4-Dimensional compact projective planes with a 5-dimensional nilradical

Geometriae Dedicata 58: 259-289, 1995. © 1995 KluwerAcademic Publishers. Printed in the Netherlands.

259

4-Dimensional Compact Projective Planes with a 5-Dimensional Nilradical Meinem Lehrer H. Salzmann zum 65. Geburtstag am 3.11.1995 in Dankbarkeit gewidmet D. BETTEN Universitiit Kiel, Mathematisches Seminar, Ludewig-Meyn-Str. 4, D-24118 Kiel, Germany (Received: 1 December 1994)

Abstract. The article is a contribution to the classification of all 4-dimensional flexible compact projective planes. We assume that the collineation group is a 6-dimensionalsolvable Lie group which fixes some flag. If, moreover, the nilradical of the collineation group is 5-dimensional,then we get 4 families of new planes which are neither translation planes nor shift planes. Mathematical Subject Classifications (1991): 51H10, 51H20, 51A35.

1. Introduction

Since the fundamental papers of Salzmann [16], [17] on 4-dimensional compact projective planes much work has been done to classify these planes. Many translation planes and several shift planes were found. The group of continuous collineations is a Lie group E with dim E _< 16 and all planes with dim E _> 7 are now explicitly known ([2], [12]). We are interested in planes for which the group has an open orbit in the space of flags. These planes are called flexible, and dim ~ > 6 in this case. From now on we assume dim E = 6. By [4] and [12] we may furthermore assume that the connected component A of P, fixes some flag ( ~ , W) E P x £, e~ E W, and that A is solvable. LetN" be the nilradical of the Lie algebra/2(A). The assumption dimN" = 6 (A is nilpotent) leads to a single shift plane ([3], [8]). Next we consider the case dim A/" = 5. In [4] several situations were worked out in which planes might exist. In the present note we study these situations in more detail and determine all 4-dimensional flexible compact projective planes which have a 6-dimensional solvable collineation group with a 5-dimensional nilradical. As the main result we get four families of new planes, which are neither translation planes (or duals thereof) nor shift planes.

260

o. BETTEN

2. Case A: ( v , a ) =

(3,2)

Besides the general assumptions (dim E = 6, A is solvable and fixes some flag (co, W), dimA/" = 5) we postulate: the group of translations with respect to the axis W has dimension r = 3 and the group of elations with respect to the center c~ has dimension a = 2. This is Case A of [4]. Let N be the connected normal subgroup of A which corresponds to A/" (and which we also call the nilradical), then by [4] the group N acts transitively on the pencil £ ~ \ {W} and transitively on the fixed line W \ {c~}. We say: N has orbit type (2, 2). 2.1. THE STRUCTURE OF N In [4] it was proved that N acts transitively on P \ W. Furthermore, the Lie algebra 2 ( was calculated there: A/" = { ( s , x , y ,

u,v)l[x,y]

= v , [s,x] = y , [s,y] = u , [ s , u ] = v } ,

(and all other commutators are = 0). The universal group N with this Lie algebra is explicitly written down in [5] (the group F5,6 on page 331): (81, Xl, Yl, Ul, Vl)(82, :g2, Y2, u2, v2) 1 2 = (82 -}- Sl,X2 q- x l , Y 2 + Yl - x281, u2 --~ Ul -~- ~x2.-q1 - Y2Sl, v2 + Vl -

1

3

1

2

1

2

-~x2s 1 + ~Y2Sl - U2~l + ~x281 - y 2 x l + X 2 S l X l ) .

By [4] we also know that the translation group T with axis W corresponds to the Lie algebra (y, u, v) and is isomorphic to ~3. We identify the element (0,0, y, u, v), y, u, v E 3~, of N with ( y , u , v ) and write T = ~3 = { ( y , u , v ) , y , u , v E ]~}. Similarly we have N[~] ~ ~2 = {(u,v)lu, v E ]~}. Let M be the subgroup of N defined by putting s = 0 in the multiplication formula for N. Then M acts regularly on P \ W, [4], and we identify this action with the action of M on itself by left multiplication: (Xl, Yl, Ul, Vl)(X, y, u, v) = (x + Xl, y + Y l , u + Ul, v + Vl -- x l y ) .

In order to get the stabilizer of N on the origin (x, y, u, v) = (0, 0, 0, 0), we have to conjugate by the parameter 8 (for formal reasons we take -8): ( x , v , u, v)" = ( x , v +

u +

2+

v +

3+

2+

-

2.2. THE STRUCTURE OF A

The plane has a 3-dimensional translation group T ~ ~3 = { (y, u, v) [y, u, v E ]~ ) and by [1] the points p E W \ {c~} correspond to the 1-dimensional subspaces

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

26 1

(y, u, v)]R,y 7k O, of IR3 . We describe these subspaces as (y, u, v)1~ = (1, ~, y)iR and introduce coordinates { = ~, u ~/= v on W \ { oc }. The action of A on W \ { oc } is induced by conjugation on T ---- 1R3 , it is therefore an affine action. In this way one calculates the action on W \ {oc} induced by N: (~, ?'/) t---+ (~ q-- 8,?'] -+- 182 -+- ~8 -'1- X).

Let £ ( A ) = (32, a) be generated by 32 and a sixth generator a. Then [a, (y, u, v)] < (y, u, v) since (y, u, v) is an ideal in £ ( A ) (corresponding to the translation group A[w,w]). Set [a, x] = z + ds, z C (x, y, u, v), d E K and apply [u, ]. Application to the left side gives

[u, [a, x]] = [[u, a], x] + [a, [u, x]] = 0 since (u, v) is an ideal in £ ( A ) (corresponding to the translation group &[~,w]) and therefore [u, a] E (u, v). Use also the fact that x centralizes < u, v >. Application to the right side gives [u, [a, x]] = [u, z] + d[u, s] = - d v since u is in the center of (x, y, u, v} and [u, s] = - v by definition. Together, this implies d = 0 and proves that (x, y, u, v} is an ideal of £(A). It follows that A is the semidirect product of the regular nilpotent subgroup M and the 2-dimensional stabilizer A0 which is generated by the parameter s and a sixth parameter a. The Lie algebra £ ( A ) = (32, a} contains the following ideals: the nilradical 32, the algebra (x, y, u, v) corresponding to M, the algebra (y, u, v} which corresponds to the translation group T and the algebra (u, v} corresponding to the group N[~o]. Since (v} is characteristic in 32 (being the third member of the descending central series of 32), this algebra is also an ideal in £(A). Since dim 32 = 5, the isotropy algebra (s, a) is not commutative and we have the commutator relation [s, a] = es, e ¢ 0. It follows that we may set:

[a, v] = av, [a, u] = bu + bay, [a, s] = es, e ¢ 0, [a, y] = c y + cl u + c2v,

[a, x] = dx + d l y + dau + d3v, where a,b, bl,C, Cl,C2,d, dl,d2, d3,e E IR. We modify the coordinates in the following way: a ~-+ a - - h i s ,

x

H

x -

au

-/3v,

where

c2 a-c

d2 b-d'

/3

d3 a-d

y

H

y

-

av

262

D. BETIEN

This change respects the commutator relations of N" and allows to set cl = c2 = d l = d2 = d3 = 0. After setting a ~+ ( 1 / e ) a we get the unique Lie algebra ~C(A) = { <X, a) l [a, v] = 5 v , [ a , u ] = 4 u , [ a , y ] = 3 y , [ a , x ] = 2x,[a,s]

=

s}.

Compare also [13] where/2(A) = G6,99 is the very last Lie algebra of a long list. For the Lie group A this means that conjugation by the parameter a leaves each of the five one-parameter subgroups (with respect to the parameter a, x, y, u or v) invariant, so we get

(s,x,y,u,v)a= (as, a2x, a3y, a4u, a5v),a > O. Summarizing we have: A is the semidirect product of the regular nilpotent normal subgroup M and the group L2 = {(a, s)ia > 0, s E ~ } where the action of L2 on M is defined by (apply first a then - s ) :

(x, y, u, v) (a,s) = (a2x, a3y + sa2x, aau + la2xs2 + a3ys, aSv .~_lg a2xs 3 q- ½a3ys2 q_ a4us -- ½a4x2s). The induced action on W \ {co} = {(~,r/)l{,r / C 1R} is (apply a, then - s , then x):

(~,T]) (x's'a) : (a~ -Jr-,-q,a2~] 'if- 182 -}-a{s + x).

