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Let r2 = (r21, ... , r2a) E Ra and r= (rl, r2).
tt
Definition (3.4.17) Suppose that AD E A' and t°° E M°A fl TA. Let
to = (tal j... , taf )5tai = ani (tail
C.. itaini )-
We define \/a12 = rja1 epi( Aa)r2i and arl
f ai+1
=
11 ( aij r-i+ltaijap-i+2 i=1 p=l p
)'rli Pjaip -jaip-1 )rlipl .. ( taijap-ltaijap 1
r) be the set of continuous functions f satisfying the condi-
Let tion
sup
(3.4.18)
If(A'k,ta)ItarlA''r2
a EA'p, ka Ena too ETaO+
fori=1,2,3. Consider elements in t°* of the form a = (al,
,
af), where
n-jpiai
ai = (ail, '
and ail,
' .
, aiai+l) ... , aiai+l )
> ail, '
, aiai+l E Rai+1 for all i. We consider such a satisfying the following
condition (3.4.19)
ail <
< aiai+l,
E(jpip - pip-i )aip = 0 for all i. P
Part I The general theory
78
Proposition (3.4.20) Let p be a path of length a satisfying Condition (3.4.16)(2). Then there exist a satisfying (3.4.19) and constants cl, , ca > 0 such that a
Op (Aa)a = ri epi(At )ci i=1
Proof. We prove this proposition by induction on l(p). Let p = (0, s) -< p' = (0', s') be paths such that p' satisfies Condition (3.4.16)(2), and l(p') = l(p) + 1 = a + 1. This implies that p satisfies Condition (3.4.16)(1). Consider ep'a+1 on M°A n MM,A. Bp, (M°A n MM,A) = M°A n Mp'A. ( 9, maps Mo to Ma.)
Suppose that Moip n Mo, = GL(lipl) x x GL(mip,ip). Let p = (pl,
GL(mipl) x
,
x GL(lipbip), and Mpip n Mp, _ pf), where lipbip
lipl
Mi = (Ail, ... , µ'iai+1) pip = aj,p -.lap-i ({A'ipl' ... µ2p1 ...
µ2p6 p
...
µipb p
and /Rips E } for all i p, s. We consider such p satisfying T7 yip' = 1 for all i +p lisµips Then any element of MIA n Mo'A is of the form pgo°, for some p and g.a, E Ma°,A. Let -y be an element in t°* of the form y = (y1i , y f), where 1ipl
lipbip
yi = (Yil, ... , yiai+l), yip = (Yipl, ... , yipl, ... yipbip, ... , yipbip and
(3.4.21)
yipl < ... < yipbip,
lips yips = 0 for all i, p. S
Also we consider an element y in t°* of the form y = (51, Mipi
, 5f ), where Mip.ip
yi = (yil ... i yiai+l)i yip = (yipl, ... , Yipl, ... , yipcip) ... I yipcip). and
(3.4.22)
yipl <
< yipcip>
mipsyips = 0 for all i, p.
Lemma (3.4.23) There exists y as above such that 9p,(p)7 = ep'a+l(A). Proof. By the definition of /3-sequences, ep'a+1(p) = p7 for some -y as in (3.4.21). Note that ppl = = p' = 1. Pp, = Bp, (Po,) is the unique subgroup of Mo which contains B n Mo and whose Levi component is Mo,. By B,', the positive eights with respect to Pa, correspond to the positive weights with respect to Pp,. Therefore, the existence of such a y follows. Q.E.D.
3 The general program
79
We continue the proof of (3.4.20). We first assume that p' also satisfies Condition (3.4.16)(1). Then AD, = Aa,. We can write AD, = da'a+i(7a'a+l)A identifying AD with fi l dai(Ao,i) Since 0,(A,) = AD 1, Bp (AD,) -_
I Op(Ai)Op'(do,a+l(X5'a+1))
l
l 0p(AD)- ep'(da'a+l(AD'a+1))
s'(a+1) =
0,
5 ' (a + 1) =
1.
By the above lemma (p = da,a+1(\a,a+l)), there exists y' satisfying the condition of the lemma for p'. This implies that 0p' (da'a+l (Aa'a+1))ry' = ep'a+l (da'a+l (Aa'a+l )) = ep'a+l (An' )ep,a (Aa,
)fl
By induction, we choose a E to* which satisfies (3.4.20) for p and Op(\a)a = , a. Let a = ma + ^y, where m is a l epi(AD)ci, where ci > 0 for i = 1,
fli
positive number. If s'(a + 1) = 0, epi(AD) = ep,i(AD), and if s'(a + 1) = 1, epi(A5) = ep,i(AD) Also Bp(Aa)5 = 1,Op,(da'a+i(A 'a+1))' = 1. Therefore, a
ep,a+1(Aa')mci.
Bp'(Aa') = ep'a+1(Aa')ep'a(Aa')+1 i=1
Hence, if m is large enough, & satisfies the condition of (3.4.20) for p'. This proves (3.4.20) for the case when p satisfies Condition (3.4.16)(1). We use the same argument again, and we obtain the case when p only satisfies Condition (3.4.16)(2). Q.E.D.
, nil for Let Ii C {1, , ni - 1} be a subset and Ti a permutation of {1, , f . Let vi be a permutation which satisfies the condition that jl < (Ti,... ,Tf), (I1,... ,If), T = j2, j2 Ii implies vi l(jl) < vi 1(j2). Let I = and v = (vl, , vf). Let I,, be as in §2.1. Let Dj,,Ti be as in (2.3.7). Let Dj,T = DIl Tl x ... x DIf,Tf. Let ai = (ail) ... ,aini-1) E kni-1 for i = 1,... f, ni(ni-i) for i = 1, , f . Let and a = (al, a f). Let ui = (uijl,j2 )jl>j2 E A 2 (gl,... (n(ui)) =< ailui2l + ... + aini-1unini-1 >. For go = , gf) E G°', we
i = 1,
define
(3.4.24)
(gi, zi) =
EBi,.i,,(gi,zi)=
J NiA/Nik EBi,«i (ginni (ui), zi) Y'.i (nni (ui))dui, t(ginni(ui)Ti)'i+,OiV).,(nni(ui))dui.
4
INcA
We assume that Ti (j + 1) < Ti (j) for j E Ii for all i, because the above integral is 0 otherwise. We define (3.4.25)
EB,«,T(go, z) = 11 EBi,oi,Ti (gi, zi),
i
0
EB,« (9, z) = 11 EBi,0i (gi, zi).
Part I The general theory
80
Also we define (3.4.26)
«(g°,
)nl+---+nf-f
1
JRe(z)=q --e, (go, w, ') )nl+...+nf-f 1 2ir
EB,«(9°, z)A(w; z)dz,
f
EB,«,T(9°, z)A(w; z)dz, e(z)=q
where we choose q as in (3.4.6) for the first integral and q E DT for the second integral. These functions are defined for Re(w) > L(q). Let (3.4.27)
i
aijEkx if jEIi °ij=o if jv!i
Wi,T,V (9°, w,
)
E E E E 1e ik i
aijEkx ifjEIi .ij=o if i97i
Since Di,T is contractible, the contour defining &«,T(g(Yi,
i
,T(9°('Y1,... ,ryf),W,0I
-YiEN
(g°, w, b) is well defined as long as we choose , ryf), w, b)'s so that q E DI,T. Of course,
l eI,T,v(9°, w,'01 < ii,T,v(9°, w, 0)
Unless the situation requires, we drop 0, and use the notation a (g°, w) etc. We estimate kj,,,v(g°, w). Let li be a function from Ii to the set of natural numbers. If Ii = {ii, ,jNi}, jl < < 3Ni, we define lip = li(jp). Let l = (ll, , l f). Let si,i, (l) be the element of the Lie algebra determined by (3.4.28)
ti''I' (c) _
I ti3 ijj+l l
1i(j)
jEIi
Let s I = s
and tsI(i) = rj ts`i(t). We fix T I. For s l (l) as above, we put q'(l) = rq - sj(l). Consider the following conditions.
E Dj T, l
Condition (3.4.29) (1) For any 6 > 0, there exists a function c6,1(z) of polynomial growth such that WT(6(a, g°), z)I <<
for {z I Re(z) E Di,T,6}. (2)M 9 (1)ivi(jf)
for (ii, j2) E Ivi.
-
q'(1)ivi(j2) >- M
c6,1(z)t(go)-s(`)
3 The general program
81
By (2.2.6), (2.4.8), (2.4.9), we get the following proposition.
Proposition (3.4.30) (1) There exists M > 0 independent of q, l such that if q, l satisfy the conditions (1), (2)M in (3.4.29) and 61 > 0, then ej,T,,,(g°,w) is bounded by a finite linear combination of functions of the form t(go)v(Tq+P-s1(d)) 11
'
11
Itij1 (9°)tij2 (g°)-'
IC:iv2
(.1i,.72)EI,,:
for L(q) + 61 < Re(w), where ci71.72 is a constant independent of q, l for all (j1, j2) E
Iv. (2) Moreover, if q E Di,,, I", (go' w) << t(g°)Tq+P
for g° in the Siegel domain and L(q) + 61 < Re(w). Note that in (1), we are not restricting ourselves to elements of the Siegel domain. Let
J
I tij1(9°)ti.92 (go)-' Ci3132 11 i (ji,72)EI,,.
This is a slowly increasing function on G°o independent of q, 1.
We have proved in (2.4.8) that there exists M1 (depending on q) such that if li, - li,+1 < M1 for all i, j then (3.4.29)(2)M is satisfied for a sufficiently large M. We are going to use (3.4.30) to estimate 9p (go, w), ip (g1, w) in this book. But our estimates are rather delicate, so the author would like to clarify the logic. We have estimates of the form (3.4.29)(1) where the range of l does not depend on q E DI,,. But for Condition (3.4.29)(2)M to be satisfied, the range of l depends on q E DI,T. When we use (3.4.30), we have two kinds of freedoms. One is the choice of 1, and the other is the choice of q. We can always fix q, and take l which satisfies (3.4.29)(1) as long as it satisfies (3.4.29)(2)M. But if we want to fix 1, and vary the choice of q, we have to be a little careful. For, if v2 # 1, the range of li could depend on q. Therefore, if vi l 1, we have to show that (3.4.29)(2)M does not depend on qj. On the other hand, if v = (1, , 1), since Condition (3.4.29)(2)M is empty, we can fix any l satisfying (3.4.29)(1) and not necessarily satisfying the condition that li9 - li7+l << 0 for all i, j, and vary the choice of q. So, for example, we can choose lij - lid+l >> 0 for all i,j. Also if some Ii consists of one element and vi = 1, we can choose an arbitrary real number lil > 1 by (2.3.24)(2). We go back to the previous situation where p is a path as before.
Theorem (3.4.31) Let CG = (9101 ...T.,)-l. Suppose that G1 = G°o and p is a path satisfying Condition (3.4.16)(2). (1) For any e > 0, there exist b = 6(e) > 0 and a finite number of points cl, t°* satisfying IIciIIo < c such that if M > L(p), I9p(go,w) - CGA(w;p)I <<
t(9a)c:
, cN E
Part I The general theory
82
on Aa(K fl Mom)To°,,7+Sta, and wo - 6 < Re(w) < M.
(2) Suppose that r = (r1, ,Tf), and r # r,. Then for any e > 0, there exist 6 = 6(e) > 0 and ri = (ril, rig) as in (3.4.17), such that rill, , ri2a > 0 for all i, Iriip8I < c for all i, p, s, and that if M > L(p), ep,T(A' ga, w) I < 1: A r2at(ga)T{1
on Apo(K fl MMA)Tb,,+fZ,, and L(p) - 6 < Re(w) < M. (3) There exists r = (rl,r2) as in (3.4.17) such that all the entries of ri (resp. r2 )
are negative (resp. positive), and that if wo < Ml < M2, 61p, (A' ga, w) I <<
Ao(Re(w)-
o)'2t(ga)(Re(w)-wo)T1
on Aa(K fl MDA)TT,7+f2o, and Ml < Re(w) < M2 for all r. , If) Let go = Aa ga Clearly, Bp (go) = Op ()4 )9p (go). ) . Consider I = (Il, such that jpip Ii for all i,p. Let o, = (Ol, ,o-f), v = (vl,... , vf) be elements of the Weyl group, where ai = fP=1 l 6ip, vi = nPl 11 vip, and o ip, vip are permutations of 3pip for all i,p and vip satisfies the condition in (2.1.10). Then, by the consideration in §2.1, Proof.
I,oT,v(0p(9D),w)
-'p,T(go,w) _ I,o,v
If (jl, j2) E
for some i, then j1i j2 E 3pip for some i,p. Also if go is in the a
Siegel domain, 0p (ga) is contained in some Siegel domain for Mp°A also. Therefore,
by (3.4.30), if q E DI,,T and 6 > 0, t,or,v(9p(go),W) <
for go in the Siegel domain and L(q) + 6 < Re(w). It is easy to see that Ei,,,,(ep (go), z) = 9p (A'
)ar=+PEI,o,,(ep (g), z)
Clearly 9p (A, )0 z+P = 9p (A, )T Z+P. We /choose an element ry in to* of the form 7 = (51, ... , 5f), where 3i = (Yip), lip = (Tip1, ... yI1pjpsp-jVsp-1 ), and
'Yip = aipl(-1,1, 0, ... 0) + ... + a?pJjp -.7pip-1-1(0, ... , 0, -1, 1), aip1, ..
, aipJp+P-.7v=r-1 -1 > 0, ryipl < ... < yiPJpip-9pip-1 .
Consider a in (3.4.20). Let h1, h2 > 0 be positive numbers and a = hla + h0Then if h1h21 >> 0, a is an interior point of t°*. If h1 is small, II&IIo can be arbitrarily small. (9p(ga),w) so that Re(z) = q We choose the contour in the definition of and q = (vr)-1(-p + ma), where m > 0 is a constant. Then Qrq + p = ma. Since -(ar)-'(p) is in the closure of DI,,, and & is an interior point of t°*, q E DI,ar.
3 The general program
83
Therefore,
eI,ar,v(Bp (ga ), w) << Op (Aa a
)m.h10Op
i mhlc
epi(At,
(ga°)mh2-1
0 nhz5 ep(ga)
i=1
By the above consideration, q E DI,-T as long as m > 0, and L(-(0'7-)-1(p)) < wo for all o, T. Clearly,
L((UT)-1(ma)) = mL((cr)-1(a)).
So, if 0 < c' < jL((orT)-1(&))j-1 for all a,T, we can choose m = c'(Re(w) - wo). Note that the choice of & does not depend on or. This proves (3). Suppose or # TG. Since L(-(UT)-1(p)) < wo, if m > 0 is sufficiently small, L(q) < wo. This proves (2). Note that if r Tp, aT # TG. Next, we consider the case aT = rG. By the above remark, r = Tp. i-1
Suppose Ii
0.
Let j E Ii. Let a' = (0 0, ai, 0,
, 0) E t°*, where
ai = (-(ni - j),
, -(ni - j), j, . . , j). Then if m' > 0 is a small number, p+m'a' is in the closure of DI,TC and L(p+m'a') < wo. If m'm-1 >> 0, then L(p + m'a' + TG(ma)) <wo and p+m'a'+TG(ma) E DI,TC. Let q = p+m'a'+TG(ma). Then TGq + p = m'TG(a') + mrG(a). Therefore, 11TGq + pilo can be arbitrarily small. Hence, we can choose q = p + m'a' + TG(ma) E DI,,, ,1). So Suppose that aT = TG and I = (0, , 0). This implies that v = (1,
eI ,rc,1(ga,w)=
)nl+...+nf-f
(2
fRe(z)=q
(z)t (O(g
))Gz+°A(w;z)dz.
Let 61,62 > 0. Let ql = (qll, "
, q1 f) E t°*, where qi1 = (giji) .. , glini ), Blip glip+l = 1 + 61 for all i,p except for i = p = 1, and q, 11 - g112 = 1- 62. We assume that 61i 62 are small and 6261 1 >> 0. Then L(q) < wo. Let q2 = (q21, , q2f) E to* . 61 for all i, p except for i = ) g2ip;) and Up g2ip+1 = 1 + such that q2i = (g2i1, p = 1 and q111 - q112 = 1. Let dz' = 1!(i,p)0(1,1) d(zip - zip+1). Then 6'I,TO,l(ga, w) is equal to ''7UU
nl+...+n/-.f
1
1
+
I
e(=)=q,
MTC(z)t(ep(ga))TC=+vA(w;z)dz
n1+...+nf-.f-1 Res
27f
IRe(z)=q2 =11-z12=1
[M,, (z)t(Bp (ga))TC=+PA(w; z)] dz'.
There exists 6 > 0 such that the first term is holomorphic for Re(w) > wo - 6. We continue this process, and eventually get CGA(w; p). This proves (1). Q.E.D.
Note that the right hand side of (3.4.31)(3) does not depend on M1i M2.
Part I The general theory
84
Definition (3.4.32) Let f, (w), f2(w) be meromorphic functions on a domain of the form {w E C I Re(w) > A}. We use the notation i - f2 if the following two conditions are satisfied.
(1) There exists a constant A' < w° such that i - f2 can be continued meromorphically to the domain {w E C I Re(w) > A'}. (2) The function f, (w) - f2(w) is holomorphic around w = w°. The following corollary follows from Theorem (3.4.31)(1).
Corollary (3.4.33) Suppose that G' = Go Aand p is a path satisfying Condition (3.4.16)(2). Suppose that f (ga) is a function on A'iM°ti/Mak, where i = 1, 2 or 3, and that there exist finitely many points cl, , c,n E t* such that If(ga)I << inf((X,t(ga))c`), x
and
I',i (KnM,A)Ta,, Sla
inf((Aat(ga))")dgo < co. i
Then
f
.f(ga)(g,w)dgCGA(w;P) f
Ani MaA/Mak
viMDAYMak
In other word, if we can prove that a function is integrable by estimating on the Siegel domain, then we can get rid of the smoothed Eisenstein series in the integral.
The statement of the next theorem does not depend on the path p, and we do not consider Condition (3.4.16) or the assumption that Gl = G.
Theorem (3.4.34) (Shintani's lemma for GL(n)) (1) For any e > 0, there exists 6 > 0 such that if M > wo,
f sup
g0 ga=kt,.n(u)E60 .e(w)<M
ni-1
I e(g°, w) - CGA(w; P)I If fl I tij (g°)tij+l (g°)-l i=1
I-E
< oo.
1
(2) There exists a constant c > 0 such that if M1 > M2 > w°i
f ni-1 sup
16' (g°, w) I If H I tij
(g°)tij+l(go)-l Ic(Re(w)-wo)
< 00.
gO=ktn(u)E60
Proof. The proof of this theorem is very similar to that of (3.4.31). The only place we used the assumption (3.4.16)(2) was the choice of a in the proof of (3.4.31). This time, we just choose & = h where y = (y1i , 5f), and
5 i =a1(-1,1,0,... 0) + ... + ai,,-1(0, ... , 0, -1,1), ail, ... aini-1 > 0, 5il < ... < 'Yini. The rest of the proof is similar to that of (3.4.31). Q.E.D.
3 The general program
85
,ni-1,
Note that ifcij is a constant for
f ni-1 f ni-1 1111 Itij(9°)tij+1(g°)-' cij(Re(w)-wo) «fJ [ i=1 j=1
Itij(9°)tij+1(g°)-1Ic(Re(w)-wo)
i=1 j=1
for go E 60, where c = infi j cij. The statement (1) of (3.4.34) implies that if f E C(G°°/Gk, r) for some r such
that rij > -j(ni - j) for all i, j, then
f
A/Gk
f (9°)8(9°, w)d9°
CGA(w;
p) J
f (9°)dg°. ,
The statement (2) of (3.4.34) implies that if f is a slowly increasing function,
fG/Gk
.f (9°)e(9°, w)d9°
is well defined, and becomes a holomorphic function in some right half plane. §3.5 The general process
In this section, we consider (G, V,Xv) such that (Me,ZZ) is a prehomogeneous vector space for all 0 = (/31, , /3a). (The character is the one which is a positive multiple of (da.) This condition is not always satisfied, for example, G = GL(n), V =
M(n, n) (the set of n x n matrices). In this section, we assume that (G, V) is an irreducible representation. Let 0, ?.' be /3-sequences such that i and 'Y2 E 9(Za'A). We define
a' and l(ti') = l(cl) + 1. Let t@1 E 9'(Z-OA),
OS,, (,P1,9D) = E 'I'1(9Dx),
(3.5.1)
xES,,k
E `y1(9ax), xE o%k
OZ,, ('Y2,9a') =
'F2(9a'x), xEZaik
for ga E G1 fl MS,t, ga, E GI fl MM'A. We also define
W1(9ax), Gs,,,st(`I'1,9a) E '1(95x),
Osa,('P1,9a)= XESa,k
xES,,,atk
E 'Fl(95'x), OY,,,st(W'1,9D') = E IF2(90'x), XEYY,k
OZa, ('1'2,95') _ E W2(9D'x), OZ,,,st(F2,9a') = xEZ,,,k
xEYa,,atk
L:
xEZo,,etk
'F2(9a'x)-
Part I The general theory
86
Let w = (wl,
, w f) E St(A < /k" )f be as before. In the following definition, we
consider the case where G = G. Definition (3.5.2) Let p = (Z,s) be a path, and T a Weyl group element which satisfies Condition (3.4.13) for Up#. For -P E .'(VA), w as above, and w E C, we define (1)
Ep,T (-D, w, w)
f
wp (ga )op (ga )E)z, (Ra'Dp, go ).p, (go, w)dga,
,,2 MaA /Mak
w, w)
(2)
f9p1 MDA/Mok -p,T+(oD, w, w)
(3)
wp (ga )Qp (ga jai (ga )E)z, (9a Ra gyp, Ba (ga )),?p,T (go, w)dga, `q,0 MaA/M k (4)
p,T#
w, w)
= Ra-bp(0)
f
wp(ga)op(go)Sp,T(go,w)dga,
AvoMaA/Mak
=p,T#(4),w,w)
(5)
_ 90Ra'Dp(0)
fno MA/Mok
wp(ga)o,p(ga)kai(ga)SpT(go,w)dga,
if these integrals are well defined for Re(w) >> 0. w, w, When we have to refer to the function z/i, we use the notation etc. We also define EP w, W)- st+(,D, W, w) etc. using '9z,. (Rasp, go) etc.
Clearly, w, w) = Cp +(4)1 w, w) + p,T,st+(,P, W, w) etc.
If the distributions in (3.5.2) are well defined, we define p ('D, w, w) _
Ep,T (4D, w, w)
etc. Equivalently, EP (-P, w, w) can be defined by replacing 6p,T(go,w) in (3.5.2) by .°p (go, w). It is easy to see that Ep,+(4), w, w) =
w, w) + Ep,st+(4D, w, w) etc.
also.
We still assume that G' = G. We consider a path p = (a, s) which satisfies Condition (3.4.16)(1), or satisfies Condition (3.4.16)(2) and there is no split torus
in the center of Ma which fixes Z,11. Let x E Z. This assumption implies that there is no split torus in the center of Ma, which fixes x, because Zak is a single Mad-orbit.
3 The general program
87
Let to, ta, ta, to be the Lie algebras of AD, A', A', T+ n M ,A respectively. Then
to=tf
ta.
Let x = (x,, , XN) E Zak. Since x is semi-stable, it is semi-stable with respect to the action of Aa also. Note that we are considering Aa as a group over R, and the notion of semi-stable points is not changed by field extensions. If p satisfies Condition (3.4.16)(1), Aa is trivial. The convex hull of {yiltii I xi # 0} contains the origin of to*. If p satisfies Condition (3.4.16)(2) and there is no split torus contained in the center of Ma which fixes ZD, the origin cannot be on the boundary of this convex hull, because that implies the existence of a split torus fixing x. Therefore, it contains a neighborhood of the origin of W.
Proposition (3.5.3) Let p = (0, s), ,r be as above. (1) The distributions Bp T w), and vp,T+(4), w, w) in (3.5.3) are w, w), well defined for Re(w) >> 0. w, w) are (2) Moreover if p satisfies Condition (3.4.16)(1), Bp,T# (4b, w, w), also well defined for Re(w) >> 0.
Proof. Let a = l(p). Then as acts on ZDA by multiplication by epa (AD). For go in the Siegel domain, we choose v(ga) E T+ so that t(ga)z = v(ga)z for any z E We choose 0 < IF E 9(ZDA) so that IRa4Pp(A gax)I << W(A v(ga)x) for g°° in the Siegel domain.
E)a,I(IF,A,v(ga)) _
(Aav(ga),x) sE ZDk
o ekx for.ef =O for i0I
We consider such I such that the convex hull of {yiIta®t$ i E I} contains the origin of td* ® to*, and the convex hull of {yilta i E I} contains a neighborhood of the origin of to*. We choose Io so that {x = (xi)i I xi = 0 for i V Io} = Z5. Then I
Ioz, (Ro bp, aaga)I << E E)a,I(IF, Av(ga)) ICIo
We fix I. We choose constants 0 < ci for i E I so that E, ciyiIt, to = 0. Since A5 acts on ZDA by multiplication by epa(A5), this implies that
(A u(ga))-Eie1
h'
= epa(A
)-Eiel°i,
Let C = EiEI ci. Clearly, C > 0. Let d = dimA5. We write elements of AD in the form as = A5Aali, where Aa E A5, 41) E A. Since the convex hull of {yilta i E I} contains a neighborhood of the origin of to*, as in the proof of (3.1.4), for any M > 0, there exist constants , h, i E I and a slowly increasing function fm (AD, v(ga)) such cM pi's for j = 1, I
that (3.5.4)
J
fM(As,v(ga))rdd,M(4
Part I The general theory
88
By (1.2.6), for any M1, M2 > 1, (eza (Rasp, )aga)I << inf(X'av(ga))-M1 E.er
CM2,j"Y'
< epa (,A')-CM1 fM2 (AD, v(ga ))rdd,M2 (dal) ).
We have the estimate (ge, w) I «A
(Re(w)-wo)r2
v(ga)(Re(w)-wo)ri
where r is as in (3.4.31)(3). We choose M2 large enough so that the function fM2 (AD, v(ga ))rdd,M2 (A(') )
is integrable with respect to A( 1). Then the resulting integral is bounded by a slowly increasing function f'(aa, v(go)) of AD, v(g°). Consider the function f '(X1 , v(ga ))epa
(Aa)-CMS Aa (Re(w)-wo)r2
v(ga)(Re(w)-wo)r1
Note that the functions epa() ), A' (Re(w)-wo)r2 depend only on AD.
If e, ()4) >_ 1, we choose Re(w) >> 0 so that the above function is integrable with respect to epi (A') for i = 1, , a - 1 and go in the Siegel domain. Then we take M1 >> 0 so that it is integrable with respect to epa (A'). If epa ()4) < 1, we fix M1 and take Re(w) >> 0. Since there are finitely many possibilities for I, this proves that =-p,T (4D) w, w), =-p,T+(4i, w, w) are well defined if Re(w) >> 0. We can estimate
aRaDp, Bp(Aaga°)) similarly except that the exponent of epa(A') is positive. Since we consider Apo for =p+(qD,w,w), the proof is similar. This proves (1). , epa) gives an We consider (2). In this case, AD = A', MMA = M°A. So (epi, isomorphism AD -* R. By (3.4.31)(3), for any slowly increasing function f(go) on ApoM,A/M'Ok, the integral Oza
JA0 Man /Mak
f (go )6p,T (ga, w)dga
converges absolutely for Re(w) >> 0. This proves (2). Q.E.D.
A similar proof to that in the above proposition shows the following.
Proposition (3.5.5) Let p = (cl, s) be as above, p -< p' = (O's') and l(p') = l(p) + 1. xi 0} Suppose that for any x = (xi) , XN) E S5'k, the convex hull of {ryiIta contains a neighborhood of the origin of to*. Then the integrals JA0 Man /Mak
f
wp (ga) kal (ga )o,p (go )esa, (go Ro Dp, 9a (ga ))ep (go, w)dga,
wp (ga )op (go )osa, (Ra gyp, ga )e, (ga, w)dgo
A'0 Man /Ma k
are well defined for Re(w) >> 0.
3 The general program
89
If p satisfies Condition (3.4.16)(1), since Aj is trivial, the condition in (3.5.5) is automatically satisfied for p'. Suppose that p satisfies Condition (3.4.16)(1). Let (3.5.6)
go) ='al(ga)
Osa, (JFaRa4'p, 00 (go)) a
t(a')=t(a)+1 E) sa, (RD D-b p, ga
a[a'
t(a')=t(a)+1
Jp(4',go) = J,(4), go) + tca1(go)-FaRa4p(0) - RD 41p (0).
If p = (0, 0), we use the notation J(4), g°), J'(4P, g°). By the Poisson summation formula, p (4', w, w) _ Ep+('D, w, w)
+
Lip (ga)op(ga)Oza(RDI)p,ga).p(go,w)dga A p0 MaA/Mak
P+ (D, L", W) + Cp+ (D, w, w) + =p# (4D, W, W) - Ep# (4), W, W)
+
JA V0 MaA /Mak
wp(go)op(go)Jp(-P, go)ep(ge,w)dgb.
By (3.5.3), (3.5.5), we get the following proposition.
Proposition (3.5.7) If p = (D, s) satisfies Condition (3.4.16)(1), then p
w, w) = up+ (p, w, w) + up+ (p, w, w) + p# ata'
t(a' )=1M+1 M+1
f
a
oMaA/Mak
fApoMaA/Mak
w, w) - Ep# (-t, w, w)
wp(go)Ka1(go)op(go)E)
wp (ga )rp (ga )es,, (Ra Dp, ga ).p (go, w)dgo
l(a')=t(a)+1
Let p be as above, p -< p', and l(p) = l(p)+1 = a+1. For go, E G°O f1 Ma,A, ua' E -01A, we define Ep'1(4', w, go,, ua') = wp (ga' )o p (ga' )OY,, (Ra bp, ga'na' (ua' )),
Ep'2('D, w,g5',ua') = wp(ga1)op(ga')ka1(ga')®Ya, (9aRaDp,ea(ga'na'(ua'))).
Definition (3.5.8) (1) If s'(a + 1) = 0,
fna Pa,A /P*'k
p, (4), w, w) is, by definition, Ep' 1(4), w, go,, ua' )f''p (ge' no, (ua' ), w)t(ga') -2p,' dga, dua,,
if this integral is well defined for Re(w) >> 0.
Part I The general theory
90
(2) If s'(a + 1) = 1, gyp' ((D, w, w) is, by definition,
f
poPa,A/Pack
Ep'2 O, W, go', u5' )fp (go, no, (u0' ),
w)t(go')-2Pp' dg0, duo,,
if this integral is well defined for Re(w) >> 0.
We consider paths in the following three classes. (1) The path p satisfies Condition (3.4.16)(1). (2) The path p satisfies Condition (3.4.16)(2) but not (1) and Ma contains a split torus in the center which acts trivially on Zo. (3) The path p satisfies Condition (3.4.16)(2) but not (1) and Ma does not contain a split torus in the center which acts trivially on Zo.
Let p' be a path such that p' = (-0', s'), and s'(a + 1) = 0. Then IA90 M,A/Mak
f
WP(go)ap(go)es,, (Ro4)p,g,)ep(go,w)dgo
ApoMA/P,k
Wp(ga)op(go)19ya, (RoDp,go)ep(go,w)dga Ep' 1(4), w, go,, uo' )e'p (ge' no' (u0, ),
w)t(ga')-2Pp' dg0' duo' .
Apo MaAnPa'A /Pa'k
The second step is because M5,,,, Ro(bp = Ra4Dp by the assumption in §3.1. Since
fa' A / U,' k rya' (RoDp,go,no,(uo,))duo' = k0'2(g5')ez,, (RD',Dp',go,), and Apo{do'a+1(Ao'a+1)} = Ap'2, the above integral is equal to
^p'(-D,W,w)+, p'(D,W,w). If p' belongs to the class (3), Ep, (4i, w, w) is well defined for Re(w) >> 0 by (3.5.2). Therefore, Ep, (1D, w, w) is well defined for Re(w) >> 0 also.
Similarly, let p' = (D', s'), and s'(a + 1) = 1. Then
f
Wp (ga)ka1(ga )op (go)Os, (go RoDp, O* (gs)).p (go, w)dga po MaA /Mak
_ =p, ($, w, w) + E"p, (4), w, w).
Therefore, p, (4k, w, w) is well defined for Re(w) >> 0 if p' belongs to the class (3). These considerations show the following proposition.
Proposition (3.5.9) Let p be a path in class (1). Then if 2p, (4i, w, w) is well defined for Re(w) >> 0 for all p', Ep=p(4),W,w) = Ep(p+(-D, W, W) + =p+((D,W,w) +
w) -
E Ep, (Gp, (-D, W, W) + E" (4), w, w)1 I a
3 The general program
91
This proposition is the basis of the iterated use of the Poisson summation formula. We now describe how we proceed.
We assume that (G/T, V, Xv) is an irreducible representation and of complete type for simplicity. This implies that G1 = G. For 4) E .'(V,1) and A E lam, we define 4)A(x) = 4)(,x).
We want to find points pi, , Pdv E C and distributions a((D, w), a2j (1b, w) for i = 1, , dv, j = 1, , dv,i with the following properties (3.5.10)
f
w(9°)J(, 9°)(9°w)d9° - a(, w)A(w, z), °e/Gk
dv dv,i
(3.5.11)
A-n' (log A)ia2j (4D, w),
a(,DA, w) i=1j=1=1
where A E R+. Clearly, 1
f A'-r; (log A)i d' A = (-1)1 j!(s -
pi)-1-1.
0
This implies that (3.5.12)
Zv(4), w, s) = Zv+(4), w, s) + Zv+(4, w-1, N - s) dv
dv,,
-13-1
+1'
giving a principal part formula. If (G, V, Xv) is of incomplete type, we need `adjusting terms' (see [86] or Chapter 4). In this book, the only such case is essentially the case of binary quadratic forms, so we do not discuss it here. For a general conjecture about prehomogeneous vector spaces of incomplete type, the reader should see [85]. In order to achieve (3.5.10) and (3.5.11), we use (3.5.9) repeatedly. If p' in (3.5.9) belongs to class (1) and Ma, is a torus or p' belongs to class (2), then we do not consider paths p' -< p". If p' belongs to class (1) and Ma, is a torus, 8p, (-D, w, w) is essentially a contour integral of a product of Dedekind zeta functions. Therefore, we can handle such distributions. If p' belongs to class (2), it is likely that we can show "p, (4), w, w), 8p' (,D, w, w) -' 0 as is the case in Parts III and IV. Suppose that p belongs to class (3). This implies that the representation of M, on Z° is reducible. We still do not have a general approach to handle all the reducible prehomogeneous vector spaces, but if the number of irreducible factors is two, we can use a technique to consider further paths. We discuss this feature of our program in Parts III and IV. So we are still obliged to show that Ep ((D, w, w) is well defined for Re(w) >> 0 for
p=
s) in classes (2), (3). We will handle these distributions when we consider individual cases. Of course, we want to ignore all the 8p (4), w, w)'s. However, it is too optimistic to expect that ,=p (4 , w, w) - 0 for all p as we will see in Part IV. For the space of pairs of ternary quadratic forms, the correct statement turns out to be Ep"p((D,w,w) - 0. p
Part I The general theory
92
Also it is too optimistic to expect that for all p, there exists a distribution by ('P, w) which satisfies the property (3.5.11) and
'p (4b, w, w) - by ((a, w)A(w; z)
This is because some cancellations have to be established between strata. Therefore, our process is quite delicate and heavily depends on the individual cases. The following proposition follows from (3.4.32).
Proposition (3.5.13) Consider a path p in class (1). Suppose that a
ap(AD) = rl epi(A,)"', i=1
and Xpi > O for i = 1,
, a - 1. Then
V+(4),w, W)
CGA(w; P) f
wp(go)ap(ga)'9z; (RoDp,9o)dgo,
Ant Man /Mak
p+(4p, w, w)
- CGA(w;P)
f
wp(ga)op(go)noI(go)19z; (goRa4,p,Oo(go))dgo
Ap0 Man /Mak
Next, we consider paths of positive length in class (2). We are still assuming that G°A = G. Therefore, we do not consider X. Suppose that p is a path which satisfies Condition (3.4.16)(2). Let Aa be as before. Let A2 be the connected component of Ker(A' -* GL(V)A) which contains 1. Then Aa = R+ for some c. We choose a subgroup Aa C Aa which is isomorphic to R+ for some d so that Aa = ADAD and A. 2 n A.3 = {1}. We write elements of Aa in the form $1) = A(2)A((3), where A(') E A, for i = 1, 2, 3. Let go = ADA(')g0 where go E G°A n
go E M°A.
a A2 We fix an identification
R+, and write x(02)
- (t(i) ... ) 11ac)
for tai), ... , 422 E R+. Let d" 42) = rli dx 4 . We choose an invariant measure dxaaz) on Aa for i = 1, 3 so that d- A(') = dx42)d-A(3), and dgo = Suppose that r is a Weyl group element which satisfies Condition (3.4.13) for M. Let 'yp,T,i(z) be a linear function for i = 1, , c such that (3.5.14)
0p(\(2))Tz = H(A((2))7p,r,i(z). i=1
We define (3.5.15)
LSpj7 = {z E * I -yp,T,i(z) = 0 for i = 1,
, c}.
3 The general program
93
Let dzI Ls,,r be the differential form such that dz = 11i=1 d'Yp,T,idzl Ls,,r
Let h E t* be the point such that 0,(
('))-Th
= op(A('))Bp(Aa2))P.
There exist numbers h1, , h, such that 0 (A (2)),h = (A(2))h; Consider I, v, v in the proof of (3.4.31). We consider the following condition.
Condition (3.5.16) For any T, a, there exists q E (LSp,T + h) n DoT such that if a E t* and IIaIIo is small, the function op(X043))Bp(A,43))T(9+«)+vep(t(9a))°T(q+«)+POZa (Ro(Dp, go)
is integrable on Ap2A3(K n MIA)T°q+S2o.
Note that OZ, (RD Dp, go) does not depend on A(2). The above condition implies that (LS,,, + h) n DoT # 0 for all 7-, a. In particular, (LSp,r + h) n DT # 0. Suppose that (3.5.16) is satisfied. We choose a subspace Hp,1. C Q so that tt = LSp,T ®Hp,r. Let z1 E (LSp,r + h), z2 E Hp,r. Let dzj be the differential form on LSp,T + h such that dz1(z-h) = dzILS,,.,. For Z2 E Hp,T, we define z2i = ryp,r,i(z2) for i = 1, c. We define dz2 = rji dz2i. Let Weput fp,T (Ao A(D3)1 901 w, Z17 Z2) Op(Ao$D3))T(Zl+z2)+PEBo,T(9p(9a),
= fp,r,I,a,v(A0A0
z1 + z2)A(w; zl + z2),
90, w, z1, z2) 9,(aoAD3))T(zl+z2)+PEB,I,°r,v(9p(ga),
z1 + z2)A(w; z1 + Z2).
Then we define the functions ep,T(A0 XD3)g , w, z2)+ 6Op,T,I,a,v(A0AD3)g , w, Z2)
by the integrals {
1
27r
(i)C
2 7r
-1
Jp,r(AoA(3),9a,w,z1,z2)dzl,
+h
Re(zl)=9
(3),
r+h.fp,T,I,a,v(AoAD90, w,zl,z2)dzl, Re(z1)=q
respectively, where we choose q in (LSP,T + h) n DT, (LSp,T + h) n DQr. The above functions are well defined if Re(w) >> 0 and IIRe(z2)IIo is small.
Proposition (3.5.17) Suppose that Condition (3.5.16) is satisfied for p. Let r be as before. Then (1) 8p (41, w, w), E`p,r(4D, w, w) are well defined for Re(w) >> 0, and
=p,T(',w,w T
Part I The general theory
94
(2) 'p,T ((D, w, w) is equal to ap(A Aa3))cp,T(Aa)a3)ga, w, 0)eza (Ra4)p, 9a)dxasdx4a3)d9a
JA,, A3 MO
-
Proof. We consider subsets of A22 like {A(2) I AO > 1 for all i}, {.X(2)
A(2)>1}etc.
I
A() 01 < -
Let Aa+ = {))(2) Aa2) >_ 1 for all i}. Since the argument is symmetric, we consider the set Aa+. We choose a E t* so that IIailo is small and I
(2))p.
0 (A((2))Ta _ (A(2))Pl ... (A DC 01
for some By assumption, .-n,T(AaA(')9a, w, z2)
ep,T,l,o,v(AaA(3)9a, w, z2),
_ I,o,v
and
Iep,,I,.,.(\ )43)ga,w,z2)I «9p(A where qa satisfies Condition (3.5.16).
It is easy to see that
(2)
))$p,T(9a,w) _
-
it v
i
T(AaA(3)9a,w,z2)fj(Aa2))zy'dz2.
c
Re(Z2)=a
Consider the function vp(AaA0 ))eza(R5-Dp,9a)e',T(AaA3)90, aw,
(3.5.18)
By Condition (3.5.16) and the choice of a, the order of the integration of the function (3.5.18) with respect to the set Aa+ and the contour {Re(z2) = a} can be changed. Therefore, p,T ((D, w, w) is well defined for all T, and (1) follows. The integral of the function (3.5.18) with respect to the set Aa+ and the contour
{Re(z2) = a} is equal to _1)c C
2
V
ap(AaAa ))oZ, (Rasp,
c
1
f i z2i
JRBc=2.>=P;
1
w, z2)
dz2.
The integral of (3.5.18) over the set {A(2) I A(2) < 11 A(2) > 1, .. , Aa2) > 1} etc. has
a similar expression.
Let e > 0 be a constant . If O(s) is holomorphic function on the set Is E C I -e < Re(s) < e} which is rapidly decreasing with respect to Im(s), fRe(8)=-
0(s)ds+J
28
e(s)=2
8
3 The general program
95
The statement (2) of the proposition follows by applying the above relation c times. Q.E.D.
When we apply our process later, we sometimes face strata Sp's such that there is a split torus in the center of Ma which fixes each point in Zp. In many cases (but not always), we can ignore such strata. This phenomenon is based on the following proposition.
Proposition (3.5.19) (The vanishing principle) Suppose that V)(P)
0,0(-7-GZ) = O(Z),
and Condition (3.5.16) is satisfied. Then if p
LSp,T + h, there exists a polynomial
P(z) such that P(p) 0 0, P(-TGZ) = P(z), and that if st' = ,b(z)P(z), then -p,T (D, w, w, 01) = 0.
Proof. We choose a linear function 1(z) so that 1(z) = 0 on LSp,,. + h and l(z) # 0. Let P(z) = l(z)l(-TGZ). Since -TGp = p, P(p) # 0 also. Since 1/)'(z) is identically zero on the contour, b(z) satisfies the condition of this proposition. Q.E.D.
Proposition (3.5.20) Let S#(w) = 6(w1) ... 8(w f). Then
fA/Gk
w(9°).'(9°, w)d9° = b#(w)A(w; P)
Proof. Let db be the measure on BA which is a product of measures we defined in § 1.1. for BA n GiA for i = 1, , f . Let BO = G°4 n BA. We choose a measure db° on BA so that
f
fBA/Bk
f(b)db=n1...nf JR1 f
f(c(A)b°)[JdxAidb°,
+BA/Bk
cn,(A1)...cnf(Af) for a1i... Af E Imo. where c(A) = Let pi = ani (I'i1' , ... , z:ini -1)' and µ = (p1i pi2µil1'
Boo -
r+1+...+nf-.t
i=1
µ f). We identify
x (Al /kx)nl+...+nf,
Then, by the choice of our and write b° = µt° where to E measure, db = P-2p fi , dxµijdxt°, where dxtO is the measure such that the volume of
(A1/kx)n,+...+nf
is 1.
Part I The general theory
96
By the Mellin inversion formula,
fG/Gk
w(g°)e(g°, w)dg°
n+.('JJ .+nf-f
_
w(g°) E t(g°'Y)z+°A(w; z)dg°dz )=4
1
JG/Gk
YEGk/Bk
fG/Bk
Re(z)=4
J e(z)=9 w(g°)t(g°7)z+PA(w; z)dg°dz
nl+...+nf-f
(2) 1
w(go)t(go)z+°A(w;z)dg°dz
7fY -1
1
= b#(w)
YEGkl Bk
n+...+nf-f
1
nl+...+n f-f
C2 7r
JR()=9
H µijj-+l-1 A ij
,
(
)
Hd x µzadz ij
= 6#(w)A(w; P) This proves the proposition. Q.E.D. Right now, (3.5.13) and (3.5.19) are the only general statements we can prove
without using properties of individual representations.
§3.6 The passing principle We have to consider various contour integrals in later chapters, and for that purpose, we prove a statement concerning a possible way to move the contour in this section. We first introduce some notation. Let p be a path, and T a Weyl group element which satisfies Condition (3.4.13) for Mp. We define ST = (STl, ... , STf )f STi =
(ST21,...
i ) ST2ni-1) E Cni -
by sril = ziT(n,) - ziT(n;-1),''' , Srini-1 = ZiT(2) - zir(1) Since the correspondence between z and sT is one-to-one, any function of z can be considered as a function of sT. For a function _f (z), let f (sT) be the corresponding function of s,.. So for example, we consider L(sr),AT(w;sr) etc. Let pT E be the element which corresponds to p. Let dsr be the differential form which corresponds to dz. Let H C C01+'.'+nf-f be a subspace of dimension d not necessarily going through the origin. We use the letter SH to express an element of H. We choose a differential form dsH on H. Let r(l),r(2) E H fl Let to be the line segment joining r(l), r(2), and D a domain containing lo. Let f(SH) be a meromorphic function which is at most of polynomial growth on any vertical strip contained in {SH I Re(SH) E D}. Let
r () w
a
=
1
27r
f (SH)V(SH) ds H fRe(sx)=r(i) W - L(SH)
3 The general program
97
fori=1,2. Suppose that there exists a polynomial P(sr) such that P(sH) f (sH) is holomorphic on {sH I Re(sH) E D fl H}.
Proposition (3.6.1) (The passing principle) Suppose P(pT) 54 0. Then there exists a polynomial P(z) such that P(-rGz) = P(z), P(p) 0, and if we define V (z) = P(z)V)(z), then E (f, r(1), w, 0') = =(.f, r(2), W, 01)
Proof. Let P(z) be the polynomial which corresponds to P(sT) by the above sub-
stitution. By assumption, P'(p) # 0. Let P"(z) = P'(-rGz). Since -rGp = p, P"(p)
0. Then P(z) = P'(z)P"(z) satisfies the condition of the proposition. Q.E.D.
Consider the following condition
0(Z)=0 (-rGZ), 1'(P):0. Consider the situation in §3.5. In later parts, we analyze the distributions
(3.6.2)
In this process, we have to study various contour integrals. What we can do by (3.6.1) is to replace the function O(z) if necessary to move the contour. By doing this, the condition (3.6.2) is not changed. Therefore, if for some L C Vks, we can prove that 8p (4), w, w) and
fA/Gk
w(91)OL(4),91)e(91,w)d91 ,., CGA(w;p)
ap(4),w), p
where ap w) is some distribution associated with the path p, then we can still conclude that a.($,w). w(g')OL($,g1)dg1 = JA/Gk
Caution (3.6.3) For the rest of this book, when we consider contour integrals of the form fRe(z)=q ' . dz or fRe(s,)=r ... ds r, we always consider w E C such that Re(w) > L(q) or Re(w) > L(r)
§3.7 Wright's principle In this section, we introduce a technique concerning the cancelation of higher order terms of distributions. We consider q,,,(s) for some n (see (1.1.2) for the definition). The use of the following proposition was suggest to the author by D. Wright after he read the first manuscript of [86] (it is used in [86]) Therefore, the author calls this proposition 'Wright's principle.'
Proposition (3.7.1) (Wright's principle) Let A E C, cl < 1 < c2, and e > 0 be constants. Let w E C be a complex variable. Let f (s) be a meromorphic function on the domain Is E C I cl - e < Re(s) < c2 + c}, having no poles along the lines Is I Re(s) = cl}, Is I Re(s) = c2}. Suppose that f (s) has a finite number of finite order poles and is rapidly decreasing with respect to the imaginary part of s. Also assume that there exists a constant A' such that 1
0' (s) .f (s)
2xRe(s)-C2 w - s - A ds - w -`q'1 - A
Part I The general theory
98
is meromorphic on a domain of the form {w I Re(w) > 1 + A - S} for some S > 0 and is holomorphic at w = 1 + A. Then f (s) is holomorphic at s = 1.
Proof. Suppose that f (s) has a pole of order j at s = 1. Since O (s) has a pole of order exactly one at s = 1, q5 (s) f (s) has a pole of order j + 1. Let 00
On (s)f (s) _
i--j-1
ai(s - 1)i
be the Laurent expansion at s = 1, where aj # 0. Then
f(s)
1
Le(s)_c2 w - 1 - Ads 1
f
-1
w-+h(w)+Eai(w-1-A)'
(S)Ads
JRe(8)=c1
i=j
for some rational function h(w) which does not have w - 1 - A in the denominator. Therefore, the proposition follows. Q.E.D.
We illustrate the use of this proposition by an easy example in the next section. Even though this proposition is very easy to prove, it does save a lot of labor for us in Chapters 12 and 13.
§3.8 Examples Before we start handling non-trivial cases, we consider two easy examples G =
GL(1) x GL(2), V = Sym3k2 and G = GL(2) x GL(2), V = k2 ® k2, applying what we have developed in this chapter. The first case was handled by Shintani in [64] and the proof in adelic language is in [83]. The prehomogeneous vector space G = GL(2) x GL(2), V = M(2, 2) has been studied by many people, but there does not seem to be any result which uses exactly the same formulation as ours. Cogdell [8] studied this case by the same method as ours (he used the group GL(2), but it is essentially the same as our case). However, his paper is written in classical language and he also chose a particular Schwartz-Bruhat function, so his analysis superficially looks very different from our analysis. Also the use of Wright's principle in the previous section simplifies the computation significantly. , wi) be a character of (Ax IV)', We first introduce a notation. Let w = (w1, and T E Y(A). For t = ( t1 , , ti) E Ai, we denote w(t) = i jj wj(tj). Let
s = (sl... si) E 0Definition (3.8.1) (1)
Oi(ww, t) = E T (tx). xE(kx )i
(2)
I tl sw(t)e(W, t)dxt.
Ei(W, w, s) = f Ax
/kx)i
3 The general program
99
If w is trivial, we drop w, and use the notation Ei(T, s) also. When i = 1, we define an analog of ZL+(-D, w, s) as follows. E1+(`I't w, s) =
I. p
/kx
ItI8w(t)ol(q1, t)dxt.
itI>1
If w is trivial, we may drop w and write E1+(,F, s) also. It is well known that E1+(WY, w, s) is an entire function, and
Ej('`,w,s)=E1+('I',w,s)+E1+(,Y,w-1,1-s)+S#(w)
(s(01
- s0))
where W is the standard Fourier transform of if. We refer to this formula as the `principal part formula for the standard L-function in one variable'.
Definition (3.8.2) For j = (jl,
so,i) E 0, we use , ji) E V j so = (so, 11 the notation Ei (j, ... ,ji)(1F, w, so) for the coefficient of 1!((s` - sl,0)ji in the Laurent expansion of Ei(W, w, s) at s = so. We also drop w in this notation when w is trivial. )
(a) The case G = GL(1) x GL(2), V = Sym3k2 Let T be the kernel of the homomorphism G -+ GL(V). We use the formulation in §3.1. Since dim G/T = dim V, Vk s = V,'. Therefore, this case is of complete type, and we use the notation Z((P, w, s), Z+ (4P, w, s) for the integrals defined in (3.1.8) for L = Vk 8.
In this case, we identify t* = to* with R so that a2(A-1, A)° = Aa for A E R. We
choose {a E R I a > 0} as ti. The Weyl group consists of two elements and the non-trivial element rG is the permutation (1, 2). We identify V with k4 by (xo, ... , x3) --+ fx(v1, v2) = x0v + xlv2 i v2 + x2v1v2 + x3v2. The weights of V are -3, -1,1,3. Clearly, 5Z \ {0} consists of two elements /31 = 3. Let 01 = (Nl),02 = ((l2). Let Pi3 = (01,51j),p2j = (D2,52j) be paths for j = 1, 2 such that sij(1) = 0 if j = 1 and sij(1) = 1 if j = 2. It is easy to see that Ma, = Mat = T and Walk = {(0) 0, x2, x3) I x2, x3 E k}, Y,,k = {(0, 0,x2, x3) I X2 E k", x3 E k},
Za1k = {(0,0,x2,0) I X2 E k}, Zalk = {(0,0,x2,0) I X2 E kX}, YD2k = Za2k = {(0, 0, 0,x3) I X3 E k}, ZO2k = {(O, 0, O, X3) I X3 E k"}.
Then by the general theory, Sark = Gk XBk YD k for i = 1, 2. We define a bilinear form [ , ] v by 1
1
[x, y]v = xoyo + 3x1y1 + 3x2y2 + X3y3
, y3). Then this bilinear form satisfies the property for x = (xo, , X3), y = (yo, [gx, tg-1]' = [x, y]', for all x, y. Let [x, y] v = [x, TGy]'. We use this bilinear form
as [,]vin§3.1.
Part I The general theory
100
We choose the constant in the definition of the smoothed Eisenstein series so
that C = C1 = 1. This implies that wo = 1. In this case, "rG = (1, (1, 2)) and CG = 921. We assume that (D E 9(VA) and MV,,I) = C By (3.5.3), (3.5.5), w, w), 8p,i w, w) are well defined for Re(w) >> 0 for i, j = 1, 2. Let - p;j J(4), g°) be as in (3.5.6). We define (3.8.3)
I0(4D,w) = f
w(g°)J((D,g°)dg°, p /C'k
I(4D, w, w) f
w(g°)J(, g°)8(g°, w)dg°.
Then by (3.4.34),
I(4i, w, w) . CGA(w; P)I
w).
By (3.5.9), (3.5.20), W, W)
EP (EP (-D, W, W) +
W, W))
P=P11,P12
+
EpEp (4), w, w) + CGA(w; P=pn,P22
By the remark after (3.4.7), "P12(4),w,w) =
P 2(4), w, w) =
Therefore, we only consider P11, 1121 Let d(A1) = (1, a2(AI 1, )1)), t ° = (t, a2(tl, t2)),
Where t,tl,t2 E A1, u E A, and to = d(A1)t°. We define d
Lemma (3.8.4) ;p (4),W, w) - 0. Proof. By definition,
e(t°,w) - 400,W) = '{1},TG,1(t°,w). By (3.4.30), there exists a constant 6 > 0 such that e(t°, w) - 8N (t°, w) is holomorphic for Re(w) > wo - b and if M > wo, l >> 0,
eN(t°,w)I «A1. By (1.2.8), for any N > 1, IeYa1 (Re,, iD, t°n2(u))I << Al Nl sup(1, Al 3).
3 The general program
101
This implies that for any/ l >> 0, N > 1, Jey,1 (Ral -P, t°n2(u))I !e(t°, w) - 6°N(t°, w)JA2 «,i N+2 SUp(1, Al 3) Therefore, the integral defining Ep11(4P, w, w) converges absolutely for Re(w) > w° S.
Q.E.D.
We fix a Weyl group element T. Let sT = zr(2) - z.,(1). Then it is easy to see that sr-1 0 2 -3 -1 2 0 Tz+ (3.8.5) apll (t) = A1A1 = Al , apse (to) = A1, (t ) P = Al Easy computations show that 1w(t °)f (tt1t)d° = b#(w) f /kx f (q)dxq,
J 1/kx)3 A
f
A1 /kx )3
w(t 0)f (tt2)dxt ° =
6(w1)6(w2)
f
1 /kx
w1(q)f (q)dxq
for any measurable function f (q) on Al /k 1. Therefore, p11,T(-D, w, w) b#(w) 27r
/- fRe(s)=r>3 E1
P21,1 OD, w/,
_
,
2)AT
T )ds T,
w)
S(w1)Slw2) 3
(Rs s, al
1
21rv'----l fRe(sr)=r>2
E1(Ra2 w1 ST
+
1)AT(w; sT)dsT.
3
w, w) - 0 If T = 1, DT = IR and L(r) _ -r. Therefore, by choosing r >> 0, for i = 1, 2. Suppose T = TG. Then AT(w; sT) = ¢(sT)A(w; si). The point p corresponds to ) are holomorphic at sT = 1. sT = 1 . The functions E1(Ra1 sT - 2), E1(Ra2 C The passing principle tells us that by changing z// if necessary, l,T (4), w, w)
27r
(V
)=r y J 0
E1(Ral 4P, sT - 2)A, (w; sT)dsT
+ CGA(w; P)6#(w)E1(Ra1 4), -1) ,., CGA(w; p)b#(w)E1(Ra1 -D, -1).
Similarly, 3
Ep21,T (,D, w, w) - CGA(w; P)
b(wl)3(w2) E1(Ra2
w1, 3
These considerations prove the following formula I(-D, w, w) - CGA(w; P)b#(w) (E1(Ra1 &, -1) - E1(Ra11), -1)) 3
+CGA(w;P)b(w1)3 (w2) (E1(R2,wi1, 3) - E1(Ra2'P,w113) + CGA(w; P)T2b#(w)($(0) - 'D (0))
Part I The general theory
102
Therefore, (3.8.6)
,-1))
E1(Ra,
1°(,D, w) = 6# (w) S(w133b(w2)
+
(Ei(&2&wi1, 3)
- E1(Ra2-D,w1i 3)
+ X125#(w)*0) - co)). It is easy to see the following relations
E1(Ra1,P),, -1) = E1(Ral,P, -1), E1(Ra1 Via, -1) = A-4E1(Ral , -1),
El(Ra2,%,w1, 3) _ A-2El(Ra24)1w11
3),
El(Ra26),wi 1 , 3) = A ,P a(0)
= -P (0), 4a(0)
3), =A-4$(0).
Therefore, by integrating .\8I°(I)A, w) over the interval [0, 1], we get the following theorem.
Theorem (3.8.7) (Shintani) Suppose P = Mv,w w, s) =
El (R,, F,-1) s-4
a + 8(wl)S(w2)
Then
+ X2126#(w)
w, s) +
+b#(w)
.
4'(0) s-4 s /
($(0)
- E1(Ral-P,-1)'
E1(Ra2$,w1 1, 2)
3s-10
J
8
3 - E1(Ra2'D,w1, 3s-2
(b) The case G = GL(2) x GL(2), V = k2 ® k2 In this case, we have a canonical measure on the group GL(2)°' x GL(2)00. Let T be the kernel of the homomorphismG -+ GL(V). We consider (GIT,V) as before. A2))(a,6) = In this case, we identify t* = t°* with R2 so that (a2(A-', A,), a2(A2 1, A A2 We choose {(a, b) E R I a, b > 0} as t* . Any Weyl group element is of the form T = (T1i T2) where Tl, 7-2 are either 1 or (1, 2). For x E V, let xij be the (i, j)-entry. We use x = (xij) as the coordinate system of V. The weights of V are (-1, -1), (-1,1), (1, -1), (1, 1). Let ((al, bi), (a2, b2)) = ala2+b1b2. Then with the metric defined by this bilinear form, the weights of V can be identified with vertices of a square. The index set 'B\{0} consists of three elements ,l3 = (1,1), ,3' _ (1, 0),)3" _ (0,1). Let = (/3). It is easy to see that Ma = T and Yak = Zak = {x I x11 = X12 = X21 = 0}. Also, Zkg = {x E Zak I X22 E k" I. Let pi = (a, si) be a path for i = 1, 2 MQ,, MQ in §3.2 is SL(2) and Zp,, Zpu such that 51(1) = 0, 52(1) = 1. For can be identified with the standard representation of SL(2). So, Z7 k, Za ,k = 0. Therefore, there is only one unstable stratum.
3 The general program
103
We define the zeta function by the integral (3.1.8) for V,,S, and use the notation Z((D, w, s), Z+(,D, w, s) where w = (w1, w2) and w1i W2 are characters of Ax /kx However, since Vks = 0, the convergence of the integrals is not covered in (3.1.4). It easily follows from the following estimate (which is a consequence of (1.2.8)).
Lemma (3.8.8) Let 4) E 9(M(2, 2)A) and g° = (91, 92) E GL(2)°o x GL(2)0. Let gi = kia2(µitil, µi 2ti2)n2(ui) be the Iwasawa decomposition of Gi fori = 1, 2. Then for any N1i N2, N3 > 1, (µlµ2)-N1
OM(2,2)k (D, go) <<
SUP(1, µi lµ2) SUP(1, /-t1p2 1) SUP(1, µ1µ2) 1152)-N3
+ (11µ2 1)-N2(/ti
SUP(1, µ1µ2)
for g° in the Siegel domain.
Let [x, y]v = Ei,j xijyi,j for x = (xij), y = (yi3 ). Then [gx, tg-1y]v = [x, y]v for all x, y. We define [x, y]v = [x, rGy]v. Then by defining g` as before, [gx, g`y]v = [x, y] v for all x, y. We use this bilinear form as [ , ] v in §3.1. We choose the constants in the definition of the smoothed Eisenstein series so
that C = (1, 1). This implies that w° = 2. In this case, rG = ((1, 2), (1, 2)) and CG = Q2 2. We assume that 4D E ?(VA) and C By (3.5.5), 3p, (oD, w, w) is well defined for Re(w) >> 0 and E p, (4, w, w) _
rp, 4),w-1, w) as before. Let d(X1, \2) _ (a2(Al
1,
o
A1), a2(A2 1, A2)), t = (a2(41, t12), a2(t21, t22))
for al,.\2 E R+, tip E Al for i, j = 1, 2. Let to = d(.1i A2)t °. We define dxt o dxto in the usual manner. We define I°(1),w), I(4), w, w) by the formula (3.8.3) (the set Vks is different of course). Then I(l), w, w) ,,, CcA(w; p)I°(d), w),
and by (3.5.9), (3.5.20), 1(4), w, w) = =-P, (4), w, w) - Ep, (4b, w, w) + CGA(w;
The element to acts on Z-OA by X22 - A1A2ti2t22x22 Therefore, Ep2 (-D, w, w) = 0
unless w is trivial. We fix a Weyl group element r = (rl, r2). Let Si r= Zirl (2) - zir, (2) for i = 1, 2.
Let dsr = dsrldsr2. It is easy to see that (3.8.9)
av, (to) = A
2A2
d(A1,.2)rz+i = A1.,-1A2,2-1.
,
We make the change of variable µ = A1A2, A2 = A2 and q = t,2t22 Then `Sr1+l`8,2+1 = `5,-2-8,1 /sr1+1 1 2
2
Since
w(t JR+ xA1 /k x
°)µ3r1+1 p1(Ra
P, pt12t22)dxµdxt ° = 6#(w)El(Ro1P, sri + 1),
Part I The general theory
104
by the Mellin inversion formula, (D, W, w) =
f
8#(w) 27rv -1
sT1 + 1)A, (w; STI, STI)dsT1.
e(s,1)=r1>1
More precisely, we are using (3.5.17).
For T = (1, 1), ((1, 2), 1), (1, (1, 2)), L(r1i ri) = -2r1, 0, 0 respectively. We choose r1 = 1 + S for a small constant 6 > 0. Then L(r1 i r1) < wo = 2. So upl (4),W, w) " 0 unless r = TG. Suppose T = TG. Then AT(w; s,1, sT1) = (sT1)2A(w; sit, sil) Let
J(`), sit) = E1(Ra$, sil + 1) - E1(Ra-D, sit + 1). These considerations show that I(-D, w, w) ^ CGA(w; P)`3 6#(w)($(0) -.p(0))
+ 6#(W)
27r/
f
I/,
e(s,l)=r1>1
J((D, ST1)4'(ST1)2A(w; sil, sil)dsii.
Therefore, by Wright's principle, E1(RD$, 2) = E1(RDID, 2), and
I(,D,w,w) ^ CGA(w;P)9326#(W)($(0)-,D(0)) + CGA(w; P)6#(w) (E1,(1) (Ra &, 2) - E1,(1) (Ra D, 2))
.
This implies that IO(,p,W) = 932b#(W)+O)
-.p(0))
+ 6#(w) (E1,(1)(Ra$, 2) - E1,(1)(RA, 2))
.
The following relations are easy to verify. E1,(1)(Ra4)A, 2) = )1-2E1,(1)(RD D, 2) - A-2(log A)E1(Ra4), 2),
E1,(1)(RaCa, 2) = X-2Ei,(1)(Ra$, 2) +A -2 (log A)E1(RD$, 2). Therefore, by integrating a3I°(4a, w) over the interval [0, 1], we get the following theorem.
Theorem (3.8.10) (Weil-Cogdell) Suppose Mv,4,oP = 4. Then w, s) = Z+(4), w, s) +
+ b#( w)
-
w 1, 4 - s) + 2?28# (w)
2E1,(O)(ROP, 2)
(s-2)2
(0) - -(0) s-4 s
+ E1,(1)(RaD, 2) _ E1,(,)(Ra(D, 2)
s-2
s-2
E1,(o)(Ra-D,2) = 0. If for some v E 9A., 4)v E Co (V, ), then Therefore, the poles of the associated Dirichlet series are all simple. Cogdell [8] spent most of his labor trying to establish the cancellations of order two terms based on the equality 2) = E1(Ra$, 2) explicitly. However, we used Wright's principle, and were able to avoid this cancellation.
Part II The Siegel-Shintani case
Chapter 4 The zeta function for the space of quadratic forms We state the result of this chapter here.
Let G = GL(1) x GL(n), and Vn = Sym2kn, where n > 1. We consider the natural action of GL(n) on V. (if n = 1, a E GL(1) acts by multiplication by a2). We define the action of a c GL(1) on V,, by the ordinary multiplication by a. This defines an action of G on V,,. Let T C G be the kernel of the homomorphism G -+ GL(V). We consider the prehomogeneous vector space (G/T, VV) in this chapter. For a character w = (w1iw2) of GA/Gk, 4) E S°(VnA), and s E C, we consider the zeta function Zv, (4), w, s) which is, by definition, the integral (3.1.8) for L = Vim" if n 2 and for L = V2k if n = 2. We show the convergence of those integrals in §4.1, because it is not covered by (3.1.4). The case n = 2 is of incomplete type and requires an `adjusting term.' We define the adjusted zeta function ZV2,ad((D, w, s) in §4.2.
Let Sn,i C V n be the set of rank n - i forms. Then Sn,l, , Sn,n-1 are unstable strata. Let Yn,i = Zn,i C V. be the corresponding subspace for S,,,i. Then Yn,i Vn_i. Let Z.,.-2,0 C Zn,n-2 be the set of points of the form {(0, x1i 0)} by the above identification. We identify Zn,n-2,o with the one dimensional affine space.
For 4) E Y'(VA), Ri4> E 9'(Yn,iA) = 9'(Zn,iA) is the restriction of 4) to Y(Yn iA). For 4 E S°(VA) we define Rn,n-2 0$ E 9'(Zn,n_2,oA) by the formula
Rn,n-2,0(x1) = fR,n_2(O,xi,x2)dx2. If n = 3, we define +026(wl)6(w2)E1(R3,24),w1, 2)
F3(4',w,s) = Z
2s-3
s
+ 5(wi)b(w2)
+ 6#(w)
ZV2,ad,(0)(R3,14', (w1,1), 3)
s-3
3E1(R3,1,0ID, 2)
2(s - 3)2
E1,(1)(Rr3,1,0-D, 2)
+
2(s-3)
)
If n > 4, we define Fn (4), w, s) = Tnb#(w)
0)
+ E (n - i)93 8(w1)6(w2) IV _iRn,i4', (w1, 1), n 2 (2s - n(n - i) i#l,n-2 + Zln_2S(wi)S(w2) ZV,,ad(Rn,n-2I, (w1, 1), n)
s-n
)
Part II The Siegel-Shintani case
106
2 + (n - 1)6(w1)b(w2)
#()
+
Zy _1,(0)(Rn,14>, (wl, 1), n
n-1) 2
2s - n(n - 1) -2b- 1) + E1,(1)(R,_2 04>, n - 1)
2(s - n)2 (nE1(R4
)
2(s - n)
The notation Zv _1i(0)(Rn,14>, (w1i 1), n (n
1) will be defined in §4.4.
In this chapter, we prove the following theorem.
Theorem (4.0.1) Let n > 3. Suppose that M ,
Zv
('D'
w, s) = Zv
+(,D,
4> = 4>. Then
w, 3) + Zv +(4', w
1,
n(n + 1) 2
- s)
- Fn($,W 1, n(n + 1) - s) - Fn(4>,w, s). 2
Therefore, the adelic zeta function has double poles in general. However, it turns and the support of 4>v is contained in Vk',' for out (see (4.7.6)) that if 4> some v E fiIO, 6#(w)E1(R',n_2,04>, n - 1) = b#(w)E1(Rn,n_2,0 , n - 1) = 0.
We consider the case k = Q. The above remark implies that the poles of the associated Dirichlet series are all simple. Consider w = 1, and drop w from the notation Zv (4), w, s) etc. Let V C VAS be the set of forms with signature (i, n i). We choose a measure on (G/T)A so that the zeta function defined by (3.1.2) coincides with the integral in (3.1.8). Let µ,,,(x) etc. be as in §0.3. Consider such that 4P is the characteristic function of Vy and 4>, E Co (Vi). 4> Then it is easy to see that
Xi(4', dn) = f D(y.)dy. = $.(0) Since Z , (4), s) has the residue Z7n4) (0) = Tn(D,, (0) at s = dn, the Dirichlet series
xEGz\Vz
µco(x)bx,oo
1
o, x)
IA(x)I00
has a simple pole at s = do with residue Zn. This means that the Dirichlet series ltoo(x)bx,oo E xEGz\Vz' 000(x) IA(x)Ion 1
with residue -. Therefore, by the Tauberian theorem, has a simple pole at s = we have the following theorem.
4 The zeta function
107
Theorem (4.0.2) (Siegel [69]) Let n > 3. Then lioo(x)bx,oo Doo (x)
:EGz\vi°nvi Io(=)I-<X
ti
n
7.-X
For discussion on the error term estimate, the reader should see [65]. In order to state the principal part formula for the case n = 2, we need the `adjusting term.' For this reason, we state the principal part formula of the case n = 2 in §4.2. In this case, the filtering process was carried out by Datskovsky, and (0.1.2) is the asymptotic formula for GQ-equivalence classes. For this, the reader should see [9].
§4.1 The space of quadratic forms Let G, Vn be as in §4.0. Let T C G be the kernel of the homomorphism G -y GL(Vn). We consider Vn as the space of quadratic forms in n variables v = (Vi,.. , vn). Let do = n 2 1 . We identify Vn with the space of column vectors of dimension do by the following map
x = (x1m)1<ji<j2
Let To be the image of GL(1) in G. For (t, g) E GL(1) x GL(n), we define Xv. (t, g) _
,
tn(det g) 2 if n is odd, and X (t, g) = t 2 (det g) if n is even. We can consider Xv as a character of GIT. The character Xv is indivisible. It is easy to see that (G/T, Vn, Xv) is a prehomogeneous vector space and Vnk' consists of nondegenerate forms. For n = 2, V2k consists of forms without rational factors. We use the formulation in §3.3. Since X*(G/T) is generated by one element, G' = GA and t = to. If there is no confusion, we identify GL(n) with its image in G by the natural inclusion map. Let Yn,i = Zn,i be the subspace of Vn spanned by the coordinate vectors of xl,n's for 1, m > i. We identify Zn,i with Vn-i. Let Sn,i be the subset of Vn consisting of rank n - i forms. Let Zn ik be the set of rank n - i forms in Zn,ikLet Zn n_2k be the set of rank 2 forms without rational factors in Zn,n_2k. Let Zn,n_2,ok = Zn,n_2k \ Zn n-2k. Let Zn n-2 ok C Zin,n_2k be the set {(0, x1i 0)} when we identify Zn,n_2 = V2. Let Z';' a_2 ok = {(0, x11 0) ( x1 E kx}. We define Sn n-2 k = Gk ' Z, ,n-2k Let Pn,i C G be the maximal Sn,n-2,stk = Gk ' ZSnnparabolic subgroup whose Levi component is GL(1) x GL(i) x GL(n - i), where GL(i) x GL(n - i) is imbedded in GL(n) diagonally in that order. Easy considerations show the following lemma.
Lemma (4.1.1) iT (1) V n k \ {0} = Vnk TTTT 1.1 lli
1 Sn,ik (2) Sn,ik = Gk XPn.,ik Zn ik for all i.
We define a bilinear form [ , ] v as follows. n [x, Y1 V.
xiiyii + 2 E xijyij.
= i=1
i<j
Part II The Siegel-Shintani case
108
Then [gx, tg-ly]v = [x, Y1' for all x, y E Vn. This can be proved easily by reducing to the case n = 2. We define [x, y] n = [x, rrGy] n. We use this bilinear form as [ , ]v in §3.1. Let JFn, g = (A, g1), g` etc. be as in Chapter 3. Let E .Y(VnA). Then the Fourier transform of the function 4P(g ) is A-dn9n4)(9 ). If n = 2, we use the notation etc. for the integrals in (3.1.8) for L = V2k. The convergence of this integral for Re(s) >> 0 is covered in (3.1.4).
