London Mathematical Society Lecture Note Series. 94
Representations of General Linear Groups G.D. JAMES Fellow of Sidney Sussex College, Cambridge
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CAMBRIDGE UNIVERSITY PRESS Cambridge London New York Melbourne
Sydn~y
New Rochelle
Published by the Press Syndicate of the Unive~sity of camb~idqe The Pitt Buildinq# Trumpinqton St~eet, camb~idge CB2 1RP 32 East 57th St~eet, New Yo~k, NY 10022, USA 296 Beaconsfield Pa~ade, Middle Pa~k, Melbou~ne 3206, Aust~alia ©Camb~idge
University Press 1984
First published 1984 Printed in Great Lib~ary
B~itain
at the University Press, cambridqe
of Conqress catalogue card number:
83-25171
British Library C&taloguinq in Publication Data James, G.D. Representations of qeneral linear qroups (London ~athematical Society lecture note series, ISBN 0076-0552: 94) 1. Linear algebraic qroups I. Title II. series ISBN
0 521 26981 4
Contents
Abstract List of Symbols
vii x
1
Introduction
1
2
Examples
7
3
Gaussian polynomials
18
4
compositions of n
20
5
Root subgroups of G
23
6
Subgroups of G
28
7
Coset representatives
30
8
Subgroups of G used for induction
35
9
Some idempotent elements of KG
10
The permutation module M A
47
11
The Submodule Theorem
56
12
A lower bound for the dimension of S
13
The Kernel Intersection Theorem for S(
14
Reordering the parts of A
80
15
The Kernel Intersection Theorem
84
16
Consequences of the Kernel Intersection Theorem
100
17
Removing the first column from [A]
109
18
Isotropic spaces
114
19
The prime divisors of Gaussian polynomials
124
20
The composition factors of S(
136
n
n
associated with compositions
n
Acknowledgements References
-
40
n
n-m,m
)
67
u
n-m,m
)
75
145
146
ABSTRACT
This essay concerns the unipotent representations of the finite general linear groups GL (q). n
An irreducible unipotent representation is,
by definition, a composition factor of the permutation representation of GL (q) on a Borel subgroup, and the ordinary irreducible unipotent n
ropresentations may be indexed by partitions A of n, as may the ordinary Lrruducible representations of the symmetric group G n• r.ature is that the representation theory of dppuurs to be the case "q
~
n
The remarkable
over an arbitrary field
= 1" ,of the subject we study here.
The most important results are undoubtedly the SUbmodule Theorem (Chapter 11) and the Kernel Intersection Theorem (Chapter 15), but there seems l.u h
lhoue whose source is quoted or which are obviously known (Chapters 3 - 8). Chapters 1 and 2 set the scene,
Gh
b¥ outlining the connection between
and representations of GLn(q) over fields of characteristic diViding q,
and by giving examples of the situation to be considered
later.
The
preliminary results which we need are derived in Chapters 3 - 8.
Thereafter,
we assume that the characteristic of our ground field K does not divide q, but otherwise K is arbitrary.
Certain idempotents of the group algebra are
defined in Chapter 9, and they are used in Chapter 10 to describe the structure of the permutation module M of GL (q) on a parabolic subgroup. A n In Chapter 11, we define a certain submodule SA of M in terms of a A
generator; SA may be regarded as the q-analogue of a Specht module.
The
t.ubmodule Theorem states that every . KGLn (q) -submodule of M,I\. either contains ~A or is contained in
st·
We proved the SUbmodule Theorem for S
n
in 1976
(James [Jll), and thereby gave the first construction of the irreducible rupresentations of
~
over an arbitrary field.
we have already published a
proof of the 'SUbmodule Theorem for GL (q) (James [J but the proof given n 9]), here is new; it is simplified by assuming initially that the ground field contains all the pth roots of unity (where q is a power
0
f p).
The Submodule
Theorem gives us an irreducible unipotent representation of GL (q) for each n
partition of n. irre~ible
In particular, the various modules SA are the ordinary
unipotent representations when the set of rational numbers is
the ground field. The aim of the next few chapters is to construct a basis for SA' and to prove the Kernel Intersection Theorem (Chapter 15), which describes SA
as the intersection of the kernels of certain KGL (q) -homomorphisms defined n
on MAo.
Here we roughly follow the approach we adopted in 1977 (James [J4] )
to prove similar results for Specht modules.
Unlike the situation for
symmetric <jroups, where bases for specht modules and the Kernel Intersection Theorem are easy for many special cases, the only partitions for which the GL (q) results are clear are (n) (when there is nothing to prove!) and n
(n - 1,1).
Even the partition (2, 2) of 4 is difficult to handle; in place
of a 2-dimensional representation of ~' We have to deal with a (q2 + q4)_ dimensional representation of GL
4(q).
Many important results (Chapter 16) follow from the Kernel Intersectiol Theorem.
For example, dim sA is shown to be independent of K, and we prove
that we have found all the irreducible unipotent representations over K. The Branching Theorem, describing the structure of sA as a KGL (q)-module, n_ l is also deduced. By combining the SUbmodule Theorem and the Kernel Intersection Theorem, it is possible to embark upon the task of finding the decomposition matrices of GL (q) for primes which do not divide q •... The problem of determining the n decomposition matrices of
~n
is still open, and we believe that the key may
well lie with the unipotent representations of GL (q). n
In Chapter 17, we prove a theorem on the decomposition matrix of
GL (q) concerning the removal of the first column from the diagram [Al, the n
corresponding
~n
result was proved only recently (James [Jal). par~s
As far as we know, only the
of the decomposition matrix of gn
corresponding to hook partitions or to two-part partitions is known (Peel [P] and James [J
2,
J
3]),
(n - m - 1, m, 1).
although work is in progress on the partitions An
analogue of Peel's results is given in Chapter 16,
ftnU in the final two chapters we determine the part of the decomposition lJIatrix of GL (q) which corresponds to two-part partitions, for all primes n which do not divide qi the evidence that the modular representation theory of G'n is just the case "q = I" is then overwhelming. Naturally, we have pondered the question why the modular representations of
ij;
n
look like representations of the group of automorphisms of an n-dimensional
voctor space over "the field of one element".
It is easy to be misled into
qivlng art unsound argument about this, and it must be noted that our proofs l10
not translate into proofs for
~.
n
More challenging still is the explanation
of the possible result that the representation theory over ~ of GLd(r} (d » n, r prime) is the case "q = 1" of our work here - see the remarks at the end of Chapter 16.
Why should the representation theory ef GL (q) over n
fields whose characteristic does not divide q throw light on the representation theory of general linear groups of different dimension over fields of the natural characteristic? Knowledge of the theory for
~n
has guided us to search for proofs to
present here which would translate immediately into proofs for the symmetric group.
We have been unsuccessful, so we cannot explain why "putting q ::: 1"
works, and entirely new techniques have had to be developed in this essay.
LIST OF SYMBOLS
Symbol
Chapter of definition
Meaning A certain subgroup of U A certain idempotent of
a C
r
JFq
(f , q
+)
9.8
5
A function from r to {l, 2, ••• , q}
9.1
A certain function from fer) to {l, 2, ••• , q}
9.4
An
e
KAr
The group of upper/lower triangular matrices
An idempotent in KG
E
8.1
n
idempotent in KG (r (r) )
9.4
An idempotent in KU
11.4
The exponent of q modulo p
19.4
A basis for V
2.4
The field of q elements
2.4
The additive group of lf q
G
A subgroup of GL (q), isomorphic to GL (q)
8.1
G*
A certain subgroup containing G
8.1
G(r)
ex ..
n
r
r
r
r
J.)
I
(i, j)6r
>
5.1
GL (q)
The group of automcrphisms of V
2.4
H
The group of diagonal matrices
5
H*
A certain subgroup contained in G*
8.1
h
The number of non-zero parts of A
4
A certain diagonal matrix
5
n
r
r
The hook length of the (i, j) node in [A] I
r
The identity r x r matrix
K
A field of characteristic coprime to q
K
K extended by a primitive p where q is a power of p
1. (m) p
20.1
th
root of unity,
The least non-negative integer i such that m < pi
8
19.1
M).,
The permutation module on P
[m]
1 + q + q
lm l
[1]
n
The dimension of V
2.4
A Gaussian polynomial
2.14
2
[2] •••
+ ••• + q
A
m-l
10.1 2.5
10.17
[m]
A binomial coefficient A
parabolic subgroup
A
prime number
6
The field of rational numbers A power of a prime number It
II.
r
A subset of {l, 2, ••• , h}
10.8
The set of subsets of {l, 2, ••• , h} of cardinality r
10.8 10.20
N* 't:
A
certain subset of
., }..