2.3. CONSTRUCTIONOF THE GEOMETRY The collineation group A acts transitively on W \ {oo} and contains the transitive translation group A[oo,w], hence A is transitive on/~ \/2[00]. Let L be the line joining (x,y,u,v) = ( 0 , 0 , 0 , 0 ) and ({,r/) = (0,0) then AL is a 2-dimensional group containing the a-parameter group and the translation group in direction ({,r/) = (0,0). Let w_,z_,w+,z+ E I~ be such that ( - 1 , 0 , w _ , z _ ) E L and (1,0, w+, z+) C L, then L contains also the a-orbits of these two points:

{ ( - a 2, 0, w_a 4, z_aS)la > 0} u {(a 2, 0, w . . b a 4 , z+a5)l g > 0}. Applying now the translation group in direction ({, r/) = (0, 0), given by the y-parameter subgroup, we get the affine part of L:

L = {(x,y,f(x),g(x))lx, y e 1R}, where

{ ~_(-~): f(*)=

• x
w+x 2 : x _ > o ' g ( x ) =

{ ~_(-~)~

: x
z+.~

: x>_o

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

263

In order to get the other lines we first apply the shift map t:

(x, y, f(x), g(x) )-t = (x - t, y, f(x), g(x) + yt) and get the line

L t = { ( x , y , f ( x q- t),g(x q- t) q- yt)lx, y C ~}. Following this we apply the s-parameter group and get the points

(x,y + sx, f(x + t) + ½xs 2 + ys,g(x + t)

Substituting ~ = y + sx and denoting ~ by y again, we get the line

L(t, s)

{ ( x , y , f ( x + t ) - ½xs 2 + ys,

g(x + t) + ( y - sx)t-

+ ½ys

+f(x + t ) s - ½x2s)lx, y E ~}. The corresponding parallels through the origin (0, 0, 0, 0) are:

L~t,~)

{ ( x , y , f ( x + t ) - f ( t ) - ½xs 2 + ya, g(x + t ) - g(t) +[f(x + t ) - f ( t ) ] s - ½x2s)lx, y e ~}.

The set of lines {L~t'S)lt, 8 ~ ~ } U {{(0,0, u, v)lu, v C ~ } } is the pencil of lines through the origin. Applying the translation group in direction ~ gives all other affine lines. Note that L~t'`) intersects ~3 = {(0, y, u, v)ly , u, v C ~} in the 1-dimensional subspace {(0, y, ys, yt + ½ys2)[y E ~}, hence the projective line L~t'~) intersects the line W at infinity in the point (~, ~) = (s, t + ~182) . We have thus constructed candidates for projective planes P(~,_,z_,~+ ,z+). 2.4.

CHECKING THE INCIDENCE AXIOMS

A necessary condition for the existence of a plane is that the origin is joined to every point (x, y, u, v) with (x, y) ~ (0, 0) by exactly one line. If x = 0 this is tree since the plane has the 3-dimensional translation group T = {(0, y, u, v)ly , u, v E ]R}. By interchanging the two points and using the action of E on the (x, y)-coordinates, we need only consider points (x, y, u, v) with x = 1, y = 0.

264

D. BETTEN

By the formula for the lines L~ t's) this condition for joining points translates to the following condition: the map h : (t, s) H (u, v) defined by

f(t + 1)- f(t)- ½~2=~ 9(t + 1) - 9(t) + [f(t + 1) - f ( t ) - t -

1 ] s - ½s 3 = v

has to be a bijection. The map h is even a homeomorphism if we suppose that the plane is topological. Denote by J the Jacobian of the map h, then det J < 0 (or _> O) everywhere, since h induces the same orientation at every point. Note that Ou Ov Ou Ov cgt Os Os Ot = - s 2 + (g'(t + 1) - g'(t))s + ( f ' ( t + 1) - f'(t)) 1), x ( f ( t + 1) - f ( t ) - t - 2

det J -

which is a quadratic polynomial in s. Let • be the discriminant, then we have the necessary condition ~(t) _< 0 for all t E lR. We calculate 49(t) = (g'(t + 1 ) - g ' ( t ) ) 2 + 4 ( f ' ( t

x ( f ( t + l) - f ( t ) - t -

+ 1 ) - if(t))

!), 2

and using the special form of f and g we get KS(t) = ( 5 z _ ) 2 ( ( - t ) 3 - ( - ( t + 1 ) ) 3 ) -8w_(2w_-l)(-t-!)

2

2 for t < - i

• (t) = (Sz+(t + 1) 3 + 5 z _ ( - t ) 3 ) 2 - 2(2w+(t + 1) + 2 w _ ( - t ) )

x((1-2w+)(t+l)

2+(2w_-

1 ) ( - t ) 2)

for - 1 < t < 0

¢5(t) = (5z+)2((t + 1)3 - t { ) 2 - - 8 w + ( 1 - - 2 w + ) ( t + k )2

for t > 0 .

We need the following LEMMA ((t+l)3-t~)2f

/~ ( 3 ) 2

for O <_ t --, oc.

P r o o f By the mean value theorem there is some ~ with t _< ~ _< t + 1 and 3 3 1 (t + 1 ) 3 - t ~ g ~ ~. This implies the convergence. To see the monotony, we check that the derivation with respect to t is positive (using the identity

(t + 1)3 - t3 = (t + 1) + (t + 1)~t~ + t). (t + 1) t - t½

4-DIMENSIONALCOMPACTPROJECTIVEPLANESWITH A 5-DIMENSIONALN1LRADICAL 265

A consequence of this lemma is: ~(t) _< 0 for all t _> 0 if and only if ($z+)5 2 < 4(32)22w+(1 - 2w+).

(R)

Similarly, ¢ ( t ) _< 0 for all t _< - 1 if and only if

(5 Z )2

,~ 4(2)22w_(2w_

1).

(L)

In order to find more necessary conditions we use the fact that each pair of lines has to have some intersection point. Using the action of the collineation group we may assume (s, t) = (0, 0) for one of the lines. Therefore the following system of equations has a solution (x, y) for given (~, t) ~ (0, 0) and m, n C ~:

f ( x + t ) - f ( x ) - lx,2 + y~ = m g(x + t) - - g(x) + (y - - sx)t - lxs3 + lys2 + f ( x + t)s - ~1X

28

~ n .

Take first s = 0, t ¢ 0. Then the first equation has a solution x for any m E R if and only if w+w_ > 0. Note that condition (L) implies w_ _< 0 or w_ _> ½ and that condition (R) implies 0 <_ w+ _< ½. Therefore we get the necessary conditions

w__> ½ and O< w+_< 1.

(W)

Next we will demonstrate that the necessary conditions (L), (R) and (W) are also sufficient. (a) det J(s, t) <_ 0 for all s, t E I~. Proof From (W), (L) and (R) it follows ~(t) _< 0 i f t _> 0 o r i f t _< - 1 . We put p_ = 2 w _ ( 2 w _ - 1 ) >_ 0 a n d p + = 2 w + ( 1 - 2 w + ) _> 0 and have for - 1 < t < 0 the estimate (use again (L) and (R)): O(t) _< /~(x/ff+-(t + 1) 3 + x/~--Z-(-t)3) 2 - 2 ( 2 w + ( t + 1) + 2 w _ ( - t ) ) ( ( a - 2w+) x ( t q- 1) 2 -'}-(2W_- 1)(--t)2).

In order to show ~(t) < 0 for all - 1 < t < 0 we divide by ( - t ) 3 and substitute z = ((1 + t ) / - t) 1/2. ~¢¢ehave then, equivalently, to verify 8 ( V / ~ Z 3 -~ V/if-if) 2 ___ 9(2w+z 2 + 2w_)((1 - 2 w + ) z 4 + (2w_ - 1))

for all z > 0. This is equivalent to showing

g(z) = p+z 6 + 2w_(1 - 2w+)9z 4 - 16v/ff~V,ffZ_z3 + ( 2 w _ - 1)2w+9z 2 + p_ _> 0

266

D. BETrEN

for all z > 0. Since p+ Z 6 q- p_ _> 0 we get

g(z) >__z2(2w_(1 - 2w+)9z 2 - 16V'~-~/P-E--z + 9 ( 2 w _ - 1)2w+) >_ 0. The last inequality is true since the quadratic polynomial in z has discriminant 162p+p_ - 4 922w_(1 - 2w+)(2w_ - 1)2w+ = 4p_p+(82 - 92) _< 0. We have therefore shown O(t) _< 0 for all t C ~ which implies det J(8, t) _< 0 for all 8, t E IR. (b) Ifw+ < ½ thendetJ < Of o r t >_ O; ifw_ > ½ thendetJ < Ofort < -1; if

both w+ < ½ and w_ > 1 then det J < Ofor all s, t C ~. Proof Assume t _ 0; by the mean value theorem there is some ~, 0 < ~ < 1, 3=3 _1 3 w i t h ( t + 1)3 - t~ ~¢2.Here,¢
with

0
and get

detJ(s,t) = - s 2 + 5z+((t + 1) 3 - t3)s + 2w+(2w+ - 1)(t + ½) = _ s 2 + ~ 5 z + ~3r~ +

1

r/)~s + 2w+(2w+

1)

(t + r/) + 2w+(2w+ - 1)( 1 - , )

= Q(s,(t + ~)½) + 2w+(2w+ - 1)(½ - r/). The discriminant of the quadratic form Q is A=(~z+.