Definition (4.1.2) Let n > 3. For (D E 9'(VnA), w = (w1, w2) as in (3.1.8), and s E C, we define (1)
Zn(",w,s) = f
(2)
Zn+(4, ")IS) =
9w(gl)Ov k( 9)d9,
Ge/C'k
f
-
GA/Gk
sw(gl)Ov;°
9)d9
A>1
We prove that if n > 3, Zv, (4), w, s) is well defined for Re(s) > dn, and Zn+(4, w, s) is an entire function. Note that this does not follow from (3.1.4).
Lemma (4.1.3) (1) Suppose n _> 3. Then there exists r = (r1, , rn_1) E Rn-1 such that rj < j(n - j) for all j, and that for any N > dn, there exists an estimate of the form n-1 119
vnk ((D, Akt) I << A-N [J (tit-1
i=1
for (2) If n = 2, there exists a similar estimate as above for Ovvk (,D, Akt). Also for any
N>3,
«A-Nt1t21
lOvzstk(4),Akt)I
for A E R+, k E St, t = a2(tl, t2) E T°+.
Proof. Let V = Vn. We choose 0 <' E S°(VA) so that I-D(akt)I << T(At) for k E S2, t E T. We consider T+ as a subset of GL(n)°A. Let ylm E t' be the element determined by the coordinate xlm. , An_1 = to-ltn 1. Then Let t = an(tl, , tn) E T+, and Al = t1t2 1, n-1
1
t1 = Al n ... An-1
_L tj=alnn=1...Aj-1
-L ...71-1 Ain
n-1
1
to=Al^...an_1 Suppose tltm = fj A"-j. Then 2 n-j n
lmj =
n2 n
n
.9
l< m <j,
l<j<m, j
4 The zeta function
109
If x E VkS, the convex hull of {rytm xi,,,, 54 0} contains the origin (but not necessarily a neighborhood of the origin). For I C {(l, m) 1 1 < 1 < m < n}, we define VI = {x E Vk I xlm # 0 if and only if (1, m) E I}. We consider I such that the convex hull Convl of {71m 1 (1, m) E I} contains the origin. We define I
CI('I',9) =
'@(gx)
xEV,,nvl
Then Ovk=(w,At) <
OI(I,At).
We fix I and estimate (3I(qf, At). Since Convl contains the origin, we choose clm > 0 so that >(l m)EI Clm Ylm = 0. We define C = E(l,m)EI Clm > 0. By (1.2.6), for any M > 0,
e1(w, At) << X-CM II t-M E(1,m)EI C'-'Y'- 11 (l,m)EI
(Atttm)-1
(l,m)EI
=A-cm-#' 11
(titm)-1.
(l,m)EI
We consider the possibilities when
E Xtmj <- -j (n - j)
(4.1.4)
(l,m)EI
Suppose that (4.1.4) is satisfied. We first consider the case n > 3.
Case 1. j < 2
Since n-2j>Oandn-j>0, Xlmj > -
Xlmj
j(n-j)(n-j+1) n
(l,m)EI
If equality happens, then (1, m) E I for j < l < m, and (1, m) V I for l < m < j. But
j(n-j)(n- j +1) > -j(n-j),
n and equality happens only when j = 1. Therefore, (4.1.4) implies that j = 1. If j = 1, n - 2j > 0. So (1, m) I for l = 1 < m. Hence, the only possibility for (4.1.4) to happen is when j = 1, and I = {(l, m) 12 < l < m}.
Case 2.j>2. Since n - 2j < 0,
E
(l,m)EI
X1mj>-j(n-j)(n-j+1)+(n-2j)j(n-j)=_j(n-j)(j+1) n
n
If equality happens, then (1, m) E I if and only if m > j. But
j(n - j)(j + 1) > -j(n - j), n
n
Part II The Siegel-Shintani case
110
and equality happens only when j = n -1. Therefore, (4.1.4) implies that j = n -1
and 1={(l,n) I1=1,...,n}. We have proved that if I#{(1,m)12<1<m
for
We consider the case n = 2. Clearly, t2 = A,, t1t2 = 1, t2 = ail. Therefore, 11(i,m)EI tltm = A for some p< 1 if and only if I = {(1, 2), (2, 2)} or I = {(2, 2)},
If n > 3 and I = {(l, m) 1 2 < l < m < n} or {(l, n) l = 1, , n}, Convj does not contain the origin. So we do not have to consider such possibilities. Since I
#I < do and the choice of M > 0 is arbitrary, for any N > dn, we can choose M > 0 so that CM + #I = N. Let rj = infl(>(l m)EI Xlmj ). Then this r satisfies the condition of (1). If n = 2, and I = {(2, 2)}, then Convl does not contain the origin. However, Convj contains the origin for I = {(1, 2), (2, 2)}. But if x11 = 0, x E V2Stk. So we do not have to consider I = {(1, 2), (2, 2)} for the first statement of (2). Since lt2 3 = t2 2 = Al 1 the second statement of (2) follows also. 11(l,m)EI(tlt'm)_1 = ti Q.E.D.
Proposition (4.1.5) If n > 3, Zv. (4D, w, s) converges absolutely and locally uniformly for Re(s) > dn, and Zv +(4), w, s) is an entire function. Proof. Let t, Aj be as in the proof of (4.1.3). We only have to show the convergence of
ARe(8)Ov°k (41,
21t)t-2Pdx Adx
t.
It is easy to see that t-2" = rjj X, j(n-j). So if A > 1, we choose N >> Re(s) in (4.1.3), and this proves the convergence of Z , +(1), w, s) for all s. If A < 1, we choose N = do + 6, where S > 0. Then the integral over the set {A I A < 1} converges absolutely if Re(s) > do + S. Since S > 0 is arbitrary, this proves the proposition. Q.E.D.
Let C(G°A/Gk, r) for r = (rl, , rn_1) E Rn-1 be as in (3.4.17) (we consider the path (0, 0)). By considering (4.1.3) on GA, we get the following lemma. Lemma (4.1.6) If n > 3, Ovnk (4), g°) E C(G°A/Gk, r) for some r such that rj > -j(n - j) for all j. Also Ov2k (4), g°) E C(G°A/Gk, r) for some r > -1.
§4.2 The case n = 2 In this section, we consider the case n = 2. We use the formulation in §3.1. In this case, Vk' consists of forms without rational factors. Let V = V2 in this section. Let t* = t°*, t°,1° etc. be as in §3.8(a). We identify V with k3 by (SO,x1,x2) - fx(vl,v2) = x0v1 + xlvlv2 + x2v2.
4 The zeta function
111
The weights of V are -2,0,2. Clearly, '$ \ {0} consists of one element ,Q = 2. Let
a = (Q). Let pi = (D,si) be paths for i = 1,2 such that sl(1) = 0,52(1) = 1. It is easy to see that M, = T and Yak = Z,k = {(0, 0,x2) I X2 E k}, Zap = {(0, 0,x2) I X2 E k" E k}.
Then by the general theory, Sak = Gk XB, Z. Let Vstk be the set of rational forms with no double factors but with rational factors. Then Vk \ {0} = Vk 11 Vstk ll S1k.
Let H C G be the subgroup generated by T and (1, TG). We define Yok = {(0, xl, x2) I x1, x2 E k}, Yok' = {(0, xl, x2) I x1 E k", x2 E k},
Zol k = {(0,x1, 0) I xl E k}, Z'os = {(0, xl, 0) I xl E k" }.
Then it is easy to see that Vtk = Gk xHk Z'ok
Consider the function a(u) in §2.2.
Proposition (4.2.1) Let F C Af be a compact set, and L C k C k. a lattice. Then there exists a constant N > 0 such that
1 > a (u) > N fi (I xI x
vEmR
2 11 (IxIy + Iuly)-l'
+ Iuly) v
vE9Rc
for of E F, x E L \ {0}. Proof. Clearly, sup (1, i Iv) = Ixly 1 sup(lxl,,, Iulv). There exists a finite set S C Off such that if v E Of \ S, xthen x, uv E ov for x E L \ {0}, u f E F. So if v E fit f \ S, then Ixly, Jul, < 1. Clearly,
[J sup(lxly,
NO
vES
is bounded by a constant M > 0. So
fl sup(l, Ixly) <_ M 11 Ixly 1= MIxJ o. vE9R f
vE9R f
Therefore,
a(x)>M-'IxIO-01 f (1+IxI2V) VE9RR
= M-'
11 vE9RR
2
J
(1+IxIy)-1
vE9Rc
(IxIy + Iuly) 2 II (IxIy + July)-1 vE9Rc
Since the inequality 1 > a(x) is clear, this proves the proposition. Q.E.D.
Part II The Siegel-Shintani case
112
Proposition (4.2.2) Let Xv = {9° = (t,92) E G°A I ltl(92)1 >_
a(tl(92)-1t2(92)u(92))}.
Then for any measurable function f (g°) on G°A/Hk,
f ,°1/Hk.f(9°)d9°
= f
.f(9°)d9°,
XV/Tk
if the right hand side converges absolutely.
Proof. Note that the above condition does not depend on the choice of the Iwasawa decomposition of 92. We fix a place v E 'YROO. The subset {9° = (t, 92) E G°A ltl(92)1 = a(t1(g2)-1t2(g2)u(g2))} is a measure zero set, because if we fix u(g2), and Itl(92)Iv for all v c 991 \ {v}, there are finitely many possibilities for Itl(92)Iv. Consider a TG-orbit {g2ig2TG}. we can assume that ltl(92)1 > It1(g2TG)I. Then I
if 92 E GL(2)°A,
92TG =k(92)TGa(t2(92), tl(92))tn(u(92))
=k(92)TGn(tl(92)-1t2(92)u(92)) . a(t2(92),t1(92))
=k(92)TGklt(tn(tl(92)-lt2(92)u(92))) - n(ul) . a(t2(92),tl(92))
for some k1 E K fl G2A, ul E A. Therefore, Itl(92TG)I = It2(92)Ia(tl(92)-1t2(92)u(92))
By assumption, ltl(92)1
It2(92)Ia(tl(92)-lt2(92)u(92)), and It2(92)I =
Itl(92)I-1
So if g° = (t,g2) and ltl(92)1 > Iti(92TG)I, then g° E Xv. Hence, Xv surjects to G°A/Hk. It is easy to see that Xv is invariant under the right action of Tk. Since a(tl(g2)1t2(g2)u(g2))} is a measure zero set, this proves the {9° ltl(92)1 = I
proposition. Q.E.D.
Definition (4.2.3) Let v be a place of k. Let IF, T, be Schwartz-Bruhat functions on A2, kv respectively. For s, sl E C and a character w of A" /k", we define (1)
(2)
TV,v('yv,w,s,s1) = f
, xk
ItvIvw(tv)av(uv)814'v(tv,tvuv)dxtvduv,
ItIsw(t)a(u)s1 (t, tu)dxtdu,
TV (IF, w, s, s1) = J x xA
(3)
Tv+(`Y, w, s, sl) =
"x xA
I tI $w(t)a(u)s1 WY(t, tu)dx tdu,
i
(4)
T(111, w, sl) =
f
w(tl)a(u)311Y(tl, tlu)dxtldu. 1xA
4 The zeta function
113
If lY = ®1Y,,, TV (IF, w, s, s1) has an Euler product as follows
s,sl) = kl - 162t k1 f TV,v('I'v,w,s,s1) V
In (4.2.4)-(4.2.6), we drop the index v from tv, uv, because the situation is obvious.
Proposition (4.2.4) Let v E JJIf. Suppose that 'v is the characteristic function of ov and w is trivial on kv . Then 1-q(s+s1)
v
Tv,v N'V' w, s, sl) _
(s+31-1)).
qv
(1 - qv s)(1 Proof.
TV, ('F"' w, s, s1) =
Jo Itly-l sup(i, jut-l lv)-3ldxtdu 00
00
_
qv
a(s-l)
(1 - gv l)qv 6 sup(1, qv
(6-a))-s1
a=0 b=0 0o
_
qv
a(s-1)(1
a-1
00
a=0
b=0 as1
(gv(s1-1)
Cqv
E qv -(s-1)(j -
qv1)
qv1-l - 1
-
vq-1)
a=O =0
(q_a(s+s1-1)(qa(s1-1)
00
E
qv1-1
a=0 00
qv
as(gv1-l
- qv
1)
4
a=0 qvs1-1
-1
- qv v1-1
a(s+s1-1)(1
-
+
a- l 1 qv - qv
+ qv asl
- qv
1)
1-qvl - 1)
s)(gv1-1
1 - qv
1)
1
-1 - qv
(1 - qv
E qv b) 6=a
00
-
a31+b(s1-1) +
- qv 1)(E qv
(1 _ qv
(3+31-1))(gv1-1
- 1)
(s+s1)
(1- qv s)(1 - qv
(s+s1-1)
). Q.E.D.
Proposition (4.2.5) Let v E'JJI f. Then Tv,v(W , w, s, sl) is a rational function of qv 3, qv 31 for all v, holomorphic for Re(s) > 0, Re(s) + Re(si) > 1.
Proof. Choose an integer l so that Tv(x1,x2) is constant on the set {(x1ix2) E A2 x11y, Ix2Iv < qv a}. Then, w(t)ItIv lav(t)311I'v(t,u)dxtdu Cqv
= Cqv
l d tdu w(t)jtjv lav (u) t1
is
(s
1 - qv (1 - qv s)(1
-
-(s+s1 )
q,-(s+"'- 1)) k>1
w(t)dx t
Part II The Siegel-Shintani case
114
for some constant C. There exist open sets U1, , UN C k,, such that Ui C qv io,x, for some mi < 1, and that if (X1,X2) is in the support of Wv and'x21v > qv-1, then x2 E Ui for some i. By replacing U1i , UN if necessary, we choose an integer l' > l so that l1 is constant on the set It I Itl < qvl' } x Ui for i = 1, , N. Then u 31
iti.,
w(t)Itl$-'a (t) F,(t,u)dxtdu
gv `
N
_
Cigv
*ni(81-1)qv
qv (8+s1-1))-1 /
w(t)dxt
ov
i=1
for some constants C1,
, CN.
Clearly, (u,-)81
w(t)ItI , lav t
qlv(t,u)dxtdu
qv duly
is an entire function. Therefore, (at)Sl
/ti
_ w(t)ItI;-lav
'F,(t,u)dxtdu
sgv
qv `
is a rational function of q,-s, qv8', holomorphic for Re(s) + Re(sl) > 1. Similarly, the integral over the set {(t,u) E k,, x kv I qvi < ItJv, Jul,,, < q,-,'l is a rational function of qv 8, qv8', holomorphic for Re(s) > 0. Since the integral over is an entire function, this proves the the set {(t, u) E kv x kv q;' < ltlv, proposition. I
Q.E.D.
Proposition (4.2.6) Suppose v E 9Jt,,,. Then for any w, Tv,v(Tv, w, s, s1) is holomorphic for Re(s) > 0, Re(s) + Re(si) > 1, and can be continued meromorphically to the entire C2. Proof. First we consider the case v E 9J1R. Since w(t) = 1 or sign(t) for t E R>1, we may consider integrals of the form
fof_oot8(1 + u2)- w(t, tu)dxtdu. C'O
The above integral is equal to the integral
t8+s1-1(t2 + u2)-zWv(t, u)dxtdu o
FOO f°° r z
= J0 J
iz
re-l cos881-2 B1Yv(r cos 0, r sin 9)drdO.
4 The zeta function
115
We define 100
fl(Wv,Z1,z2)=
r2
Trz1COS129Wv(rcos0,rsinO)drdO,
J
2
too
2
f2(1Yv,z1,z2)=J
,rZlcosz20sin9AYv(rcos0,rsin9)drd9. 2
Then 1
fl(Wv, z1, z2)
(fl(a1'Fv, z1 + 1, z2 + 1) + f2(82Wv, z1 + 1, z2)),
zl + 1
f2zl, z2) = Z2 11
(f2(a1wv, z1 + 1, z2 + 1) - fl(a2wv, z1 + 1, z2 + 2)),
where a1i a2 are partial derivatives with respect to the two coordinates. The meromorphic continuation of Tv(Wv,w, 5, Si) for a real place v follows from the above formula. w(re2nV---1e) = e2""v----io. Next we consider an imaginary place. Suppose that
Let t = r cos 9e2i, u = r sin
9e2"1/'- 102. Then
d"tdu = 4r tan9drd9d¢1dq52.
Therefore, oo
T, (IF,., w, s, sl) = 4
p2
010 Jo
2n
p2"
F(r, e, 01, 02, s, sl)drd9d0ld02,
0
where 9e2"V/--1nk1
=r2s-l
F(r, 9, 01) 02, s, SO
Costs+2s1-3 9 sin
9e2"N---1"0',
x 11v(r cos
r cos
9e2"Vr-l"O2 ).
The meromorphic continuation of TT (tII0, w, s, Si) is similar to the real case, and it is holomorphic for Re(s) > 0, Re(s) + Re(si) > 1. Q.E.D.
Suppose IF = OT,. If w is a character of A' /kx, w is trivial for almost all v. Let 9)1oo C P C ))t be a finite set such that w is trivial on kv and 4f, is the characteristic function of the set ov for v 0 P. We define (k,P(z) = jjv0P(1-qv Z)-1, andTv,P(W,w,s,sl) = IOkI 2ttk111vEPTv(1I'v w s,51). By (4.2.4), TV (`I', w, s, Si) = Tv,P(`I') w> S, Si)
(k,P(3)Ck,P(8 + 81
(k,P(S + Si)
Definition (4.2.7) For SY, w, s, sl, and P as above, we define (1)
Tv(q', w, s) _
(2) Tv,P(IF,w, s) _
d
d dsl s1=0
TV (IF, L.), S, S 1), TV' (IF, U))
d dsl s1=0
Tv,P(W,
Si).
ds1 s1=o
TV (`I', w, Si),
Part II The Siegel-Shintani case
116
By (4.2.4)-(4.2.6), we get the following proposition.
Proposition (4.2.8) For any w, T('Y, w, s, s1) can be continued meromorphically to the entire C2 and is holomorphic for Re(s) > 1, Re(s) + Re(sl) > 2
Let b ° = t °(1, n2(uo)) E B00, and b° = d(\,)-b° E B. We define measures db °, db° on BA, BO in the usual manner (db° = dx Aldx t 0duo)
Proposition (4.2.9) Letw= (W1,W2) be a character of (Ax/kx)2. Then (1)
i/T,.
J1
asw(b °)ezo(4i, Ab 0)a(uo)sldx.db 0
f3A
converges absolutely and locally uniformly for all s, si E C, f1
(2)
J0
f
Xsw(b °)Oz,
°)a(uo)37d Adb °
0/ Tk
converges absolutely and locally uniformly for Re(s) > 2 + e, Re(si) > -e, where e > 0 is a constant, and w(b0)ezo(-D,b0)a(uo)sldb0
(3)
converges absolutely and locally uniformly for all s1 E C.
Proof. Let o, = Re(s), of = Re(si). There exist Schwartz-Bruhat functions 4)1i (D2 > 0 on A such that E) z,, (4), A0) << 1 41(AX1)4)2(AXluo) x1Ekx
Since the order of the integration and the summation can be changed, we can make the change of variable ul = Axlu0. Then (1)-(3) are bounded by constant multiples of the integrals
f Ilf J
'Dl(Ax1)(D2(ul)a(A-1xllul)"d"\dul,
Au-1
A
x1Ekx
E 4)1(AX1)4)2(u1)a(A-lxllul)6ldxAdul, x1Ekx
JAx1Ekx Y'
4)1(xlA2(ul)a(xllul)"Idol,
respectively. There exist a lattice L C k and a compact set F C A f such that 'D1(Ax1)4?2(ul) _
0 unless x E L, ul f E F. Since the finite part of J.ul is the same as the finite part of ul, by (4.2.1),
axlw + IA-lull)-1
a(A-1x11211) ? N II (Ix1w + IA-lull) vEmR
vEfilc
4 The zeta function
117
for some N > 0. If o1 > 0, then a(A-lxl lul)°1 < 1. If of < 0, since (Ixlly + IA-lu'lly) 2 < At k:Q av(Ax1)°lav(ul)o1 for
v E 9A98,
2°,
(Ixlly + IA-lulls)-°1 < A[k:Q[av(Ax1)°'av(ul)°1 for v E
,
we have the following inequality
aWixl lul)°1
A°' 11 av(Ax1)°1av(ul)°1. v EM
Therefore, the proposition follows from (1.2.6) (consider A = 1 for Q.E.D.
For later purposes, we fix some notation. Definition (4.2.10) For (D E VA, we define Rod E Y(YOA), R''
E 9'(ZZA) by
Rol)(x, y) = P(0, x1, x2), Ro,D(xl) = f(0,xi,x2)dx2. The proof of the following lemma is an easy consequence of the definition.
Lemma (4.2.11)
(1) F (2)
f
f
a sw(b o)ezb (41, Ab o)a(uo)sldx Adb o =
b(w1w21)7,(Ro(P,
A/Tk
w(bo)Ozo(41,bo)a(uo)sldbo = A/Tk
Definition (4.2.12) For 4), w as above, and s E C, we define Est(4)>w>s) = 6(w1w2 1)T1(Ro4),w1i
Let
1-s ). 2
00
Est(4),w, Z) = E Est,(i) (4),w,1)(z - 1) i=o
be the Taylor expansion around z = 1. Then (
-s`w12 2
1
_
Definition (4.2.13) ust(4', w, w) = f
w(go)Dv,st(4), go)8(go, w)dgo. A IGk
wl> s, sl),
Part II The Siegel-Shintani case
118
Then by (3.4.31),
,
t(4), w, w) is well defined for Re(w) >> 0. As in §3.8,
-pi (1), w, w) is well defined for Re(w) >> 0 for i = 1, 2, and
We define
J8(4', g°) = E ($((g°)`x) - -b(g°x)) , XEGk\Vk
w(g°)J8(,g°)dg°,
I°(-D,W) = f A/Gk
I(4), w, w) = f A/Gk
w(g°)J8(4), g°)e(go, w)dgo.
Then 1(4), W, w) ,,, CG (W; P)I°(1,W),
and by (3.5.21) and the argument of (3.5.9), I (d), w, w) = Epi ($, w-1, w) - Epl (4), W, w) + CGA(w; P)9J2S#(W)($(0) - 4)(0)) _1
Definition (4.2.14) Let 1b, w, s be as before. We define lS(W1W2 1)
ZV,ad(C W, 8) = Zv(D,W, s) -
TV(ItA,Wl, s)
2
We call ZV,ad(1b,W, s) the adjusted zeta function, and b("12 21)Tv(R°4), w1i s) the adjusting term. We devote the rest of this section to the proof of the following theorem.
Theorem (4.2.15) (Shintani) Suppose P = M ,,D. Then ZV,ad(C W, S) = ZV+(qb,W, S) + Zv+(4', w-1, 3 - s)
- 8(wl2 w2 1) (Tv+(Ro(b, wl, s) + Tv+(Ro$, wi 1, 3 - s)) (U3 (0)\ +28#(W) b(Wl)S(W2)
+
2
Wl 1, 1)
(s-2)2
6(W1)6(W2) 2
+ 1)
(s-1)2 (E',)(R,w1,
+
s-2
E1,(o)(R1) s-1
1, 1)
f .
Since the first four terms of (4.2.15) are entire functions and we proved the meromorphic continuation of Tv (R°4), wl, s) in (4.2.8), Zv (-D, w, s) can be continued meromorphically everywhere.
4 The zeta function
119
Corollary (4.2.16) The adjusted zeta function ZV,ad(4), w, s) satisfies a functional equation ZV,ad(', w, s) = ZV,ad(D, w _ 1,3-S).
We first consider
p, (4), w, w). The element to acts on ZSA by x2 -* )1b2tt2x2.
Let T be a Weyl group element. Let sT be as in §3.8 (a). It is easy to see that apl (to) = A2 (to)TZ+P = A '
'
Also if f (q) is a function on Al /kx,
f
w(t°)f(ttz)dxio
A1 /kX)3
= 6(co1)6(w2) f 1 /kX
wl(q).f(q)d"q.
Therefore,
w, 'w)
p1,7-
=
6(W2
1)6(W2) 2 27r
1
1
f
E1 (Ray w1
sT + 11
Re(s,-)=r>1
2
l AT(wjsT)dsT.
Ifrr=1,wecanchoose r>> 0andL(r)=-r<<0. So Bp,T(4D,w,w)-0. Next, we consider °st(-D, W, W).
Proposition (4.2.17)
w, w) - E
1
T
-1
J e(sr)=r>1
w, z)Ar(wi sT)dsT.
Proof. By the standard consideration,
w(g°)ev,(, g°)e(9°, w)dg°
=st(4), w, w) = f A/Gk
fGl/Hk JXV/Tk
w(g°)Oza
g°)e(go,
w)dgo
w(g°)Ozo (iD, g°)e(g°, w)dg° w(b°)Ozo (4), b°)e(b°, w)db°.
'XvnBA/Tk
Lemma (4.2.18) The integral fXy nBA/Tk
w(b°)Ozo (,D, b°)(oo(b°, w) - eN(t°, w))db°
is holomorphic for Re(w) > 0.
Proof. As in §3.8 (a), there exists a constant 6 > 0 such that e(t°, w) - &N (to, w) is holomorphic for Re(w) > w° - 6 and if M > w°, l >> 0, I .-(to, w) - eN (t°, w) I << All.
Part II The Siegel-Shintani case
120
Since tl,t2 E A' IV which is a compact set, by (1.2.3), there exist SchwartzBruhat functions 1D1i 4D2 > 0 on A such that
Oz., (4)fb1)e1(bofw) << E 4D1(xl)4'2(xluo)Ai x1Ekx
Then the above integral is bounded by a constant multiple of 2duo
JAx1Ekx E =
xlx
fi(xi)2(xiuo)a(uo)duo
f
=
2dul,
where ul = x1uo. Therefore, the convergence of the above integral follows from (4.2.9). This proves the lemma. Q.E.D.
If f (q) isf a function on A1/k<, wl(q)f(q)dxq.
f1/kx)aw(t0)f(tt1t2)dxt0 = 6(wl)6(w2) J1/kx
Also, `3T_ldx/1\1
I,>
^ Ck(up)
=
a(uo) 2 ST - 1
Therefore, by the above lemma and (4.2.11), 1
st((Df U), w) T
ST)dST.
ST -
27rV -1 fRe(sr)=*>1
This proves Proposition (4.2.17). Q.E.D.
It is easy to see that
A('wfs)ds,. ^ 0
1
27r
IRe(s4r> 1
sT
if r Let T = TG, and J(-D, w, sT) -
6(wl 21) (E1
(R&wif 2J - E1 (RwlT + 1 2
1
+ ESt(sfw )s ) ST-1
- F-St(-D, wf'ST) ST-1
4 The zeta function
121
By the above considerations, I0(oD, w, w) ^' CGA(w;
4D(o))
J(,D, w, s,)O(s,)A(w; sT )dsT. 1 J 27rV -1 Re(s,)=r>1
+
By Wright's principle, J(4P,w, sT) must be holomorphic at sT = 1. Since
El(IF,w1i
2E1,(-1)(W, w1,1)
sT + 1 2
sT - 1
)
+ E1,(o)(W,wl,1) +O(sT - 1),
we may conclude that w, w) ' CGA(w; 6(w1 6(w2)
( E 1,(°)( Ra
p)
(
w-1 ,
1
,
1
) - E 1,(0)( Ra
,
w1, 1 ))
w-1, 1) - Est,(1)(4), w,1)).
+ CGA(w;
Hence,
I0(4,w) _ 9326#(w)($(0) - 4)(0)) +
6(wl)6(W2)
1
2
,1) - E1,(o)(Ra41,wl,1))
+ Est,(1)(41, w-1, 1) - Est,(1)(It, W, 1).
It is easy to see that qla(o) _ (P(O), oa(O) = , wl,1) _ A-1E1,(o)(Ra1, w1i 1) - ))-1(1og A)E1,(_l)(R,P, w1,1), E1,(o)(RD
A-2E1,(o)(Rah,wl1,1)+A-2(logA)E1,(_1)(RD$,wi 1, 1). Also,
1
f
As,Da(O)dxA _
1)s4)a(t1)dXA
_
0
1
f AsEl,(o)(R, A,wl,1)d'A =
E1,(o)(RDP,wl,1)
+ E1(R5 , '1,
o
1
J
AsE 1,(0)( Ra a,
w-1 1 dx = El,(o)(RD$,w1 1, 1) ) 1
,
s-2
0 1
A8Est,(1)(I A, w,1)dx.A _ - 6(w12
1
f
0
1
A_-
21)
-
1),
_ E1,(-1)(Ra_,wl 1,1) (s - 2)2
(7'v(Ro41, wl, s) - 7'v+(RooD, w, s)),
6(wlw2 1) Tv+(Ro$, wi 1, 3 - s), 2
Part II The Siegel-Shintani case
122
because the order of the integration with respect to A and the differentiation with respect to sl can be changed by (4.2.9). This finishes the proof of (4.2.15).
§4.3 3-sequences In this section, we introduce some notations related to /3-sequences. The unstable strata of V. are Sn,,1, , S,,,,.-1, and Y,,,,i = Vn-i for all i. So we identify /3-sequences with subsets 0 _ 0 1 ,--- , ja} C {1, , n} where 1 < j1 < < ja < n. With this identification, YD = Vn-ia. If 0 and l(cl') = l(a) + 1, Pa, C Ma is a maximal parabolic subgroup. Therefore, all the ,3-sequences satisfy Condition (3.4.16) (1). Hence, Ep ((D, w, w) etc. are well defined if Re(w) >> 0. We choose da1(Aa1), , daa(Aaa) in the following manner da1(Aa1) =
da l Ail
a
In
in
-ji
\- n-i2) r.
da2(Aa2) =
'D 2
Iia-1
do,, (AD.)
'a
_
via-ia- In-ia
(n-ia)Iia-ia-1
-aa Let
AD = do (Aa1,...,Aaa) = dal (Aa1)...daa(Aaa),
AD = {A5 = d(Aal, ... , AD.) I Aal, ... , AD,, E R8 }.
Let go E Mam fl G. Then go can be written uniquely as go = Aaga°, where A5EAa,9a°EMA. We define (4.3.1)
aa1 = jl(j2 - j i )
( j a - j . -
-
ja),
t at = 2aj1(j2 jl) "' U. - j.-1), trla3 = trf51t352 = 2-a(n - j1) ... (n - ja). Let dx Aa = dx Aa1 ... dx AD,. We define an invariant measure dga on MDA fl G°°
by dga =,'ldxaadga. This measure satisfies the condition after (3.3.11). We define 'XJa = 331Qji2-il .. Tia-ia-1. Let S2a etc. be as in Chapter 3. If 0 = {i}, we also use the notation Fn,i instead of go. , ja} be a /3-sequence of length a > 0. Let jo = 0, ja+1 = n. We Let i) = {ji, define constants fai, hai in the following manner (4.3.2)
(n - 9i-1)(n - ji)
(n - ji)(n - ji + 1) fai =
2
,
hai =
2
for all i = 1, , a. Let Is = faa, ha = has We define a constant cp for each path p of positive length inductively in the following manner.
4 The zeta function
123
Suppose 1(p) = 1. Then (4.3.3)
cp
1i-za. 2
s(1)=1.
If p -< p' and 1(p') = 1(p) + 1, cp, =
(4.3.4)
j
2(ha ha,) cp
5'(a + 1) = 0,
2(ha+ha,) fa) cp
s-'(a + 1) = 1.
For w = (W1iW2) and a path p = (0,s), we define Sp(w) = 6(w2Db(w2). Let wp = (w1, 1) if #{i I s(i) = 1} is even, and wp = (wj 1, 1) otherwise. Let Ti be the set of p's such that Z = {1, , a -1, i} for some i # a, n - 2, and l(0) > 0. Let 312 be the set of p's such that 0 _ {1,... , a} for some a < n-3, , and
l(z) > 0. Let 3 be the set of p's such that 0 = { 1, ,a-1,n-2} where a < n-3. 933 is the empty set if n = 3. Let 3 be the set of p's such that a = {1, 2,- , n-2}.
§4.4 An inductive formulation We will formulate an inductive way of proving Theorem (4.0.1) in this section. We first have to introduce some notations. Let
Zv (4i, w, s) = E ai(s - so)
(4.4.1)
be the Laurent expansion of Z n (4i, w, s) at s = so. Then we define Zv,(_) (ID, w, so) = ai.
Of course, when we consider these values for a particular n, the meromorphic continuation of Zv. (4), w, s) has to be known. However, since we prove the principal part formula by induction on n, we do not logically depend on the meromorphic continuation proved in [60], [64]. We use a similar notation ZVZ,ad,(i) ('D, w, so) for the adjusted zeta function for n = 2. Suppose that 0 = {jl, , jQ} is a ,3-sequence such that ja = n - 2. Then we identify Za with V2.
Definition (4.4.2) (1) We define Z' ,o c Za to be the subspace which corresponds to Zo in §4.2 by the
above identification. (2) For T E Y(ZDA ), Ra,OW E 9(A), R°'Y E .'(A2) are functions which correspond to R'V2 ' OWY E 9'(A), ROW E Y(A2) in §4.2 by the above identification. We define Z'asok similarly.
Let J(4 D, g°) be as in (3.5.6). Let 9(g°, w) be the smoothed Eisenstein series which we defined in (3.4.7). We define (4.4.3)
IO(4),w) = f
GA/Gk
1(4), w, w) = f
w(g°)J(4,g°)dg°, w(g°)J(D, g°)e(g°, w)dg°.
A /Gk
Part II The Siegel-Shintani case
124
We remind the reader that 1
Zv (4b, w, s) = Zv+(,D, w, s) + Zv+($, w-1, do - s) + f A'I0(4>A, w)dx A. 0
Since
_ g°) - °v k ($, (g°)`), by (4.1.3), J(lk, g°) E C(G°°/Gk, r) for some r = (r1, , rn_1) such that rj > -j(n - j). Therefore, we can use Shintani's lemma for GL(n). The statement of J(-P, g°)
(3.4.34) implies that I(41, w, w) - CGA(w; p)I°(1, w).
By (3.5.9) and (3.5.20), we get the following proposition.
Proposition (4.4.4)
I
E
w, w) _
4)(0))
Ep,rp (4', w, w) + b#(w)A(w;
pE'a31Ual3U`a34
+E E
E Ep(2p#(4),w,w)-w,w))
pE`.az
Note that since all the paths satisfy Condition (3.4.16)(1), ^p($,w,w) is well defined for Re(w) >> 0 for all paths p.
Definition (4.4.5) (1) Let p E'Pi and l(p) = a. We define Ip(-P,w) = CpUabp(w)Zv
(Ra4)p,Lop, ha)
(2) Let P E'432 and 1(p) = a. We define 1p+ (41, w) = cpbp (w) (Zv a+(Ra,Dp, wp, .fo) + Zv-a+(ARo Dp, wp1, 0))
,
qP.fa0)
Ip# (4', w) = Cpb#(w)
Ipl (1)' w) = - c p'JJ n-a- 2 b#(w) + Cp93n-a-2b#(w) w)
Cp'un
(n -
a - 1) 2(n - a)2
E1,(1)(Ra, oRa,go4bp, n - a - 1) 2(n - a)
(n - a)E1(R0,,oRa,4)p,n - a - 1) a-2b#(w) 2(.fa - (n - a))2 E1,(1)(Ra,,oRa,-bp, n - a -
c
p'XTn_a_2b#(w)
2(fa-(n-a))
where a' _ { jl, , ja+1} is the unique ,3-sequence such that t and ja+1 = n - 2.
1)
,
-
a', l(0') = l(0) + 1
4 The zeta function
125
(3) Let P E X33 and l(p) = a. We define Ip (4), w) = cpJn-a_2bp (W)ZV2,ad(RaI)p, wp, n - a + 1).
(4) Let P E 4+4 and l(p) = a. We define Ip(4,w) = cpbp(w)ZV2,ad,(0)(Ra,Dp,wp,3).
Proposition (4.4.6) Suppose My ,,,4D = 4). Then
I°(1,w) =
4)(0)) + E 10(C w) pETi
+
2 n-2b#(w) ( Fl,(l)(Rn,n-2,0 , n - 1)
+
- E 1 (1)( Rn,n-2,0 n-1) ) ,
(Ip+(1), W) - Ip# (4, w) + Ip1(-D, L11) + Ip2(D, w)) PET2
+
Ip(,D,w)+ PET3
PET4
We devote §§4.5-4.7 to the proof of (4.4.6). For the rest of this section, we prove that (4.4.6) implies Theorem (4.0.1) by induction on n. The following lemma is the basis of cancellations of various distributions in this chapter.
Lemma (4.4.7) Let dx = fv dx be the ordinary measure on An, and dxt = H d" t, the ordinary measure on Ax. Suppose that I E Y(An) is invariant under the action of the standard maximal compact subgroup of GL(n)A. Then
f
4)(x)dx = Z (n) n
f
x
ItITD(t,0,... ,0)dxt.
Proof. By the choice of our measure, IA"'D(x)dx = I
kl
f
2
4b(xv)dxv, v
, 0)dx t.
fAX tIn(t, 0, ... , 0)dx t = it- l 1 f Itv l(ty, v
Let v E 9)l. It is easy to see that fk= (1 - q1)n v
Let fn(i) = (1 - qv 1)n
q(xii
it
qv
in
Then
00
fn(i) = (1 _ qv 1)n E
qvi-i2-..._in
i2 ,... in =i 00
+ (1 - qv 1)n
qv ii =i+1 inf(i2, ,in)=i
_ (1 - qv 1)q, ni + qv (i+l)fn-1(Z)
ii-i2-..._in
Part II The Siegel-Shintani case
126
By induction, fn(i) _ (1 - qv n)qv n. Therefore,
f,,.(i)4'(,rv 0 ... 0) _ (1 - qv n)
fk X
Itvly,P (t,,, 0, ... 0)dxt2,.
Next, we consider the real place. Let Dn be the unit ball in R. Then vol(Dn) _
it2F(2 +1)-1. So, ( xR)dxR =
fR ^
f
x+ + xR, 0,
,
0)dxR
R
r = nvol(Dn) J
(r, 0,
,
0)rndxr
= 2vol(Dn) fR ItRIR4)(tR,0,0)dxtR x 0, ... , 0)dxtR.
n r- 21(2) ftx
Finally, we consider the imaginary place. Note that dxc is two times the usual Lebesgue measure. As in the real case, Ixnclc, 0, ... , 0)dx,
-D(xc)dxc = f ( = 2nvol(D2n)2n f
r2n-P(r, 0,
(7r)
( ) fcx (27r)nln
,
0)dxr
tcInD(tc,0,... ,0)dxtc. I
Therefore,
(x)dx = Zk(n)1(27r)T2 IA"
9k
0)dxtv
J
Itln.p(t,0,... ,0)dxt,
Zk(n) fAx
because
(27r)T29tk.
Q.E.D.
Lemma (4.4.7) is essentially the relation between integrals with respect to the cartesian coordinate and the polar coordinate. Easy considerations show that (4.4.8) E1,(1)(Rn,n_2,o6a, n - 1) _
n_2 0
+
A-" 2
n - 1)
(log A)E1(R' ,n_2,A, n - 1),
El,(1)(Rn,n-2,o4)a, n - 1) _ A E1,(1)(R.,n_2,oob, n - 1) - A-n( logA)E1(R.,n_2,04, n - 1),
4 The zeta function
Zv (Rn,iCA, (W1 1, 1), n
_A
2
1+
2
(Rn,i,'),
Zv _i
127
n(n - i) 2
)
ZV _i (Rn,i-D, (WI
1, 1), n(n2
i)
),
n(n - i) ) (W 1, 1), 2
=A-
Zv, (Rn,i D, (Wl, 1), n(n2
2)
),
for i# 1,n-2.
Ifn=3,
(4.4.9) ZV2,ad,(o)(R3,1ia, (W1 1 1), 3) = A-3ZV2,ad,(O)(R3,1$, (W1 J),3) + )-3 (log A)9326# (W ).° 3,1R3,1$(0), Zv2,ad,(o)(R3,lDa, (W1, 1), 3) = A-3ZV2,ad,(O)(R3,1'D, (WI, 1), 3)
- A-3 (log A)'Z726# (W ).
3,1R3,1D(0).
Ifn>4, (4.4.10)
A-dn4(0), DA(0) = qD (0), (0) _ _1 ZV2,ad(Rn,n-4a, (W1 1), n)
n(n-,)
_
2
-1
ZV2iad(Rn,n_21, (Wl
1),n),
ZV2,ad(Rn,n-2'a, (w1, 1), n)
= A-"ZV2,ad(Rn,n-2', (01, 1), n), 1 n(n - 1)
Zn_,,(o)(Rn,l4)a, (Wl
-A-nZVn-,
,
1),
2
(0)(Rnl, (W1 _1, 1)) ,
)
n(n -1) 2
+ A-n(log A),Zn-lb#(W)9n,1Rn,1$(0), n(n - 1) 2 ) Zn-,,(o)(Rn,lca, (1)1,1), _ n(. -1) n(n - 1) 2 Z n-,,(0)(Rn,l I, (W1, 1), = 2
- A-
nc2 ,)
(log A)Tn-lb#(W)9n,1Rn,14D(0).
Lemma (4.4.11) If $ is K-invariant, (1)
Zn_2F_,1(Rn n-2 0'p, n - 1) =
(2)
21n_2E1(Rn,n-2,0$, n - 1) = 93n_19n,1Rn,1(D(0).
Proof. Note that 9n-2
'1(x12, ...
_1 = zk(1 1). We define `F1 E Y (An-1) by
,
x1n) =
JA 4)(x11, x12, ...
, x1n, 0, ... , 0 dx11.
Part II The Siegel-Shintani case
128
Let W2(x) = 1Y1(x, 0, , 0). By assumption, W1 is invariant under the action of the standard maximal compact subgroup of GL(n - 1)A. Easy considerations show
that E1(R71
n-
,9I'n,1Rn,1$(O)
=f
n - 1), W1(x12,"' ,xln)dx12...dxln.-
An-1
Then the first relation follows from (4.4.7). The second relation is similar. Q.E.D.
Definition (4.4.12) For n = 3, we define 3
{{/ J3(I',w) = Q36#(w)D+ (0)
2)
+ 6(W21)6(W2)ZV2,ad,(O)(R3,1(D, (w1, 1), 3)
+
#(w) E1,(1)(R3 1 0IP, 2).
For n > 4, we define fn(1D,W) = 9n8#(w)'D (0)
+
(n - i)1i8(wi)6(w2)
ZV
-i (Rn,iob, (w1,1),
n(no
i#1,n-2
+
Zn-26(wl)6(W2)ZV2,ad(Rn,n-2', (wl, 1), n)
+ (n - 1)S2 -1
(w2)
ZVn-1,(o)(Rn,lD, (w1,1), n(n2 1) )
+ Zn-26#(w) E1,(1)(Rn 2
n - 1).
Also we define (4.4.13)
1, 1(4,w)
E Ip(4''w)+ 1: Ip(F,w)+ E Ep1
PEP3
1(p)>1
1(p)>1 I a(1)=0
s(1)=0
PE'p4
s(1)=0
+ > (Ip+('D,w)-Ip#('D,w)+Ipl(1',w)+Ip2(4,w)), pEP2
s(1)=0
E Ip('D+w)+ pEP1 1(p)>1 s(1)=1
+
E
IEP3
1(p)>1
IpN,w)+ E Ip(4),w) pET4 s(1)=1
s(1)=1
(Ip+('D, W) - Ip#(D, L") + Ipl(4), W) + Ip2('D, w)) IEP2
s(1)=1
a) )
4 The zeta function
129
We consider three statements (An), (B.), (Cn). The statement An is as follows.
I°(-D,w) = fn($,w-1) - fn
(An)
The statement (B3) is as follows. I3,1(P, w) = -b(wi)b(W2)7'V2,ad,(0)(R3,1'P, (wi, 1), 3),
(B3)
I3,1(.p,w)
=6
1,
(wi 1), 3).
The statement (Bn) for n > 4 is as follows.
(Bn)
In,1(-D, w) _ -
In 1(4', w) -
(n - 1)b(wl)b(L02)
(n -
Zv, -,,(o)(Rn,l41, (w1,1), n(n2 1) ),
Z n-1,(o)(Rn,i
n(n2
1) )
, (w1 1, 1),
The statement (C3) is the empty statement. The statement (Cn) for n > 4 is the statement of Theorem (4.0.1) for Vn_1.
Proposition (4.4.14) The statements (An), (Bn), (Cn) are true for all n.
Proof. Clearly, (4.4.6) and the statement (Bn) imply the statement (An). So we only have to deduce the statements (Bn+l), (Cn+i) assuming the statements (An), (Bn), (Cn), and (4.4.6). Consider the case n = 3. In this case, T2i 33 are the empty set. Therefore, (B3) is Definition (4.4.5)(4). Now we consider the step from n to n + 1. Let 4 be as in Chapter 3. We multiply as to I°(4) a, w) in (An) and integrate it over A E [0, 1]. Then (Cn+1) follows by the relations (4.4.8), (4.4.9), (4.4.10). 1 Let 4) E Y(Vn+1,e), and W1 = Rn+1,1'D,'F2 = Rn+1,i . Let fn = n , hn,i = n n-i . By applying (Cn+1) to `I'1 and using (Bn), we get the following lemma. 2
2
Lemma (4.4.15) If n > 4, b(Wl)b(w2)Zn,(0)('1, (w1, 1), fn)
=
6(w2
flb(w2)(Zn +0I'1, w, fn) + Zn+(Wi, w, 0))
- 93n6#(w)
f
(0) n
+ E (n - i)
ib(wi)S(w2)
i961,n-2
(n - i)Zj6(wi)b(w2) i541,n-2
(wl 2hn
1
1), hn,i)
Zn (Rn,iW1) (w1, 1), hn,i) 2(fn - hn,i)
Part II The Siegel-Shintani case
130
(w1 1, 1), n)
+n-26(W1)6(W2) 2
- Tn_26(W1)b(W2)
+
In,1('F 1,W)
hn,n-2 ZV2,ad(Rn,n-2W1, (Wl, 1), n)
f, - hn,n-2
- In,1(W1,W) fn - hn,l
hn,l
+n_2/5#O W
-
nEl(Rn,n-2,o1, n - 1) + El,(1)(R',n-2,o1, n - 1) 2h 2
2hn,n_2
n ,n-2
-_2S #(w) (nE'(R2i,n 2(fn - hn,n-2)2
n,n-,oq1- 1)
+ E1,(1)(Rn-2,oW1, n - 1) 2(fn - hn,n-2))
)
/
For a path p = (a, s) for Vn, we associate a path p' = (D', s') for Vn+l so that
0'={1}U{i+1IiED},s'(1)=0,ands'(i+l)=s(i)fori=1,...,l(p). Any
path p' for Vn+1 such that 1(p') > 1 and s'(1) = 0 can be obtained in this way. By definition, cp' = - z cp.
Let pi = (0',s') be a path for Vn+l where 0' = 1}, s'(1) = 0. If we multiply (4.4.13) by - z , the sum of the first two lines and the last two lines is Ipi+(4),W) - Ip1#(4),W) + Ipi1(b,W) + Ipi2(D, W)'
The sum of the third and fourth lines is equal to
E
W),
p ,E'J1 1(p')=2 s(1)=0
where we consider paths for Vn+1. The sum of the fifth and the sixth lines is equal to Ip' (-P' L`')' p'E'V3
1(p')=2 a(1)=0
where we consider paths for Vn+1. Finally,
n In,l(Fl,w) 2
hn,l
n In,1('1,W) + 2 fn - hn,l
gives the rest of the terms in (Bn+l) for In+1,1(,D, w). This proves the step from (Bn) w) is similar using to (Bn+1) for n > 4 and In+i,i(-P, W). The argument for WY2. The step from (B3) to (B4) is similar using F3(1, w, s). This proves (4.4.14). Q.E.D.
Now we begin our analysis, and study the distributions ,=p ((P, w, w) in the next three sections.
4 The zeta function
131
§4.5 Paths in 131 We mainly consider paths in X31 in this section. We first determine o'p for an arbitrary path p. Lemma (4.5.1) Suppose up (AD) = fa 1 epi (Xa )x"' . Then
_
Xpi -
fori=
hai - hai+l
s(i + 1) = 0,
hai + hai+1 - fai
s(i + 1) = 1,
and Xpa =ha.
Proof. We prove this lemma by induction with respect to 1(p). Suppose that p = (-0,s) -< p' = (-0',s') and 1(p') = l(p) + 1 = a + 1. Let Xa' = do,(Xa'1, 1 Xa' a+l) The statement (3.3.10) says X2(ia.+1-ia)
(4.5.2)
D 'a+1
-
J l ea'a+1(Xa')ea'a(Xa')
ea'a+1(Xa')ea'a(Xa')-1
s'(a+l) = 0, s'(a + 1) = 1.
Therefore,
da'a+1(Xa'a+1)
(4.5.3)
2p _
(ea'a+1(Xa')ea'a(Xa,)-1)ha'
(ea'a+1(Xa')ea'a(Xa'))h°'
s'(a + 1) = 0, s'(a + 1) = 1.
In all the cases, Ya = Zo. So r,02(AD)
Suppose s'(a + 1) = 0. Then a-1 Qp'(An') = (IT ep'i(Xa')XCi)ep'a(Xa')ha i=1
(ea'a+l(Xo')ea'a(Xa')-l)h
, a - 1, Xp'a = ha - ha' , and Xp'a+l = ha' . Therefore, Xp'i = Xpi for i = 1, Suppose s'(a + 1) = 1. Then OD (AD,) = A,-,', and ep'a(Xa')-fa
ka1(ea(Xa')) = epa(Xa')fa =
Therefore, a-1
ep'i(A0')XO )ep'a(Xa')ha(ea'a+1(Xa')ea'a(Xa'))ha ep,a°.
Qp'(.X') _ i=1
Hence, Xp'i = Xpi for i = 1,
, a - 1, Xp'a = ha + ha' - fa, and Xp'a+l = ha' . Q.E.D.
, a - 1. Therefore, Xpi > 0 for all i. X14, hai = fai for i = 1, If P E 4'1e The following lemma is an easy consequence of (4.5.1). Lemma (4.5.4) Epaa3 f i=1 Xpil = ep The following lemma follows from (4.1.3).
Part II The Siegel-Shintani case
132
Lemma (4.5.5) Let p = (a,s) E 31 or T2 Suppose a = {1,
a - 1,i}.
Then
(n-i)(n-i+1)
for any N1i N2 >
2
op(Ab)IOza (Ra4)p, Xaga)l
(1)
a-1
<< fj
epi(Ab)Xn'inf(epa(A,)h.-Ni
epa(Aa)ha-N2)t(gO)r, l /
i=1
ap(Aa)ra1(Aa)IOza (Rasp, 00(.aga))j a-1 << 11 epi(Aa)Xeiepa(Aa)ha-d°-i+Nlt(go),.
(2)
i=1
for Aa E Ap2 and go E (K n MaA)ToO,+52a, where r = (r1,
rn_i_1) E R'-1,
rj < j(n - i - j) for all j, and t(g°)r is as in (3.4.17). Proposition (4.5.6) (1) Suppose p = (a,s) E X31. Then
ep=p(,D, w, w)
(2) Suppose P
E 132.
CGA(w;p)cp`2abp(w)Zv
(RsIp,wp,ha)
Then
Ep =p+(,D, w, w) "' CGA(w; P)cpZa6p (w)ZV.-is+(Ro Dp7 wp, ha), Ep
p+(-D, w, w) - C'GA(w; p)cpTabp (w)ZV.-a+(9a Ra 4D p, wp
1
0),
Ep=p# (11, w, w) - CGA(w; P)ep jp# (,D, w)'
Proof. Consider (1). We choose N1i N2 in (4.5.5) so that N1 >> 0 and N2 is close to (n-i)(2-i+1) Then ha - N1 < 0 and ha - N2 > 0. This is possible because ha > (n-i)(n-i+1) if p E'131. 2 This implies that the right hand side of (4.5.5)(1) is integrable on the Siegel domain. So we can apply (3.4.32), and ep-p(4),w,w)'"epC'GA(w;p) J 2
Mas/Mak
wp()()e(R0p,\ag°)dg
An easy consideration shows that the right hand side is equal to the right hand side of (1). The first two statements of (2) follow from (3.5.13). Consider the third statement of (2). Since ha > 0, ][li=1 epi(Ab)X"iepa(AD)ha is integrable on Apo also. Therefore, we can apply (3.4.32) to =-,# (-D, w, w). Thus, the proposition follows. Q.E.D.
§4.6 Paths in '133i'134 Suppose that p = (a, s) E '133 or '4i and l(p) = a.
Let a'-< a,and1(a')=a-1. ForWE9(ZZ'A),wedefine (4.6.1)
IF(gax) xEZ'°D°,Ok
4 The zeta function
133
Let p11 = (01, 511), p12 = (01,S12) be paths such that p - P11412, a1 = aU{n-1},
and 511(a + 1) = 0, 512(a + 1) = 1. Clearly, 93 = Ta, _ 93n.-a-1 By (3.5.9),
W,w)+ "p+(4),W, w)+ p#W,w)p#(4b) W, w)
W, W)
+ P12 (4), W, w) -'='Pii (-D, W, W).
Also
p+(D,W, w) =
(4), W, w) +
(D, W, w),
gyp+(4),W, w) _
(4), W, w) +
(
The following proposition follows from (3.5.13).
Proposition (4.6.2) (1)
Ep"p+(
(2)
fp,rp+(4,,W,w) ^' CGA(w>P)cpZabp(W)Zv2+(gaRa4Dp,Wn 1,3- ha).
We consider the contribution from Zastk. W e define Ma = f 1 Ma`, and Xa = Al x Man x Xv2. Let X.}. = {(uo, µ) E Ax R+ I t > a(uo)}. Let Ha = GL(1) x Ma x Hv2 and La = GL(1) x Ma x TGL(2), where TGL(2) is the set of diagonal matrices in GL(2), and Ma, TGL(2) are imbedded in GL(n) diagonally in that order. Then Zasstk = Mary xHak Z'ag0 .
Any element ga of Xa can be written in the form ga = (t, g'a, ba), where go E Man, ba = n2(uo)a2(it1, A-1t2), uo E A, µ E IlB+, and t, t1i t2 E A'. Let dba = duodX_µdxtldxt2i and d go = d"tdg°°dba. If f(go) is a measurable function on Ma°A/Hak, then (4.6.3)
fm.n/Hak
f(9°)dg° =
f
f(9a)d
Xa/Lak
if the right hand side converges absolutely. Let ga = Raga, and dga = dxAad g. Let Mp, Up, Up*, Tp etc. be as in Chapter 3. Also let Oz,.,t (RD Dp, go) be as in (3.5.1). By (4.6.3), Wp (9a )ap (Aa )E)za,st (Ra Dp, 9a).1p (9a, w)dx Aad9a
Apt Man/Mak
J
wp (9a )ap (Aa )Gz;,o (Ro-Dp, 9a )ep (9a, w)dx Aa d9a pi Man/Hak
fpl Xa /Lak
f
Wp(9a)op(Aa)Gz;.o(RIDp,9a).p(9a,w)
a,
Wp (9a )ap (AI)-a1(AD )Gza,,t (go Ra gyp, ea (9a ))-p (9a, w)dx Ao d9a
Ao Man/Mak
_
Jpo Man/Hak _
fpo Xa /Lak
g.
Part II The Siegel-Shintani case
134
The group Mp is either one of the following forms Case 1 GL(1) x GL(1)m x GL(n - a - 1) x GL(2) x Case 2 GL(1) x GL(1)m x GL(2) x GL(n - a - 1) x
GL(1)a-m-1 GL(1)a-m-1
Since it, E Pa1A, we can consider ga as an element of G°A n Pa1A. Let
9a = Aa(t,ga,a2(/Itl,µ 1121))
If we write ga = )talga°1 for aal E A,,,ga°1 E Ma A, then Aal = )adala+l(µ-1) Consider the function -'pll (ga, w) It is easy to see that n+m-a-1 a-m-1 ep(dala+l(µ-1)) =an( 1,... ,1,µ,Il-1,1,...
for Case 1 and m
n-m-2
Op(dala+l(/-1)) =an 1,...,1,µ,N-1,1,...,1 for Case 2.
Lemma (4.6.4) (1)
J
1Xa/Lak
wp (a)op
(AD )Oz, o (Rio 4p, ga )1p (go, w)d9a
^'0
fApj Xa /Lak (2)
_
wp () o p (A
(RD Dp, 9a )e 11,Ta11 (ga, w)d9a,
CJp
Apo Xa /Lak
Apo Xa /Lak
(gp, w)dga.
wp
Proof. We first compare ap(ga,w) with 0p11,'rpll (g0', w). By definition, -'p (ga, w) - -'P (go" w) = ip11(ga, w)
Lemma (4.6.5) For any e > 0 there exists S = S(e) > 0 and cl,
, Ch E t* = t°*
such that IIcjlb° < e for all i and for any l >> 0, M > w°i
for w° - 5 < Re(w) < M, and go in some Siegel domain.
Proof. Let I = In - m - a} for Case 1, and I = {m + 1} for Case 2. Then e,,', (op (ga ), w)
8pll (g-D, w) _ I,T
4 The zeta function
135
(see (3.4.27) etc. for the definition). Since v = 1 in this case, for any 6 > 0 and l >> 0, there exists a function ct,6(z) of polynomial growth such that I EB,I,T,1(9p(9a), z)I «Ct,6(z)t(Op(9a))TZ+P,-1.
Since I # 0, by the same argument as in the proof of (3.4.31), for any e > 0, we can choose q E DI, such that L(q) < wo, and IITq+PIIo < E. Therefore, there exist 6 = 6(e) and c E to* = t* such that IIcIIo < e and for any l >> 0, M > wo, 91,T,1(Bp(9a),'0) << t(ep(9a))cP-1
for wo - 6 < Re(w) < M. Note that there is no condition on p in this estimate (see the remark after (3.4.30)). We can write t(Bp(ga))c = t(ga))` for some c' E t*, and if the norm of c is small, the norm of c' is small also. Q.E.D.
- e'pi(9a,
Next, we consider the difference e,,, (ga, w) w). By definition, ep,li(90') = epi(Aa) for i = 1, a, and N'-Zepa(Aa) Therefore, (3.4.31)(2) implies that for any e > 0, there exist
6=6(e)>0,
andcEt*
such that ri < e for all i, IIcIIo < e, and if M > wo, a I&p11 (9a, w) - &pii,rp (9a> w)I <<
µ-r°+i t(9a )c
IIepi (Aa )r' i=1
for wo - 6 < Re(w) < M. Therefore, for any e > 0, there exists 6 = 6(e) > 0 such that if M > wo, ep (ga, w)-(go, w) is bounded by a finite linear combination of functions of the form a A-ra+1 t(9a)c
I'
epi(Aa)r i=1
for wo - 6 < Re(w) < M, and ga° in some Siegel domain, where c E t*, Irfl, IIcIIo < e
fori=1, ,a,andra+1>0.
We choose 0 < W E Y(A2) so that
Iez;,o(RaDp,9a)I << E T(epa(_a)x,epa(_a)xuo), XEkx
Ieza,j9-oRa,'p,ea(9a))I4 E W(epa(ta)-1x,epa(Aa)-1xuo) sEkx
These functions do not depend on ga. The function t(ga)' is integrable on the Siegel domain; therefore, we can ignore this factor. By (4.5.1), a
a
'i
-ra+1 i=1
-r.+1
Xv.+ri i=1
Part II The Siegel-Shintani case
136 Since ra+1 > 0,
µ-*a +ldxµ = 1_a(uo) - '11 2
µ>
ra+1
--(v.p)
Since #cai (7a) = epa (Aa)-3, multiplying i ,1(aa) affects only epa (.a ). Since ri can
be arbitrarily small, we can choose such ri's that Xpi + ri > 0 for i = 1, Let µi = epi (AD) for i = 1, , a. Then the integration with respect to µi, converges. So we only have to check that µaa+r°a(u0)_ -
1
iP(µx,Ixuo)d"µaduo
1
, a - 1. , µa_ 1
< oo,
xEkX 1 1
'F(N'a 1x, µa lx2lp)d"µa d260 < 00.
2
0
A
xEkEX
This follows from the fact that Tv2+(`P, w, s, Si) converges absolutely and locally uniformly for all s, s1. This proves (4.6.4). Q.E.D.
Let z = (z1,... , zn,) E C , z1 + ... + zn = 0. Let yl = za-m, y2 = za-m+1 for Case 1 and y1 = zn_m,_1, y2 = z,,-,n for Case 2. Let y = yl - y2. We define
T1=(a-m,a-m+1) for Case landT1=(n-m-1,n-m) for Case 2. In Case 1,
Tp(1) =n, ,,r,(m) =n-m+1,
Tp(m+1)=a-m+2,...,-rp(n+m-a-1)=n-m, Tp(n+m-a) =a - m, rp(n+m-a+1) =a-m+1, Tp(n+m-a+2)=a-m-1...... p(n)=1. In Case 2,
Tp(1) =n, ,Tp(m) =n-m+1, Tp(m+1) =n-m-1, Tp(m+2)=n-m,
Tp(m+3)a-m...... -(n+m-a+1)n-m-2, Tp(n+m-a+2) = a - m - 1, ,Tp(n) = 1. It is easy to see that Tp11 = T1Tp. Let U = U 11#, T = 7-p11. Then
MT(z) =
MTl
(Tpz)M" (z) = c(y)M1- (z).
Clearly, Bp(dal,a+l(µ-1))Tz+P = µl-y. Let I1 = {(i, j)
In-m+1<j<
n, 1< i < j}, 2= {(i,j) 1 1< i < a-m-a - m < j< n - m}, and I3={(i,j) I 1
n-m n-m m(Za-m. - zj) q (zi - zj) 11 O(za-m+1 - zj) H j=a-m+l j=a-m+2 (i,j)EIIUI2UI3
4 The zeta function
137
for Case 1, and
f
n-m-2
n-m-2
11 q5(zi - zn-m-1) [J q5(zi - zn-m)
O(zi - zj)
i=a-m
(i,j)EI1UI2UI3
i=a-m
for Case 2.
Let
9p(AD)rz+P = 9p(.a)T"z+P = ep1(Aa) i i(z) ... epa(Aa)7 (z)
Lemma (4.6.6) Suppose that ypi(z) =
cijzj+cio, where c, c0 are constants.
Then
(1) Ej cij + cio = 0, (2) caa_m = 1 for Case 1 and can-m-1 = z for Case 2.
Proof. In Case 1, n-a-1
m
a-m-1
-----------------
a
9p(d0a(ADa)) = an(
0.
a o- ia da
e
>
>
Therefore, 9p(daa(Aaa))T"z
-
AOa
(n-a-1)(Y1+Y22(za--+2+...+zn-,n
(n-a-1)Y+2((Y2-z-+2)+...+(Y2-zn-,n))
AOa
In Case 2, n-a-1
,m
Op (daa (Aaa)) = an(lr
' Aa (n_a_1)'
-
(n_a-1)'
A2
a-m-1
' 1 ... ,1
e
Aa
2
.
Therefore, (n-a-1)(Yl+Y2)+2(za-,n+...+zn---2) 0p(daa(ADa))T"z = A
_ So
a-1
ep(Aaa)T"z = fj ep(dai(Azi))T"zep(daa(Aaa))T"z, i=1
and the first factor is a product of powers of epi (.\a )'s for i = 1, (2) follows from (4.5.2). If z1i
, a - 1. Thus,
1, then Op(Asa)T"z+P = 9p(Aaa)TCz+P = 1. Thus, (1) follows. Q.E.D.
We define
Qp={zIzi-zi+l=lfori=a-m+2,,n-m-1}
Part II The Siegel-Shintani case
138
for Case 1, and
Qp ={zI zi-zi+1 for Case 2.
Let CO be a contour of the form {z E Qp I Re(z) = q}. Let
dzlcp=
1
l li=1 1 d,j Un-m dzj a
7'7a-m-1
lli=1
7'' n-1
dxi l In-m-2 dzi
Case 1, Case 2.
By the Mellin inversion formula, w2(det a)}1eS,T(ep(Aa(9a>ba)),w)dga a+1 MT(z)p1-IOp(Ao)T=+PA(w;z)dzlcp.
=6#(w2) (2v)
IC'
Let F0 1(4D, t, Aa, ba, z), Fp2 (D, t, Aa, b°°, z) be the following two functions ap(Aa)p1-IOp(Aa)7Z+P8Za.o (R5IDp, (Aot,1, bo)), ap (Aa)na1(Aa )pl-y9p (Aa )FZ+Ppza Al so
a
Ra,Dp, 9(t, 1, ba )).
let Ep,st+
7pa(z) + ha, 2 )
z)
&2(Y - 1) lli1 (7pi(z) + Xpi)
3 - 7pa(z) a-1
2p,st+(D, z) =
f
)
t a2 (y - 1) fi=1 (7pi (z) + Xpi )
, a. Then d" As = aa2 d" pl
Let pi = epi (AD) for i = 1,
ha,
1/kx XApi XX+X(Al/kx)2
dX Pa. Therefore,
w1(tt1t2)t1Fp1(-D, t, Xa, b°>
= 6#(w1)Ep,st+(41, z), wl(ttit2)}1Fp2('D, t, A1, ba, z)dxtd) Aadbo Al /kx xApo XX+X(Ah /kx)2
= 6#(wl)2p,st+(D' z). Hence,
=
1
a+1
6#(w) (2) J
z)MT(z)A(w; z)dzl cp ,
p
wp
w)dga
fA,,,Xa /Lak
=
1
_
a+1
b#(w) (2) f
p
Ep,st+('D,z)MT(z)A(w;z)dzlcp.
4 The zeta function
Let
[
hp(z) =
139
cb(zi - zi).
(ia)EI1u12uI3
In Case 1, we define a function fpo(y, za-m+i) by
fpo (y, za- +1) =z,_1=1 Res
.
Res
Res
za_m_1=1
Res hp(z)IQ,.
z1=1
In Case 2, we define a function fpo(y, zn-,n-2) by the same formula. Let fill(y) _ fpo(y,1) in both cases. Let 9tk
fp2(Y) =
Zk(y)
Zk(n - a) Zk(y + n - a)
Then Res z,_1=1
.. Res
. Res MT(z)IQ, = fpl(y) fp2(y) Res z,._m=1Resza_-+1=1z1=1
for Case 1, and
Res .. Res
Res
z._1=1
Res
za_,,_1=1
.. ResMT(z)IQ, = fpl(Y)fp2(Y) z1=1
for Case 2. Also Resy=1 fpl(y)fp2(y) = %DCG = 3n-a-1 n1
Let Qp = {z E Qp l zi = l fori = 1,... ,a-m-l,a-m+l,n-m,.. ,n-1} for
Case 1, and Qp = {z E Qp I zi = 1 for i = 1, , a-m-1, n-m-2, n-m, for Case 2. Any function on Q' can be considered as a function of y. Let 7i(y) = -Ypi(z)IQ,
,
E,8+ (D,y) _
z)IQp, Ep,at+(4), Y) =
, n-1}
z)IQp
Because of (4.6.6), -ypa(y) = y21
Lemma (4.6.7) 1a+1
1
Ep,st+('k,
z)MT(z)A(w; z)dzIcp
(1)
n
- 27r- J 1
(2i)
(2)
1
y)fp1(y)fp2(y) [A(w; z)] IQdy,
e(y)=1+b
a+1
f Ep,st+(4), z)MT(z)A(w; z)dzIc,
Le(y)1+S
,st+(
)fl(y)fp2(y) [A(w; z)]
dy,
where 6 > 0 is a small number.
Proof. Let Qpl = {z E Qp zl = 1}. Let Cpl be the contour of the form {z E Qpl Re(z) = q}. Let Cpl be the contour of the form {z E Qp I Re(z) = q} I
I
where gl-Q2=1-61,gi-qi+1=1 +62
foriV{1,a-m+2,,n-m-1}or
Part II The Siegel-Shintani case
140
{1, a - m, , n - m - 3}, and 61621 >> 0. Let dzI c,1 be the differential form on Cpl such that dzlcp = dzldzlcn1. Then i
(2) 1
a+1
fP
(2) f I (2) a
1
Res [Ep>st+(-D, z)MT(z)A(w; z)] dzI c,1
v/- 1
1
a+1
1
+
Ep,st+(iD, z)MT(z)A(w; z)dzIc,. 1
The second term is holomorphic for Re(w) > wo - 6 for some 6 > 0. If we continue this process, we get the lemma. Q.E.D.
These considerations show the following proposition.
Proposition (4.6.8) Suppose P ET3 U T4 Then (1)
Ep°p+(4),W,w) CcA(w;p)cpWabp(w)Zv2+(RaIPp,wp,h0) alb#(W)
+ Ep 27r
(2)
Re(y)=1+b
Ep st+(D, y)fpl(y)fp2(y) [A(w; Z)] IQ; dy,
ep=P+(4), W, W)
- CGA(w; p)cpZabp(W)Zv2+(gaRa4)p, Wp 1, 3 - ha)
+ EpValb#(W) 21N
1
JRe(y)=1+6
Ep,st+(4),y)fp1(y)fp2(y)[A(w;z)] IQ,dy,
where 6 > 0 is a small number. Next, we consider , Epl,
W, W)) Ep12 (p, W, W).
Lemma (4.6.9) Let T be as before (7 = Tp12 also). Then W, 211)
wp1. (ga1)aa1 (A01)9z11 (Ra1 p1. , go,)
t71a11 J
4v1.0 xMa1A/Mo1k
for i = 1, 2.
Proof. Suppose r # T. By (3.4.31)(2), for any e > 0, there exist
b=b(e)>0,
andcEt*
such that ri, llcllo < e for all i and if M > wo, << IS ,,,.(ga1,w)I
a+1
ri epl.j( j=1
1)r't(9a1)c
4 The zeta function
141
for wo - b < Re(w) < M, and gal in the Siegel domain. By definition, ha1 = 1. Since t1+T+1e1(Ro, fAX
I
p1i,
t)dxt
converges absolutely for i = 1, 2, the lemma follows. Q.E.D.
By (4.5.2), 16
Ao1a+1 = ep1ia+1(Xot)2eptta(Pat)
2 = ept2a+1(Asi)2
Therefore,
TZ+P
=
TZ+P
TZ+ P
ep (d,,.+,
a
_
epi(Ao)7Pi(z)XY-1 AY-11 a
= [Iep11i(Ao)'1pi(z)AY-i i=1
-eptta(Ao1)ryp(Z)-
' =i
2 ep11a+1(Ao1)
Y 2
a-1 Hep11i(Aot)'1pi(z) i=1
Also 0p12(Ao1)7Z+P
=
Bpeo(Xo)TZ+Pepeo(do1a+1(Aa1a+l))TZ+p
= ep(AD1)rz+POp(dala+l(ASla+1))7z+P a\\
_
fjepi (Ao)-'1pi (z)A5 i=1
as+l
a-1 =ep12a(AD,)'1111(z)+
1ep12a+1(Ao1)
2
ep12i(A01)'1pi(z) i=1
By definition, Xptia = ha define
- 1, Xpita+l = 1, Xp12a = ho - 2, Xp12a+1 = 1
We
bp(W)E1(Rot,DplltWpt y21) Ep11 (4)+ W, Z) =
Ep12 (fit Wt z) =
2to2(ho-1+ypa(z)- y21)I]'= (Ypi(z)+Xpi) 6p(W)F-1(Ro11)p12IWp 1' 2 ' 2%o2(ha - 2 + ypa(z) + y21)Ia=1(7pi(z) + Xpi)'
where we consider c.Jp, wp 1 as characters of AX /k < by restricting to elements of the
form (t,1). Let µj = ep1tij (ao1) for i = 1, 2, j = 1,
dxA, = ad-pi ... dXµa+1 = 2-1 a2
, a + 1. Then dxu,a+i
Part II The Siegel-Shintani case
142
Also , a1 = ,aa,1. Therefore, we get the following proposition.
Proposition (4.6.10) up1.
)a+11
1
w, w) - aa1
(.D,
1
w, z)MT(z)A(w; z)dzI c,
Ep,. n
for i = 1, 2. We define Ep w, y) = Ep1i (4, w, z) IQ, for i = 1, 2. We can get the following lemma as in (4.6.7) and the proof is left to the reader.
Lemma (4.6.11) =p1. (', w, w)
2
1 Re(y)=1+6
aV
Ep,, (-b, w, y)fp1(y)fp2(y) [A(w; Z)] I Q'p dy
for i = 1, 2, where b > 0 is a small number. The proof of the following proposition is similar to that of (4.5.6)(2), and is left to the reader.