.'
A
certain submodule of M A
e;,
The symmetric group on n symbols
n
'1'
A
±
u
± 1I).
V
\
11.11
The initial A-tableau
4.2
The group of upper/lower unitriangular matrices
5.6
± A certain subgroup of U
6.1
The n-dimensional vector space over If of which GL (q) is the group of automorphisms q
2.4
The group of permutation matrices
5
A root subgroup
5
n
w
An element of X, ,
5
~J
The ring of integers
Q 1(,
s. y,
Elements of f'q
15
closed subset of
5.1
I'
A
r
The "commutator" subset of r
I' (r)
{(i, j)
II
A KG
n
In;;; i
> j
-homomorphism
i;Il
S;
r
S;
n}
5 9.4
An
K
A,
ll,
\)
element of K
COmpositions of n
4
v (m)
The largest integer i such that p
11",
Permutations
p
0, T
i
divides m
19.1
11">..
A certain permutation, depending on >..
11.3
: "'~11" .
A certain permutation, depending on R
12.1
R
s
r;Il
{(i, j)
1 ~ it' j
.+
{(i, j)
1
~ i
<
j :li n}
5.6
-
{(i, j)
1
s
<
i
s nl
5.6
r;Il
j
5
n}
4l 1 , 4l 2 ,
Linear i-characters of
A
Xl' X2 '
Linear K-characters of
(If , +)
9.6
r
9.1
q
linear K-character of G(f)
9.1
Xc
A
X>..
The ordinary character of SA
lPd,i
A certain KGn-homomcrphism defined on M
1 A
~
A transitive relation on the set of compositions of n
e-
~
<••• >
The group, or vector space, generated by •••
but not
=
15.1 4.1 4.1
A bilinear form on M
A
1l.1
INTRODUCTION
1
Among all classes of groups, it is arguable that the symmetric groups
a: n
have the richest representation theory.
Not only are the symmetric groups
interesting in their own right - the theory of their representations is extremely elegant, and still contains many mysteries - but they can also be mind in several ways to shed light on the representations of other groups,
"n(l their theory can be applied in fields as diverse as quantum mechanics and
polynomial
identity algebras.
r~presentation theory
We hope to convince the reader that the
of symmetric groups is just a special case of a far
11cu.1per, but equally interesting., topic I namely the theory of unipotent n-'Llr.esentations of the finite general linear groups GL
n
(q).
It is well-known that there is a close connection between representations
or '"n
over a field F and the representations of GL (F) over the same field F. n
Hut this is not what we shall explore here; instead, we open up a new avenue by considering representations of GL (q) over a field K whose characteristic n
~O.8
not divide q. The ordinary irreducible characters of GL (q) have been determined by n
ureen [G
I]
I
but earlier work of Steinberg [S] produced one ordinary
irreducible character X for each partition A of n, A
It appears that no
provious work has been done on constructing the representation modules for
or,
n
(q),
and we shall deal here entirely with the modules corresponding to the
characters obtained by Steinberg, the so-called unipotent representations of GL (q). n
We shall be working, then, with representations of GL (q) which n
ttre indexed by partitions A of n , U
f
The ordinary irreducible representations
e'n are also inqexed this way. Two striking results about the characters
X, of GL (q) have been proved which already indicate an analogy with G' :
A
n
n
deg X = q A
}:(k k
1)\ 1 (qn _ 1) (qn-l _ 1) _.•. (q _ 1) II ~~- h (qh _ 1) ,
where the product is over the hook lengths h
=
j) in the
the diagram [A]. (Olsson gives a similar formula for all the character degrees. of GL (q).) n
1.2 not q
e
THEOREM
(Fong and Srinivasan [FSJ).
Assume that p is an odd prime
dividing q, and e is the least positive integer such that p divides - 1.
Then
[A] and
[~]
XA~X~
are in the same p-block of GLn(q) if and only if
have the same e-core.
Campare these results with two theorems from the representation theO% of G;"n (We shall also denote by X the ordinary irreducible character of A
«.
corresponding to the partition A of n.)
1.3
THEOREM (Frame, Robinson and Thrall [FRT]).
deg X). = n(n - 1)
1.4
••• 1
~ [~}
•
THEOREM (Brauer [B] and Robinson [R]).
p-block of
~n
For the symmetric group
if and only if [A 1 and
[~]
X ~ X~ belong to the same A
have the same p-core
What is more, there are similar results in the theory of Weyl modules We shall not use Weyl modules
here, so we shall not give a formal definitio
of them, but refer the reader to the relevant literature (for example, Green [G or James and Kerber [.ne]). 2]
SUffice it to say that if F is a
sufficiently large field, and V is the d-dimensional vector space over F on which GL (F) acts in the natural way, then for each non-negative' integer n d
and for every partition A of n having at most d non-zero parts, there is lPn a GLd(F)-Submodule W~, called a Weyl module, of V •
We emphasize that W
A
is a representation module for GLd(F) over the "natural" field F.
1.5
THEOREM
The dimension of W is independent of F, and A
dimW . A=
n
+
(d
j - L)
(i,j)E[A]
Two weyl modules W and
A
oequence A :: A l
I
A I 2
•••
,
\
W~
~ (~)
•
are said to be connected if there exists a
= II of
partitions such that for each i ,
W have a common composition factor. ~i+l 1.6
THEOREM
If F has characteristic p and A,
are partitions of the same
~
integer, then W and WIJ are connected if and only if [A] and [lll have the same A p-core. In fact, the theory of Weyl modules is much closer to the symmetric group than one might expect.
For each A, W has a unique maximal submodule; A
we denote the quotient by FA' whereupon every irreducible polynomial representation of GLd(F) is isomorphic to some FA" n, there is a Specht module, defined over F I for
For each partition A of
G(.
n
Most Specht modules
have a unique maximal submodule J the various quotients give all the irreducible F~ -modules.
n
To illustrate the connection between F~ -modules
n
and FGLd(F)-modules, and also for future reference, we now give an example.
1.7
EXAMPLE.
The following matrices describe the composition factors of
aome Weyl modules.
The entry in row A and column II is the number of
composition factors of W which are isomorphic to Fu ' A (1,1)
1
l·
(2)
if char F:: 2
(1,1,1)
r
(1,1,1) 1
(2,1) 0
01
l
0
1
0
1
0
IJ
(1,1,1)
(2,1)
(3)
1
0
0
1
1
0
1
I
(2,1) (3)
I
(1,1,1)
l
(2,1) (3)
( 3)
I
if char F = 2
I
1 I 0 i I IJ
if char F = 3
(11+)
(2,1 2 )
(2,2)
(3,1)
( 4)
(1 4 )
1
0
0
0
0
(2,1 2 )
1
1
0
0
0
(2,2)
0
1
1
0
0
(3,1)
1
1
1
1
1
0
1
1
(1 4)
(2,1 2 )
(2,2)
(3,1)
1
0
0
0
a
1
0
0
:J :1
1
0
1
0
0
0
0
0
1
0
0
a
1
0
1
L
(4)
I
(11+)
II
(2,1 2 )
I
(2,2) (3,1)
L
(4)
if char F = 2
(4)
if char F = 3
Remember that all these matrices give information about every general linear group GLd(F) over F. d.
The partitions are not partitions of the integer
On the other hand, the matrix involving partitions of n contains the
decomposition matrix of
~n
this decomposition matrix.
over F {James [J
7
]) i
the underlined entries give
It can also be proved (James [J that every 10])
column of the matrix for n corresponds in a natural way to an indecomposable module for
F~
n
•
The results which are illustrated in the example above are true in general.