)2+4.2w+(2w+_l)<0

by(R).

It follows that Q _< 0 and for w+ < ½ we get

detJ(s,t) < 2 w + ( 2 w + - 1)(½ - ~) < 0. The second statement is proved similarly. If w+ < ½ and w_ > ½ then the discriminant at the end of(a) is < 0 since p_ > 0 and p+ > 0 and 4p_p+(8 2 - 9 z) < 0. Therefore also q~(t) < 0 and det J ( s , t ) < 0 f o r - 1 < t < 0. (c) Intersection of two different lines: Since w+w_ > 0 two distinct lines intersect if s = 0, t ¢ 0. Now take 8 ¢ 0. Then we can eliminate y and get one equation for x. This equation has the form

wl2

t,

(

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

267

where we have set w = w+ for x, x + t _> 1, w = w_ for x, x + t _< 0 and r C is some constant. Note that we do not need the formula for the middle part of this function in the following. If w+ # I (w_ # ½) then the growth of the function for x -+ oo (x ~ - o o ) is determined by the quadratic term in x. This follows from the fact that the term g(x + t) - g(x) has growth of exponent -3 2 only (mean value theorem). Thus, if both w_ and w+ are different from ½, then we get a solution of the equation since w+ - 21-and w_ - ½ have different sign by (W). If, say, w+ = ½, then (R) implies z+=O

and

-1

(184+2w+t2)x

~-sign(s)oo

for x - - , o o .

8

Since w_ - ½ _> 0, the function converges to sign (8)00 for x --+ - o o and the equation has a solution x. (d) The map h is a proper map, i.e. if I(t, 8)1 -+ oo then ](u, v)l --+ oo. Proof. I f t _< S remains bounded for I(t, s)l -+ oo then u -+ - o o and I(u, v)[ --+ oo. Therefore we may assume 0 < t --+ oo in the sequel. In this case the map h has the form

u = 2w+(t+½)-½s

2

v = z+(t + 1) 5 / 2 - z+t 5/2 + (2w+ - 1)(t + ½)s - 183 whose Jacobian matrix

j =

3u Ou ) at os O..~v O._2v Ot 0s 2w+

=

-8

)

z+S(t + 1) 3/2 - z,~_2 ~5-'3/2 + (2w+ - 1)s (2w+ - 1)(t + ½) - s 2

has determinant det J ( s , t ) = - s 2 + (z+5(t +

1)3/2- z+5t3/2)s +

2 w + ( 2 w + - 1)(t + ½).

If w+ = ½ then z+ = 0 by (R) and the map h restricted to t > 0 reduces to: = t + ½ - ½8 2

v = -½8 3

and h is obviously proper. Therefore we may assume w+ < ½. It follows by (b) that det J < 0 and that the restriction hit>0 is a local diffeomorphism. We want

268

D. BETTEN

to use the theorem of Hadamard, see [14, 5.3.10], and check if IIJ-1 II is bounded above for t > 0. Replacing ( t + 1)3 - t3 by 3(t +r/)~,0 <~ r j < ½, in J wehave

J=

z+iS.~(t+r/)½+(2w+-l)s

(2w+-l)(t+r/+½-T/)-s

2

"

Note that det J < 0 for t __ 0. Now, itJ_ll I = (2w+) 2 + , 2 + ( z + ~ . -~(t + r/)~ + (2w+ - 1)s) 2 1) (--82 --}- ~5 Z + ~3( t "1- ?'1)½8 "-}-2w+(2w+ +((2w+ - 1)(t + ~ + ½ - ~/)- s2) 2 ×(t + r/)+ 2w+(2w+ - 1)(½ - r/)) 2" The discriminant of the quadratic form Q(s, (t + rl)½) in the denominator is A = (-~z+ .3)2 + 4 . 2 w + ( 2 w + - 1) _< 0 by (R). We suppose first A < 0. Then the quotient is essentially

iij_~l I ~

( ( 2 w + - 1)(t + ~ ) - s2) 2

(_~2 + ~z+~(t 5 3 + ~)½~ + 2w+(2w+ _ 1)(t + ~))2 and remains bounded for (t, s) --+ oo. By the theorem of Hadamard hit>0 is a diffeomorphism and in particular proper. If A = 0 then we cannot apply the theorem of Hadamard. But in this case we will see the properness directly from the form of the map h. From the first equation we have 1

u

t+2-

2w--7 + ~s2"4w+ 1

Since by the mean value theorem (t + 1)~ - t~

(t+ ~)z 1

3_

5 --+- for t--, o0 2 5

.

we replace in the second equation (t + 1)~ - t~ by ~-(t + 1)3 Inserting t + gives v~

z+- - - +

s2

2 2w+

+(2w+-l)

+ 1-Ls2~s-~,~. 4w+

If u remains bounded for (s, t) + 0% then Is[ --+ c~ and 5

1

-3

v "~ z+-~(~w+)=s3 + (2w + - 1 ) 4 1 + s 3 - ~ s 3 = a s 3 .

/

1

4-DIMENSIONAL COMPACT PROJECTIVE PLANES W I T H A 5-DIMENSIONAL NILRADICAL

269

To prove the properness of h it suffices to check that the coefficient a is # 0. Assume by contradiction that a = 0. Then also 5 2 1 3 / 2 w + -- 1 ( Z2+) - ( 4- w- + ) = |\ ~w+

1)2.

Since we are in the case A = 0 we may substitute

and get 2 . 4 ( 1 - 2w+) = (3(2w+ - ~ - 4 ~ ÷ ) ~ 32(4w+)2 \ 3.4w+ which leads to 4w 2 + 4 w + + 1

=0

or w+ = - ½ .

This is a contradiction to w+ > 0. (e) Joining two distinct points. We have to show that the map h" (t, 8) ~ (u, v) is a homeomorphism. Forw_ > I andw+ < ½ we have proved in (b) that det J ( s , t ) < 0everywhere. This means that the map h is a local homeomorphism. Since the map h • ]~2 __+ R2 is also a proper map, it follows that h is a homeomorphism, see for instance [6]. In the mixed case, say w+ = ½, w_ > ½, it follows from (L) that ~(t) < 0 for all t < - 1 . For the function g(z) in (a) we get g(z) = p+z 6 + p _ > O, and therefore ~ ( t ) < 0 also f o r - 1 < t < 0. Thus, ~(t) = 0 may only occur for t _> 0. But then d e t J = - 8 2 = 0 only i f s = 0. Let N := {(s, t) [ t > 0, s = 0} be the null set for det J. Then

hIN " N ~ h( N ) , u = t + ½, v = O is a bijection. Next we state that h ( ~ 2 \ N ) C IR2 \ h(N): if (~, t) ¢ N then either t < 0 or t _> 0 and ~ # 0. In the first case

f(t+l)-f(t)<

f(0+l)-f(0)=w+

=

1

1 since f is planar [19, 74.1-3] which implies u(a, t) = f ( t + 1) - f(t) - ½~2 < 5" In the second case we get f ( t -4- 1) - f ( t ) = t + ½ since w+ = ½. Therefore

v(8, t) = ( f ( t + 1) - f(t) - t -

~1 ) s - ½J

In both cases we have proved h(~, t) • h(N).

=

-½s 3 # 0 .

270

D. BETTEN

Now we know by (a) and the definition of N that det J < 0 on ~2 \ _/V. Furthermore the restriction of h to ~ 2\ N is again a proper map (note that properness may also be described by the property: the preimages of compact subsets are compact). Therefore the restriction h : ~2 \ N ~_ ]~2 ___+]R2 ~ ]R2 \ h(N) is a local homeomorphism which is also a proper map and therefore a homeomorphism. Since h: U --+ h(N) is a bijection, we see that h : (s,t) ~ (u, v) is a continuous bijection of ~2 to ]~2 and therefore a homeomorphism. The other mixed case is proved similarly. In the special situation w_ = w+ = ½ it follows z_ = z+ = 0 and the map h reduces to t + 1 _ ½+=

u,

v.

This is obviously a homeomorphism. We have thus shown that the incidence axioms of a projective plane are fulfilled. That the plane is a topological projective plane follows by using Criterion 3.5. 3. Case C with

Lie Algebra

G6,94

We consider next Case C of [4], which can be described by '(v, a) = (3, 3); N, A : (1,2)'. This means that the translation group with respect to W has dimension 7- = 3 and the group of elations with respect to the point ~ has dimension a = 3. Furthermore one orbit in P \ W under the translation group is fixed by A and N acts transitively on W \ { ~ } . We will first determine the structure of N and prove then that only two Lie algebras £ ( A ) are possible. The geometries for these two Lie algebras are determined in this and the next section. 3.1. THE STRUCTUREOF N The Lie algebra Af is uniquely determined and has the structure H=

{(y, u , v , s , n ) ] [ s , y ] = u , [ s , u ] = v , [ n , y ] = v}

(and all other commutators are = 0). To prove this let R 3 = {(y, u, v)[y, u, v C ~} be the translation group which we identify with the fixed orbit "x = 0". Since the translation group acts regularly on this fixed orbit and is a normal subgroup of A, the group A is a semidirect product of ~3 by a 3-dimensional linear subgroup (the stabilizer). Similarly N is a semidirect product of ~3 and the 2-dimensional stabilizer No. We show that, up to conjugation, the Lie algebra of No is generated by the two endomorphisms

s=

(0) 10

010

,

n=

(0) 00

100

.