Proposition (4.6.12) (1) Suppose P E 33 U X34. Then Ep p# ((D, w, w) - CGA(w; p)cp91a6#(w) TP(0) . (2) Suppose p E X33. Then
(0) epEp# ((D, w, w) ,r CGA(w; p)cpZab#(w) ha - 3
Now we are going to establish a cancellation of distributions. This cancellation basically comes from the case n = 2. In this process, we reconstruct the order two pole of the adjusted zeta function for the case n = 2. Let Ep1)
y) = Tv2+(Ra,oRa,Dp, Y 2
I
+ ha,
+Tv2+(Ra,o9aRa4)p, 3 -
2 Y)
y-1
bp(w)E1(Ra141P12,wp 1> 2 )
2(ha - 2 + y - 1)
1
2
- ha,
1-y 2
),
- bp(w)E1(Ra1Dp11,wp) 2(h, - 1)
b#(w)E1(Ra oRa-Dp, v21 +ha - 1)
y-1 +
En2)(.D,
y-1 y)
We consider the Laurent expansion of EP(3) (,D, w, y) at y = 1. Since Tv2+(Ra,oRa(bp, s, 0) = E1+(R',oRa4,p, s - 1) etc.,
4 The zeta function 1
Tv2+(Ra,oRa l)p,
2
143
+ ha, 0) + Tv2+(Ra,o " DRa(Dp, 3 - y 2
- E1(Ra oRa-Pp, y - 1 + ha - 1) + 2
2
R,,0Ra4'p(0)
+
1
- ha, 0)
Ra,ORa-917a4)p(0)
1
- 1-ha
2-
Clearly, 1
y1 2 + ha -1
2-y2ha
_
1
1
2(ha - 1)2 (y - 1) + O((y - 1)2),
ha - 1 2
lha+2(2 lha)2(y1)+O((y-1)2),
and
bp(w)E1(Ra1(-bp11,wp)1)
b#(w)El,(-1)(Ra1,Dp11,1)
2(ha - 1)
(ha - 1)(y - 1)
+
bp(w)E1,(o)(Ra1 -Dp11, wp) 1)
+ O(y
2(ha - 1) 1, 1L) 2(ha - 2 + y - 1)
- 1),
b#(w)E1,(-1)(Ra14)P12'1)
(ha - 2)(y - 1) _ 6#(w)E1,(-1)(RD1 '12,1) (ha - 2)2
+
by(w)E1,(o)(Ra14p12,WP 1, 1)
2(ha - 2)
+O(y - 1).
If b#(w) = 1, 4) is K-invariant. This implies that Ra,oRa-Pp(0) =
(Ra14Dp1111)1
Therefore, E(3) (4), w, y) is holomorphic at y = 1 and (4.6.13)
Ep3)(1),w, 1)
_ -b#(w) 2
ha) +Tv2+(Ra,o.3'iRs p, 3 - ha))
_ b#(w)E1,(-1)(Ra1Dp12'1) +
bp(w)E1,(o)(Ra1-Pp12,WP 1,1)
2(ha - 2)2
2(ha - 2)
_ b#(w)E1,(-1)(Ro14)p11' 1) _ 2(ha - 1)2
6p(w)E1,(o)(Ra14)p11Iwp,1)
2(ha - 1)
Let (4.6.14)
w, w) =
b#(w)E,st+(,p,
w) +
+Ep12(4,w,y)-Ep11(4,w,y) Also let °,p
8t (4),w,
w) be the following integral
b#(w)Ep,st+(.p, w)
Part II The Siegel-Shintani case
144
(4.6.15)
2 + ha - 1) fp1(y)fp2(y)[A(w;
E1(R
t71a36#(W)
Le(y)=l+ (y - 1) l1i_1 (1pi(y) + Xpi)
z)]I Q,, dy.
Then (4.6.16) tea
Y-1
(4)
JRe
aa3
Ep (W, W, W)fpl(y)fp2(y)[A(W;z)JIQ,,ay
Ep3)(
I
J
2irV -1 Re(y)=1+6
flid_-11(7pi(y)
y)
+Xpi)
fpl (y)fp2(y)[A(w; z)] I Q,, dy
+ p,st((D, W, W).
Since E(3) (4>, W, y) is holomorphic at y = 1, and Xpi > 0 for i = 1,
, a - 1,
(4.6.17) 2ir
^
Re(y)=1+6
7 llia=-11(7'pi(y)
+ Xpi)
fp1(y)fp2(y)[A(w;z)]IQ,,dy
C, Z' E (3) (4,' LL,'
+
Eptrla3
27rV '-1
f
dEp3)(i(,W,y)
fp1(y)fp2(y)[A(w;z)]jQ,,dy
e(y)=1-6 IJi=1 (7pi(y) + Xpi)
CpIjaE(3)(4p,W,1)A(w; z).
Note that Epaa3 f1a=1 Xpil = Cp. By (4.2.15), (4.6.8), (4.6.12), (4.6.17), we get the following proposition.
Proposition (4.6.18) (1) If P E f 33, Ep °v
w, w)
IiGA(w; p)Ip
w) + Ep =p,st
w, w)
(2) If P E 34, Epp ((D, w, w) N CGA(w; P)Ip (4),w) + Epp st (4>, w, w) + Ep ='p# (4D, W, w)'
§4.7 The cancellations In the last section, we established the cancellation of distributions based on the case n = 2. In this section, we will establish another kind of cancellation of distributions. This cancellation is established between -',1,,t (4), w, w) and °p2# w, w) for Pi E X33, P2 E X32 or P1, P2 E 134
; a - 1,nLet pi = (0l, s1), P2 = (D2,S2) be paths such that a1 = { 1 , 2}, 02 = {1, , a}, s1(i) = 52(i) for i = 1, , a - 1, and sl(a) + S2(a) = 1.
4 The zeta function
145
The correspondence between pl and p2 is one-to-one. Therefore, if pl is of Case 1 (resp. Case 2) in §4.6, then we say p2 is of Case 1 (resp. Case 2). If p2 is of Case 1, Mpg = GL(1) x GL(1)m x GL(n - a) x GL(1)' m. If p2 is of Case 2, GL(1)a-m-1 Mpg = GL(1) x GL(1)m+l x GL(n - a) x
Proposition (4.7.1) If a > 1, let p3 be the unique path such that p3 - P1, p2 and l(p3) = a - 1. (1) If a = 1, Epl
'1
(
/
WI w) +
W))
CGA(wiP)EpiOn26#(W)Z1(1)(R',,°Ral4)p,,n-1).
(2) If a > 1 and si(a) = 1, pi,st(''' W, w) + Ep2-p2#(, W,
Ep1
w) - CGA(wi P)Ipsl (I
W)
(3) If a > 1 and si (a) = 0, W, W) + 6P2;42# ('DI w, w)
Epl
CGA(wi P)Ip32(I, W)
We devote the rest of this section to the proof of this proposition. This will complete the proof of (4.4.6). Let I1, 12, I3 be the subsets of {1, definition,
, n -1} which we defined for pl in §4.6. By
n-m M1-,2 (z) =
H
O(zi - zj)
(i,j)EI1UI2UI3
11
Y'(za-m - zj )
j=a-m+1
for Case 1, and M'b2 (z) _
[
n-m-2
O(zi - zj) J O(zi - zn-m) i=a-m
(i,j)EI1U12UI3
for Case 2. Also by definition, Xp2a = hat = fat, and t021(A,2) = ep2a(AD2)-fat Therefore, opz(A )ND21(A12) = IIi=1
ep2i(A12)XU2i.
Lemma (4.7.2) "P2#(4', W, w)
aa21 f1,20 M,2A/M,2k W 2 (9°)a2P2 (A12)K121(A2 )2,r12 (0p2 (9a2 ), w)d A02 dg°2. Proof. Suppose r
rl,
r,2. By (3.4.31)(2), for any e > 0, there exist 6 = S(e) > 0
, ra+1 > 0, c E t* such that ri, IIcIIo < e for all i and if M > wo, a
Ip2,r (902 f w) I << [1 ep2j (A12 )T't(9az )c, j=1
Part II The Siegel-Shintani case
146
for w - S < Re(w) < M and ga°2 in the Siegel domain. Since /Caz (Aaz) = epza
(Aaz)-ho
ri + Xp2 > 0 for i = 1,
tr , aa2 (.a2 )Ka2 (A , 2 ) = 11a=1 e.2i (Aaz )Xe2 i . Since
, a - 1, and re > 0, the lemma follows. Q.E.D.
Let
Qp2 ={zIzi-zi+i=1fori=a-m+1,
,n-m-1}
for Case 1, and Qp2 for Case 2.
Let Cpl be a contour of the form {z E Qp2
I
Re(z) = q}. We use the same
definition of y1i y2i y as in §4.6. We define Q'2 similarly as in §4.6.
Lemma (4.7.3) Let a (Aaz)7-12 Z+P =
eazi(Aa2 )"12'(z).
epz
i=1
Then ryaza(z) =
(1-a2
on Qp,.
Proof. In Case 1, \
BP2(daza(AD2a))T2
Z+
\\(n-a)za-m-(za-m+1+...+Zn-m)-(n-a+2)(n-a)
P = ^a2a
, n - m - 1. Therefore,
On Qp2, z i = 1 for i = a - m + 1,
(n-a)za-m - (za-m+1
=(n-a)y+
n-a-1 i=1
zn-,t )
_
(n-a+1)(n-a) 2
- (n-a+1)(n-a) i=(n-a)y } (n-a-1)(na) 2 2
_ (n - a)(y - 1). Since Aa2a = epza (a02)ep2a_1(A02 )'1, we get the lemma for Case 1. Case 2 is similar. Q.E.D.
Zk (y)
fa21(y) = f i (y), fp22(y) = JJk(y + n - a)
fa22(y) =
w, z) =
Zk(91k
26#
a)fp12(y)-
a2 4'P2 (0)
(n - a)(y - 1) lli=1(lazi(z) + X020
4 The zeta function
We define Ep2#
w, y) = Epz#
147
w, Z) 1Q02 . By (4.7.3) and a similar computation
to that in §4.6, p2# (D, w, w) ^' 1aa23
(27rv/1
/ a Jc Ep2# (.Dw, z)MT2 (z) [A(w; z)dz] I c, 12
a0,3
,,
27 r V -i Re(y)=1+6
Ep2# (4, w, y)fpz l (y)fp22 (y) [A(w; z)JIQ' 2 dy.
Clearly, epl = -ep2. If a = 1, we define Elan.(4), Y)
E1(R1,oRa1 p1 >
y-1
y-1
2
+ n - a)
Zk (n - a) 9az Rax'PP2 (0)
y-1
9ik
If a > 1, we define y21
+n - a) E1(Ra1,ORa1,tp1, (Y - 1)(-Ynia-1(y) + Xp2a-1)
Zk(n-a) Nk
gal
R1e-bp2(0)
(y - 1)('Yp2a-1(y) +Xpza-1)
Lemma (4.7.4) We have the equality rypli (y) = ryp2i (y) for i = 1, (1) Yp1a-1(y) = 7p2a-1(y) - (n-a+2)(y-1) ifsi(a) = 0, (2) 'Ypia-1(y) ='Yp2a-1(y) +
(n-a2
, a - 2, and
if -,,(a) = 1.
Proof. If sl (a) = 0, Aa1a = epla(^al)2epla-1(\a1) 16
2,
1
'aza = ep2a (^a2) 2 ep2a-1(Aa2) 2
The lemma follows by the proofs of (4.6.5) and (4.7.3). The argument is similar for
the case sl(a) = 1. Q.E.D.
Sincen-aaaz3 ? = trlal3,
^
2
Ep1 ='p1,st
w, w) + Epz'='pz #
c''J, w )
a- E...' (-D' y) Ep1 9-0136#(w) f fp11(y)fp12(y)[A(w; z)] IQ,, dy. JRe(y)=1+b ii-1(Yp1i(y) + Xplti) (If a = 1, we do not consider the denominator in the above integral.) We consider Ecanc(4), y) only when w is trivial. This implies that 4) is K-invariant. Therefore, 4Pp, is invariant under the action of the standard maximal compact
subgroup of GL(n - a)A. So by (4.4.7), Ecanc(), y) is holomorphic at y = 1. We consider the value Ecanc(',1). If a = 1, Ecanc((P, 1) = 1 E1,(1)(Ra10OR514)p,, n - 1).
Part II The Siegel-Shintani case
148
Therefore, in this case, Ep,
..
past ('D, W, w) + Epa
pa# ('D, w, w)
P)Ep193n-26#(w)
CGA(w;
2
E 1 (1)(a1,O R' Ra , p1, n - 1).
Suppose a > 1. There are two possibilities for s1i52, namely, 51(a) = 0,52(a) = 1 or s1 (a) = 1,52(a) = 0. The following lemma is an easy consequence of (4.7.4).
Lemma (4.7.5) Suppose a > 1. Then (1) If si(a) = 0, (n - a + 1)E1(R'01,0R11,Dp1, n - a) Ecanc(4), 1) =
2 2Xpla-1
+
E1,1(Ra1,ORa1,Dp1 , n - a) 2Xp1a-i
(2) If si(a) = 1,
1) _ -
(n-a+1)E1(R'1,ORa,4)pi,n-a)
Clearly, aai3 =
2
+
E1,1(Ra1,OR51
2Xpja-1
2Xpla-1 1 30,,3
1,n-a)
and Ta, = `fin-a-1 Also a-2
a-2
-1
Xp,i =Ep3 aa33
Ep3trla13
i=1
Xp3i = Cp3.
i=1
Clearly, cp, = -Ep3 if si(a) = 0, and Ep, = Ep3 if sl(a) = 1. Therefore, Ep1 A,36#(W)
=
Ep1
E canc( ', y)
f e(y)=1+ lli=1 (1i(y) + Xp,i) 2
27r
'7a-2
Ya1393n-a-1b#(W)
fp11(y)fp12(y)[A(w; z)]IQ,l dy
Ecanc(4, 1)A(w; p)
l ii=1 Xp3i
+
Ep19'O136#(W) 27r
f
Re(y)=1-6 l1i=1 (rypli (y) + Xp1i) Re(y)=1-6
fp11(y)fp12(y)[A(w;z)]IQ,,dy
_1)o1(a)+l Cp3 Tn-a-16# (W) F'canc (-D, 1)A(w; p).
By definition, (n-a-}-1)(n -a)
Xp,a-1 =
n-a+1
= fa3 - (n - a + 1)
si(a) = 0,
2
sl(a) = 1.
Let a' = a-1. Then l(p3) = a', and n-a+l = n-a' etc. Therefore, this completes the proof of (4.7.2), and hence (4.4.6).
Remark (4.7.6) The order two terms of the poles of the zeta function are constant multiples of 6#(W)E1(R'n n_2
n - 1), 6#(W)E1(Rn n-2,o$, n - 1).
4 The zeta function
149
These distributions are zero if w is not trivial. If w is is trivial and 4D = Mv,,,,D,'D is K-invariant. Therefore, by (4.4.7),
E1(Rn,n-2,A)n - 1) = So, if
Zk(n - 1) k
- n,1R.,1ID (0).
and 4)v is supported on Vk'y for some v E 93
,
n - 1) _ -`Fn, 1Rn,1ID (0) = 0.
This implies that in the case k = Q, all the poles the associated Dirichlet series are simple as was proved by Shintani [65].
§4.8 The work of Siegel and Shintani Both Siegel and Shintani worked on the space of quadratic forms. We consider their work from our viewpoint. Even though I was an undergraduate student at the University of Tokyo and took some courses from Shintani, I was not personally acquainted with him nor was I aware of the theory of prehomogeneous vector spaces at that time. So the greater part of this section is my speculations based on papers by Siegel and Shintani. Siegel did at least two kinds of work on quadratic forms. One is the proof of what we call Siegel-Weil formula [67], and the other is the proof of the average density theorem for the equivalence classes of integral quadratic forms as in this chapter [69].
His argument in [69] was clever, and essentially determined the rightmost pole of the zeta function even though he used the argument based on his Siegel-Weil formula and did not use the zeta function explicitly. Shintani was apparently influenced by Siegel's paper and tried to interpret the content of Siegel's paper from the viewpoint
of prehomogeneous vector spaces. Here, we try to trace what Shintani might have thought in pursuing this goal. We assume that k = Q for simplicity. One difficulty of the space of quadratic forms comes from the fact that we have to adjust the zeta function for the case n = 2 as we saw in §4.2. If we use this formulation, the case n = 2 is more or less included in the general case. However, Siegel and Shintani handled the cases n = 2 and n > 3 quite differently. We used the smoothed Eisenstein series for all the cases, but Shintani used the ordinary Eisenstein series for the case n = 2. The function he considered was the following integral (4.8.1)
JGA/Gk IXv(
)8Ov, (, )E(g°, z)d,
where V is the space of binary quadratic forms, (t, g°) E A" x GL(2)11, and E(g°, z) is the Eisenstein series for the Borel subgroup of GL(2). This function can be considered as a two variable zeta function associated with the action of the Borel subgroup on V. As in the case of the smoothed Eisenstein series, the pole of (4.8.1) at z = p gives us the one variable zeta function. Shintani himself was aware of the approach in this book, but chose the above formulation. He might have chosen the
150
Part II The Siegel-Shintani case
ordinary Eisenstein series, maybe because he wanted to achieve the cancellation in §4.2 explicitly. Then why didn't he consider the generalization of the above function for the general case? Let us consider the following integral (4.8.2)
GA/Gk
jXv(7jsOvka(1b, g)E(g°, z)dg,
where V is the space of quadratic forms in n > 3 variables, g = (t, g°) E Ax x GL(n)°n, and E(g°, z) is the Eisenstein series for the Borel subgroup of GL(n). This would be the natural generalization of the integral (4.8.1). However, there is a slight difference here, i.e. we have to consider all the points in Vk". The effect of this is that the theta series Ovka (1), g°) is not rapidly decreasing with respect to the diagonal part of g° any more, whereas it is true for Ov,, (4), g°) for the case n = 2. In fact the function Ovk. (,D, g°) is barely integrable on G°n/Gk, and it is already non-trivial to show the convergence of the integral (4.8.2) in a certain domain. Moreover, we do not know if (4.8.2) can be continued meromorphically to a domain which contains p. This means that we cannot recover the one variable zeta function from (4.8.2). Shintani must have considered integrals of the form (4.8.2), and faced this difficulty. My guess is that this is the reason why he chose a different approach for the case n > 3. However, he could not compute the residue of the associated Dirichlet series for indefinite quadratic forms in general. So it was natural to look at Siegel's original approach and see how he handled this difficulty. Siegel's approach for n > 3 was quite different from the case n = 2 also. Roughly speaking we can summarize his argument as follows. First, he used his Mass formula to describe µ,,,(x) (see §0.3 for the definition) for this case. Then in order to estimate Iloo(x)bx,oo o CO
(4.8.3)
'
he considered integral quadratic forms with two different signatures (m, n-m), (m2, n - m + 2) such that the discriminant is less than or equal to Q
TIPg[s]. log p
= P
For 0 < S < N, let C,,,,(S) be the number of residue classes modulo Q which contain an integral quadratic form with the signature (m, n - m). Let D(S) be the number of modulo Q congruent classes such that detx = (-1)"-'S for x in such classes. Siegel related (4.8.3) with C,,,(S) by the Mass formula, and showed that Cm(S) + Cm_2(S) < D(S). By estimating D(S), he obtained an upper bound for Cm(S) + Cm_2(S). The lower bound is relatively easy, and this argument was the point of his argument. But why two different signatures (m, n-m), (m-2, n-m+2)? His argument was very clever, and I spent a long time trying to see from our viewpoint what he was trying to accomplish. My conclusion was that Siegel was trying to accomplish cancellations of divergent integrals one way or another. After investigating which distributions Ep (ot, w, w) may give such cancellations for some time, I reached the cancellations of distributions as in (4.6.18), (4.7.5). In a way, his
4 The zeta function
151
two different methods for n = 2 and n > 3 correspond to two different cancellations (4.6.18), (4.7.5) in this chapter. Siegel did not have a fully developed theory of Eisenstein series or adelic language to hand, and the fact that he was able to handle the case n > 3 is due to his genius. We now go back to Shintani's approach for the case n > 3. Recognizing the difficulty of using (4.8.2), Shintani used the approach essentially based on microlocal analysis. If we summarize his approach, we may say, that he considered the relative invariant polynomial as a hyperfunction, and tried to compute its Fourier transform. As I look back, he was unlucky in choosing this approach. Shintani was basically considering the associated Dirichlet series i (4), s) (see §0.3 for the definition). However, while the adelic zeta function has order two poles, the associated Dirichlet series only have simple poles. Nevertheless, the difficulty based on the order two poles of the adelic zeta function existed. In our approach, we indirectly reconstructed the order two part of the adelic zeta function in the inductive computation as in (4.7.5). But, since the associated Dirichlet series only have simple poles, Shintani was, in a sense, fighting against an invisible enemy. It is an interesting question to see whether it is possible to reconstruct the proof of Theorem (4.0.1) by micro-local analysis. Shintani passed away about three years before equivariant Morse theory was established. Nevertheless, he managed to produce all the essential ideas in this book in his single paper [64]. He had almost all the right tools except for the uniform estimate of Whittaker functions at infinite places. In this sense, there is no doubt that the content of this book is what Shintani would have done by himself. The purpose of this section is not to subordinate the contribution of Shintani concerning his work on the space of quadratic forms, but rather to point out how a true pioneer sometimes has to go through a tremendous struggle and yet may not achieve all his goals. And Shintani was a true pioneer of the subject.
Part III Preliminaries for the quartic case In the next three chapters, we consider the following three prehomogeneous vector spaces (1) G = GL(2) x GL(2), V = Sym2k2 ® k2, (2) G = GL(2) x GL(1)2, V = Sym2k2 ® k, (3) G = GL(2) x GL(1)2, V = Sym2k2 ® k2.
Case (3) was handled by F. Sato [55] in a slightly different formulation. These are rather easy cases, and are essentially the same as the space of binary quadratic forms. However, we need the principal part formula for the zeta function for these cases, because these representations appear as unstable strata for the prehomogeneous vector space G = GL(3) X GL(2), V = Sym2k3 ®k2 which we call the quartic case.
Coefficients of the Laurent expansions of the zeta functions for (2) and (3) appear in the Laurent expansion at the rightmost pole of the zeta function for the quartic case, and therefore they are particularly important. The principal part formulas which we are going to prove are (5.6.4), (6.3.11), (7.3.7).
Chapter 5 The case G=GL(2) x GL(2), V=Sym2k2 k2 §5.1 The space Sym2k2 0 k2 We consider the prehomogeneous vector space G = GL (2) x GL(2), V = Sym2k2®
k2 in this chapter. We will use a new technique concerning the choice of the constants in the definition of the smoothed Eisenstein series in this chapter. Generally speaking, if G is a product of more than one GL(n), it sometimes makes the computation easier to choose the constants in the definition of the smoothed Eisenstein series in a certain way. The reader can see the effect of this technique in the proofs of (5.3.9), (5.4.10), and (5.5.6). For the rest of this book, W is the space of binary quadratic forms in two variables v = (v1i v2). We can identify W with k3 by the map fx(v) = xovi + xlV1V2 + x2v2 _ (xo, x1, x2).
The vector space W is V2 in §4.2. Let V = Sym2k2 ® k2 = W ® W. Any element of V is of the form
f = (fl, f2), fl = fxl, f2 = fx2, where xl = (x10, x11, x12), x2 = (x20, x21, x22). We use x = (x1, X2) = (xlo, x111 x12, X20, X21, x22)
as the coordinate system of V. We define Gl = G2 = GL(2), and G = Gl x G2. The group G acts on V by the formula b) /lgl g)( f ) = (afglxl + b fgixz cfgixi + df9ixz) for g2 = (a \ dJ 2 fxl xz C
5 The case G = GL(2) x GL(2), V = Sym2k2 ®k2
153
Let T c G be the kernel of the homomorphism G -+ GL(V),and Xv(91,92) = (det gl)2 det 92. Then Xv can be considered as a character of G/T and it is indivisible.
Let
Let U = {x E V I pi(x) # 0 for some i}. Consider the map Vk D x - (p1(x),p2(x),p3(x)) E V2 Then G2k \ Uk = ]En, and the action of Glk can be identified with the action of
GL(2)k on ]P(Sym2k2). Therefore, this is a prehomogeneous vector space which is essentially the same as Sym2k2.
By the above consideration, Vk 0 0. Let G°,A etc. be as in §3.1. In this case, G' = GA, so t = V. We identify t with {(zl, z2) = (Z11, z12, z21, z22) E R4 z11 + z12 = z21 + Z22 = 0}. We define a Weyl group invariant inner product by (z, z') = Ei j zijzi'j for z = (zip), z' = (z2j). Let yij E t* be the weight of xij for all i, j. We use the notation yij (t) for the rational character of T determined by yij. This should not be confused with t7' which is a positive real number. We identify yij's with elements of R2 as follows.
Let f8 be the parametrizing set of the Morse stratification. Elements of B \ {0} are as follows. Cl 0
C2
Part III Preliminaries for the quartic case
154
C3
Let 01,
C4
, /34 be the closest points to the origin from the convex hulls C1i
, C4
respectively. We choose G" = SL(2) x SL(2) as G" in §3.1. Then Zpi etc. for i = 1, 2, 3 are as follows. Table (5.1.1)
RR
131 =
414)
4
4)
Zo
WO
X12, X21
X22
02 = (0)0i-2f 2) 03 = (-1, 1; -2, 2)
x20,X21,X22
/34 = (-1,1; 0, 0)
x12, x22
M1161
-
X22
SL(2) x {1} {(a2(t-1, t), a2(t2, t-2))} SL(2) x {1}
Let ai = ()3i) for i = 1, 2, 3. By the above table, Zaik = {x E Zp1k I x12, x21 E k" }. We identify Za2 with W and consider Za2k, Z'a2 ok etc. Since M,3 acts trivially
on Z,3, Zaik = Zaik \ {0}. The vector space Z-04 is a standard representation of Mao. Therefore, Za4k = 0. There is one /3-sequence of length 2 which is a4 = (/32,/33 - /32). Clearly, Za, is the subspace spanned by {e22}, and Z14k = Za4k \ {0}. It is easy to see that Vk is the set of x E Vk which are not conjugate to elements of the form (0, 0, x12, x20, 0, 0). This set corresponds to the set of forms without rational factors by the map U -* V. Let Yv,o (resp. Zv,o) be the subspace spanned be the subspace spanned by by {e12, e20, e21, e22} (resp. {e12, e20, e21}). Let {e12, e2o}. Let Z'vok = {x E Z' ,ok X12, X20 E k" }. Then VStk = GkZ'vok Therefore,
11 Vk \ {0} = Vk
3
Vstk
J__L
ll Spik. i=1
Let Za2,o C Z.O. (resp. Z-02,0) be the subspace spanned by {e21} (resp. {e21, e22}). Let Z'a2iok = Za2,ok \ {0}. Then Za2,stk = Ma2kZ'02,ok We write elements of TAO in the form (5.1.2)
t ° = (a2(tll,t12), a2(t21,t22)),
A = d(A1, A2) = (a2(A1 1, A1), a2(A2 1,A2)),
to = At °,
where A1i A2 E IIB+, tij E Al for i, j = 1, 2. We define
dxt° = fl dxtij, dx,A = dxA1dxA2, dxto = i,2
dxAdxt°.
5 The case G = GL(2) x GL(2), V = Sym2k2 0 k2
155
The inductive structures of VBik and Zt, ,stk can be described in the following manner. Lemma (5.1.3) Let Hat C G be the subgroup generated by T and (TGL(2),1). Let Hv C G be the subgroup generated by T and TG. Then (1) Vst,k = Gk XHvk Z'V,ok, (2) Za2,stk = Ma2k XHa2k Z'ja,ok.
The proof of the above lemma is easy and is left to the reader. Let w = (w1i w2) be as in §3.1. We use the alternative definition (3.1.8) of the zeta function using L = V,', and use the notation Zv (4i, w, s), w, s). Let [ , ]' be the bilinear form such that 1
1
2
2
[x, y] v = X1OY10 + X12Y12 + X20Y20 + X22Y22 + 1 X11y11 + 1 X21y21
for x = (x2j), y = (yjj) E V. Then it is easy to see that this bilinear form satisfies the property [gx, tg-ly]', = [x, y]'v. We define [x, y]v = [x, TGy]'I, and use this bilinear form as [, ]v in §3.1. Let (5.1.4)
b((g°)`x) -
Js(D,g°) = xEVk\V,
I°((D,w) = f
Cg°x), xEVk\Vk°
w(g°)J8(4',g°)dg°. A /Gk
For A E R+, let 4ba be as in §3.5. By the Poisson summation formula,
Zv(4', w, s) = Zv+(I,, w, s) +
w-1, 6 - s) + JR+ A8I°((>a, w)dx A.
We will study the third term in §§5.2-5.5. Throughout this chapter we assume that li = Mv,u,4) (see (3.1.11)). Let 8(g°, w) be the smoothed Eisenstein series which we introduced in §3.4. We choose the constants in (3.4.3) so that C1 = 1, C = C2 > 4. This choice of constants C1i C2 will have some effect later as we mentioned at the beginning of this chapter. We define (5.1.5)
I ((D, w, w) = J A/Gk
Since J8(4>, g°) = ev,. (,D, g°)
w(g°)J(, g°)e(g°, w, )dg°.
- ev,: ($, (g°)`), by (3.1.6), (3.4.34),
I(4, w, w) ,., CGA(w; p)I°(4), w).
We define (5.1.6)
V,st(4), W, w) = JCGOIGk w(g°) ev,k (4), g°)-O(g°, w)dg°;
=v,st(4), w, w) = JG/Gk w(g° )ev(&,
(g°)t)(g°, w)dg°.
Part III Preliminaries for the quartic case
156
The functions evik (,D, go), Cvtk (', (go)`) are slowly increasing by (1.2.6). Therefore, by (3.4.34), the above distributions are well defined for Re(w) >> 0. Since e((go)`, w) _ 6°(go, w), "v,st(4p, w, w) = Ev,st(4) , w-1, w).
Let 6#(w) be as in §3.6. By (3.5.9) and (3.5.20), (5.1.7)
I(-P,w,w) = 6#(w)A(w; p)+0) - -P(0)) +
ep(vp(1p,w,w)+Ep(4p,w,w)) p,l(p)=l
+ °v,st(4D, w, w) - EV'st(-D, w, w).
If p = (tI's), and 0 = tIi, p belongs to class (3) in §3.5. Therefore, °p (4b, w, w) is well defined for Re(w) >> 0. This implies that °,p (4), w, w) is well defined for Re(w) >> 0 also. If tI = a2, t3, °p (4, w, w) = 0, and Ep (I, w, w) is well defined for Re(w) >> 0 by (3.5.5). Therefore, all the distributions in (5.1.7) are well defined for Re(w) >> 0. If t = 04, tI satisfies Condition (3.4.16)(1), and therefore, ,3p(,P,w, w) is well defined for Re(w) >> 0 also.
Let i- be a Weyl group element as in (3.4.13). We use the same notation as in §3.6 in this chapter. So sT = (Sri, Sr2) E C2 where s,i = ZiT1(2) - ZiT1(1) for i = 1, 2.
We define ds, = ds,lds,2. For a similar reason to that above, Bp,T(4D,w,w) is well defined for Re(w) >> 0 in the same cases as 3p (,D, w, w)
§5.2 The adjusting term In this section, we define the adjusting term Tv(W, w, s, Si). This distribution is required in order to describe the contribution from V. Let ' E 9'(Zv,oA). Let q = (ql, q2) E (A1)2, uo E A, µ = (µl,µ2) E R+, dx q = dxgldxg2, and dxµ = dxµldxp2. We consider the function a(uo) in §2.2. Let wl, W2 be characters of Ax /kx, and w = (Wl, W2). We define w(q) = wl(gl)w2(g2). Let fv('F, It, q, uo, Sl, 82) = µ2l a(uo)32 ww(µlµ2 lql, µ1µ2q2, 2p1L2g2uo)
Note `2' in the third coordinate. Definition (5.2.1) For complex variables s, S1, 82 and q1, w as above, we define (1)
Tv(W,w,S,sl,S2) =JR2
x(Al)2xA
w(q)µifv('y,µ,q,uo,sl,32)dxpdxgduo,
(2) Tv+('I, w, S, S1, S2) = JR2 x(A1 )2 xA w(q)pifv('y, p, q, uo, S1, S2)dxidxgduo, µ1>1
(3)
TV' (12, W, Sl, S2) = f
+x(A')2xA
w(q).fv('I',1, µ2, q, uo, sl, s2)dx p2dx gduo.
Proposition (5.2.2) The distributionTv(1Y, w, s, sl, S2) can be continued meromorphically everywhere, and is holomorphic for Re 2 s1 > 1, Re si > 1, Re si + ti Re(s2) > 2. The functions (2), (3) are entire functions.
2
5 The case G = GL(2) x GL(2), V = Sym2k2 0 k2
157
Proof. The convergence of (2), (3) is easy, so we briefly discuss (1). Suppose that there exist Schwartz-Bruhat functions 'y1 E 9'(A), ty2 E 9'(A2), such that 'F (Y1, y2, 2y3) ='I'i(yi)'P 2(y2, ya). Then
TV(W,w,5,s1,52) = 2E1(IFI,wl,
s
2s1)Tw(W2,w2,
s
2s1,s2),
where the last factor is as in §4.2. So TV (W, w, s, s1 i 52) can be continued meromorphically everywhere. In general, we can assume that IF = and the finite part W f of T has a similar decomposition. So in order to consider the meromorphic continuation of TT(q', w, s, s1i $2), we can assume that Tv(W, w, s, s1i s2) has an Euler product and its finite part can be continued meromorphically everywhere and is holomorphic for Re 2 sl > 1, Re si > 1, Re sl + Re(s2) > 2. One can prove that the infinite part of Tv(W, w, s, s1i 82) can be continued meromorphically everysi Re sl > Re + Re(s2) > 1 where and is holomorphic for Re 2 sl > using the polar coordinate as in §4.2.
2
2
a
a
Q.E.D.
Definition (5.2.3) For 'P, w, s, s1i s2 as above, we define (1)
Tv (iY, w, s, 1 s )= d - ds2
TV ('y, w, s) s1) $2), 82=0
d
(2)
TV+('y, w, s, s1) = ds 2 TV' (IF, w, s1) =
(3)
TV+('y, w, s, s1, s2), 82=0
d
1
ds2 82=0
Definition (5.2.4) (1) Let 'y1 E Y(Zv,oA). Let p, q, no be as in the beginning of this section. We define eZv,o ('I'1, µ, q, u0) _ E 1y1(1!1µ2 181x, µ1µ'242y, 2µ1µ2g2yu0) x,yEk"
Note `2' in the third coordinate. (2) Let 'y2 E Y(Za2,oA). Fort' E Ax, uo E A, we define 'y2(t'x, t'xu0)
eZa2,0('y2i t', n0) _ xEkx
(3) For ty3 E 9'(Za2A) and gal E G°A fl Ma2A, we define ®z;2,o ('y3, 9a2) =
1:
'ya(9a2x.
xEZa2,ok
The following lemma is clear from the above definition.
Part III Preliminaries for the quartic case
158
Lemma (5.2.5) Letw1iw2 be characters of A' 10, andw = (w1,w2). Lets1is2,q, p be as in (5.2.1). Then for kk E So(Zv,oA), 1
Ty (T, w, 81,S2) fR+x(Al/kx)ZxA
w(q)l4a a(uo)82eZVo ('Y,1, i 2, q, uo)dxµ2dxgduo.
§5.3 Contributions from 01, 03 In this section, we consider paths p = (0, s), p' = (a, s') such that D = a1 or 03 w-1, w). So we only consider p. For 0-sequences a = a1, 03, Ma = Tk. Since l(0) = 1, we do not have to worry too much about Ape etc. for these 3-sequences. So, instead of the the general notation go E G°A n MaA, we just use t° in (5.1.2) to describe elements of G°A n MaA. When we consider A, Ax in the next three sections, we consider the standard measures on them. We start with some definitions. and s(1) = 0, s'(1) = 1. Since e(g°, w) _ 9((g°)`, w), -Ep, (4D, w, w) = SEp
Definition (5.3.1) Letw = (W1,W2) be as before. We define (1) bv,st(w) = b(wiw2 2), wv,st = (w2,w2), (2) Sai (w) = b(wlw2 1)6(w2), wai = (w2,w2), (3) ba2(w) = 6(w2), wag = (1,w1), (4) 602,-t GO) = bas(w) = 6D4 (w) = b#(w) = 6(w1)6(w2)
Definition (5.3.2) For 4)1 E .9(VA), 42 E .9(Z02A), we defineRv,o4)l E.(ZV,oA) and Ra2,A2 E 9'(Za2,OA) by Rv,o,D(x12, x20, x21) = f 4(0, 0, x12, x20, x21, x22)dx22, JA
(1)
R02,A(x21,X22) = -P(0,X21,X22)
(2)
Definition (5.3.3) For 4 E .9(VA), and s = (s1i s2) E C2, we define
s1-1 s1 + 282 - 1
bal (w)
2
2
,
2
where the first (resp. second) coordinate of Rol4D corresponds to x12 (resp. x21).
For s E C, we define Eal (4, w, s) = Eal (4), w, s,1). Let 00
(5.3.4)
Ear (1), w, s) = E Ea"(j) (4', w, So)(s - so)i i--2
be the Laurent expansion at s = so. For the rest of this chapter, r = (r1, r2) is a point in R2. First we consider the case 0 = 01.
Lemma (5.3.5)
(-ib, w, w) - 0.
5 The case G = GL(2) x GL(2), V = Sym2k2 ® k2
159
Proof. Let a = (al, a2) E V. For u = (ul, u2) E A2, we define
< an(u) > = < alul >< a2u2 > . Let
fl,a(t°) = f
®y,1 (4', t°n(u)) < au > du,
NA/N,
f2,a(w, t°) =
JNA/N,
9(t°n(u), w) < au > du.
Then by the Parseval formula,
fl,(t°)f2,-(w,t°)(t°)-2Pdxt°
'=p(-D+w, w) = L2/T,. w(t°) aEk2\{0}
Lemma (5.3.6) For any N1, N2, N3 > 1, Ifl,a(t°)I < A 2A-'(A2a21)-N1A2N2(A2 12)N3. aEk2\{0}
Proof. Let ui = u'l (x, u) = x21u1 + x12u2i u2 = u'2 (x, u) = X12U2. Then u1 X2_1 (ui - u2), u2 =x12 u2. Since A/k is compact, f1,a(t°) is equal to l
fA2/k2 x12,:21 Ekx
( 712(t°)x12,
721(t°)x2i, 722(t°)(x22 + ui)) < au > du
x22Ek
X12,
kx
A2/k2
Ra1D(712(t°)x12>721(t°)x21,722(1°)(x22 + ul)) x22E:k
x < au > du. Since
< au >=< aix21 u1 >< (a2x12 - alx21 )u2 >,
we only have to consider terms which satisfy the condition a2xr2 = a1x21. This implies that a1i a2 are both non-zero. _ Let 1) 1 be the partial Fourier transform of Rat 1' with respect to the third coordinate and the character < >. Then it is easy to see that fl,a(t°) = Al 2A2 1 E 411(712(t°)x12, 721(1°)ala2 'X12, 722(t°)-1a2x12) x12Ekx
Therefore, if we choose '2 E 9(A3) so that I'D1I < (D2i I fl,a(t°)I
Al 2'x2 103(D2, 712(t°), 721(t°), 722(t°)-1)
aEk2\{0}
Hence, the lemma follows from (1.2.6). Q.E.D.
Part III Preliminaries for the quartic case
160
Let M > 1 + C. Clearly, If2,«(w,to)I = i1,TG,l(t0,w)) aE(kx )2
where I = ({1}, {1}). Let 6 > 0. By (2.4.10), for any 11, l2 >> 0, <
Ilig jTG l(t0, Z)I
for Re(z) E DI,TG,6i where c1,6(z) is a function of polynomial growth. We can choose q E Dr,T,6 so that L(q) < wo. Therefore, there exists S' > 0 such that for any11,12>>0, Oi,TG,1(t0, w) << A A 12
forwo - S' < Re(w) < M. This implies that for any N1, N2, N3 > 1 and ll, 12 >> 0,
Ifl,a(t°)f2,a(w,t°)I «al'a2A 2a21(a1A21)-NIA N2 (A
2)N3.
«Ek2\{0}
Since the convex hull of {(2, -1)j(0j1)j(-2, -1)} contains a neighborhood of the origin of 1R2, the above function is bounded by a constant multiple of rd2,N(Al,.\2)
for any N > 0. This proves (5.3.5). Q.E.D.
Lemma (5.3.7) w)
_
(2)7r11
2
fRe rl>13,rr2>1
a
('p,
w, sT)'+T (w, sT)dsT.
Proof. Suppose that f (q) is a function of q = (ql, q2) e (A/ k' )2. Then J
(Al /k x )4
w(t0)f(t12t21,t11t12t22)dxto = sa(w) J
wo(q)f(q)dxq (Al /k x)2
after the change of variable Ql = t12t21, q2 = t11t12t22. Let t' = (ti, t2) _ (1q1,µ242) E (Ax)2 where p1,a2 E R+, gl)g2 E A1. Let dxt' = dxtidxt2. We make the change of variable µa = Al2A2i µ2 = A21. Then dxµldxµ2 = 2dxaldxA2i and Al = jtl 2 112 2, A2 = µ2
It is easy to see that Qp(to)(to)TZ+P =
T1-1
P1 2
2
112
Therefore,
sa2w)
f
W,sr)
(Ax /kx )2 (Ax
Hence, Ea(4),w,sr)AT(w;sr)dsr. _
Q.E.D.
1
1>3, r2>1
5 The case G = GL(2) x GL(2), V = Sym2k2 ®k2
161
Lemma (5.3.8) Cp,T ($, w, w) - 0 unless r = TG. Proof. We choose rl, r2 >> 0 for r = (1, 1), r1 fixed and r2 >> 0 for r = (TG1,(2),1),
and r2 >> 0 and r2 fixed for T = (1, TGL(2)) Then L(r) < w°. This proves the lemma. Q.E.D. If r = TG, then srl = Zii - Z12 and Sr2 = Z21 - Z22
Proposition (5.3.9) Let r = TG and s = sTl. Then (OD, w, W)) 2,7f
0-1
Re( >3
1
2
1)0(s)A(w; s, 1)ds.
rl
Proof. By (5.3.8), ED(-P, W,ST)AT(w;Sr)dsT
((D,W,w)
rl>3, r2>1 2
27r
+
f
Re(.
l>3,
ST)'+T(w;ST)` s, 1>>r=2r
>O
60
27
-1
Re(1>3r1
If rl is close to 3 and r2 is close to 0, then r1 + Cr2 < 1 + C, because C > 4. Therefore, we can ignore the first term. This proves (5.3.9). Q.E.D.
Next, we consider a path p = (0, s) such that 0 = Z. Proposition (5.3.10) By changing V) if necessary, 1z p(4D,W,w) - 0.
Proof. If f (q) is a function of q E X41 /k x ,
f(A1/kX)4
w(t °)f
(t12t22)dx t °
= 6# (w) f
f (q)dq
Consider the situation in (3.1.18). In this case, AD = {d(pi, pi) µl E ]f } and Aa = Aa = {d(p2i µ2 2) µ2 E III. }. We make the change of variable Al = Then dxA = 5dxµldxp2. Let r, sT be as before. It is easy to 91/p2, A2 = see that epl(A) = y5 and and ATZ-p = Misr1 Esr2-3/le2ry-2sry-1. Also LS,,, + h in I
1
(3.5.17) is {sr I sr1 = 2sT2 + 1}. Therefore, by (3.5.17), 'Ep,T (qD, w, w) is well defined for Re(w) >> 0 for all T, and
6a '-'(7,T
W1 w) =
2ir
(w)
JR(.2)3 r2>1
E1(Ra4l, sr2 + 1)Ar(w; 28T2 + 1, ST2)dsT2
Part III Preliminaries for the quartic case
162
Let pT be the point in C2 which corresponds to p by the substitution in §3.6.
Then it is easy to see that pT = (1, -1) for T = (TGL(2), 1), p,. = (-1, 1) for T = (1, TGL(2)), and pT = (1, 1) for T = TG. For these cases, pT is not on the line s,1 = 2sT2 + 1. Therefore, by (3.5.19), we can change t/i if necessary and assume that Bp,T (4), w, w) = 0. For r = (1,1), we choose r1 = 2r2 + 1 >> 0. Then LT (r) = -r1 - Cr2 < w°. Therefore, p T ((D, w, w) - 0 for this case also. This proves the proposition. Q.E.D.
§5.4 Contributions from D2,04 In this section, we consider paths which start with /32. We do not consider paths
p = (cz,s) such that s(1) = 1 for a similar reason as in §5.3. Let p2 = (02i52) be
a path such that 52(1) = 0. Let Ni = 04,s4i) be a path for i = 1,2 such that 541,(1) = 0 and 541(2) = 0,$42(2) = 1.
We use the notation Eat (Ii, w, s), Ea2,ad(T, w, s)
for the zeta function, the adjusted zeta function for the space Za respectively. Also we use the notation Ta(Ra2,OW1w1s) etc. for the adjusting term etc. We define Ea2,ad,(j) (W, W, s°) similarly as in (5.3.4).
Since D2,04 satisfy the condition (3.4.16) (1), (5.4.1)
°p2 (D, w, w) =
w, w) + Ep2+(4, w, w)
p2#(4?,w,w)=p2#(4)1W1W) + yP42 (4?, w, w) - up41 (cb, w, w).
Also w),
P2+('P1 w, w) _ 12+(4D, w, w) +
w, w) =
w, w)
w, w).
Let f be a function on Za2A. Then J(A1/kx)z
= 6(w2)
w2(t21t22)w1(det91)f(t2291x)d"t21d
f
w1(detgl)f(t2291x)d"t22,
where gl E G1A, X E Za2A. It is easy to see that ape (a2(A1 1, Therefore, by (3.5.13), (5.4.2)
-1),1) = all
(R, 2),
ypz+(,D, w, w) ^ CGA(w; PS2 wED2+D2 I, wp2
.02+((k> w, w) - CGA(w; P)ba2 (w)Ea2+(-Fa2 Ra2 "P, wpzl, 1)'
Let gal = (n2(u°), 1)t°. We define X-02 = {kga2 I k E K, Al < a(u°)-2
5 The case G = GL(2) x GL(2), V = Sym2k2 0 k2
163
As in the case W = Sym2k2, Xa2/Tk is the fundamental domain for G°A/Ha2k. Let dg-02 = dxt°duo. Then p2,st+(4p1 W, w) = IX02 naA/z.k W(9-02 )E)Z0'2 (Ra2'P+9a2)A2 282 ( 2, 049021
1-'2 nA1aA/Tk <1
W (9az
w)dg
(Ra2'Pl 0.02 (9D2
Lemma (5.4.3)
f
(1)
^ (2)
a2 nBA/T,,
J J
2
(Ra2-D, 902)A22eN(t°, w)d9a2 a2 nBA/Tk
!C2 nBA/Tk ,.
W(a2)®z2(Ra21)) 2g (2,w)d
a2
w(2(Ra2 , Ba2 (2 ))A2 S,2 (21
fBA/Tk
W(9,2)eZa2
Proof. Let I = ({1}, 0). Then I8p2 (902 1 W) - -N (t°, W) I << E 41-,1(9021
w).
T
Let M > wo. By (3.4.30), for any e > 0, there exist 6 = 6(e) > 0 and c E t* such that Ilcllo < e and for any l >> 0, 0
(t)c)"1
4,73 (W121 w) <<
forwo-5
.+xAx /kx xA
Ai It'I`
where c' is some constant, and l' can be arbitrarily large. Since T0e+(`f'1 s, s1) converges absolutely for all s, i, the above integral converges absolutely. Q.E.D.
Let r, sT be as before. We define (5.4.4)
Ep2,st+((D, sr) =
T0e+(R02,0R024" 1 + Sr2, 12T1) Tae+(Ra2 °a2 Ra2 41, 2 - 9T2,
Sr) _
,
STl -1
Sr1-1
1
2 2
Part III Preliminaries for the quartic case
164
Lemma (5.4.5) (1)
J
nBA/Tk X1>1
(2) f 1
st(W) dal (2)
2
Ep2,st+(d),
sr)AT(w; ST)dsT.
R`(°*)=* 1>1, r2>1
fX02 nBA/zk W(ga2 )0Zaz.o (RaAf ea2 (go. ))A21eN,r(t0f w)d-jaz X1<1 2
daz,st(W)
Ep2,St+(4),ST)Ar(w;Sr)dST.
rI>', '2>1
In particular, these integrals are well defined.
Proof. Suppose that f (q) is a function of q E (Al /k")2. Then
f f
W(t°)f(tiit12t22)d"t° =
A'/kX)4
d2,st(W)
f
l /kx
Therefore, (1) is equal to da2,st(w)
6z02,0(. 52,0R524) t',UO).2eN,T(to
w)dx.2dxt'duo,
R+XAX /kX XA
where t' = )(2t11t12t22 Since (t°)'=+P = jt'js.2+1A1r1-1 the lemma follows for (1). The proof of (2) is similar. Q.E.D.
Lemma (5.4.6) If r 54 TG, 2 1
(1)
C27r / 1
(2)
C27rv/---l
f
E2,5+(, ST)XT(wj Sr)dST ^ 0,
rl>1, r2>1
)f 2
Re(ar)=* 1>1, r2>1
Ep2,st+(4)1 sr)AT(wisT)dsT .v 0.
Proof. Since Tae+( , S, Si) is an entire function for IF = Ra2,°Raz 4)> Ra2,oga2 Ra2 4)f we can choose rl, r2 >> 0 for T = 1, r1 fixed and r2 >> 0 for T = (TGL(2), 1), and r1 >> 0 and r2 fixed for T = (1, TGL(2)). Then L(r) < wo. This proves the lemma. Q.E.D.
By moving the contour in the above integrals so that r2 < 1, we get the following proposition.
Proposition (5.4.7) Let r = rG, and s = srl. Then (1)
(2)
WI w)
W, w)
da2,st (W ) 27f
R`(°)=*1 Ep2,st+(1)1 Sf 1WS)A(wi s, 1)ds, rl>1
2,P2,st+(, ()
27r
fRe(.)1 rl>1
L'p2
s, 1)¢(s)A(w; s, 1)ds.
5 The case G = GL(2) x GL(2), V = Sym2k2 0 k2
165
Next, we consider P41, P42. Let r, sT be as before. We consider the domain DT (2.4.7) for r = Re(sT) also.
Lemma (5.4.8) (1)
w7 w)
`-'p41 ,T
_
Qaa4(w)
E1(Ra4"Pp417
J
R`(")=r
-ST1 + 2ST2 + 1
R.("r)-r rEDr, -1>3-2r2
ST1 + 2ST2 - 3
rEDr, r1<2r2+1
(2)
p42,T ('D,
w7 w)
_
s.22t)..,
A,(w; ST)ds.r.
Pba4 (w) 27r
1/-
AT (wj ST)dST.
Proof. Consider sT such that r = Re(sr) satisfies the conditions in (1) and (2). It is easy to see that Ap412TA° = {t° 1 A2 < 1}, and Ap422TA = {t° 1\2 > 1}. Easy computations show that 1
0
2
2
0
2
O`p41 (t) = A1A27 O'p42 (t ) = AlA27 (t0)Tz+P = As"-'As 2-1,
1-3,2
A1f
Suppose that f (q) is a function of q E Ax IV. Then
(ti2t22)dxt° = ba4 (w) f/kx f(q)dxq fA1/kx)4 (
after the change of variable q = t212t22. Let t' = pq where µ E R+. We make the change of variable p = A A2, A2 = A2. Then dx pdx A2 = 2dx A1dx A2i and
A1=µ2A22. So o,P41
(t°)(t°)TZ+P
=p
',.1+2a,.2+1
1}a
2r1 A2
2
°P41 (t°)Ba2 (to)Z+P
= µ
l+s 2
-s
+2s
-3
A2
Since 1
2 JR2 xA1 /kx
,I1+ar, \ fA 2 ^2
a2<1
+1_ 2
2
OZ04 (R a41)p41 Y
II
` q)dJC ^2dx dx
q
)
-srl + 2ST2 + 1 1
fR+XA1/kx f'I 1+er7 A2 e1,+2ar2-3 OZ04 2
2
` 'I x x (Ra44)p42,pq)dA2d
/a] q
a2 >1
E1(Ra4'DP42'
Sr1+1 2
)
ST1 + 2ST2 - 3
the lemma follows. Q.E.D.
Lemma (5.4.9) If T TG, p4i,T ('D, w,w) - 0 for' = 1, 2. Proof. If T = 1 or (TGL(2),1), we fix r1 and choose r2 >> 0 for P41, P42. Then L(r) < wo and r1 < 2r2 + 1, r1 > 3 - 2r2. If r1, r2 are close to 1, then L(r) < wo
Part III Preliminaries for the quartic case
166
and rl < 2r2 + 1 for T = (1, TGL(2)) and P41 If rl >> 0, then rl > 3 - 2r2, and L(r) < wo for T = (1, TGL(2)) and P42. This proves the lemma. Q.E.D.
Proposition (5.4.10) Let r =To, and s = sTl. Then r (1)
1
3-s
A 1
s -1
a4 (w)
T
"V42 (`r, w, w)
(2)
21r
2) cb(s)A(w; s, 1)ds,
I/'
eva4 21r -1 J
w, w)
)
El(Ra44Dp42''
1v
,
¢(s)A(w; s,1)ds.
Proof. P41
- fr1=2, r2=i Re(ar)=r
+
Pty
AT(w, sT)dST
i)
2ST212
rl =r2=2
yy
E1(R04iDP41' 9 2AT(w;sT)dsT !l -STl + 2S-r2 + 1 Jn1(Ra4
(w) 1
ON (w)
27r J
p41,
(s)A(w; s,1)ds
3-s
2e(.>=rl
E1(R04
0(s)A(w; s,1)ds,
3-s
R<*i<3
because L(2, 42) = 2 + AC < 1 + C (C > 4). This proves the first statement. 4 El(R041)P41,-2-)AT(w;s,)dsT
ti
$T1 + 2ST2 - 3
JRe(sr)-(2,4)
+
Paa4(w)
2
J 2aV-1 L-(.)=j 1
S- 1
Pba4 (w) /' 27rVr--1 1
1
)
)
0(s)A(w; s, 1)ds
q,(s)A(w; s, 1)ds
for the same reason. Q.E.D.
Proposition (5.4.11) p2#((D, w, w) - A(w; P) Pb#(w)1b(0)
(1)
2
'p2# ((D, w, w) " -A(w; P)96#(w)`7'02 Rat Y (0).
(2)
Proof. Let 7-2 be a permutation of {1, 2}. By the Mellin inversion formula, _
2
(27rV
+) I
1A/G1kXT2A/T2k
b#(w) f
a2<1
A(1,T2)(w; -1, s) e
rl>2
s-2
ds.
w(91,t2)A21.I{l}xN2,(1,T2)((91,t2),w)d91dxt2
5 The case G = GL(2) x GL(2), V = Sym2k2 ®k2
167
If T2 # TGL(2), we can choose r >> 0, and ignore this function. 11 72 = TGL(2), 1
27r-,/
A(,,,2) (w;
1f
-1,
s-2
*1>2
s) ds
- -Pb#(w)A('w; p).
Similarly, 2
(27r) fGO1A/G1kXT2A/T2k w(gl,t2)A2r'11}xN2,(1,T2)((g1,t2),w)dgldxt2 a2S1
= b#(w) 2ir
JRl >1
A(1T2)(w;-1,s)ds
8+1
Bb#(w)A(w; p) 2
This proves (5.4.11). Q.E.D.
§5.5 The contribution from Vsik In this section, we consider the contribution from V. Let b = (n2(uo),1)t°(1, n2(u2)), db = ))2 dxt°duodu2,
where t° E TT°, uo, u2 E A. We define
Xv={kbI kEK,A1
Proposition (5.5.1) Let T =TG, and s = sT1. Then v,st(
w w)
TV' (Ry,o(D, wy,st,1, Pbv,st (w) f J R<(<>=r1 s-1 2ir
2 0(s)A(w; s, 1)ds.
*1>1
Proof. By (5.1.3), =V,st(w, 4p, w) = f
A/Gk
w(g°)Ov(, g°)6(g°, w)dg°
JA/Hv k
w(g°)Ozc o ('b, g°)e(g°, w)dg°
JX1 fBA/Tk w(b)Ozv
0
(41,
w)db.
Lemma (5.5.2) Let c1i c2 be any constants such that c1 +2c2 > 0. Then the integral JAil x v nBA/Tk
A22Ozv.u (4, b)db
Part III Preliminaries for the quartic case
168 converges absolutely.
Proof. Let µ = a12A2, and IF = Rz,,,o4D. It is easy to see that
f
Ozv (4), b)du2 = .112A-'6zvo (91, 1, µ, q, uo),
where q = (qi, q2), qi = t12t2l, q2 = ti1t22 Therefore, the above integral is equal to A11-2Ac2+1 eZV
fRx(A1/kx)2xA a':5
o (`1',1, µ, q, uo)dx aid x µdx gduo
-(UO)-1
1 +2C2 µC2+10
=
1
zv,o (',1,µ,q,u0)dxA1dx/tdxqdu o
R+x(A1 /kx )2 xA
a(-0)-1
a1
- I2 x(A1/kx)2xA _
C2+1
µ
< _ 1+2c2
a(uo) Cl +2c2
2
pzv,u('I',1,µ,q,uo)dx µd xt°duo,
which converges absolutely.
Q.E.D.
Lemma (5.5.3)
f
(b)O, (, b)e(b, w)db v nBA/Tk
f
w(b)Czo (, b)eN(t°, w)db.
XV n BA/Tk
Proof. Let M > w°. By (3.4.30), there exists a slowly increasing function h(A1, A2) and a constant 6 > 0 such that for any 11, l2 >> 0, 1-1(b, w)
- -IN (t', w) I << h(\1, A2) sup(A
,
a2, A A2 )
for w° - 6 < Re(w) < M. Therefore this lemma follows from (5.5.2). Q.E.D.
Suppose that f (q) is a function of q E Ax /kx. Then
J
A1 /k x )4
f(ti2t21,ti1t22)dxt°=6v,st(w)f A1 /kx )2 wv,st(q)f(q)dxq
2 q2 = ti1t12 We make the change of variable after the change of variable q1 = t2 µ = A-2A2, Al = Al. Then dx µdx Al = dx Aldx A2, and A2 = .1iµ Clearly,
IA ez; ,o
(n2(uo),1)t°(1, n2(u1))duo = A 1
2
A2
1ezv,O (I',1, µ, q, uo).
Therefore, 8v,gt('D, w, w)
^ 6v,st(w) f
R+x(A1/kx)2XA A1<
a(up)-1
wv,st(q)6zv,o('P,1,µ,q,uo)µ.'N(t°,w)dxaldxµdxgduo.
5 The case G = GL(2) x GL(2), V = Sym2k2 0 k2
Let r, sr be as before. We define
169
Ev,St,r (Z , w, w) by the following integral
(5.5.4) bv,st (w) R+x(A1/kx)2xA
wv,st (9)ezvo (`I',1, p, q, uo),a.N,T(to, w)dxa1dx µdxgduo,
al< a(u0 )-1
if this integral is well defined. Let (5.5.5)
Ev,st(p,w,sT) = bv,st(w)
TV1 (RyoD, wy,st, Sr21
-srl-23.2+31 2
Sr1 + 2Sr2 -3
It is easy to see that (to)rt+a = Ai,1+2sr2-3µ3r2-1 Therefore, we get the following proposition.
Proposition (5.5.6) The distribution E'vt,T(4>, w, w) is well defined for all r and
vst,r('p, w, w) - (
2
Re (ar>=r
27r
rE Dr, rl>3-2r2
If r # rG, we can choose r1 or r2 >> 0, and v,st,r(4), w, w) r 0. Let r = TG. Then C27r
+
2
-i
f TV' (Rv,o4, wv,st,1, 123)
Pby,st(w) 21r
1
S-1
Re(e)=r1
rl>1
¢(s)A(w, s, 1)ds.
Since C > 4, 2 + s C < 1 + C. Therefore, we can ignore the first term. This completes the proof of (5.5.1). Q.E.D.
§5.6 The principal part formula In this section, we prove the principal part formula for the adjusted zeta funcThis implies that
tion. Throughout this section, we assume that 4) = MD,u,, Ra 4) = RD 4) for all the path p. We define
J1(4),w,S) =
bal(es) 2
E2(Ral-D,wal,
s-1 s+1 2
2
+ ba2,st (w) T0e+(Ra2,oR02 'b, 2,
1 2
s-
Ta2+(R02,0 g-02 Ra2 -P,1,
+ ba2,st (w)
+ba3(w)
12 s)
s-1
(El(Rs4 42 s-1
s-3 1_8)
1
+
8)
bv,st(w)Tv(Ry,oP, wv,st,1,
s-1
2
Part III Preliminaries for the quartic case
170
J2(1b, w) = Sae (w) (Eaz+(Raz , w02, 2) + Eat+(-97a2 Ra2
wa2, 1))
(0)
Then, by (5.3.9), (5.3.10), (5.4.2), (5.4.7), (5.4.10), (5.4.11), (5.5.1),
w, w) -
0
27rV /-l IR.(j )1''1
(Ji( &, w-1, s) - J1(), w,
s,1)ds
+ CGA(w; p)(J2($, w-1) - J2(4,, w)).
We define
J3(P, w) = 80 1(w)E0 1,(o) (C w, 1) Sa2,st (w) 2
(Ta2+(Ra2,oRa2 1D, 2) + Tae+(Ra2,oRa2 gal , l))
ba42w)
+ _
El,(1)(R04(Pp42,1) sa4 (w) 2
(El,(-1)(Ra44PD41, 1) + El,(o)(Ra4(I)p41, 1))
by,st(w) 11V 2
wv,st, l)
Tv k
l))
+ Sae (w)(Ea22+(Ra2'P) wag) 2) +
-'2728#(w)
(2o
) + 24)(0)) +
Then, by Wright's principle, Jl w-1, s) - Jl (4), w, s) must be holomorphic at s = 1, and the following proposition follows.
Proposition (5.6.1) I°((b,w) = J3(&,w-1) - J3(-D, w). Let IF = Ra2 41). Then, by the principal part formula (4.2.15), Ea2,ad,(o) (
)wag, 2) =Eat+('1') L')02)2) + Eat+0, wa2', 1)
8(1) 2
(Ta2+(Raz,o'y, 2) + Tae+(Ra2,o'I',1))
- 9326(w1) 8(w1) 2
(o2wo) +
F(0))
1)(Ra44)p4111) + El,(o)(Ra44Dp41,1))
Therefore,
8a42w)
J3(,b, w) = 8a 1(w)Ea1,(o)(11, w,1) +
+8a 2 (w)E 0 2,ad, (o)(Raz
E1,(1)(Ra44)p42, 1) +X28# (w) -P (0)
, waz ,
2)-8v,st(w)Tv(Rv,o-P,wv,st,1).
The following relations are easy to prove, and the proof is left to the reader.
5 The case G = GL(2) x GL(2), V = Sym2k2 ®k2
171
Lemma (5.6.2) W,1)
(1)
=
w,1)
2
-,\-2(log.1)Eal,(-1)(,b,w,1) + A-2EO,,(o)(4b,w,1). w-1, 1) =
(2)
1A-4(logA)2E0,,(-2)*
w-1, 1)
+ A-4(1og.A)Ea1,(-1)(4',w-1, 1) 1).
2A-2(log
(3)
E02,ad,(o)(R021'a,w02,2) =
A)2E02,(-2) (Ra24P,w02,2)
- A-2(log A)Ea2,(-1) (Ra2
W0212)
+ ^-2Ea2,ad,(o) (Ra2 `l', W021 2).
-4(log )2Ea2,ad,(-2) (Ra2 $, wa2 2)
E02,ad,(o) (Ra2 'a, wa21, 2) =
(4)
+2 \-4(log A)F'a2,ad,(-1) (Ra2 $, wa21 2) + A-4Ea2,ad,(o) 0, w0-21, 2).
E1,(1)(Ra4")Ap42,1) = IA-2(logA)2E1,(-1)(Ra4Dp42, 1)
(5)
+ A-2(1ogA)El,(o)(R04')p42,1)
+A-
2E1,(1)(R04'p42,1).
E1,(1)(Ra4Cp42,1) = 2a-4(1ogX)2E1,(-1)(Ra3-9702$,1)
(6)
+-4E1(1)(Ra3
1).
Since
-6(2wl)E1,(-1)(Ra4''p42,
2) _ Ea2,ad,(-1)
w02, 2) =
6(wl) 2
1),
E1,(o)(Ra4 'bp42,1),
etc. and 604 (w) = E(wl)6a2 (w), 6a,2w) 602 (w)Ea2,ad,(o) (Ra2
(w)A-ZE02,ad,(o) 602
, w0212)
+
E1,(1) (Ra4 Ap42 ,1}
(Ra2 (I), W02,2)+ 6a42w) A -2E1,(l)(Ra4 DP42,1), 1
602 (w)Ea2,ad,(o)(6'\p42,w02 ,2) +
= 6a2 (w)A-4Ea2,ad,(o)(R 12
f
6a4 (w) 2
w-1) +
,T
El,(1)(R04Y'Ap42) 1)
6042w}A-4E1,(1)(R04p42,1).
Part III Preliminaries for the quartic case
172
Definition (5.6.3) wv,st, s - 1,1).
w, s) -
Zv,ad(4D, w, s) =
We call Zv,ad(4, w, s) the adjusted zeta function, and 6"2 " )Tv (Rv,o4), wv,st, 81, 1) the adjusting term. It is easy to see that 1
f AsTv(Rv,o(Pa, wv,st,1)d"A = Tv(Rv,o-P, wv,st, s - 1,1) 0
- Tv+(Rv,o I, wv,st, s - 1, 1),
_
1
wvgt,1)d"A = Tv+(Rv,o$,wv,gt, 5 - s,1). 10
We define w) = bat (w)Ea2,ad,(o) (Ra2 C w02 2) +
aa (w) 2
E1,(1) (Ra, P42' 1).
Then, by the above considerations, Zv,ad(4,w,s) satisfies the following principal part formula.
Theorem (5.6.4) Zv,ad(4, w, s) = Zv+(4D, w, s) + Zv+($, w-1, 6 - s)
- 6v5(w) 2 dy,st(w) 2
wv,st, s - 1,1)
Tv+(Rvo1P, wvst) 5-8 ,1)
+ Z2s#$(0) 2 (w)
s-6
- 4'(0)
+ E($, W-1)
s
s-4
3
+
(-1)i+1 Ea1,(-j+1) ( w-1) E (s - 4)j j-1
- E('D, w) s-2
_ Ea1,(-j+1) (4, w) (s - 2)j
The following functional equation follows from the above formula.
Corollary (5.6.5) Zv,ad(4, w, s) = ZV,ad($, w-1, 6 - s). Also it is easy to see that Ea1,(-2)
w) = 26#(w)E2,(-1,-1)(Ro, ID, 0, 1), (C w) = 601(w) (E2,(-1,o) (Ra1(p, wa1, 0,1) + E2,(-o,-1) (Ra14 , wa1, 0, 1)),
Ea1,(o) ('P, w)
=dal (w) (E21(o10)(Ra1 2
+
da1(w)
Lao, 107 1) + E2,(-1,1) (Ra1(), wa1, 0, 1)
wa1, 0,1)
Chapter 6 The case G=GL(2) x GL(1)2, V=Sym2 k2® k §6.1 Reducible prehomogeneous vector spaces with two irreducible factors We consider two prehomogeneous vector spaces G = GL(2) x GL(1)2, V = Sym2k2 ® k,Sym2k2 ® k2 in this chapter and the next chapter. In both these cases, V is of the form V = V1 ® V2 for some irreducible representations V1, V2. Therefore, we consider such representations in general. We assume that Vks ¢ 0. We choose T+ as in §3.1. If we restrict the weights ryi to T+, the convex hull should contain a neighborhood of the origin. This can happen only if dim T+ < 1. If dimT+ = 0, G1 = G°A, and we considered such cases. Therefore, we assume that
dim T+ = 1. We choose an isomorphism a : R+ - T+. Then any element of G1 is of the form gl = a(A1)g°, where Al E R+, go E G. We define kv1, icv2 to be the rational characters defined by the determinants of GL(V1), GL(V2) respectively. Let rcv. (gl) = Iicv (gl)I-1 for i = 1, 2. We assume that a(X1) acts on V1, V2 by multiplication by A', Ai for some a > 0, b < 0. This means that a dim VI
iy1 (a(A1)) = Al
,
Icv2 (a(A1)) =
Ai dim V2
For i = 1, 2, let [, ]v, be a bilinear form on V such that [gxi, 9Lyi]v; = [xi, yi]V, for all x, y, where g` = TGtg-'TG as before. Let [ , ]v be a bilinear form on V defined by the formula [(xl, x2), (yl, y2)]v = [xl, y1Jv1 + [x2, y2]v2.. For P E S°(VA),
let Ry'I E .Y(V A) be the restriction for i = 1, 2. For P E 9'(VlA) (resp..So(V2A)), we define a Fourier transform 9v1 4) (resp.." v1 4)) by [ , ] v1 (resp. [ , ] v2 ). We also define partial Fourier transforms gv1 CF, gv24) for AD E .O(VA). Since the restriction
to Vl or V2 gives the same result, we use the same notation. If g' E GA,
(6.1.1) 6v,--(4),g1) = icv(9')ev,,. (-q'v4', (9l)`) + iv(91)9v4)(0)
+ E (iv(g1)es, (-q'V4, (g')`) - E s
- D(0) g1))
SO0V1,V2
+E
(iv(g')®s (9v
,
(g')`) - OSp (,D, g'))
S1CV1 or V2
Lemma (6.1.2) (gl)`) - esp(,D,gl)) + tv(9')9v4'(0) -,D(0) SoCV1 or V2
Gsp(gv1gv(D,9')+lcv(9')
=1cv2(9')
SpCV2
SpCV1
esa(4,g')-iv2(91) E @Sp('Fv24'1(9')L) S/CV1
SpCV2
Part III Preliminaries for the quartic case
174
=kV (91) > 19Sp(-q17V4,(91)1)+kV,(9') 1: es'(9v29vp,g1) spcv1
S19cv2
-kvJ91) >2
as"(b,91)
s1cv1
spcv2
Proof. This lemma follows from the following relation (6.1.3)
kv(91) > -v4((91)`xl, 0) + kV(91) > -07v4'(0, (91)`x2) x1EVlk\{0}
E
x1EV1k\{0}
40, 91x2) + kV(91)9V4D(0) -'b(0)
'D (91x1, 0)
x1EV1k\{0}
x1EV2k\{0}
= kv2(91) > 9v1Fv'D(91x,,0)+kV(91)
-
9vD(0,(91)'x2) x1EVlk\{0}
x1EV1k\{0}
> ID(91x1, 0)
>2
- sV2(91)
x1EV1k\{0}
.v24D(0, (91)`x2)
x1EV2k\{0}
= kV (91) > -Fv4)((91)`x1, 0) + kv1 (91) > - kv1 (91)
>
91x2)
x2E V2k\{0}
x1 E V1k \{0}
'Fv1'((91)1x1, 0) - > t(0, 91x2)
x1EV1k\{0}
x2EV2k\{0} Q.E.D.
Consider a 3-sequence 0 such that 1(a) = 1 and So
V1, V2.
Lemma (6.1.4) Let g1 = a()1)g° be as before, and X as in §3.1. Then the integral
fA/Gk
w(91)X(91)es, (4,,
91)8(90,
w)dxgl
is well defined for Re(w) >> 0.
Proof. Since Sa St V1i V2, for any N1, N2 >> 0, there exists a slowly increasing function hN1 iN2 (g°) of g° such that es, (4, 91) << hNl,N2 (90) inf(Ai 1 A 1 N2 ).
W e choose N1, N 2 large enough so that the function inf(A tai N2)X(a(.X1)) is integrable on T+. We fix such N1i N2. If Re(w) >> 0, the function hN1,N2 is integrable on the Siegel domain. Therefore, the convergence of the above integral (9°)9(90, w)
follows. Q.E.D.
The following lemma follows from the same argument as the above lemma.
Lemma (6.1.5) Let X be as in §3.1. Then the integral v,8t(D,w,X,w) = JG/Gk w(91
)X(91)ev(,91)g(9°,w)dxl
6 The case G = GL(2) x GL(1)2, V = Sym2k2 ® k
175
is well defined for Re(w) >> 0.
Definition (6.1.6) Let 0 = (a) be a (3-sequence such that Sp C V1 or V2. Let X be as in §3.1. We define J GA/Gk
w, X, w) =
ai>1
w(91)X(91)es5 (D, 91)f
(9°,
w)dg'
foA/Gk w(91)X(91)esa (41, (91))8(9°, w)dg1
S,6 CV1i
SOCV2i
ai>1
<(D, w, X, w)=
fGA/Gk w(91)X(91)esp (4), (91)`)x(9°, w)dg1
SoCVI,
J GI/Gk w(g1)X(91)esp (-D, g')e(g°, w)dg1
SpCV2-
ai<1
X1<1
If Sp C V1, we define (6.1.7)
Va,1((D, w, X, w) = --- 0, >- (9V, 90b, U), r. V2X, W), 77
0,2 (D, LO, X, W) = Ea,> (b, w, X, w), 0,3(.D,
X, w) = =-a,
0,4 (-P, w, X, w) = -=D,< (gv, -P, w, Kv1 X, W).