If we had complete information about the components of permutation
modules of symmetric groups, then we could construct all the matrices which
describe the composition factors of Weyl modules, and conversely, since these matrices contain the decomposition matrices of sYmmetric groups, knowledge of the composition factors of Weyl modules would give the decomposition matrices of symmetric groups.
we hope now to have given sufficient support to the claim that the representation theory of
~
n
is very closely tied to that of GL (q) over fields
n
whose characteristic divides q.
How then does the representation theory of
GLn(q) over fields of characteristic
~
dividing q fit into the picture?
We aim to show that the representation theory of Gt must be "the case q n
uf this theory.
It is well-known that
f;;
n
= 1"
looks like "the general linear
lJroup over the field of one element", but the results go through in an unexpected and beautiful way.
We emphasise that the methods used in this
l!Ssay do not apply to the symmetric groups; everything we achieve is ~nalogous
to the theory of symmetric groups, but whenever we suggest that q
IIhould be put equal to 1, we do so for the results but not for the proofs. Wf! do not know why this works. If K is a field whose characteristic does not divide q, we shall produce an irreducible KGL '(q)-module for each partition of n. n
I
here are fewer irreducible
I
ho re is a Weyl module for each partition of n ,
K~
n
In general,
-modules than there are partitions of n, but
fhp theory of Weyl modules, too?
Could our new theory subsume
The evidence is flimsy, but consistent
(Iwe Chapter 16). A light-hearted
observation concerns the tendency of theorems from
'he representation theory of symmetric groups to take a form which does not ,ply on the primeness of the field characteristic (for example, several
theorems involve the p-core of a diagram, which exists whether or not p is I" i me.) dIP
It appears that this phenomenon occurs because, for example, there
primes which divide 1 + q + q2 + q3 but not 1 + qi when we put q
=
1 the
'/I·neral linear group. r.Pflults for these primes turn into empty theorems about
those primes dividinq 4 and not 2. It will be clear by the end of this essay that there are still many interesting open questions. ~
A recent volume on the representation theory
0
by the present author and Adalbert Kerber [JK] contained nigh on a
thousand references t.o works on that subject. the tip (q == 1) of a very big iceberg I
we believe that this is just
2
EXAMPLES
Few of the results from this section will be used later on, but it' is enlightening to look at some. special cases of the problems we shall consider, before plunging into the general situation. First we consider a representation of GL (q), denoted by (n - 2,2) , n which looks relatively straightforward.
It is already difficult, though, to
complete the relevant calculations for this representation. The symmetric group Gr
is the group of all permutations of {1, 2, ••• , n
n
We define a representation of
fi{
n
which is indexed by (n - 2,2), and then show
what happens for the general linear group. Consider a vector space, over a field K, whose basis elements are the unordered pairs {i, j} from {i, 2, ••. , n}.
belongs
to
Since S
our vector space.
space may be regarded as a
K~
n
n
-module.
Thus, for example,
permutes the pairs {i, j}, our vector The dimension of the space is simply
t.he number of unordered pairs, n(n -
1)
2
Let S
= S(n-2,2)
be the subspace consisting of those vectors satisfying
t.he following two conditions: (0) ~
1 +
K
2
The sum of the coefficients is zero.
(Thus we require that
= 0 in our example above, for this condition to hold.) (1)
For each 1-e1ement subset {u} of {1, 2, ••• , n}, the sum of the
('oofficients of the pairs containing u is zero. The subspace S is a I,reserved under the action <'onditions (0) and (1)
is:
~
n
-module, since conditions
of It:n •
An
(0)
and (1)
are
example of a vector satisfying
(2.1)
{12} - {23} + {34} - {41} ,
for certainly the sum of the coefficients (+1 - 1 + 1 we look at u
= 2,
1)
is zero, and if
for example, the pairs containing u are {12} and {23}, and
the sum of the coefficients (+1 - 1) of these pairs is zero.
That condition
(0) and (1) are fulfilled is seen most easily by drawing a picture to
describe the vector (2.1): 1
+1
(2.2) -1
-1
4
3
The following are facts about the space S: (2.3)
(i)
S is non-zero if and only if n
(ii)
S
(iii)
Dim
(iv)
If K has characteristic zero, then S is an irreducible
~
4.
has a basis consisting of vectors of the form (2.2).
The first
S
= n(n - 3)/2.
three
of
KS-mo~
n
'
these results are true regardless of the field : 1
,I
we now turn to the general linear group, and choose some notation
j •
which will remain in force throughout this work:
i
2.4
DEFINITIONS.
space over
~'
Let F be the field of q elements, and let V be a vectj q
with basis e
l,
e
2,
•.• , en.
We denote the general linear
By definition, G is the group of all automorphisms of n
V.
We shall freely identify G with the group of non-singular n x n matric n
over 1F , the automorphism given by the matrix (g .. ) being the one for which q
1)
(1 S i S n)
If v
l'
v
2'
•.• , v
••• ,
k
vk >
•
are vectors in V, we let
I
v
denote the subspace of
2.5
DEFINITION. [m]
Ln
=1
l'
v
2'
•.• , v k•
If m is a non-negative integer, let
+ q + q
particular, [OJ
spanned by v
2
= 0,
+ ••• + q
1, and [m]
[1]
Note that if we put q
m-l
=
(qm - l)/(q - 1).
1 in this definition, we get [m]
~
= m. Bear
Lhis in mind while we go through a "q-analogue" of our statements about C5. • n
Consider a vector space over K whose basis elements are the 2-dimensional aubapacee of V.
Thus, for example,
hclongs to our vector space.
Since G permutes the 2-dimensional subspaces n
uE V, our vector space may be viewed as a KG -module. n
The dimension of this
lIj,Jace is simply the number of 2-dimensional subspaces of V, which, by a lIimple calculation (see Theorem 3.1) is In] [n -
1]
[2] Consider the subspace S the
S(n-2,2) consisting of those vectors satisfying
following two conditions:
( ~~ .6)
(0)
+ K
2
=a (1)
I 'lIt'
~
The sum
of the coeffic ients is zero.
(Thus, we reqUire that
in our example above, for this condition to hold.) For each l-dimensional subspace U of V, the sum of the
fficients of the 2-dimensional subspaces containing U is zero. The subspace S is a KG -module, since conditions (0) and (1) are pren
"" rved under the action of Gn • We recommend at this stage that the reader attempts for himself to , "rlstruct a non-zoro clement of S.
It is conceivable (and after some effort,
the reader might believe likely1) that S is the zero subspace; this would not be very interesting. To exhibit a non-zero element of S, we resort to the notation of
projective geometry.
A line will denote a 2-dimensional subspace U of V,
and the points on the line will be the l-dimensional subspaces of U.
EXAMPLE.
If q
= 2,
<e
<e >, and <e + e >. 2 2 l <e >, 2
<e
l
l,
e > contains three l-dimensional subspaces <e >, 2 l
In the picture below, the line through the points <e
1
+ e > is labelled +1; this denotes the fact that in our linear 2
combination of 2-dimensional subspaces, <e
l,
e > occurs with coefficient +1 2
<eo +1 -1
(2.7)
-1
-)
<e,-fe" fe3+e..>
<e,-te..>
+1
<e..;> The picture corresponds to
an element of the vector space over 2-dimensional subspaces of V.
K
whose basis eleme~ts are the
A glance at the picture shows that condition
2.6 hold, so we have constructed an element of S.
EXAMPLE.
If q
= 3,
<e
l,
e > contains four l-dimensional subspaces, <e >, 2 l
<e >, <e + e >, <e - e 2 l 2 1 2>·
-" 10 -
+1 -1
-1
-]
-1
+1
8)
+1 +1
leave
the reader to label the four points in the middle of the picture;
example, the top left-hand point is <e fmensional intersection of <e
l
+ e
4,
e
l
2
+ e
2
+ e
3
+ e
4
>, being
the
+ e > and <e + e e + e >. 3 l 4 2, 3
in, we have an element of S. Compare (2.8; q = 3), and (2.7, q = 2), with (2.2, q
= I?).
Notice that both the above Examples used four linearly independent tors e
lJ)
l,
e
2,
e
3,
e
4•
In fact, all the following statements are true for
Assume that the characteristic of K does not divide q.