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

271

Proof By [11] each4-dimensional, connected commutative collineation group of a 4-dimensional projective plane is isomorphic to ll~4 , R3 x 5'02 or JR2 × (5'02) 2. By [8] the first case leads to translation planes, dual translation planes or shift planes, and all these flexible planes are known. The second case leads to planes with two fixed points and two fixed lines, which were studied in [10]. The group R 2 x (5'02) 2 fixes three non-collinear points, and in this situation the full group A is only 4-dimensional, [11], a contradiction. Hence we may assume that the Lie algebra of N does not contain a subalgebra isomorphic to R4. The group No fixes the line 0 V ee and therefore the corresponding 3 by 3 matrix has zeros at positions (1,2) and (1, 3). Assume that No acts transitively on the set of 1-dimensional subspaces of the space {(0, u, v)lu , v E IR} then the group induced by N on 0 V eo would contain the Euclidean motion group. Since this group is not nilpotent we get a contradiction. Choosing the coordinates suitably we may therefore suppose that the matrix group No consists of lower triangular matrices. Again by nilpotency these matrices have 1 on the diagonal. It follows that the Lie algebra of No is a 2-dimensional subalgebra of

{(0) m 0 nsO

]m,n, s E ~

} •

If all endomorphisms had s = 0, then the group N would have a subgroup of the form

{(1 )uvmnC} 0 1

urn1 vnO1

This group is isomorphic to ~4, a situation excluded above. It follows that we may generate the 2-dimensional Lie algebra by the two endomorphisms

a 0

blO

,

0 O0

,a,b,q, rC~.

Since a 2-dimensional nilpotent Lie algebra is commutative, commutation shows that q = 0, which allows to set b = 0 and r = 1. If a = 0 then the one-parameter group

{(1) } 0 1 Osl

Jsc~

272

D. BETTEN

fixes the space { (y, 0, v) lY, v E IR} which contains quadrangles, a contradiction to the lemma on quadrangles. Therefore a ~ 0 and by conjugation

(a)(o 01 001

a0 010

0 0

1 01

) (o) =

10 010

we may take a = 1. If we denote by y, u and v the elements of the Lie algebra which generate the translations in the directions y, u and v, then we get the Lie algebra defined above. This is the Lie algebra Gs,3 in the list [5] and G5,5 in the list [13]. 3.2. THE STRUCTURE OF/~(A) There are two possible Lie algebras £ ( A ) = (A/', a), defined by the commuting relations: ~6,94 : [a,v] -- ()~-{- 2)v, [a,u] -- ()~q- 1)u, [a,y] = Ay,

[a,n]=2n, [a, sl=s, ~6,98 : [ a , v ] = v, [a,u] ~-~ hv--~ u, [a,y] = y , [a, n] = O, [a, s] = hn, h e 1R. P r o o f We may choose the one-parameter group corresponding to the sixth gene-

rator a E £ ( A 0 ) ) in such a way that it fixes the point at infinity (~, r/) --- (0, 0). Since (s, n) is an ideal in the linear stabilizer £ (A0) = (s, n, a), the endomorphism a must have the following shape: a=

o)

0 c + k 0 f c+2k

, c , k , f E IR.

From this we get the commuting relations [a,y] = cy, [a,u] = (c + k)u + f v [a, v] = (c + 2k)v, [a, s] = ks + f n , [a, n] = 2kn. If k = 0 then c ¢ 0 otherwise £ ( A ) would b e nilpotent. Taking lc a instead of a we get the Lie algebra ~6,98 with h = ( f / c ) . For k • 0 we substitute s ~ s - ( f / k ) n , u ~-+ u - ( f / k ) v and we may suppose f=0. After the substitution a ~ ( 1 / k ) a we get the Lie algebra G6,94 with )~ = ( c / k ) . (The numbering is taken from [13].) 3.3. CONSTRUCTIONOF THE GEOMETRY FOR G6,94 We consider the Lie algebra G6,94 and its corresponding universal Lie group A. We define the following 'standard action' of A on the affine point space P \

4-DIMENSIONALCOMPACTPROJECTIVEPLANESWITHA 5-DIMENSIONALNILRADICAL

273

W coordinatized by ]R4 = { ( x , y , u , v ) l x , y,u,v E ]R}: A is generated by the translation group ]R3 (which we identify with the fixed orbit 'z = 0') and the following three linear one-parameter groups:

s2

I~ E ~

{{er

s 1

'

t

-ffTsl

It E IR

1 t

'

1

e(l+k)r e(l+2k)r e(l+3k)r

Here, k is some real parameter and the structure of ~6,94 implies k ¢ 0. It is easily checked that the group A so defined has the Lie algebra G6,94 (3~ = (1 + k)/k). Now we propose that each collineation group A with £ ( A ) ~ G6,94 acts on P \ W equivalently to the standard action just defined.

Proof This equivalence is clear for the restriction to the fixed orbit 'x = 0' since this restriction is the semidirect product of the translation group and the stabilizer of the origin. We have still to show that the globalization of this action to P \ W is uniquely determined (up to equivalence). The 3-dimensional group E ,of elations with center cx) has the form 01

0 u v

1 n

1]~3.

[ u,v,n E R 1

n

1

By identifying the vertical line L(z,v ) E £ ~ \ W with the 1-dimensional subspace of E which has the line L as axis of shears we get the horizontal coordinates

x =-u/n,

y =-v/n.

Now we start with the action of A on the fixed orbit 'x = 0'. Letting A act on E via conjugation, this allows us to calculate the horizontal action of A:

( (

e(l+k)~ e(l+2k)~ e(l+3k)r

1 O1 uO1 vlO1

1

e-(l+k)~ e-(l+2k)r e-(l+3k)r

)

×

274

1

0 1 e(l+2k)ru 0 1 e(l+3k)rv e 2kr 0 1

D. BETTEN

)

hence ( x , y ) -~ ( - x , - y , 1)~ r (_e(l+2~)~x, _e(l+3k)~y ' e2k~)~ = ( - C x , --e(l+k)~y, 1)II~ -~ ( C x , e(l+k)~y). Similarly, 1 Yl 1 001 0001

0 1 uO1 v 1 0 1

)(1 )(, ) -Yl 1 0 O1 0 O0 1

--

0 u v-y1

1 O1 O0 1

implies

(x,y)-~ (--x,--y, 1)R Y'> ( - x , - y - yl,1)I~ (x,y+ yl). This shows that the horizontal coordinates are the right ones. But in the geometric context the topology on P \ W is uniquely determined by the horizontal coordinates. It is the direct product of one horizontal line and the translation group {(u, v)lu , v E ~ } in direction oc. In order to construct the geometry we first note that all orbits under the translation group A[oo,w] must be lines, therefore the geometry contains all 'vertical' lines

s(~,~) = {(~,y,~,v)lu, v e ~ } ,

x, y e ~ .

Let L(°,°) be the line which joins the origin (0, O, O, O) and the point (~, 7?) = (0, O) at infinity and assume that this line contains the points ( - 1, O, w_, z_) and (1, O, w+, z+). If we set

-l:x =

~(x)

=

O:x

<0 =

l:x>O

0

(w_'E=-I

z_:c=

-1

w+'¢=l

Z+:E:

1

, wE =

then these two points can be described by (c, 0, wE, zE), c = 4-1, and the following calculations hold for x _< 0 and for x _> 0. The line L(°,°) can be constructed from (c, 0, wE, z~) by applying the r-parameter group (which fixes the origin) and the group of translations in direction (~, r/) = (0, 0) (the y-parameter group): L (°'°) = {(x, y, (ex)l+2kwc, (ex)l+3kz~)]x, y E I~}.

4-DIMENSIONAL COMPACTPROJECTIVE PLANES WITH A 5-DIMENSIONALNILRADICAL 275 Application of an element of the s-parameter group gives the points: (X, y Ar 8X, (~x)l+2kWe ~- 8y ~- 182X, (£x)l+3kze

Substituting Y = y + 8x and denoting Y afterwards by y again we get the line

L(°, ") = {(x, y, (~.)1+2%. + 8 y - 182x, (,~)1+3~z,

+(~x)1+2%~8 + ½82y - ~8~x)lx, y ~ ~}. Finally we apply an element of the t-parameter group (the shears with respect to S(o,o)) and get the line L (t's) :

{ ( z , y , ( ~ * ) l + 2 k W e AF [t-- 182]X ~- 8 y , ( ' x ) l + 3 k Z e +(6-x)l-k2kWe8 -- 183X + [182 "~ + t]Y)lX, Y C R}.