Also if Sp C V2, we define (6.1.8)
Ea,1(b, W, X, w) = za,>
w) 'vX,w),
0,2 (4), w, X, w) = ca,> (6 V2 Y', w, K'V2 X, w),
=a,3(,,, w, X, w) =0,4 (p, w, X, w) _ ---D,< OD, w, X, w)-
We also define =S, at C W, X, w ,
a,at,i N', W, X, w
similarly by considering SQk, Sf,$tk.
Proposition (6.1.9) The distribution 'Ea,;,(4), w, X, w) is well defined for all i if Re(w) >> 0.
Proof. Since the argument is similar, we prove this proposition for =a,1(4p, w, X, W)
for Sa C V1. Let g1 = a(X1)g°. By (1.2.6), for any N >> 0, there exists a slowly increasing function hN(g°) such that Osa (4D, g1) << Al NhN(g°)
for g° E 60. We fix such N so that AINX(a(.ll)) is integrable on the set {Xl A, > 1}. Since hN(9°)&(g°,w) is integrable on 67° if Re(w) >> 0, this proves the I
proposition. Q.E.D.
An easy consideration shows that
-1X-1 ,w). 0,I(C w,X,w) _ °0,4(9VC w -1 kV
Part III Preliminaries for the quartic case
176
Therefore, (,D, w, X, w) for i = 1, , 4 can be constructed from any one of them. Let Sa C V1, and a = (,3). Suppose that YO = Za for all Sa C V1 or V2. Notice that this condition is satisfied by our two representations.
It is easy to see that (6.1.10)
X(9a)t(9a)-2pa Gza (RaID, 9a).ua (9a, w)dg
W, X, W)
.
cAJInMDA/Mok a1>_1
We define Js(4),gl) similarly as in (5.1.4) Let 4
W) =
(6.1.11)
totOD, W, X,
W, X, W), SacV1,V2 i=1
I(-I), W,X,w)= J A IGk
W(91)(91)Js(,91)e(90,w)dg1.
We have proved the following proposition.
Proposition (6.1.12) Suppose that YD = Za for all Sa C Vl or V2. Then X, W) =
E f i(a)=1,SagV1,V2 -
A/Gk
W(91)X(91)Kv(91)Gsa (, (91)(90
fA/Gk 1(a)=1,S5QV1,v2
W(91)X(91)Gsa(-D,91).(90
w)dxgl
w)dxgl
+ tot(,', W, X, w) + "V,st* W-1, /£V'X-1, w)
W, X, w).
§6.2 The spaces Sym2k2 ® k, Sym2k2 ® k2 Let V1 = W = Sym2k2, V2 = k or V2 = k2 and V = V1 ® V2. We consider these cases in this chapter and the next chapter. We write elements of V in the form x = (x1, x2) = (x10, x11, x12) x20), where xl E Vi, X2 E V2, or
x = (Xl,X2) = (x1o, x11, x12, x20, x21), where x1 E V1i X2 E V2.
Let G1 = GL(2), Gi = Ti = GL(1) for i = 2, 3, and G = G1 x G2 x G3. The group G acts on V by (91,t2,t3)x = (t291x1,t3x2) or (t291x1,t391x2) for g = (g1,t2it3) E G.
Let T C G be the kernel of the homomorphism G -+ GL(V). Consider a rational
character Xv(91,t2it3) = (detgl)°'t22t33 of G. This character is trivial on T if
6 The case G = GL(2) x GL(1)2, V = Sym2k2 ® k
177
and only if al = a2 or 2a1 - 2a2 - a3 = 0. We choose such a,,a2ia3 > 0 so that they are coprime integers. Then we can consider Xv as a character of G/T
and it is indivisible. We consider (GIT,V). We fix al = a2 = 5,a3 = 3 for V = Sym2k2 ® k, and al = 5, a2 = 3, a3 = 4 for V = Sym2k2 ® k2, because these are the normalizations which we will use in later chapters.
We have to choose T+ C T+ (see §3.1) so that T+ = T+/T+. We can choose T+ _ {(1, t2, t3) E T2+ X T3+ I t2 t3 = 1} or T+ = {(1, t2, t3) E T2+ X T3+ I t2 t3 = 1}.
In Chapters 6 and 7, t11, t12, t2, t3 are elements of Al. We define t1(A1) _ a2(A1 1, Al) and t2(.\2) = (a2, A2 5) or (A2, 2 3) for 1i 2 E IIB+. Let
d(A1,\2) = (t1(A,),t2(\2)) Then T+ = {(1, t2(A2)) I A2 E 1R.+}. Let t1 = a2(tll, t12). We define (6.2.1)
t ° = (tl, t2, t3), to _ (t, (A,), 1)t °, tl = d(\1, A2)t °.
We use this notation in Chapters 6 and 7. Let dxto - dxtlldxt12dxt2dxt3, dxt° = dxaldxt°, dxtl = dxAldxA2dxt°. Let GA, be as in Chapter 3. Any element of G' is of the form gl = t2(\2)9° where g° E G. We use dxt1 for the diagonal part to define a measure dgl on
G. Let dg = dxAdgl. We define an action of GA on VA by gx = Aglx, where we consider the ordinary multiplication by A. Let eij be the coordinate vector of xij. Let yij be the weight of ei3 i.e. teij _ yij (t)eij for t E T. We identify t with {(zl, z2) E R2 I zl + Z2 = 0} x R by d(A1, ).2)(-a'a;b) = A2a.>,2
Let
((a, -a; b) (a, -a ; b'))
-
r 2aa' + lbb' s
Sl
2aa' + 14 bb'
Sym2k2 ®k,
Sym2k2 ® k2.
Then this is a Weyl group invariant inner product on V. We use this particular inner product, because it is the inner product which we will consider in Part IV. We can identify yij's with elements of R2 as follows.
(/ 3/)
(v2,
1
(7
14
3
14)
Note that the origin is above the line segment joining the two points (f, 14)
and (- 1f , -
3 ) 14
Part III Preliminaries for the quartic case
178
Let be the parametrizing set of the Morse stratification. Then elements in S \ {0} for V = Sym2k2 S k are as follows. Cl
C2
C3
C4
I Elements in 93 \ {0} for V = Sym2k2 S k2 are as follows.
C3
C4
6 The case G = GL(2) x GL(1)2, V = Sym2k2 ® k
179
C7
We choose SL(2) x {(t3,t-5)} or SL(2) x {(t4,t-3)} as G" in §3.2. Then Zp etc. are as follows. Table (6.2.2) (for V = Sym2k2 (D k) ZO R N1. = 2 (-1, 1; -2)
x12, x20
,32 = (0, 0; 3)
x10, x11, x12
03 = (-1, 1; 3)
x12
,34 = (0, 0; 5)
x20
WO
Moll
-
{(a2(t-l, t), t3, t-5)} SL(2) x {1} (t-3, t3),
{(a2
t-6, t10)}
SL(2) x {1}
Table (6.2.3) (for V = Sym2k2 (D k2) (3
Nl = 16 (-1, 1; -6) RR
Q2 =
q
(-1, 1; 2)
Z13
W13
Moll
X12, x20
X21
{(a2(t-3, t3), t4, t-3)}
x11, x21
x12
{(a2(t-l, t), t-4, t3)}
Q3 = 10 -L(-7,7; -2)
x12, x21
X34 = (0, 0; 4)
x10, x11, x12
05 = (-1, 1; 4)
x12
,66 = (- 2,
-3)
2; X37 = (0, 0; -3)
X21
X20, x21
-
-
t2), t28, t-21)}
{(a2(t-2
SL(2) x {1} {(a2(t2, t-2), t4, t-3)} {(a2(t3, t-3), t-4, t3)}
SL(2) x {1}
Consider V = Sym2k2 ® k. It is easy to see that x E V is k-stable if and only if (x10, x11, x12) is k-stable with respect to the action of GL(2) and x20 # 0. In particular, Vk 0 0. Let Di = ((31) for i = 1, , 4. By Table (6.2.2), the weights of x12i x20 with respect to M'1 are t5, t-6 respectively. Since there are both positive and negative weights, Z1 1k = {x E Za,k I x12,x20 0 0}. We identify Z.O. with W, and consider Zazk, Z02 O etc. Let Eat (IF, w, s), Ea2i (W, w, s) be the zeta function, the adjusted zeta function respectively. We also use the notation Tae (R02iOT, w, s) etc. for the adjusting term etc. Let Zero (resp. Z'1,,0) be the subspace spanned
0}. Let by {ell, e12, e20} (resp. {ell, e20}). Let Z'i. Ok = {x E Zc,Ok x11, x20 Hv = H.O. be the subgroup generated by rG and Tk. Then Vgtk = Gk X H,,, Z'i, ok, I
and Sa2,stk 2-'- Gk XHa2,, Z'aZ,ok Since Ma acts trivially on Z?, for D = D3, 04i Zak _
180
Part III Preliminaries for the quartic case
Zak \ {0} for 0 = 03, N. We do not consider /3-sequences of length > 2. Let Rv,o4>
be the restriction to Zv,o Consider V = Sym2k2®k2. Note that if a convex hull of a subset of ryij's contains
the origin, it contains a neighborhood of the origin. Therefore, Vk8 = Vk. Let Q be the binary quadratic form which corresponds to (x10, x11, x12), and l the linear function which corresponds to (X20, X21). Then it is easy to see that x is semistable if and only 1 does not divide Q. In particular, Vk8 # 0. Let Di = (/3i) for i = 1, , 6. By the above table, the weights of x12, x20 with respect to M0"1 are t10, t-6, the weights of x11i X21 with respect to Mat are t'4, t4, and the weights of x12, x21 with respect to MQ3 are t32, t-19. For these cases, there are both positive and negative weights. Therefore, Talk = {x E Za1k I x12, x20 ss
752k = x E Za2k I x11, x21 54
0}, ,
ss
Za3k = x E Zask I x12, x21 54 0}.
We identify Z54 with W, and use similar notations to D2 of the previous case. Let Ha4 be the subgroup generated by TG and Tk. Then Sa4,stk - Gk XHs4k Z'a,,ok Since Ma acts trivially on Za for 0 = 05, D6, Zak = Zak \ {0} for 0 = c05, c06. As in Chapter 5, ZR1k = 0. We do not consider /3-sequences of length > 2. In both cases,
Vk\{0}=Vk8llflSe,k. We define ez,.o u0) etc. for cl = D2 in the first case and for 0 = Z4 in the second case as in (5.2.4). /kx)3. We define w(g') as in §3.1. It is Let w = (wl, w2, w3) be a character of (Ax easy to see that rcv(gl) = XZ 4 rcvl (g') = X2 9 and 1v2 (gl) = X2, for V = Sym2k2®k. Also ICv(gl) = X2 6, r,Vi (gl) = A 2 12' and K y2 (gl) = a2 for V = Sym2k2 ®k2. For V2 = k2 or k, let [ , ]' be the standard inner product, and [x, y]v2 = [x, rGy]',2. We defined a bilinear form [ , ]v for V = Sym2k2 in §4.1. We use this bilinear form as [, ]v1 in §6.1. We define J8(4),g1) similarly as in (5.1.4). Let (6.2.4)
w(gl)(gl)J8(,gl)dg',
Il((D, W, X) = J AIL,.
I(4), w, X, w) = f
A/Gk
w(gl)(gl)J8(,gl)e(g°,w)dgl.
(J$ (,D, g1) = J(41, g1) for Sym2k2 (9 k2.) We define the zeta function using (3.1.8) for L = Vk for the first case, and L = Vk8
for the second case, and use the notation Zv(,D, w, X, s) etc. Let X(t2(1\2)) = \2 for some c E C. By the Poisson summation formula, Z01, w, s) = Zv+(-D, w, X, s) + Zv+($, w-1, rcv1X-1, N - s)
+ JR+ ABIl(`I'a,w,X)d'A,
6 The case G = GL(2) x GL(1)2, V = Sym2k2 ® k
181
where N = 4 for the first case, and N = 5 for the second case. By (3.4.34), - CGA(w; p)I1(4, w, X)
1(4),
As in Chapter 5, we assume that 4P = Mv,4,-P in this chapter and the next chapter.
§6.3 The principal part formula In this section, we prove a principal part formula for the zeta function for the space Sym2k2 ® k. Since the zeta function is almost a product of the Riemann zeta function and the zeta function for the space of binary quadratic forms, we only outline the proof.
We assume that c # -5, -2,0,1,3,4,9. Let T E 9(Zv,oA). Suppose that q = (ql, q2) E (Al )2, uo E A, and p = (91,µ2) E 1R+. Let a(uo) be as in §2.2. Let dxq = dxgldxg2i and dxµ = dx pjdx p2. Let w1, W2 be characters of Ax/kx, and w = (w1, w2). We define w(q) = wl(gl)w2(g2). Let fV (`L) p, qi uo, S1, S2) = µ23 a(uo)$2
(1µ2g1, Alµ2gluo, p1/t2 5q2)
Definition (6.3.1) For complex variables s, sl, 82 and tI1, w as above, we define (1)
TV(1,w,s,S1,S2)
=f
w(q) '1fv(IF,p,q,uo,Sl,s2)dxpdxgduo, 2 x (A1) 2 x A
(2) Tv+(`F, w, s, S1, S2) = JR2 X(A1)2 XA w(q)µifv(IF, µ, q, no, S1, S2)dxpdxgduo, 1>1 TV' (W, w, S1, s2)
(3)
= J +x(Al)2xA w(q)fv('I,1, P2, q, no, sl, s2)dxl.12dxgduo.
As in (5.2.2), Tv(1Y, w, s, s1, s2) has a meromorphic continuation. We define TV ('F,
si)
Tv+('W,
Si)
d ds2 s2=0
TV('F, W, S, S1, S2),
d ds2 s2=0
d
T' (W, w, Si)
Tv+(W, LO, S, S1, S2), 1
ds2 92=0
In our case, Ya = Za for all 0. Therefore, all the distributions in (6.1.12) are well defined. Let 1)6(w3)TV (IZVOl), (wl, 1), c, 2 (1 - s)) EV,st(4), w, X, S) = 6(wlw2
s-1
Then by considering r = rG, s = ST, (6.3.2)
1
°,v,st(-P, w, x, w) -
fRe(s)=r>i
w,
,
s)(s)A(w; s)ds.
Part III Preliminaries for the quartic case
182
Let b(wl)b(w2)
Ear 0') w, X, s) =
1
E2(Ra,11 (W2, W3),
10
2
1
(S+1), 10 (3s + 3 - 2c)).
Then similarly as in Chapter 5, by considering r = rG, s = sr, w(g')X(g1)OSal ('p, 91)8(9°, w)dxgl
(6.3.3)
FGA'/Gk 1
2Ir
fRe(s)=r>1
Ear (-P, w, X, s)O(s)A(w; s)ds.
Let
Jz(,D,w,X)=
w2 a2 W (( 21,
3
11),
3)
By (3.4.31), (6.3.4)
02,2 (P, w, X, w) + "02,4(4P1w, X, w) - CGA(w; p)
w, X).
Let b(w3)b(w1w2 1) Ta2+(Ra2,oRa2I,w1, 3, L01 X, s) =
S-1
3
2(1 - s))
+ b(w3)b(wlwz 1) Ta2+(Ra2,oga2Ra2'P,wi 1,3- 3, 2(1 - s)) 3 s-1 Then similarly as in Chapter 5, by considering T = TG, s = sT, (6.3.5)
, w, X, w) +a2,st,4 (`P, w, X) W)
02,st,2 ( 1
27rV -1 Re(s)-r>1
Ea2,st (D, U)' X, s)¢(s)A(w; s)ds.
Let ED, (4D, w, X, s) be the following function
b(wl)b(w 2b(w3) (
E1(Ra,,P,w2, 2(s+1)) + E1(Ra,.
3s-2c+3
v%
41,w21, 2(s+ 1))
3s+2c-15
Then similarly as in Chapter 5, by considering r =TG, s = sT, (6.3.6)
w, X, w) + 'y a3,4
a3,2
N
1
-1 Refs)-r>1
w, X, W)
Ea3 ('b, w, X, s)O(s)A(w; s)ds.
Let
J4(D,w,X) =
Q32b(w1)b(w2)
5
+
C
E1+(Ra4Cw3,-5)
J2b(w1)b(w2) 1 E1+(9o,Ra,4),w3 5
5(c+5)).
1
6 The case G = GL(2) x GL(1)2, V = Sym2k2 G k
183
Then by using the Mellin inversion formula, (6.3.7)
-=D,,2 (-D, W, X, W) + Ea4,4(1b, w, x, w) ^' CGA(w; P)J4(,D, w, X),
where 26(w1)6(w2) 5
c
E1(Ra4D,w3,-5) (--D(O)
+ !I 12b(wl)b(w2)6(w3)
JIFc 50)
+
All these computations are valid replacing (1) by $, X by'cj1X-1 etc. by the assumption on c. We define Ea1i(j) (1D, w, x, so) etc. similarly as before.
Easy considerations show that (6 3 8 ) .
.
E vsc, (o) ( , w, X, 1) _ E al,(O) ( 41 , w, X, 1 ) =
- b(wl w 2 )5(w z
3 ) T% ( R vo , P , (w1 , 1 ),
c),
,P , (w2i w3),
1,
(w2, w3),
1,
z
3b(W
50 (w2) E 2 (1,-1) ( Ra1 a(wl)b(w2)
10
E 2,(o,o) (R al
,
5 ( 3 - c)) 1
5
(3
- c)).
Also
(6.3.9)
J2($, w-1, I£V'X-1) + J4($, w-1, kV1X-1)
w-1, v1X-1, 1) + Ea3,(o) (I, w-1, kV1X-1, 1) - J2(4', W, X) - J4(I, W, X) + Ea2,$t,(o)
- Ea2,st,(o) (b, W, X, 1) - Ea3,(o) (P, W, X, 1) 6(W3)
3
+
(ED2,ad (Ra2
212b w1)b(w2
$) (w2 1 , wl 1) ,
) (El(RD4&w1,
4
3
c 5
c) - Ea2,ad (Ra2 -P, (w27 wl), 3))
)-
E1
c (Ra4 1b w3 - 5)
Definition (6.3.10) ZV,ad(', W, X, S) = Zv(4, w, X, s) - b(w1w22 )s(w3)Tv(Rv,o4), (w1,1), s, c).
We use the terminology `adjusted zeta function', `adjusting term' for this case also. Let F(4i, w, X, s) =
b(wl)b(w2)E2,(o,o)(Ral , (w2, W3),1, 1(3 - c)) 2(5s + c - 8)
36(wl)b(w2)E2,(l,-1)(Rol' , (w2, w3),1,1(3 - c))
+
+
2(25s + 5c - 40)
4b(wl)b(w2)E2,(o,-1)(Ra1 , (w2, w3), 1, 1(3 (5s + c - 8)2
- c))
184
Part III Preliminaries for the quartic case + b(W3)
Ea2,ad (Ra2 , (W2, W1), 3 )
3s-c
+ 32b(W1)6(W2)
E1(Ra,',W3i-5) 5s + C
Then by considering relations as in (5.6.2), we get the following proposition.
Proposition (6.3.11) Suppose b = Mv,,4). We also assume that
c0 -5, -2,0,1,3,4,9. Then Zv,ad(4',w, X, s) satisfies the following principal part formula W, X, s) = Zv+('D,W, X, S) +
W-1, /£V1X-1, 4 - s)
6(W1W2 1)b(W3)Tv+(RVo4), 2
6(W1W21)b(W3) 2
(wi, 1), s, c)
Tv+(Rvo(D, (W1
1
1),4-s,4-c)
- F(,D, w-1, '4 1X-1, 4 - s) - F(-D, w, X, s).
When we apply this result in Chapter 13, we consider the situation s = -3, c =
-1.
Chapter 7 The case G=GL(2) x GL (1)2, V=Sym2k2® k2 In this chapter, we consider the case G = GL(2) x GL(1)2, V = Sym2k2 ® k2 using the formulation in Chapter 6. We use the notation in §6.2.
§7.1 Unstable distributions In this section, we define distributions which arise from unstable strata.
Definition (7.1.1) Let w be as before. We define (1) bal (w) = 6(w1w2 2)b(wlw3 1), wal = (w2, w3), (2) 6a2 (w) = 6(w1w2 1)b(w3), wag = (w2, 1), (3) 6,3(w) = b(wl)b(w2w3), w03 = (w2,w3), (4) 604(w) = b(w3), bay,st(w) = 6(w3)6(wlw21), way = (w2,w1), wa,5t =w2, (5) 6as(w) = 6(w1)b(w2)6(w3), was = w2,
(6) 6#(w) = bas(w) = 6(w1)6(w2)6(w3)
Definition (7.1.2) Let 4), w, X be as in §6.1. For a complex variable s E C, we define X, s) =
(1)
2 bat (w)
E2(Ral , wal ,
3s-3-c , 2s - 3 - c), 2
3s - 7 + c
(2)
Eat (D, W, X, s )
(3)
Ea3(C w,X,s) =
(4)
Eay (1), w, X, s) = 6 4 (w) Ea4,3,d (Ray D, way, + s 4 4
(5)
Eas,l (D, W, X, s) = bas (w)
>
=
bal(w)
4
63(w) 10
E2(Ra2$1 wag,
E2(Ra3 C , w a3,
4
3s + 3 + c 2s + 2 - c),
E1(Ra54i7 w55,
4(s + 1)
-
s - 1),
10
5
1)
2c2
E1(Ra5YVl-D,wa51, 8+1 )
X, s) = bas (w)
(6)
4(s - 5) + 2c E1(Ras 4), s + 1)
Eas,l (4), X> w> s) =bas (w) Ea6,2 (4), X, w, s) = bas (w)
3s + 3 + c E1(Rae9v24), s + 1)
3s - 3 - c
We define Eay,ad (4), w, X, s), E04+(", w, X, s) similarly. The distributions Ea; ((D, w, X, s) for i = 1, , 3 and Eay (4), w, X, s) have at most
a double pole at s = 1. The distribution Eas,1((k, w, X, s) has at most a double pole at s = 1 for l = 1, 2. The distribution Eae i, (4),w, X, s) has at most a simple pole at
s=1for1=1,2.
We define E*,,(j) (,P, w, X, 1) etc. similarly as in (5.3.4).
Part III Preliminaries for the quartic case
186
§7.2 Contributions from unstable strata We consider I(,D, w, X, w) in this section. Let pi = (ai, si) be a path such that si(1) = 0, and pz = (0j,si) a path such that si(1) = 1 for i = 1,2,3. Let 0 = 01, 02 or a3. It is easy to see that w(g1)X(g1)kv(g1)eS, (&
fA/Gk
fA/Gk
(g1)`)-,(go, w)dxgl
w-1(g1)X-1(g1)rv1(g1)esa($,gl)S(g° w)dxgl
because g((g°)°, w) = g(g°, w). Therefore, we only consider fG/G,.
w(gl)X(g1)©s, (-P, gl)S"(g°, w)dxgl.
Lemma (7.2.1) For i = 01, 02, 03, JA/Gk
w(g1)X(g1)OSa (4', g1).I(g° w)dxgl
JAnTA/Tk
w(t1)X(t1)(t°)-2"ic i(tl)eZa (R,4), tl)&N(t°, w)dxtl.
Proof. Clearly,
fG/G,.
w(g1)X(g1)esa (,D, gl)8(g° w)dxgl w(t1)X(t1)(t°)-2Pe ,(Ra(D,t1n(u))8(t°n(u),w)dxtldu.
fAnBA/Bk
For a E k, we define
fl,a(tl) =
JNA/Nk
f2,a(tl) = f
9y, (-D, tl(n2u), 1)) < au > du, Oy,2 (,(D, tl(n2u), 1)) < au > du,
NA//N,
f3,a(w, t1) =
JNA/Nk
8'(t°n2(u), W) < au > du.
Then, by the Parseval formula, we only have to show that
fe/Tk
w(t1)Ac E fi,a(tl)f3,-a(W,tl))2 dxtl ,.. 0 aEkx
for i = 1, 2. We fix a constant M > w°. By (3.4.3), there exists 6 > 0 such that if l >> 0,
lf3,-a(w,tl)l « Vi
7 The case G = GL(2) x GL(1)2, V = Sym2k2 ® k2
187
for w° - S < Re(w) < M. Let 41 be the partial Fourier transform of Rat 4 with respect to x21 and the character <>, and 1)2 the partial Fourier transform of R,2 4) with respect to x12 and the character <>. Then
f1,-W)
f
/k Z12,x2oEkx,x21Ek
u
Jru 'D(712(tl)xi2, 72o(tl)x2o, 721(tl)(x21 + ux20)) < au > du
xiz,xzoEkx
_Ef X12,xzoEkx
_
'D(712(tl)x12, 72o(tl)x2o, y21(tl)(x21 + ux20)) < au > du
/k x21Ek
(712(tl)xi2,72o(tl)x2o,721(t1)x2ou) < au > du
A
i(712(tl)xi2, 72o(tl)x2o, 721(11)-1x20 a)
)11 1A2
x12,x2oEkx
Similarly,
f2,,(tl) = Al2A24
'P2(711(tl)x11,721(11)x21,712(tl)-1x12 a) x11,x21Ekx
Therefore, by (1.2.6), for any N1i N2, N3 > 1, (7.2.2)
«AiiA201A2)-N1(A11A23)-N2(A-lA3)-N3,
Ifi,a(tl)j aEkx
If2,a(tl)I «A12A24A24N1(AiA23)
N2(Al2)24)-N3
aEkx
Since the convex hull of each of 1(2,4),(-l,-3),(-1,3)1,1(0,4),(l,-3),(-2,-4)1 contains a neighborhood of the origin of R2, for any N >> 0,
E Ifi,a(t1)I, E l f2,a(tl)I << rd2,N(A1, A2)aEkx
aEkx
We fix 1. Then JA/Tk
w(tl).2
fi,a(t1)f3,-a(w,11)A1dx11
aEkx
is bounded by a constant multiple of
f Rrd2,N()1, A2))2+lAcdxAldx)2 2 for i = 1, 2. This proves (7.2.1). Q.E.D.
Let r be any Weyl group element. For i = 1, 2, 3, we define (7.2.3)
Epi,T(,Ib,w,X,w)= J p/Tk w(tl)opi(tl)Oz.(RD ,t1)eN,T(t°,w)dxtl.
Part III Preliminaries for the quartic case
188
By a similar proof as in (6.1.4), Bpi (4D, w, X, w) is well defined for Re(w) >> 0. By (7.2.1), the first two terms of (6.1.12) 3
w) - =pi,T
VX i=1
w, X, +u
T
Easy computations show that apl (t1) = XiX2+3 Qp2 (t1) =X24, and Op3 (t1) _ A2Xc 1 2
Proposition (7.2.4) Let r, s = sT be as before. Then pi,T (I), w, X, w) -
27f fR)=
Eoi (4), w, X, s)AT(w; s)ds.
Proof. Let t' = (ti, t2) = (µ1q1, µ2q2) E (AX )z, where Al, µ2 E ]R+, q1, q2 E A1. Let
dxt' = dxtidxt2.
(/J(A1
Suppose that f is a function on (A1/kx)2. It is easy to see that
f
Al/kx)4 w(t °)f (ti2t2, tllt3)(tx t ° = 6" (w) J Al
///
/kx)4 w(t °)f (t11t12t2, t12t3)dxt ° = 62 (w)
f
Al
J
\A1
w11(q)f (q)dx 4,
/kX)2 w02 (q)f (q)dxq,
(w(t °)f (t2tz, t12t3)dx t ° = 613 (w) fAl /kx)2 w13
/kx )4
after the change of variable (ql,q2) = (ti2t2, t11t3), (t11t12t2,t12t3), (t12t2, t12t3) respectively. We make the change of variable µ1 = A2A4, 92 = X11Xz 3 for p1, µ1 = X2, µz = X1123 for P2, and µ1 =XiX2, µ2 = X1X23 for P3. Then dx µldx µ2 = 2dx X1dx X2, X1 = µi N'2, A2 = µ1 2µ2 1 for pi,
dxµldxµ2 = 4dxA1dxA2, X1 =µiµ2, X2 = µi for p2, and dxµldxp2 = 10dxA1dxA2, X1 = µ'1 N'2 , X2 = Al µ2 10 for p3.
Clearly, £pi,T(t°, w) = &N, (t', w). It is easy to see that 3a-3-c
Al op,
2
(tl)(to)TZ+n =
2s-3-c
µ2
z = 1,
A2-1
2 = 2,
° µ1
4
3a+3+c U
10
4e+4-2c N'2
10
i=3.
7 The case G = GL(2) x GL(1)2, V = Sym2k2 $ k2
189
Therefore, w(tl)(t1)Ozo(Ra,t1)N,T(to,w)(to)-2pdxtl
J
= E(,w) X,A
/Tk
for i = 1, 2, 3. This proves the proposition. Q.E.D.
Proposition (7.2.5) Ep;,T (1b, w, x, w) - 0 if T 54 TG for i = 1, 2, 3.
Proof. If T # TG, we can choose r >> 0. Then Z(r) < wo. Q.E.D.
Next, we consider So, for i = 4,5,6. Since
-1 -1
-1 Ea.,1w-
w, X, W) =
a,,4
)
r--V1 X- 1
1
,
,
w),
we only consider Ea w, X, w) for j = 2, 4. The following proposition is a direct consequence of (3.4.31)(1).
Proposition (7.2.6)
w w-
(1)
a4,2(
(2)
14,4 (D , w, X, W)
,
, X,
c).
G
)
(
4
,
CGA(wi P)
w 4 as+(R as, a4,
Eaa+ (ga4 Rio, 4), aal, 3 - 4
Let c
Eaa,st,2(D) w, X, s) = 604,st(w) Ea4,st,4(4), w, X, s) = ba4,st(w)
1-s
7a4+(Raa,oRa4wa4,st, 2 )
4(s-1)
,
Taa+(Ra4) o`aaRaa'b ,wa41st,3- 41 123) 4(s- 1)
The proof of the following proposition is similar to (5.4.3)-(5.4.7), and is left to the reader.
Proposition (7.2.7) Let r = TG, and s = sT0. Then (1)
Ea4,st,2
1Xs)(s)A(w;
w, X, W)
e,->1
s)ds,
>1
f
(2)
aa,st,4 (4), w, X, w)
27f
-- J
R ,->)1
Eaa,st,4 ('D, w, X, s)0(s)A(w; s)ds.
Since YO, = Za; and M-6; T for i = 5, 6, E0;,2(4), w, X, W) =
J T/Tk
w(t1)X(t1)()za; (.b, t1)c"'N(t°, w)(to)-2pdxtl,
A1>1
a.,4(4b,w,X,w)
_f
T A/Tk A1<1
w(t1)X(tl)eza;(,D,tl)f°'N(t°,w)(t°)-2pdxtl,
Part III Preliminaries for the quartic case
190
for i = 5, 6. Let r be a Weyl group element. We define
i
w, Xf w) =
(1 .2.18)
f
T.l/Z.k A
w(tl)X(tl)e
w)(to)-2Pdx
,5 (D,
t'.
al>1
i = 1,
We define E S,i,T (4i, w, X, w) for i # 2, and 8a5 larly.
, 4 simi-
Proposition (7.2.9) The distribution ^aSiT(-P,w,X,w) is well defined for i =
1, ,4 if Re(w) > 0, and (1)
.a5,2,TR, w, X, W)
w, X, w) ti
(2)
2r
Refs)=r
vas (w)
27rv/---l
E a5,1Rf w, x, s)AT(w; s)ds, E05,2 (C w, X, s)AT(w; s)ds.
/ yo
Proof. Let t' = µq, where p E R+, q E A'. Suppose that f is a function on Al /k x . It is easy to see that
f
Al
/k5)4
w(t o).f (tzt2)dx t ° =b(w) Al f /kX
wp
after the change of variable q = t212t2- We make the change of variable p = X1A2, A2 = A2. Then dxudxA2 = 2dxaldxA2, Al = µA22 It is easy to see that X(tl)(to)T -P = Y" -A2 2(s+l)+C Therefore,
fl
w(tl)X(tl)ez,s (Ra54,f
tl)(to)TZ-Pdxt1
=
w, Xf s)
T /Tk a2>1
This proves the first statement. By the remark after (6.1.9), the second statement can be obtained by replacing (D by 9v I), wa by wa 1, and c by 6 - c. Q.E.D.
Proposition (7.2.10) .^aeijiT (w, 4), X, w) is well defined for i = 1,
, 4 if Re(w) >>
0, and as,4,T(, (1) , X, w) -29r
(1 )
(2 )
1z
06,2,,r (w, 4) ,
X, w) ^
2 7r
Aer>1 r 1
->1
ae,1( 4, f Xf s )A T ( w; s )ds,
fR)=a5,2
(
, X) s )A T ( w; s )ds.
Proof. Let t' = pq, where p E R+, q E Al. Suppose that f is a function on Al/kx. It is easy to see that
JM /k-)4 w(t e).f (t12t22)dx t e = b#(w)
Jl
/kx
.f (q)dxq
7 The case G = GL(2) x GL(1)2, V = Sym2k2 ®k2
191
after the change of variable q = t12t22. We make the change of variable µ = A1.\23, A2 = .\2. Then dxpdxA2 = dxAldx\2, Al = µA2 It is easy to see that X(tl)(to)rz-' = us+1X23+3+c
Therefore,
fTp /Tk
w(tl)X(tl)Oz 6 (Ra6,D,
t1)(t0)'
-Pdtl = Eb6,1(oD, X, s).
a2 <1
This proves the first statement. The second statement can be obtained similarly. Q.E.D.
Proposition (7.2.11) If r O rG, `-'a6,i,r (D, LO, X, w),
(-P, w, X, w) ". 0
Proof. We only have to choose r >> 0. Q.E.D.
§7.3 The principal part formula We define 3
Jl
w, X, S) _ E 1"a, 0, w, X, S) i=1
+
c (w) Ta4+(Ra4,oRa4'P, w04,$t, 4 4(s-1)
1-S 2
+
c
4(s-1) + +
J2w,X) J3
(
U), X) =
4
1 - s) 2
W, X, S) + Ea6,2 (4, w, X, S)
w, X, S) +
w, X, S),
Sa44' (Ea4+(Ra44),wp4,4)+Ea4+(ga4R044"LON,3
4))'
c
- ba4,st8 (w)
T04+(Ra4,oR04D1 wa4,-t,
4)
1 44,-t (U)) Ta4+(Ra4io'a,Ra4Cwa4 8
st,3- 4c
Let T = TG, and s = sr. Then by (7.2.1), (7.2.4)-(7.2.11),
I(4,w,X,w)
2 seta>-r (J1(
,
w-1, Kv1X-1, s) - J1(D, W, X, s))O(s)A(w; s)ds
.moo
+ CGA(w; P)(J2($, w-1, r.V1X-1) - J2(4), w, x)).
Part III Preliminaries for the quartic case
192
Nv1X 1, s)-Jl(,D,w, X, s) must be holomorphic
By Wright's principle,
at s = 1. Therefore, 27r 1
-1
Recs>=.
(J1(
,
w-1, 1V1X-1, s) - Ji(lb, w, x, s))O(s)A(w s) ds
.moo
,., CGA(w, p)
(Ea.,(o) ($, w-1, rcv1X-1, 1) -
w, X> 1))
i=1,2,3
+ CGA(w; p)
(4', w-1, v1X-1> 1) -
w, X, 1))
i=5.6 1=1,2
+CGA(w,p)(J3($,w-1,, 1X-1) - J3(D,w,X)) This implies that
Il (4', w, X) _
w-1
(Ea ,(o)
v1X-1,1) - Ea,,(o)
w, X, 1))
i=1,,2,3
+ L,
Ea,,d,(o)
w, X, 1))
=1,2
i=2,3
Since the reason for studying this case is to apply it to the quartic case, we restrict ourselves to X satisfying the condition 4, 64c # 0, 2, 3. In Part IV, we will consider the situation c = 2, so this condition is satisfied. By assumption, Ea6 w, X, s), Ea6 w-1, v1X-1, s) for i = 1,2 are holomorphic at s = 1, and their values are E1(Ra641, 2) 606 (w) sa6 (w)
6+c E1(Ra6,D, 2)
12-c
-Sa6 (w) -bah (w)
E1(Ra69y2ID, 2) c
E1(Ra62) 6-c
respectively. The following lemma is an easy consequence of (4.4.7).
Lemma (7.3.1) If -D is K-invariant, Z
E,(Ra64D,2) = Z k)g04$(O)l
).D (0).
The above relations hold by replacing 4) by . Also by assumption, bas (w) El,(-1)(Ra5..v14), wa5 1,1)
E05,2,(O) (D, W, X, 1)
8
+
(4
- 2)2
1 sD5(w) E1,(o)(Ra5. y1',was ,1)
8
(4 - 2)
The above relation holds by replacing 4b, w, X by $, w-1, rcV1X-1 respectively.
7 The case G = GL(2) x GL(1)2, V = Sym2k2 ® k2
193
If c#4, bas (W)
E1,(-1)(R,5 )w05,1) + E1,(o)(Ras4),waa,1)
8
(4 -1)
(4 - 1)2
/
If c = 4, x, 1) =
6as8W)
E1,(1)(Ra54i, Wa5,1).
If c ¢ 4, we define b54 (W)
J4('P,W, X)
Wa4, C). 4).
4
If c = 4, we define bag8w)
6a44W)
J4('p, W, X)
Ea4,ad,(o) (Ra44', W041 1) +
E1,(1)(Rab 4), WD" 1).
By the principal part formula for Ea4,ad (Ra4 -P, w1,,, s) and the above considerations, we get following proposition.
Proposition (7.3.2) X) _
(Ea;,(o)(4', W-1, v1X-1, 1) i=1,2,3
-
W,
X,1))
+ J4* W-1, icv1X-1) - J4(P,W, X) If c 4, as in §1.7.
J4W,X) =
If c = 4, J4(4)x,W,X) = A-1J4(4',W,X)
Definition (7.3.3) Zc, P2(X) = 4, P3(X) = 1 10 c P4(X) =
P1(X)
(1)
7
4c ,
7
q1= 2, q2=4, q4=1.
(2)
Easy computations show the following relations and the proof is left to the reader.
Lemma (7.3.4) For i = 1, 2, 3, (1)
E
(,pa,
W, X, s) =
-q;(s-1)-p;(x)Ea; (4p, W, X, s),
E1 (Ex, W-1, kV/X-1, s) = \q;(s-1)-(5-p;(e 'x-'))ED ($, W-1,
(2)
KV1X-1
The following proposition is an easy consequence of (7.3.4).
Proposition (7.3.5) For i = 1, 2, 3, (1)
Ea;,(o)(%, w, X, 1) 0
1
(_qi logA)`A-p:(x)Ea:,(i)(4), W, X, 1), (2)
ED,, (0) (4'x, W-1, Kv1X-1, 1) 0
1
(qi
log X)-iXv'x
v1X-1, 1).
s).
Part III Preliminaries for the quartic case
194
Definition (7.3.6)
F,(-P,w,X,s)-
(q,)-jba;(S)Eai,(j)(q),U,X,1) pi)1-j
(
i=1 j=-2 X)
F2(F,w,X,s)=
3-4
These considerations imply the following principal part formula.
Theorem (7.3.7) (F. Sato) Suppose 4D = M,4P. Also assume that 4, 64` 0 0, 2, 3. Then Zv(1b, w, X, s) satisfies the following principal part formula Zv (4', w, X, s) = Zv+(,D, w, X, s) + Zv+(&, w-1, kv X-1, 5 - s) i=1,2
Because of the above formula, the following functional equation follows.
Corollary (7.3.8) Zv(4D, w, x, s) = Zv(&, w-1, r.v1X-1, 5 - s). Since we want to use (7.3.7) for the quartic case, we describe F, ((b, w, X, s) explicitly here. We assume that c = 2. So 6 - c = 4. We define Fa1 (4b w s) = w, s) _
Fat
1 E2(Ra1 4(D, wa1, -1, -3)
s+3
2
1 E2,(o,o)(Ra21', wag, -2, 0)
s-2 1
4
3 E2,(1,-1)(Ra2 , wa2, -2,0)
+
s-2 1
16
7 E2,(o,-1)(Ra24',wa2,-2, 0)
+16 F03(4',w,s) _
(s-2)2
1 E2(Ra3'D,wa3, 45 > 25)
s-56
10
1
Fa1
( w s) =
F02(
w s) =
1 E2(Ra14, wa1 , -2, -5)
11-s
2
1 E2,(o,o)(Ra2 , wa21, 0, 0) + 3 E2,(1,-1) (Ra2 4 , wa21, 0, 0) 16 4 wa2', 1 E2,(-1 1)(Ra2$, 7 E2,(o,-1)(Ra2$1 wa21, 0, 0) 0, 0)
4-s
4-s
+3
4-s
7 E2,(-1,0) (Ra2 $, wa21, 0, 0)
+ 12 P03 (
w s) -
(4- s)2
1 E2,(o,o) (Ra3 $, wa31, 1, 0)
4-s
10
+ 3 E2,(1,-1) (Ra3 $, U)0311110)
4-s
7 E2,(-1,0) (Ra3 $, wa31,1, 0)
+ 30
(4-S)2
4-s
40
2 Ez,(-1,1) (Ra3 , wa31 1, 0)
+15
(4-s)2 49 E2(Ra2 $, wa21, 0, 0) + 48 (4-S)3
+16
7 E2,(o,-1) (R03 , u'-111, 0)
+40
(4-s)2 1111
49 E2,(-1,-1) (Ra3 $, W+ 120 (4-S)3
0)
7 The case G = GL(2) X GL(1)2, V = Sym2k2 ® k2
195
Then (7.3.9) i=1,2,3
Therefore, the poles of ZV(4, w, X, s) are s = -3, 2, 54, In Chapter 12,13, we show that the Laurent expansion of Zv w, X, s) at s = -3 will contribute to the rightmost pole of the zeta function for the quartic case. The function ZV (C W, X, s) has a simple pole at s = -3, and the coefficient of the order two term of the principal part of the quartic case turns out to be a constant multiple of E2(Ral lb, wal, -1, -3).
Part IV The quartic case In the next six chapters, we consider the quartic case G = GL(3) x GL(2), V = Sym2k3 ® k2. In Chapter 8, we study the stability and the Morse stratification of our case. In particular, we explicitly describe all the ,Q-sequences we need. In Chapter 3, we proved that the distributions associated with certain paths are well defined. However, we have some paths which are not covered there. The representations Zpe, Zp8 of MQ6, Mp8 (in §8.2) are reducible and require a handling similar to Chapters 6 and 7. We prove that certain distributions associated with N6, ,as are well defined in §9.1. In §9.2, we prove some special estimates of the smoothed Eisenstein series which are required in Chapter 10. In Chapter 10, we prove that we can ignore the non-constant terms associated with unstable strata. We will prove that the distribution Ep (oD, w, w) is well defined for all p and Ep'='p (OD, W, W) " 0. p
For this purpose, some cancellations between different paths have to established as in §10.4, 10.7. In Chapters 11-13, we compute the constant terms EEp(-D, W, W) associated with paths explicitly, and prove a principal part formula for the zeta function of our case. In our case, we can fortunately use Wright's principle. This case is of complete type. Let Zv(4D,w,s) be the zeta function defined by (3.1.8) for L = Vks = Vk. The principal part formula for the zeta function for this case is (13.2.2). The location of the poles is s = 0, 2, 3, 9, 10, 12. The orders of the poles at s = 2, 10 do not coincide with the multiplicities of the corresponding roots of the b-function for this case.
Chapter 8 Invariant theory of pairs of ternary quadratic forms §8.1 The space of pairs of ternary quadratic forms Let V_be the space of quadratic forms in three variables v = (vl, V2, v3). We identify V with k6 as follows: 2
2
Qx(U) = x11v1 + x12V1V2 + X13V1V3 + x22,2 + X23V2V3 + x33,3
-* x = (x11, x12, x13, x22, x23, x33)
The group Gl = GL(3) acts on V in the following way: Q91..(v) = Qx(vgl), for g1 E G1. _ Consider V = Sym2k3 ® k2 = V ® V. Any element of V is of the form
Q = (QS1, Q,), where x1 = (x1,111 x1,12, x1,13, x1,22, X1,23, x1,33),
X2 = (x2,11) x2,12, x2,13, x2,22, X2,23, x2,33)
8 Invariant theory of pairs of ternary quadratic forms
197
We choose x = (x1i x2) as the coordinate system of V.
Let G2 = GL(2), and G = Gl x G2. The group G acts on V as follows. Let g = (91, 92) where g1 E G1i and 92 = I a d
)
E G2. We define
g ' Q = (aQ91'x1 + bQ91'X2, cQ91.x, + dQ91'x2 ).
Let T be the kernel of the homomorphism G -+ GL(V). We define
Xv(9) = (detgl)4(detg2)3. Then Xv can be considered as a character of GIT, and it is indivisible. In the next six chapters, we consider (G/T, V). Throughout the next six chapters, we use the notation (8.1.1)
a(t11, t12, t13; t21, t22) = (a3(tll, 42, t13), a2(t2l, t22))
Let T C G, G', t*, t' etc. be as in Chapter 3. In our case, G°A = GA, and therefore, t* = t°*. We identify t* with {z = (z11, Z12, Z13; Z21, Z22) E W I z11 + z12 + Z13 = 0, Z21 + Z22 = 0}.
We use the notation z1 = (z11) z12, z13)/, z2 = (z21, z22), and write z = (z1i z2). For z = (z11, Z12, z13i z21, z22), z' = (z11, z12, z13; z21, z22), we define (Z, Z) = z11zi1 + z12z12 + Z13Z13 + z21z21 + z22z22'
This inner product is Weyl group invariant, and we use this inner product to determine the Morse stratification in the next section. Let 1111 be the metric defined by this bilinear form. We recall that G' = Ker(Xv). We choose G" = SL(3) x SL(2) for G" in §3.1. The weights of xl,ij, x2,ij for 1 < i < j < 3 are as follows: 4 1
(13 3 1
(3'
2
2 -2.1 -1)
X2,11
3' 2 -2)
X2,12
1
2
1. 1
3' 3' 2'
2, 1
4
1 1
1. 1
2
(-3,3'3'2'2 4, 1 2
-3'2
1
2
3' 3' 2'
X2,13
2
(-3' 3' 3' 2,
1
X2,22
x2,23 1
2
X2,33
() (3+-3'-3+-2,2 2._1 1
1
3' 3' (1
2
1
3'
1,
3' 3'
3'
(2 4 (2
-2.
2' 2 1 1 2' 2 ) -1 1)
1 1._1 1) 2' 2)
31 3 ' 3 '
(2 -3,-3'2 4-11) ' '
Therefore, with our metric, the weights of coordinates look as in the picture on the next page. Let 7l,ij,'Y2,ij be the weights of xl,ij, X2,ij respectively. If t E T, we denote the value of the rational character determined by ryl,ij (resp. 'Y2,ij) by yl,ij (t) (resp. 72,ij(t)). This should not be confused with ty-j etc. which is the adelic absolute value of ry1,ij(t) etc. We use this notation throughout the next six chapters.
Past IV The quartic case
198
§8.2 The Morse stratification We study the stability over k. Consider a point x = (x1i x2) E Vk as before. Let Q1 = Qxl, Q2 = Qx2. The quadratic forms Q1, Q2 define subschemes V(Q1), V(Q2)
of Pk respectively. We define Zero(x) = V(Q1) fl V(Q2) and call it the zero set of x. The stability with respect to G', G" are the same. We proved the following proposition in [84] ((1) was proved in [59]). Here, we give a proof based on geometric invariant theory.
Proposition (8.2.1) (1) P(V)k = P(V)80)k and is a single Gk orbit. (2) If X E Vk \ {0}, ir(x) E P(V)7 if and only if V(Q1) n V(Q2) consists of four points.
Proof. Let fB be the set of minimal combination of weights. Table (8.2.2)-(8.2.4) are the list We first show that V, 0. It is easy to verify that if x E Vk and x E Sp, for some 1 < i < 10, then Zero(x) has at least a double point or contains a line or a conic. Therefore, if we can show that Sp; = 0 for i = 11, 12, any x E Vk such that Zero(x) consists of four distinct points should be semi-stable. Such x clearly exists, so Vss 0 0. So we consider SO,l, Sp12.
Consider Spl,. Let ap,, (a) = a(a-1, a-1, a2; a, a-1) for a E GL(1). Then by Table (8.2.3),
MPI, =
{((9112 1) I2)
ap11(a) 91,12 E SL(2), a E GL(1) } .
Elements of the form aQ11 (a) act trivially on Zp11, and the action of 91,12 on Z01,
can be identified with the standard representation of SL(2). Therefore, ZQ11k = 0. Consider S012 By Table (8.2.3),
Mp12=I
(
\91,12
1),92)
91,12, 92 E SL(2) I .
Elements of the form (( 91,12 1) ,1 ) act on ZQ12 trivially, and the action of 92 on Z1912 can be identified with the standard representation of SL(2). Therefore, Zp12k = 0. This proves that V88 0. Note that we did not use a relative invariant polynomial.
8 Invariant theory of pairs of ternary quadratic forms
Table (8.2.2)
Strata
Convex hull
Conics
Sol
2 identical non-singular conics
i i
SRZ
X 2 identical reducible conics
Sa3
040
K
2 identical double lines
SR4
1 common component
199
Part IV The quartic case
200
Strata
Convex hull
Conics
SP5
1 common double line
Sas
2 multiplicity 2 points
Sol
C-) The local ring = k[e]/(E4)
Sag
1 multiplicity 2 point
8 Invariant theory of pairs of ternary quadratic forms
Strata
Convex hull
Conics
Sp9
1 multiplicity 3 point
Sp1o
The local ring = k[Ei, E2]/(Ei, E2)
Sp11
Sp12
201
Part IV The quartic case
202
Table (8.2.3) ((71,11 ' ) (71,12
13 1
Q1 = (0, 0, 0; - 12, 12)
,132 = (-3, 3, 3;-2, 21)
('12,33 ' N))
-1, -1,1,1,1,1,1,1)
2(-1,
s(-11,-5,-5,1,1,1,-5,1,1,7,7,7) !(-11, -11 ,1 -11, 1 11, - 5-5,77 - 5 7, 17)
1 1) /j3=(-37-231 4. 3,-2, 2)
g
/34 = (- s, - s, 3. 0, 0)
2 1 7. 1) a5-(-3,12,12,-4,4
6
06 =
1(-5 -5 -1, -57, -17, 1, ,1,, 1,, 5, 1, 5) 24 , , ,
2
1
(-1 _1 1.-18,8 1)
/37 = (
24,
4,
241 121
O,
4,
4, 4)
1) 14,14 211 - 2 1211-L.--L 1 (39 = (-1 5, 0, 1' 3, - 1 10 , 10)
NR8
_
,Qlo =
2, 3, 0, 0) 1 _ 13; 1 _ 1 1) g, 61 2, 2
/iii = (_q, _2, 4 _j 1) 1312 = 333, 2, 2)
1(-2, -2,1, -2,1, 4, -2, -2,1, -2,1, 4)
1(-19, -10 -4, -1, 5 11,-13, -4,2,5,11,17 ) 12 4
(-3,-2,-l,-1,0,1, -1,0,1,1,2,3)
1(-11 -5 -5,1, 1 ,1 ,-5,1,1,7,7,7) 10(-5, -3,, -1,, -1,1,3, -3,-1,1,113, 1)
42
,
,
1(-4, -1, -1, 2, 2, 2, -4, -1, -1, 2, 2, 2)
1(-5 -5, -2,-5,-27 1 1 1 4 1 4 17) g 1(_4 -4 2 -4 2 8, -4, -4 2 -4 2 8) 3 >
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
Table (8.2.4) Zp
Wp
a
I1,31I2
,Ql
2
x2,j,j2 for 31, j2 = 1, 2, 3
02
g
x2,22, x2,23, x2,33
,Q3
6
x2,33
04
6
,Q5
12
X1,33, x2,23
-L
x 1,33, x 2,ji,j2
06
24
-
x1,13, x1,23, x2,13, x2,23
X1,33, x2,33 X2,33
for ilj 2 ,
=
1, 2,, ,
3
x2,13, x 2,23, x 2,33
x1,33, x2,13, x2,22
X2,23, x2,33
42
x1,22, x1,23, x1,33, x2,12, x2,13
X2,22, x2,23, x2,33
10
x1,23, x2,13, x2,22
x1,33, x2,23, x2,33
2
3
xi,jij2 for i = 1, 2, il, j2 = 2,3
Nil
3
X2,13, x2,23
012
3
x1,33, x2,33
Q7
4
,Q8
09 a1o
-
X2,33
8 Invariant theory of pairs of ternary quadratic forms
203
Let w = (Q1, Q2) where Q1 = V2V3 - V03, Q2 = V1v2 - v2v3.
Easy computations show that Zero(w) = {(1, 0, 0), (0, 1, 0), (o, 0, 1), (1,1,1)}.
If a conic in p2 contains three points on the same line, it contains the line. Therefore, if Zero(x) consists of four points, they are in general position. There exists an element of GL(3)k which sends Zero(x) to Zero(w). Therefore, we may assume that x = (Q1i Q2) where Q1 = x1,12v1v2 + x1,23V2V3 + x1,1003, Q2 = X2,12V1V2 + X2,23V2V3 + x2,1003, and
x1,12 + x1,23 + x1,13 = X2,12 + X2,23 + X2,13 = 0
Without loss of generality, we can assume that x1,12 0. By applying a lower triangular matrix in GL(2)k,, we can assume that x2,12 = 0. After a coordinate change, we can assume that Q2 is a constant multiple of v2v3 - V1V3. Then by applying an upper triangular matrix, we can assume that Q1 = V1V2 - v2v3. Thus, the set of x such that Zero(x) consists of four points is a single Gk-orbit. Since the set of such points is an open set, this proves (8.2.1). Q.E.D.
A straightforward argument shows that x E Sp, for some 1 < i < 10 if and only if Zero(x) has the corresponding geometric property. However, we do not logically depend on this statement. All we have to know is the fact that So, # 0 for i = 1, , 10 and the inductive structure of Sp; for i = 1, , 10. So the verification of the geometric interpretation of Table (8.2.2) is left to the reader. We now consider the strata Sp;. In our situation, there are only a few possibilities for parabolic subgroups of G. Let
Pl = {
91 , 12
0
P2 = { (t01
0
t13) 9 023)
91,12 E GL(2), t13 E GL(1)
l
t11 E GL(1), 91,23 E GL(2) } .
Standard parabolic subgroups of G1 are G1, P1, P2, and B1. Standard parabolic subgroups of G2 are G2 and B2. We consider MR in §3.2. (a) Sp, We can identify the vector space Zp, with the space of quadratic forms in three variables, and Mpl SL(3) = {(g1i I2) 91 E SL(3)}. We discussed this case in §4.1, and Zplk consists of non-degenerate quadratic
forms.
Part IV The quartic case
204 (b) SR2
We can identify the vector space Zp2 with the space of binary quadratic forms. Let ap2 (a) = a(a2, a-1, a-1; a-2, a2) for a E GL(1). Then
MQ2
1
0
0
91,23
'
I2
91,23 E SL(2), a E GL(1) } .
a192 (a)
Since elements of the form ape (a) act trivially on Zp, a point in Zp is semi-stable if and only if it is semi-stable under the action of 0
SL(2) '= {((;
det 81,23 E SL(2) } .
91,23
I2/
Therefore, ZQzk consists of non-degenerate forms again. (c) SR3
In this case, we can identify Zp, with the one dimensional affine space. Let aR3 (a) = a(a, a, a-2; a-2, a2 )
for a E GL(1). Then
Ol
MQ3=!((91,12
12
a,3,
Since MO3 acts trivially on Z 3, ZQ3k = Zp,k \ 101(d) Sp4 We can identify the vector space Z$4 with M(2, 2), i.e. the vector space of 2 x 2
matrices. It is easy to see that Mp4 = SL(2) x SL(2)
{((912
01)
, 92 f
(
91,12, 92 E SL(2) } .
Therefore, ZZk is the set of rank two matrices. /
J
(e) SRS
The vector space Zp5 is spanned by coordinate vectors of x1,33, x2,23. Let
ap5 (al, a2) = a(al, a2, al 2a2 l; al Sat 1, al a2) for al, a2 E GL(1). Then MR5 = {ap5 (al, a2) al, a2 E GL(1)}.
The weights of the coordinates x1,33, x2,23 determine the characters al 9a2 3a3 of ap5 (al, a2) respectively. These characters depend only on a3 = a3ja2i and there
are both positive and negative weights of a3. Therefore, Z7 k = {(x1,33, x2,23) E (kx )2}. (f) SRe
Let ape (a) = a(a, a, a-2; a-1, a) for a E GL(1). Then Mrs Re
91,12
_
0
0 1
92 )
a p, Re (
)
91,12, 92 E SL(2), a E GL(1)
.
8 Invariant theory of pairs of ternary quadratic forms
205
Let V1 (resp. V2) be the subspaces spanned by coordinate vectors of {x2,11, x2,12, x2,22} (resp. {x1,33}).
Clearly Zp6 = V1 V2. Elements of the form ap6 (a) act on V1, V2 by multiplication
by a3, a-5 respectively. Therefore, we have already discussed this case in Chapter 6. So Za6k 0. Let ZZ6k be the set of k-stable points, and ZR0 Stk = Zpek \ Zpek. As we defined in Chapter 3, SQ6k = GkZp6k, Sp6,stk = GkZQ6,stk' (g) So,
Let ap, (al, a2) = a(al, al 2a2 2, aia2; a2i a21) for al, a2 E GL(1). Then Mp,7 = lap, (al, a2) I al, a2 E GL(1)}. The coordinates x1,33 x2,13, x2,22 determine characters aia2, aia3, ai 4a2 3 respectively. The convex hull of {(2, 5), (2, 1), (-4, -3)} contains a neighborhood of the origin of R2. Therefore, Zpk = Zp,k = {(x1,33 x2,13 x2,22) E (V )3}. (h) Spe
Let ape (a) = a(a-2, a, a; a2, a-2) for a E GL(1). Then Mpe
=
{(@ 91,23)I2) aps(es)
91,23 E SL(2), a E
GL(1)}.
Let V1 (resp. V2) be the subspace spanned by coordinate vectors of {x1,22, x1,23 x1,33} (resp. {x2,12, x2,13})
Clearly, Zp8 = V1 ® V2. Elements of the form aRs (a) act on V1, V2 by multipli-
cation by a4, a-3 respectively. Therefore, we have already discussed this case in Chapter 6. In this case, ZQsk # 0, and Zaek = Zpek. (i) Sp9
Let a,% (a1, a2) = a(al, a2, al a2; a1 2a21, aia2) for al, a2 E GL(1). Then MR0 =laps (a1, a2) I al, a2 E GL(1)}. The coordinates x1,23, x2,13, x2,22 determine characters a1 3a2i ai, aia2 respectively. The convex hull of {(-3, -1), (2, 0), (2, 2)} contains a neighborhood of the origin of 2 . Therefore, Z7.k = Zp9k = {(x1,23,x2,13x2,22) E (kx)3}. G) Sp1U
We can identify Zp10 with Sym2k2 0 k2. It is easy to see that
M' = SL(2) x SL(2) - { I
I
0
9 023)
,
92) 91,23,92 E SL(2) }
Therefore, we have already discussed this case in Chapter 5. So ZQ10k Let Zp10,k be the set of k-stable points, and ZQ10,stk = ZR10k \ defined in Chapter 3, Sp1o,k = GkZp10,k, Splo,stk = GkZalu,stk'
.
0 Zp10,k.
As we
Part IV The quartic case
206
§8.3 /3-sequences of lengths > 2 In this section, we describe /3-sequences of lengths > 2 for our representation explicitly. Let 81, , )31o be as in §8.2. Let (31,1 = 02 - /31, /1,2 = /33 - /31. Then (/31i 01, 1), (01, /31,2) are the /3-sequences
of length 2 which start with /31. Let 31,1,1 = (33 - 02. Then only /3-sequence of length 3 which starts with /31.
is the
Let /32,1 = ,33 -(32. Then (/32i /31,2) is the only /-sequence of length 2 which starts with ,32.
There is no /3-sequence of length > 2 which starts with /33. Let /34,1 = /32 -/34. Then (34i /34,2) is the only /3-sequence of length 2 which starts with 04. We do not consider /3-sequences of lengths > 2 which start with /35. We can write any element of T+ in the form vp6 (A1)ape (A2)(a3(-
1,
A3,1),1),
where vpe is a 1PS proportional to )36 and A1, A2, A3 E R+. If we identify Pp. (A1), ap.(A2), (a3(\ 1, A3, 1), 1) with 1PS's, these are orthogonal to each other. Let tp8 = {z E t 1 (/36i x) = 0}. We identify t% with {(-a, a; b) I a, b E R} = 30 so that 1))(-a,a,b)
(vo,(A1)aRe(A2)(as(A 1,A3, 1),
= AbA2a 2 3
Since 11(1,1, -2; -1,1)112 = 8 and 11(-1,1, 0; 0, 0)112 = 2, 11(-a, a; b)112 = 2a2 + 8b2.
If (/t6i /3) is a /3-sequence, /3 is an element of tts. Since the metric on t;. is the same metric as we considered in §6.2, /3 corresponds to /3-sequences which we considered in §6.2. Let /36,1, , (36,4 be the elements which correspond to /31, . . , /34 in §6.2. Then /3-sequences of length 2 which start with ,Q6 are ((36, /6,1), , 036036,4) We do not have to know /36,1 for our purpose. We can describe /36,2, , /36,4 explicitly as follows: 1
(36,2=( 3,
1
3,24 33 2
3
2
1
3;- 2)
1
2)-/3s,
2
1
3
22
241 332
1
1
2
We do not consider /3-sequences of length 3 which start with /36. We do not consider /3-sequences of length > 2 which start with /37.
We define t% similarly as in the case S6. Since II(-2,1,1; 2, -2)112 = 14, the metric on tQ8 coincides with the metric we considered in §6.2. , /36 in §6.2. Then 08,6 be the elements which correspond to /31i Let /-sequences of length 2 which start with /38 are (/38, /38,1), , ((3s, /38,6). We do not have to know /38,1, /38,2, /38,3 for our purpose. We can describe /38,4i /38,6 explicitly as follows:
21
1
1
1
,38,4 = (-3, 3, 3; 2, - 2) - Qa, (38,5
2
=
3
'_2
4 1
3 3;
1
2
3
33
1
-1
2, 1
2) - /38, 1
22
q
8 Invariant theory of pairs of ternary quadratic forms
207
We do not consider ,3-sequences of length 3 which start with X38. We do not consider ,3-sequences of lengths > 2 which start with ,39. We define 1
1
1310,1 = (0,4, 4,
1
1
4, 4),
1310,2 = 132 - 1310,
,310,3 = 133 -,610-
Then (13io,131o,1), (010, 010,2), 0310, 010,3) are the ,3-sequences of length 2 which start with ,310. Let 310,2,1 = 1310,3. Then (310, 010,20310,2j) is the only 3-sequence of
length 3 which starts with ,310.
Chapter 9 Preliminary estimates §9.1 Distributions associated with paths Let (P, w, s, GA, g etc. be as in §3.1. In our case, the condition dim G/T = dim V is satisfied. Therefore, (G/T/V) is of complete type. We consider the integrals in (3.1.8) and use the notation Zv(4), w, s) etc. for L = V. Note that since G1 = GA, we have a canonical measure on G. _ Let [ , ]i, be the bilinear form for V = Sym2k3 which we defined in §4.1. We define a bilinear form on V by [(X1, X2), (yl, y2)]V = [x1, y2]V + [x2, y1]v
Then this bilinear form satisfies the property [gx, g`y]V = [x, y]v for all x, y E V where g` = TGtg-1,rG. We use this bilinear form as [, ]v in §3.1. Let J(oD, g°) be as in (3.5.6). We define (9.1.1)
1°(`li, w)
= J s/Gk w(g°)J(, g°)dg°;
1(4D, w, w) = f
w(g°)J(, g°)(g°, w)dg°. /Gk
The Poisson summation formula implies that _ 1 (9.1.2) Zv(OD, w, s) = Zv+(4), w, s) + Zv+(&, w-1,12 - s) + f A sI°(,
w)dx.
Since the first two terms are entire functions, the last term is the issue. As in previous parts, we assume that P = Mv,,,D. Also we assume that V)(z) = and V)(p) 0 0, so (g°, w) = 9((g°)`, w) By (3.4.34), (9.1.3)
1(-b,w,w)
CGA(w;p)I°(-D,w)
(CG=T2-1T31).
So we study I(4, w, w) as a function of w. By (3.5.9) and (3.5.20), (9.1.4)
I(4), w, w) =S#(w)A(w;
4D(o))
p,l(p)=1
We know that 8p (4), w, w) +-":"-, ((b, w, w) is well defined for Re(w) >> 0. If p = (r), s)
and 0 = (i31), (/32), (03) or ((310), Ep (-D, w, w) = 0. If p belongs to class (1) or (3) in §3.5, cp (,D, w, w) is well defined for Re(w) >> 0. This applies to p = (0, s) where
D = (04), (as), (07), (08) or (f39). Therefore, all the distributions in (9.1.4) are well defined for Re(w) >> 0 except for p = (D, s) such that i) = (,35). We prove that (,D, w, w) is well defined for Re(w) >> 0 and 8p (,D, w, w) - 0 for this case in §10.3.
Since the rank of our group is 3, we have to consider paths of lengths up to 3. It turns out that if a is a /3-sequence of length > 2 which starts with /3k, (32, /34, ,3lo,
9 Preliminary estimates
209
we can apply our results in Chapter 3. However, if a is a 3-sequence of length > 2 which starts with 36 or /38i we have to use a method similar to Chapters 6 and 7. For the rest of this section, we define certain distributions associated with /36i /38, and prove that they are well defined. Let p = (a, s) be a path such that a = (/36) or (/38). In both cases, Za = Va1® Va2 where Valk = Sym2k2, Va2k = k for /36, and -01k = Sym2k2, Va2k = k2 for /38. We define a bilinear form [, ]v,; on Vai for i = 1, 2 similarly as in §6.1, using the longest element of the Weyl group of M.D. Let ° v,; be the partial Fourier transform with respect to [ , ] v,, for i = 1, 2. In both cases, Ap2 = Aa (see (3.3.11)). By Table (8.2.3), AD = {da(.\1) I Al E R+}, where 6
3,
3
2
1
R) ,
2
a = (as
-1)
da(A1) _
1-1) -1 1-1
Let kv,; be the rational character of Ma defined by the determinant of GL(Vai) for i = 1, 2. For ga E G°O fl MsA, we define rv,1 (go) = Ikv,, (9a)I-1 for i = 1, 2. By (6.1.1), (6.1.2),
Oz, (Rasp, 9a) = ral(9a)Oz, (A-sRa4)p, ea(9a)) +AaRsDp(6) - Rs4p(o)
+
(ica1(9a)Os,, (AaRa-Dp,ea(9a)) - Os,, (RalDp,9a)) Z,, QVa1,Va2
+
(Ial(9a)es,,("aRDDp,Ba(9a))-Os,,(Ra-Dp,9a)) Z,, CV01
+
(ral(9a)Os,,
es,, (Ra4Dp,9a)) ,
Z,, C Va 2
and (ic
Os,,
(9a)Os,,
Z,,CV,1,Va2
+ .`'aRa,Dp (6) - Ra,Dp (6)
= Kv,2 (9a) E Os,, (9v,1 9aRo Dp, 9a) Z,, C Va 1
Os,,(9sRa4'p,Ba(9a)) Z,, C Va 2
T,
OS,, (Ra,Dp,9a) - 'cv,2(9a)
OS,, (` V,2Ra4'p,ea(9a))
Z,, C V02
Z,, C V01
_k01(90) E Os,,("aRoDp,ea(9a)) Z,, CV01
+acv,,(9a)
p,9a) Z,,CV,2
-' v 1 (ga)
Os,Z,,CV,1
Rasp+9,(9a))
-
es,, (Rasp, 9a), Z,,CV,2
where a' runs through /3-sequences of length 2 and a -< D.
Part IV The quartic case
210
Let X be a principal quasi-character of go E G°A f1 MSA/Mak, and W E .©(ZDA). We define JAaoM,'A/Mat X(9a)esa,(IF,9D).p(go,w)dga
sD' C Vol,
J AaoMOS/Mat X(9a)©sa, ('y,9a(9a))ep(go,w)dgD
So' C Vat,
("vat (sa) -
"Va1 00-1?1
So, C Vol,
X(9D)©s,, ('y, 0a(9a)).p(go,w)dga f AaoM' /Mat 1 W, X, W) _
"Va1 (go)- <1
J AaoMOA/Mat
Sa, C Vat.
X(9a)Gsa, ('y,9a).p(9a,w)dgo
"V,1 00-1 <1
Lemma (9.1.6) The distributions
are well
defined if Re(w) >> 0.
Proof. Let go = Raga, where A5 = do(A1) is as in (9.1.5), and ga E MMA. Let ao(X2) = ape(A2) or ap8(22) for A2 E R+. Then we can write ga = aa(A2)ga, where
9a E M. The element Aaaa(A2) acts on VD1A, V02A by multiplication by ep1(As)Aa, ep1(As)A ',
respectively, where a, b > 0 are constants. Then'v01 (go)-1 = By (1.2.6), for any N >> 0, there exists a slowly increasing function hN (go) of go such that (epl(As)Aa)dim V02
go) < (epl(Ab)Aa)-NhN(9a) 0sa, ('1'e 9a(9a)) << (ep1(Aa)-lA1)-NhN(9o) Os,, ('F,95(95)) << (epl(A,)Aa)NhN(9o) es,, (IF, go) << (ep1(A5)-lA2)NhN(9a)
esa,
for Zo, C Vol, for Z5, C V52, for Za' C Vat, for Za, C Va2.
Therefore if
Let p1 = ep1(Xa), µ2 = IV,, (9D)-l. Then A2 = Ill N >> 0, there exists a slowly increasing function h'N(pl,ga) such that go) <
for Zo, C Vol,
esa, ('1`, 05(g5)) << /12 NhN(Al, ga)
for Zo, C VD2,
e5,, ('I',95(go)) «p2
for Zo, C Val,
esa,('1',9a) <
for Zo, C Va2.
Let 9
o _ (( 91,12
-
43
) Ia2(t21,t22)) or
((t11
91,23)
,
a2(t21, t22))
for d = ()36), ()38) respectively, where tll, t13, t21, t22 E A', 91,12, 91,23 E GL(2)A. By
(3.4.31), there exist constants rl, r2 > 0 such that if Ml, M2 > w0i µ11(Re(w)-wo)Itll(91,12)tl2(91,12)_lI-r2(Re(w)-wo)
e ( 9a, w ) « 1j 1,11(Re(w)-wo) l It12(91,23)t13(91,23) p
1
I-*2 (Re(w)-wo)
a = (Q6), a =
('38),
9 Preliminary estimates
211
for M1 < Re(w) _< M2 and ga in the Siegel domain. We choose Re(w) >> N >> 0. Then the function Itll(91,12)t12(91,12)-1I-r2(Re(w)-wo)
g,)lt'1(Re(w)-wo)
X(9a)µ2 NhN(fti,
is integrable for p, < 1, µ2 > 1, and ga in the Siegel domain. Other cases are
similar. This proves the lemma. Q.E.D.
Let' = Ra4bp. If Si,' C V 1, we define (9.1.7)
w, w)
w, op, w),
p,a',3 (b, w, w)
(go IF, w, Ic010'p, w),
w, w) _ p,a',<
w, /CVal Qp, w).
Also if Si,' C Va2, we define Ep,a',1(C w, w) = =p,a',> (90 IF, w, Ka1op, w),
(9.1.8)
0 ,a',2
(', w, w)
(Va2 'I, W1 KV02 Qp, w),
=p,a',3 (), w, w)
w,'Vai op, w),
(..
p ,a',4 (-t, w, w) _ Ep,a',< (IF, w, op, w)
We define 4
(9.1.9)
p,tot('D,w,w) _
(-1) iFlC
W, W).
Za'CV1,Va2 a=1
For a' such that Za' ¢ Vat, Va2, we define rp' (Ra' gyp', w, w), gyp' (Ra gyp', w, w) in
exactly the same manner as in (3.5.8). Then by (3.5.5),
8p'(Re'(P,',w,w)+,=p'w,w) is well defined for Re(w) >> 0.
These considerations show that (9.1.10)
"p+(-D, w, w)
-,p(op,w,w) = E
+
(2p'(R,'4p',w,w) +'p'(Ra',Pp',w,w)) p-
'(p')2
Zp' a Val Va2
+ =p,tot (4D, w, w)
We will prove in Chapter 10 that the distributions =p' (Ra' gyp' , w, w), 'p' (Ra' gyp' , w, w)
are well defined for Re(w) >> 0.
Part IV The quartic case
212
§9.2 The smoothed Eisenstein series In this section, we prove some estimates of the smoothed Eisenstein series which are not covered in Chapter 3. We will use these estimates in Chapter 10. We first fix some notation. For u = (ul, u2) E A2, let (9.2.1)
((I2
np, (u) = nP2(u) =
U
\ \tu
For u = (u1, u2i u3) E A3, let 1
nB, (u) =
(9.2.2)
u1
((U2
J2)
1
U3
1
For u E A, let 1 (9.2.3)
n'(u) =
u
)I2)n"U)=
1
1
1
I2
1
U
1
j1
nB2 (u) _ (I3, (U
For) ER,t11, ,t22EA1,let 1
(9.2.4)
dpi (A,) = a(A1 1 , .\1 , A2 1, 1),
dp2(A1)=a(A 1 t0
For u = (u1i
,
2
Al,1i1,1), A
= Cl(tll,t12,t13it21,t22)
u4), let nB(u) = (nB, (u1i u2, u3), nB2(u4)). Also let dxt o =
ndxti;.