Then
(i)
S is non-zero if and only if dim V
(ii)
S has a basis consisting of vectors of the type illustrated by
(2.7) and (2.8). (iii)
dim S = q2 [n] [n - 3] / [2) •
(iv)
If K has characteristic zero, then S is an irreducible XG -module. n
Compare these statements with (2.3). expected one, but still, if we put q
= 1,
The formula for dim S may not be then ••••
The first three of the statements above are hard to prove from the Inition of S.
One of the reasons for the difficulty is that the results
false - unless we
U8e
the hypothesis that the characteristic of K does
2.10
EXAMPLE.
Consider the following linear combinations of 2-dimensional
subspaces of V:
q '" 2
<e.s>
q '" 3
(The
<e
l
remaining points are labelled <e
- e
2
l
+ e
2
+ e >, <e - e + e >, l 3 2 3
- e >, <e + e - e >, readinq clockwise from the top left.) 2 3 l 3
These vectors satisfy condition 2.6(0), and for all I-dimensional subspaces U of V, except U '" <e > or U '" <e >, the sum of the coefficients 2 3 of the 2-dimensional subspaces containing U is zero.
But if we let our
coefficient field K have characteristic q, then condition 2.6(1) holds for U
= <e2> and
U '" <e >, also. 3
Thus, if we allow the characteristic of K to
divide q, we can construct a non-zero element of S when dim V'" 3, contradictinq
•
2.9 (i.) - (iii). So
far, we have concentrated on the 2-dimensional subspaces of V.
But there is an obvious way of qeneralising the ideas to other cases. - 12 -
Let us
see what happens. Define M(
n-m,m
) to be a vector space over K whose basis elements are
the m-dimensional subspaces of V.
Thus, M(
n-m,m
of the set of m-dimensional subspaces of V. M(
n-m,m
) is a " c oordinatized" version
Formally, we could regard
) as the vector space of K-valued functions on the set of m-dimensional
subspaces of V, but we find our description conceptually easier.
In fact,
we shall later perform all our calculations with a right ideal of the group algebra KG
n
which is isomorphic to M(
n-m,m
).
The plan is to construct
inside M a KG -module S( ) which is the q-analogue of the Specht (n-m,m) n n-m,m module for
~
n
corresponding to (n-m,m).
In Chapter 11, we shall define a
module SA for each partition A of n. To provide examples of what we are about, we need a notation for the m-dimensional subspaces of V.
Although we were able to draw lines for the
2-dimensional subspaces, something more elaborate is required for a general m. An element of V may be regarded as a n-tuple (a row vector of length n)
of elements from
IF. q
Explicitly,
+ a e
n n
~
V
(a . .; IF) ~
q
corresponds to the row vector
Given m elements of V, viewed this way, we construct an m x n matrix by placing the row vectors one after the other.
If the m elements of V are
linearly independent, then the matrix has rank m, and corresponds to an m-dimensional subspace of V.
This SUbspace is unchanged by applying each of
the following operations to the matrix:
(2.11)
(i)
Divide all entries in a row by a non-zero element of
(ii)
Subtract
(iii)
HeiJrrange the order of the rows.
a multiple of one row from another.
f. q
By applying (i) and (ii) we can arrange that the last non-zero entry in each row is a 1 and that the other entries in the column containing this 1 are zero.
Suppose, having done this, that the last non-zero entry in row
i is a 1 in column a
m
s n.
i
6
Using 2.11(iii), we may assume that 1
~
a
1
<
A
2
< •••
Thus, we can get our matrix in the form:
a ** **
**
a
1
a
2
*100
000
*0** *0**
100 0**
...
3
a
0 ••
a
0 •• 1 ••
0
(2.12)
m
J n
It is clear that there is
bijection between the set of matrices in
A
this "canonical form" and the set of m-dimensional suspaces of V.
2.13
EXAMPLE.
n
= 4,
m
= 2.
Bach
2-dimensional subspace of V corresponds
to one of the following matrices:
] . (1OafS 000]1 (a100 (o1000) 100 (o1000) a 1 0 fS 0 1 0 a l O O) ( fS 0 y 1
(a,
a,
Y,
sE
0)
a 6 1 ( Y 001
F). q
Thus there are 1 + q + 2q2 + q3 + qlt 2-dimensional subspaces
•
of V.
2.14
DEFINrrION
(see also 2.5).
In] [n [m1 [m -
Let
11 11
[n - m [1}
+ 11
,
if n
~
m
~
a
= 0, otherwise. The number [:] is a q-analogue of the binomial coefficient [:), when we
put q
= 1,
we qat
- 14 -
[:] = (:]
It is easy to show (cf. Theorem 3.1) that the number of m-d1mensional subspaces of V is [:].
EXAMPLE.
=
4] [2
(4) (3) [2][1]
=
(1 + q
+ q2. + q3) (1 + q + q2.) (1 + q)
From Example 2.13, we see that this is, indeed, the number of 2-d1mensional
••
subspaces of a 4-dimensional space.
Note that the arrangement of *·s in (2.12) corresponds to a partition, naDl8ly (am - m, ••• , a
2
- 2, a
l
- 1) •
?:'his partition has at most m non-zero parts, and each part is at most n - m. Since there is a bijection between the set of such partitions and the set of sequences
,~
the two different ways of counting the m-dimensional subspaces of V give:
2.15
THEOREM
(cf. Andrews [A] p. 33)
m(n-m)
I [:] = i=O where a
n,m
.
a
(1) q1
,
n,m
(i) equals the number of partitions of i, each of which has at most
m non-zero' parts and largest N t at most n - m.
J is
In particular, [:
a polynomial
polynomial.
- 15 -
•
in q, it is called a Gaussian
2.16 from
DEFINITION. M(
n-m,m
)
Assume that 0 s i s m.
into
M(
.. )
n-:1.,:1.
~l,i
Define
to be the linear map
which sends each m-dimensional subspace of
V
to
the sum of the i-dimensional subspaces contained in it. The reason for the first subscript 1 in to define more general maps.
2.17
EXAMPLE5
(i)
m
= 2,
~l
,:1..
m
=3
(11)
i
~l,l
It is clear that
~l
. will emerge when we come ,:1.
~l
. is a KG -homomorphism. ,:1. n
= 1.
<e
l,
e
2,
e > 3
~
<e > + l
L
a€/f'
l
+ e > + 2
q
~1,2
<e l, e 2, e 3>
*
[~
0 0 1 0
~)
+
L
a, ~£IF
~e2
+ e 3>
q
L
aE/F
q
o0
(~ a
1
~)
+
L
a, ~e IF
q
In the case where m = 2, a vector in M (n-2,2) belongs toKer and only if it satisfies Condition 2.6(0), and it belongs
(~
1 0 0) 0 1
oj
~1,0
tOKer~l,l
if
if and
only if it satisfies Conditon 2.6(1) • In general, the subspace of
M(
n-m,m
)
on which we shall focus our
attention is m-l S
(n-m,m)
=
n Ker ~l ' i
i=O
We must warn that we shall not define s( n-m,m ) this way, and it is only after some hard work that we can prove that our module is a kernel intersection, as above. The reader who wishes to explore some of the intricacies of the subject for himself is urged to construct a non-zero element of S( n-m,m ) when - 16 -
3
=m S
n - m, and to prove that s(
characteristic
n-m,m
) is zero if m > n - m and the
of K does not divide q.
As
a hint to the generator of S
(n-m,m)
which we shall define later, draw the lines corresponding to
and compare the answer with Example 2.10.
'\ {[o 1
0 0) _ LaO 0 1
af.F
(lla
Then consider
1 0 0) _ [0 1 0 0) 001 a 0 1 1 +
(1a 10 01 1o)}
q
For
a
generalisation of this idea for m arbitrary, see Example 11.17(v); to
generalize (2.7) and (2.8), see Chapter 18.
- 17 -
3
GAUSSIAN POLYNOMIALS
The Gaussian polynomials, which we introduced in the last section, behave in many ways like binomial coefficients.