Note that L(t,~) intersects W in the point (~, r/) with ~ = s, r] = ½82 + t. We have thus constructed the pencil £0 = {L (t'~) 18, t C R} U {L(0,0)} of lines through the origin. The other lines are the images under the translation group in direction oo. 3.4. CHECKING THE INCIDENCE AXIOMS If w_ = z_ = w+ = z+ = 0 then the plane is a translation plane and the full collineation group has dimension > 6 (and does not fix the orbit x = 0). Since we want to construct a topological projective plane, a necessary condition is: k > -½ if z_ = z+ = 0 and k > -½ if (z_,z+) 7£ (0,0).

Proof The functions u(x, y) and v(z, y) in the formula for L(t,~) have to be continuous. If z+ = z_ = 0 then w+ ¢ 0 or w_ ¢ 0 and from the continuity o f ( e x ) l + 2 k w ~ i t f o l l o w s l + 2 k > 0 o r k > 1 g. I f z + ¢ 0 0 r z _ ¢ 0 t h e n t h e continuity of (ex)l+3kz~ implies 1 + 3k > 0, i.e. k > _1. We will now check whether two points (xl, Yl, ul, vl) ¢ (x2, Y2, u2, v2) are on a unique line. This is true if the following system of equations has a unique solution #, u, 8, t: [~(Xl)Xl]l+2kWe(xl) Ar (t -- ½82)Xl -]- 8y 1 @ ]A -~ U 1 [e(X2)x2]l+2kWe(x2) q- (t -- 21-82).2 -1- sy2 q- # = u2 [~(Xl)Xl]l+3kZe(xl) q-[E(Xl)Xl]I+2kWe(xl)8 -- lX183 q- (182 "~ t)yl q- P = Vl

[~(.:)x:]1+3%(~:) + [~(~:)x:]~+2'%(~,)8 - }x:.~ + (½8 2 + t)y: + ,. = ,~2

276

D. B E T T E N

Equivalently, the following system of equations must have a unique solution 8. for given a, b C II~:

We consider first the case xl = x2: If also ffl = Y2 then the two points are on the vertical line S(xl,yl). If yl ~ y2 then it is obvious that there exists a unique solution s, t. Let now be Xl ~ x2, say xl > x2. Then by eliminating t we get the following polynomial equation of third degree in s:

. v -

v - z

A unique solution 8 exists if and only if the derived function is semidefinite. For this we must check if its discriminant is _< 0: Up to the positive factor 4(xl - x2) this discriminant is

I f 0 _< x2 < xl then A = e(x 1l + 2 k - x 2l + 2 k x )w+ < 0 if and only if w+ _< 0. Similarly, for x2 < xl _< 0 we have A = ( ( - x l ) 1+2k - ( - x 2 ) l + 2 k ) w _ _< 0 if and only if w_ _> 0. Conversely, if one assumes w+ _< 0 and w_ _> 0 then for x2 < 0 < xl we get A = x]+Zkw+ -- (--x2)l+2kw_ _< 0. Thus under the assumptions 1 + 2k > 0, w_ _> 0 and w+ _< 0 two distinct points are uniquely joined by a line. In order to check if two different lines intersect, we may assume that one line has the parameters (s, t) = (0, 0) and the other one (8, t) ¢ (0, 0). This leads to the system of equations

For s = 0 we have the solution x = ~ , y = ~, hence we may assume s 7~ 0 and even s = 1 by using the action of A on W. Inserting y = a - (t - ½)x from the first equation into the second one we get f(x)=

(t 2 + ~ ) 1x + B ,

BEIR.

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

277

Since f is decreasing by the condition k > -½, w_ _> 0, w+ _> 0 and since t 2 + ~ > 0, we get a solution x. It follows that two different lines intersect. The intersection point is unique since we have already proved that two different points are on a unique line. Inspection of the lines L (t'~) = { ( x , y , u ( x , y , s , t ) , v ( x , y , s , t ) l s ,

tE ~},s,t E

through the origin shows that the functions u and v depend continuously on x, y, and t. Therefore the projective planes are topological by the following criterion. 3.5. CRITERION Let 79 = (P, £ ) be a projective plane with line at infinity W, a point oo E W and coordinates P \ W = {(x, y, u, v)lx, y, u, v E IR} on the affine point set. Let

s(x,y) = { ( x , y , u , v ) l u , v e

E

be lines through {c~} and let y, u , v )

(x, y, u + re, v +

E

be translations with axis W and center { ~ } . Suppose that W \ {oc} is coordinatized by {(s, t)ls , t E R} and that the lines ~ S(0,0) through the origin (0, 0, 0, 0) are described by L (~,t) = { ( x , y , u ( x , y , s , t ) , v ( x , y , s , t ) ) l x ,

y E JR}

including the lines E = { ( x , y , x , y ) [ x , y E ~ } and X = { ( x , y , O , O ) l x , y , E ]~}. If the functions u and v depend continuously on x, y, s and t then the plane is a topological projective plane. Proof We introduce coordinates suitably and calculate the ternary ring. The assumptions imply that the ternary operation is continuous and by the theorem of Knarr and Weigand [9] the ternary ring is topological. See also [19, 43.8]. Remark. This criterion also applies to 4-dimensional translation planes. If the spread in ~4 = { ( x , y , u , v ) l x , y , u , v E ]~} is defined by the vertical line S(0,0) = { (0, 0, u, v)lu, v E ]R} and the system of 2-dimensional subspaces,

v

=

c(s,t) d(s,t)

y

'

then it is sufficient to check whether the four entries a(~, t), b(8, t), c(8, t), d(8, t) depend continuously on s and t. Remark. Note that in [1] two cases for planes with a 3-dimensional translation group were established: the space of orbits on P \ W under the translation group

278

D. BET~N

is homeomorphic to ~ and A is either transitive on this space ('transitivity') or A fixes some element of this space ('fixed orbit'). Case C is a mixed situation, where both cases occur at the same time. For the plane itself there exists a fixed orbit and for the dual plane we have the situation 'transitivity'. 4.

C a s e C with Lie A l g e b r a

~6,98

We consider now the second Lie algebra which is possible in Case C, see Section 3.2. This Lie algebra may be generated by ~3 = {(y, u, v)ly, u, v E I1~}and the three endomorphisms s=

(0) ( 0 ) ( , ) 10

,

n=

00

010

,

a=

01

100

,hEir.

Ohl

The map u ~ - u , s ~ - s , v ~ v, y ~ y, n ~ n, a ~ a defines an isomorphism of the Lie algebra with parameter h to the Lie algebra with parameter - h . Hence we may suppose h >_ 0 in the following. For h = 0 the group {exp ta]t C ~ } is a group of homologies and we are in a situation already studied in [1]. Therefore we consider only h > 0 . 4.1. THE COLLINEATIONGROUP We prefer to work in the dual situation. Here we have orbit type (2,1) which means that N and A act transitively on £ ~ \ {W} and N and A have a 1-dimensional orbit on W \ { ~ } . The collineation group acts transitively on the space of points W (formerly lines) and this is an RNS-action. There is a regular nilpotent normal subgroup (Lie algebra (s, n, u, v)) and a 2-dimensional non-commutative stabilizer (Lie algebra (y, a)). We will now describe this action in more detail. We introduce coordinates ~4 = {(8, n, u, v)[8, n, u, v C ]~} on the affine point space. The group ll~3 corresponding to the Lie algebra (n, u, v) is the translation group with respect to the line W at infinity. The action of the fourth parameter s on the affine point space can be calculated by the following matrix multiplication: (1 0 1 0 o 1 0 -2- a 1

1 0 u

( 1 s

v -£+n

1 s 1

=

1 0 u v+{ru

) 1 8+(7 2

+n

1 s+a

" 1

Hence the regular nilpotent normal subgroup acts on itself by left translations in the following way (81,nl,Ul,Vl).(O,n,U,V)

= (S+sl,n+nl-l-slsW-~,uq-ul, v q- vl q- 81u)~81~nl~ Ul~ Vl~8~n~ u~ v E IR.

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

279

By conjugating (in the matrix representation of the group given at the beginning of this section) we get the action of the stabilizer:

( s , n , u , v ) ~ = (s,n + has, eau, haeau +eav),a > O, and (~,n,u,v)y = (~,~,u-y~,~-y~),

yea.

We have thus determined the action A on P \ W. From this action we see that the Lie algebra £ ( A ) - G6,98 may be generated by the following vector fields

o

o

o s'

On~Ou~Ov~

:

o-~+s£+u

0--~- a'

Ov~

:

hso-~+uo-~+(hu+v)o~,

y'---- - - 8 ~ u -- n0--~ .