In our situation, the Eisenstein series EB (g°, z) is a function of g° E GA, z = (z1, z2), where z1 = (z11, z12, z13) E C3, z2 = (z21, z22) E C2 and Z11 + Z12 + Z13 =
0, z21 + z22 = 0. Let e(g°, w) be the smoothed Eisenstein series. We choose the constants (Cl, C2) in (3.4.3) so that C1 = 1, C2 = C > 100. Let I = (I1i I2), T = (Ti, T2), v = (v1i v2), sI(l) = (s 1,i, (1), 52,12 (l)) be as in (3.4.24)-(3.4.28). Note that v2 is always 1. Consider an element of the form A = dpi (A1)a(p1, µ1 1, 1iµ'2,µ2 1),
where A1, µ1,µ2 E R+- We can write any g° E G°A in the form g° = kAt °nB(u), where k E K and u = (u1i , u4) E A4. We write at ° = (t1i t2), where t1 E (g°, w) for various T°A, t2 E T20A. For the rest of this section, we estimate cases. We recall that by (3.4.30), if q E t*, for any 6 > 0, M > w°i
9 Preliminary estimates
213
for L(q) + 5 < Re(w) < M, where h(A) is a slowly increasing function independent of 5, M. Moreover, if v = 1, h(A) = 1. Let q = (ql, q2) _ (qil, q12, q13; q21, q22) E Dj,r. We define rl = qll - q12, r2 = q12 - q13, and r3 = q21 - q22 Then 2r1 + r2 ( 9.2.5 )
qll =
-r1 + r2 3
qlz =
3
,
q13 = -
rl + 2r2 3
The following table shows the condition for q to belong to the domain Dj,,. Table (9.2.6) Il = 0
1
no condition
(1,2)
rl > 1
rl > 0
(1,3)
rl>1
rl>1
(2,3)
r2 > 1
(1,2,3)
rl>1 rl+r2>1
(1,3,2)
rl+r2>1
-
--
rl>0
r2 > 0 rl>1 rl+r2>0
-
r2>1
r2>1
rl>O r2>1
-
r2>0
r2>1
-
I1 = {1,2}
I1 = {2}
I1 = {1}
Tl
rl+r2>0
-
-
--
r2>0
It is easy to see that Al rl-2T2 idrl
t11g1+P1 =
(9.2.7)
Ai 3µ X
Al rl-2r2µ1 rl 2r1+r2 -r2 A 1
Al
Pi rl+r2 rl+r2
µl µl
2r1+r2 r2
Al
A T1+r2µl -Tl-r2 1
Also 1 µi 2111 (9.2.8)
-$1,1111 -
-
tl
t2S2,12
2
(1)
-
3
t
µl
1 -111 )
3111
_ 1
111
µ'1 µ12111(A -3p1 -')-112 - A1112µ112-2111
-212. /2 2
Suppose vi = (1, 2). Then (9.2.9)
t1
1o
3µl)-111 = A1111µ1111
2111 (Al 3/11)-112 = A1112µ1111-112
Ii = {2}, I1 = {1, 2}.
Part IV The quartic case
214
Proposition (9.2.10) Suppose I # (0, 0). (1) Suppose that r1 (1, 3), (1, 3, 2), or T = (1, 3) and Il = {1, 2}, or r = (1, 3, 2) and I1 = {1}. Then for any N > 0, e > 0, there exists a constant b > 0 such that if
M>w0, ej,T,1(9°, w) <<) I,42µ23
for w° - S < Re(w) < M, where cl > N, C2 < -N, C3 < f(2) Suppose that 12 = {1}, and ri = (1, 3) or (1, 3, 2). Then for any e > 0 there exists a constant S > 0 such that if M > w°i &r,T, l (9°, w) << )i' Itµc3
for w° - S < Re(w) < M, where c1 > -2 +8 C, C2 < -(-2 + sC), and c3 < f-
(3) Suppose that (I1i 7-1) = ({2}, (1, 3)) and 12 = 0. Then for any e > 0 there exists a constant S > 0 such that if M > w°i 2+c1 p1: c2
0
1"'i,T,l(9,w) K Al
c3
µ2
for w° - S < Re(w) < M, where Ic1I, IC2I, IC3I < e.
Proof. If 12 = 0, tT2 Z2+P2 2
-1
1+r3
ll
µ'2-r3
T2=1, T2 = (1,2).
If r2 = 1, we choose r3 = -1, and if r2 = (1, 2), we choose r3 > 1 close to 1. If 12 = {1}, tT2s2+P2-82,12(t) 2
1-r3-212.
=2
We choose r3 = a for this case. In all the cases, Lj(q2) < 1 + e1i where e1 > 0 is a small number. Also t22z2+P2 = µ2'3 for some c3i where C3 < e in all the cases and
IC3I
T1=(1,2,3), wechoose rl= 2-r2>>0. Then L(q)=3+(1+e1)C<w°=4+C if el is sufficiently small. Then ti1g1+P' satisfies the condition of (1). Since v = 1, we can fix 1. This proves (1) for r1 # (1, 3), (1, 3, 2). If I1 = {1, 2}, we choose r1 = r2 = 2, and 11, >> l12 >> 0 in (9.2.8). Note that we can do this because v = 1. Consider the case Ti = (1, 3, 2), Il = {1}. We choose r2 >> 0 and ri = 1 - r2. Then L(q) < w°, and
ti
- -A A
1
2r2-1 -1-211' 1
P'1
tC
Therefore, if we choose 11, >> 0, we have an estimate of the form (1).
Next, we consider (2). If Ti = (1, 3), we choose r1 = r2 = 1 + 16C and r3 = 2. Then L(q) = 4 + 2C < w°i and T191+P1
-
Al16
µl-16
9 Preliminary estimates
215
We have already considered the case I1 = {1, 2} in (1). If I1 = {1} or {2}, we choose 11, = 2 in (9.2.8). Note that we can choose any 11, > 1 in this case. This consideration shows that we have an estimate of the form (2). If rl = (1, 3, 2), we choose r1 = 0, r2 = 1 + 8C, and r3 = 2. Then L(q) < w°i and
-2+8C -IC
= Al
t1
Al
We have already considered the case I1 = {1} in (1). Therefore, we have an estimate of the form (2). Finally, we consider (3). We choose r1 = E2, r2 = 2 - e3, 111 = 1 + E4, where E1,' , E4 > 0. Then L(q) = w° + 2(E2 - E3) + CE1, and "I,r,I(9°, w)
<< A
2+2E2-E3+3E4µ13+E4pC31
where Ic3I < E.
, E4 so that L(q) < w°i we have an estimate of the
Since we can choose El, form (3).
Q.E.D.
Proposition (9.2.11) (1) Suppose I2 = {1}. Then there exist a constant 6 > 0, a slowly increasing function
h(a1i µ2), and a constant c > 1 such that if M > w°i 'I,r,v(9°, w) << P'1 `h(A1, P2)
for w° - 6 < Re(w) < M. (2) Suppose that I1 E) 2 and v1 = (1, 2). Then there exist a constant 6 > 0 and a slowly increasing function h(A) such that if M > w°i for any 1 >> 0, ei,r,v(9°, w) << h(A)A11µ-1
for w° - 6 < Re(w) < M. Proof. Since we assumed that C > 100, we covered the case when v = 1 in (9.2.9). Therefore, we assume that v1 = (1, 2) (v2 = 1) and prove (2). If I1 = {1, 2}, by (3.4.30), 4'I,r,v(9°,
w)
11-112 << A3`12µ21 1
A(rq+P)h(A),
where h(.\) is a slowly increasing function. So we choose any q E Dj,r so that L(q) < wo, fix 111, and choose 112 >> 0.
If I1 = {2}, by (3.4.30), I,r,v (9°s w) <<
1111 p-111 Av(rq+P)h(A),
where h(A) is a slowly increasing function. So we choose any q E DI,, so that L(q) < w°, and choose 11, >> 0. Q.E.D.
Chapter 10 The non-constant terms associated with unstable strata In this chapter, we prove the following two theorems.
Theorem (10.0.1) Let p = (D, s) be a path. Then 8p (4D, w, w) - 0 for (1) (2) (3)
= ()35),
(4)
_ (08),
(134), (137),
(5) r) = (138,138,1), and s(2) = 1, (6) a = ()38,138,2), and s(2) = 1, (7) -0 = ()3io, /31o,1)
Theorem (10.0.2) Let p1 = (131, 51), p2 = (132, s2) be paths. Then Epi
W, W) + Epz 'p2
=p1
(1) 01 = (136), 02 = (138,138,1), (2) 01 = (N8, 13s,2), 02 = (,39),
for
W, W) - 0
and sl(1) = 52(1), 52(2) = 0, and 5i(1) = 52(1), s1(2) = 0.
We will prove (10.0.1) (1) in §10.1, (2) and (7) in §10.2, (3) in §10.4, (4) in §10.5, (5) in §10.3, and (6) in §10.6. Also we will prove (10.0.2) (1) in §10.3, and (2) in §10.6. Corollary (10.0.3) Ep'= p (4), W, W) - 0. p
In this chapter, g8 always denotes an element of G°A f1 M0A. We use the measure dgo on G°A fl MDA which we defined in §3.3 (just after (3.3.11)). Let dxtO be the measure on TA° such that dg° = dkdx t°du, where dk, du are the measures on K, NA which we defined before.
§10.1 The case 0 = (,34) In this section, we prove the following proposition. Proposition (10.1.1) Let p = (0,s) be a path such that D = 134. Then yp (4', w, w) 0.
Proof. Let n(ut,) = npl (ua) for ua = (u2i u3) E A2. Consider E )y, (`I', 9an(ua)),9(9on(uo), w)du8
(10.1.2)
J(A/k)2
for IF = Ra4,p.
Let da(A1) = dpl(A1) for Al E I1 (see (9.2.4)). Let 913 = (91, 92), 91 = ( 91,12
t13)
'
10 The non-constant terms associated with unstable strata
217
where I det 91,121 = It13I = I det 921 = 1. Let go = d(A1)g°00.
For a = (al, a2) E k2, let < a ua >=< alu2 + a2u3 >. We define (10.1.3)
J1,«(9a) = I(A/k)2 OYa (IF,9an(ua)) < a ua > duo,
,fz,a(90,W) = fA/k)2 e(9an(ua), w) < a ua > duo. (
By the Parseval formula, (10.1.2) is equal to
8,«(91).2,-«(9a,w) aEk2
We consider 91,12, 92 in the Siegel domain as follows: (10.1.4)
9112 = k1
F(tltll
1
1)'
µ11t12) (u1 (P2t21
92 = k2
1
u4
µ21t22
1
,
where tll etc. are chosen from a fixed compact set in A', and P1,µ2 are bounded below.
We can write any element y E Yjk' in the form y = (yl, y2) where y1 EZak, Y2 E Wak and
yl
X1,13
X1,23
x2,13
X2,23
)
X
'
y2 = ( x2,33 )
Then 9an(ua)y = (A,t13g2y1tg1,12,)it1392(y2 + yltua)) Therefore,
fl,a(9a) = f A2
W(A1t1392y1t91,12,
A4't12
g2y1tua) < a ua > dua
ylEZak
392yltua) < a ua > duo.
_ yl EZnk
JA2
For y2 = t(x1,33, x2,33), y2 = t(x1,33+x2,33), we define [y2, y2] = x1,33x1,33 +
x2,33x2,33 Let IF, be the partial Fourier transform of IF with respect to y2 and the above bilinear form. Then t
fl,«(9a) = 1 8
4 2t -1t -1t 1 t13 92 ty1 a)
yl EZak
Let (10.1.5)
f3,a(9a) = A1
8
W1(A1t1392y1t91,12, A1 4t 13 t92
lty1 lta),
v1 EZak
-1,13#0
14 ,«(90)
_
Alb
A14t13t92 lty1 lta ) v1 EZak -1,13=0
Part IV The quartic case
218
Then fl,a(9a) = f3,a(9a) + f4,a(9a) Lemma (10.1.6) Suppose that µl, µ2 are bounded below. Then for any N > 1,
E If3,a(9a)I << rd3,N(Ai, pl, p2) aEk2\{0}
Proof. Let a' = (a', a2) = ayi 1. Then T1(A1t1392y1t91,12, A1 4t13 t92 lta )
f3,a(9a) _ A1 $ aEk2\{0}
v1 E ZDk
«'Ek2\{0}
x1,13#0
An easy consideration shows that Al 4t21 t9 2
ltai
= tk2 l \ 4t 12
il-163x2) )
/62 lt2i
A2t22 a2
J
.
Let
f5(90) = E
If3,a'(9a)I, f6(91) = E
If3,a'(91)I
a2=O,aiEkx
aiEk,a2Ekx
We choose 0 < \Ir2 E .9(ZaA), 0 < `P3 E 9(WDA) so that IWl(yl, y2)I << W2(yl)'F3(y2)
By (1.2.8), for any N > 1,
E
XF 3(-1
4t 13
t92
ltx)
<< sup(1,)lp2)(Al 4P2)-N,
«4Ek «2Ekx
3(A1 4t13 t92
lta) « (Aiµ2)N,
«i Ekx 1
«2'4 =0
E W2(\1tl392yit91,12) <
.1,13#0
where hl(A1, /61, P2) = sup(1, Al'Ptl 1tt2) sup(1, A1 lftlft2
1)
sup(1, )1 l/t1ft2).
So for any N1, N2 > 1, f5(9a) << Al shl(\1, /tl, /t2) sup(1, A4 1p2)(\l/1i/A2)-N1(Al 4U2)-N2,
f6(9a) << Al shl(A1, Al, /t2)(A1p1Et2)-Nl (Al 4µ2l)-N2.
Therefore, for any N1, N2, N3 > 1, f3,a(9a) << h2(Al, Ml, /62)('\1µlµ2)-Nl ((Al 4p2)-N2 + (Al 4/62 aEk2\{0}
l)-N3),
10 The non-constant terms associated with unstable strata
219
where
h2(A1, µl, µ2) = Al 8 sup(1, \1µ2)hl (X1, µl, µ2).
If A1 > 1, we choose Nl >> 0, and if Al < 1 , we choose Nl = 2N2 = 2N3 >> 0. Since Al, 92 are bounded below, we get the proposition (10.1.6). Q.E.D. Next, we consider f4,a(ga). Consider yi E Zak such that x1,13 = 0. This implies that x1,23 0 0 and x2,13 0 0. An easy computation shows that t
-ltyi-lt a = det x-
92
1
L 1t21 ((x2,23 + x2,13u3)al - x1,23«2) _1 _E2 22 x2,13a1
Let
f7(9a) =
Y,
1.f4,a(9a)I, .f8(9a) = E
If4,a(90)I
a1=0,o2Ekx
aiEkx,a2Ek
Lemma (10.1.7) Suppose that µl, µ2 are bounded below. Then we have the following estimates. (1) For any N > 1, f7(ga) << rd3,N(A1, Al, A2)-
(2) For any N > 0, f8(9a) «
12 Al-2N
sup(l, Al lµl/t2)µi µ2 Proof. By (1.2.8), for any N1, N2, N3 > 1, f7(9a)
sup(1, A4A2) (1µ1µ21)-N1(A1µ1µ2)-N2(A14µ2)_N3
X
We choose N1 = N4 + N3i and N2 = N4, where N3, N4 > 1. Then 2
f7(9a) « A-' sup(l, A-1µ4µ2) sup(l, AW2)A12N4 (A1 2 µi
/J-N3
Since 91, µ2 are bounded below, we get (1). Consider f8(ga). By (1.2.8), for any Ni, N2, N3 > 1, f8(9a) « Al 8 sup(1, Al 1µ1µ2)(A1µ1µ2 1)-N1(A1µ1
1µ2)-N2 (a14µ21)-N3.
We choose N1 = 1 + N, N2 = 2 + N, N3 = 1, where N > 0. Then f8(9a) << Ar
2
sup(l, ai 1µ1µ2)µi µ2 .
This proves (2). Q.E.D.
We continue the proof of (10.1.1). By the consideration as in §2.1,
E 1f2,a(91,w)I <EiI,7,v(9a,w), aEk2`\{0}
L, a1=0,a2Ekx
jf2,a(9a,w)j «
Part IV The quartic case
220
where I = (11,12) and either Il or 12 is non-empty. Therefore, there exists 6 > 0 such that if M > wo, EEk2\{o} If2,a(90, W)I is bounded by a slowly increasing function for wo - 6 < Re(w) < M. This implies that If4,a(9o)f2,a(9a)It(9o)-2p,
If3,«(9o)f2,a(9o)It(9a)-2pa,
«Ek2\{0}
«iEkx,a2Ek
are integrable for go in the Siegel domain. By (9.2.10), there exists 6 > 0 such that if M > wo, (go,w) «sup(Ail µ i2µz3,A
2+C
1µ12µ23),
where cl > 8, c2 < -4 and Ic3I, Icil, Ic2I, Ic3I are small. Note that we are restricting ourselves to the Siegel domain, so we can assume that Ic3I, Ic3I are small. t(ga)-2p = It is easy to see that A1912µ22. Therefore, for any N > 0, t(go)-2pfs(9o)
E
If2,-a(9o, W) I
a1=°,a2Ekx
is bounded by a constant multiple of 1
sup(1, A1 µ1µ2) 3llp(A 1
2-2N+c1 -2+c2 -2+c3 µl µ2 , A1 ^1
2-1
-2+cs -2+cs µ1
µ2
)
If A, > 1, we choose N >> 0, and if Al < 1, we choose N = 0. Hence,
f
a M,A/Mak
f1,a(9o)f2,-a(9a)t(9o)-2pdga _ 0.
wp(go)
aEk2\{0}
This proves (10.1.1).
Q.E.D.
§10.2 The cases a = (,Qs), and 03 = (01o) We consider paths pi = Let 01 = ()35), 02 = (01,51)4P2 = (02,52), and p3 = 03,53) such that 51(1) = 52(1) = 53(1) (i.e. P3 P2)- For 0 = 01,D2, Ma = T. Therefore, we use the notation go = to = (tl, t2) where tl E T°A, t2 E T20A.
We first consider pl. Let p = p1, 0 = 01.
Proposition (10.2.1) The distribution 8p(ID,w, w) is well defined for Re(w) >> 0, and , p (4), w, w) - 0 .
Proof. Let W = RaPp, and n(uo) = nB(uo) for uo = (u1,u2,u3,u4) E A4. It is easy to see that Oy, (W, t°n(uo)) does not depend on u1i u2. For a = (al, a2) E k2, let
< a, uo >=< a1u3 + a2u4 >. We define (10.2.2)
fl,«(t°) = J
OYa (IF,t°n(ua)) < a uo > duo, A/k)4
f2,a(to, w) = J A/k)4
ea(t°n(uo), w) <
uo > duo.
10 The non-constant terms associated with unstable strata
221
By the Parseval formula,
I
fin (t°)f2,-a(t°, w).
y.('W,t°n(uo))8(t°n(uo),w)duo = aEk2
We can write any element y E Yak as y = (yl, y2) where yi = (x1,33, x2,23) E Zak, y2 = x2,33 E Wok. Then
`1(t°n(uo)y) < a uo > duo.
fl,a(t°) = (A/k)4 Y2Ek
YiEZak
An easy computation shows that t°n(ua)y = (71,33(t°)x1,33, 72,23(t°)x2,23, 72,33(t°)(x2,33 + X1,33U4 + x2,23u3)) Let u4 = x1,33u4 +x2,23u3, ui = ui for i = 1, 2, 3. Then u4 = x1,33(u4 - x2>23u3) We define
(10.2.3)'(t°, u4, y) = i'(ry1,33(t°)x1,33, 72,23(t°)x2,23, 72,33(t°)(x2,33 + u4)).
Since t°n(uo)y does not depend on ui,u2,
fl,a(t°) _ YiEZak
f(A/k)2
T V/(t°, u4, y) < alu3 + a2x1,33(u4 - x2,23u3) > du3du4. Y2Ek
Since < alu3 + a2x1,33(u4 - X2,23, u3) > _ < (al - a2x1,33x2,23)u3 + a2x1,33U4 >,
the above integral is 0 unless x1,33 = al 1a2x2,23. Therefore, /
fl,a(t°) =
Ek -1
ZA V (t
0
-1
/
/
, u4, yl) < a2x1 33u4 > du4.
-1,33-°1 -2-2,23
Let W1 be the partial Fourier transform of W with respect to the coordinate x2,33
and < >. Then fl,a(t°) is equal to (72,33(1°)
(-1
W1(71,33(t°)x1,33, 72,23(t°)x2,23, 72,33(t°)-1a2x1,33) a1 E ZOk
-1 -1,33=°1 -2-2,23
Since X1,33, X2,23 54 0 for yl E Zak, the above integral is 0 unless al, a2 # 0 or al = a2 = 0. We choose 0 < q'2 E .9(YSA) so that 011 <''2. Then
E If1,a(t°)I C aEk2\{0}
I72,33(t°)I-1e3('P2,71,33(t°),72,23(t°),72,33(t°)-1),
Part IV The quartic case
222
Let Pl = 171,33(0°)I, P2 = I'Y2,23(t°)I,µ3 = I'Y2,33(t°)I-1, and (10.2.4)
d(µ1, P2, µ3) =a(µ1 21
_1 _12
a
µi1
,
,E
µ2
1 _ 41
µ121
11
i
µl
1 21
2 ).
Then to can be written in the form to = d(pi, P2, µ3)t °. By (1.2.6), for any N1i N2, N3 > 1, Ifl,a(t°)I << IL-NjIL-N2 A1-N3
(10.2.5) aEk2\{0}
By the consideration as in §2.1, 1f2,a(t°, w)I E iI,r,1(t°, w)
r
aE(kx )2
where Il = {2}, 12 = {1}.
Lemma (10.2.6) Suppose that Il = {2} and I2 = {1}. (1) If Ti = (2, 3) or (1, 2, 3), there exists a constant S > 0 such that for any N > 0, M > w° there exist cl, c2i c3 > N satisfying ei,r,l(to,w)
<< Pi1P'22P33
for w° - 6 < Re(w) < M. (2) If Tl = (1, 3), there exists a constant S > 0 such that if M > wo, «P11µ2P3
forty°-S
<<
(P2P3)-111(P1P3)2-2r3-ia1 x
1(911,913,912)+P1
Ti = (2,3),
1(912,913,911)+P1
Ti = (1, 2, 3),
1(913,912,911)+P1
T1 = (1, 3)
for L(q) + S < Re(w) < M. An easy computation shows that ti' = Pl ° p2 lp3 ' . We fix r3 = 2, 111 = 121 = 2. Then Pl I,r,1(tO,w)
X
2*1
4r2 P2 r1-r2
P3
2r1-4r2
Ir1-qT2 -T2 -qrl-q r2
P1
P2
P3
Irl+Zr2 r2 -4r1+lr2
Pl
P2 P3
In the first case, we fix r2 and choose rl << 0. In the second case, we choose r1 = -r2 + S >> 0 where b > 0 is a small number. Then it is easy to see that the statement of (1) is satisfied. In the third case, By choosing q so that r2 = 10, rl = 1,
10 The non-constant terms associated with unstable strata
we get the estimate of (2). Since L(q) = 22 + L(q) < w°. This proves (2).
223
2C, w° = 4 + C, and C > 100, Q.E.D.
By (10.2.5), (10.2.6), for any N, Nl, N2, N3 > 1, 0
0 -2p «rd3,N(/tl,/t2,/-13)+µla -Nl 9-N2 4-N3 P2 µ3
0
Ifl,a (t )f2,-«(t ,w)I(t) aEk2\{0} Therefore,
0 JA/Tk
wp(t°) E fl,a(t°)f2,-a(t°,w)(t°)-2pdxt° - 0. aEk2\{0}
This proves (10.2.1) (We have also showed the absolute convergence for Re(w) >> 0). Q.E.D.
Next, we consider P2 such that 52(2) = 0.
Proposition (10.2.7) ;42 ((DI WI W) - 0' Proof. Let tY = Rae 4)p2. Let nag (u3i u4) = (n" (U3), nB2 (u4)) for u3i u4 E A. Then 43 (t°na2 (u3, u4), w) < 0lu3 + a4u4 > du3du4 = f2,a(t°, w). J(A/k)2
Since ep21(t°) = (t°)010 = Al /.t2µ3 and 52(1) = sl(1), we are considering the integral
J
wp2(t°)
fl,a(t°)f2,-a(to,,W)(t°)-2pdxt°
El/ aEk2\{0}
TA/Tk
µ1D2 µ3 <1
But we have already estimated this integral in a larger domain. Therefore, p2 (4i, w,
W) - 0. Q.E.D.
Next we consider P2 such that 52(2) = 1.
Proposition (10.2.8)
p2 (4i, w, w) - 0.
Proof. Let eYa2 (`11, t°Ba3 (na2 (u3, u4))) < l3 + 2U4 > du3du4,
(10.2.9) fl,a(t°) = f A/k)2
f2,a(t°, w) = J
r'p3 (9D3 (t°)na2 (u3, u4), w) < a1u3 + a4U4 > du3du4.
A/k)2
Then, by the Parseval formula, Ep2 ($, w, w) is equal to
f
wp2 (t°) E TA Tk
;,,;.2 3>1
aEk2\{01
fl,a(t°)f2,-a(t0
W)((t°)-2Pa3 IGa31(t°))-1(t°)-2pa2 dxt°.
Part IV The quartic case
224
We can estimate EaEk2\{0} I fi,a(t°)I in exactly the same way as in (10.2.5). Let µ4 = µ1µ2µ3. Then µ1
Ni
µ2 N2µ3
NS
= µl2
-N1 -f'2 µ32 fl -N3 P4
Therefore, for any N > 0, ,«(t°)I << 4 Nrd2,N(µl, µ2) aEkx
Since EaEkx I f2,_a(t°, w)I is a slowly increasing function, well defined for Re(w) > wo - 6 for some 6 > 0, °p2 (4>, w, w) - 0. Q.E.D.
§10.3 The cases D = (/36), In this section, we consider the cases al = (/36), D2 = (38,,38,j). Let (l3 = (08)-
such that si(1) _
Consider paths pl = (D1,51)42 = (22,52), and p3 = 52(1) = 53(1).
We first consider D = t)1, s = si. In this case, Pa = P1 x B2. The proof of the following lemma is easy, and is left to the reader.
Lemma (10.3.1) The convex hull of each of the sets (1) {71,33, 72,11, 72,22, -72,33}, (2) {71,33, 72,11, 72,22, -72,13},
(3) 171,33, f 12,111 72,22, -72,23},
(4) 01,33, 72,121 -72,13, -72,23}
contains a neighborhood of the origin. For ua = (u2i u3, u4) E A3, let n(ua) = (npl (u2i u3), nB2 (u4)). Let 91
091,12
0
0 t13
,
t2
t21
=1\ 0
0 t22
where I det gi,12I = It13I = It21I = It22I = 1. Let ga = dpl (,i)a36 (µ2)(g1, t2), where a36(µ2) is as in §1.2. Let 91,12, ui etc. be as in (10.1.4), and go = kt°n the Iwasawa decomposition of ga. We can write any element y E Yakl in the form y = (yi,Y2) where y1 = (x1,33, x2,11, x2,12, x2,22) E Zak and y2 = (x2,13, x2,23, x2,33) E Wak. Let
IF = Ra4)p. Then DYa('P,9an(ua)) =CY; (1I',9an(uo))+ey,
(11,9an(ua))
We recall that these are theta series associated with subsets of {(y1, y2) E Zak, Y2 E Wak}, where y1 E Zak, Za Stk respectively. By assumption,
ey, ('1', 9an(ua)) = w}1(k)CYa (W, t°nn(ua))
So when we estimate various functions, we assume that k = 1.
10 The non-constant terms associated with unstable strata
Let
fl,«(90) = J
e; (, 9n(uo)) < a u> duo, A/k)3
e(9n(u), w) < a u> duo.
f2,. (90, w) = f A/k)3
Proposition (10.3.3)
J
f1,«(9o)f2,-«(9a,w)t(9a)-2PDd9a ^ 0.
wp(9o) n M,A/Mok
«Ek
0}
Proof. We define 2
2
Q,, (u2, u3) = x2,11u2 + x2,12u2u3 + x2,22u3, lye (u2, u3) = x2,13u2 + x2,23u3
Then n(uo)y = (yl, x2,13 + au2Qy1, x2,23 + au3Qy,, x2,33 + lye + Qy, + x1,33u4),
where 8u2 etc. are the partial derivatives. Let u'2 = 8u2Q,1, u3 = 8u3Qy1. Then
-
C
2X2,11 X2,12
X2,12
U2
2x2,22) ( u3
So if A(yl) is the determinant of the above matrix, U2
( u3) = 0(91)-1 Note that if yl E ZaSk, A(VI)
2X2,22
-X2,12
2x2,11)
u3
)
0.
As before,
fl,«(90) = ylEZok
I I'(9an(uo)yl) < a uo > duo. A3
Let A2,13 = A2,13(9a, y1, u1, u5) = 72,13(t0)au2Qy1,
A2,23 = A2,13(9a, Y1, u1, uo) = 72,23(t0)(au3Qy1 + ul(x2,13 + au2Qy1)),
A2,33 = A2,13(90, yl, u1, ua) = 72,33(t0)(42 + Qyl + xl,33u4).
Then
fl,«(go) = yl EZak
f
IF (9o y1, A2,13, A2,23, A2,33) 3
< a uo > duo.
225
Part IV The quartic case
226
Lemma (10.3.4) There exists a constant 6 > 0 such that for any N > 0, M > wo,
E Ifl,a(9s)f2,a(9s, w)I < rd3,N(Al, 111, P2) aEk3\{0}
forwo-S
dinate x2,33 and < >. Then 1F(gon(uo)yl) < a3u4 > duet JA y2,33(to)-1«3x1,33)
IF1(goy1, A2,13, A2,23,
< -a3(1 ,2 + Qy1) >
We choose 0 < "2 E 9(Z-OA x A) so that
fA2
IW
1(yl,x2,13,x1,33)Idx2,13dx2,23 «W2(y1,x1,33)
Then `I'2(9ay1,y2,33(t0)-1«3x1,33)
Ifl,a(9a)I << X 6N'2 T, a3Ekx
yi EZok
Note that this bound does not depend on al, a2. By (10.3.1), for any N > 0,
E Ifl,a(9a)I <
By the consideration as in §2.1,
E If2,a(91,w)I << E
(go, W),
n1,a2Ek
a3Ekx
where I=(I1i12)and12={1}. Therefore, there exists 6 > 0 such that if M > wo, 1,a2Ek,a3Ekx If2,a(gs, w)I is bounded by a slowly increasing function for wo - 6 < Re(w) < M. So for any
N>0,
E If1,a(gs)f2,a(9a,w)I <
(10.3.5)
a1,a2Ek a3Ekx
for wo - 6 < Re(w) < M. We consider a such that a3 = 0. Let
al = A(yl)-1(2«1x2,22 - «2x2,12), a2 = A(yl)-1(-alx2,12 +
2a2x2,11)
Then alu2+a2u3 = aiu'2+a2u3. The condition a 54 0 is equivalent to the condition a'=(a',a')0.
10 The non-constant terms associated with unstable strata
227
We define 'I'3(yl, x2,13, x2,23) fA2 `1'(yl, x2,13, x2,23, x2,33) < x2,13x2,13 + x2,23x2,23 > dx2,13dx2,23dx2,33
Then
fl,a(90)=I72,33(t°)I-1 E 'F3(9ay1,72,13(t°)-la' -a2u1,72,23(t°)-lag yi E Zo k
We choose 0 < 'P4 E So(ZDA),0 <'I's E So(A2) so that I
3(y1, x2,13, x2,23)I <'I'4(y1)W5(x2,13, x2,23)
By (1.2.6), for any N1, N2, N3 > 1,
E T4(9ay1) <<
I71,33(t°)I-N1I72,11(t°)I-N2I72,22(t°)I-N3,
V1 EZnk
'k5(72,13(t°)-la' - a2u1, 72,23(t°)-la2) << sup(1, I72,13(t°)I)I72,23(t°)INI, -1 Ek .Z Ekx
E 'I's(72,13(t°)-1x1,0) <<
I72,13(t°)IN1.
aiEkx
Therefore, by (10.3.1), for any N > 0,
E
(10.3.6)
Ifl,a(9a)f2,-a(9a, w)I << rd3,N(.Al, µ1, P2)-
(-1--2 )Ek2\fo)
°3=0
Hence, (10.3.4) follows from (10.3.5) and (10.3.6). Q.E.D.
Now Proposition (10.3.3) is an easy consequence of (10.3.4). Q.E.D.
Next, we consider
(A/k)3
©y.,., (`F, 9an(ua))e(9an(ua), w)dua.
Let (10.3.7)
d(Al, µl,µ2) = a(Aj 1µ1,'l
lµi 1
, Air L 2, µ21),
and t° = d(A1i pi, µ2)E °. Let go = n'(u°)t° for u° E A. Let Ha be the subgroup generated by T and the element ((1, 2), 1). We identify ZOk with the space Sym2k2 ® k in Chapter 6. We define Za °k to be the subspace
Part IV The quartic case
228
which corresponds to Z,,ok in Chapter 6. We define Z'a3ok similarly. Then Zl,.tk Mak XHak Z'asok. We define
Xb = {kn (uo)to I k E K fl MbA, Al >
(10.3.8)
a(uo)}.
Then as in §4.2, if f (go) is a function on GO fl Maw/Hak,
f °1fMae/Hak f(9)d9=
f(9)(to)-2dxodo.
f
ar1Be/Tk
We define (10.3.9)
1,
6z;,0 (W, 9an(ua)) =
T(9an(ua)(yl, y2)),
51=(-1,33,0,-2,12,0), Y2 E Wak -1,33,-2,12 Ek x
f3,«(96) = f(A/k)3 6z6,a (W, 9an(ua)) < a ub > dub.
Then fA,MA/M,k (A/k)3
eya,., (`f', 9an(ua))e(9an(ua), w)dgbdua 0
f3,«(9a)f2,-a(9a,w)(t) Xa f1BA/Tk ,,Ek3\{0}
2p 2 x t0duo. Al
By a similar consideration as before,
E 51(13
f3,a(90) =
'(9an(ua)yl)dub.
fA
-1,33,-2,12Ekx
As in the proof of (10.3.4), If3,a(9a)I << A 6µ2
'1'2(901,
72,33(t0)-1x1,33a3)
51=(-1,33,0,-2,12,0)
-1,33,-2,12 Ekx
This bound does not depend on a1, a2. We choose 0 < 9'6i T7 E 9'(A2) so that '1'2 (9ay1, 72,33(t0)-1x1,33a3) 12µ21x2,12, A12µ21x2,12u0)q'7(A1i
<< '6(A
For it El
,uoEA, let hi(µ,uo) =
W6(µx2,12,ux2,12uo) x2,i2Ek5
We also define 'F7(A4iE2X1,33, AI 4A-2X-,I
h2 (Al, P2) X1,33,a3Ekx
10 The non-constant terms associated with unstable strata
229
Then If3,a(ga)l «A16/12h1(A12121,uo)h2(.A1,µ2)
(10.3.10) a3Ekx
Lemma (10.3.11) For any 81i S2,83 E C, the integral As' A12 2 a(uo)s3hl(t12u21,uo)h2(A1,IU2)d-Ajdxµ2duo
LXA converges absolutely.
Proof. Let µ3 = A 1 2µ2 1, µ4 = A1/12 Then A1 = pi µ4 , /12 = 113 2µ4 1 By (1.2.6), for any N1, N2 > 1, h2()1iµ2) << Therefore, our integral is bounded by a constant multiple of µ4N'(µ34µ43)-N2.
µ4N2+siµ33N2-N1+s2 18+
a(u°)
Re(83 )
X
x
hl(P3,uo)d µ3d µ4du°,
xA
where si = Re(2s1 - 282), s2 = Re(2s1 - 82). Let r1i r2 E R, and c > 0. Then the integral µs1 a(uo)r2 h1(µ3, uo)dX p3duo
converges absolutely and locally uniformly on the domain of the form {(rl, r2) E R2 I r1 > 2+e,r2 > -e} by (4.2.9). So if µ4> 1, we choose N1>>N2>>0, and if µ4 < 1, we choose N2 >> N1 >> 0. This proves (10.3.11). Q.E.D.
By the consideration as in §2.1,
If2,a(g1,w)I «I: 4,Ty(ga,'w), a3Ekx '11°2Ek
where 12 = {1}. By (3.4.30), and (9.2.11)(1), (10.3.10), (10.3.11), we get the following proposition.
Proposition (10.3.12) Xa nBA/Tk
wp(t°) E
a1,a2Ek -3Ekx
Note that
E f3,a(n(uo)t°)f2,-a(n(uo)t°,w) '1,°2Ek -3Ekx
is invariant under the action of Tk. We consider a such that a3 = 0.
f3,a(n(uo)t°)f2,-a(n(uo)t°,w)(t°)-2pµidXt°duo
0.
Part IV The quartic case
230
Proposition (10.3.13) X, nBk/Tk
Wp(9a)
E
f3,a(9a)f2,a(9a,w)(t°)-2P1A2ld"t°duo
0.
a1 EkX
a2Ek a3=0
Proof. It is easy to see that (t°)-2Pµ1 = A61/12 2, r'02 (to) = A l sµ2. Let c4 = -a2x2 i2, as = -a1x2 i2. Then f3,a(9a) is equal to
-6 3 1 µ'2
T3(9ay1,'Y2,13(t°)-la1 -'Y2,23(t°)-1a2u0, 72,23(t°)-la2) -1,33,-2,12 Ek X
We choose 0 < T8 E .'(A2 ), 0 < IF9 E .'(A3) so that 1W3(90Y1, x2,13, x2,23) <<
2µ2 1x2,12, A 1 2µ2 1x2,12u0)
9(A1 u2x1,33, x2,13, x2,23)
Let h3 (p, uo), h4(t°, u°) be the following functions
E T8 (µx2,12, AX2,12u0) x2,12Ekx
E q'9(A411!2x1,33, 72,13(t°)-1a1 - 72,23(t°)-la2uo, 72,23(t°)-la2) -1,33 Ek x ai Ek
a2Ek%
The condition al # 0 is equivalent to the condition a'2 54 0. So If3,a(9a)I << A1 6A2h3(.1 212 1, u0)h4(t°, 460) a1Ekx a2Ek
By (1.2.8), for any N1i N2 > 1, h4(t°,uo) «sup(1,A1/.i1µ21)(A4N2)-Ni(Al
lµ1112)_N2
By the consideration as in §2.1,
E If2,a(91,w)I < E 1FI,T,V(9a,'w), I'T'V
EkX a2Ek *1 a3=0
where I1 3) 2, vl = (1, 2). Therefore, by (3.4.30), there exists a slowly increasing function h5(t°) and a constant b > 0 such that if M > w0i
> If2,a(9a,w)I << h5(t°) a1Ekx a2Ek a3=0
10 The non-constant terms associated with unstable strata
231
forwo-b
1)h5(t°).
Then, for any N1, N2 > 1,
E If3,a(90)f2,-a(90,w)I '1 EkX a2Ek
3=0
<< h3(A1 2/.12 1, uo)h6(t°)A3N2-2N1(A1 2/12 1)N1+N2µ1 -N2
If Al > 1, we choose N1 >> N2 >> 0, and if al < 1, we choose N2 >> N1 >> 0. Then the right hand side is integrable on Xa f1 BAITk. This proves the proposition. Q.E.D.
We consider a such that al = a3 = 0. This implies that a2 = 0. Let h7(t°) _
19(A1/_12x1,33, 72,13(1°)-1a1A31,33,aiEkx
Then for any N1, N2 > 1, h7(t°) << (A
p2)-Nl (Al lµl lµ2)-N2,
and
If3,«(9a)I << A16µ2h3(A12µ21,uo)h7(t°). '1='3=0 a2Ekx
By the consideration as in §2.1,
E If2,a(91)I C £I,,,1(9a, w), °1=a3=0 .2Ekx
where Il D 2. Therefore, by (3.4.30), there exists a slowly increasing function h8 (t°),
and a constant b > 0 such that if M > wo and 1 >> 0, If2,a(9a)I << h8(t°)(Aiµ1)1 a1=°3v0 a2Ekx
for wo - 6 < Re(w) < M. Let h9(t°) = A1 -6µ4h8(t°). Then for any N1i N2 > 1, l >> 0, (10.3.14)
E If3,«(90) f2,-«(9a) I '1='3-0 .2Ekx « h9(t°)h3(A12µ21, uo)(Ai/12)-Nl (A1 lµ1 lµ2)-N2 (AlA1)i = h9(t°)h3(A12u21 u0)Ail-2N1+3N2 1 N2(\j2µ21)N1+N2.
Part IV The quartic case
232
Lemma (10.3.15) The integral
f
u'p(t°) A/TkxA
(uo)t°,w)(to)-2Ppidxtoduo
f3,a(n (uo)t°)f2,-,(n °1=-3=0 -2Ekx
is well defined if Re(w) >> 0.
Proof. By (3.4.30), there exists a slowly increasing function h10(t°) such that
An+r2 1
T1+r2
N1
Airl+r2µ2
it,-r,I(9a, w) << hlo(t°) { Il
r, +r2
Tl = (2, 3), T1 = (1, 2, 3),
r2
This is possible because v in (3.4.30) is 1 for our case. If Ti = (2, 3), we fix r2 and consider rI << 0. If Ti = (1, 2, 3), we fix r1 + r2 and consider r1 >> 0. If Ti = (1, 3), we fix ri and consider r2 >> 0. It is easy to see that the convex hull of the set
j(-2,0,-l),(4, 0, 1), (-1, -1, 1), (-1,1, 0)} contains a neighborhood of the origin.
Let to = (t1, t2) where tl E T°A, t2 E TiA. In all the cases, there are a finite number of points ql,
-
, qj E D11,1.1 such that
If3,«(n(uo)t°)I inf((tl)Tlq,)(t°)-2puihlo(t°) °1=°3=0 -2Ekx
is integrable on TA°lTk x A. Then our integral is well defined if Re(w) > L(qi) for
i=1...,j.
Q.E.D.
Lemma (10.3.16) -Pi
w)
L2IT.xA
-2Ekx
Proof. Consider the integral
TA/TkxA Cl1
fdxt°du0. f3,u(9a )f2,-a(90)(t°)-2Pµ2
Wp1 (t°)
( 0)
f3,«(9a)f2,-.(9a)(t°)-2Pµidxt°du0.
Wp1 (t°)
.1 =.3 =O
-2Ekx
We use the estimate (10.3.14). If Al > 1, we choose Ni >> N2, 1, and if Al < 1, we choose 1 >> Nl >> N2. Then the above integral converges absolutely. Therefore, f3'.(gD)f2'_.(gD)(t1)-2p[12
WP' (to) TA/TkxA µl< a(u0)
a1=°3 e0 -2Ekx
Q.E.D.
10 The non-constant terms associated with unstable strata
233
We define
f4,a2,a4(t°) = J f3,(°,a2,°)(t°n'(ul)) < a4u1 > du1, A
f5,
2,4(t°,w) =
fT2/TkxA
ITVTk
IA/k
f2,(°,2,°)(t°1Z (ul),W) < 4u1 >
f3,a(go)f2,-a(ga)(t°)-2pµ2ldxt°duo
Wa1 (t°)
'2Ekx
Wp1 (t°) T f4,a2,a4(t°)f5,a2,a4(t°, w)(t°)-2pµdxt° .2Ek x a4Ek
By (10.3.1)(4), for any N > 0,
If4,a2,a4(t°)I «rd3,N(A1,91,µ2)
T,
a2,a4Ekx
Since we have an estimate of Eat a4Ekx If5,1112,a4 (t°, w) I by a slowly increasing function as before, we get the following lemma.
Lemma (10.3.17) JTA/Tk
W pl
(t°)
f4,a2,a4(to)f5,a2,a4(t°,w)(t°)-2pdx to - 0.
2Ex '4Ekx
Let f6,a2 (t°) = f4,a2,0(t°),, f7,a2 (t°, w) = f5,a2,o(t°, w). Then
&-2 (to) =
JA/k
f7,a2 (t°, w) =
/k
OY02 (Ra2(Dp2,ton" (u3)) < a2u3 > du3,
J3
(t°n"(u3), w) < au3 > du3.
Consider P2 such that 52(2) = 0.
It is easy to see that ep21(t°) = (t°)"8 =
Al µ'1 µ2 ' . So Ap2°Ma2A = {t° E TXX I ,1µi lµ21 < 1}. Therefore, p2 (1 , W,
W) = f
Wp2 (t°) E fs,a2 (t°)f7,-a(t°,
al
TAO
/Tk
µll
w)(t°)-2pdx
to.
a2Ekx
µ2k1
Hence, (10.3.18)
Ep1 E 1("'' , W, w) + EpJ42 ((D, W, w) Ep1
J
Tk/Tk a1µ11µ21>1
f6,a2(t°)f7,-a2(tO,w)(t°)-2pdxt°.
Wp1(t°)
a2Ekx
Part IV The quartic case
234
Note that E2 = -flLemma (10.3.19) There exists a constant S > 0 such that for any N > 0, M > wo, I f6,a2(t°)f7,-a2(t°,w)I << (Alpl lµ21)-Nrd2,N(/-1l,P2) a2Ekx
forw°-6
>1, l>1,
E
Ifs,a2(t°)f7,-a2(t°,w)I
°1=°3=O a2Ekx «hll(t°)(A1/1'2)-N1(A11/111/12)
N2(A12µ21)-N3(A1/11)1
Let µ3 = \lpl lµ21 Then the right hand side of the above inequality is (10.3.20)
hll(t°)111
(4N1-2N2-2N3-41) µZ (5N1-3N3-31)µ3 (4N1-N2-2N3-31)
We choose N1, N2, N3,1 in the following manner:
N2i N3,1 fixed, and Nl >> 0 N2i l fixed, and N1 = 1 N3 >> 0
Al > 1, 92 > 1, Al > 1, 92 < 1,
N2iN3fixed, and l=4N1>>0
µk1,µ2>1,
N2 fixed, and N3 = 41, Nl = 3l >> 0 8
Al < 1, µ2 < 1-
This gives us the bound of the lemma. Q.E.D.
The following proposition follows from (10.3.18), (10.3.19).
Proposition (10.3.21) If 51(1) = 52(1) and 52(2) = 0, Ep1"p1(4),w,w)+EP2yP2(4)'w,w)
0.
Consider P2 such that 52 (2) = 1. Then -P2 (4), w, w) is equal to
f
TA Tk
a1M1 1µZ 1>1
Wp2(t°) T,
fs,a2(t°)f7,-a(t°,W)((t°)-2hD3ra31(t°))-1(t°)-2P°2d"t°.
a2Ekx
By (10.3.19), we get the following proposition.
Proposition (10.3.22) If 52 (2) = 1, `^'p2 (D, ca, W) - 0
§10.4 The case 0 = (/37) In this section, we consider a path p = (a, s) such that 0 = (07) An easy consideration shows the following lemma and we omit the proof.
10 The non-constant terms associated with unstable strata
235
Lemma (10.4.1) The convex hull of each of the set (1) 171,33, 72,22, 72,13, -72,23)1, (2) 171,33, 72,22,'Y2,13, -72,33} contains a neighborhood of the origin.
Let = Ra4p. In this case, Pa = B, so we write go = to _ (tl, t2) where t1 E T 10A, t2 E T 20A. We choose an isomorphism d : R+ -* T. Therefore, we can write to = d(A1, A2, A3)t °, where Al, A2, .\3 E R+. Let n(uo) = (nP, (u2, u3), nB2 (u4))
for uo = (U2, u3, u4) E A3. For fixed to and ul E A, Oy, (IF, t°n'(ul)n(uo )), e(t°n'(u1)n(uo),w) are functions on (A/k)3.
For a = (a1i a2, a3) E k3, let < a uo >=< alu2 + a2u3 + a3u4 >. By the Parseval formula,
f
OYY (%F, t°n (ul)n(uo))e (t°n'(ul)n(uo), w)duo
A/k)3
_
fl,a(tOn'(ul))f2,-a(t°n'(ul),w), aEk3
where
©y, (P, t°n (ul)n(uo)) < a uD > duo,
.f1,a(t°n'(ul )) = f A/k)3
f2,a(t°n'(ul), w) = J
e(t°n'(ul)n(uo), w) <
u> duo.
A/k)3
Proposition (10.4.2) There exists a constant 6 > 0 such that for any N > 0, M > wo,
E I fl,a(ton
(ul))f2,-a(ton (ul), w)I <
aEk3\(0)
for w0-6
By definition, n (ul)n(uo)y = (y1, x2,23 + x2,13U1 + 2x2,22u3, x2,33 + x2,22u3 + X2,13U2 + x1,33u4)
Let
2
u2 = x2,13u2 + x2,22u3 + x1,33u4, ui3 = x2,13u1 + 2x2,22u3,
ui¢ = u¢.
Then
u2 = x 2,132 u 1
u3 = 2 i
U4 = u¢.
4-1x-1 x-1 u' 2+ 2-1x-1 ul u 4-1x-1 x-1 xX 33u4, 2,22x2 13 u21 - X2,13 32,22 2,13X2,22 3
-1 / 2 22x2,101, x2,22u3 - 2-1X-1
Part IV The quartic case
236
Clearly, n'(ul)n(uao)y does not depend// on u/4. Therefore,
fl,a(t°n (ul)) _
J W(t°n'(ul)n(no)yl)du2du3. yiEZ'k
A2
Let V), (a, ul, u') = < aix2,13uI + 2-1a2x2,22u3 > x < ai(-4-1x2,13x2,22u32 + 2-1x2,22uiu'3) >
X < (a3 - alx 2,l3X1,33)u4 >, 02(a, u1) = < -4-1aix2,-122x2,13u1 - a2x2,22x2,13u1 -1 >, 1x-1 x-1 u' 2+ 2 2,22 3 u+ al( -4 03( a ,u 1, u'3) = < 2,13 2,22 3 1
2-lax-1
2-1x-1
2,22
u u3) 1
>
.
Then fi,a(t°n'(ui)) is equal to T(61, 72,23(t°)u3, 72,33(t°)u2) 1(a, u1, u)du2du' ,b2(a, ul) y1EZak
AZ
The above integral is zero unless a3 - a1x2,13x1,33 = 0, because n'(ui)n(u,)y does not depend on u4. Let W1 be the partial Fourier transform of T with respect to the coordinate x2,33
and < >. Then u, u)du' du3
fA2 W (t°y1,72,23(t°)u,
=
We choose 0 < T2 E .'(Z,A x A) so that IA Since I02(a,u1)I =
I03(a,u1,u3)I = 1,
Ifl,a(t°n (ui)) «i02(t°)
`W2(t°y1,0'1x2,1372,33(t°) Y1 EZOk
s2,13=-1 a3 T1,33
If the condition x2,13 = alai 1x1,33 is satisfied, ai $ 0 implies a3 # 0. So
1f1,a(t°)I <
Y'EZDk
o,Ekx
This bound does not depend on a2. By (10.4.1), for any N > 0,
E Q1 a3Ek'
Ifl,a(t°)I << rd3,N(Al, A2, A3)-
10 The non-constant terms associated with unstable strata
237
There exist a slowly increasing function h(t°) and a constant 6 > 0 such that if
M>w0,
I f2,a(t°, w)I << h(t°) Ekx .2,3Ek
for w0 - 6 < Re(w) < M. Therefore, for any N > 0,
E 1f1,a(t°)f2,-a(t°, w)I << rd3,N(Al,'2, A3)ulEkx
°2,3Ek
Next, we consider terms such that al = a3 = 0, a2 # 0. Let 'F3(yl, X2,23) _
T (yl, x223, x2,33) < x2,23x223 > dx2,23dx2,33
Then fl,a(t°n'(ul)) is equal to K02(0) E 'F3(t°y1, 2-1a2x2,2272,23(t°)-1) < -2-1a2x2,22xz,13ul > yiEZnk
By (10.4.1), for any N > 0,
E I fl,a(t°n (ul))I <
2Ekx
al=°3=O
The rest of the argument is similar. This proves (10.4.2). Q.E.D.
The following proposition follows from (10.4.2).
Proposition (10.4.3) Ep (1), w, w) - 0. §10.5 The case 0 = (/3s)
In this section, we consider a path p = (D, s) such that a = (/38).
Proposition (10.5.1) cp(4P,w,w) - 0. We devote the rest of this section to the proof of (10.5.1). For A1, 91, µ2 E Rte, we define (10.5.2)
d(A1, µl, µ2) = a(A1 2, 1u1, lµl
µ2, µ2
We write ga = kd(Ai, µi, µ2)t °n"(u3) where k E K fl MOA. As in §3.4, when we estimate various functions, we assume that k = 1. Let to = d(A1, µl, µ2)t °. Let n(ua) = (np2 (u1, u2), nB2 (u4)) for ua = (ul, u2, u4) E A3. For N = (Nl, N2i N3), we define (10.5.3)
h1,N(9a) =
71,22(t°)I-N"I'i2,13(t°)I-N2I72,23(t°)IN3,
h2,N(9a) = 71,22(t°)I-Nl I72,13(t°)I-N2I72,33(to)IN3,
h3,N(9a) = h4,N(9a) =
71,23(t°)I-Nl 172,12(t°)I-N2I72,23(to)IN3,
I71,23(t°)I-"I72,12(t°)I-N2I72,33(t°)IN3
Part IV The quartic case
238
Then
h1,N(ga) = A (2N,-N2-2N3)A1 -(2N1-N2)P2-(Ni-N2+N3) 1
h2,N(ga) = Al (2Ni-N2-2N3)µ1 -(2N1-N2+2N3)µ -(Nl-N2+N3) 2 (2N1-N2-2N3)µl-N2 -(Nl-N2+N3) h3,N(9a) _ Al N2 (2N1-N2-2N3)µ= (N2+2N3)µ2 (Nl-N2+N3)
h4,N(9a) = Al
Consider the condition (10.5.4)= For any N = (Nl, N27 N3) satisfying N1i N2, N3 > 1, If (9a)I << hi,N(ga). Lemma (10.5.5) Suppose that f (go) is a junction on G°Af1MMA/Mek which satisfies
Condition (10.5.4)= for some 1 < i < 4. Then for any N > 0, If (9a)I << pi Nrd62, N(Al, p2)
for go in the Siegel domain. Proof. Since Iy2,23(t°)I = P I72,33(t°)I >> 1%,33(0°)I, we only have to consider i = 1, 3.
Suppose that (10.5.4)1 is satisfied. Then we choose N1, N2, N3 in the following manner:
N2,N3fixedand Nl>>0,>>0
ifal>1,µ2>1,
N3 fixed and N1 = 2N2 >> 0 N3 >> Nl » N2
if Al > 1, µ2 < 1,
N1=N2,N3=4N2>>0 3
ifAl<1,µ2 _ 1-
if Al < 1, µ2 1 1,
Suppose that (10.5.4)3 is satisfied. Then we choose N1, N2, N3 in the following manner:
N3 fixed and Nl »N2>>0 N3 fixed and Nl=2N2>>0 Nl fixed and N3 » N2 Nl fixed and N3 = 33N2
ifal> 1,µ2> 1, if Al>1,µ2<1, if Al < 1, µ2 > 1, if )1 < 1, µ2 < 1-
Then we have the estimate of the lemma. Q.E.D.
Let T = Reap. As usual, we can write any element y E Yak' in the form y = (y1, y2) where yl = (x1,22, x1,23) x1,331 X2,12, x2,13) E Zak and Y2 = (x2,22, x2,23, X2,33) E Wok
For a = (al, a2i a3) E k3, we define < a ua >=< alul + a2u2 + a3u4 >. Let (10.5.6)
fl,.(ga) = f
DY, (W,9an(ua)) < a ua > due,
A/k)3
f2,«(9a, w) = fA/k)3
(gt n(ue ), w) < a u-0 > dua.
(
The proof of the following lemma is easy, and is left to the reader.
10 The non-constant terms associated with unstable strata
239
Lemma (10.5.7) Let P(y1) = x2,12x1,33 - x2,13x2,12x1,23 +x2,13x1,22 Then P(y1) is a relative invariant polynomial under the action of M.D. Let
A(y1) =
2x2,12
0
x2,22
x2,13 0
x2,12 2x2,13
x1,23
u'1 ,
x1,33
u2
u2
= A(y1)
( u2 )
.
u4
u4
Then n(ua)y = (yl, x2,22 + ui, x2,23 + u2, x2,33 + ui)
Therefore,
fl,.(9a) = E f yiEZak
iP(9an(ua)y1) < a ue > due.
3
A
Let (ai, az, as) = (a1, a2, a3)A(y1)-1 (det A(y1) = P(y1)
0). Then
< a1'u'1 + a2 u2 + a3u4 >=< alu1 + a2u2 + a3u4 > .
It is easy to see that 9an(ua )y1 = (9ay1, ui, u2 + 2u1u1', u4 + uiui + u1u2).
For u' = (u'1, u2, u4), we define
'I"(9a, y1, u) _ IF(9ay1, ui, u2 + 2uiu'1, u4 + uiui + u1u2). Also we define 11 U3 = u2 + 2u1u1)
l
2
'u4 = u4 + u1u1 + u1u2.
Then
u2 = u3 - 2u3ui, ) = u"4+ u32u'1 u 4= u"4 - u2u 32u3u13 1 - u3(u"
-
- u3u'' 3
Therefore,
< a' u >=< (a' - 2a2u3 + au3)ui + (a2 - a3u3)u2 + a3u'4 > . Let
V)a' (ul, u2) =< (a' - 2a2u3 +a3, u3)ui +(a2 - a3u3)u2 > .
Let If1 be the partial Fourier transform of W with respect to the coordinate x2,33
and < >. Then
J(9n(u)y1) < a ue > due yiEZ3
= 112,33(t°)1'
yiEZa k JA2
1(9ay1, u1u3, a3-Y2,33(t0)(ui, u2)duidu2.
Part IV The quartic case
240
We choose 0 < 4f2 E 9(ZZA x A) so that
f
I'F1(Y1, ul, u3, x2,23)Idu' du" < qf2(y1, x2,23)-
2
A
T hen If1,o (9a)I << K
(go) E 'P2(9ay1, a3
Y2,33(t0)-1),
yi EZak
By (1.2.8), (10.5.5), for any N > 0, Ifi,a(9a)I << Pi Nrd2,N(Al, µ2) a3Ekx
for g, in the Siegel domain. This bound does not depend on a', a2. Therefore, by a similar argument as before, there exists a constant 6 > 0 such that if M > wo, (10.5.8)
If1,a(9a)f2,-a(9a,w)I << P1 Nrd2,N(Al, P2) «i.n2 Ek a3 EkX
for go in the Siegel domain and wo - 6 < Re(w) < M. Next, we consider a such that a3 = 0. Let P3(y1) x2,22, x2,23) _ JA2 'Y(yl, x2,22, x2,23, x2,33) < x2,23x2,23 > dx2,23dx2 33
We choose 0 < q14 E 9'(ZOA x A) so that 1.
IW3(y1,x2,22,x2,23)Idx2,22 < T4(y1,X2,23)
Then If1,a(ge)I «!a2(9a2) -o °3n3EkX
W4(9ay1,a272,23(t0)-1),
-3=o y1EZok n2 EkX
By (1.2.8), (10.5.5), for any N > 0, Ifi,a(go)I < P1 Nrd2,N(Al, P2)
3- 0 o3EkX
for go in the Siegel domain.
Therefore, there exists a constant 6 > 0 such that if
M>w0, (10.5.9)
E
a3=0 a2Ekx ai Ek
I fl,a(9a)f2,-a(ga, w)I «N'1 Nrd2,N(A,, /22)
10 The non-constant terms associated with unstable strata
241
for go in the Siegel domain and w° - 6 < Re(w) < M. Finally, we consider a' such that a'2 = a'3 = 0. Let 'y5(y1, x2,22) =
Jf3
"' (yi, x2,22, x2,23x2,33) < x2,22x2,22 > dx2 22dx2 23dx2 33
We choose 0 < is E Y(ZZA x A) so that I'y5(yi,x2,22)I <_ 'P6(y1,x2,22) Then If1,0(9a)I <<'
'I'6(g0y1,'Y2,22(t°)-1a11)
(90) yiEZn%
ai Ekx
Clearly, a = (c4, 0, 0)A(yl) = (2x2,12ai, 0, x1,22ai). If Y1 E Za8k, x2,12 # 0 or x1,22 0 0.
Suppose a1 i4 0. Then If2,0(90, w)I
Z3
1(9,,W) I,T
.i Ekx
°1#0
where I = (11i 12) and I1 3 2. So by (3.4.30), there exists a constant 6 > 0 and a slowly increasing function h(t°) such that if 1 >> 0, M > wo, If2,n(9a,w)I << h(t°)(Xiµl)l.
(10.5.10) 2alEkx °1*0
Consider the following conditions (10.5.11)1 For any N1, N2 > 1,1 >> 0,
f(9a)
«I71,22(t°)I
N1I72,13(t°)I-V2(A13ul)`;
(10.5.11)2 For any N1, N2 > 1,1 >> 0, f(90) <<
Iy1,23(t°)I-N1I72,12(t°)I-N2('A1p1)t.
Lemma (10.5.12) Suppose that f (go) is a function on G°, fl MJA/Mak which satisfies either (10.5.11)1 or (10.5.11)2. Then for any N > 0, If(9d)I<
for g, in the Siegel domain. Proof. If f (go) satisfies (10.5.11)1i we choose
N1>> N2>>l>>0
Al>1,1L2?1,
N1 = 2N2 »0, lfixed
Al
1,µ2
1,
N1 = 2 N2, 1 = N2, N2 >> 0
A,
1, µ2
1,
N1=3N2, 1= 1N2, N2>>0 Al<1,µ2<_1.
Part IV The quartic case
242
If f (go) satisfie s (10.5.11)2, we choose
>1,µ2?1,
Nl»N2>>l>>0
Al
Nl=4N2>>0, (fixed
Al < 1, µ2 > 1,
Z N1=N2,l=2N2,N2»0 Al >1,µ2<1,
N1=3N2,l=4N2,N2>>0 Al <1,µ2<1. Then we have the estimate of the lemma. Q.E.D.
By (10.5.10), (10.5.12), for any N > 0, (10.5.13)
If1,.(ga)f2,-.(ga, w)I << µ1 Nrd2,N(A,, µ2) #2 - 3 ai Ekx
n1#0
for go in the Siegel domain.
Suppose that al = 0, a3 0 0. If a2 = a3 = 0, a2 = 0. So If2,.(go,w)I S 1: I°I,,,1(go,W) .1_°2_°3=0
I,T
a1 Ekx '300
where I = (11,12) and 12 = {1}, I1
1. Consider 8j,1.,1(go, w) for such I. Let 113 = I72,13(t°)I,µ4 = 172,22(0°)I. Then 03 = Al lµ1 1µ2 1, µ4 = A2 A2'U 1 and
3 3 A1 = µl lµ3 3 µ4 , A2 = µ3 µ4 . Let
(10.5.14)
d(µ1,µ3,µ4)=(µ13µ43,p3A3,p 2A33A43;
aA3gµ3
Then to = d(µ1,µ3, µ4)t °. By (1.2.6), for any Ni, N2, N3 > 1, (10.5.15)
I'W6(gay1,72,22(t°)-lai)I «µ3N1-N2µ4N` v1 E ZDk
-1,22.x2 13Ek%
Ekx
Lemma (10.5.16) Suppose that I1 ; 1, 12 = {1}. Then there exist a constant b > 0 and a slowly increasing function h(µ3, µ4) such that if M > wo, l > 1, ,fi,r,l(ga, w) « µl 5h(µ3, µ4)(µ33 µ43 )'
for go in the Siegel domain and w° - 6 < Re(w) < M. Proof. Let q E DI,,, rl, r2i r3 be as in (9.2.5). There exists a slowly increasing function hr,r(P3,µ4) for each T such that Al(rl+r2)hr,, (µ3)
4
T , 1(ga , w)
P4)
µ1r2hT,r(µ3, µ4) 2(rl+r2)hr,r(µ3,
2
<< µ i (µ3 µ4 ) x
µl
1
µlr1 hr,r(µ3, µ4)
µl
I µu
2,1
h,.,, (p3, A4) 1r
r (µ3, µ4)
T1 = 1,
rl = (1, 2), µ4)
Ti = (1, 3), Ti = (2, 3), Ti = (1, 2, Ti = (1, 3, 2).
10 The non-constant terms associated with unstable strata
243
Note that if II = {2}, we have an extra 11 21' factor for some 1' << 0, but since µl is bounded below, for ga in the Siegel domain, we can ignore this factor. We choose r3 = If Ti = 1, we choose rl, r2 << 0. If -r1 = (1, 2), we fix r1 and choose r2 << 0. If Ti = (2, 3), we fix r2 and choose r1 << 0. If T1 = (1, 2, 3), 1.
we choose r1 = 1 - r2 >> 0. If TI = (1, 3, 2), we choose r2 = 1 - r1 >> 0. Then L(q) < w°i and the estimate of the lemma is satisfied. If Tl = (1, 3), we can choose
r1=r2=2. Then L(q)=8+2C<4+C since C> 100. Q.E.D.
It is easy to see that (t°)-2P = µi sµ3 31.t4 3 and K02(90) = µ6µ43. Therefore, by (10.5.15), (10.5.16), for any N > 0,
E
(10.5.17) (t°)-2'lc02(90)
W)
v1 EZab
s1,22,x2,13Ek5
-i Ekx
<< µl 8rd2,N(µ3, µ4)
Proposition (10.5.1) now follows from (10.5.8), (10.5.9), (10.5.13), (10.5.17).
§10.6 The cases Z = (38, 08,2), (09) Let 01 = (08,88,2), a2 = (09), and 03 = (08). Consider paths pi = (0i,5i)42 = (Z)2, Z2), and p3 = (03,63) such that sl(1) = 52(1) = 53(1). This means that p3 -< pl. Let (10.6.1)
d(.\i, Al, µ2) = a(A14p1, iµ2 Aiµl 1 µ2
3µl 1 ,
A3
and to = d(A1, µl, µ2)t °. For u2, u3, u4 E A, a = (al, a2, a3) E k3, Let n(u , ), <
a- u8>beasin§10.4.
We first consider pl such that sl(2) = 1. It is easy to see that ep11(t°) _ 2 (IC132(t°)(t°)-2Pa3)-1(to)-2p°1 (to)-Rs = A-1 and Clearly, 43 (t° w) = As µ12µ2 is the constant term with respect to np2(u2i u2).
Let %F = R,1 Pp1. We consider
IA/k
Oy,1 (W, ton "(u3))3 (t°n'(u3), w)du3.
For a E k, we define (10.6.2)
f1,a(to) =
f
Oya1 (W,t°n'(u3)) < au3 > du3, /kf
f2,a(t°, w) = J /k
-ep3
(toni(u3), w)
< au3 > du3.
Lemma (10.6.3) Let f (t°) be a function on TA/Tk. Suppose that for any N1, N2, N3 > 1, f(to) << Iy1,23(to)I-N1I72,13(t°)I-N2I-Y1,33(to)IN3
Part IV The quartic case
244
Then for any N > 0, f (to) «.Al Nrd2,N(µ1, p2). Proof. Easy computations show that I71,23(to) I-N, I72,13(to) I-N2 I71,33(to) I Na (N1+N2-N3)ll-l1 (-2N1+N2+3N3) lt2-(-N2+2N3)
- Al
We just have to choose N1, N2, N3 in the following manner:
N1=N2=N3»0
Al _ 1,/-t2
N1iN3fixed and N2>>0
hl>1,tt2<1,
1,
N2 fixed and Nl>>N3>>0 /L1<1,lt2>1,
N3 fixed and Nl»N2>>0 pl<1,µ2<1. Q.E.D.
The following proposition is an easy consequence of the above lemma.
Proposition (10.6.4) Suppose sl(2) = 1. Then pl (-D, W, w) =
fa
2/Tk
wp1(to) E A8111 2112 2f1,a(to)f2,-a(t°, w)dxto - 0. aEk x
,>1
Next, we consider P2 and pl such that sl(2) = 0. The proof of the following lemma is easy, and is left to the reader.
Lemma (10.6.5) The convex hull of each of the sets (1) (111,23, 72,22, 72,13, -72,33}l, (2) 171,23, 12,22, 72,13, -72,23}
contains a neighborhood of the origin.
Proposition (10.6.6) Epl
pl("),W,w)+Ep2p2((1),W,w) - 0.
We devote the rest of this section to the proof of the above proposition. Let W = Ra2Dp2. Any element y E Ya.k can be written in the form y = (yi, y2) where yl = (x1,23, x2,13, x2,22) E Za2k and y2 = (x1,33, x2,23) x2,33) E Walk Then n (ul)n(ua)y = (y1) x1,33 + u3, x2,13 + u4 + x2,13u1, x2,33 + u2),
where 7lr3 = x1,23u3,
u4 = 2x2,22u3 + x1,23u4, /
2
/
u2 = x2,13u2 +x2,22 U3 + u3u4.
Let
(10.6.7) f3,a(ton (u1)) =
f IF (ton'(ul)n(ua)(yl, 0)) < a ua > duo; 3
Y1EZa2k
A
I(ton (ul)n(u), w) <
f4,a(t°n (ul), w) = A/k)3
u> due.
10 The non-constant terms associated with unstable strata
245
An easy computation shows that
< a ua >= < alx1 u 2,13 - a l(x2,13x1,23x2,22u32 - x2,13u3(u4 - 2x1 23x2,22u3)) > 2 X < a2x1,23u3 + a3(x1,23u4 - 2x-1,23x2,22u3) >
Let Y 01(yl, u3, u4) = < -a, (x2,13x1,23x2,222132 - x2,13u3(1b4 - 2x1 23x2,22u3)) > /1
X < a2x1,23u3 + a3(x1,23u4 - 2x1 23x2,22"3) > .
We also define (to) yl, ul, u2, u3, u4)
= W (t°n'(ul )yl, -Y1,33(t°)u3, -Y2,23(t°)(u4 + x2,13"1),'Y2,33(t°)u2)
Then f3,a(t°n'(ul)) is equal to I WY'(t°, yl, u1, u'2) u'3, u4) < a1x2-,1 13u2 > W ,,/'a (y1l u'3+ u4 )du'. 3
yi E Za2 k
Let WW1 be the partial Fourier transform of IF with respect to the coordinate X2,33 and < >. We choose 0 < W2 E 9(Z-02A x A) so that
fA2
I ''1(yl, x1,33, x2,23, x2,33)Idxl,33dx2,23 < 'P2(y1, x2,33)
Then
If1,a(t°n (ul))I <<' 22(t°)
'1'2(t°y1,72,33(t°)-161x2,13) Y iEZazk
This bound does not depend on a2i a3. By (10.6.6), for any N > 0, E I f3,a(t°n'(u1))I <
Therefore, there exists a constant 8 > 0 such that if M > wo, (10.6.8)
I f3,a(t°1t (ul))f4,-«(t°n'(ul), w)I <
-2,n3Ek
for wo - 6 < Re(w) < M. Next, we consider a such that a1 = 0, (a2i a3) # 0. Let W3(yl, x1,33, x2,23) = 1A2
(yl, x1,33, x,23, x,33) < x2,23x,23 > dx23dX,33.
We choose 0 < qt4 E So(ZOA x A) so that IA I W3(yl,x1,33,x2,23)Idx1,33
'4(Y1,x2,23)
Part IV The quartic case
246
Then If3,a(t°n (ul))I << 6;022(t°) E 'W4(t°y1,"Y2,23(t°)-1a3x1,23). YlEZa2k
By (10.6.5), for any N > 0, I I f3,a(t°n (ul))I <
Therefore, there exists a constant 6 > 0 such that if M > wo, (10.6.9)
E I f3,a(t°n (ul))f4,-a(t°n'(ul), w)I << rd3,N(.\1, µl, /22) a3Ekx 1=0 2Ek
for wo - 6 < Re(w) < M. Finally, we consider a such that al = a3 = 0, a2 E k". Let `W5 be the partial Fourier transform of T3 with respect to the coordinate x1,33 and < >. Then W5 (t°y1,y1,33(t°)-1a2x1,23).
f3,a(t°n (u1)) = r-022(t°) y 1 EZ12 k
Thus f3,a(t°n'(ul)) does not depend on ul any more. Therefore,
E(t°n'(ul))f4,-(t°n (u1),J
/k '1=°3=0 a2EkX
f3,a(t°) LIk f4,-(t°n'(ul), w)dul, 2Ekx
and
(10.6.10)
J /k f4,-a(t°n (ul ), w)du1 = f2,-a2 (t°, w).