We wish to record some of
their properties, referring to Andrews [A) for a fuller account.;'
3.1
.!m!..Yl!l-:!2
...
a (m-b) q
Proof:
~1L-Y2
THEOREM.
o.
be subspaces of V with dim,V
l
... a, dim,V
2
... b,
The number of m-dimensional subspaces W of V such that
rn - a ~ -
b
bl
J
Given a basis of V , the number of ways of choosing a sequence of 2
m - b further linearly independent vectors in such a way that no non-zero linear combination of them lies 1n V1 •
V 2 is .~
,1
n - qa+b) (n q - qa+b+l) ••• (n q - qa+m-l) (q ... (q -
I
l)m-b q(m-b) (m+b+2a-l) /2 [n _ a - b] [n - a - b - 1)
[n -
a - m+
• ~
If 1,
But an m-dimensional
subspace containing V2 has the following number of bases
extending the given basis of V 2: (qm - qb) (m q -qb+l) ••• (m q -qm-l)
=
(q _ 1) m-b q (m-b) (m+b-1) /2 [m _ b) [m - b - 1] ••• [1] •
The quotient of these two numbers is the required answer. Taking V ... V ... 0, we deduce that the number of m-dimensiona1 1 2 of V is [:].
3.2
COROLLARY.
dimU=c.
Let U, V be sub.paces of V with V 2 2
S;
U, dim..Y2
= b,
Then the number of m-dimensional subspaces W of V such that
- 18 -
•
.~
(c-b) (m-b) q
Proof:
~ ~
Cl bJ
Let V1 be a complementary subspace to V2 in U, and apply the Theorem •
• 3.3
COROLLARY.
(i)
= c.
Then
The number of m-d1mensional subspaces W of V such that W ;! U is
The number of m-dimensional subspaces W of V such that W ~ U
(ii)
Proof:
Let U be a subspace of V with dim U
=0
is
For part (i) put V2 = U, and for part (d.L) put V2 = 0 in Corollary 3.2 •
• Either by direct calculation, or using Corollary 3.3 to count the number of m-dimensional subspaces of V which do or do not contain <e >, we get l
3.4
THE~
[:] =
=
~-
r
a r~ ~ + qm
~~
+ qn-m
~
-
fl .
Applyinq this result, and using induction on n, one easily proves a q-analogue of the Binomial Theorem:
3.5
THEOREM.
For all x, n
(1 + x) (1 + xq) ••• (1 + xqn-l )
=
~ L
j=O
[n] qj (j-l) /2 x j j
/'
- 19 -
4
COMPOSITIONS OF n In section 2, we demonstrated how to construct a representation of
G for each m with n
a
S m S n, by considering the vector space over K whose
basis elements are the. m-dimensional subspaces of V (the n-dimensional space over
IFq on which Gn acts in the natural way).
A way of generalis1ng this
idea uses the concept of a composition of n , by which we mean a sequence
of non-negative integers, whose sum is n, Consider subspaces
Va' Vl, ••. of V with the properties that
SUch a set of subspaces is called a A-flag.
Clearly, the action of G
n
on V
induces a permutation of the A-flags, and as a result we may construct a permutation representation of G by taking the vector space over K whose n basis elements are all the A-flags. was that where A
= (n
The situation discussed in Section 2
- m, m, 0,0, ••• ).
It is important to point out that it is unclear whether or not reordering the parts of A gives rise to an isomorphic permutation representation of G. n
(The result
;La
easy enough if K has characteristic zero, but is
not true for arbitrary fields It i see Section 14.)
For this reason,
it is
necessary to work for some time with compositions of n , instead of the more £aDdliar concept of partitions of n. ). for which A l
~
A 2
C!: •••
•)
(A partition of n is a composition
This Chapter is devoted to adapting various
ideas concerning' partitions to suit our needs. The last non-zero part of the composition A will always be denoted
by A (h standing for "height"), and we shall write h ( Al' A2, ,••• ' A 0, 0, ••• ) = (A A .. ···, A • h, l, 2, h) - 20 -
Given a composition A, its diagram [A] is defined to be the set of ordered pairs (i, j) with 1
i
5
hand 1
5
~
j
Ai' and we always regard
5
[A] as an array of "nodes" as in the example:
xxxx =XX
[A]
(4, 2, 3)
X X X
2, ... )
Let Aj equal the number of nodes in column j of [A]. is a partition of n.
4.1
For instance, if A
DEFINITIONS.
If A and
1
i=l
A'
5
~
I
(3, 3, 2, 1).
are compositions of n, we write A ~
~
~
~
~
if
but u ~ A.
Notice that the relations possible to have A !:::
~
and
~ !:::
~
and c> are transitive, but that it is
A without A
=~.
In fact, A ~
is obtained from A by rearranging the parts.
~
and
~ ~
A if and
For example, if
A ='" (4, 2 I 3), u = (3 I 2, 4), v = (3, 3, 3), then A !::: u , u A
=
for all j.
~~
i=l
We write A e- u if A !:::
~
2, 3) then AI
j
j
only if
= (4,
Then A' = (Ai, A
~
A !>v.
A-tableau is an array of integers obtained by replacing the nodes
in [A] by the numbers 1, 2, ••• , n in some order (no repetitions being allowed).
For example, 124 5
3 6 8 5
3 6
2 7
789
1 4 9
are two (4, 2, 3)-tableaux.
The symmetric group
~
n
acts on the set of
A-tableaux in the natural way (e.g. T (1 3 2 6 7) (4 8) = T above). l 2, If T is a A-tableau, then row i of T is the set of numbers replacing the nodes (i, j) in [A] (1
~
j
~
Ai)' and column j of T is the set of numbers
replacing the nodes (i, j) in [A] (1 T
l
~
i
above.
- 21 -
~
A!); e.g. {4, 9} is column 3 of J
Given a tableau T, we define the functions row I colT: T
{l, 2, •.•
I
n}
-+
u,
2I
••• }
by
4.2
row (b) T
=i
if b belonqs to row i of T I
cOlT(b)
=j
if b belongs to column j of T.
DEFINITION.
The initial A-tableau will always be denoted by
it is defined to be the A-tableau in which the numbers 1, 2, ••• , n in order when readinq from left to riqht in successive rows.
e.q.
If A = (4, 2, 3)
I
then T
A
= 1 2 3 4 •
5 6
789
- 22 -
5
ROOT SUBGROUPS OF G
n
It is useful to adopt some standard terminology from the theory of root systems. Let ~ group act on
(See, for example, Carter [C].)
= {(i, ~
(i, j) 'IT
u,
If
j)
I
= j,
i
1 ::;; i ::;; n, 1 ::;; j ::;; n}, and let the symmetric
accordinq to the rule:
=
(i'IT, j'IT) ~
j) c
«i, j)
E:~,
and a. Elf, let Xi' q
)
'IT
E~)
n
(u) C! G
n
•
be the matrix with (i, j)
entry equal to a., all diagonal entries equal to 1, and zeros elsewhere: j
1
Xij(a.)
=i
Ct.
·1
Then
Therefore,
is a group, isomorphic to
+), the additive group of If. q
(X,.
J.)
is known
as a root subgroup of G . ) n
5.1
DEFINITIONS.
( j, k)
£,
r,
A subset
we have ( i,
k)
6
r of
~
is said to be closed if, for all (i, j)
r.
If r is a closed subset of
~,
let G(r) be the following subgroup of
G : n G (f)
=
<x..
J.)
I (i,
j) 6
r> •
The next result is very easy, but it is of fundamental importance in the later Chapters: - 23 -
5.2
THEOREM.
If f is a closed subset of It, then
G(f)
I
X j
U:,j)ef
i
~
,
where the product can be taken in any order.
The group G (r) consists of all
• diagonal and zeros in those places (i, j) those matrices which have lis on tlie for which (1, j)
e
\
~
each element of G(r) has a I
x
(i,j)ef
Proof:
Once the order of the product has been chosen «
f.
unique expression of the form
(a.)
ij
l.j
This is iDmediate from the followinq commutation formulae, which can
be deduced from the definition of x
(5.3)
(a) :
SUppose that (1, j), (j, k), (k, R,) € t.
In particular, IG(f)l= qlfl.