The points of W \ {ec} correspond to the l-dimensional subspaces (n, u, v)IR, n 0 of the translation group T. Therefore, by letting A act on T via conjugation we get the induced action of A on W \ {oc} = {(5, rl)]5, rl E 1~}:

(~,rl) ~) ( e ~ ,haea ~ + earl), (~,rl) y) (~,rl_ y) , (~,rl) s)

(~, 4 + rl). The translation group T with Lie algebra < n, u, v > acts, of course, trivially on W. Note that the line ~ = 0 is invariant under the action in accordance with the orbit situation (2,1). 4.2. CONSTRUCTION OF THE GEOMETRY We need only construct three special lines, passing through the points at infinity (~,rl) = ( - 1 , 0 ) , ( 0 , 0 ) and (1,0), the other lines can then be determined by applying the group. From the induced action on W \ {c~} it follows that the stabilizer of the point (~, rl) = (e,0), e = - 1 , +1 is given by a = 1 and s = ~y. The connected component of the stabilizer on a line through (e, 0) may be generated by two vector fields

0 0 0 x = s' + Ey' + q' 0-~n + W'~u + Z'~v 0 0 Y = o---~+ E -Uu -

with real numbers q,, w,, z~ C ~. By switching from X to X - q,Y if necessary one may assume that q, -- 0. The vector field X generates a one-parameter group which leaves the line invariant and acts transitively on the set of orbits under the translation group. The vector field Y generates the 1-dimensional translation group with axis W and center (~, rl) = (e, 0). By inserting s' and y ' we get

0

x = ~

+ ~

0

+ (w~ - ~ )

+ (u + z~ - ~n)

.

280

D. BETTEN

By integration we get the one-parameter group (8,~'t,U,V)

exptX

-:

t2

t2

( 8 + t , n + st + - ~ , u + ( w , - e s ) t v + (u + & - en)t + ( w , -

- e-~,

2e8)~- - 2e~).

The one-parameter translation group is (8,?'t,U,V)

exprY

= ( 8 , n '1-

r , u + er, v).

We now apply these two groups to the origin (s, n, u, v) = (0, O, O, O) and replace t and r by 8 and n, respectively. This gives the points

82

By substituting N = T q- n and writing n again we get the lines L(~,°) =

a , n , W e S - e 8 2 q-en, Z e S - k w e - ~ - e - ~

la, rtE ]R .

which join the origin and the points (~, y) = (e, 0) at infinity. We now apply the stabilizer of the origin which is generated by the a-parameter subgroup and the y parameter subgroup and get the lines through the points at infinity with coordinates (~, ~1), ~ ~ 0. In this way we get two orbits consisting of the following lines: L(¢,o) = { ( s , n , ea(wcs - es 2 + e(n - 2has)) + rlS , haea(weS

-

f.8 2 -

82

83

ehas) 4- ea(ze8 q- we y - e T ) +

tin)Is, n c R} where ( = eae and e = q-l, - 1 . The lines which join the origin with the points of the fixed 1-dimensional orbit on W can be determined by the limit process ( ~ 0. Since for a ~ - c ~ it follows ae a --+ 0 and a2e a --+ 0 we get L (°'~) = lim L (¢'v) = { ( s , n, r/s , y n ) l s , n E ~ } , rl C ~.

¢-.o

4.3. C H E C K I N G T H E INCIDENCE AXIOMS

We want to show that the origin can be uniquely joined to each point (s, n, u, v) (0, 0, 0, 0) by a line. This is tree for a = 0 since the lines intersect the set {(0, n, u, v)ln , u, v E ~ ) in 1-dimensional affine subspaces. Therefore we may suppose 8 ~ 0. Moreover, using the action of the stabilizer on the origin, it suffices to consider n = 0. This leads to the following system of equations:

*DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL ea(wes

-- ¢8 2 --

281

2hase) + ~]s = u

h a e a ( w , s - es 2 - hase) + e a

&s + w , - ~ - e-~

= v.

After dividing by s we can write the second equation as f ( a ) = ea ( - e h 2 ) a 2 + h ( w e - C s ) a + z e + w e

i8 - - 6 T

=- 8

Let us first assume v ~ 0. Since f ( a ) ---+ ( - e ) o o for a -+ + c ~ and f ( a ) ---+0 for a ---+ - o o there is at least one solution a (take e = - 1 if sV > 0 and e = 1 for < 0). This solution is unique if the derivative 8

"1

+ h ( w , - es)]

is definite. Let A be the discriminant, then this is equivalent to A h2

__

l s 2 + w e +2 4 h 3

2+4e&<0

for all s ¢ 0. Hence, we get the condition W2

q- 4h 2 + 4e& < 0

for the uniqueness of the value a. The value r / c a n be calculated from the first equation. N o w assume v = 0. Then the quadratic polynomial f ( a ) • e - a has discriminant

h-5 = (w~ - es) 2 + 4 e =

(

& + w~2

s2)

We2 q_ 4e& - le2823 < W,2 + 4e& < We2 + 4h 2 + 4e& <_ 0.

It follows that f ( a ) = 0 has no solution. This means that there exists no joining line of the form L((,~), ~ > 0, or L((,~), ~ < 0. But the point (s, 0, u, 0) lies on the unique line L (°''), 77 = u / s , of the 1-dimensional orbit. Explicitly, the condition for existence consists of two inequalities, namely e=+l =-1

• w2+4h 2+4z+_<0 • w2 +4h 2-4z_

<0.

282

D. BETTEN

Note that this implies z+ < 0 < z _ . Let us check that two different lines intersect. If one of the two lines has the form L(°,~) we get calculations similar to those for joining points. Thus we may suppose that both lines intersect W in { # 0. Let the first line have 'slope' al, ~l and side E1 and let the second line have 'slope' a2, ~]2 and side e2. If e=lcl = ea2e2 then it is easy to see that the system of linear equations for the intersection point has a unique solution. In the general case ealc1 ~ ea2e2 this system of equations has the form dn

+

Q(s)

6(8) + (T]I- /2)n

= A = B

where d # 0 , A, B E ~ , Q(8) is a quadratic polynomial in 8 and C ( s ) is a cubic polynomial in 8. If 7?1 = r/2 then we get 8 from the second equation and afterwards d from the first one. If Yl # ~/2 then we insert n from the first equation into the second one. The resulting cubic equation has a solution s and from the first equation we get n. In order to prove that these projective planes are topological, we use Criterion 3.5. The lines # S(o,o) through the origin have the form

L (~'') : { ( s , n , u ( s , n , ¢ , r l ) , v ( s , n , ( , ~ l ) ) ]

s , n 6 ~},(,~/6/i~.

The functions u and v determined above are continuous since aC' ---+ 0 and a2e a ---¢"0 for a ~ - o o .

5. Case D In Case D of [4] we have (7-, a) = (3, 3), N : (1, 2) and A : (2, 2). This means that the group of translations with axis W and the group of elations with center oo are 3-dimensional. Furthermore, the nilpotent normal subgroup N has a 1-dimensional orbit in/2oo \ { W ) and acts transitively on W \ {co}. Furthermore, the group A acts transitively on L2oo \ { W ) and transitively on W \ {c~). We will prove that this situation does not lead to a projective plane. From the orbit type N : (1, 2) it follows as in 3.1 that the 5-dimensional nilradical has the structure A f = {(y, u , v , s , n)l[s,y] = u , [ s , u ] = v , [ n , y ] = v ) , where (y, u, v) corresponds to the 3-dimensional translation group A [ w , w ]. Let R' be the space of orbits of N on £oo \ { W ) , then R" is homeomorphic to ~ [1], and A acts transitively on A"by the assumption A : (2, 2). It follows that the induced action is either isomorphic to the L2-action {x ~ ax + b[ a > 0, b E I1~)or is isomorphic to the R-action {x ~-+ x + b [ b E ]R}. In the first case the commutator A / of A would also act transitively on ,-V. Since the commutator A' of the solvable group A

4-DIMENSIONALCOMPACTPROJECTIVEPLANESWITH A 5-DIMENSIONALNILRADICAL 283

is nilpotent, we have A I _< N and A / fixes an element of A" (the fixed orbit unter N). This contradiction implies that the induced action of A on ,V is an k-action. We introduce horizontal coordinates £ ~ \ {W} = {(x, y) I x, y E II~) and note that the horizontal action is an affine action (since a = 3). The stabilizer on L(0,0) fixes all lines L(0,y), y E ~ , and all orbits x = const., hence the transitive horizontal action is the affine action N i l = {(x, y) ~ (x + m, y + cx + n) I c, ra, n E R). Therefore the group A must be an extension o f ~ 3 ~- A[~,oo] by the group Nil. We now use the table No. 8 of [13] where all 6-dimensional real solvable Lie algebras are written down which have the 5-dimensional nilradical A/'" CI~C2~c3~e4~C5)~C 3 o c 4 = CI~C 2 o e 5 = Cl~e 3 0 e 5 = C2

9

c 1 oc 6

e 2oC 6

e 3 oc 6

c 4oe 6

e5 oe 6

96,94

(~ + 2 ) e l

(/~ + 1)e2

/~e3

2e4

e5

96,95

2Cl

e2

0

c 1 + 2e4

e5

96,96

3el

2e2

e3

e2 + 2e4

e3 + e5

96,97

4el

3e2

2e 3 or_ e4

2e4

e5

96,98

el

h e 1 + e2

e3

0

he4.