Let W6 be the partial Fourier transform of q/5 with respect to the coordinate x2,22 and < >. Let W7i `W8 be restrictions of %F5, `1'6 to the first three coordinates. Let f1(t0, x1,23, x2,13, x2,22) a2)
('Y1,23(t°)x1,23, Y2,13(t°)x2,13,-Y2,22(t°)-1x2,22, Y1,33(t°)-1a2X1,23)
We define 'f'5(A(t°,x1,23, x2,13, X2,22, a2)),
f5,-2(to) = lc,22(t°)I-Y2,22(t°)I-1 Y1EZ12k
f6,a2(t°) =
r-122(t°)I72,22(t°)I-l
'F8(A(t°, x1,23, x2,13, 0, a2)), X1 ,23,52,13 Ek x
f7,a2 (t°) = ic,22 (t°)
WY7(A(t°, x1,23) X2,13, 0, a2)) 51,23,x2,13Ekx
10 The non-constant terms associated with unstable strata
247
Then f3,(o,a2,o)(t°) = f5,a2(t°) + f6,a2(t°) - f7,a2(t°). We choose Schwartz-Bruhat functions 0 < W9i Wio on A4, A3 so that IT5I <-'I9, IT6I, I''7I < W10 Then
E If5,a2(t°)I, T, If6,a2(t°)k E If7,a2(t°)I a2Ekx
012Ekx
a2Ekx
are bounded by /ca22(t°)Iy2,22(t°)I-1'J4('y9, -Y1,23(t°),'Y2,13(t°), Y2,22(t°)-', Y1,33(t°)-1), '522 (t°) I-Y2,22(t°) I-1''3('y10,'Y1,23(t°),'Y2,13(t°), 'Y1,33(t°)-1)' ICD22
(t°)' 3('y10, 'Y1,23(t°),'Y2,13(t°),
-Y1,33(t°)-1)
respectively.
Proposition (10.6.11) (1) Suppose that f (t°) is a function such that for any N1, N2, N3, N4 > 1, f(t°) «I"Y1,23(t°)I-N1I72,22(t°)I-N2I-Y2,13(t°)I-N3I-Y1,33(t°)IN4
Then for any N > 0, If (t°)I << Al Nrd2,N(µl, µ2) for A1 > 1. (2) Suppose that f (t°) is a function such that for any N1, N2, N3, N4 > 1,
f(t°) <<
I"Y1,23(t°)I-'
I'Y2,22(to)IN2I-Y2,13(t°)I-N3I.Y1,33(t°)IN4
Then for any N > 0, If (t°)I << ANrd2,N(µl, µ2) for Al < 1.
Proof. Easy computations show that I"Y1,23(t°) I -N, I-Y2,22 (t°) I-N2 I'Y2,13(t°) I-N3 I-Y1,33(t°) IN4 (N1+7N2+N3-N4) N'-1 (-2N1+N2+N3+3N4) -(2N1 -N3+2N4)
Al
N'2 I'Y1,23(t°)I-N,
1-Y2,22(t°)IN2
(-Y2,13(t°)I-N3I`Y1,33(t0)IN4
Al (N1-7N2+N3-N4) N-1 (-2N1-N2+N3+3N4) /22-(2N1-N3+2N4)
We choose N1i
, N4 in the following manner.
Nl, N3 fixed, and N2>>N4>>0 µ1>1,µ2>1, N1, N2, N4 fixed, and N3 >> 0
Al > 1, p2 < 1,
N3iN4fixed, and N1>>N2>>0 µk1,/2>1, N2,N4fixed,and N1>>N3>>0 µ1<1,µ2<1. This proves (1).
We choose N1,
, N4 in the following manner.
N1iN4fixed, and N3=2N2»0 µb1,µ2>1, Nl, N2i N4 fixed, and N3 >> 0
Al > 1, µ2 < 1,
N3,N4fixed,and N2>>N1>>0 µk1,µ2>1, N1iN4fixed, and N2>>N3>>0
µ1<1,/2<1.
Part IV The quartic case
248
This proves (2). Q.E.D.
By (10.6.11)(1), (10.6.12)
Wp2 (t°) E f3,(O,a2,0) (t°)f2,-a2 (t°,
w)(t°)-2Pdxto
a2Ekx
- 0,
and by (10.6.11)(2), f5,a2(t°)f2,-a2(t°,
wp2(t°)
(10.6.13) JT'O /
Tk
A1<1
w)(t°)-2pdxt°
- 0.
a2Ekx
The following lemma is easy to prove, and we omit the proof.
Lemma (10.6.14) The convex hull of each of the sets (1) {(1, -2, 0), (1, 1, -1), (-1, 3, 2), (-4,1, 0)}, (2) {(1, -2, 0), (1, 1, -1), (-1, 3, 2), (-2,1,1)} contains a neighborhood of the origin.
Now we want to cancel out °p1 (4i, w, w) with "p2 (-D, w, w). However, we are still obliged to prove that 8pl (,D, w, w) is well defined for Re(w) >> 0. Consider el,,,l(t°,w) such that I = (1,,12) and 12 = 0, 1, = {2}.
Lemma (10.6.15) (1) If Tl = (2,3) or (1,2,3), there exists a constant 6 > 0 and a slowly increasing function h(t°) such that for any N >> 0, M > wo, ,rj,r,l(t°, w) << h(t°)(Ai/-L1 1)N
for w° - 6 < Re(w) < M. (2) If r1 = (1, 3), there exists a slowly increasing function h(t°) and a constant c > 0 such that if M2 > Ml > wo,
ij,r,l (t°, w) << h(t°)
(A1,,1 l µ2 1)cRe(w)
for Ml < Re(w) < M2. Proof. Let q E DI,,. Let r = (r1, r2, r3) be as in (9.2.5). By (3.4.30), there exists a constant 6 > 0 and a slowly increasing function h(t°) such that if M > w°i
1 4r1 -2r2 Alrlµ'2-r2 u I,r,l(t°, w) << 1,h( t°) x
-(r1+r2) Al2r1-2r2-rl µ1 µ2
12rl+r2 Al-(rl+r2)µ2-r1 for L(q) + 6 < Re(w) < M where 6 > 0, M > w° are constants. If Ti = (2, 3), we fix r2 and choose rl << 0. If Tl = (1, 2, 3), we choose rl = 1 - r2 >> 0. Then we get an estimate of the form (1). If T1 = (2,3), we fix r2 and choose rl >> 0. Then L(q) depends linearly on rl. Therefore, we get an estimate of the form (2). Q.E.D.
10 The non-constant terms associated with unstable strata
249
Since ep11(t°) = (t°)a8 = al, by the Parseval formula,
jpl (4), w, w) _
wp1 (t°) E f6,a2 (t°)f2,-a2 (t°, w)(t°)-2Pdx t°.
fA/Tk
a2EkX
a1 <1
By the consideration as in §2.1,
E
(t°, w),
If2,a2(t°,w)I << T
a2EkX
where I1 = {2},12 = 0. Therefore, the following proposition follows from (10.6.14), (10.6.15).
Lemma (10.6.16) The distribution 22p, (,D, w, w) is well defined if Re(w) >> 0. Therefore, by (10.6.10), (10.6.16), Ep1
w, w)
Pi (ID, w, w) + 1P2P2
Wp1 (t°) E f7,a2(t°)f2,-a2(t°,w)(t°)-2PdxtO.
N EP,
1T0ITk A1<1
It is easy to see that
a2EkX (t°)-2p
= A18Al 21'22 and rD22(t°) = A1 15µiµ2. Therefore,
for any N1, N2, N3 > 1, 2)-N1
If7,a2 (to) l << A3
X131'1 2A2 2
011'21'2 1)-N2
a2EkX
We have already estimated E I f7,a2 (t°) I (°I,T,I (t°, w) a2Ekx
for T1 # (1, 3), so we assume that T1 = (1, 3). By (9.2.10), for any E > 0, there exists a constant 6 > 0 such that if M > wo, 4I,T,1(t°, w) << ai+011'121'23
for w° - 6 < Re(w) < M where IC11, Ic21, IC31 < E. Therefore, (t°)-2P
(10.6.17)
«
1f7,112(t°)1'k1,1,1(t°,w) a2EkX A5+c1tt2+C2A2+C3
(A,t-tl 2)-N1(A1/-tll-t2 1)-N2 (A1 1tt3tt2)-N3.
We fix N1, N2 and choose N3 >> 0. Then there exists a slowly increasing function
h2(t°) such that for any N > 0, (10.6.17) is bounded by a constant multiple of h2(t°)(A11'1 31'22)"'. (10.6.17) is also bounded by a constant multiple of 2+C1 1'1
$+C3
1'2
3+C1
,1
1+02 -1+C3
1'1
1'2
, 12 +C1 1'12 +C3 1'22 +C3 )
by choosing (Ni, N2, N3) = (1, 3,1), (3,1, 2), (3, 2,1). Therefore,
wp2(t°) E f7,a2(t°)f2,-a2(t°)(t°)-2Pdxt° " 0. f""-ITk X1<1
a2EkX
This completes the proof of (10.6.6). Hence, we finished the proof of Theorem (10.0.1), (10.0.2). This implies that we can ignore the non-constant terms associated
with unstable strata.
Chapter 11 Unstable distributions In the next three chapters, we consider distributions ^ p (4i, w, w) and compute the principal part of the zeta function for the quartic case.
§11.1 Unstable distributions In this section, we define certain distributions associated with unstable strata. We remind the reader that we use the notation in previous chapters like n'(u), n"(u) in (9.2.3) or F' in (9.2.4). Definition (11.1.1) For a character w = (wl, w2) of (Ax /kx )2, we define (1) S#(w) = 6(wl)6(w2), (2) 6-0 (W) = s(w2), wa = (1,w1) fort = (i1), (3) sa(w) = s(wl), wa = (1,w2) ford = (04), (4) Sa (w) = S/(wi), wa = (wl, w2, w2) for (l = (or,), (5) Sa(w) = Slwi)S(w2), wa = (W2,w1,w1 lW2)fort = (37 ), (6) Sa(w) = 6(w1), wa = (W1,W2,W2) ford = (08), (7) 58(w) = S(w1')6(w2), wa = (1,w1 1, w1) for cl = (Qs),
(8) Sa(w) = S(wl),S8,st(w) = b(wl)S(w2), wa = (1,w2),wa,.t = (W2,W2) for Z _ (N1o)
Definition (11.1.2) We define subsets 01i 02 of permutations of {1, 2, 3} as follows. (1) 111 = {1, (2, 3), (1, 2, 3)}.
(2) 02 = {1, (1, 2), (1, 3, 2)}.
Let 0 be a (3-sequence for which Pa# = Pl x B2 or P2 x B2. The double coset
Pa#k \ Gk/Bk is represented by 01 x {1, (1, 2)} if Pa# = Pl x B2, and by 02 X {1, (1, 2)} if Po# = P2 x B2. The reductive part Ma of Pa# is of the form Ma = Ma1 X Mat and Mal = Mall X M812 where Mall = GL(2), Ma12 = GL(1) if
Pa#= P1xB2andMoll 'GL(1), Ma12=GL(2)ifPa#= P2xB2. By a standard argument, PlkrlBlk/Blk, P2kTlBlk/Blk are represented by the sets {7172 I yl E Ma11k/Mallk n B1k, Y2 E N 8}, {''1''2 -'1 E Ma12k/Ma12k n Blk, 72 E N,k}, I
respectively. Let dp (al) be as in (9.2.4), and di(\1i A2) = (dpi (.Al), a2(.\2, A21)) for i = 1, 2.
Let Aa = dl(A1, A2) if Pa# = Pl x B2, and as = d2(.\1, X2) if Pa# = P2 x B2. Any element ga can be written in the form ga = gaga where go E MDA. The element go is of the form (11.1.3)
t13) (t21
t22)) or
((tii
91,23) (t21
t22))
I
where 81,12, 81,23 E GL(2)00 and 1431 = It21I = It22I = Itiil = 1. The following lemma can be obtained by a standard argument and the proof is left to the reader.
11 Unstable distributions
251
Lemma (11.1.4) In the following two statements, Tr = (r,, r2).
(1) IfPo#= P1xB2, EB(gsn(us),z)dus fUs#A/UD#k MT(z)A/DZ+//EMa11nB1(91,12)z1T1(1),z1T1(2)). 'i E$11
,2=1,(1,2)
(2) If Po# = P2 x B2,
IUD#A/U,#k
=
E
EB(9an(u,),z)dua TZ+P
MT(z)A/D
EMc12nB1 (91,23, z1T1(2)) Z1T1(3))
'1E202
,2=1,(1,2)
The following lemma is a consequence of the Mellin inversion formula.
Lemma (11.1.5) Suppose that f (s) is a holomorphic function which is rapidly decreasing on any vertical strip contained in the domain Is Re(s) > 0}. Let I
E(g°, s) be the Eisenstein series on GL(2)00, and w a character of Ax /kx. Then 1
w(det g°) f (s)E(g°, s)dg°ds = 6(w) f (1). 1
GL(2)A/ GL(2)k
Refs)=r>0
Let W = Sym2k2, i.e. the space of binary quadratic forms. For IF E so(WA), w = (w1i W2) a character of (lax /k>1)2, and s E C, let Ew(W, w, s), Ew+('1', w, S), E W,ad(T ,W, s)
be the functions which correspond to ZV2 (',W, S), ZV2 (T, W, S), ZV2,ad (9', W, S) in
C W be the subspaces which correspond to Zo, ZO, in §4.2 respectively. We define 4,0k etc. in the same manner as in §4.2. Let Rw,O, R/w,0 be the operators which correspond to RO, Ro in §4.2. Let Hyl. be the group which corresponds to H in §4.2. Let XW C Al x GL(2)00 be the subset which corresponds to Xv in §4.2. Let TW(Rw,OiW, w, s, sl) etc. be the adjusting term etc. We use the same function a(u) as in §2.2. Let g E A x GL(2)11 = Rh x Al x GL(2)0A. For IF E .Y(WA), we define §4.2. Let Zw,o,
(11.1.6)
xEZ'IW,Ok
Gw,st('I', 9) _ 5 '1'(9x) xEZW,stk
Then (11.1.7)
Ow,st('1', 9) =
GZw,o (`F, N. 7EGklHw,k
Part IV The quartic case
252
Next, we consider distributions associated with unstable strata. (a) 'a = (,31). We can identify ZD with Sym2k3, i.e. the space of ternary quadratic forms. For T E 5°(ZDA), w = (W1i W2) a character of (Ax /kx )2, and s E C, let ED (IF, w, s), E'+ (q" w, s)
be the functions which correspond to Z,,3 (W, w, s), Zv,+(WY, w, s) in Chapter 4.
(b)D=(,34)
We can identify ZD with M(2, 2), i.e. the space of 2x2 matrices. For IF E 9(ZDA), w = (wl, w2) a character of (Ax /k x )2, and s E C, let ED (W, w, s) be the integral (3.1.8) for L = ZaZ. We discussed this case in §3.8. (c) D = 06), (D8)-
We can identify ZD with Sym2k2 ® k or Sym2k2 We considered these representations in Chapters 6 and 7. For IF E Y(ZDA), w = (w1iw2,w3) a character of (Ax /kx )3, X a principal quasi-character of GA /Gk, and s E C, let V.
ED (W', W) X, s), ED+(W) W) X, s)
be the functions which correspond to Zv (qD, w, X)s), Zv+(4,, w, X, s) for the case V = Sym2k2 ® k or Sym2k2 ® k2 in Chapters 5 and 6. Let t2(A2) be as in §6.2. Let XD be the principal quasi-character of Gj/Gk such that XD(t2(A2)) = A2-' for (06), and XD(t2(X2))= r.Tv,2A2 for (/38). We use the notation 'v1,nv2,gv1,gv2 instead Of'v,1,kv,2,.9v,1, as in Chapter 9. We identify ape ()t2), ape (.\2) in §8.2 with t2(\2). Then no, (t2(A2)) = X24 for (/36),
and /c,l(t2(A2)) _ A26 for (/38). We consider r,01 as a principal quasi-character of the group in §6.2. For D = (/36), let ED,ad (`Y, w, X, s) be the adjusted zeta function. We denote by RD,O, R" O the operators which correspond to Rv,o, R',,0 in §6.2. We define To (RD,OIF, w, s, s1) etc. similarly. Let HD C MD be the subgroup generated by T and for the element ((1, 2), 1). Let to = a(X11, \12, \13; A21, A22)t ° E TA° where Xzj E
all i, j. We define (11.1.8)
XD = {kn'(uo)t° I k E K fl MDA, uO E A, IA11A12 I ? a(uo)}.
Then GA fl MDA/HDk = XD/Tk. (d) D = (i31o) We can identify ZD with Sym2k2 ®k2. For 'P E 9'(ZDA), w = (wl, w2) a character
of (Ax /kx )2, and s E C, let ED('I'+w, X, S), ED+(1', W, X"5)
be the functions which correspond to Zv (ob, w, s), Zv+(4), w, s) for the case V = Sym2k2 ® k2 in Chapter 5. Let YD,o, ZD,o, ZD',0 C ZD be the subspaces which correspond to Yv,o, Zv,o, Z(,,0 in Chapter 5 respectively. We define RD,O etc. similarly. Let ED,ad (W, w, s) be the adjusted zeta function and TO (RD,01W, w, s, Si) etc.
the adjusting term. Let HD C MD be the subgroup generated T and the element ((2, 3), (1, 2)). Let to = a(Alli A12, A13; A21, A22)t ° E T. We define
(11.1.9)
XD = {kn"(uo)t°nB2(ul) I k E K fl MDA, u° E A, IA12A13 I ? a(u0)}.
11 Unstable distributions
253
Then GAf1MsA/Hak = Xa /Tk. Let h = (µl, µ2, ql, q2, uo) and hl = (1, µ2)ql, q27 uo) where Al, 92 E R+, q1i q2 E Al, uo E A. We define dh in the obvious manner. For lp E .9(Zs,OA), we define (11.1.10)
OZa,o
h) lY(µlµ2-1g1x1,337 µ1µ242X2,22> 2i 1µ24zx2,z2uo)
X1,33,52,22 Ekx
(e) As before, we use the notation Ew,aa,(2) ('I', w, so), Ea (2) (`I', w, s°) etc. for the
i-th coefficient of the Laurent expansion of Ew,ad(T,w, s), Ea(IF,w, s) etc. In the next three chapters, ga denotes an element of G°A fl MDA, and g°° an element of MaA.
Let dga, dga be the measures which we defined in §3.3. Let d"t° be the measure on TA such that dg° = dkd" t°du where dk, du are the canonical measures on K, NA. (f) Next, we define some distributions associated with the space W of binary quadratic forms, and distributions which are variations of the standard L-functions. Let T = (T,,T2) be a Weyl group element. Let x = xl, x2 = x117 x12> x137 x217 x22 E Q .
Consider the following substitution (11.1.11)
ST11 = x111(3) - Zlri(2), S7-12 = x111(2) - x111(1)7 sT2 = x212(2) - x212(1)
Let sT = (ST1l, 5112, ST2), and sr, = (ST117 ST12) We define ds7-1 = dsTlldsr12, dsT = dsTlds.2
_
Then dz = dsT.
We define functions L,(sTl), M1 (STl) of sT1 E C2 by the following table. Table (11.1.12)
Ll(srl)
Ti
'i T1(ST1)
-2(5111 + ST12)
1
(1,2)
-2sT1l
O(Srl2)
(2,3)
-2S 12
O
(1,2,3)
2sT11
7 f(ST11)/(ST11+
(1,3,2)
2ST 12
1
ST12)
0(ST12)0(STr 1 + ST12) .
1 + ST12)
2(5111
(1,3)
(Srii)
f1
0(ST11)0(ST12)w(ST11 + ST12)
We also define (11.1.13) L2(sT2) _ { -sT2
T2
ST2
MT2(sT2) =
1
T2 = 1,
_ (1, 2),
1
T2=1,
O(ST2)
T2 = (172)7
L(ST) = Ll(STl) +CL2(ST2)7 MT(ST)
M1(ST1)MT2(3T2),
fi(x)
A(w,sT) _
W - L(ST) Al(w7STl) =
(
AT(w;sT) = MT(sT)A(w; sT),
x17 III
w-LiT1(ST1)-V
)
A111(w7S11) = Al(w7STl)MT1(ST1)7
Part IV The quartic case
254
where z and sr are related as in (11.1.11). It is easy to see that L(z) = L(sr). We remind the reader that w° = 4 + C. We also remind the reader that we are still assuming that b(-TGZ) = O(z), O(P) 0 0.
We consider the domain which corresponds to Dr for Re(sr) and use the same notation D. For example, if T = ((1, 2, 3), (1, 2)), a point r = (ri, r2i r3) E Il83 belongs to Dr if and only if r1 > 1, ri + r2 > 1, r3 > 1. Let pr E R3 be the element which corresponds to p by the substitution (11.1.11). Similarly, let pi, E R2, P2r2 E R be the elements which correspond to Pt, P2 by the
substitution (11.1.11). It is easy to see that P2r2 = -1 if T2 = 1, and p2r2 = 1 if r2 = (1, 2). Also Pir1 = (-1, -1), (-2,1),(l, -2), (2, -1),(-1,2), (1, 1) for ri = 1, (1, 2), (2, 3), (1, 2, 3), (1, 3, 2), (1, 3), respectively.
Definition (11.1.14) Let i be a positive integer and IF E Y(Ai). Let 11(s,), li(sr) be linear functions of sr, and li+1(sr) a polynomial of sr. We define
Ei,sub(W, 1, sr) = li+1(sn)-1Ei(1F, 11(°Sr), ... , li(sr)).
(`sub' stands for `substitution.') Let G = GL(2), and G°o = GL(2)°A. Let lt(sr) be a linear function, and 12(sn) a polynomial. We assume that li(-8,111 8,11 + 8,12, 5T2) = li(STi11 S7-121 S.2) for i = 1, 2. Let g° E GL(2)00, t E All. Definition (11.1.15) LetT = (Ti,-r2) be a Weyl group element such that r1 E 232. For ' E 9(WA), we define (1)
Itli1(sr)
o
Esub (l , t , g, ST ) =
12(8T)
EB (t , go ,
- 8,11
),
_
3
(2)
4 .b,, (1 , t , g ° , w ) = ( 2nV -l f2w,T('I'11,w)=
(3)
Re(sr)=r
/k X XG O /Gk iti>1
jtj>1 7, QW, st ,r(Y',
Gwk:<(y.,tg°)esub,T(1,t,g°,w)dxtdg°,
A
Qw,T('I'11, w) = fX /kx XGO/G k
(4) (5)
)f
E 5b (l , t , g° , sr )An ( w;
Ows OF, tg°)19sub,T(11
t)
9°,
w)dxtdgo,
A
1, w) = f X /kX XGOA/Gk Gw, st('y1 t9 0 )4sub , r(l, t, g0 , w)dx td9 0 , t1>1
(6)
3
Ew,SUb+('F,1, ST) = fAX /kx XGOA/Gk Ow. ('I', t9°)Esub,T(l, t, g°, sr)dx
tdg°,
It1>1
where in (2), we choose r = (ri, r2i r3) E R3 so that r E Dr, l2(r) > 0, and r1 < -1. For the rest of this book, when we consider contour integrals with respect to sr, r is of the form r = (r1, r2, r3) E R.
11 Unstable distributions
255
The distribution Ew,sub+(q', s,) is well defined as long as Re(s,ll) < -1, and
_ (2)f 1
S2n,'1,w) =
E.W,SUb+(q'11, ST)AT(wl ST)ds,. Re(ar)=
,Y <-1, l2(*)>0
Clearly,
w) _ cv ('k,1, w) + 1lw,sc,,(W,1, w).
Since we have to deal with integrals in (11.1.15) many times in Chapter 12, we consider them simultaneously in this section and the next section. Consider (11.1.15)(5). Let ESUb(l, t, g°, s,) (resp. Esu6(l, t, g°, s,)) be the constant (resp. non-constant) term of Esub(l, t, g°, s,). It is easy to see that the nonconstant term EEdb(1, t, g°, s,) is holomorphic as long as Re(s,ll) < 0.
Definition (11.1.16) ST) =
(1)
f
OW,st(`
,
tg°)Esub(ll t, g°, s,)dxtdg°.
Ax /kxxGA/Gk
(2)
ntW,st,T('y,1,
w) = f
E"w,st,sUb+(`k,1, s,)A,.(w; s,)ds,.
Re(ar)= *1<0, 12(")>s
s,) is well defined as long as Re(s,ll) < 0. The distribution We consider the constant term. Let to = a2(µt1, µ-1t2) where µ E R+, tl, t2 E A1. It is well known that t 11 (3+)
Esub(1, tl t°l 8T)
(-Srll)Nl+sr11 ).
I12(sT) (µl-gr11
Let r' = (Ti, T2) be a Weyl group element such that Ti = 1 or (2, 3)Tl. We consider s,, as in (11.1.11) for T' also. If Ti = (2, 3)71, then s,,ll = -s,ll, STt12 = 8T11+ST12, and q(-S,ll)A,(w; S,)= A,, (w; s,,). Therefore, by the assumption on 1,
JRe(s)=r
Esub(l) t, t°, s,)A,(w; s,)ds, Itl1 (s,-,)N,1-s,,11 ,.,
T'=r,((2,3),1)T
Re(s,.,)=r'
12(8T')
A,, (w; S,, )ds,,.
Let B be the Borel subgroup of GL(1) x GL(2), b = (t,n2(u°)a2(ptl, P-%)), and db = dxl-tdxtdxtldxt2du°. We identify the maximal torus of GL(1) x GL(2) with GL(1)3. By a similar argument as in §4.2,
f
XWnBA/(k")3 ItI>1
12(ST')
OZ,W 0 (lY b)db =
Tw+(RW,°1k111(s,,)' 2(i - S,tll))
(8r'll - 1)12(8,') l
Part IV The quartic case
256
Let Tw+(Rw,o'L, 11(s"), 2(1 - ST'11))
(11.1.17) E W,st,sub+(LY, 1, sT')
(sT 11 - 1)12(sT') 1
F, w W,st,T OF, W) =
(2)7f Le(sr,)=r
W,st,sub+(`Y, 1, s7, )' ,, (w, ST' )dsT'.
Then (11.1.10)
fW,st,T (IF, 1, w) =
E
Q
w,st,T' (i , 1, w) + cl'w.t,T(w, 1, w).
T'=T,((2,3),1)T
§11.2 Technical lemmas Consider a Weyl group element T = (Ti,T2) such that T1 E 02. First we consider the distributions Styy T(gf,1, w) etc. in §11.1. We consider the following three possibilities for 12: (1) 12(ST) = -ST11 - 2sT12 + 2ST2 - 1, (2) 12(ST) = sT11 + 2sT12 + 2sr2 - 7, (3) 12(ST) = ST11 + 2ST12 + 2ST2 - 9.
If r2 = 1, we fix r1 = -2, r2 = 4, and choose r3 >> 0. Then L(r) < 8 - Cr3 << 0. w), S2'W'St,T(IQ,1, w) - 0 if TZ = 1. So we assume that r2 = Therefore, Q8 (1, 2).
Lemma (11.2.1) Suppose that T1 E 02 and T2 = (1, 2). Then by changing b if necessary, w) - 0 unless Ti = (1, 3, 2), (1) f2fv,T('y,1, w), (2) If 7-1 = (1, 3, 2), n'w,st,T('P,1, w) - 0, and f2ft,,T('p,1,w)
-CGA(w;P)Ew+((,ll(-1,)2,1))
Proof. If r2 = (1, 2), l2(PT) # 0 for all the cases. Therefore, by the passing principle (3.6.1), we can ignore the set 12(sT) = 0 when we move the contour. So for the case (1), we choose r so that 12(r) < 0. We move the contour so that r3 < 1 as follows. 1w,T('I`,1, w)
(2) 1
1rV-1
+0
3
fRe(s.)=(-2,6,2)
1
(2)7rVl'-'-l
3 £W,sub+('@, 1, ST)A'* ST)dST
2
JRe(.FW,sub+( ,1, ST1, 1)A1T1(w; ST1)ds
1.
*1+2r2 >7
Note that l2(-2, 6, 2) # 0 for all the cases. Since C > 100, Z(-2,6, < 12 + < 4 + C. So, we can ignore the first term. 2C 1, w) also. The same argument works for2)SE,,,St,T (IF,
257
11 Unstable distributions
Consider the second term. If T1 = 1 or (1, 2), we fix r1 = - 2, and choose r2 >> 0Then i(r) << 0 or i(r) = 3 + C. Therefore, we can ignore these cases. This proves (1).
Consider Tl = (1, 3, 2). We move the contour to Re(sT1) _ (-1 - 6, 2 + 6) where
6 > 0 is a small number. For (2), (3), we can use (3.6.1), and ignore the line l2(sri, 1) = 0 Therefore, in all the cases, 2
1
0
27r
=0
JR.(1)=(12) r1+2r2>7
1
EW,sub+('', 1, ST1, 1)A1T1 (w, ST1)dsrl
2
JRe(si)=(-i-6,2+5)
(2)7r 1
= [J
E
\
1, ST1, 1)A1T1(w, STi)dsTi
2
EW,sub+('lf, 1, ST1, 1)A1T1(w, STi)dsTl P
Res JRC(sr12)=2+6 Sr11=-1
27r V
EWsub+(I',1, ST1, 1)AiT1(wi -1, ST12)dsT12
Note that since we move the contour from the left to the right, the sign of the second term is negative.
Since L1(-2,2) = 3, we can ignore the first term. Since M-1(-1,5T12) _ 02(ST12 - 1), and
Res E Sub (l t , g°, s T1, 1)=-
s+ll=-1
,
VIt111(-i,s,12,1)
12(-1, ST12, 1)
we only have to move the contour to 1 < Re(sT12) < 2. Q.E.D.
Lemma (11.2.2) Let T' = (T1,T2) be a Weyl group element and 6 > 0 a small number. Then by changing 7/i if necessary, (1) S2v,st,,, ('I`,1, w) - 0 unless r' = TG,
(2) If r' = TG, QW,st,T,
('I`,1, w)
27rv -1 Re(sr,11)=1+6
EW,st,sub+('y, 1, ST,11, 1, 1)Yi2(sr,11)A1(w; ST,11, 1)dsT'ii.
Proof. Similarly as in (11.2.1), we can ignore the case T2 = 1, and assume that T2 = (1, 2). Then Qw,st,T,
_0
(
('I',1, w) 1
2
EW,St,sub+(W, 1, sT,i, 1)A 11 w, sT'1)dsT'i. 2 7r V1----1
*1+2,2>7
(We have to use the passing principle (3.6.1) for the case (1).)
258
Part IV The quartic case
If 1.1' = 1, (1, 2) or (1, 3), we choose rl, r2 >> 0. Then L(rl, r2,1) << 0. If Ti = (1, 2, 3), we fix r1 = 2, and choose r2 >> 0, and if Ti = (1, 3, 2), we fix r2 = 2 and choose r1 >> 0. In both cases, L(rl, r2, 1) = 3 + C < 4 + C. This proves (1).
Suppose rl = (1, 3). Then p, = (1,1,1). Clearly, 12(1,1,1) : 0 for all the cases. Therefore, we can move the contour crossing this line. So 1
2
) f('i)("i2) EW,st,sub+('I')1, 81.'1) 1)A1,'1(w; ST'1)ds1'1
(21rV
2
, 1+2r2>7
1
1
)'I
+ 2n V 1
Ew,st,sub+(W') 1, S7-'I) 1)A11.(w; ST'1)dsT'1
e(sT'1)=(1,5)
1) ST'll) 1) 1)
fRe(srii)=1+6
X 02(s1'11)A1(w)8T'11) 1)dsT,ll)
where 6 > 0 is a small number. We can ignore the first term by the usual argument. This proves the lemma. Q.E.D. The following lemma follows from these considerations.
Lemma (11.2.3) Suppose that r' = TG and b > 0 is a small number. Then by changing 7/i if necessary
E nw,st,T('I')l)w) *1EW2
r2=1,(1 2)
CGA(w; p) P2
+
27r
Ew+(`I',1, (-1, 2,1)) 12(-1, 2,1) JRe(s11)=i+8 EW,t,sub+( T, 1, sr'11, 1, 1)2(ST'11)A1(w; sr11, 1)ds111
Now we consider slightly different contour integrals. Let l = (2 (3,11 + 1), 12) where l2 is one of the following: (1) 12(s1.) = (-sT11 - 23,12 + 2s1.2 - 1)(s,12 + 1),
(2) 12(8,) = (-ST11 - 28,12 + 2ST2 - 1)(8,11 +'ST12 - 1), (3) 12(51.) _ (sr11 + 2sT12 - 2sT2 + 1)(-sill + 28,2 + 1), (4) 12(8,) = (ST11 + 231.12 - 2ST2 + 1)(3,11 + 28,2 - 3), (5) 12(8,) = (s1.11 + 281.12 + 281.2 - 7)(8T12 + 1), (6) 12(8,-) = (31.11 + 231.12 + 281.2 - 7)(sTll + 8,12 - 1))
(7) 12(ST) = (s1.11 + 251.12 + 2ST2 - 7)(5sr11 + 4s,-12 - 2ST2 + 1), (8) 12(51.) = (`81.11 + 2s,-12 + 2ST2 - 7)(s,-11 - 4sT12 + 2ST2 - 13),
(9) 12(ST) = (8,-11 + 2s,12 + 2ST2 - 7)(8,11 - 28,12 + 281.2 + 1), (10) 12(8,-) = (sill + 2s,12 + 25,2 - 7)(3s1.11 + 2s,-12 - 2sT2 - 9),
(11)
12(81.) = (5,-11 + 2s1.12 + 2ST2 - 7)(-s,11 + 2ST2 + 1),
(12) 12(3,-) = (5,-11 + 25,-12 + 281.2 - 7)(sT11 + 251.2 - 3),
(13) 12(81.) = (8,-11 + 28,-12 + 28,2 - 9)(5s,-11 + 43,12 - 2sT2 - 9),
(14)
12(8,-) = (5,11 + 28,-12 + 28,-2 - 9)(8,11 - 431.12 + 25,2 - 3),
11 Unstable distributions
259
(15) 12(ST) = (sill + 2sT12 + 28,2 - 9)(s711 - 2s.12 + 2ST2 - 5), (16) 12(87) = (s.11 + 2s.12 + 2sT2 - 9)(3sT11 + 26.12 - 28,2 - 3).
For these cases, let 121 (resp. 122) be the first factor (resp. second factor) of 12. Let 41 E 9(A). We consider 1
(2)3fR(S,- E1,sub(
, 1, s.)AT(w; sT)ds.,
where we choose the contour so that r = (rl, r2, r3) E DT, r2 > 1, and l21(r), 122(r) > 0.
Lemma (11.2.4) Suppose that T' = TG and 6 > 0 is a small number. Then by changing V7 if necessary, 1)31
1
2
N 27rV
E e(sr)=r
l,sub(, 1, sT)AT(wi 87)(187
1, sT'11, 1, 1)W2(sT'11)Al(w; 37'll, 1)dsr'ii.
-1 JRe(s11)=1+6
We devote the rest of this section to the proof of this lemma.
Lemma (11.2.5) Suppose that l(s.) = (2(s,11 + 1),12(57)) where l2(sT) is one of (1)-(16). Then by changing 7/J if necessary, (1) if T2 = 1,
)31
1
.(',)=r
27r
El,sub(i, 1, sT)AT(w; 5r)d5r - 0, and
(2) if T2 = (1, 2),
)3j 27rV -1
e(s,)=r
1
E1,sub(W, 1, sT)AT(w; sr)dsr
)21
e(s.,l)=(ri,r2)
27r
1,sub(I, 1, sTl, 1)AiT1(w; 8T1, 1)dsT1,
where we choose the contour so that (rl, r2) E DTl, r1 > 1, 121(rl, r2, 1) < 0, 122(rl, r2, 1) > 0 for cases (1), (2), and
(rl, r2) E DTJ, rl > 1, 12i(rl, r2, 1) > 0 for i = 1, 2
for cases (3)-(16). Proof. If r2 = 1, we can choose r3 >> 0, and Z(r) << 0. Therefore, (1) is clear.
260
Part IV The quartic case
We assume that r2 = (1, 2). Then P21-2 = 1. For cases (1) and (2), it is easy to check that the point pill is not on the lines 121(s1-i,1) = 0,122(81-1,1) = 0. Therefore, by the passing principle, we can move the contour crossing these lines. So we can assume that 121(ri, r2, 1) < 0,122(ri, r2, 1) > 0 for these cases. Consider the following values for (ri, r2).
(rl, r2)
(3,5)
(1), (2), (4) - (7), (10), (12), (13), (16),
(2, 20)
(3), (11), (8), (9), (14), (15).
(15,2)
Then for cases (1) and (2), 121(ri, r2, r3) < 0, 122(ri, r2) r3) > 0 for r3 = 2,1, 2, and for cases (3)-(16), 12i(ri, r2, r3) > 0 for i = 1, 2 and r3 = 2,1, 2. Therefore, 1
27f
)3f El,sub(W, 1, s1-)AT(w; ST)ds1-
e(sr)=(ri,T2,2)
)31
Re(s.,)=(ri,r2,2)
21r N/- 1
1,sub(W, 1, s1-)A1-(w; s1)ds1-
2
1
+N
E
( 27rV -1
J Re(srl)=(rl,r2)
E1,SUb(W, 1, s1-i, 1)X1,1 (w; srl)dsri.
In all the cases, L(rl, r2, 2) < 43 + 2C < 4 + C since C > 100. Therefore, we can ignore the first term. This proves statement (2). Q.E.D.
If r1 = 1, (1, 2), (2, 3), Li (ri, r2) < 0 for the above choices of ri, r2. Therefore,
)21
1
El,sub(W, 1, sT1, 1)AiT1 (w; 51-1, 1)ds1-i - 0
( 2r
e(s,l)=(rl,r2)
for these cases. Suppose that ri = (1, 2, 3), (1, 3, 2) or (1, 3). Let 6, 61 > 0 be small numbers such that 616-1 >> 0.
Lemma (11.2.6) By changing -i if necessary,
Fl,sub(W, 1,'STl) 1)A1T1 (w; sTl, 1)dsTl
( 27r N
)2j(w;
)21
1
e(sr1)=(r1,r2)
1
2r
sTl,1)ds1-t.
e(s,l)=(1+6,1+6)
Proof. For cases (1) and (2), the domain {(ri, r2) 1121(rl, T2, 1) < 0, l22(rl, r2, 1) > 0} contains (1+6,1+6). For cases (3) and (4), the domain {(ri, r2) 1121(ri,r2, 1) > 0,122(ri, r2, 1) > 0} contains (1 + 6,1 + 6). For ri = (1, 2, 3), (1, 3, 2), (1, 3), Pill = (2, -1), (-1, 2), (1, 1). For cases (6)-(11), (13)-(15), 12i(P1T1,1) # 0 for i = 1, 2. For cases (5), (12), (16), 121 (P1,1, l) # 0,
and the line segment joining (ri, r2) and (1 + 6, 1 + 6) does not meet the line
11 Unstable distributions
261
122(81.11, S,-12, 1) = 0. Therefore, the lemma follows from the passing principle (3.6.1)
for these cases. Q.E.D.
If Ti 54 (1, 3), Ll(1 + 6,1 + 6) < 4 as long as 6 > 0 is small. So
)2j
1
2iri
-1
e(s.,,)=(1+6,1 -6)
El,sub( ) 1, sr1, 1)Alrl (w; srl)dsrl - 0.
Suppose rl = (1, 3). Then
)21
1 271.
-1
)2
1 (2'i)71.
+
--
21r J
e(s.l)=(1+6,1+6)
E1,sub(W, 1, srl)
fRe(sj)=(i+5,1-5i)
(w; srl)dsrl
+l,sub(W, 1, srl, 1)Alrl (w, srl)dsrl
El,sub(W, 1, 81.11, 1, 1)02(sr11)Al(wi 81.11, 1)dsrll e (sr11)=1+6
27r fRe(sii)1+6
Fil,sub(W) 1, Sr11, 1, 1)02(sr11)Al(wi 81.11, 1)dsr11,
because the first term can be ignored by the assumption on 6, 61. This proves (11.2.4).
Q.E.D.
Chapter 12 Contributions from unstable strata §12.1 The case 0 = ((31) We now start the analysis of the constant terms 8p (4', w, w). We remind the reader that we are still assuming that 4' = Mv, 4, so if 6# (w) = 1, ID is K-invariant. In this section, we consider paths p = (a, s) whose ,3-sequences start with /31.
Since e((g°),w) = e(g°,w) and the characters epi do not depend on s(1), we only consider p = (0,s) such that s(1) = 0. For paths with s(1) = 1, all the results are valid replacing 4' by $ and w by w-1, and changing the sign. All the /3-sequences which start with Nl satisfy Condition (3.4.16)(1). Let = (ao,so) where 0 = (f31). Let pll = (01, 511), P12 = (D1, 512) where (i1 = (31, i1,1), and RRPo
511(2) = 0, 512(2) = 1. Let p21 = (02, 521), P22 = (D2, 522) where 02 = (/31,,31,2), and 521(2) = 0,522(2) = 1. Let 03 = ((31.,,(311,/31,1,1), and p3i = (03, 53i) for i = 1,2,3,4
where (s3i(2), S& (3)) = (0, 0), (0, 1), (1, 0), (1,1) for i = 1, 2, 3, 4 in that order. We consider contributions from these /3-sequences. Throughout this section, we assume that IF = Rao 4'po Let
(12.1.1)
do(A1) = a(1,1,1; ail, A1), d1(Al, A2) = a(A22, A2, A2; A11, A1), d2 (A,, A2) = a(A2 1, A2 1, A22; Al
1,
i),
d3(A1, A2, A3) = a(A2 2 -2'13 1+ 2 i-i 1, A1),
for A1, A2, As E R+.
Easy computations show the following two lemmas.
Lemma (12.1.2) A1a2. A1, (1) (2) ep121 (dl(A1,A2)) _X11, ep,22(dl(A1,A2)) = A1a2. (3) ep2i1(d2(Ai, A2)) = A1, ep221(d2(Ai, A2)) = Al 1. A2, A3)) = A1, ep3,2(d3(Ai, A2, A3)) = AjA2. (4) (5) ep321(d3(Ai, A2, A3)) = Al 1, ep322(d3(Ai, A2, A3)) _ (AiA2)-1 1 ep332(d3(A1, A2, A3)) = ala2. (6) ep331(d3(Ai, A2, A3)) _ A (7) ep341(d3(Ai, A2, A3)) = A1, ep342(d3(A1, A2, A3)) = (A1A2)-1
Lemma (12.1.3) A1-3 A2-6.
(1) op,,(dl(A,,A2)) = A2A2, o'12(d1(A1,.2)) = A'A', ,a,1(d1(A1,A2)) _ 21 (d2(Al, A2)) = A2A2, o' 22 (d2 (A,, A2)) = A4A2. (3) Qps, (d3(A1, A2, A3)) = A1'A3, 0P32 (d3(Al, A2, A3)) _ A1A3, spas (d3(A1, A2, A3)) = A1A1Am3, ap34 (d3(A1, A2, A3)) = Al 1A3
(2) o'
Let T, = {po, p11, P12}, P2 = {P21, P22, P31, P32, P33, P34}. By (3.5.9),
(12.1.4)
Epo
P. (4',
w,w) _
Ep
P+ (C w, w) + °p+OP, w, w)
PET,
pE`p,
+
Ep p(4',W,W) PEP2
12 Contributions from unstable strata
263
Po+N' w, w), w, w) The element do(A1) acts on ZaoA by multiplication by A1. It is easy to see that -po(do(A1)) = )s , Kao1(do(\1)) _ Al s Therefore, the following proposition follows from (3.5.13). (a)
Proposition (12.1.5) (1)
"po+(ID, w, w) - CGA(w; P)sao (w)Eao+(q', wao, 2).
(2)
=po+(1D, w, w) ' CGA(w; P)aao
(b) =po#(-1),w,w),
(w)Eao+(j;-ao W, wa 1, 4).
w)
Proposition (12.1.6) (1)
po# ( , w, w) - CGA(w; P)
(2)
po# ( , Lo, w)
2736# (w)'y(0) 2
CGA(w;P)2136#(w)9ao'F(0) 4
Proof. Let 0 = Do, p = po. Let go E Ma, 9a = do(A1)gao Then dga = d
cp#(',w,w) = `I'(0) f
w(go)A
p(ga,w)dga,
Ri. X MDA /Mak
p#('b,w, w) = a'0 T '(0)
f
ai <1
w(ga)A14e(ga,w)dga
R+x MOA /Mak 1<1
Let T = (1, T2) and r2 = 1 or (1, 2). Let sr2 be as in (11.1.11). By a similar consideration as in (3.5.20), 6#(w)
p#( D ,w, w) _
11
72
f
2vJ
1-e(8 T 2)=*' 3 >1
MT2(s-2)
A(w;P1,z2)
S T2
+1
d sTz,
where z2 and sit are related as in (11.1.11). If T2 = 1, L2(Re(z2)) = -r3 < -1 on the above contour. Therefore, we only have to consider the case 72 = (1, 2). We then move the contour to 0 < r3 < 1, and (1) follows. The proof of (2) is similar. Q.E.D.
w, w), E w, w) In (c), (d), let a = 01,p = p11, and W1 = Ra14)p,,. Let go = di(A1,A2)ga° where ga E MA is as in the second element of (11.1.3). Also in this case, M ,1A = MA. It is easy to see that dga = 2dxA1dXA2dga. Let p = A1A2. Then 2d \1dxA2 = (c)
dXA1dxp. Let T = (7-1iT2) be a Weyl group element, and sT as in (11.1.11).
Definition (12.1.7) (1) 1p11 = (1p11,1,1p11,2) where 1p11i1(ST) = 2(sT11+2s-12+3), lp,,,2(S) = -ST11 2sT12 + 28,2 - 1,
Part IV The quartic case
264
(2) "pll = (lpll,1, 1p11,2) where 1p11,1(ST) = 3 - lp11,1(ST), 1p11,2(ST) = 1p11,2(ST)
Note that 1,11 (-ST11, ST11 + ST12, ST2) = lpii (ST11, ST12, ST2) etc.
By (12.1.3)(1), \2)TZ+P = A1T2-1A2r11+28+12-3 _
(12.1.8)
o,p(d1(A1,A2))d1(A1,A2)TZ+P,a1(d1(A,,A2)) =
Ai`n2(Sr)µi
l(er)-3
The elements dl(A1, A2), 05 (d, (A,, A2)) act on ZDA by multiplication by µ, µ 1 respectively. Also the integrals defining =pll+(4), w, w), upll+(4), w, w) do not depend on tll, t21. Therefore, by (12.1.8), ll+(,D,w,w)=26#(w)
Qw,T(`I'1,1p111w)1 r1 EID2
r2=1,(1,2)
w, w) = 26#(w)
QW,T(gW'F1, 1p11, w). r1 E 2112
T2=1,(1,2)
The following proposition follows from (11.2.3).
Proposition (12.1.9) Suppose that r = 'rG and 6 > 0 is a small number. Then by changing 7/i if necessary,
- CGA(w; P)6#(w)Ew+('Y1, 3)
=pll+('D, w, w)
+ 2026#(w)J f 27rV -1
EW,st,sub+(``1) lpll, 5T11, 1, 1)
Re(sT11)=1+6
X 02(ST11)A1(w, ST11, 1)dST11,
w) - - CGA(w; p)6#(w)EW+(9W'F1, 0)
+ 2026#(w) 27rV
i
f
e(s,ii)=1+6
1W,st,sub+(gW 1, 1p11 , ST11, 1, 1 )
X g52(ST11)A1(w, si11, 1)dsi11.
(d)
We define w, w, ST12, ST2) =
W) W, ST12,
(0)
ST2)-
A T(w; -1, ST12, ST2) (ST12 + 1)(-ST12 + ST2) 1(O)
A. ( w; -1, ST12,ST2) (ST12 - 2)(-Sr12 +ST2) .
Then by (11.1.4), (11.1.5), (12.1.8), and the Mellin inversion formula, the distributions p11#(D,W,w),upll#(D W,W) are equal to 2
(12.1.10) E
CO,
T1
r2=1,(1,2)
T1E'ID2
r2=1,(1,2)
2 (_1 1r
1
/
(1\21) 27r v
fRe(, r12.ar2)=('2, r2>2, r3>r2
Ep11#,T((D, w, w, ST12, ST2)dsT12, ST2, r3)
f(arl2,ar2)=(r2,T3) r2>2, r3>"2
Gpll#.T O', W, w, ST 12, ST2)dST12, ST2
12 Contributions from unstable strata
265
Proposition (12.1.11) Suppose that T = TG and 6 > 0 is a small number. Then by changing ?,b if necessary, CGA('w;p)126#(w)`I'1(0)
(1) ( 2)
Ep11#(41,w,w)
(
"p11#
3
,W,w )
Jr
27r\ -1
2(ST11)Al(w; sr11, 1)
Re(srll)=1+6
s,.n(1 - srll)
d sr11
Proof. Consider the term in (12.1.10) which corresponds to r. If r2 = 1, we can ignore such a term by the usual argument. So, we assume that 7-2 = 1. If Ti = 1, (1, 2) PT = (-1, -1, 1), (-2, 1, 1). Therefore, by (3.6.1), we can move the contour crossing the line sri2 = 2. So in both cases, we can move the contour so that
r2 = 1 + 6, r3 = 1 + 26 where 6 > 0 is a small number. Then L(-1, r2i r3) < 2 + (1 + 26) C < 4 + C if 6 is sufficiently small. Therefore, we can ignore these cases.
Suppose Ti = (1, 3, 2). The point pr = (-1, 2,1) is not on the lines sri2 + 1 =
0, -sri2 + sr2 = 0. So we can move the contour crossing these lines by (3.6.1) so that r2 < r3. Then for (1), the term in (12.1.10) which corresponds to T = ((1, 3, 2), (1, 2)) is equal to
(2) 2 fR(.,22)=(2,3) rs>r2>2
)1
1
C2
+
Ep11 #,T
2
V -1
27r
W, w, 5T12) sr2)dsrl2dsr2
Ep11#,T(4), w) w, 8,12,sT2)ds,12dsT2
Re(srl2,sr2)=(3, 2 )
Res Ep11#,T(4,w,w,sr12,sr2)dsr12
Re(s,.12)=r2>2 sr2=1
We can ignore the first term as usual. If T1 = 1 or (1, 2), L(-1, 3,1) < 2 + C < 4 + C. Therefore, we can ignore both these terms. Suppose T1 = (1, 3, 2). Since M1.1 (-1, sr12) = 02(sr12-1) and this function has a simple pole at sr12 = 2, we only have to move the contour to 1 < Re(sTii) < 2. This proves (1). The proof of (2) is similar except that the integrand of the second term has a pole of order 2 at sr12 = 2,
and we leave as it is. If T' = TG, 5r'12 = sr12 - 1. We consider the substitution z -+ -TGZ. Then sr)11i Sr'12 are exchanged, and A1(w; sr)11i 1) = Al(w;1, sr'11) by the assumption 7,b(-TGz) = i,b(z). Thus, after these substitutions, we get the expression of (2) in terms of Q.E.D. (e) E 12+(4)w,'w), "p12+(4''w,w).
In (e)-(g), 0 = 01, P = P12, and W2 = Rat'I
12
Let go = d1(.X1, A2)9° where
goo E M OA is as in the second element of (11.1.3). By (12.1.2)(2), (12.1.3)(1),
"P12+ (D) w, w)
=2f R+x MaA /Mak
Al>1, ala2>1
w(9a)-1A14
A2
(`y2,9aA12(9a)w)dxAldxa2(1900)
Part IV The quartic case
266
W) w)
w(go)-1AlOz,(9W`Y2,00(ga))gp12(gD1w)d'AldxA2dga
=2 1.2 +x MDA/MO k
J.111, A1a2<1
Let µ = A1a2. Then A' A2 = a1µ3. The functions Oz, (W,go),Oz,(` 0Y,8s(ga)), and ep12 (gD, w) do not depend on tll, t21. Therefore, these distributions are 0 unless w is trivial. Let r, sr be as before. Easy computations show the following lemma. Lemma (12.1.12) Bp12(d2(A1, 2)) r=+v = Al 2(-srll-zsrl2-zsr2+s) µ 2(sr11+29,12-3)
Definition (12.1.13) (1)
1P12 = (lp12,1,lp1z,2)
where lp12,1 (Sr) = 2 (s111 + 2.s ,12 + 3),
lp12,2(Sr) = Sr11 +
2sr12 + 2sr2 - 7,
(2) 1P12 = (1p12,1,1p12,2) where lp12,1(sr) = 3 -1p12,1(Sr), 1p12,2(Sr) = lp12,2(sr). Note that 1p12(-sr11,sr11 + Sr12,sr2) =
lp12(sr11,Sr12,Sr2) etc.
By (11.1.4), (11.1.5), (12.1.12), SlW,r('1'2,1P12' w), T1 EQ2
*2=1,(1, 2)
p12+(4P,w,w) = 26# (w)
E
QWr(9WT2, lp121 W) -
r1E'h12
r2=1,(1, 2)
The following proposition follows from (11.2.3).
Proposition (12.1.14) Suppose that r = rG and b > 0 is a small number. Then by changing 0 if necessary, W, w) - -CGA(w; P)b#(w)EW+(%'2, 3)
f
2,o 2
+
JRe(sr11)=1+b
2a
EW,st,sub+(P2,1Piz) Sr11, 1, 1) X 02(Sr11)A1(w;1) sr12)dsr12,
p12+(41), W, w) " -CG A(w; P)b#(w)EW+(-g'-W'P 2, 0) 2g2 _ EW,st,sub+(`W W2, lp12, 5r11, 1, 1) + 27rV
1
fRe(sii)1+6
X 02(5r11)A1(W; 1,5r12)dsr12
(f) "p12#(4',w,w), up12#(4,w,w) These distributions are 0 unless w is trivial for a similar reason. We define A7(w; -1, Sr12, Sr2)
+sr2 -4)(Sr12+1)1 Ep12#,r(
,
w,w,sr12,sr2) = b#(w) `W1Y2(0)
Ar( w; -1, Sr12, Sr2) (Sriz + Sr2 - 4)(sriz - 2)
12 Contributions from unstable strata
267
Then by (11.1.4), (11.1.5), (12.1.12), and the Mellin inversion formula, the disWI w) are equal to
tributions ` P12# ("'I WI w)) "P12#
)2j..
1
(IF,
,1
-
2
r2=1,(1,2)
w) `ST121 S r2)ds,.12dsT2.)
(.j2,er2)=(*2,*3) r2>2, r2+r3>4
21 )
27r
r12
'
> P12#,T('f') W) W) ST12, Sr2)dSrl2dsr2 R<('r12,'r2)=(r2,r3)
r2>2, r2+'3 >4
,2=1,(1,2)
Proposition (12.1.16) Suppose that T = TG and S > 0 is a small number. Then by changing 0 if necessary,
-C G A (wp) 932S#( 3 )' 2(0)
( 1)
= p12#( p ,w,w)-
( 2)
2 p12#( 4 ,w,w ) ^' QS#(w)." w'F2(0) 27r
J
12(STII)AI(w; STIl, 1)d ST1I e(s,II)=1+6
(STIl - 1)(sTI1 - 2)
Proof. As usual we can ignore the case r2 = 1, and assume that T2 = (1, 2). We can ignore the cases rr1 = 1, (1, 2) by the same argument as in (12.1.11). Suppose Tl = (1, 3, 2). Consider (1). We move the contour in the following manner: 1
(2)2JR(l2.2)(2r3) Ep12#,
(WI) w, w, Sl2) 32)d12dS2
r2>2, r2+'3>4
2
p12#,T (TI) W, W, ST12, ST2)dsrl2dsT2
(2)7r-,/
+
fRe(si2,s2)=(4,)
f
1
27r
Res Epi2#,T( 1, W, W) ST12) 8T2)dsr12
e(sr12)=rz>3 Sr2=1
We can ignore the first term because L(r) < 8 + 1C < 4 + C. The point pT = (-1, 2, 1) does not satisfy the condition ST12 = 3. Therefore, we can ignore the pole of Ressr2=1 Ep12#,T('yl, w, w, sTl2, sT2) at ST12 = 3 by (3.6.1). Hence, Res E'p12#,T (W1, w, W, ST12, ST2)d5T12
1
JRe(si2)=r2>3 3,2=1
27r 1
27r
Res Ep12#,T(W1,w,w,5r12,sr2)dsT12
R'(' 12)=r2 8,2=1 1
- CGA(w; P) 02S#(w)y1(0) 3
We can ignore the first term in the usual manner. This proves (1), and the proof of (2) is similar to (12.1.11). Q.E.D.
Part IV The quartic case
268 (g) = p21
w, w), =P22 ( w, w)
Let 0 = 02. Let go = d2(A1, A2)9a where ga is as in the first element of (11.1.3). It is easy to see that dg, = 2dx )1dx )2dg°°. Let 'd = Rae Rp2i for i = 1, 2. In this case, Pa# = P1 x B2. By (12.1.2), (12.1.3),
P21 (D, w, w) = 2 f.2
w)d"Aldx.2dga,
w(9a)A2A2ez, (`I`1, 9sWp21(9a, x M ak /Mak
al<1
f
w(9a)-1A
R+x Mak /Mak
a1>1
Then A?.\ = Ai µz, a1 2 = Al µ2, and dxA1dxA2 = Let p = Let r = (r1, r2) be a Weyl group element such that rl E 01. Easy computations show the following lemma.
Lemma (12.1.17) Suppose Sr12 = -1. Then A2))Tz+P = A (-8x11+29r2)µ2(8,11-2)
(1) 0P21 (d2(\1,
(2) ep22 (d2(A1, a2))
TZ+P
-=
i (-s,11-28,2+4)µ2cw 11-2)
Definition (12.1.18) Let 42i = (2 (sill + 1), lp2i,2) for i = 1, 2 where (1) lpz1,2(Sr) = -s-11 + 2Sr2 + 1, (2) lp22,2(Sr)
= srll + 2ST2 - 9.
w) do not depend on g1,12, t21 for Note that the functions Oz, (lWe, ga), i = 1, 2. Therefore, the distributions .p21 (`Y, w) w), °p22 (I, w, w) are 0 unless w is trivial. We define Ep2i (D, W, Srll, Sr2) = [E1,sub(Wi, 42i, 8,)A,-(w; ST)I8,12=-1
for i = 1, 2. Then Ep2i ($, W, W) is equal to 1
2
6#(w) E (2ci) f
Ep2i ('P, w, sr)dsi11dsr2
R.e(s 11 ,8r2)=(r1 ,r3)
,1 E,d,l
for i = 1, 2, where we choose the contour so that r1 > 2, 2r3 > rl - 1 for i = 1 and
r1>2,2r3>9-r1 for i = 2. Proposition (12.1.19) P21(45,w,w),,, CaA(w;P)`V26#(w)E1(Ra2(Dp21,
(1) (2)
(4), w, w) - -CGA(w; P)
P22
2).
X1265 (w) E1(R02OPP221
2
Proof. Let ri = a, r3 = a for i = 1 and ri = 9, r3 = 2 for i = 2. Then 1
2
(2) fRe(sii,s2)(ri,r3) Ep2i (4), w, s,)dsT11dST2
(2) 1
=
7r V
+
i
2
fRe(aii,sr2)=(rl,r9)
4e
(s,11)=r1 sRes
Ep2i
(1)1 W1 Sr)dST11dgT2
E12i ('P, w, sT)dsrll
12 Contributions from unstable strata
269
Since L(ri, -1,r3') < 18+ 2C < wo for all the cases, we can ignore the first term. The rest of the argument is similar as before. Q.E.D.
(h) ,-p3i (41, w, w) for i = 1,
, 4.
Let a = D3. Let to = d3(A1,A2,A3)t°. Then dxt° = 2dxA1dxa2dxA3dxt°. By (12.1.2)(4)-(7) and (12.1.3)(3), / w(t°)A2A ®Z, (R1 4'p33) 1 2A 3
(-D, w, w) = j
(t°, w)dx to,
TO /Tk Al<1, ala2<1
P32
J
w, w) =
w(t°)-1AiA3OZa (R133 to)4p32 (t°)
w)dxto,
1
Tp/Tk
l>1, ala2>1
P33
w(t°)-'A4 A6 A3OZa
((D, w, w) =
J
(Ra4)p33) to)4p33 (t°, w)`"xt°>
TA /Tk
al>1, \1A2<1
paa (-D, w, w)
_f
w(t°)A1'A20Z,
(Ra.Dp34) t°Ap34 (t°, w)dxt°.
Tp/Tk
a1<1, N1),2>1
For a similar reason as before, these distributions are 0 unless w is trivial Let µl = Al A 2, and µ2 = ala2A3. Then A1A2A3
= A11µ1µ2, A1A3 = Alµl 1112)
2 A 4 A2A3
= A111i112, and Al 1A3 = Al 1µ11µ2
It is easy to see that dxto = 2dxAldxµldxµ2dxt o Easy computations show the following lemma.
Lemma (12.1.20) (1) Bp31
A2, A3))
(2) OP3z (d3(A1) A2, A3))
rz+P == 1A(-sr11-25,12-2Sr2+1)tts1 ,12-1µ2(s,11-1) Tz+P = A 2 (s -11+2s -12+2sr2-1) -s,11-x,12+2 z (sr11-1) 111 112
rz+p = z(-s,11-2s,12-2ST2+5) Sr12-1 z(srll-1) A1 µ1 112 (3) (s, 11 F2sr12+2s,z-5) -sr11-sr12+2 z (Sr11-1) (4) Bp3a (d3(A1, A2) A3))Tz+1 = A21 111 112
Definition (12.1.21) Let lp4i = (2 (sill +/ 1), lp4i,2) where (1) 1p31,2(ST) = (-STll - 2ST12 + 2ST2 - 1)(ST12 + 1), (2) 1p32,2(ST) = (-ST11 - 2sr12 + 2ST2 - 1)(ST11 + ST12 - 1)) (3) 1p33,2(ST) = (ST11 + 2sT12 + 2ST2 - 7)((ST12 + 1), (4)
1p34,2(ST) = (ST11 + 2ST12 + 2ST2 - 7)(sT11 + ST12 - 1).
By (12.1.20), (-D, w, w) = S#(w)
(2) 1
f
Psi
T T
3
E1,SUb(4),
1P30
ST)AT(w, ST)ds.r
for i = 1, , 4 where we choose the contour so that r1, r2, r3 > 1 and both factors of lp3i (r) are positive.
Part IV The quartic case
270
Let
i
E1(Ra3 p31, `Z (`SILL+1))
E P31 (4) , ST11 )
2(sTLL + 1)
EP32 (4) , S TL1 ) = ST 11) -
Ep33
E p39
(
, SILL) -
(ST11 +1)) sy11(8T11 + 1)
E1(Ra3 4)P32 , 2
E- 1 (Ra3 p33, 2 (3,11 + 1))
2(sTll - 3) E1(R53'p34, 2(s,11 + 1)) 3,11(3,11 - 3)
The following proposition follows from cases (1), (2), (5), (6) of (11.2.4).
Proposition (12.1.22) Suppose that 7- = TG and 6 > 0 is a small number. Then by changing b if necessary, 26#(W)
_p-3i (S , w, w) -
27rV1
fRe(sii)=i+5
8T11)02(8T11)A1(w; ST11, 1)dsrii
Ep3,
fori = 1,2,3,4. (i) Now we combine the computations in this section. Let i) = Do, P = Po. We define Jl((D,w) =sno(w) (Eao+(Ra0Pp.,wao,2)+Eao+(9aoR30 Po,wao ,4)) (0) - XTgs# (w) ( Rao ,Ppo 2
+
`_Do Raa -Ppo (0)
1
4
- X726#(w) (ElRD2P2l1 2) + 5E1(Ra24)p22, 2)) J2 (,D, w) = (Ew+(Ra1,Dp11 , 3) + Ew+(gwRa1(Dp11 , 0))
- (Ew+(Ra1IPp12, 3) + Ew+(,FwRa14 12, 0)) 02
+
(-Ra, lpii(0) + Rat p12 (0)) ,
Ta,1(P, W) = J1(', w) + 6#(w)J2(1D), 2Tw+(Rw,oRa1
(sill + 5), 2 (1 - s,11))
2 (sT11 - 1)(ST11 + 1)
2(1 - s,.11), 2(1 - ST11))
+ J(4D e T11 )_ 4
(STL1 - 1)(sT11 + 1)
2Tw+(Rw,oRa1 ,1Dp1, 2 (sTll + 5), 2 (1 - 3,11)) (sT11 - 1)(3,11 - 3)
2Tw+(Rw,o9wRa14'p12, 2(1
+
- ST11), 2(1 - S'11))
(3,11 - 1)(s,11 - 3)
12 Contributions from unstable strata
271
W, sill) = b#(W)(J3(I, s1-11) + J4('D, STll))
7
wRD, IDpii (0) + +Z26#(w)
+ +
s,ll(sTll - 1)
r
9wRa14)p12 (0)
l
(sT11 - 1)(s,-11 --2))
S1-ll) - 13P32(4)) ST11) 6#(W)(-E 33 (ID, ST11) + E 34
(4), STll)).
Then by (12.1.4)-(12.1.6), (12.1.9), (12.1.11), (12.1.14), (12.1.16), (12.1.19), (12.1.22), and by changing 0 if necessary, Epo (4),W, w) r., CGA(w;
W)
_
2 10
+ Let sion. We define
J5(4)) =
J6(D) =
J 21iV -1 Re(s,,,)=1+6
TO,2('D, W, ST 11) 02(ST 11)Al(w; S, 11, 1)dsT11.
W,1), J3,(2)(,D,1) etc. be the i-th coefficient of the Laurent expan-
-12(Tw+(Rw,oRo,
2-1
(Dpli, 3) +Tw+(Rw,oJIwRal4)pll, 0)),
(Tw+(Rw,o " wR51'Dp11, 3) + Tw+(Rw,o9wRa14P12' 0))
Then J3,(o)(4),1) = J5 ((P) + 2
+2
Tw+(Rw,oRa1 ,Pp11, 2(81-11 + 5), 0) d
sill + 1 srll=l Tw+(Rw,o9wR,i,Dp11, 2(1 - srll), 0)
ds1-11
d
ds,.11lsill-l
J o (4) 1) = J,(4)) + 2
d
sill +1
Tw+(Rw,oRal (Pp., 2(8.11 +5),0)
I
6,11 I3'11=1
d
81-11 -3
Tw+(RwoJFwRo1'Pp121
+2 dsTll
2(1- 8,11),0)
81-11 -3
3,-11=1
By the principal part formula for the standard L-function in one variable, if 4P is K-invariant, Tw+(Rw,oRo1 4)p1,, 2 (31-11 + 5), 0) + Tw+(Rw,o-q'w Ra1 (Ppll ) 2 (1 - sT11), 0) /
El Rw
1
2
(81-11 +3))+
2
D3
Ra3Ip31 (0)
sill + 3
2g'a3 Ra3 4p32 (0)
sill + 1
Tw+(Rw,oRa1 p12, 2(s1-ll + 5), 0) +Tw+(Rw,oFwRa1 Dp12, 2(1 - STIl), 0)
=
1
2(81-11 + 3)) +
2ga3 Ra3 -Dp33 (0)
2ga3 Ra3 p34 (0)
sT11 + 3
81-11 + 1
Part IV The quartic case
272
Therefore,
J3,(o)(D,1) = JA(D) + 2 (-Ei(Rw,2) + RD3 4)p31 (O) +
3
J4,(o)(,D, 1) = J6 (,D) 8
2
2))
R (O), 02 (0)
(E1(R'w,o-Dp12, 2) + E1,(1)(Rw,o41p12, 2))
FZ'3 RD3 4Pp33 (O)'
It is easy to see that _
1
ST11(ST11 - 1)
1
--
1
(s,11 - 1)(ST11 - 2)
- 1 + O(sl - 1)
,
ST11 - 1
- 1 + O(S'11 - 1).
1
sT11 - 1
Also 1)
Ep31,(o)
Ep32,(O)(4), 1) = 1)
Ep34,(o)( ), 1)
IE1,(-1)(Re3DP3111)IE1,(o)(Ra34p31,1), 4 4 2E1,(-1)(Ra3Dp321 1)
2E1,(o)(Rlo3'Dp32, 1),
4E1,(-1)(Ra3'Dp331 1)
1 E1,(o)(Ra3-pp33, 1),
1 E1,(o)(Ra34p34, 1).
1 El,(-1)(Ra34Pp3411) 2
2
We define J7((D)
-
1)
-2
4E1,(o)(Ra3'p31) 1)
El,(-,)(Ra3'Dp32, 1) + 2 E1,(o)(Ra31)p32, 1)
+ 8 E1,(-1)(Ra3N33, 1) + 4 E1,(O)(Ra3')p33, 1)
+ 2 E1,(-l)(Rb34N3411)
6
E 1,(o)(RO31)p34, 1),
J8(4D) = 2 (-3E1,(o)(R'wo4Dp11) 2)+El,(1)(R'w,o'Dp11,2))
- 2 (3E1,(o)(Rw,Ap12, 2) + E1,(1)(Rw,01)p1212)) Clearly, ga3 R534)13i (0) = E1,(-1)(Ra3443i 11)
for all i. Also if 4 is K-invariant, by (4.4.11), `323wRa143p11(0) = E1(R'w,o'1)p12, 2), 9329wRa1
(0) =
2).
12 Contributions from unstable strata
273
Therefore, S
w, 1) = 6#(w)
Ta,2,(o)
i=5
Ji(b)
By the principal part formula (4.2.15), J2(4') + J50) + J6(4) + J7(4) = EW,ad,(0)(Ra1,Pp11, 3) -
EW,ad,(o)(Ra1,Pp12, 3).
Hence, Ta,l (4),w) + Ta,2,(o) O, w, 1)
= Ji((D,w) +6#(w)Js(4) + 6#(w) (EW,ad,(o)(Ra14)p11, 3) - Ew,ad,(0)(Ra1'DP12, 3))
By Theorem (4.0.1), we get the following proposition.
Proposition (12.1.23) T,,i (,P, w) + T,,2,(o) (D, w,1) = ba (w)ED (RD gyp, wa, 2).
§12.2 The case a = (,32) We prove the following proposition in this section.
Proposition (12.2.1) Let p = (t, s) be a path such that Z = (,32). Then by changing V) if necessary, Ep (4), w, w) - 0.
Proof. Since .$((g°)`, w) = 6°(g°, w), we only consider p such that s(1) = 0. Let g° be as in the second element of (1.1.3). Let r = (T1,-r2) be a Weyl group element
such that r1 E 02Consider the situation in (3.5.17). In this case, we choose Aa = a(-1 4, _l, 1
i,
11
31
-1),
A,2) = a(A2 2, A2, A2, .2, A2 2),
where A1, A2 E 1R+ (A 3) = 1). Then ap(Aa)ATz+P
=
`26,11+4x.12+3s.2+9
ap(A((2))(A((2))TZ+P - A2r1+2x.12-2s,2+1
Therefore, LSp,r + h in (3.5.17) is Is, 1 + Sr11 + 2sT12 - 2sT2 = 0}. Let 91,23 = ka2(pt1i µ-1t2)n2(u) be the Iwasawa decomposition of 91,23 where p E l
1[8+, tl, t2
Then t(g°)urZ+P = p 8r11+1 if or = 1, and t(g5)aTZ+P = /-,S,11+1 if
v = (2, 3).
We choose rl = 3, r2 = 2, r3 = 4 if a = 1, and r1 = -3, r2 = 5, r3 = 4
if
o, = (2, 3). Then op(AD)A'Z+Pt(9a)aTZ+Pl
= A,A-2.
If q = Re(z), q E Dar for the above choice of r. Moreover, if T2 = 1, L(r) < 22 - 4C < 4+ C since C > 100.
Part IV The quartic case
274
It is easy to see that ) acts on ZDA by multiplication by ail. Hence, by (4.1.3)(2), the condition (3.5.16) is satisfied for all v, r, and p,T (4), w, w) - 0 unless r2 = (1, 2).
If T2 = (1, 2), pT = (-1, -1, 1), (-2,1,1) or (-1, 2, 1). These points do not belong to the set {sT 11 + sr11 + 2sr12 - 2sT2 = 0}. Therefore, by (3.5.19), we can replace V) if necessary and assume that "Ep,T (4D, w, w) = 0. This proves the proposition. Q.E.D.
§12.3 The case x = (03) We prove the following proposition in this section.
Proposition (12.3.1) Let p = (x, s) be a path such that x = (/33). Then by changing 0 if necessary, 8p (4), w, w) - 0. Proof. As in §2.2, we only consider p such that s(1) = 0. For )( = (A,, a2) E R+, let ' , ' , A ; 1 , 2 ) . Let g5 = d(A)g° where g°° is as in the first element of (1.1.3). Then dgo = 2d".\1dx\2dga. Let T = (T1iT2) be a Weyl group element such that Tl E 01. Let sT be as before. Consider the substitution p = A \2. Then d(X) = a (
P - 2s,.11+5,12+3 a1 2s,2+1 = 125t h +5,12-45,2-1µ8r2+1
(12.3.2)
Also dx,\1dxA2 = dxXldxp. Let W = Ra(Pp. The functions Oz,(`w,gD),ep(ga,w) do not depend on t13,t21. So E :7p (4P, w, w) = 0 unless w2 is trivial. By (11.1.4), (11.1.5), W, W)
E
E20 1
#(w) 27r
fRe(s2)=r3>i
E1(F, ST2 + 1)AT(w; 28T2 + 1, -1, 8T2)d5T2
,2=1,(1,2)
Let 1 w,
w) =
1)A (w; 2s+ 1, -1, ST2)dsr2. JRe(s2)=r3>1 E1( , 5T2 +
As usual, we can ignore the case r2 = 1. Suppose 72 = (1, 2). If T1 = 1, (2, 3), (1,2,3), pT = (-1,-1, 1), (1, -2,1), (2, -1,1). So in all the cases, pT does not belong to the set {Sr I sill = 2sT2+1}. Therefore, by (3.5.19), Ep,T (c, w, w) - 0 Q.E.D.