(5.4)
ij
If
r , r l' r 2
we
also have:
are closed subsets of
e,
with r
~
r1 u r2
and
r 1 f'\ r 2
empty, then G (f) =- G (f1) G( r 2) and every element q of G(f) can be expressed in a unique way as 9
If
r
f'
= {(i,
~
9
192
with
is a closed subset of t, define k)
I
(i, j)
and
(j, k)
c r for some j} •
It is easy to see that f' is a closed subset of It contained in f; indeed,
r'
is strictly contained in runless f is empty.
- 24 -
From (5.3) we obtain:
(5.5)
The commutator subgroup of G(T) is G (f I)
5.6
DEFINITIONS (i)
~+
= {(i,
( i i)
~
= {(i
ThenU
•
j)
I
, j)
G (~
1 s j < i s n} and U
i.
+ (respect1vely, . U ) is the group of upper (respectively, lower)
unitriangular matrices:
U+
"" the set of matrices like
U
= the
set of matrices like
The commutator subgroup (U-)
I
of U
[>.:] [~. <].
consists of those lower unitriangular
matrices which have zero immediately below the diagonal: (1 ({j)
I
o
= the
set of matrices like
* If 1 s i matrix in
G
n
:S
0)
1 0 1
0 1)
n and 6 is a non-zero element of
with (i,
~,
let hi (6) be the
entry equal to 6, all other diagonal entries equal
L)
to 1 and zeros elsewhere: i 1
1
o h.(6) = i 1.
o
Let H
=
1 ~ i :5 n,
s
ElF' {OJ>. q
matrices: - 25 -
Thus, H is the group of diagonal
H
= the
For c , 13 E: f1:", 13 q
(5.7)
-1
~
[':'.~.].
set of matrices like
~
0, we have
if k ::: j
(13) x i j (u) ~ (13) ::: Xij (a.B) , ::: x
ij
(a.f3
-1
if k ::: i
),
if k Thus, H normalizes each root subqroup. qroups U+H ::: HU+ and U H
i, j •
~
Hence H normalizes u+ and U •
= HU are denoted by B+ and B-, respectively
are called Borel subgroups.
B+
B
= the
set of matrices like
= the
set of matrices like
By (5.7), if 1
~
**.••
rlO
*] *
[:*...:]
d < n, then
and h
-1
d
(f3}x, 1 . (a.)h ::: x. 1 ,(a.) d(f3) ]+ ,J J+ ,J
if j > d.
Hence:
(5.8)
Suppose that 1 s d < n, and that 13 is a non-zero element of
there exists h h h
-1 -1
£
H, such that for all a. e ~,
xd+l,d (a.) h
= xd+l,d(a. B} ,
x , 1 . (c) h ::: x , 1 ,( o) ]+ ,] J+ ,J
and for all j
- 26 -
~
d.
~
If
'11' E.
c;'n' we shall identify
which acts on the basis e (I
~
l,
'11'
with the permutation matrix in Gn
e , ••• , en of V by
2
i :Ii n)
•
Let W be the subgroup of G obtained by this identification: n
W = the set of permutation matrices.
For calculations, it is useful to notice that if q 'Ii
-1
£ G
n
and '11' E W, then
g (respectively, g'li) is obtained from q by applying the permutation
the rows (respectively, columns) of q. (5.9)
For example,
1 0I 0] 0l O 0] a [1 10 0 0] 0 _ [Cia y 0 a y I 0 1 0 0 0 I 0 0 0 [ 000 1 6 £ t 1 6 &t I
o0 01
The followinq results are elementary: (S.IO)
(i)
For each '11'
-1
-1 '11'
(ii)
'11' E
W, a E'
IF':, (i , q
xi'J (a) 'Ii = Xi 'Ii, j '11' (a) , and Xi j 'l1'
= Xi 'l1' , j 'l1'
•
W normalizes B.
- 27 -
j)
~
t,
'Ii
to
SUBGROUPS OF G ASSOCIATED WITH COMPOSITIONS
6
n
we
shall study the KG -module which consists of the vector space over n
K whose basis elements are the )'-flaqs.
It is often best, in practice, to
view this module as a riqht ideal of the qroup alqebra KG. n
To do this,
we consider the stabilizer of the )"-flaq
where • ••
e n>
I
(0 S i S h)
,
and
The stabilizer of this )'-flaq is
"1
p).,
= the
set of matrices like
>'2.
.
.. .
• .- *" •
0
•
Recallinq the definition of the initial ).-tableau T)., we may write P). (a so-Qalled parabolic subgroup of G ) as n row
T ).
we
(i) S row (j» T ).
shall be amply rewarded later for usinq this apparently clumsy notation
involvinq T). ..
6 .. 1
DBFINITIafS (1)
Let u+
(ii)
Let U). = <Xi j
).
-
= <X i j
row (i) < row (j» T T ).
x
I
row
T
(i) > row (j» T
x
- 28 -
x
Then
+
U
~* o
the set of matrices like
A
I
A~
).,1
UA
•••
~o
the set of matrices like
•
*" ..... ) ,
For example, when A = (1, 1, 1, If
r
p
A
= B+ ,
is any closed subset of ell and
r1
{(i, j) €
r2
= {(i, j) €
r r
I
row T
I
row T
1T
(i)
row T
S;
A
1T
A
(i) > row T
r1 =
r =
{(i, j)
u
=
(U
i
=
1.
-1 f)
1T
,
+, U
U
1T
1T
1T
-
A
r.
Therefore, applying
(i)
-1 I)
'IT
UATI)
- 29 -
-.
(j)}
> j and row (i) > row (j)} T 1T T 1T A A
PATr)(U
U
(j) }
A
That is
A
A
we have
i > j and row T
{(i, j)
and G(r ) () G(r ) l 2
~
A
then c:
1T
are disjoint closed subsets of ell whose union is to the case where
+
U
7
COSET REPRESENTATIVES Since we shall consider the permutation representation of G on the n
riqht cosets of P A' we wish to determine a set of riqht coset representatives for P
A
(By a riqht coset, we mean a set of the form PAq with q E G . ) n
in G • n
We shall illustrate with several examples of these coset representatives, since they will be used frequently in later chapters. First, we deal with the special case where A
7.1
Proof:
Every right coset of B
THEOREM.
It is known that G
n
But B+ and B
=
in G
n
(1, 1, i., ••• ):
+
has the form B
is a union of double
CO sets
where
B 'll"B , where
'II'
c: W.
Therefore,
U B+'ll"B'll"IW
(we shall call this lithe Bruhat Decomposition").
Now, every element of B-
has the form hu with h E: H, u £ U- , and W normalizes H
riqht coset of B+ in G has the form B+'II"U, with u n
we may write u
u' e
+ +
'IrU,
are conjugate by the element of W which reverses the numbers
1, 2, ••• , n.
Gn
+
=
-
U
= u 'u", "
'II"
€
u.
~
+ B.
Thus, every
Further, by (6 .2) ,
where
-1 +
u" ~
B 'II" ,
U
-lu- 11"
t1 'II'
•
Then
+
Bw
= B+Wl'll"-1 nu"
=
+ B nu" ,
•
as stated.
we
remark that it is simple enouqh to construct an alternative proof,
and at the same time prove the BrUhat Decomposition.
For, qiven a riqht
coset B+q of B+ in G , then we can pre-lIDlltiply q by elements of B+ in such n
a way as to"rcw reduce q, starting' at the bottom" (cf. (2.11) and the argument follcwinq it).
We leave the details to the reader.
- 30 -
7.2
If n
EXAMPLE.
= 3,
then each right coset of 8 + in G has one of the n
following forms:
[ ~ ~r o~] , [~ ; ~] , [~~~] , [~ ~ ~] «
100
ala
100
(a, B, y e F ) • q
The element
la 1]
(B 0 1 0
100
[g100 ~ ~] ,
is in the right coset 8+ \
since we can "row reduce it starting from the bottom".
EXplicitly,
1°1 -a-6] [6a 01 11OJ = [00 01 oj o [0 0 1 1 0 0 100
1)
It is interesting to verify that q3 + 2q2 + 2q
we should like to comment here on our notation.
+ 1
=
(1 + q) (l + q
+ q2)
•
In Theorem 7.1 we
have committed ourselves to working with right cosets; there wae also a choice of signs ± in the double coset decomposition Gn '" The alternatives involve thinking about column vectors instead of row vectors, and reversing the parts in partitions.