The correspondence to o u r notation is: e 1 ~-+ v~ c 2 F-+ u~ e 3 ~ y~ e4 ~-+ n~ e 5 s. The Lie algebra of A[oo,~] is generated by u, v and n corresponding to e2, el and e4 and we have to check if the quotient algebra of 96,, modulo (el, e2, e4) is isomorphic to Nil. But by the table in all five cases this quotient algebra has a subalgebra isomorphic to L2, a contradiction. 6. Case B: (r, a) = (3, 3);

N,A:(2,2)

In this case N (and A) acts transitively on £oo \ { W ) and transitively on W \ {oo). For the group A[oo,oo] of elations with center c~ and the group A[w,w ] of translations with axis W we have dim A[oo,oo] = dim A[w,w ] = 3. 6.1. THE LIE ALGEBRA/~(A) The Lie algebra is isomorphic to 96,94 with )~ = 2, in our notation: £ ( A ) = (A/',a)

with

N ' = {(y, u , v , s , n ) [ [ s , y ] = u , [ s , u ] = v , [ n , y ] -- v ) , [a,y] = 2 y , [ a , u] = 3u, [a,v] = 4v, [a,s] = s , [ a , n ] = 2n} and all other commutators are = 0. This Lie algebra may be written as a semidirect product oflR 3 ---- (y, u, v) and (s, n , a ) , where s=

(Z) 0 10

,n=

O0 100

,a=

3

. 4

284

D. BETTEN

The embeddings of the translation group with axis W and the elation group with center c~ are defined by £(A[w,w]) = (y, u, v) and £(A[oo,oo]) = (u, v, y + n). Proof The group N/(A[oo,o~]) is 2-dimensional and nilpotent, hence commutative. This implies N ~ C A[~o,~o]. Analogously, N ~ C A[w,w] and therefore N ~ C A[oo,w], i.e. d i m N ~ < 2. If we again assume that the Lie algebra .M of N does not contain a subalgebra ~4, we see from the list of 5-dimensional nilpotent Lie algebras ([5], [11]) that there are only two possibilities: d i m N ' = 2 : [ n , y ] = v , [s,y] = u, [s, u] = v , dim N r = 1: [n,y] = v , [ s , u ] = v, and all other commutators are = 0. Assume first dim N r - 2, i.e. .Af-- {(y, u , v , s , n ) l [ s , y ] = u,[s, u] = v , [ n , y ] - - v } . This Lie algebra already led to projective planes in Section 3 where A[w,w ] and A[~,oo] corresponded to (y, u, v) and to (u, v, n), respectively. We will now determine the subalgebras of Af which correspond to the groups A[w,w] and A[oo,oo] in Case B. They must contain the commutator algebra AP = (u, v) of Af and have therefore the form (u, v, a y + bn + cs), a, b, c C ~. From the commutativity it follows that [ay + bn + cs, u] = cv = 0, which implies c = 0. Hence £(A[w,w]) and L;(A[c~,oo]) are both contained in the 4-dimensional Lie algebra (u, v, y, n), which is the direct product ofI~ ~ (u) and Nil TM (v, y, n). The Lie algebra of A[oo,c¢] has the form ( u , v , ay + bn), a, b E ~. Assume by contradiction that a = 0 (and b # 0). Then (v, n) is an ideal of Af and the adjoint representation of N on (u, v, n) does not act transitively on the set of 1dimensional subspaces of (u, v, n) which are not contained in (u, v). Since these subspaces correspond to the lines of L;oo \ {W}, the group N cannot be transitive on L;oo \ {W} if a = 0, a contradiction to the assumptions of Case B. Similarly, £(A[w,w]) # (u, v, n) since N acts transitively on W \ {oc}. Hence we have £(A[oo,oo]) = (u, v , y + b2n) L:(A[w,w]) = (u, v, y + bin)

with bl 56 32, and we may assume bl < b2 since Case B is selfdual. Set a ---x/~-blthens~ is, u~u, v ~ Ia v , y + b l n ~ - + a y , y + b 2 n ~ t x ( y + n ) is an automorphism of Af which maps £(A[w,w]) and £ ( A [ ~ , ~ ] ) to (u, v, y) and (u, v, y + n) >, respectively. In order to determine the 6-dimensional Lie algebra £ ( A ) we note that. £ ( A [ w , w ]) and £(A[oo,oo]) are ideals of £ ( A ) by geometric reasons. This implies that in table No. 6 of Mubarakzjanov [13] only the Lie algebra ~6,94 with A = 2 is possible. Now we assume dim N p -- 1, i.e.

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

285

and all other commutators are = 0. The Lie algebra £,(A[w,w]) contains v and has the form ~3 __ (v, fi, ~'). In order to determine the action of A on W \ {oo} we look at the adjoint respresentation of N on (v, fi, ,9). Since dim N ~ = 1, each 2-dimensional subspace containing v is fixed, and therefore the action of N on W \ {oc} cannot be transitive, a contradiction. 6.2.

THE ACTION OF A

By construction, the group M = exp(y, u, v, s) acts transitively on P \ W. Since (y, u, v, s) is an ideal i n / : ( A ) we have an RNS-action of A on P \ W. After renaming s by x the regular normal nilpotent subgroup is

M = { ( x , y , u , v ) [ x , y , u , v e ~}, ( ~ , ~ l , ~ t , O ) . ( x , y , u , v ) = x + 2, y + 9, u + ~t+ 2y, v + ~ +~2u + -~y . The stabilizer on 0 is generated by (a, y + n). We describe the corresponding one-parameter groups by the parameters a E ~ and b 6 ~ respectively. The action of the stabilizer on P \ W can be calculated by conjugating M by a and b. In this way we get

(x, y, u, v) a = (eax, e2ay, e3~u, e4~v), x2

(x, y, u, v) b = (x,v,u + bx, v - by + b y )

Remark. The horizontal action is the affine group generated by the mappings (x, y) ~ (z + ~, y + if), ~, ~ E ~

and (x, y) ~ (cx, c2y), c > O.

The induced action on W \ {oc} = {(~, ~)1~, Y E R} is generated by the mappings

(the shears), and (~, y) ~ (c~, c2~]), c > 0. 6.3.

CONSTRUCTION OF THE GEOMETRY

We first construct the line L (°,°) which joins the origin and the point (~, y) = (0, 0) at infinity. Let w+, z+, w_, z_ be real parameters such that L(°,°) contains the points (1,0, w+, z+) and ( - 1 , 0 , - w _ , z_), then one gets all other points of the

286

D. BETTEN

line L (°,°) by applying the a-parameter group and the y-parameter group to those two points (and to 0). In this way we get the line L (°'°) = {(x, y, w¢(x)x 3, z4x)xa)lx, y E ~ } , where we(z) = w+, z~(x) = z+ for x > 0, wc(x) = w_, zc(x) = z_ for x < 0 and w~(x) = z~(x) = 0 for x = 0. We then get the other lines of the plane by shifting L(°,°) in the direction of x (apply the parameter ~" = - t ) and then using the shears exp(y + n) (parameter b). After a suitable translation in the direction of c~ we then get the following pencil of lines through the origin 0:

L(t, b) = u(x,y)

{(x,y,u(x,y),v(x,y))lx,ye

~ } , t , b c 11~, with

=

w~(x+t)(x + t) 3 - w~(t)t 3 - ty + bx

=

z~(x+t)(x -1- t) 4 -- zc(t)t 4 -- t(w~(x+t)(x q- t) 3 -- wc(t)t 3) + -t2f y + b ~ - by.

6.4.

CHECKING THE INCIDENCE AXIOMS

We have to check whether the origin 0 is joined to each point (x, y, u, v) # 0 by exactly one line. This is easy to see if x = 0 , so we consider the general case x # 0. From the horizontal action of A we see that it suffices to take x = 1. From the two equations for the pencil £0 we eliminate b and get one equation for t:

f(t) r(t) a(t)

= F(t)(½-Y+t)-G(t)-~y+ty ---- W~(l+t)(1 q- t) 3 -- W~(t)t 3 and

t2

2-Yy=C•

with

= Z,(l+t)(1 + t) 4 - z~(t)t 4.