§12.4 The case x = (04) Let d1(A1) = dp1(A1) and d2(A1, A2, )3) = a(A 1A21, Ai 1A2, Ai; ', A3) for )11, )(2, A3 E R+ We consider paths po = (xo, so), pll = (x1, 521), P12 = (x1, 522)
such that Do = (i34),x1 = (/34,,34,1), and 521(2) = 0,x22(2) = 1. As in previous sections, we only consider the case so(1) = 521(1) = 522(1) = 0.
12 Contributions from unstable strata
275
Since YO = Za for 0 = 01, ,=pl. (,P, w, w) is well defined for Re(w) >> 0 for i = 1, 2.
Therefore, by (3.5.9), (12.4.1)
po
w, w)
w, w) =
po+(, w, w)
+ po# (D, w, w) - Epo# (4, w, w) + P12 (D) w, w) - PHi (4), w, w).
Easy computations show the following two lemmas. Lemma (12.4.2) ep111(d2(Ai, A2, A3)) = A1, ep,21(d2(A1, A2, A3)) = Al 1.
Lemma (12.4.3) (1) opo(dl(A1)) = Al 2, Kpo1(dl(A1)) = a1 (2) orp 11 (d2(A1, A2, A3)) = A1-2 A22
o'P12 (d2(A1, A2, A3)) = a1 2a3.
w, w), 2po+(4), w, w) In (a), (b), 0 = DDo,p = po, and T = Ra(Dp. Let ga = d1(A1)ga where go is as in the first element of (11.1.3). Then dgo = 2d'AldgO. The element d, (A,) acts on ZSA by multiplication by A1. Therefore, the following proposition follows from (3.4.14) and (12.4.3)(1).
(a)
Proposition (12.4.4) Po+(4b, w, w) - 2CGA(w; p)bao (w)EDo+(Roo'po, wao, -2).
(1)
°po+(4P, w, w) " 2CGA(w; P)bao (w)Eao+(31oo Roo 41po, w0-0 , 6).
(2) (b)
"po#(41,W,w)
Consider r = (r1,1) such that 'r1 E l1. It is easy to see that d1(A1)TZ+P =
A2s.,11+s.,12-3
By (11.1.3), (11.1.4), (12.4.3)(1), (12.4.5)
Epo#(4',w,w) = E T,E`Q,
b#(w)T(0)
27r1
A,T, (w; si11, -1)
fRe(s,ii)=rj>3
6#(w)-FOT(0)
dsTu,
sill - 3
A1T1(w ;8T11,1)dT1E1
JRe(sii )=r1>2
s'11-5
Note that A1T, (w; s,11, -1) 4-1,10; 5T111 -1, -1). Proposition (12.4.6) By changing zV if necessary, (1)
=po# (4), w, w) - -CGA(w; P)T2b#(w)Rao 4'po (0),
(2)
"po# (4), w, w) - -CGA(w;
Rao,D po (0)
3
Proof. Since the proof is similar, we only prove (1). If T1 = 1 or (2, 3), we choose
r1 >> 0 in (12.4.5). Then, L1(rl, -1) << 0 if 7-1 = 1, and Ll(rl, -1) = 2 < 4 if ri = (2, 3). Therefore, we can ignore these cases. If 7.1 = (1, 2, 3), MTl (sill, -1) =
Part IV The quartic case
276
02(sr11 - 1). The point pr = (2, -1, -1) does not satisfy the condition sill = 3. Therefore, we can move the contour crossing this point. So, A1T1 (w, sr11, -1)
/ Re ca,,
C1Sr11
sill -3
rl >3
=
f
J
Re
AiT1(w, sT11, -1) =''1
1<,l <2
dsTll
81-11 - 3
- CGA(w; p)X2
If rl < 2, Ll(rl, -1) < 4. So we can ignore the first term. Therefore, this proves the proposition. Q.E.D.
w, w),-'p12 _ (c) "pll w, w). Let Z) = D1, Ti = Rat 0p1, for i = 1, 2. Let to = d2(A1, A2, A3)t °. Then dxt° _ 2dxAldx)2dxA3dxt°. By (12.4.2), (12.4.3)(2), w(t°)a12A2A3Oz,
(12.4.7) Epll (-P, w, w) = f
(,@,, t°)ep11 (t°, w)dxt°,
TA/1-k
w, w) =
Pie
f
Aj<1
w(t°)-1A A A (`12, t°)ep12 (t°, w)dxt°. 2 OZ, 3
TA/Tk Al >1 2
Let tc = A1A2A3. Then
= A4µ2 and )16A2A =
A4/,t2
Also
dxAldxA2dxA3 = dxAldx.A2dxp. The functions Oz, (Ra4)p11 , t°), f°°p1, (t°, w) etc. do not depend on t11, t21. Therefore,
these distributions are 0 unless w is trivial. Easy computations show the following lemma.
Lemma (12.4.8) (1) ep11 (11201, A2, 3))
-1
2
Tz+p - 2sr11+s.12-sr2-2 Sr 12-3+2 M s.r2-1
A3))Tz+P = A 2s,11-sr12-sr2+4A2,12-8+21Ls,-2-l.
(2) 8P12(d2(A1, "2,
Definition (12.4.9) Let 42i = (lp2i,1, lp2i,2)
where
(1) 1p11,1(ST) = ST2 + 1, 1p11,2(8T) = 8711 - 3, (2) 1p12,1(81) = sT2 + 1, 1p12,2(s1) = ST11 + ST12 - 4.
The function Oz, (Ti, t°) does not depend on A,, A2. So by the Mellin inversion w, w) is equal to
formula, EP2i
(2)
b#(w)
7r
1
2
ffte(s1)j,2) r1,'2>>0
l,sub(
i, 1P2' S)ST12)AT(w; ST1)ST12)dsTl
for i = 1, 2.
Proposition (12.4.10) By changing % if necessary, P12 (4), w,
w) - Ep11(4), w, w)
- CGA(w; p)S#(w) (- E1,(0)(' 2, 2) + El,(l)(I'1, 2) _ El,(1)(T212) 4
2
2
1
J
.
12 Contributions from unstable strata
277
Proof. We define E(1)(4', 8,1) = E1,sub(Wi, 4221 Sr11 Sr12) - E1,sub(Wi) 1P21, 5r1, sT12), E(2)(4,, 5r12)
= E(1)(C 1, sr12)
Then E1(W1, 8,12 +
E(2)( 4i, 5r12) =
1)
+ E1(W2, sr12 + 1)
sr12-3
2
and P12
(), w, w)
=6#(w)
- :Pi, (4), W, w)
)
1
--
27r,/ -l
2
B.(er1)=(T1.'2)
r'(1) (.I,, sr1)Ar(w, 5r1, s-12)ds11.
11,'2>0
We can ignore the case r2 = 1 as usual, and we assume that r2 = (1, 2). If ri = 1, (1, 2), (2, 3), Then L(r) << 0. Therefore, we can ignore these cases, and we assume
that rl = (1, 2, 3), (1, 3, 2) or (1, 3). The point pr does not satisfy the conditions sT11 = 3, srll + ST12 = 4. Therefore, by (3.6.1), we can move the contour crossing these lines by replacing z,b if necessary so that (r1, r2) = (1 + 6,1 + 6) where 6 > 0 is
a small number. If r1 = (1, 2, 3), (1, 3, 2), L(rl, r2i r2) = (2+6)(1+C) < wo = 4+C if 6 is sufficiently small. Therefore, we can ignore these cases also. Suppose that Ti = (1, 3). Then
)21
1
21rV-1 1
e(,,l)=(1+a,1+a) r1,r2>0
ST1)AT (w, 5r1, 5r12)dsrl
E(1)
)21)=(1-61,1+6)
(2) +
ST1)AT(w, ST1, ST12)dsT1
C
27r
E(2) (,D, 5T12)0(ST12)02(Sr12)A(w11, sr12, Srl2)dsT12 Re(sr12)=1+6
JRe(sj2)=i+6
27r
E(2)sr12)O(Sr12)W2(Sr12)A(wi 1, Sr12, Sr12)dSrl2,
where we choose 6, 61 > 0 small and 616-1 >> 0. We proved that E1(W1i 2) = E1(W2i 2) in §3.8. But if w is trivial, 4) is K-invariant
by assumption. Therefore, this is a consequence of (4.4.7) also. This implies that E(1)(4),1) = 0. Since 1
1
8T12 - 3
= -1 -
1(512 4
- 1) + O((sr12 - 1)2),
E(2)(4), 5,12) = E(3)(')(sr12 - 1) + O((sT12 - 1)2),
where El,(0)( 2, 2)
E1,(1)Nf 212)
E1,(1)(IF 1, 2)
4
2
2
Therefore, "a12
W, w) -
p11 (D1 w, w)
^ CGA(w;
Q.E.D.
Part IV The quartic case
278
The following proposition follows from (12.4.4), (12.4.6), (12.4.10), (3.8.10).
Proposition (12.4.11) By changing z/b if necessary, upo (4), w, w) - 2CGA(w; p)sao (w)Eao (Rao 4)po , woo, -2).
§12.5 The case r) = (/35) We prove the following proposition in this section. Proposition (12.5.1) Let p = (Z,s) be a path such that 0 = (/35). Then by changing 0 if necessary, E p (4), w, w) - 0.
Proof. As in previous sections, we only consider a path such that s(1) = 0. Let
IF = R.
Suppose that f (q) is a function of q = (q1, q2) E (A1 /kx )2. Let FO be as before. Let dxq, dxt ° be the usual measures. An easy computation shows that
f
(w(t °)f (71,33(1 °), 72,23(1 °))dxt ° = #(w) fAi
A 1 /k x)5
/kx)2
For µ =0-11, µ2, p3) E R+, we define
d(µ) =d(µ1, 92,93) =a(µ 2 1µa, u1-12
u2/13-2
,µ1iµ3 ;
t,-1,
4
)
Let to = d(p)t°. Then dxt° = dxµldxµ2dxµ3dxt°, and µ1 = 171,33(t°)I, P2 = (t°)-p = 1 1 _ µ1 µ2µ3 , K02 (t°) = p11µ3 2 and I72,23(t°)J. Let r, sT = (ST12, sT11, sT2) be as before. It is easy to see that
(12.5.2)
(t0)Tz P
16
= µl
(5,11+1)L2
(sr11-2sr12+2sr2+1)
Since OZ, (WY, t°) does not depend on /13i by (12.5.2), w(t°)ia2(to)(to)TZ-POZ, (IF,
t°)dpldµ2dxt o
fnx(A1/kX)5 S#(w)µ3
2
-S-12+s,2 -3
E2(
,
2
(sT11 - 1), S,12 T12 + 1).
By the Mellin inversion formula, 1
21rV -1 JRe(8T2)=r3 >0
_
AT(w, ST12,'ST11,
µ3 5r11
2
- S r n+s+2 - 2
AT(w;sT)dsTz
3
+ ST12 + ^ ).
Let EON , w, 5T1) = E2 (x', 2 (ST11 - 1), ST12 + 1)AT(w, ST12) ST11,
5x11
2
3
+ ST12 + -).
12 Contributions from unstable strata
279
Then w,w)
(w)E (2)
2f ///
Eo(IF,w,srl)ds,l.
7r'
ll)=crl ,r2>
rl>3, r2>1, 2r2+1>rl
We can ignore the case r2 = 1 as usual, and assume that T2 = (1, 2).
If T1 = 1, we choose 2r1 = r2 >> 0. Then L(rl, r2i -1121 + r2 + 2) = -2(rl + r2) + 2C << 0, so we can ignore this case. If r1 1, pT is not on the hyperplane + 8r12 + 2. Therefore, we can ignore the rest of the cases by (3.5.19). ST2 = Q.E.D.
§12.6 The case a = (,36) Let Do = (N6), ai = (06, B6,i) for i = 1, ... , 4. Let po = (ao, so), pll = (a1i511)412 = (a1is12) be paths such that 511(2) = 0, 512(2) = 1. As in previous sections, we only consider such paths satisfying so(1) = s11(1) = 512(1) = 0. Throughout this section, T = Rao Dpo Let dl (A 1, A2) = a(A1 1A2, A1
1-2, -1-2
2,
d2(A1, n2, A3) = CL(Al l 2 13 1, Al
-1 3L-2 11 AI 2 ),
12 l)'1-2
for A1, A2, A3 E ]R+.
Let Zao = Vl ® V2 where Vlk = Sym2k2, V2k = k. Then dl(A1i A2) acts on VIA, V2A by multiplication by A1A2,A1A2 5 respectively. Let Xao, kvl etc. be as in §11.1. (a) Epo+(,D, w, w) etc.
Let p = po, a = Do. Let ga = d, (A,, A2)g° where g° is as in the first element of (11.1.3). Then dga = 8d" A1d" A2ds°, and dl(A1i A2)-2P = A12Az 4.
Easy computations show the following lemma.
Lemma (12.6.1) (1) Opo (dl(Al, A2)) _ (2) kaol (d1(X1, A2)) _ Al 4A2 4 "115"2 (3) kao2(d1(y1, X2)) _ "13"21.
The following proposition follows from (3.5.13).
Proposition (12.6.2) po+(,', w, w - 8CGA w p)bao(w)Eao+ tY woo) Xao, -3). "po+(qb, w, w) ^' 8CGA(w; p)Sao
IF, wao , kaolXao 7).
Next, we consider Ep,st+(-D, w, w).
Let to = d2(A1iA2,A3)t°. Then dxt° = 8dxAjdxA2d)A3dxt°. Let Ha,X5 be as in §11.1. Let ga = n'(uo)t°. Then ga E Xa if and only if A3 < a(u0)-2. Let dg-.o = dxt°du0.
Part IV The quartic case
280 By (12.6.1),
Zp,st+(4),w,w) JXo nBA/Tk
w(9a)A13A21ez;o(`y,9a)4 (9a,w)d9a,
=p,st+(4),w, w)= JXDflBA/Tk w()If f (q/) is a function of q = (ql, q2) E (A1 /kx )2,
J(Al/kX)a w(t °)f (72,12(1 °), 71,33(1 o))dxt ° = 6#(w) JAl/kx)2 f (q)dxq. Therefore, these distributions are 0 unless w is trivial. We will show that we can replace fp (go, w) by eN(t°, w) in the above integrals. It is easy to see that 60p(ge,w) is the constant term with respect to Pl x B2.
Lemma (12.6.3) 3A2 1e zo
(1) =p,st+N,w, w) - b#(w) LflBA/Tk
w, w) - 6#(w)
(2)
f
Xa fBA/Tk
(W, 9)N(t°, w)dxt°duo.
Al 7A2 5eza,o (go T, 6a (9a))6ON(t°, w)dxt°duo.
Proof. Fix a compact set C C Al so that C surjects to Al/kx. Clearly, I'y2,12(t°)I = A1A2, I71,33(1°)I = AlAz 5 Therefore, there exist Schwartz-Bruhat functions 0 < %Fj E .5"(A2), 0 < W2 E 9(A) such that W (9ax) << T1(A1A2X2,12, AJA2X2,12uo)'F2(A1A2 5x1,33),
W (ea(9a)x) << pl((AJA3)-1x2,12, (A1A2)-1x2,12uo)T2((AJA2
5)_1x1,33)
for x E Za ok, tip E C. Therefore, for any N > 1, Ozn.o (W, 9a) « (A1A2 5)-N E W1()1A2x, A1A3xuo), xEkX
,9z,,,, (go IF, 9 (go)) <<
5)N
(A1A2)-lxuo) xEkX
By the consideration as in §2.1,
Iep(9a,w) - 8N(t°,w)I «EiI,T,1(90,w), where the summation is over I = (11,12) such that I1 = {1}. So by (3.4.30), there exists a slowly increasing function h(t°) and a constant 6 > 0 such that if M > wo and l >> 0,
I $p(go, w) - 8N(t°, w)I < h(t°)As
for<wo-6
12 Contributions from unstable strata
281
Let X+ = {(A1) A2, A3, u0) E R+ x A I )13 < a(uo)
2
6}-
We substitute µ = A1 A2. Then the differences of the left hand side and the right hand side of (1), (2) are bounded by constant multiples of the following integrals h(t0)A (-N-1)µ3(5N-1)A3
fx+ al
T1(µx,µxuo)d"Ald"µdu0,
1
xEkx >1
`I'1(µ-'x, µ-lxuo)d"Ald µduo.
JX+
xEkx
al<1
These integrals converge absolutely if N >> I. This proves the lemma. Q.E.D.
Definition (12.6.4) (1) lpo,st(si) = (Sill + 2si12 + 3ST2 - 9, -ST11 - 2si12 + si2 + 1, 2 (1 - STll)), (2) lpo,st(sT) = (-sill - 2sT12 - 3ST2 + 13, sTll + 2sT12 - ST2 + 3, 2(1 - ST11))' We define T5+(R5,oIF, lpo,st(si)) ST11 - 1
1po,st(s'r))
sT) _
Sill - 1
It is easy to see that (t0)TZ+P - %2sr11+Sr12+3sr2-6 `-2sr11-sr12+sr2+2 \s,-12-1.
(12.6.5)
If
2
1
3
((1, 3),1)T((1, 3), 1),
A., (w; si),
AT (w; STl21 ST11) ST2)
because z/'(-TGz) _ V)(z). Therefore, by (12.6.5), and exchanging Sill, ST12, p,st+ ('P, W, w)
_
_ =
3 1
86#(W) E (2) frEDr, r1>1 1
86#(W) T
2 7r
Ep,st+(P, ST12) 5T11, si12)Ai(w, si)dsT
3
V) f
Ep,st+(,D)
(w; s1)ds,.
Re (
rEDr, rl>1
Similarly,
(2) f 3
p,st+(,D, w, w) - 86#(W)
Ep,st+((D, Si)Ai(w; si)dsi. Re(ar)=r
rEDr, rl>1
sT) are holomorphic for Re(si11) > 1, the following Since Ep,st+(-P, si), proposition follows by the usual argument.
Part IV The quartic case
282
Proposition (12.6.6) Suppose that r = TG and 6 > 0 is a small number. Then (1)
w) W)
8026#(w)
2xY-1 JRe(sjj)=1+5
Epp,st+(4'f ST11, 1, 1)02(STll)Al(wi srll) 1)dSTl1,
(2) =po,st+(4), w, w) 8p26#(w)
post+(
Le(sii )=1+6
s T11) 1f 1 )2 T11 (s)A
f
(w s T11) 1)ds T11 1
)
(b) uPli (,P, w, w)
Let i) = D1, and Wi =
Ra145P1i for i = 1, 2. Let
32
Al = I-Y2,22(d2(Ai, A2, A3)I = A1A2A3, 12 = I-Y1,33(d2(Ai, A2, A3)I = A1A2
-4
1
5.
3
Then A2 = Al it2 5) A3 = a1 bµi µ2 and dxAldxA2dxA3 = 110 Easy computations show the following lemma.
Lemma (12.6.7)
_ 24
-d"Ald"µld"µ2.
4
(1) opll (d2(A1, A2) A3)) = Al 3A2 'A3 =32Al 1 µ1u2 2 2
(2) op12 (d2 (A,, `A22, `3A3)) = A7A2A3 = A /p1p2 5. (8,12-1) p (43,11+58,12-28,2-7) (3) Boll (d2(^) A2, ^))TZ+P = ^51 (s,11+28,2-3) Al2 µ2 (-3,11 -sr12-23,2+4) 2 (8,12-1) (4) 9P12 (d2(A1, A2f A3))TZ+P 1 µ Xµ1° (-4s,11+8,12+2s,2+1)
-
Definition (12.6.8) (1) lpii,l (sr) = z ST11 + 1),
L10
(5sr11 + 48-12 - 2ST2 + 1), ST12 + 2sT2 - 6).
(2) 1p12,1(ST) = (2(ST11 + 1),10(ST11 - 4ST12 + 2ST2 - 3))ST11 + ST12 + 2ST2 -8).
By (12.6.7), and exchanging STil, sT12,
(2)
Pli (4), w, w) = 6#2w)
3
R.(.,)=
2,sub( i) lpli) ST)AT(w, ST)dsT
rl>jr2, r3>1
for i = 1,2 where the first (resp. second) coordinate of Wi corresponds to
X2,22
(resp. X1,33)- We define 1
1
cE +'P11 (D, STll) = - E2(Ral Dpii) (Srll 6 2
LP12 (D) sill) =
E2(Ral'DP12)
2
+ 1),
1
10
(5sr11 + 3)),
(Srll + 1), !-(Srll - 5)) 2(sll - 5)
Proposition (12.6.9) Suppose that r = TG and 6 > 0 is a small number. Then by changing t/i if necessary, 26
w, w)
2x #
fRe(sii)=l+6
l
ST11, 1)dsi11
12 Contributions from unstable strata
283
fori=1,2. Proof. We can ignore the case T2 = 1 as usual, and assume that r2 = (1, 2). Let r = (rl, r2, r3) = (20, 2, 2). Then by the usual argument, 1
(2) fRe(s)=r 1
2,aub(
i,lplt+ ,)A(w;s,)ds,
2
fRe(si)=(rj,T2)
(2)7r
2,SUb(Wa) 1pli, sri, 1)AiT1 (w; sTl)dsT1.
If 7-1 = 1, (1, 2), (2, 3), we choose rl >> r2 >> 0. Then L(rl, r2, 1) << 0. Therefore, we can ignore these cases. The point p1,1 is (2, -1), (-1, 2), (1,1) for -rl = (1, 2, 3), (1, 3, 2), (1, 3). So p1T1 is not on the lines 5,12 = 4, 5sT11 + 4s,12 = 11 for these cases. Therefore, we can move the contour crossing these lines by (3.6.1) so that (ri, r2) = (1+6,1+6). Then we only have to move the contour so that (ri, r2) = (1 - 61i1 + 6) where 6, 61 > 0 are small numbers and 616-1 >> 0. Q.E.D. W, W)
(c)
Let Z = Zo,p = po, and ZZ' = c02. Let tll = .1a2. Then dxAldx.\2 = Easy computations show the following lemma.
Lemma (12.6.10) (1) d (\ \ Tz+P = 1\^ll /12) \\1
sdxAldxµl.
3(2s..ii+8-12+25-2-5)4 (-2sr11-sr12+s,.2+2)
-
1
(2) ap(d1(A1,A2)) = Al 3µl 3 (3) nD 1(dl(A1, )2)) = Al 3 /.t1 3 .
(4) tv1(dl(A,, A2))) = (A,A3)-3 = tbl 3. 5)-l = Al $tti (5) kv2(dl(Ai, A2)) = Definition (12.6.11) Let 12i = (12i, 1, 12i,2) for i = 1, 2, 3, 4 where (1) 121,1(8,) = g(-s,ll - 2sn12 + ST2 + 6), 121,2(ST) = sr11 + 28,12 + 2sr2 - 9, (2) 122,1(8,) = 3 (-ST11 - 28,12 + ST2 + 1), 122,2(ST) = ST11 + 28,12 + 2ST2 - 7, (3) 123,1(Sr) = 3 - 121,1(8,), 123,2(8,) = 121,2(8,), (4) 124,1(8,) = 3 - 122,1(8,), 124,2(5,) = 122,2(3,)-
Note that 12i(-8.11, sill + 5,12,',2) = 12i(5T11, 5,12, ST2) for i = 1, 2, 3, 4. Let W1
W,' IF,
IF, T4=Ro,9v1IY.
By (12.6.10), and exchanging ST11, sr12, S2w,12i,T('yi, w)
1=P'O"i ((.D+W 1 w) = 26#(w) T1 E!W2
*2=1,(1,2)
for i = 1,2,3,4. The following proposition follows from (11.2.3).
Part IV The quartic case
284
Proposition (12.6.12) Suppose that r = Tc and 6 > 0 is a small number. Then by changing zb if necessary, p,a,,i (41, w, w) - 2CcA(w, p)
6#(w)Ew+(W i, 12i,l(-1, 2,1)) 12i,2(-1, 2, 1)
27r+b
2 + 2P 6#(W)
fRe(srii)=1
Fw,st,sb+( 'P, 12ii
1) 1)
X Y'2(s-11)Al(w, sr11, 1)dsT11
for i = 1,2,3,4. (d) PO,03,i
W, W)
Let 0' = c03. For the cases i = 1, 2, we make the change of variable µ1 =
A1A2, µ2 = A1A2A3 = µ1A3 Then A3 = µ1 2µ2, and (12.6.13)
d2(A1,X2,A3)TZ+n
_ Al3(2sr11+sr12+2s,-2-5) Al
6(-4sr11-5sr12+2s,2+7) 1(sr12-1)
µ2
For the cases i = 3,4, we make the change of variable µi = \i)2, µ2 =
(A1.2)-'A2 3 = µ1'A3 . Then )3 = µi µz , and (12.6.14)
d2(A1,X2,X3)T=+a
(-4sr11+3712+2sr2+1) 1 (s, 12-1) g (28-11 +Sr 12+2972-5) P2 Pi6 I1
In both cases, dXA = 1dXAid"µidXµ2 It is easy to see from (12.6.1) that vp(d1(A1, )2))A3 = Al 3µi 3 µ2 for the cases 8
2
i = 1, 2, and 0 p (dl (A1, A2))A 3 = Ai 3µ1µ2 for the cases i = 3, 4.
Definition (12.6.15) Let 132 = (2 (8,11 + 1),131,2) where (1) 131,2(Sr) = (5,-11 + 2Sr12 + 2sr2 - 9)(5ST11 + 4sr12 - 2sr2 - 9), (2) 132,2(S,-) = (s,-11 + 25r12 + 2sr2 - 7)(55-11 + 45-12 - 2sr2 + 1), (3) 133,2(5,-) _ (s,-ll + 2Sr12 + 2sr2 - 9)(5.11 - 4sr12 + 2sr2 - 3), (4) 134,2(5,-) = (5-11 + 25r12 + 25-2 - 7)(5,-11 - 4sr12 + 2sr2 - 13).
Let
IF1 = Ra' Fv1 Fa'I','Y2 = R1,'Y, T3 = Ra,.`'a'IJ,'F4 = Ra,.rv1 T. By (12.6.1), (12.6.12), (12.6.13), and exchanging 8711, 8712, W, w)
_
- 66#(W) for i = 1,2,3,4.
(
\2n
1
3
/
f
Re(sr)r1>'2, r3>1
sr)AT(w, ST)dsT
12 Contributions from unstable strata
285
We define sT11)
EP31
Yyjga'1', 1(ST11 + 1)) - 6E1(Ra3 = (ST11 - 5)(5sT11 - 7)
6E1(Ra3'I,
(,`i', 7
ST11) -
EP32
2(sTll + 1))
(sTn - 3)(5sT11 + 3) 6El(Ra39-o 'I',1(sr11 + 1)) , EP33 (), ST11) (sT11 - 5)2
6E1(Ra3gyiIF, 2(sr11 + 1)) P34
,
T11 (sr11 - 3)(sT11 - 15)
The following proposition follows from cases (7), (8), (13), (14) of (11.2.4).
Proposition (12.6.16) Suppose that r = TG and 6 > 0 is a small number. Then by changing 0 if necessary, 26#(w)
(I, w, w) -
ko
21rN
1
3T11)02(3T11)A1(W, 8r11,1)dsT11
fRe(sii)=i+fi
fori = 1,2,3,4. (e) °Po,a4,i (4), w, w)
Let 0'= N. Let 112 = Al , p3 = (A,A )-3. Then dxAldxA2 = 40dx p2dx p3. Let 11 = (A2, A3), and dxµ = dxp2d"p3
Let T = (Tl, T2) be a Weyl group element such that Ti E 01. The following lemma is easy to prove and the proof is left to the reader.
Lemma (12.6.17) (1) ap(d1(A,, A2)) = 112
356
(2) ial(dl(A,,A2)) = µ21µ3 9
(3) kv1(dl(Ai,A2)) = 113 (4) kv2(dl(A1,A2)) = µz 1µ3
It is easy to see that (12.6.18)
[dl(Al, A2)
Tz+P
s,ll+s,2-3 113 1(2x,12-sr2-3)
13,12=-1 = 112
Let `Y1 = RD,goIF, W2 = Rio, 9v2IF, T3 = Ra,.Fv29a'y, W4 = RD'W. We define distributions Ep,a,,i (C w, s,.ll, 5T2)'S for i = 1,2,3,4 by the following integrals in that order. 36#(w) ,11+s,2-5113 (23,11-3,2+2)el(1y1, pdxt, µz
5
JR2 XAl/kx
-
A2!51, 113 <1
36#(w) 5
112,11+s,2-5P3(2s,11-s,2-7)e1('y2,(
1RXAh/kX
-23
µ2G1, µ3<1
36#(w) 5
112,11+8,2-4113(2x,11-s,2+7)61(W3,µ
fRxA1/kX
11 t)dxpdxt, -2-3
µ2G1, µ3>1
36#(w) 5
112,11+s,2-4113(2x,11-s,z-2)01(4,112µ3t)d"11d"t.
fRxAh/kX 112<1, 113>1
-
µ)-lt)dxpdxt,
Part IV The quartic case
286
We define w, w, ST11, sT2) _ p,a',i
F'p,a',i
w, ST11f ST2)AT(w; sill, -1f sT2)
Let s'. = (STii, ST2) and dsT = dsr11dsT2. By (12.6.17), the Mellin inversion formula, rY-1)2f
p,e',i (4), w, w) _
e(s)=(r,.r2)
27
T1 E 9]7 1
40"i
w, wf
s/T)dsITf
T2=1.(1,2)
where we choose the contour so that (rl, -1, r3) E DT, and r1 >> r3 >> 0. For A E ] and i = 1,2,3,4, we define via (x) = WYi (,fix). Clearly, 113 (resp. qf4) is the Fourier transform of 11 (resp. W2) with respect to the character < >. Therefore, for any c1i c2 E C, 1
0 0
1
1
Act-1Ej+(1f'ia-1,1 - c2)dXA
A`1E1+('Fi+2a,c2)dxA+1 0
1 C1 - C2
E 1 (Ti+2A, C2) +
'Pi+2(0)
_
q1i(0) (Cl - 1)(c2 -
C1C2
1).
Hence,
p,a',1
w, 8r11, ST2) + p,a',3 w, sill, ST2) = b#(w) E1(W3, 5 (2ST11 - ST2 + 7)) sri1 + 2sT2 - 9
3(0)
+3(J#(w)
(ST11 + ST2 - 4)(2sT11 - ST2 +
7)'
q/ (0)
- 36#(w)(ST11 + sT2 - 5)(2ST11 - 8T2 + 2)' p,a',2
w, ST11, ST2) + Ep,a',4 (4, w, ST11, ST2)
E1( 4, 1(2sT11 - sT2 - 2)) ST11 + 2sT2 - 6
'1'4(0) (ST11 + ST2 - 4)(2sT11 - 8-,2 -
- 3b#(w)
2)'
gi2(0) sT2 - 7).
(ST11 + ST2 - 5)(2sr11
Proposition (12.6.19) By changing 0 if necessary, (1)
w, w) + "p,a',3 (1), w, w) ^ -CGA(w; p)
2d#(w)E1(Ra' IF1, -1) 5
+ 3CGA(w; P)b#(w) (2)
ap a ,2 ($f w, w) +
p,a 4 ('Dww)
--CGA(w, P)
(_T3(0) 10
+'F
100)
IZ26#(w)E 1 (Ra,T4 , - g1) 2
- 3CGA(w; p)b#(w) (w4(0)
- _28(0)
I .
12 Contributions from unstable strata
287
Proof. As usual, we can ignore the case T2 = 1, and assume that r2 = (1, 2). By the usual argument, 1
27rV---i P
f
w, w, sT) + Ep,a',i+2 (4', w, w, ST)) dsT
e(sr)=(rl,r3)
Res (Fip,a',i (4p, w, w, sT) + Ep,a',i+2 ((D) w, w, sT)) dsT11 J e(s,ll)=r1 sr2=1
21r
for i = 1, 2. The poles of the function ST11, ST2) +
ST11) 81-2)18
2=1
for i = 1, 2 are si11 = 2, 2, 3, 4, 6, 7. But pi, does not satisfy these conditions. Therefore, by (3.6.1), we can move the contour so that r1 = 1 + 6 where S > 0 is a small number. Then L(1 + 8,1,1) < wo if ri = 1 or (2, 3). So we can ignore these cases. Suppose 7.1 = (1, 2, 3). Then by the usual argument, g
27r\ -1
Res (Ep,a',i J e(srll)=1+b a*z=1
^' CiGA(wi P)932
w) w) sT) + Fp,a',i+2 w) sT
w, sT) + Ep,a',i+2
w, w, sT)) ds111
=(2+1)
Q.E.D.
(f) Now we combine the computations in this section. Let Z = 00, p = po We define J1(4), w) = 8ba(w) (Ea+('1`)w,,X,-3)+Ea+('a4')wa 1,ka11Xa 1,7)) +'x126#(w) 8#2w)
J2(4', w)
(_E1(R4, -1) + 2E1(Ra4'Y)-5))
(Ew+(Rw 62
+ 8# (w)
)
3)
3
1
(Ew+(R2I' ) - 3) + Ew+(JlwRa2'1',
+3'2128# (w)
+3'x128#(w)
(D4R4W(o)
3
))
R04 ."DWY(0))
+
8
10
(gNRa4 -q 10 aW(0)
10
+ R14 'F(O)
TO,1(1))w) = Ji('P,w)+J2(4),w), J3(`) s,11) _
8Ta+(Ra,o'P, sill - 41 -sT11) 2 (1 - ST11)) ST11 - I
8Ta+(Ra,o-'a'I', 8 - si11) sTil + 4, 2(1 - ST11))
+
J,4 (4k 8T.l )_
ST11 - 1
2Tw+(Rw,oRa2ga'Y) 1(s,-11 + 4), 2 (1 -s,11 ))
)
(sill - 1)(sTll - 5)
+ 27'w+(RwogwRa2. 'a'Y) 3(5 - sill)) 2(1 - ST11)) (si11 - 1)(sii - 5)
Part IV The quartic case
288
J (ffi)- - - 2Tw+(Rw,oRa2'y, -1, a (1 - sr11)) s
(sr11 - 1)(sT11 - 3)
2 (1 - sTii)) (sT11 - 1)(sT11 - 3) 5 Ta,2(-D)w,ST11)
i=3
4 6#(w)\-1)z+1F'V3i
+
sT11) i=1
The following proposition follows from (9.1.10), (12.6.2), (12.6.6), (12.6.9), (12.6.12), (12.6.16), (12.6.19).
Proposition (12.6.20) Suppose that r = rG and S > 0 is a small number. Then by changing zb if necessary,
O (4i, w, w) -CGA(w; p)To,l (,D, w) N
+
27w-1 Re(srll)=1+6
Ts,2(4),w, sT11)02(sT11)A1(w; ST11, 1)dsT11
Let Ta,2,(i) N, w, 1) etc. be the i-th coefficient of the Laurent expansion. We define
J6(4i) = -4 (Ta+(Ra,o1F, -3, -1) + Ta+(Ra,oFF'I', 7,5)),
J7(f) =
4
4
(TW
5
owR52W, 3) + Tw+(Rw,o9sRa2`y, 3)
-1(
3-))-
2
It is easy to see that d
J3,(o)(c,1) = J6(4)) + 8
+8
d
dsr11 S.11=1
J4,(o)(4),1) = J7(4)) + +2
+2
To+(R1,o'I', sr11 - 4, -sr11) 0)
d d
d dsr11
J5,(o)(41,1) = J8(4)) - 2
To+(Ra,o9a1W,8-sr11,sr11+4)0)),
3(5 - s,-11), 0)
d
dsrii
sr11 - 5
s+11=1
Tw+(Rw,oRa2 ga `I', (8rii + 4), 0) s,11-5s sr11=1 d dsT11 sr11=1
d
- 2dsr11 Is+11=1
,
Tw+(Rw,oRa2'I'1 - iLu 3 ) 0)
s,,11-3 s (si11 + 9), 0)
si11 - 3
12 Contributions from unstable strata
289
The distributions To+(RD,o'I', s111 - 4, -sr11, 0), Tt+(Ro,o_IIo'1', 8 - 5rii, `ST11 + 4, 0)
are equal to the following integrals A ''1-5A2 `11-3e2(R',o',AlA241, A1A2
'Rx(A' /kX)2 al>1
ai-sr11 AZr11+1@2(R1,o
JRx(A1/kX)2
"o'y, 1A241, A1A2
542)d" A1dxA2d
a1>1
where the first (resp. second) coordinate of R',oxF,R',o.3'0' corresponds to x2,12 (resp. x1,33). For IF E .Y(Za oA), we define ,Fo,p'1'(x2,12, X1,33) =
JA, (y2,12, Y1,33) < x2,12y2,12 + X1,301,33 > 42,1241,33-
Then if (b is K-invariant, R',ogoW = 9,',oR',oY. For W E 9'(Za oA), let R1, R2 be the restrictions to the first coordinate and the second coordinate respectively. Also let (12.6.21)
FMx2,12) = f 'y (y2,12, 0) < X2,12Y2,12 > dy2,12, A
F2'y(x1,33) = f i (0, y1,33) < x1,33y1,33 > dy1,33 A
Lemma (12.6.22) Suppose that 4D is K-invariant. Then 8To+(Ro,o'I', 5.u - 4, -sr11, 0) + 8To+(Ro,oA-o'1', 8 - sr11, sr11 + 4,0) = E2(R',o'1'+ 2- 2 (s,11 - 7), 2(5.11 - 3))
2E1(R1Ra,oAoIF,3(3.11+1)) 2E1(92Rao 0IF,s(s.ll+6)) 8Th - 9 3.11 - 5 2E1(FiRa,o'I', 3(s.11 + 6)) 2E1(R2Ra0o'1`, 5(srll + 3)) + s,11-7 S,11-3 + Proof. By (6.1.2), e2(R',o'I', A1A241,
542) - Al 2A2e2('` a,0Ra,0W, Al 1A2 391 1, Al 1A242 1)
_ A1 1Aa01(F1R',o9o'1', A1A291) + a1 2A201(R2R',o` oW, A11'X242 1)
- O1(RlRa0o'I', a1a241) - A11A581(92R',o'f','1 1Az42 1)
=A 2A2e1(R1R',o9oW,'l 1'2 3q1 1) + Al 1A2 391(92R',o.o'It, A1A2 542) A-'A 3e1(91R',o'I', A'A2 3q1 -1) - e1(R2Ra,o'I', A1A2 5q2),
-
Part IV The quartic case
290
We divide the integral according as A2 > 1 or A2 < 1. If A2 > 1, we use the first equation, and if A2 < 1, we use the second equation. Then (12.6.22) is equal to the summation of 1 1 E2(Ra,o'y, 2 (s1-u - 7), 2 (8,11 - 3))
and the integral of the following function . sj'3 -6 asT1 -7
-81-3 +2)
El+
1Ra,o
(
,
E1+ (R2Ra,o9a)a1 1'
-s1-5 - 11
(
+ ai'31-5
J
-sin - 3al
E1+
-s.-
,
___ -6
+2
5
81-7
11
A31T1
s'11
E1+ ((R1Ra o
3
s,11+6)
_____6
1
E1+
5
+
_11-3 -6 E1
5 I
s113+ 6\
91R' T
sT15+3\
over Al < 1.
By the principal part formula for the standard L-function in one variable, _8,11-6
-81-11 +2)
E1+
3
3
s 11-7
+
A1T3
E1+ (R1Ra o9a1')a 1
81-13+ 1)
J
4er1 ] -20
81-13+ 1)
E1
(R1Ra 0'Qa 3 A r" 691Ra 0. 5 '(6)
-8r11+2
+
A1,1-7R1Ra 0g5W(0)
+
8r11+1
-sr 5 - 1) srl1-6 1
+
5
8r11 + 6l
E1+
S
4sr1, -36
Al
5
/
(2R,0w,81-15+6)
E1
girl'
Airy'-7R2Ra ova (6) 81-11 + 1
+
69J2Ra ST11 +6
12 Contributions from unstable strata A81'11-5
291
-sT 1-3
Ei+ (RlR,o'')A1
J ST113+ 6)
A4s,11-12 1
sT11 +6 E1
3
3
As-"-SRiR'
A1iR',o'y(o)
oW(0)
+
s,11+3
sTii +6
-sT5 +2)
5
s,ll-5 1
+
E1+
+
srii +3
(R2Ra,o`I'),\,
A4sr11-28
i
l/
E1 (R2Ra o'M, sT11 5
5
+3
5
o'y(o) +
A
A1r1l-5R2Ra oy(0)
-s,,11+2
sT11 + 3
is the Fourier transform of Note that standard bilinear form on A etc.
with respect to the
If 4) is K-invariant, 1Ra,0JF,1@(0) = 92Ra,oY(0),
R1Ra,o9a''(0), RiRa,o'I'(0) = R2Ra,o'I'(0),
o(0). Now (12.6.22) easily follows from these considerations. Q.E.D.
If P is K-invariant, by the principal part formula for the standard L-function in one variable, Tw+(Rw,oJFwRa2 JFa iY, 3 (5 - sy11), 0) + Tw+(Rw,oRa2 9-a`y,
3
(8T1i +4),0)
+ 1)) + 3E1,(-1)(Ra3YwRa2.F'a'I',1) = E1(R1Ra,o9a'Y, 1(s-r11 3 2 - sill
+ 3E1,(-i)(Ra3ga'I',1)
3
8,11 + 1
Tw+(Rw,oRa2'I', 1
= E1(J91Ra,o`I',
3
s
0) + Tw+(Rw,ogwRa2 `F, 3 (srii + 9), 0)
(s,11 + 6)) +
3E1,(-1)(Ra3W,1)
s,11+3
,
3E,,(-,)(R D39wRD2 'F' 1) ST11 +6
Part IV The quartic case
292
Note that if oD is K-invariant, 5(ST11 + 6)) = Res E2(Ra, (P12) S') S'=1
--(Sr11 + 1)),
5
5(Sru + 3)) = Res E2(Ra,,kp,,, s', 5(s,11+3)).
E1(R2Ra,o4',
Therefore, 9
Ji,(0)(b) _ i=3
Ji(.D) i=6
-
E
1, 5)
5) +
E2,(-l,o)(Rs, p,2' 1, -5)
32
- 3El,(-1)(Ra3IF,1) + 98 15
E1,(-1)(Ra3..WRa2'y,1)
15
+
3E1,(-1)(Ra3-gra4',1) -
E1,(-1)(Ra39wRa2ga'y,1),
8
where 2E2,(l,o)(R'a OWY, -3, -1) + 1 E2,(o,1)(Ra,OWY, -3, -1).
Also,
1, 5) - 6E2,(o,o)(Ra,4'p,,5), Ep,,,(0)(41,1) = - 1E2,(-1,1)(Ra,Dp,,) 6 40E2,(-1,1)(Ra,4Dp,211) --)
EP12,(0)(D,1) =
- 8E2,(o,o)(Ra,
16E2,(-1,o)(Rol iDP121 1, -5)
11 -5),
EP31,(0)(C 1) =38 El,(-1)(Ra3J"wR02ga`I',1) + 4E1,(o)(Ra:,-FWRa2g0I',1),
EP32,(0)(4,1) =3E1,(-1)(Ra34',1) - 3El,(o)(R53W,1), 3
Ep33,(o)(p, 1) _ 8E1,(-1)(Raa.
3
W, 1) + 8 E 1,(0)(Ra3 ga'1', 1),
E133,(0)(b, 1) =42 E1,(-1)(R039wRa2'y, 1) +
4 E1,(o)(Ra39wRa2'1', 1).
We define 5) - 8E2,(o,o)(Ra,4p12, 1, -5)
+ 1E2,(-1,1)(Ra,4tp,,) 1,4)
40E2,(-1,1)(Ra,obp,2'1,-5
- 8E2,(-l,o)(Ra,4)p,,,1, 5) + 3E2,(-l)o)(Rs, P12,11-5),
12 Contributions from unstable strata 9
3
J11(I)) =16
+
293
E1,(-1) (Ra3 J'a `I', 1) + 8 E1,(o) (Ra3 -11'a T, 1) 9E1,(-1)(Ra3J"'wRa2.a`I',1) + 3E1,(o)(Ro39w Ro2goT,1) 4
4
- 2 E1,(-1) (Ra3 W,1) +8 E 1,(o) (Ra3 IF, 1) - 98E1'(-1)(R039wRa2I1`,1) - 14 E1,(o)(Ro3.` wRa2W,1). By the above considerations, we get 11
JJ(-D)
Ta,2,(o)(4),w,1) = 6#(w) -s
By the principal part formula (4.2.15), J2(4)) + J7(4)) + J8 (,D) +
Ew,ad(Ra2 IF, - 3) -
-Ew,ad(Ra29a IF, 3 ).
Therefore, Ta,l (b, w) + Ta,2,(o) (4), w,1)
= J1(4, w) + 6#(w)(Js('D) + JA (f) + +6#(W)(EW,ad(Ra2IF,-3) - 2EW,ad(Ra29 W,
3))
By (6.3.11), we get the following proposition.
Proposition (12.6.23) Ta,1('', W) + Ta,2,(o) (4),w) = 86a (W)Ea,ad(RaIp, wa, Xa, -3) + 6#(w)J9((D)
§12.7 The case t = (37) Let p = (0, s) be a path where 0 = (37). As in previous sections, we only consider such path such that s(1) = 0. Let W = Ra,Dp throughout this section. Suppose that f (q) is a function of q = (ql, q2, q3) E (Al /kx )3 An easy consideration shows that
I
A1/kx)5
w(1 0)f (71,33(1 °), 72,13(1 °), 72,22(1 0))dxlo
= 6a(w) f
A1 1kX)3
w(q)f(q)dxq.
Let
d(p) =d(µ1, µ2, 93) =
-3
z , µ2
1
µ2 µ3 u2
Let to = d(µ)1°. Then dxto = sdxµdxl0 , and Al = 171,33(t°)I, µ2 = I72,13(t°)I, µ3 = h2,22(t°)I
2
)
Part IV The quartic case
294
Let r, sT be as before. Easy computations show the following lemma.
Lemma (12.7.1) (1) K02(t°) = P1 2 µ2 213 2. (srll+sr12+2)A2 (srii+2sr2+3)µ3 (srii+3sr12+2sr2+6)
(2) (to)-'-P = p1
Definition (12.7.2) Let lp = (lp,1i lp,2i lp,3i 1) where 2(ST11 + ST12 - 1), lp,2(Sr) = 3(srll + 2ST2 - 3), lp,3(ST) = 6(Sr11 + lp,l(ST) 3s112 + 2ST2 - 3).
By (12.7.1),
(12.7.3) f
w(to)IGa2(to)(to)TZ-peZ, (,I, to)dxto =
ba(w) 6
TA /Tk
LetD={rIr1+r2>3,rl+2r3>4, s + 2 + 3 >
( ap (`I', w, w) = 6a (w) E
1
'-
By(12.7.3),
3
27r v"- 1
T
2}.
E3,sub(1I') lp, wa, s.)
JRe(s)=r
E3,sub(WI', lp, LID, ST)AT (wi ST)dsT,
where we choose the contour so that r E D fl DT. We define Ta,2(4>5 w, ST11) =
ba (w) STll E3(`I', WO) , 1 6 2 3
(S,11+2)). (Srll - 1), 1 6
Proposition (12.7.4) Suppose that r = TG and 6 > 0 is a small number. Then by changing V) if necessary,
P26a(w) f
U), W)
27r
Ta,2((D,w,ST11)02(STl0Al(w,8T11,1)ds,11.
Re(s,ii)=1+6
Proof. We can ignore the case r2 = 1 as usual, and assume that rl = (1, 2). The pole of E3,sub(`P, lp, wa, Sr) in the set is, I Re(srl) = (6, 3), Re(s12) > 0} is Sr2 = 1. So 1
27
e(sr)=(6,3,2)
)3j
1
F+3,sub(W, lp, wa, ST)AT(w, s1)ds1
27r-1 +0
E3,sub('I')lp, wa, s1)AT(wi ST)ds,
e(sr)=(6,3,2)
)21 E3,sub(lI, lp, wa, sr1, 1)A1T1(w; s11)ds1l.
(2) 1
Since C > 100, Z(6,3, < 18 + C < 4 + C. Therefore, we can ignore the first 2) a term. If rl = 1, (1, 2), or (1, 3), we choose r1i r2 >> 0. If Ti = (1, 2, 3), we fix r1 = z and choose r2 >> 0. If rl = (1, 3, 2), we fix r2 = z and choose rl >> 0. For these cases, L(rl, r2, 1) < wo. Therefore, we only have to consider the case r = TG.
12 Contributions from unstable strata
295
The pole structure of E3,sub('F, lp, wD, sTl,1) is as follows.
The point plr, is not on the lines sill = 4, 3sr12 + Srl1 = 1, 7, 8T12 + 8T11 = 3. Therefore, by (1.2.1), we can move the contour crossing these lines so that (ri, r2) = (1+6,1+S). Then we only have to move the contour so that Re(sT1) _ (1+6,1-61) where S, 6l > 0 are small and 616-1 >> 0. Since E3,sub(W, lp, wD, Sr11) 1, 1) = T0,2
w, ST11),
the proposition follows. Q.E.D. Let T8,2,(i) (b, w, 1) be the i-th coefficient of the Laurent expansion. The following proposition is clear.
Proposition (12.7.5) 2,0, 2)
T0,2,(o)(4,w)
+ +
S(w1)6(w2) 4
S(wl)S(w2) 12
1
E3,(l,-i,o)(`1', (w2, 1, w2), Z, 0, 1
1
2 1
E3,(o,-l,l)(F,(w2,1,W2), 2,0, 2).
§12.8 The case 0 = (,138)
Let Do = (08), and Di = (,38,,38,i) for i = 1, ... , 6. Let Po = (Do, 5o). Let Pil = (Zli, 5il ),Pit = Oi, si2) where sil (2) = 0, si2(2) = 1 for i = 1,2,3. As in previous sections, we only consider such paths such that so(1) = sil(1) = Si2(1) = 0. Let' = RDo4)po throughout this section. Let -1-2i l dl(Al,.A2) = a(A1-4A2 2 2) -1-2,
-13-2,
Cl2(A1, A2, A3) = Cl(A1
4-2 2) -1-2-3 1, -i'2L13i
a1A22).
Part IV The quartic case
296
Let Z-0 = V1 ® V2 where Vlk = Sym2k2, V2k = k2. Then d1 (A1, A2) acts on VlA, V2A by multiplication by A1 A2, A1A2 3 respectively. Let Xa,'w1 etc. be as in §11.1. (a) Epo (41, w, w)
Let ga = dl(A1, X2)ga where ga is as in the second element of (11.1.3). It is easy to see that dga = 14d
Let r, sT be as before. An easy computation shows that (12.8.1)
d2(Al,A2,X3)
TZ+P
2s*11+4sr12+33,-2-9X23,-11+23x12-23,-2-183"' -1 a
= Al
Easy computations show the following lemma.
Lemma (12.8.2) (1) o,,. (d, (A,, A2)) Al A2 A15A26. (2) Kao1(d1(A1, X2)) = A121 (3) kao2(d1(A,, A2)) _
Since Za'k = Zak, the following proposition follows from (3.5.13) and (12.8.1).
Proposition (12.8.3) (1)
=p,,+ (1), w, w) - 14CGA(w; p)6,0 (w)Eao+('1', wao, Xao, -3),
(2)
=p,)+ ((P, w, w) - 14CGA(w; P)bao (w)Eao+(9ao'I', wao , Kaa1x
(b) '
, 8).
p1t (1D, w, w)
Let Z = D1. In (b)-(d), t° = d2(A1,'2, A3)t °. Let µl = I'Y1,33(t°)I = A1A2 A3 , µ2 = I72,12(t°)I = A1A23A-'. 3
Then A2 = Al µl 2µ2
A3 = X1 2µiµ2 It is easy to see that
d"t° =
7dxAjdx121dx122d"to.
Easy computations show the following two lemmas.
Lemma (12.8.4) (1) ep;1(d2(X1, A2, A3)) = A1, ep:2 (d2(A1, A2, A3)) = Al 1 for i = 1, 2, 3.
Lemma (12.8.5) (1) ap11 (d2(A1, A2, A3)) = Al 4A A3 = 121 2 (2) ap12 (d2 (A,, A2, A3)) = A1A2A3 = A IL1 2122 5 A3))TZ+P= \73,-12-7 s.ll-S,-12+S,-2-1 3x11-23,-12+2sT2-1. (3) O,,, (d2(A1, A2, 122 121 1 (4) eplz
A2, 3)) Tz+P
3s,-11+2x,-12-2s,-2-3 = 1-78T11-73,12+14 2s,-11+s,-12-s,-2-2 122 121
Definition (12.8.6) Letlpli =(lpli,1ilp1i,2,lp1i,3) where (1) 1p11,1(ST) = -ST11 + ST12 + ST2 - 2, lpi,,2(ST) = -2ST11 + ST12 + 2sT2 4, 1p11,3(ST) = ST11 - 1,
(2) 'p12,1(ST) = ST11+2ST12-ST2-4, lp12,2(ST) = 2s111+3sT12-2sT2-8, lp12,3(s1) = ST11
+ ST12 - 4.
12 Contributions from unstable strata
297
By (12.8.1), (12.8.4), (12.8.5), and exchanging sr11, sr12, pli(4, W, w)
3
- 6#(W)
1
27rV
R.(-)=r
1
_ E2,sub(Ral-Ppli) lpii, sT)A'r(wi sT)dsr,
'1>'2, r3>1
where the first (resp. second) coordinate of R51 4)p1i corresponds to x1,33 (resp. x2,12) We define E2(Ra1,Dpii, -51-11, -2sr11 - 1) Sr11 - 1
Srll)
Ep11
where the first (resp. second) coordinate of Rat qDpll corresponds to x1,33 (resp. x2,12)
Proposition (12.8.7) Suppose that T = TG and 6 > 0 is a small number. Then by changing z/i if necessary, (1) =p11 (p, W, W) V (2) =p12
10 26#(W)
271-/
,
/Epi1 Re(s.,11)=1+6
(D, srll)02(sr11)A1(wi 8,11, 1)dsr11,
6#(W)£'2(Ro1'Pp12, -2, -5)
w, w) ^' -CGA(w; P)
2
where in (2) the first (resp. second) coordinate of Rat dopii corresponds to x1,33 (resp. x2,12).
Proof. The proof of this proposition is similar to (12.6.9). The only difference is that we have to check that the points (2, -1), (-1,2), (1, 1) are not on the lines 5T11-5712 = 1, 2, ST11-2ST12 = 2, 3, 25r11+5T12 = 6, 3Sr11+25x12 = 10, 11, 5x11 = 3
in order to use (3.6.1). Q.E.D.
(c) 'P2i ('P, W, w)
Let Z = 02. Let Al = I71,23(t°)I = .\i-4, µ2 = I72,13(t°)I =.1)23A3.
dxto = 7dxAldxµldxµ2dxto. 2
Easy computations show the following lemma.
Lemma (12.8.8) (1) up., (d2(\1, A2, \3)) = Al 4A2 2 = al 2µl 2 (2) up. (d2 (A,, )2, .\3)) = A1 (3) OP21 (d2(A1, A2, A3))
TZ+P = A z (5,.12+s, 2-2) z (2sr11+S, 12-8.1-2-2) 31-11-1 P2 1 N1
Part IV The quartic case
298
(4) Bp22 (d2 (A,, A2, A3)) T2+p == Al
2(-ST11 -8,12-S,-2+3) 21 (3,-11-3,12+3,2-1)
Al
3,-11-1
µ2
Definition (12.8.9) Let 42i = (lp2i,li 42i,2i lp2i,3) where (1) lpzi,1(ST) = 2 (2ST11 + ST12 -sT2 - 3)f lp21,2(ST) = ST11 - 1) lp21,3(sr) = S,-12+
S,2-3, (2) lpzz,l(ST) = z(ST11 - ST12 + ST2 - 1)f lpzz,2(ST) =sill - 1, lp2z,3(ST) = ST11 5T12 + ST2 - 5-
By (12.8.8), and a similar argument as in previous sections,
)1
F(
_
3
1
= 6#(W)
lp2i'sr)AT(w; sr)dsr,
>'2, ,-) 1
T3>1
where the first (resp. second) coordinate of Ra2 4DP2i corresponds to x1,23 (resp. X2,13) for i = 1, 2
Let 4)(21 E 9'(A) be a function defined by the formula 4)p21 (x1,23) = f W(O) x1,23, x1,33, 0, x2,13)dx1,33dx2,13
If 4) is K-invariant, D(1) is the Fourier transform of e22 with respect to the character < >. We define 4)p22 similarly for Ra21DP22
We define 1
Epzi (4)f sT11) = E2(Ra2 (bp21) 2 (2sr11 EP2 2 1
-3),s,-11 - 1),
(,D, sTll) = r12(Ra2 ADp21 I ST11 - 1, ST11 - 1)) E2(Ra24)V2z) 2(sT11 - 1))sT11 - 1)
E P22 (4)7 ST11) =
sill -3
Let '.217' = {(1, 2, 3), (1, 3, 2), (1, 3)}.
Proposition (12.8.10) Suppose that T = TG and 6 > 0 is a small number. Then by changing V if necessary, (1)
:P21(rf WI w)
-
26# w
()
27rVw-1 JRe(sii)=1+b 6#
+ T'=(T1 .(1.2))
Ep21 (D, ST11)02(ST11)A1(w; sf11, 1)dsr11
W
2
fRe(s,11)=ri
21r V'
X121 ((b) ST'11)A17i (w) sr'11, 2)dsT'11
,-1 E'ID'
E1('Dp211 22.
06#(W)
+
T'=('-
1,2)) ,-i EW
2
1
fRe(s,12)=r2
5T'12 - 2
A1T1(w) 2, ST)12)d8r'12,
12 Contributions from unstable strata
299
(2) ''P22('P)W)w)fRe(sii)=1 X26#(W)
() s711 )2(sT11)A1(w) sll) 1)ds
11
+6 P221
+
2
2
A171(w;2,s7'12)ds7i12
rl en,
Proof. We can ignore the case r2 = 1 as usual, and assume that T2 = (1, 2). Let r = (rl, r2i r3) = (10, 2, 2). By the usual argument,
(2) f5= 1
-p
5u(2
2,T2) 1
P2,' 1P2i) 8T )AT (w; s7)ds7
'r
2
fRe(si)=(ri
E2,sub(R52
P2i)1P2iIsr1,1)A171(w;sr1)dsrl
fori=1,2. The pole structure of E2,5Ub (R52 -DP2i )1P2ti) 871,1) for i = 1, 2 is as follows.
(1+S,1+S)
H
am.
If Tl = 1, (1, 2), (2, 3), we choose rl >> r2 >> 0. Then L(rl, r2, 1) << 0. So we can ignore these cases. The point pi, is (2, -1), (-1, 2), (1,1) for Tl = (1, 2, 3), (1, 3, 2), (1, 3). The point p171 is not on the lines 2s711 + ST12 = 4, 6, srll = sr12 + 2, 8711 + 5712 = 4 for these cases. So we can move the contour crossing these lines by (3.6.1). We move the contour to Re(sr1) = (1+6,1+6) for i = 1, and Re(s71) = (1+6,1+ s ) for i = 2. In this process, we have to cross the lines s711 = 2, s712 = 2, and the condition of (3.6.1) is not satisfied for these lines. When we cross these lines, we pick up the third term of (1), and the second term of (2). If Tl = (1, 2, 3), (1, 3, 2), 1
(2)7r V
2
-1
JRe(si)(i+6,1+6)
E2,SUb(Ra2'1 P21 ) 1P 211
5T1)1)A1r, (w; ST1)dsT1 - 0,
Part IV The quartic case
300
because L(1 + 6,1 + 6, 1) < wo for these cases. Similarly, we can ignore the cases Ti = (1, 2, 3), (1, 3, 2) for i = 2. Then we only have to move the contour so that (ri, r2) = (1 + 6,1- 61) where 61 > 0 is a small number such that 616-1 >> 0. This proves the proposition. Q.E.D.
(d) yp3i (,D, w, w)
Let 0 = 03. Let Al = I71,33(t6)I =.1)1 23, 'X /22 = Iry2,13(t6)I = A1A23A3-
Then A2 = .1i µr /22 g, A3 = "1 °/2 °/22 It is easy to see that dxto = ?dx.Xldxµld"122dxto. 5
Easy computations show the following lemma.
Lemma (12.8.11)
-21 4 2 2 (1) 0p31 (d2(A1, A2, A3)) _ Al 3A 22 A 3 =A 1 5/21 /25
(2) op32(d2(A1,A2,A3)) = AgA A3 = 11/21 µ (2sr11+8.12-8,2-2) Tz+{)-- A (8T31+38,12+2sr2-6)i (3) Op., (d2(A1, A2, A3)) Xµ2
(s,l1 -28,12+2sr2 -1)
(4) ep32 (d2N, A2, A3))
5(8,11-8,12+8,2-1)
Tz+ F
= All -28,11-38,12-2sr2+7) Al
Xµ2 (38,11+28,12-28,2-3)
Definition (12.8.12) 43i = (lp3i,1, Ip3i,2i lp3i,3) where (1) 1p31,1(Sr) = 5(2ST11 + ST12 - 8r2 + 2), 1p31,2(Sr) = 1(s,11 - 2sr12 + 2ST2 + 1), 1p31,3(ST) = si11 + 381-12 + 2ST2 - 9, ST2 + 4), lp32,2 (S r) = 5 (3sT11 + 2s112 - 2ST2 -3 ), (2) 432,1(ST) = 5 (81-11 - 51-12 1p32,3(ST) = 281-11 + 381-12 + 2ST2 - 12.
By (12.8.11), and a similar argument as in previous sections, w, w)
=p3i
3
6#(w) T
( 21rV1
1
)
Re(sr)=r
F12,SU11(Ra3
b13i 1
lp3i,
_ ST)AT(w; ST)dsT,
where the first (resp. second) coordinate of Raqbp3i corresponds to x1,33 (resp. X2,23)
for i = 1, 2. Let
srll) _
Ep32
F.,2(Ra341p32, 1(s111 + 4), b (3sT11 - 3))
28,11-7
Proposition (12.8.13) Suppose that T = TG and 6 > 0 is a small number. Then by changing 0 if necessary, PM ($, w, w) - -CGA(w; p) ` P32
w, w) '.,
2 6#(w)E2(Ra3'p31, 45 , 5)
3
X26#(w) 27f
fe(s,11)=1+6
Ep32 (-F, s111)02(s111)A1(w; sT11, 1)dsrij.
12 Contributions from unstable strata
301
Proof. The proof of this proposition is similar to (12.6.9). The only difference is that we have to check that the points (2, -1), (-1, 2), (1, 1) are not on lines s,11 = 2sr12+2, sill+3sr12 = 7, sT11+2s,12 = 4, 2sr11+3sr12 = 10, 3sT11+2s,12 = 10 in order to use (3.6.1). Q.E.D.
(e)
(b, w, w)
Let0=cDo,a'=c04,andp=po Let 9a = dl(A1) Az)dga, and µl = A1A 2. Then dg, = 2d
A 1'µi, and dl(A1) A2)-.+p = A1
(12.8.14)
(6.11+26.12+2s+2-5)
(5,ii+26,12-2sT2-1)
Easy computations show the following lemma.
Lemma (12.8.15) (1) vpl (d1(A1, A2)) = A 3A = Al 2µi I
_2
(2) Ka1(dl(A1, A2)) = A15A2 6 = Al 2µl 2
- /213. (4) Kv2(dl(A1,A2)) = (A1A = Al 2µi Definition (12.8.16) Let l4i = (14i,1,14i,2) where (A1A2)-3
(3) kvl(dl(A1,A2)) =
3)_2
(1) 141,1(8,) = 4(6,11 + 28,12 - 2sT2 + 7), 141,2(6,) = sr11 + 26,12 + 28T2 - 9, (2) 142,1(6,) = 4(sT11 + 2ST12 - 2sT2 + 1), 142,2(6,) = 8,11 + 28,12 + 2ST2 - 7, (3) 143,1(ST) = 3 - 141,1(8,), 143,2(ST) = 141,2(6,),
(4) 144,1(5,) = 3 - 142,1(6,)) 144,2(6,) = 142,2(8,).
Note that 14i(-ST11, S,11 + ST12, 8,2) = 142(6,11) 6,121 s,2) for i = 1, 2, 3, 4. Let
IF1=R14-Fv1 'e W,' 1'2=RD,T, `f'3=R54', 'F4=R149v1q,. By (12.8.14), (12.8.15), and a similar argument as before, =p,aI,i (,P, w, w) = 26#(w)
S2W,141,r(Pi, W)
for i = 1, 2,3,4. The following proposition follows from (11.2.3).
Proposition (12.8.17) Suppose that r = TG and 6 > 0 is a small number. Then by changing 0 if necessary, w) w)
2Ew+( i,14i,1(-1, 2,1)) 141,2(-1, 2, 1)
+ 2v2b#(w) 27fV
1
f
Re(s,jj)=1+6
CGA(w; p) 1W,st,sub-{-(Ti, 14i) sill, 1, 1 ) X 02(sTll)Al(w) sT11, 1)dsT11
for i = 1, 2, 3, 4.
Part IV The quartic case
302
w, w)
(f) up,a5,i
Let 0' = c05. Let to be as in (b). When we consider the cases i = 1, 2, we make the change of variable p, = A1A2, P2 = a,a2a3 = P1A3. Then (srii-1)
A1(s.,ii+2sriz+2srz-5)P
d3(,)--+P =
(12.8.18)
K v, (d2(A1, A2, A3)) etc. are the same as in (12.8.15). When we consider the cases i = 3,4, we make the change of variable P1 (A1a2)-'A3 2
A1A2, P2 =
(12.8.19)
2
= Pll\3. Then
dg(A)TZ+P = Al
4(sri,+2sriz+2srz-5) 16(3s,ii+2s,12-2S,2-3) )
Pi
P2
i(srii-1)
It is easy to see that 2
i = 1, 2,
r A l 2 Pl 2 P2
Qp(d2(A1,A2,A3))A 3 = Sl
(12.8.20)
r
3
i=3,4.
A 12P192
In both cases, dxto = 4d>A1d)p1d>P2dxto Definition (12.8.21) Let l5i = (2(s:1i + 1),15i,2) where (1) 151,1(8,) _ (ST11 + 28,12 + 2ST2 - 9)(8-11 - 26,12 + 28-2 - 5), (2) 152,1(8-) = (s-11 + 25:12 + 2ST2 - 7)(8,11 - 25:12 + 25,2 + 1), (3) 153,1(5,) _ (8,11 + 2sT12 + 28T2 - 9)(3s,11 + 28,12 - 2sT2 - 3), (4) 154,1(5,) = (5-11 + 2s-12 + 25,2 - 7)(38,11 + 25-12 - 2sT2 - 9).
Let Wi be as in (e). Then by (12.8.18)-(12.8.20), and a similar argument as before, w, w) \113
415#(w) 1
/
E1,sub('Fx, 152, 5:)A:(w, S,)dST r3>1, *1 Ar2
for i = 1,2,3,4. We define
Em"i (4), 5,11) -
4E1(R,5
4E1(RD,IF, 2(s-11 + 1 ))
Ep ,a5,2
(s-u - 3)(5:11 + 1) 4E1(Ra5.
p
s
E p,55,4
2(5,11 + 1))
(s-11 -5)2
'I',
2(5,11 + 1))
3(Srll - 5)(s,11 - 1)
(
,
8 ,11
)=
4E1(R,S9y,IF, 3
2(5:11 + 1))
(s,ll - 3 ) 2
The following proposition follows from cases (9), (10), (15), (16) of (11.2.4).
12 Contributions from unstable strata
303
Proposition (12.8.22) Suppose that T = TG and b > 0 is a small number. Then by changing 0 if necessary,
27rb
026#(W) OD, W, w)
JRe(s,-ji)=1+ V()S111)dST11
for i = 1, 2, 3, 4. (g) "P,a6,i (`b, W, W)
Let to be as in (b). We make the change of variable µi = A142, {12 = (A1A2 3)-'A3 = Al 1a2 A3 1
-.1
for the cases i = 1, 2. Then A3 = AI µ'1 ' 92, and (12.8.23)
d3(A)Tz+P = Al
(S,ii+s,iz+s,z-3)µi
-12
We make the change of variable µ1 = A1A2, µ2 = A1A23A3 = A1A23A3 1
4
for the cases i = 3, 4. Then A3 = Al ' µ1 11'2, and d3(A)Tz+P = A (s,i2+s,z-2)i (2s,11+s,12-e,z-2)µ2,h-1
(12.8.24)
It is easy to see that ( 12.8.25 )
o p(d2( 1, A2, A3))A 3 =
A
i=1 '2
2 2
-7 2 1
2
µ1µ'2
,
i = 3,4.
In both cases dxto = 7d)Aid1µldxµ2d1t1 Definition (12.8.26) Let 16i = (STii + 1,16i,2) where (1) 161,1(ST) = (ST11 + ST12 + ST2 - 4)(ST11 - ST12 + ST2 + 4), (2) 162,1(ST) = (ST11 + ST12 + ST2 - 4)(ST11 - ST12 + ST2 - 2), (3) 163,1(ST) = (ST12 + ST2 - 4)(2s,11 + ST12 - ST2 - 4), (4) 164,1(ST) = (ST12 + ST2 - 4)(2ST11 + ST12 - ST2 + 2).
Let
'113 = Ra69v29aIF, T4 = Ra6T.
IF1 = Rasga`y, W2 =
Then by (12.8.23)-(12.8.25), and a similar argument as before,
MO (41), W, W) 3
f
E1,sub('Pi, 16i, S)AT(w; S,r)dsr 2b#(w) E (2) f Re(,-)=r »''2. T3>2
for i = 1,2,3,4.
Part IV The quartic case
304
Proposition (12.8.27) By changing 0 if necessary, p,ae,i (1), w, w) - 2CGA(w; p)6#(w)E1,sub('Fi, 16i, 1, 1, 1)
for i = 1,2,3,4. Proof. We can ignore the case T2 = 1 as usual, and assume that r2 = 1. Let r = (ri, r2i r3) = (10, 5, 3). By the usual argument,
)31
1
27r V -1
(27r 1 0
31
V- 1
e(sr)=r
El,sub( i, 16i, s -)AT(w; Sr)C1Sr
) JRe(sri)=(ri,r2) E1,sub(Wi, 16i, Srl, 1)Alrl (w; sr1)dsT1. 2
The points (2, -1), (-1, 2), (1, 1) are not on the lines srii + Sr12 = 3, Sr12 = 5111 = Sr12 + 1, 2srii + Sr12 = 5. Therefore, by (3.6.1), we can move the
contour crossing these lines so that (rl, r2) = (1 + S,1 + S). Since El,sub(%Fi, lsi, sr) is holomorphic at (1, 1, 1), the proposition follows by a similar argument as before. Q.E.D.
(h) Now we combine the computations in this section. Let 0 = Do, p = P0. We define Jl(,D, w) = 148, (w) (Ea+(W, Loo, Xo, -3) + Eo+(-Fo'I', Lob 1, rIai Xa 1 8))
-
b#2w)
E2(Ro,'p12, -2, -5) +
6#3W)
E2(Ro.Dp31 5, ), 5
J2('D) = -2Ew+(Ro,-91o4', 1) - 1
2)
+ Ew+(Ro4AF, 2) + Ew+(JFwRa4IF, 2)
+
E1(J"a6Rae_R'a11,2) 2
+
E1(Ra6W,2) 4
2
- 2 El (RD, 9a IF, 2) -
RD, IF, 2),
Ta,l ((D, W) = J1 OD, w) + 6#(w)J2(4'),
2Tw+(Rw,oRo4-'a', 4(5 - sr11), z(1 - 8111)) (8111 - 1)(s,11 - 5) 2Tw+(Rw,o9wRa4g04', 4(8111 + 7), 2(1 - srll)) (sr11 - 1)(s111 - 5) + 2Tw+(Rw,oRa4`y, 4(8111 + 1), 2(1 - sril)) J4 (a
,
S r11 )
(8111 - 1)(srll - 3)
2Tw+(Rw,0 wRo4`W,4(11-8111),2(1-8111)) (Sr11 - 1)(sr11 - 3) TD,2 (D, w, 3,11) = b#(w)(-Ep11(4', sri1) + EP(12)1 (b, srii))
+ 6#(w)(Ep22 ('P, Srii) + Ep32 (4, Srii)) + 6#(w)(J3(4, 8111) + J4(D, 8111)) 4
+ 6#(W)
(-1)i+1Ep,oe,i (b, Sr11). i=1
12 Contributions from unstable strata
305
Since the third term of (12.8.10)(1) and the second term of (12.8.10)(2) cancel out, we get the following proposition by (9.1.10), (12.8.3), (12.8.7), (12.8.10), (12.8.13), (12.8.17), (12.8.22), (12.8.27).