All our later results could be amended
to cope with whatever form of Theorem 7.1 we cared to take, but we feel that the development runs most smoothly with the choice we have made.
Consider now the right cosets of the parabolic subgroup Since 8 IT
+
~ P~,
ea~h
riqht coset representative of P
A
P~
has the form
of Gno ~u
with
E W, u € U •
. LEMMA o
7.3
-1
1Tl~2
, with lT t...!"2 I f P 1T u '" P 1T l A l 1 A 2U2
lfWIlPAo
Proof:
- 31 -
e
W, and ul~2 S U , then
u
-1 -1
-1
w has a 1 in the (i, i) place, and so w lulu2 2
1u 2
-1
where w 1w2
has a 1.
There fore,
-1
~
1T 1 1T 2
has a 1 in each place
•
W ('I P A•
The Lemma illustrates nicely the analogy between our parabolic subgroup of G and Young subgroups of n
~
n
•
We know from (6.2) that, given w
~
W, every u € U can be written in
the form
Therefore, (7.4)
-1
Remembering that (w
7.5 of P
THEOREM A
where
1, we have proved:
PAW)
suppose that A is a composition of n.
Then every right coset
in G has the form n
7l" €
r
W, and u E: U = {(i, j)
/l
Ii>
W-
j
\J~ 7l"
= G(r), and
and
row
T
W
(i) > row
A
T
W
(j)}.
A
Furthermore, if
•
It is useful to understand clearly the form of the right coset representatives wu of P so each row and each
A•
Suppose W is given; this is a permutation matrix,
co~umn
contains precisely one 1.
The elements
lTU,
-
with
u € U , are obtained by replacing each 0 Which lies to the left of a 1 by an arbitrary element of IF (cf. (5.9». q
In order to explain Which of these
- 32 -
elements
'I1U
-~~ 'Ir I we
have u C U- "
imagine that the rows of the matrix
'Ir
'Ir
are divided into batches, batch b consisting of rows b-l 1 + ~ ~i i=l
b
to
L
A. •• 3..
i=l
Then the elements in
'Ir (U
arbitrary element of
Fq
-lu~ 'Ir)"
1\
'Ir
each 0 in
'Ir
are obtained by replacing by an
which lies both tb the left of .a·1 J.n the
same row and below a 1 which lies in a sUctly higher batch of rows.
(cf.
Example 7.2, there each individual row forms a batch.) 7 .6
EXAMPLES.
If n = B, A.
(1)
(2, 3, 3) and
•
I I I
'Ir :::
•
I
then every element of
-1 -
n
(U
1T
U)"1T)
'Ir
has the form.
I I
. .
I I
··· ·• ..... ··
•
•
I
I I
. .
I
where each. is replaced by an element of
IF. q
(Empty places denote zero~)
In this example,
1 2 T)"
=3
5 1 2 7
=3
4 5
6 7 B
and U-
1\
-~~ 'Ir
'Ir
6 4 B
is the set of matrices of the form I
·•
•
I I
. .
,
·. · .. ·•• ·· ·
I
I
•
I
- 33 - .
In this last matrix, there is a • in the (i, j) place if i > j and row
T A'11"
(i) > row
If n = 4 and h = (2, 2), then the right coset representatives
(ii)
• IX
y
,
,
(j).
T x"
I
,
l
6:
I
IS
(ll,
I
I
oc y
I
I
I
I I
I I
~
I
I
I
I
I
Ol
"
I
,
I 0<
I
I
B, y, 15 £ IF ). q
Whenever A
=
(h
l,
A has two non-zero parts, a coset representative 2)
is determ1ned by the entries in the last A rows (compare Example 7.6(i1) 2 with Example 2.13).
We recall that each right coset of P A in
G
n
always
corresponds to a A-flag; all that Theorem 7.5 achieves, therefore, is a "coding II for the collection of A-flags.
For example, if A
=
(2, 3, 3), the
right coset representatives of Ph in G correspond to flags n
o = 8,
where dim V
dim V l
= 6,
dim V 3
= 3.
Notice that the last A rows of the coset representative matrix h correspond to a subspace V _ of V, the last A _ + hh rows to a subspace h l h l _ .2 V - , and so forth. h l h 2
V
The fact that the batch of hh_l rows correspond
to a quotient space V and so on, illustrates our point about h-2!Vh_ l,
ll
row
reducing from the bottom". To summarize, there is nothing mysterious about the coset representati constructed in Theorem 7.5; they merely represent a translation of the concep of A-flags into a language which allows us to work with the group algebra of G • n
- 34 -
8
SUBGROUPS OF G
n
USED FOR INDUCTION
Unlike the. situation for 'the 8ymmetric qroups, where r; 1 is clearly
n-
the subqroup of
~
n
which shoUld be u8ed for induction on n , there is some
choice of "obvious" subgroups of G n,
We can do better than restricting our
attention to a subgroup isomorphic to GL
n-
subqroups we shall use.
l(q), and we introduce here the
In the next section we shall derive, using them,
some results on induced representations.
(Bence the ambiguous title of this
chapter! ) There are four types of subgroup which we wish to name G , G*, ft*, r r r A , respectively, and these are most clearly described in terms of pictures:
r
r, G :
o "*
r
G*: r r
II
...
A : r
"*" I·
0:
I.-
f
~
To
r
Ih-r c:::::>
~
.. ~
j
r
In... c::::::> H*: r
~
f
avoid any possible ambiquity, we give formal definitions of these
qroups.
8.1
DEFINITION •
Assume.O
:S
r
:S
n.
(i)
G = <Xi"' hi(B) r J
i, j > n - r, B £ ~ ..... {oj>
(i1)
G* = <X , hi (~) r ij
i > n - r, ~ E
(iii)
H* = <X r
i j, ~(l3)
q
n=q
..... {a}>
i, ,k > n - r + 1, ~ £ or i :: n - r + 1 and j
(iv)
A
r
... <X
i,n-r
I
i > n - r>.
f q ..... :S
{OJ,
n - r>
In particular, GO
= GO = He = AO = the
identity subgroup.
We have (see Definition 6.1 and (5.3»: (B.2)
IAr I = q r • G*
r-l
4 B* C G*.
r
r
and U
U
B.3
(n-r,r)
is an abelian normal subgroup of G*. r
(n-r,r)" Gr
:=
THEOREM ..!-r~;'
and
1
G*
r
=U
(n-r ,r) Gr •
and given any two nan-identity elements
alt...i'-2~r
there eXists q £ G such that r
In view of (8.2) and the fact that Ar -C U (n-r ,r) , it is sufficient
Proof:
to consider the effect of conjugating an element of A by an element of G • r r Write elements a £ A , g (: G in the form r r
a
=
9 =
II'
o
where y is a column vector of height rand m is a non-singular r x r matrix. Then direct calculation reveals that
t., c:::> gag
-1
=
0... v
Ir
where y is the column vector mu. N
The result now follows. - 36 -
•
From now on (except where otherwise explicitly stated in one or two examples), we assume that K is a field whose characteristic does not divide q.
-
If q is a power of the prime p, let K be the field obtained from K by adjoining a primitive pth root of unity.
8.4
DEFINITION.
-
If A is a subgroup of G , then a linear K-character X n
of A is a mapping
X:
-
A+K \ {a}
such that x(a
l•
a ) 2
= x(a l)x(a 2 )
for all a
X of A is said to be non-trivial if X(a)
Now, every element a of A
r
7t-
l,
a
2
e
A linear K-character
A.
1 for some a
f;
A.
is determined by the column vector
~
of
height r in the description
a =
and the effect of multiplying two elements of A is to add the corresponding r column vectors.
Hence the proof of Theorem 8.3 shows that the action of G
r
on A , by conjugation, is the same as the action of G r
(viewed as the group
r
of r x r non-singular matrices over height r and entries from IF.
q
r
t
=P ,
If q
q
IF) on the set V of column vectors with then V is a vector space (of r
dimension tr) over the field IF of p elements. p
the set of non-zero elements of the
Irp -vector
Since G
acts transitively on
r
space V , G r
r
also acts
transitively on the set of non-zero elements of the IF -vector space V* dual p
to V • r
(As usual, v« is the set of r
IFp -linear
the G action on this set is defined by v(8g) r
This proves:
- 37 -
r
mappings 8 from V to f ' and r p g(v8}
for v € V , g E G .) r r
8.5
+It..!.2
If
THEOREM.
then for some q
(!
r
G ,
r
=.