A unique line exists if and only if for every C C I~ and every y C ~ there exists a unique solution t of this equation, i.e. the map t ~ f ( t ) is a bijection. It remains to find out for which parameters w+, w _ , z+, z_ this is true. A necessary condition is if(t) >_ 0 (or _< 0) for all t. Since

f ' ( t ) = y2 + y ( - F ' ( t ) - t - ½) + i1F , (t) + tF'(t) + F(t) - G'(t) is a quadratic polynomial in y, the discriminant of this polynomial has to be < 0 for all t E ~ :

O(t) = (F'(t) - (½ + t)) 2 + 4 G ' ( t ) - 4 F ( t ) < 0

for all t E ~ .

4-DIMENSIONAL COMPACT PROJECTIVE PLANES WITH A 5-DIMENSIONAL NILRADICAL

287

Inserting F(t) and G(t) we get

¢(t) = ( ( 6 w + - 1)(t + 1))2 + ( 1 6 z + - 4 w + ) ( 1 + 3t + 3t 2) ift > 0, = (3w+(1 + t) 2 - 3w_t 2 - (t + 1))2 +(16z+-4w+)(l+0 3-(16z_-4w_)t 3 for-1 0 it follows that t 2 has a coefficient _< 0, therefore we get the necessary condition (6w+ - 1) 2 + 3(16z+ - 4w+) _< 0.

(R)

If, conversely, the condition (R) is fulfilled then 16z+ - 4w+ _< 0 and

• (t) <_ - 3 ( 1 6 z + - 4 w + ) ( t + 1)2 + ( 1 6 z + - 4 w + ) ( 1 = (16z+-4w+)¼<0

+ 3 t + 3t 2)

for allt>_0.

Similarly we get the necessary condition (6w_ - 1) 2 + 3(16z_ - 4w_) _< O,

(L)

which is equivalent to ~(t) _< 0 for all t _< - 1 . Note that the conditions (R) and (L) may also be written as

(~+ - ~): + 4(z+ - ½) _< o,

( ~ _ - ~): + %~ _ - ~ ) _< o.

We will now show that the conditions (L) and (R) also imply if(t) _< 0 for all t with - 1 < t < 0. By (R) and (L) we have 16z+ _< -12w~_ + 8w+ - ½ and 16z_ _< - 1 2 w 2- + 8w+ - 1 which allows the estimate < ( 3 w + ( l + t ) 2 - 3 w _ t 2 - ( t + 1))2 +(-12w 2 +4w+::

½)(1 + t)3 - ( - 1 2 w 2 + 4 w _ - ½)t 3

Q(w+,w_,t).

For fixed t, - 1 < t < 0, Q(w+, w_, t) is a quadratic form in w+ and w_ with the following quadratic terms: (9(1 + t) 4 - 12(1 + t)3)w~_ - 18(1 -t- t)2t2w+w_ + (9t 4 q- 12t3)w 2. Since the discriminant is 144(1 q-t)3t 3 < 0, the quadratic form is strictly definite and has a unique extremal point. We get the coordinates of this point by calculating

OQ

OQ

Ow+

Ow_

~0,

D. BETTEN

288 This leads to the system of equations

3(1 + t)(3t - 1)w+ - 9t:w_ - t + ½ = 0 - 9 ( 1 + t):w+ + 3t(3t + 4 ) w _ + t + 3 = 0 with the unique solution w+ = w_ = I' Since Q(~, 1 , t ) = 0, we have proved O(t) _< Q ( w + , w _ , t ) < 0 for all t w i t h - 1 < t < 0, and therefore ~ ( t ) _< 0 for all t C I~. As a consequence, if(t) >__0 for all t C ~ . Moreover, the above calculations show: ~ ( t ) = 0 is only possible for 6w+ - 1 = 0 and 16z+ - 4w+ = 0, i.e. w+ = ~ , z + = , or w_ = ,z_ = , or both. In all cases we calculate f ' ( t ) = (y - (t + ½))2. This means that f'(t) = 0 can occur for only one t (y E ]R fixed) and this implies the strict monotonicity of f. It remains to check that f behaves correctly for Itl --+ o~. For t ~ oo the function f has the form

f(t) =

t3(3w+ - 4 z + ) + terms of lower order in t if 3w+ - 4z+ # 0 t(y 2 + ~1) + const. If 3w+ - 4z+ -- 0, i.e. if (w+, z+) = ( - I , - ½ ) "

For t ~ - c ~ one has similar expressions, replace the indices + by - . If 3w+ 4z+ # 0 then by the condition (R) this number is > 0. If 3w+ - 4z+ = 0 then the behavior for t ~ oo is determined by the term (y2 + ~ ) t which has the positive coefficient y2 + ~ . A similar statement holds for t ~ - o o . In any case we get limt~co f ( t ) = c~ and limt--.-oo f ( t ) = -oo. Let us now prove that each pair of different lines intersects. Using the collineation group we may suppose that one line has parameters (t, b) = (0, 0) and the other one (t, b) # (0, 0). This leads to the following system of equations:

w~(x+t)(x + t) 3 - w~(z)x 3 - ty + bx = m

We have to show that for m, n C ~ and (t, b) # (0, 0) there exists a solution (x, y) C I~ 2. Since this is obvious for t = 0 we may suppose t # 0. Eliminating y we get one equation for z. For x --+ c~ this equation has the form f ( x ) = 0 where

x3t2(4z+ - w+) + terms of lower order in x if 4z+ - w+ ~ 0 f ( x ) = { x ( - ~ t 4 - b2) + const, if 4z+ - w+ = 0, i.e. if (w+, z+) = (~, ~4).

4-DIMENSIONALCOMPACTPROJECTIVEPLANESWITH A 5-DIMENSIONALNILRADICAL 289 F o r x ~ - oo w e get similar expressions, replace the indices + b y - . F r o m (R) and (L) it f o l l o w s that 4 z + - w + _< 0 and 4 z _ - w _ _< 0. F u r t h e r m o r e the a s s u m p t i o n t ~ 0 implies - ~ t 4 - b2 < 0. T h e r e f o r e in any case we h a v e limx__+~ f ( z ) = - ~ and l i m x ~ _ ~ f ( z ) = c~. H e n c e there exists a solution o f f ( x ) = 0 and the t w o lines intersect.

References 1. Betten, D.: 4-dimensionale projektive Ebenen mit 3-dimensionaler Translationsgruppe, Geom. Dedicata 16 (1984), 179-193. 2. Betten, D.: 4-Dimensional compact projective planes with a 7-dimensional collineation group, Geom. Dedicata 36 (1990), 151-170. 3. Betten, D.: 4-Dimensional compact projective planes with a nilpotent collineation group, Mitt. Math. Gesellsch. Hamburg 12 (1991), 741-747. 4. Betten, D.: Orbits in 4-dimensional compact projective planes, J. Geom. 42 (1991), 30-40. 5. Dixmier, J.: Sur les repr6sentations unitaires des groupes de Lie nilpotentes lIl, Canad. J. Math. 10 (1958), 321-348. 6. Browder, F. E.: Covering spaces, fiber spaces and local homeomorphisms, Duke Math. J. 21 329-336. 7. H~ihl, H.: Homologies and elations in compact, connected projective planes, TopologyAppL 12 (1981), 49-63. 8. Knarr, N.: Topologische Differenzenfltichenebenen, Kiel, 1983. 9. Knarr, N. and Weigand, C.: Ein Kriterium fiir topologische Tem~rk6rper, Arch. Math. 46 (1986), 368-370. 10. Knarr, N.: 4-dimensionale projektive Ebenen yon Lenz-Baflotti - Typ II.2, Results Math. 12 (1987), 134-147. 11. Knarr, N.:4-dimensionaleprojektiveEbenenmitgrol]erabelscherKollineationsgmppe, J. Geom. 31 (1988), 114-124. 12. Lfwen, R.: Four-dimensional compact projective planes with a nonsolvable automorphism group, Geom. Dedicata 36 (1990), 225-234. 13. Mubarakzjanov, G.M.: Classification of solvable Lie algebras of sixth order with one nonnilpotent basis element, Izv. Vyssh. Uchebn. Zaved. Matematica 4 (1963), 104-116. 14. Ortega, J. M. and Rheinboldt, W. C.: lterative Solution of Nonlinear Equations in Several Variables, Academic Press New York: (1970). 15. Salzmann, H. R.: Topological planes, Adv. Math. 2, (1967), 1-60. 16. Salzmann, H. R.: Kollineationsgmppen kompakter vier-dimensionaler Ebenen, Math. Z. 117 (1970), 112-124. 17. Salzmann, H. R.: Kollineationsgruppen kompakter 4-dimensionaler Ebenen II, Math. Z. 121 (1971), 104-110. 18. Salzmann, H. R.: Elations in four-dimensional planes, TopologyAppL 3 (1973), 121-124. 19. Salzmann, H. R., Betten, D., Grundh6fer, T., H~ihl, H., L6wen, R., and Stroppel, M.: Compact Projective Planes, de Gmyter, Berlin, 1995.