Proposition (12.8.28) Suppose that 7 = TG and 6 > 0 is a small number. Then by changing 0 if necessary, Po (4, W, w)
,., Ta,1(,D,w)CGA(w; p)
+
P
2
27rV -1
f
_ e (srll)=1+6
ra,2(4), W, sTii)ni(w; sTii,1)dsT11
s#(W) f J
+
1
l z)) 27rV
2
Re01r1 l)=r2
A )"'1Ti (w; sr'11, 2)ds T'11
p21
r1 EM,
i=1,2
Note that we have proved in Chapter 10 that Mp;j (4), w, w) is well defined for
i,j=1,2and pi2(4),W,w)-0fori=1,2. Let Ta,2,(i) ((b, w,1) etc. be the i-th coefficient of the Laurent expansion as before.
It is easy to see that E2,(1,o)(R51,Ppll1 -1, -3) - 2E2,(o,1)(Ro1 cpil' -1, -3),
Ep11,(o)(1',1) Ep21 (o)
(4), 1) =E2,(1,-1) (Ro2 0P21 1 - 2, 0) + E2,(o,0) (Ra2 Dp211 -
Ep22,(0)(4),1)
2 2,(o,0)(Ra2'DP22, E 0,
0)_ 1
2, 0),
E2,(-l,o)(Ra24'P22, 0, 0)
4E2,(o,-1)(Ra2'P22,0,0) - 4E2,(-1,-1)(Ra2'DP22'01 0) E2,(1,-1)(Ro2'P22, 0, 0) -
Ep32,(0)(4), 1) _ -
-E2,(o,o) (Ra3
,1, 0) 5E2,(0,-1)(Ra34DP32'
15 E
2,(1,-1)
(Ra3 41) p32 ,
5
E2,(-1,1)(Ro2(DP22, 0, 0),
E2,(-1,0) (Ra3 4)P32
1,0) 1,
E
,1, 0)
52,(-1,-1)(Ra34DP32'
0) - 5 E 2,(-1,1) (Ra3
P32'
1 ' 0 ))
We define
4(Tw+(Rwo9wRN "o'p,2)+Tw+(RwoRa4go IF,1)), J6(4,)
2 (Tw+(Rw,oRa4'I', 2) + Tw+(Rw,oJ'wRo4'I',
2
))
1,0) .
Part IV The quartic case
306
Then
J3,(o)
,
1) =J5
2
4(srll + 7),0)
d I
dsr11
sr11 -5
=1
d Tw+(RwoRo4 JFo lF, (5 - sr11), 0) +2 sr11 -54 dsr11 Is.,,,=1 I
Tw+(Rw oRa4 `y , (sll + 1) , 0)
d
J4,(o)(4),1) =J6(41) - 2dsrll
,
srll - 43 Tw+(RwogwR04q,4(11-srll),0) d - 2dsru s,11=1 sr11 -3 s,
=1
Easy computations show that
= 4E1,(-1)(RD,
fa `y, 4 (srll + 7),0) + Tw+(Rw,oR04 -1'-O'Y, 4 (5 - srll ), 0) o%F,1) + 4E1,(-1)(Ro5JFwRo4JF,`F,1) sr11 +3
1-s- ll
- RessE2(Ra2,Dp22,
Tw+(Rw,oR04 +,
1
4(1 - sr11), s'),
1(sr11 + 1), 0) + Tw+(Rw,o. 'wRa4 WY, 4 (11 - srll), 0)
4E(Ra5'F,1)
4E1,(-1)(Ra59wR04W,1) 7-s,11
s,11-3 - Ress E2 (Ra2 4)p,,,
1
4
(sr11 - 3), s').
It is easy to see that
- Res
4(1
- srll),s) + 4E1,(-1) Ra5 ,a D 1) Ill
_ -E2,(o,-l)
+ O((srll -
(Ra24)p22, 0,
0) +
4E2,(o,-l)(Re2Dp220, 0)(srll - 1)
1))2).
Therefore,
J3,(o) (4),1) = J s (4) + 8 E2,(0,-1) (Ra2 DP22, 0, 0) - 8 E2,(1,-1) (Ra2 41) p22, 0, 0),
J4,(o)(4),1) = J6(4)) -
2
E2,(o,-l) (Ra2(l)p21, -
2
,
0) - 4E2,(l,-l)(Ra2Dp21, - 2 , 0) 4
4
- 2E1,(-1)(Ra5`T',1) +
4E1,(-l)(Rb59WRa4W,1)
12 Contributions from unstable strata
307
Also it is easy to see that 1
1
4E1,(-1)(Ra5
Ep,a5,1,(0)
4E1,(o)(Ra5
wRa4
Ep,a5,2,(0) (4),1) = E1,(o)(Ra5 `F,1),
Ep,a5,3,(0)(b,1) = 124E1,(-1)(RaS"''a`I',1) - 1 12 E1,(o)(Ra59a1y,1)
Ep,a5,4,(0)(4',1) = 3(RD,9wRa4I`,1) +
3E1,(o)(Ra59WRa4W,1).
We define 3
1
J7(P)
1
4E2,(1,-1)(Ra24p21,-2,0)
=E2,(o,o)(Ra2'Dp21,-2, 0)+
- 2E2,(o,-1)(RaAp21) -1- ,0) - 2E2,(o,0)(Ra2")P22,0,0)-
8E2,(1,-1)(R02"P22,0,0)
-5 - 5 E2,(o,o)(Ra3 'Dp32 ,1, 0) -
2E2,(-1,1)(Ra24'p22,0,0)-
1E2,(o,-1)(Ra24'p22,0,0)
8
0E2,(-1,0)(Ra2")P22,0,0) -
20
J8(f)
4E2,(-1,-1)(Ra2')p22,0,0)
20 E
2,(1,-1) (Ra3 4)p32, 1, 0)
E2,(O,-1) (Ra3 4)p32 , 1, O),
=14E1,(o)(RaS,FwRa4ga'',1)
1
+ 1E1,(-1)(RD,9wRa4gbW,1)
+ E1,(o)(Ra5 W,1) - 2E1,(-1)(Ra5'y,1)
-
2El,(-1)(RD,9wRa41Y,1).
1E1,(o)(Ra59WRa4IF,1)-
If 4) is K-invariant, E2,(-1,-1)(Ra2qDp22) 0, 0) = E1,(-1)(Ra5
z
,1),
E2,(i,-1)(Ra3'PP32) 1, 0) = -E1,(i)(Ra59a'P,1), E2,(-1,i)(Ra34'p32, 1, 0) = -E2,(-1,i)(Ra2'bp22, 010)
for all i. Therefore, these considerations show that T ,2,(0)(4), W,1) =6#(W)(J7(D) + J8(P) - 1 E1,(1)(Ra5,Fa`I',1))
+
-1, -3)
+ 2b#(W)E2,(0,1)(Ra1 (Pp11, -1, -3).
Part IV The quartic case
308
By the principal part formula (4.2.15) for E W,ad,(o) (Ra4'a IF, 1) etc.,
J2(D) + Js(D) = -2EW,ad,(o)(Ra4. aIF,1) + EW,ad(Ra4IF, 2
Note that by (4.4.7), if 4D is K-invariant,
E1(go,Ra69aW',2) ='b2"aF (0), E1(Ra6 `I', 2) = T2----wR14 A'F(0) EI(Ra69a`Y, 2) = tZ729WR14'Y(0) EI(9a6Ra61Y, 2) = 932T(0). So
T0,1('P,w) +Ta,2,(o)(D,w) =J1(4),w) +6#(W)(J7(f) - 4E1,(1)(Ra59a`y,1)) 6# (W)
2
EW,ad,(0)(Ra4'q'aW,1)
+ 6#(W)EW,ad(Ra4'I', 2 ) + 6#(W)E2,(1,o)(Ra1 lbpii, -1, -3) + 26#(W)E2,(o,l)(Ra1,Dp11, -1, -3)).
By (7.3.7), we get the following proposition.
Proposition (12.8.29) Ta, l (,D, w) + Ta,2,(o) (OD, w) =146a (w) Es,(o) (Ra 4Dp'b, wa, Xa, -3)
+ 6#(w)E2,(l,o)(Ra, 4)p11, -1, -3)
+ 26#(w)E2,(o,l)(Ra1Ip11, -1, -3).
§12.9 The case 0 = (09) Let p = (z, s) be a path such that a = (,39). As in previous sections, we only consider such path such that s(1) = 0. Let IP = Ra-Dp throughout this section. Let 223' be as in§12.8. Suppose that f (q) is a function of q = (qi, q2, q3) E (A' /kI)'. An easy consideration shows that w(t 0)f 0'1,23(1 0), 72,13(1 0), -Y2,22(t
0))dxt 0
=6a (w) fA' /kx)3 wa(t0)f(q)dxq. Let 1
d(µ) = d(µ1,µ2, µ3) = a(p 11%2 31,33 ,
-i3 AA , A1µ2 µ2-23
2
16
3,
µ3
µ2
0 ).
12 Contributions from unstable strata
309
Let to = d(µ)1°. Then dxto = 3dxµdxto. It is easy to see that 91 = 171,23(t°)I, 92 = 172,13(0°)I, P3 = I72,22(t°)I
Let T, sr be as before. Then r-02 (to) =µ1Sµ2'µ3 3, and
d(µ)" =
µlrii+sri2+2µ2 (3x.,11+2sr12+2sr2+7)µ3(s,.12+s,.2+2) 3
Definition (12.9.2) Let 1p = (lp,li 1p,2, 1p,3i 1) where 1
lp,l(Sr) = srll + Sr12 - 3, lp,2(sr) = 3 (3sr11 + 2sr12 + 2Sr2 - 9), lp,3(Sr) =
3
(8r12 + 8r2)
Let D = {r I r1+r2 > 4, rl+2(32+r3) -3, r 3r > 1}. Then a similar consideration as in §12.7 shows that
(2)
W ,W)
7rV
3
1
_
fRe(s)=r
E3,sb(
W,sr)Ar*sr)dsr,
where we choose r from D (1 Dr. As usual, we can ignore the case T2 = 1, and assume that T2 = (1, 2). The point (3, 3, 2) belongs to the domain D (1 Dr. By moving the contour to Re(sr) = (3, 3, ), a 1
)
3
'-
(,-,)=(3,3,2)
C2 7r V- 1
E3,sub(W, lp, Wa, Sr)Ar(w, Sr)dsr
3
1
2
+
E3,sub(T, 1p, Wa, sr)Ar(w; sr)dsr
Re(s*)=(3,3,
(2)7r
27r
2
)
fRe(si)=(3,3)
E3,sub(, lp, Wa, Srl, 1)nlrl(wi srl)dSr1.
Since C > 100, Lr (3, 3, _< 12 + C < 4 + C. Therefore, we can ignore the first 2) z term.
If Ti = 1, (1, 2) or (2, 3), we choose rl, r2 >> 0. Then L1(r1i r2) < 0 for these cases. The function E3,sub(W, 1p) We, Sri, 1) has the following pole structure.
Part IV The quartic case
310
The points (-1, 2), (1, 1) are not on any of the lines except for the line sT12 = 2. Therefore, by the passing principle (3.6.1), we can move the contour from the point (rl, r2) = (3, 3) to (1 + 6,1 + 6) where 6 > 0 is a small number crossing other lines. We define Cpl) E .9(A) by the integral '(x1,23, x2,13, x2,22)dx2,22
(x1,23, x2,13) = J
(Dp
A
Let f (C W, ST'11) = E2('D p
1
, sr'll - 1, s -'11 - 1)A,,,, (w; sT'11, 2),
and r = (1 + 6,1 + 6). Then by the above consideration,
P6a(w )
2
1
r
fRe(srj)=r E3,sb( , lp) Wi, s-rl, 1)Ai 1(w; .rl)d- j
2
P6#(W)
+
w, s
fRe(s11)=ri>1
*'=(,-' (1,z))
11)dS
'll
T1' Effi'
If r1 = (1, 2, 3), (1, 3, 2), L1(1 + 6,1 + 6) < 4. Therefore, we can ignore these cases in the first term. Since the function E3,sub(F, 1P, WD, S,1, 1) is holomorphic at
s,j = (1, 1), we get the following proposition.
Proposition (12.9.3) By changing V) if necessary, (4), w, w) ^' CGA(w; P)
6a
3W) E3(RaDp, Wa,
e6#(W)
+
*'=(+1,(l,z)) 27r V -1 ,. E'm'
-1, -3, 3)
f Re(sr'ii)=ri>1
f
w, sr'11)dsT'11.
12 Contributions from unstable strata
311
§12.10 The case Z = (,31o) In this section, 00 = (31o), 01 = (/31o, /310,1), 02 = (010, /310,2), a3 = (010, /310,3), and 04 = (31o, 810,2, N10,2,1). Let po = (ao, so), pll = (al, 511), p12 = (Z1, 512), p21 = ('02,s21),P22 = 02,522)) p31 = (03,531),P32 = (D3,s32) where (5il(2),si2(2)) = (0,1)
for i = 1,
Let Ni = 04,540 for i = 1,
, 4.
, 4 where (54i(2), 54i(3)) =
(0, 0), (0,1), (1, 0), (1,1) for i = 1,2,3,4 in that order. As in previous sections, we only consider such paths such that po(1) = 0 etc. Let tY = Ra04Dp0 throughout this section. Let di (A,) = dp2(Ai), d2(Al, A2) = a(A1 2, A1, Ali
d3(Al, A2, A3) =
A2), A
-A-',
Easy computations show the following two lemmas.
Lemma (12.10.1) (1) epi, l (d3(A1, )'2, X3)) = A1, epi21(d3(Al, A2, A3)) = Al 1 for i = 1, 3. (2) ep2,1(d2(A1,A2)) = A1, ep221(d2(A,,A2)) = A11. (3) epa11(d3(Al, A2, A3)) = A1, epa,2(d3(A1, A2, A3)) = A A2. epa21(d3(Al, A2, A3)) =All, ep4,2(d3(Al, A2, A3)) = (A A2)-1 1 (5) (5) epasl (d3(Al, A2, A3)) _ A1 epa12(d0l, A2, A3)) = A A2. (a1A2)-1
(6) epa41(d3(Al,A2,A3) _ A1, epa12(d3(Ai,A2,A3)) =
Lemma (12.10.2) (1) opo(d1(Al)) = A6 (2) op11 (d3(A1, A2, A3)) = Qp12 (d3(A1, A2, A3)) = A A2. (3) ap21 (d2(Al, A2)) = O'p22 (d2(Al,, A2)) = A1A2.
/
(4) 0P31(d3(A1,A2,A3)) = 0P32(d3(A1,A2,A3)) = Al A2.
(5) opal (d3(Ai, A2, A3)) = o 43 (d3(A1, A2, A3)) = A?A2A3. (6) o 42 (d3(Al, A2, A3)) =ttapama (d3(Al, A2, A3)) = A2A3. Let Pi = {po,p21,p22J, P2 = {P11,P12}, T3 = 431,P32,P41,P42,P43,P44} The
paths Po, p21, p22, p4i for i = 1,2,3,4 satisfy Condition (3.4.16)(1). The paths P11412 satisfy Condition (3.4.16)(2). For the paths p = P31432, =p ((D, w, w) = 0. Therefore, by (3.5.9), (12.10.3)
EP(='p+(41,w,w)+C.p+(-D, w, w))
Ep0E 0(-',w,w) = PET,
+ E EP 0-'P# (.D, W, W) - =P# (41, W, W)) PEP,
+ pE12
+ PEP3
and all the distributions in the above formula are well defined for Re(w) >> 0.
Part IV The quartic case
312
(a) _po(,D,W,w)
Let a = Do,p = po. Let ao = dl(A1)g°° where go E M°A. Then dg-0 = 2d
Proposition (12.10.4) so+(4,, w, w) - CGA(w; P)bao (w)Eao+(IF, Loa., 3).
(1)
w, w) - CGA(w; P)boo (w)Eao+(,Foo T, wao , 3).
(2)
Let to = d3(.X1, )'2,.\3)t °, and ga = n"(uo)t°nB2 (u4) for uo, u4 E A. It is easy to see that dxt° = 2dxA1dxA2dx.3dxto Let dg, = A32dxt°duodu4. Clearly,
olp(g,,w) is the constant term with respect to P2. Let OON(t°, w) be the constant term with respect to the Borel subgroup as before. We show that we can replace .'p (go, w) by N (to, w).
Lemma (12.10.5) w(t°)ap(to)8Zn,o ('F, 9o)eN(t°, w)dga
(1) Ep,st+("D, W, w) r., f Xa fBA/Tk
(2) =p,st+(ID,w,w) -
f
A1>1
(9a))8N(to,w)dgo.
X1 n2A/Tk Al<1
Proof. Consider (1). By the consideration as in §2.1, -'p (9a, w) - 4°0(t°, w)I < E 4,r,1(t°, w),
where the summation is over I = (Il, 12) such that Il = {2} or 12 = {1}. By (3.4.30), there exist a slowly increasing function h(t°) and a constant b > 0 such that if M > Wo, l1i 12 >> 0,
1,,1(t°, w) << h(t°)(A2 + A3 )
for wo - 6 < Re(w) < M. We choose Schwartz-Bruhat functions 0 < 411 E 9(A), 0 < W12 E 9(A2) so that
fI
o
(IF, 9o)Idu4
I'Y2,33(t°)I-1e1(W1,'Y1,33(t°)) E 'F2(_Y2,22(t°)x, y2,22(t°)xuo). xEkx
Then by (1.2.6), for any N > 1, 11i 12 >> 0, the difference of the left hand side and the right hand side of (1) is bounded by a constant multiple of the following integral
f
h'(t°)(AA 'A 3)-N(A + A
)
TAO /Tk xA
Al>1, a2<
a(up)-1
T 2('2,22(1°)x, -Y2,22(t°)xuo)dxt°duo,
X
xEkx
12 Contributions from unstable strata
313
where h'(t°) is a slowly increasing function independent of N, 11, 12.
Let p = ai,2A-2. Then 3 (A2A2
2)-N (A1 +
12)
= ANA
4N((/-'AI
2X2)11
12).
+ A3
If N >> 11, 12 >> 0, the above integral converges absolutely. This proves (1). Since ea(d3(Al, A2,,A3)) = d3(Al 1,.2iA3), and the integral is for Al < 1, the proof of (2) is similar. Q.E.D.
Let T, sT be as before. Let p = A2A3 2. Then (12.10.6) (t0)Tz-p = AsT1+2sT12+3A3.2+1As,11+1 = )S-11+2s.12+3/,,S.2 +1A8.11+25,2+3 l 2 1 3 1 3
Definition (12.10.7) (1) lst,l(ST) = (2 (ST11 + 2s112 + 1), Sr2,
2
(-sill - 2sr2 + 3).
(2) 13t,1(ST) = 21(7 - STll - 2ST12)) ST2) 2 (-sT11 - 2ST2 + 3)).
(3) 13t,2(sT) = lst,2(sT) = sTll + 2ST2 - 3.
We define Ep,st+(OD) W, ST) = 8a,5t(w)
wf ST) = sa,st(w)
7a+ (Ro,o xF, wa,st ) lst, l (s,) ) 13t,2(8T) 1 7a+(RaoJFaW, wa,st, lst,2(ST)) ,
lst,2(ST)
Since I12,33(t°)I = A A2A2 = A2/tA3, by (12.10.6) and a similar consideration as before, _
3
s)A(w;s)ds,
p,st+(4>,w,w) - E (2) JRe(s)=r T
w, w) r TE T
( (2)7f
3
1
__
p,st+(, w, s)A(w, s)ds
fRe(s)=r
,
where we choose the contour so that r E DT, r1 + r2 > 3. w, 5r1, 1) are holomorphic for Re(sr11) > 1, Since Ep w, STI) 1), we get the following proposition by the usual argument.
Proposition (12.10.8) Suppose that T = TG and 8 > 0 is a small number. Then (1)
W, W)
P28a,5t(w) Ep,st+( >w) sTll, 1, 1)02(sT71)A1(w srll, 1)dTll) 27rVr-1 Je(sr11)=1+b (2)
P28a,st(w)
fRe(s,,)=1+5
() w
1 1)2(s II )AI (w sTII1)ds f II' f
)
Part IV The quartic case
314
(b) Epo#(,P,W,w),
Let r = (Ti, (1,2)). Let sT be as before. It is easy to see that
d 1( A 1) Tz+P ]9x11=-1,9rz=1 = 29r12-4 fcao(dl(A1)) is as in (a). By (11.1.4), (11.1.5), (12.10.2), and the Mellin inversion formula,
upo# (
, W,
b#(W) IY ( 0 )
w) =
Z-1
[
-(T1.(1,2))
A1T1 (w; -1, ST12)
2ir/
-
dST2,
ST12 + 1
e(9r1z)-T2> 2
+i E'ID2
('D' W, w)
PO
=
A11( w,1,ST12)
S,12-5
fRe(s .1 z)=''2>2
=(,(12)) r1 E'N2
ds2.
Proposition (12.10.9) By changing t/' if necessary, qj2S#(W)T(0)
po#(4', w, w) - CGA(w; P)
3
CGA(w; p) Z2d#(W)3
po# ((D, w, w)
,
so
(0)
Proof. If 7-1 = 1, (1, 2), we choose r2 >> 0, and we can ignore these cases. Suppose Tl = (1, 3, 2). Then 9rz+1, 9rx-4 are holomorphic at p1T1 = (-1, 2). Therefore, the proposition follows from the passing principle (3.6.1) and the usual argument. Q.E.D.
w, w) P12 Let 0 = (l l. Let t° = d3(A1, A2, A40 . We make the change of variables
(C) ":'P11 (), W, w),
2
2
/11 = I'y1,33(t°)I = A1 A21A3, /A2 = I72,23(t°)I = ).X2.
Then it is easy to see that dxt° = dxAjdXµldxp2dxt°. Also A4A2 = Aiµ2. Easy computations show the following lemma.
Lemma (12.10.1 0) (1) ep11
\ A2, 13))TZ+P = -9T31+23x12-2sT2+1 1
(2) 9P12
A2,
2 (9*11-1)
A3))Tz+P = A_38.11_2312_28.2+7 i
P22
(9x11+2sT2-3)
(9x11+2sT2-3)
Definition (12.10.11) Let 42i = (2 (8111 - 1), 2(ST11 + 2S,-2 - 1), lpxi 3) where (1) lp11,3(ST) = -ST11 + 2ST12 - 28T2 + 3. (2) 1p12,3(S1) = 3ST11 + 2ST12 + 2sT2 - 9.
By (12.10.1), (12.10.2), (12.10.10) and a similar consideration as before, P. (', W, w) 3
b#(W)
1 T
2 -7r
_ F+2,sub(Ra1,Dp1., lpli, s1)AT(w; s1)dST,
12 Contributions from unstable strata
315
where we choose the contour so that r E DT, r1 > 3, rl + 2r3 > 3, and lpl; (r) > 0 for i = 1, 2. Also the first (resp. second) coordinate of Ra1,kp1i corresponds to x1,33 (resp. X2,23). We define E p11 (
f
E2(R61Dp11,
8T11) =
2(8'11 - 1), (8,11 + 1)) f
3 -sill
/ / _ E2(Ra4)p12 f 2(s'11 - 1), 2(s'11 + 1))
Ep12 OD, 3'11) -
3sT11 -5
Proposition (12.10.12) Suppose that r = TG and 6 > 0 is a small number. Then by changing V) if necessary, 926#(W)
(1) "p11 ('D,Wf w) ,v (2)
p12
(Df W f w) `
I
Ep11
5'11, 1)d3T11,
926# (W) Epi2 27rV 1 fRe(s111)=1+6
sT11)I2(sT11)A1(wf s'11f 1)dsT11
21rV -1
Re(srll)=1+6
Proof. The proof of this proposition is similar to (12.6.9). The only difference is that we have to check that the points (2, -1), (-1, 2), (1, 1) are not on lines sT11 = 2sr12 + 1, 2sr12 + 3sT11 = 7 in order to use (3.6.1). Q.E.D.
(d)
P21+(1'1W f w), '='p21+(4), w, w) etc.
In (d), (e), 0 = (l2. Let ga = d2 (A,, A2)90. Then dga = 2d' Ald' A2dga . Let A _ d2(A1, A2), and dxA = dxAid)
P22+(4)1L"1W)1
P22+
Lo' W)
are equal to
IRxM
W (9a )A1A2oz, (Ra P21 f 9a)"'p21 (go, w)d" Adga, /Msk
Al <1, ),2 X2>1
W (9a )A210z, (ga Ro 4)p22 f ea (9a) )p21 (9a, w)d" Adga f
IRxMA/Msk a1<1, a1 a2<1
fRXM/Mk
w(9a)-'A 6A2ez,
(Ra4)p21
f go)-422 (go, w)dx Adga f
X1>1, ).1)2>1
fRxM
w(9a)-1A 'ez,
(9aRa'Dp22f 00
/Msk
al>1, X2P2:51
respectively. Easy computations show the following lemma.
(90Ap22(901
w)d"Adgaf
Part IV The quartic case
316
Lemma (12.10.13)
-
Tz+p = As,"' +2s,12-2s,2-1µs,2-1 (1) Bp21 (d2(A1) A2)) )2))Tz+p = Al1 ,11-2x,12-2sr2+5µs,2-1. (2) OP22 (d2()1,
Let
Definition (12.10.14) Let 102, = (lp2i,l) lp2i,2), 1P2i = (lp2i,l,lp2i,2) be linear functions for i = 1, 2 where (1) 1p21,1(ST) = ST2 + 1, 1p21,2(Sr) = 3r11 + 2ST12 - 2Sr2 + 1, (2) 1p21,1(Sr) = 3 - lp21,1(ST), 1p21,2 (ST) = lp21,2(ST), (3) lp22,1(ST) = ST2 + 1, lp22,2(ST) = sT11 + 2sT12 + 2ST2 - 7, (4) 1022,1(ST) = 3 - 1P22,1(Sr)1 1022,2(ST) = 1022,2(S,)-
Note that 102i (-ST11) ST11 + ST121 ST2) = 1021 (ST11) ST12, ST2) etc.
By (12.10.1), (12.10.2) and a similar consideration as before, =p2i+((D, w, w) = 6#(w)
E
j,,, ,T(Ra'Pp2i) w),
r1EW2 r2=1,(1,2)
"p2i+('D, w, w) = 6#(w)
1:
W142{,T(9-0 Ra4)p2;,w),
,1 E202
,2=1,(1,2)
for i = 1, 2. We define
E2+(, ST11) =
)102,, ST11)1)1)1
E2+(, sill) = 2F+W,st,sub+(ga2 R02 4'p2i I 42i 1 ST11) 1, 1)
for i = 1, 2. The following proposition follows from (11.2.3).
Proposition (12.10.15) Suppose that r = TG and S > 0 is a small number. Then by changing z// if necessary, (1)
=p2i+(4),w,w)
2CcA(w; P)b#(w)
C # fRe(si (2)
Ew+(RDDp21 , lp2i,1(-1) 2, 1)) lp2i,2 (-112) 1)
1)=1+b
1)dSTll)
p2+ (,
021+(4),w,W)
^ 2CGA(w; P)s#(w) Ew+(ga2
Rat 'Dp21 ,
lp2i,1(-1, 2,1))
lp2;,2(-1, 2,1)
+
P26#(W #() JRe(sii)=1+5
p2i+(, sT1 1)2(s111)Al(w; s111,1)ds111
fori=1,2. (e) "p21#(')'w,w), ' 022#(4),w, w) etc. Let p = A 2. Then ai. 2, a21 =dip-1
12 Contributions from unstable strata
317
By (12.10.1), (12.10.2), p21 # (4D) w, w) = Ra2
JR2
X MO A /MDk
w(g0
1
2S,21(ga, w)d" A1dx /.cdga,
a<1, µ(1
"p21#(41fwfw) =
J
w(g0')A21421(go, w)d"A1dx µdga f + X MIA/Mak
A1<1, µ<1
'p22#(,D7 wf w) = R021)p22 (0)1
µdga°,
w(ga)-1A1 26'22 (go,
R+x MIA /Mak
al>1, µ<1
=P22#(4>, w, w) = aRa2 p22 (°) f
22
R2 x MIA / Ma k R+
(go, w)d" Ald"pdga
A1>1, A<1
Easy computations show the following lemma.
Lemma (12.10.16) A2))Tz+P =
(1) 0P21 (d2(A1,
(2) 0P22 (d2(A1, A2))T +P = a1
= \ sr11+2sriz-2sr2-1µ5r2-1 sr11-2sr12+3As -2-1 = Al srll-2sT12-2872+5µw2-1. A8T11+25712-3A3T2-1
Let sT = (S,12,8,2) and ds'T = dsT12ds,2 We define AT(w; 8712, -1, ST2)
ST) = 26
Ep21#,7(`Y
(ST12 - ST2)(ST2 + 1)
Ep21#,7 ((D, w, w, sT) = 2b#(w)9-DRa2 4)p2, (0)
Ep22#,T(
f
wf
Ep22#, (Yfwf
f
W,
S'7 = 26## ((w)Ra2
p22 (
0)
AT (w; ST12, -1, Sr 2) (8712 + ST2 - 4 `)('ST2 T 1)
Ra2 P22( 0 )
8T') =
, 8T2) ' A. ( w; sr12, (ST12 - ST2)(ST2 - 2)'
A7
f
(w; w; ST12f -1, ST2)
(8712 + ST2
- 4)(872 - 2)'
Then by (12.10.15) and a similar consideration as before,
)21E2#,T((A),
'-'p2i#(-'f w, w)
w,
'1E2
,'r3)
T2=1,(1,2)
P2i# (4), w, w)
E(
rl EW2
)2f 2%rV1
1
.e(sr)-(T2,'r3)
Ep2i#,T (4k, w, w, sT)d4T,
T2=1,(1,2)
where we choose the contour so that (-1, r2, r3) E DT and r2 > r3 etc.
Proposition (12.10.17) By changing O if necessary,
(2)
p21#((D,w,w) - CGA(w;P)T2b#(w)Ra2`'p21(0), -2CGA(w; P)`Z126#(w)gaRa2 `DP21(0)1 ' jP21# ((D, w, w)
(3)
E P22# (ID, w, w)
(4)
Ep22#(ID, w, w) - 2CGA(w; P)Z2b#(w)9bRa2 DP22 (0)
(1)
-CGA(w; P) 2b#(w)Ra2 DP22 (0)f
Part IV The quartic case
318
Proof. We can ignore the case T2 = 1 as usual, and assume that T2 = (1, 2). If TI = 1, (1, 2), we choose r2 » 0, r3 = 1 + 6 where S > 0 is a small number. Then L(-1, r2, r3) < 2 + (1 + 6)C < w° if 6 is sufficiently small. Therefore, we can ignore these cases. Suppose TI = (1, 3, 2). Then the point pr = (-1, 2, 1) does not satisfy the conditions ST12 = sr2) sr12 + sr2 = 4, sr2 = 2, -1. Therefore, we can move the contour crossing these lines by the passing principle (3.6.1). Since Ep21#,rN, w, w, sT) etc. are holomorphic at s' = (2, 1), the proposition follows. Q.E.D.
(f)
w, w).
w, w), '^--'P32
= 03. Let to be as in (c). Let is = Iy2 33(t°)l = A1.A2A3. Then A1A2A3 =
Let
AIA2 t, and d1t° = dxA1dxA2dxµdxt°. By (12.10.1), (12.10.2), P31 (,D, w, W)
w(t°)-'A
1A2A28Z.
t°Ap3 (t°) w)dxt°)
T2 /Tk a1<1 P32
(4)1 w, w)
w(t°)A1A2A29Z, (R5DP32 f t°)4032 (t°, w)dxt°. Tp /Tk Al >1
Easy computations show the following lemma.
Lemma (12.10.18) A2, 3)) Tz+P =
(1) ep31
(2) 9
32
(d3( 1, A2,
3))Tz
1s*
-1
12-1
z (-s*11 +2srz+1)
A2
µ 2 (s.,ll-1)
+P - -13+11 -sr12+4 2A2(-8T11+2sr2+1) 16 (srll -1) 2
Let
Ep31,7
(4if w) wf sT12) sT2)
Ep32,T(4i)
w)
W, sT12)
sT2)
2S# (w)E1 (RZI 4> p31 f
sT2 + 1)
AT(w; 2sr2 + 1, Sr12, ST2) ,
ST12+1 AT(w; 28,-2 + 1, ST12, ST2)
28#(w)E1(Ra
4?p32) sT2 +
1)
sT12 + 2sr2 - 3
Then by (12.10.7) and a similar consideration as before, =p3: (4>, w) w)
_
(
1
)21 e(sr12,s.2)=(rl,r3)
w, w) ST12, Sr2)dsr12dsr2,
where we choose the contour so that (2r2 + 1, r2, r3) E Dr, r2 > 0, and r2 > -1, r2 + 2r3 > 3 for i = 1, 2 respectively. Proposition (12.10.19) By changing t/i if necessary, 8p3i (4>, w, w)
0 for i = 1, 2.
Proof. We can ignore the case r2 = 1 as usual and assume that r2 = (1, 2). Then the point pT is not on the line sril = 2sr2 + 1 for all the cases. Therefore, the proposition follows from (3.5.19). Q.E.D.
12 Contributions from unstable strata
(g) Ep4i (c, w, w) for i = 1,
319
, 4.
Let 0 = 04. Let to be as in (c). Let pi = .1\2i and p2 = A A2A1. Then A1 2 3 = ip1p2, A2A3 = Al 2µ2 It is easy to see that dxt° = dxA,dxpldx p2dxt °.
By (12.10.1), (12.10.2),
t°)1041 (to W)d"t°
w(t°)A1A2A3eZ,
w, w) _
p41
TA/Tk a1 2 <1 w(t°)-1)t2A3OZa
p42 ('D' w, w) _
(RD,P042, t°)4042
(to w)dx x1dx t°,
TA/Tk f)`1>1, X2P2>1
w, w) =
.1043
w(t°)-'A 6A2A3eZa (Ra4'p43, t°)4pa3 (t°, w)dxt0,
J
TA/Tk 1 a2<1
II/a1>1,
044 (4), w, w)
=
w(t°)A2.eZ,
J
t°)4044 (t°, w)dxA,dxt°.
TA / Tk
a1<1, aia2>1
Easy computations show the following lemma.
Lemma (12.10.20) (1) 0P41
(2) 9P42
(3) ep43 (4) ep44
(t0)TZ+P
= 1r11+2sT12-2s,z-lpl (-3T11+2srz-1)02 (3T11-1)
(t0)TZ+P = Al 8111-28112+2srz+lp (-8111-2srz+3)02 (8111-1) (t0)TZ+P = al 8111-23x12-23x2+spi (-8111+2srz-1)02(3.,11-1) As,"' +28,12+28.,2-5Alz(2
0 TZ+P
(-sT11-23.,2+3)
z (s,11-1) 02
Definition (12.10.21) Let 44i = (2 (sTi1 + 1), lp4;,2) where (1) lp41,2(ST) = (sT11 + 2ST12 - 2ST2 + 1)(-ST11 + 2ST2 + 1), (2) 1p42i2(&T) = (sT11 + 2sT12 - 2sT2 + 1)(sT11 + 2ST2 - 3),
(3) 1043,2(ST) = (sT11 + 2sT12 + 2ST2 -
7)(-ST11 + 2ST2 + 1),
(4) 1p44,2(ST) = (ST11 + 2ST12 + 2ST2 - )(ST11 + 2Sr2 - 3).
By (12.10.20) and a similar consideration as before,
Ni (D, w, w) _
3
=2b#(w) T
(2)1r
fRe(s)r
E1,sub(RDipa1,1p4:>ST)AT(w;sT)dsr,
where we choose the contour so that r E DT, rl > 1 and both factors of lp4.,2 are positive for i = 1, 2, 3, 4. We define Epal
STll) _
EP42
(D, ST11) -
E043 Nf ST11) = E044 (Df ST11) =
2E1(R14'PP41' 2(s,11 + 1)) (ST11 + 1)(3 - sT11) + 1)) 2E1(R04'DP42' 2(ST11
(ST11 + 1)(ST11 - 1) 2E1(R14 Dp43, 2 (ST11 + 1))
(sT11 - 3)(3 - 8,11) 2E1(RD4 4044 , 2 (ST11 + 1))
(ST11 - 3)(ST11 - 1)
Part IV The quartic case
320
The following proposition follows from cases (3), (4), (11), (12) of (11.2.4).
Proposition (12.10.22) By changing 7/i if necessary, 2826#(W)
p4i (, LO, w) -
21r
JRe(s,ii=1+6 Epi (, ST11)AiT1(w; Sr111 1)dsrll
fori = 1,2,3,4. (h) Now we combine the computations in this section. Let 0 = 00, p = po We define
Ji(,W) =ba(W) (ED+(`F,WD,3)+Ea+( ,IF,Wa if 3)) 26#(W) 3
(RD 41p (0) + 9, RD -Pp (0)),
-
J2(4))
(Ew+(R,24'p21,
2) + Ew+ (-9'wRa2-Pp21 f 1))
- (Ew+(Ra2 P22 f 2) + Ew+(9wR02 4P22' 1)) + Q32 (Ra2 (I)P21 (0) + 2.rwR,2 (I)p21 (0)) + 02 (Ra2 4)}f22
(0) + 29w Ra2 DP22 (0)) f
TL, 1(4),W) =Ji(,W)+5#(W)J2(4), J3(4f W) sT11) =
To+(RD,oIF,wo,3t, 2(srll + 3), 1, 2(1 - sT11)) ST11 - 1
Wa t, 2(5 - ST11)f 1, 2(1 - ST11))
+
J(4) s T11 )_4
ST11 - 1
2Tw+(Rw,oRa24)P21, 2, 2 (1 - 8T11))
f
(sT11 + 1)(s"11 - 1)
2Tw+(Rw,o9wRa24 211 1, 2(1 - ST11)) (S T11 + 1)(STll - 1)
J5(, Srll)
=2Tu +(RK,ORaz'Dp22 ) 2f 2(1 - ST11))
(sTll - 3)(sTil - 1) +
2Tw+(Rw,o9wRa24'p22f 1, 2(1 - ST1l)) (STll - 3)(sTll - 1)
7a,2(, Wf ST11) -ba,st(W)J3(, Wf ST11) + 6#(W)(J4(, ST11) + J5(, ST11))
+ b#(W) E (-1)iEpli (, si11) i=1,2
+ 8#(W)(Ep41 (, STll) - EP42 (, ST11)) + 5#(W)(-Ep43 (, Sl1) + EP44 (4)1 S"11))
By Theorem (10.0.1) and (12.10.3), (12.10.4), (12.10.8), (12.10.9), (12.10.12), (12.10.15), (12.10.17), (12.10.19), (12.10.22), we get the following proposition.
Proposition (12.10.23) Suppose that T = TG and 6 > 0 is a small number. Then by changing 7/i if necessary,
'po (, w, w) -CGA(w, p)Ti,l (, w, w)
+
g
_
2
JRe(sii)=h+5
W,STl 1)2(sT1l)A1(wfST11,1)dsii.
12 Contributions from unstable strata
321
Let Ta,2,(o) (P, w, 1) etc. be the i-th coefficient of the Laurent expansion as before. We define
J6(P, w) = -1(Ta+(Ra,o`F, wa,st, 2,1) + ZTa+(Ra,o9-aW, wa, t, 4,1)).
Then it is easy to see that d Ta+(RD,0W, wa,st, ds-11 Sri1=1
J3, (0) (ID, w,1) =Js(4 , w) +
Ta+( Ra,o
a IF, wa,9t, -1
1(sill + 3), 1, 0)
2 (5 - sTll),1, 0).
S, 11=1
Also, Ta+(Ro,o'W, wo,st,
2
(sT11 + 3),1, 0),
ar11+1 2
e2(Ra,ow,1L1E2 141, µ1µ242)d µ1d",L2dXg1d"g2,
2 x (A1 /kx )2 µ1 >1
7a+(Ro,oAoIP, wat, 1(5 - ST11),1, 0) 3-erll p1 62(Ra,o9-a`I',t1Ft2141,µ1µ24z)d"p1d"µzd 4id"42 2 .2 x (A' /k X )2 c it
For IF E Y'(Za,oA), We define 9a,0T(x1,33, x2,22) _ f (y1,33, Y2,12) < x2,12y2,12 + x1,33y1,33 > dy2,12dy1,33 A2
Then if (k is K-invariant, Ro,O-5FaRa(D(x1,33, x2,22) =.a,oR',oRa-P(X2,22, x1,33)
For T E 9(Za oA), let R1, R2 be the restrictions to the first coordinate and the second coordinate respectively. Also let (12.10.24)
91(x1,33) = f (yi,a, 0) < X1,301,33 > dy1,33,
.2(X2,22) =
//
J
'F (0, Y2,22) < x2,22Y2,22 > dy2,22
The proof of the following lemma is easy, and is left to the reader.
Lemma (12.10.25) Suppose that 4) is K-invariant. Then -QriRa,0W(0), 92R0 ',09a`y(0) = F2Ra,o`y(0), E2,(-1,-1)(Ra1 p11, 0, 1) = -E1,(-1) (Ra445P42' 1), E2,(-1,-1)(Ra1 DP12) 0,1) =
-E1,(-1)(Ra4N44)
1),
E1,(i)(91Ra,o`W,1) = (-1)tE2,(i,-1)(Ra1 iDp11 , 0,1) for all i, E11(i)(91Ra o a1Y,1) _ (-1)iE2,(i,-1)(Ra1 4(Dp12, 0,1) for all i.
Part IV The quartic case
322
Lemma (12.10.26) bo,st (w)T1+(Ra,o4y, wo,st, 2 (S'11 + 3), 1, 0)
+ ba,st (w)T1+(Ra,o9-a IF, wa ba,st(w)
=
2
+ b#(W)
+ b#(w)
/
Ez(Ra,o'I`,wa,st,
sill + l
ST11 + 1 4
4
,
(4E2(_l_l)(Rll4'p11,0, 1)
+
2E2,(o,-1)(Ra1P1201)) 2E1,(o)(Ra44p42, 1)
+
(ST11 + 1)2
4E1,(-1)(Ra4'P1441 1)
2
/
ST11 -3
4E1,(-1)(Ra4')p42, 1)
Ta+(Ro,o41, wa,st,
)
ST11 + 1
(4E2(_l_l)(RDi p12,0,1) (sT11 - 3)2
(srll - 3)2
)
2E2,(o,-1)(Ra14)p11,0,1)
(ST11 + 1)2
+ b#(w)
- 6#(w)
t,12 (5 - ST11),1, 0)
)
sT11 + 1
2E1,(o)(Ra4'bp44, 1)
s,11-3
+
(ST11 + 3), 1, 0), Ta+(Ra,oJFa 4y, wa,5t, (5 - ST11), 1, 0)) 2
are equal to the following integrals "" 2+1
fRx(A1/kx)2
wa,st(41,4z)A1
_
z(Ra,o`I`,-
1,
1 A2)d"Ald A 2dxgldxg2,
a1>1
fRX(A1/kx)2
wa,1St(41,42)A
z .. Q2(Ra,O-9,4I`,AjA 1>A1A2)dxA dxA2dx41dxg2.
a1>1
Note that wa,st(41,g2) = we,st(42,41) By (6.1.2),
A12
A1r12
O2(Ra,0W, -A1-A2 141,
E)2(ga,ORa,OF,A 1241
erll-1 = Al 2 A2
-
A1A2 141) + Al 2 A1r,2-1
1O1(R1Ra,oq,,
A1",2+1
A1'a 141) -
A2
+li-3
= Al
1)
A21e1(R2Ra,o9a4Y,
2
- Al
191(R1R/0,0
W, Al 1'2 142
1)
1)
A-1e1(92Ra,o`I',A 1A 2'g2 1) srll-1 2 261 (9R',09, T, A1 Az4z) + Al °r11+1
er 11-1 2
1,
a+11
A2E)1(91Ra,oqf, Al 1' 241 1) - Al
2
A201(R2Ra,04y, 1A242)
Note that each term is independent of either ql or 42. We divide the integral according to A2 < 1 or A2 > 1. If A2 < 1, we use the first equation, and if A2 > 1, we use the second equation. Then (12.10.26) is equal to the sum of the function ba,st(w) 2
/
Ez(Rao4y,wa,st,
ST11 + 1 ST11 + 1 4 , 4
12 Contributions from unstable strata
323
and the integral of the function
-
2
E1+((92Ra,o9a1If)a1,1) + Al
+ 6#(w)(A1 2 E1+((R1R',O1F)a1, 0) + Al r 1,-3
- 6#(w)(A
2
r -1
+ 6#(w)(A1
2
E1+((RlRa,o9a'F)a 1-1, 0))
2
E1+((1F2Ra,o'F),-1,1))
2
r11 2
E1+((R2Ra,o9a`F),\ 1, 0) +.
1))
D
"'+1
E1+((91Ra,o'F),\ 1, 1) + Al
E1+((R2Ra,o'F)A1, 0))
2
over 0
r -3
.,, -1 Al
2
= Al
E1+((" 2Ra,0JFa 2
2
E1+((R2Ra,oFa)a 1, 0)
2
E1,(o)((92Ra,o9a'F)a1,1) + Al rl I -1
rl 1 -3 A1
1) + Al
El+((RlR',oca'F)a 1, 0) + Al
E1+((91Ra,o9a'F)A 1,1)
2
."'-1
r11-1
Al A1
2
Al Al
2
_ Al
2
E1,(o)((FlR0',o9a'F)A1,1) + Al
EI+((RIR',o'F)A1, 0) + Al 2
2
`'
1R1,O9o1F(0)
E1+((" 1Ra,o1Y)a 1,1
2
E1,(o)((91Ra,o'F),\ 1, 1) + Al
E1+((JT2Ra,o'F), 1,1) + Al 2
92Ra,oJgaW(0),
2
2
E1,(o)((92Ra,o'F)a 1, 1) + Al
2
91Ra,o'If(0),
E1+((R2Ra,o1F)al, O) 2
Y2Ra,o'I'(0).
is the Fourier transform of R2Ra o i a'F with respect to the standard bilinear form on A etc. r11-1 By (12.10.25), terms like Al 22,9Ka1F(0) cancel out. It is easy to see that Note that _0'2 R'
A-' (log Aj)Ej,(_j)(92R',O9-o IF, 1) + Aj 1E1,(o)(92Ra,o `' a1F,1),
E1,(o)((FlRa,o9a'F)A1,1) = - Aj 1(log +A1 1E1,(o)(91Ra,o9a'F,1), E1,(o)((91Ra,o`F)a-1,1) =Ai(logA1)E1,(-1)(_F1Ra,o'F,1) + AlE1,(o)(91Ra,o'F,1),
E1,(o)((92Ra,o'F)a1) =A,(log Al)E1,(o)(92Ra,o'F,1) +AlE1,(o)(F2R'a ,OF,1).
If P is K-invariant, 92Ra,o9a'F = Ra4 n44, 92Ra,o'F = Ra4Dd42
Part IV The quartic case
324
Therefore, integrating over Al < 1, and using (12.10.25), we get the proposition. Q.E.D. We define
J7('F, W) =8(E2,(l,o)(Ra,o4,ws,st) 2, 2) + E2,(o,1)(Ra,oI,wa,st, 2, 2)), Js((D) = - E2,(-1,-1)(Ra1 p11, 0,1)
1
- 1 E2,(o,-1)(Ra1
0,1)
1
0,1)
+ E2,(-1,-1)(Ra, 4'P12, 0, 1) + 1 E2,(o,-1)(Ra1 1
+ E1,(-1)(R044)p42,1) - 21,(o)(RaJp42,1) E E 21,(o)(Ra4'Pp441 1),
Js(,P) =2(Tw+(Rw,oRa2()P21,2) +Tw+(Rw,o9wRa2'DP21,1)),
J1o(p) = 2 (Tw+(Rw,oRa2
4p22,
2) + Tw+(Rw,o9wRb2'bP22, 1))
Then by (12.10.26), oa,3t(w)J3,(o)('P, w,1) = sa,st(w)(JO), w) + J7(4,w)) + 5#(w)J8('D).
also, J4,(o)(4',1)
= J9(.') +
1(Tw+(Rw,oRa2'PP21, 2, 0) +
Tw+(Rw,o9wRa24)P211 11 0)),
J5,(o)(4),1) 1
= J10(-P) -
2(Tw+(Rw,oR024P22, 2, 0) +Tw+(Rw,o-11wRa2'DP22,1, 0))
By the principal part formula for the standard L-function in one variable, Tw+(Rw,oR021'P21, 2, 0) + Tw+(RwoJFwR02'DP21 , 1, 0) = -E2,(-1,0) (Ra1 p11, 0,1) + E1,(-1) (Ra4 'bp41 , 1),
Tw+(Rw,oR02 4P22, 2, 0) +
Tw+(Rw,o-11wRa2'kP22, 110)
= -E2,(-1,o)(Ra1 4DP12, 0, 1) +
E1,(-1)(R544)p43,1).
Therefore,
J4,(o)('D, 1) = 6#(w)(Js('P) - 2E2,(-l,o)(Ra1,Dp11, 0,1) +
A'(0)(4), 1) = b#(w)(J1o(1k) +
0,1) - 2E1,(-1)(R041)p43,1))
12 Contributions from unstable strata
325
It is easy to see that Ep11,(o) (1p, 1) =1 E2,(o,o) (Ra, 4)p1, , 0, 1) + 1 E2,(1,-1) (Ra1 2 2
0,1)
+ 2E2,(-1,1)(Ra,(Dp11,0,1)+ 2E2,(o,-1)(Ra1,Dp11,0,1) 1
1
+ 2E2,(-1,o)(Ra1 4)p1,, 0, 1) + 2E2,(-1,-1)(Ra1 4p11, 0, 1),
0,1) - 2E2,(1,-1)(Ra1
2E2,(o,o)(Ra1
0,1)
- 2E2,(-1,1)(Ra14)p12,0,1) 9
3
- 2E2,(-1,0)(RD, 4DP12, 0,1) - 2E2,(-1,-1)(Ra1 qtP12, 0,1).
Also 1
Ep42,(o)((,), 1) = 2 E1,(1)(Ra4 Dp421 1) - 2 E1,(o)(Ra4-pp42, 1)
+ Ep43,(0) ('p , 1) = 2- E 1)
1 E1,(1)
l,(o)(R544)p43,
1) - El,(-1)(Ra4 p43, 1), 2 E1,(o)(Ra4'Dp44, 1)
(Ra4Dp4411)
-
1).
We define J11(4)
2E2,(o,o)(Ra,-Dp11,0,1) 0, 1) - E2,(o,-1)(Ra14bp11, 0, 1)
2E2,(-1,1)(Ra1
- E2,(-l,o)(Ra1 4bp11, 0, 1) - 2E2,(-1,-1)(Ra1 bp11, 0, 1)
-1
2E2,(1,-1)(Re,4DP1210,1) E2,(-1,1)(Ra1 Pp, 0, 1) - E2,(o,-1)(Ra1 DP12, 0, 1)
- E2,(-1,o)(Ra1 Dp12, 0, 1) - 2E2,(-1,-1)(Ra1 Dp12, 0, 1)
-
2
E1,(1) (Ra4 OPP421 1) -
2
E1,(1)(Ra4'Dp44, 1),
1
J12(1) =2E1,(o)(Ra4'p41,1) +
+
2E1,(-1)(Ra4'Dp41, 1)
1
1
2E1,(-1)(Ra4'Dp43,1).
Then by the above considerations, 6# (W)
Ji(4),W)
To,2,00D, W) = ba,.t(W) E i=6,7
+
12
E Ji(.D) i=9
Part IV The quartic case
326
By the principal part formula (4.2.15), Ji(41) = -Ew,ad,(0) (Ra2,Dp2i , 2) - EW,ad,(o) (Ra2 4)P22, 2)* i=2,9,10,12 Therefore,
T ,,1(4', W) + Ta,2,o (4 , W )
= SD,st(w) T, Ji(4),W) + 6#(w)Jii(-D) i=6,7
- 6#(w)(Ew,ad,(0)(Ra24)P21, 2) + EW,ad,(o)(Ra2'DP22, 2)).
By (5.6.4), we get the following proposition.
Proposition (12.10.27) T
w) + Te,2,(o) (4),w)
= bb (w)Ea,ad (RD Pp, wa, 3)
+ sa,8t(w) (E2(lO)(ROwwO,St,
) + E2,(o,
2,
2
Wa,st,
2
,
2)
Chapter 13 The main theorem §13.1 The cancellations of distributions In this chapter, 0 = (/3i) for i = 1,
,10, Ds,i = and 88,2 = (,38,08,2)Let P8,2 = 08,2, $8,2), p9 = (D9, s9) where 88,2 = (138, 08,2), a9 = (09) and 58,2(1) _ 59(1), 58,2(2) = 0. Then EP8 2 = -Ep9. An easy consideration shows that R814Dpl = qDp9) Therefore, the third term of (12.8.28) and the second term of (12.9.2) cancel out. We define
i=1,6,8,10
+ 2604 (w) (Ea4 (R54 $, wa 1, -2) - Ea4 (R04 , wa, -2))
+
sa93w)
I E3(R89
22
,wa91, -1, -3
-
+ 9329J36#(w)(4(0)
3) -
-1, -3, 3)
I
4)(0)),
and
E
T2(4), w, s-11)
w-1) sT11) - TO ,2(4D, w, sT11)).
i=1,6,7,8,1°
Suppose that r = TG and b > 0 is a small number. Then by (10.1.1), (10.1.2), and the results in Chapter 12, by changing if necessary, I(4), w, w) '.'Ccn(w; p)T1(4),w) p2
+
-
fRe(ii)=1+S
siii)2(sT11)A1(w, sTll, 1)ds,11.
We get the following lemma by Wright's principle.
Lemma (13.1.1) The distribution T20), LO, ST11) is holomorphic at si11 = 1. Therefore, the cancellations of higher order terms of T2((D,w,si11) are guaranteed, and we are not obliged to check them. Hence, we get the following proposition.
Proposition (13.1.2) I°(4), w) =
w) + T2('D, w,1).
Clearly,
T2(`I', w,1) = E (Ta.,2,(0) ($, w-1, 1) -
(D, W,1))
i=1,6,7,8,10
Let ps,1 = 08,1,55,1), p6 = (x6,56) be paths such that 68,1(1) = 56(1), 58,1(2) = 0.
Then it is easy to see that E2,(i,9)(Ra8.1 `Dp8,1) -1, -3) = E2,(7,i)(Ras
-3, -1)
Part IV The quartic case
328
for all i, j. Also if p7 = (D7,57)410 = (aio, sio) are paths such that 57(1) = slo(t), E3,(1,-1,0)(Ra7I 7,w57, (w2, 1) W2), E3,(0,-1,1)(Ra7Op7, Wa7, (W2,1) W2),
2
0,
1
2) =
21
),
1, 0, 1) _ -E2,(o,1)(Ra1o,o4'p1o,wo,st, 1, 1)
22
2
2
We define (13.1.3)
f(4), w) =6a, (w)Ea1(R51'p, wa1, 2) + 2654 (w)Ea4 (R5, 4), wa4, -2) + 8656 (W)Ea6,ad(Ro6'D, was, X56, -3)
-3, -1)
+ +
+
26#(w)E2,(0,1)(Ra6io-b, -3, -1) 1 1 657(w) 6 E3,(0,o,o)(R574,w57, 2,0' 2)
+ 1465, (w)Eas,(o) (Ro,,P, woe, Xas, -3)
22 33
bag (w) 3
+ 65io (W)F'alo,ad (R51o 4), W,,o, 3)
_ +
balo,8 (w) E2
(1,o)(Ralo,o,D,walo,st)
balo,st (w) 24
E2,(0,1)(Ra1o,O
2, 2)
, walo,st, 12 , 21 )
By (13.1.2), we get the following proposition.
Proposition (13.1.4) I0(t1 w) = f * w-1) - f (C w)
§13.2 The principal part formula We are finally ready to prove the principal part formula for the zeta function for the space of pairs of ternary quadratic forms. For 4D E .'(VA), X E lam, let 4)5(x) = D(Ax) as before. The following relations are easy to verify, and the proof is left to the reader.
Lemma (13.2.1) 4)),(0) _ (D(o).
(1) (2)
(3) (4) (5)
E51(R51,Da,w51,2) = E54 (R54
)' , Wa4, -2) = E54 (R54
, w54, -2).
Ea6,ad(RD6(1) A,W56,X56,-3) = E56,ad(Ro64),w56,X56,-3). E2,(1,o)(Ra6 O4)A, -3, -1) = E2,(1,0)(Ra6,04), -3, -1)
- (log A)E2(R'a6,o4), -3, -1). (6)
E2,(o,1)(Ra6i04)a, -3, -1) = E2,(0,1)(Ra6i0-D, -3, -1)
- (log A)E2(RD6,0 4)' -3,-1).
13 The main theorem
329
2, 0,21
(7) 6D,
wa7, 2, 0,2)
1, 1
+ A -3 tog (logXA)SD 1ost( W) E2(R'Dio,o cj, walost, 2, 2
(8) 6ae (W)ED8,(0) (Rae A, was, Xae, -3) = 8 x8 (w)ED8,(o) (Ras 4), WD8, X08, -3)
+
2A(log.)6#(w)E2(R0g,oRag,D, -3, -1).
2,2
(9)
2
3)
(10) (11)
3)
EDio,ad(RD1o4)a,WD1o,3) = A-3ED1o,ad (R,1oID,Wa1o,3). E2,(1,0)(R0'
X-3E2,(1,o)(Ra1o,oID,WD1o,st,1,1) 2 2
2 2
1
1
- A-3 (log.)E2(Ra1o,0-D, w,1o,st, 2, 2 )
Note that (8) of the above lemma follows from (7.3.7). Similar relations hold for also. We define F(-1) (1, w, 0) =9J29T36# (w)4)(0) + 26D, (w)ED, (Ra4 D, W041 -2)) + 86Dg (W )Eag,aa (Rag
, was, Xo6 , -3)
+ 146x8 (W)EaB,(o) (Rae,D, Loa, XD81 -3)
+
6#(U)) (5E21(1,0)(Ra,,,0
, -3, -1) + 3E2,(o,1)(Ra6,0 (1), -3, -1)),
36#(w)E2(R0' g,oD, -3, -1),
F(-2)(C W, 0)
6
F(-1)((D, w, 2) =6a1(w)ED,(RD,Ob,wD,,2)+ F(-1)(4), w, 3)
(w)E3(Ra94),wa9,-1,-3, 3),
Wa 2' 0, 2 + 601o (w)Ealo,ad(RD1o D) wD1o, 3))
_
(w)
6D 1o,
8
412,(1,0) (R'610'0 4) , W01o,st, 2,
+ halo,st(w)E2 24
6alo,st(w)
F(-2)(1,w,3)
4
,
(o ,
2)
1), 1)(R' 10, 04),wD lo,st, 1, 2 2 ,
1
E2(R1o,0(D,wojo,st, 2>
1
2)
We define
F(
w)s) = F(-1)(4),w,2) s-2 +
F(-j) 03 (s - i)i j=1,2
By replacing
Theorem (13.2.2) Zv(t1, w, s) = Zv+(-D, w, s) + Zv+($, w-1,12 - s) - F($, w-1,12 - s) - F(,D, w, s).
330
Part IV The quartic case
§13.3 Concluding remarks We used the passing principle (3.6.1) in various places to move contours. Instead of establishing cancellations of various distributions, it enabled us to ignore them as long as we can check some easy conditions. We also used Wright's principle to ignore higher order terms of certain distributions. However, it is possible to study the poles without these two methods, and in fact, the author's original approach was to establish cancellations explicitly. It is an interesting exercise to establish cancellations of various contour integrals without using the passing principle (3.6.1) or Wright's principle, because one can see that the cancellations between different paths are happening. They correspond to cancellations between divergent integrals, and our two methods and the choice of the constants in the definition of the smoothed Eisenstein series enabled us to handle most of them without establishing the cancellations. In Chapter 12, we computed contributions from unstable strata, and obtained more or less special values of the zeta functions for the corresponding representations. Those zeta functions have higher order poles, so we had to recover the higher order terms. When we analyzed the constant terms in Chapter 12, they came not only from the unstable strata, but from the adjusting terms also. The author does not know if this is a special case of something more general. But for this reason, we had to compute the contributions from unstable strata explicitly, and compare with principal part formulas of the corresponding zeta functions. This was rather annoying, and the author hopes to find a better interpretation of this phenomenon in the future. If the center of the stabilizer of a point in Zap contains a split torus, we could ignore jp (,D, w, w) in most cases. However, this is not always the case, for example, in the case G = GL(2) x GL(2), V = k2 ® k2. This is another reason why we still cannot handle all the cases simultaneously. We restricted ourselves to groups which are products of GL(n)'s, mainly because we do not have the uniform estimates of Whittaker functions. As far as the formulation of equivariant Morse theory is concerned, there does not seem to be any major problem in generalizing our formulation. However, for other groups, we still do not know how we can describe the Eisenstein series in terms of generalized Whittaker functions, and until we find out the answer to this problem, we cannot handle other groups. On the other hand, there is no doubt that the cases where the group is a product of GL(n)'s contain the most interesting cases (see [84]), and we hope our approach enables us to handle more cases. It is clear from the computations in this book that the complexity of our computations increases exponentially with the rank of the group. Therefore, the number of the distributions we have to handle can be quite large, and may require a computer calculation. However, we still do not know when the significant cancellations as in §10.3, §10.6 and §13.1 will occur. Also, in order to do such a task, our theory has to be generalized so that we can handle reducible representations. However, these are questions to be answered in the future.
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List of symbols f (x) << g(x) §0.1 VR5 §0.2
o(x), oco(x), µ(x), µ,,,(x), b,,,,, §0.3 ZZ,,,(4p) s), Xx,,,(4D) s),
s) §0.3 b(s), b*(s) (0.3.4)
an(tl) ... , tn), nn (U) (1.1.1)
K=
§1.1
p (half the sum of positive weights) §1.1 O(s), On(s) (1.1.2) P after (1.1.2)
93n (1.1.3)
NT, NT (2.1.4) IT after (2.1.5) (2.2.1)
a(u), a
(2.2.2)
Ea(g°, z) (2.2.3) M(z) (2.2.8), (2.2.10) TG before (2.3.1), after (3.1.8)
Wn(a, z) etc. (2.3.1) An,,, (t, z) (2.3.7) Qn,,,(a, z) (2.3.12)
D(n, r) (2.3.16) K, (x) (2.3.18) 6(a,t) (2.3.25) W, (a, z) (2.4.2) El,,,,(g°, z) (2.4.6) Dr, DI,T etc. (2.4.7) T+ before (3.1.1) OL(4),g) (3.1.1) ZLN, W, X, S), ZL+(D, W, X, s) (3.1.2) VkS (3.1.3)
rdi,M(µ) (3.1.5) w(g°) (3.1.7) [x,y]v (3.1.9), (3.1.10)
g° after (3.1.10) Mv,u, (3.1.11) t, ( , ), 11 11 §3.2 (3.2.3)
0 = (01, ... , 13a) (3.2.6) Pa, Ma after (3.2.7) Ma the end of §3.2
go after (3.3.1) RD IF,RaI1 (3.3.2) ODp, Op (3.3.3), (3.3.4), (3.3.5) Po#, Ua#, Pa (3.3.6)
List of symbols
336
P'DA , M'A D
etc. (3.3.7)
S4, Tai after (3.3.7) epi (3.3.9), (3.3.10)
Apo etc. (3.3.11) ga, dgo after (3.3.11), before (3.3.13) ep (3.3.12) wp after (3.3.12) °p (3.3.13) to the beginning of §3.4 (, )o, II 11° after (3.4.1) L(z), A(w; z), A, (w; z) (3.4.2), (3.4.3), (3.4.6) MT(z) (3.4.5) e(g°, w, V)) (3.4.7) ep (ge, w), Sp' (ge' n(ua' ), w) (3.4.8), (3.4.9), (3.4.10)
8u(g°, w), eu,r(g°, w) (3.4.12), (3.4.14) rp etc. after (3.4.13) (A ga)T (3.4.17) C(A'oMM°A/M k, r) (3.4.18) eI,T,v(g°, w) etc. (3.4.27) CG (3.4.31)
fi(w)
12 (w) (3.4.32)
es, ('1'i, 9a) etc. (3.5.1) p (-D, w, w) etc. (3.5.2) Jo(41,go),Ja(`D,go) (3.5.6)
p(,P,w,w) (3.5.8) 4a _after (3.5.9)
s, A(w; sT) §3.6 pT §3.6
Ei('yi, S), Ei,(j .....
so)
(3.8.1), (3.8.2)
Sn,i etc. §4.0.
W (the space of binary quadratic forms) the beginning of Chapter 5. Xv (4.2.2), §5.5 Tv2 (W, w, s, sl) etc. (4.2.3) Ro, Ro, Rv,o, Ra2,o etc. (4.2.10), (5.3.2) ZV,ad(D, w, s) (4.2.14), (5.6.3) a,i etc. (4.3.1)
.fo, ho (4.3.2)
cp (4.3.3), (4.3.4) '131 etc. after (4.3.4) ZV2,ad,(i) ('I', W, so) etc. (4.4.1)
Ra,o, Ro,o (4.4.2)
Ip(I,w) etc. (4.4.5) ez;,o (Ro Dp, go) etc. (4.6.1)
Xo etc. after (4.6.2) t°, to (5.1.2), (9.2.4)
TV (c w, s, Si, s2) etc. (5.2.1), (6.3.1) Ozv,0 (wi)t', uo) etc. (5.2.4)
List of symbols
Xa2,Xa §5.4, (11.1.8) =a,i (4),w, X, w) (6.1.7), (6.1.8)
Zero(x) §8.2 =p,al,i (,', w, w) (9.1.7), (9.1.8) Ep,tot(4,W,W) (9.1.9) n'(u),n"(u) (9.2.3) 01,02 (11.1.2)
L(sT), A(w; sT) etc. (11.1.3) RW,o etc. before (11.1.6) ®W,st(W, 9) etc. (11.1.6), (11.1.7) OZ,,o (IF, h) (11.1.10) Ei,sab(W,1, sT) (11.1.14) Esub(1, t, 9°, Sr), 4sub(1, t, 9°, w) (11.1.15) cW,T(`y,1, w), EW,sub+(,P, 1, sT) etc. (11.1.15), (11.1.16), (11.1.17)
91' before (12.8.10)
337
Index
adjusting term, 91, 105, 118, 156, 162, 172, 179, 183,252 adjusted zeta function, 105, 118, 172, 179, 183 asymptotic formula, 1, 18 ,B-sequence, 69, 122, 154, 180, 209 Bessel function, 49, 50 b-function, 8 bilinear form, 12, 65, 71, 99, 103, 108, 155, 173, 180, 209
binary quadratic form, 1, 12, 18, 149, 180, 251 binary cubic forms, 2, 18 Borel subgroup, 23, 29, 34, 66, 149 Bump, D., 29, 49 Casselman, W., 42, 43 characteristic function, 12, 15, 106, 113 class number, 1 classification, 4, 14 Cogdell, J., 98, 104 complete type, 6, 8, 64, 91, 196 constant term, 34, 196, 262 contour, 80, 94, 96, 138, 146 contragredient representation, 7, 12 convex hull, 63, 67, 87, 109, 199, 224 cubic equation, 3
Gamma function, 49 Gauss, C. F., 1 geometric invariant theory, 21, 67 Goldfeld, D., 2, 18 Grassmann variety, 4 half the sum of positive weights, 23, 34, 62, 65, 72 Heilbronn, H., 18, 19 Hilbert, D., 67 Hoffstein, J., 2, 18 hyperfunction, 151 ideal class, 1 Igusa, J., 8, 13, 20 incomplete type, 6, 12, 20, 64, 91, 105 infinite place, 7, 29, 49, 50 integral orbits, 13 integral test, 24, 35 invariant measure, 23 involution, 12, 65, 71 irreducible, 4, 5, 20, 76, 85 Iwasawa decomposition, 23, 34, 65, 112, 224
Kempf, G., 21, 68
Khayyam, 0., 3 Kimura, T., 4, 10, 20, 21, 70 Kirwan, F., 21, 68
Datskovsky, B., 2, 15, 18, 107 Davenport, H., 6, 18, 19 density theorem, 2, 20 differential operator, 8 Dirichlet series, 12, 17, 104, 106, 149 discriminant, 1, 18, 19, 150
k-stable, 63, 67, 69, 179
Eisenstein series, 22, 29, 149 equivalence class, 1, 13, 107 equivariant Morse theory, 21, 66, 151 error term, 1, 13 Euler factor, 16, 19 explicit formula for p-adic Whittaker functions, 29, 42
Lipschitz, R., 1
Laurent expansion, 99, 123, 142, 152, 195 length, 31, 69, Levi component, 5, 68, 70, 75 Liouville's theorem, 12 Lindel5f's theorem, 12 local theory, 7, 15, 18, 20 longest element, 40, 65, 71, 76 Mass formula, 150 maximal compact subgroup, 29 (see 23 also),41, 65, 125
filtering process, 15, 20 finite place, 8, 29, 42 Fourier analysis, 1 Fourier expansion, 29, 30, 55 Fourier transform, 8, 65, 71, 108, 173, 209 functional equation, 5, 8, 119, 173, 194 fundamental domain, 167
measurable function, 24, 112, 133 measure (Haar measure), 5, 8, 12, 19, 23, 38, 65, 73, 95, 100, 122, 125, 158 Mellin transform, 49 meromorphic continuation, 5, 20, 114, 123, 156, 181 micro-local analysis, 151 minimal combination of weights, 67, 198 Mumford, D., 67
Calois cohomology, 13 Calois group, 13
Ness, L., 21, 68 non-constant term, 75, 196, 216
Index
1 PS, 23, 67, 72
orbit, 2, 7, 13, 18, 21, 72, 112 orbit decomposition, 21, 70 Ozeki, I., 10, 21, 70
pairs of ternary quadratic forms, 3, 11, 19, 196 parabola, 3 parabolic subgroup, 5, 62, 68, 70, 71, 75, 122, 203 parabolic type, 5 passing principle, 96 path, 71, 125, 131, 132, 158, 196, 216, 262 permutation, 14, 23, 31, 35, 55, 76, 250 Pjateckij-Shapiro, I., 29 Poisson summation formula, 9, 28, 89, 155, 180, 208 pole, 10, 18, 21, 97, 104, 142, 149, 185, 196 prehomogeneous vector space, 2, 29, 62, 67, 70, 85, 91, 98, 105, 152, 173 principal part, 11, 18, 20, 62, 91, 99, 107, 152, 169, 181, 191, 196, 273, 293, 308, 326 principal quasi-character, 62 product form, 8, 15 properly-stable (point), 66 quartic case, 93, 152, 196 quadratic field, 1 quadratic form, 5, 9, 14, 19, 105, 196 quartic equation, 3
rational character, 2, 23, 69, 73, 173, 197 rationality, 21, 67
rational orbits, 13 reduced, 4 reducible, 8, 91, 173, 196 reductive group, 2, 66 regular, 5 regular elements, 69 regulator, 2 relative invariant polynomial, 3, 5, 198 representation, 2, 18, 65, 69, 91, 152 residue, 15, 106 Riemann zeta function, 12 Rubenthaler, H., 5 Sato, M., 2, 4, 8 Sato, F., 8, 20, 152
339
semi-stable (point), 3, 66, 87, 180, 198 Shalika, J., 42 Shintani, T., 2, 6, 8, 20, 29, 42, 50, 98, 105, 118, 149 Shintani's lemma, 62, 73, 84, 124 Siegel, C. L., 1, 105, 149 Siegel domain, 24, 60, 63, 72, 82, 103, 135, 141, 174, 211
Siegel parabolic subgroup, 5 Siegel-Weil formula, 14, 149 single orbit cases, 14, 20 smoothed Eisenstein series, 22, 73, 75, 84, 100, 149, 152, 196, 212 split torus, 5, 23, 62, 66, 87 square free integer, 1 stabilizer, 6, 19, 67 Stade, E., 40 Stirling's formula, 49 (Morse) stratification, 21, 66, 153, 198
Tauberian theorem, 13, 17, 106 tempered distribution, 15 ternary quadratic form, 3 theta series, 21, 24
unipotent radical, 5, 68, 75 unstable (point), 66, 102, 122, 152, 186, 262 unstable distributions, 185, 250 vanishing principle, 95 Vinberg, E. B., 5 volume, 6, 12, 24
Watson, C. N., 49 Weil, A., 25, 104, 149 weighting, 75 Weyl chamber, 66, 69 Weyl group, 23, 34, 65, 69, 96, 99, 254
Whittaker, E. T., 49 Whittaker function, 29, 40, 50 Wright, D., 15, 97 Wright's principle, 97, 104, 121, 170, 192, 196, 327, 330
Zariski open, 2, 5, 13 zero set, 3, 198 zeta function, 5, 20, 62, 105, 155, 180, 196 Zhu, X: W., 20