1 1 ('gag-1 )
are any two non-trivial linear i-characters of A
2 .;. (a.. ;,}__f_or_a_1_1_a_f_Ar•
Since (iF, +) is an elementary alielian p-group, it has q distinct q
linear K-characters, by which we mean mappings
F
x:
K ,
+
q
{a}
such that X (a • (J2) = X (a ) X (a ) for all a , a ~~. l l 2 l 2 Theorem 8.5 (arising from taking r
8.6
J.fq '
~1t..J(.2
COROLLARY.
= 1)
A
particular case of
is:
are any two non-trivial linear K-characters of
+), then for some 6 E ~ \ {O}, for all a ElF.
Xl (a6) = X (a) 2
,;;;.--------_....q
Gi.ven any linear i-character X of (~, +), we obtain from it a linear i-character
+ of I.. . r
A , by defining
r
C'::)
Qj
0p:.··· 8.7
LEMMA.
B* r
Proof:
= tg
I~
If the character. of A described above, is non-trivial, th r, .
+(gag-1 ) = +(a)
€ G* r
for all a E A} r
Since
I_ e::::> h
I~
0·•·
i., c:=:> !Ii
0...*.
Ir
«..
- 38 -
I..
for all h E B*, We certainly have r
H* ~ {q E G*
r
r
I
• (gaq
-1
)
= • (a)
But, by Theorem 8.5, G* is transitive on the qr - 1 non-trivial linear r
K-characterB of Are
Since
IG;:B;I
c:
qr - 1, we deduce the result.
- 39
9
SOME IDEMPCYI'ENT ELEMENTS OF
KGn
With the preliminaries now behind us, we commence our study of modules in this chapter.
For each closed subset
several idempotent elements of
KGn ,
r
KGn -
of t we shall define
and investigate the properties of these
idempotents.
9.1
DEFINITIONS.
(fq , +),
Let Xl' ••• , X be the distinct linear K-characters of q
with Xl equal to the t~ivial character (that is, xl(a)
=1
for all
a elF). q
r
Given a closed subset
... ,
of t, let c be any function from
r to
{l, 2,
q} such that ~i ,
j) c
=1
if (i, j)
e
f' •
Define the function X from err) to It by c X: c
x
II (i, j)e I'
(a ij
)
l+
ij
II (i,j).r
X(i,j)c
(ex •• ) ~J
,
(a
ij
€
.IFq ) •
In view of Theorem 5.2, and the fact that we have taken c to have value 1 on f', Xc is a well-defined linear i-character of G(f).
Bence
is an idempotent element of KG • n
9.2
THEOREM.
If
r
is a closed subset of
~,
and flt-I2, ••• , r
closed subsets of • whose union is f, then
where the product can be taken in any order. Proof:
Making repeated use of Theorem 5.2, we have
- 40 -
k
are disj
1 =rrT
\
L
q
=
Xc(g)g
g€G (r)
+
l
f
ql11
j»]
II
{X [ xi.(f(i, c (i,j)€r J
r
the sum being over all functions f from
II
(i,j)€r
x .. (f(i, ~]
j»} ,
to IF , and the product being in any q
order,
=:Tn=
r L
IT
l(U~
(i,j)Er
q
II
(i,j),r
[1
Xij(O:)J~
(0:)
X g] .
l
qgEX
X(i,j)c
(g)
ij
c
Since the product here can be taken in any order we please, the
•
Theorem follows.
9.3
EXAMPLE. c:
(2, 1) (3, 2) (3, 1)
Let n ~ 3, q = 2, 1+ 1+ 1+
r
= ~-, and c be defined by
2 2 1 .
Then
1 00l} [1o 10 1OJ
{II 00] 0 1 0 0 0 1
IIf10 001 10OJ} {(Ilu0 100011)0'
-
-
fllO 10 OJ} 0 . 0 1 1
The result is the same if we take the product in any other order. notice that the last two terms in the product do not camnute.
IJ. 4
DEFINITIONS. (L)
SUppose that I
S;
Let r(r) ~ {(i, j) c ~-
r
:S
n.
I
j
:S
- 41 -
r} •
However,
Define the function
(ii)
(j + 1, j)c (i, j)c (iii)
If n
EXAMPLE.
E
from r(r) to Ii., 2, ."., q} by
r
=2
r
=1
if j < r
for all other (i, j) e fer) •
Let
=
Er
9.5
r
C
1 == -
1
:rr7rrT q
= 3,
L Xc (q)q • qCG(f(r» r
then
[1 00]
L
a 1 0 601
q2 a,6E/f q
rl
E
0 01 X2 (a) a .1 0 2 == -q3 ~ Y 1 a , 6, ve IF'q
E
1 x (y ) 3 == -q3 2(a}x2 a,S,Yf!fF
1
l
L
J
[1 00]
L
a 1 0 S Y 1
q
We emphasize that these are elements of KG).
(We never consider matrix sums.)
From (5.8) and Corollary 8.6 we have the following useful result: (9.6)
Suppose that 1
S
d < n.
•
Let +d' +a. be non-trivial linear K-characters
of Xd+l,d' and for each j with 1
S j
< n and j ~ d, let
~.
J
be an arbitrary
linear K-character of Xj+l,j" Then, for some h E H, h-
l(
L
I
'd(g)q)h
qliXd+l,d
L
1
h- (
+d(q)q, and
q,xd+l,d +j(q)q)h
=
g,"xj+l,j Applying this, we obtain: (9.7)
Suppose that c' is any function from fer) to {l, 2, ••• , q} such that r
(j + 1, j)c 1 > 1 if j < r r
(i, j) c
I
r
=1
for all other (i, j) 6: I' (r) • -
42 -
Then, for some h E H,
For this reason, the exact nature of c
r
does not matter for our
purposes - any function c' satisfying the conditions of (9.7>. would do equally r
well in the definition of E • r
Now, G(r(r»
= An _ l
A-
••• A -
n 2
(see Definition 8.1).
n r
with 0 S s s n - 1, A is an elementary abelian s
For each s
p qroup (q ... a power of p) ,
s
of order q •
9.8
DEFINITIONS.
Suppose that 0 s s S n - 1.
q
s
be the linear X-characters of A , the s
notation beinq chosen so that for 1 S i
~
q, i .is the character qiven by
t., c::::::>
0 '·"••
Is
Gfr
(ii)
s
For 1 S i S q , let
Theorem 9.2 gives (9.9)
E r
= A(2)
n-l
In particular, E
A(2 ) .•• A(2) n- 2 n-r+l
A (1)
n-r
(1) and E ... A(2) ... An-l A (2 ) n n-l ••• 1 • l
(i) I 1 ~ i S qn-l} is a set Now , {An_l""l sum is 1.
0
f orth oqona1 idempotents wuuse ~-
The results from the last section show that for each
- 43 -
KGn-module
H,
q
M=MA(l).( n-l
n-l
l
MA (i)
i=2
)
,
n-l
as a KG* l-module. nGiven an irreducible KGn-modUle 0, the cOmposition multiplicity of 0 in M means the number of factors isomorphic to 0 in a composition series of M.
The above decanposition of M as a KG
I-module, together with the fact
n-
that A 1 has order coprime to the characteristic of K shows: n(9.10) The composition multiplicity of the trivial KG -module in arbitrary n
KGn-modUle M is at most the composition multiplicity of the trivial KG n_lmodule in MA (1) n-l
= MEl.
We next investigate more fully the decomposition of M when it is viewed as a KG*
-module:
9.11
If M is any KG -module, then n
n-l
THEOREM.
n
(i)
(I r=2
M '" MEl Ee
n
(ii) dim M =
L
ME
r
G* 1) , and
n-
(qn-l _ 1) (qn-2 _ 1) ••• (qn-r+l - 1) dim (ME ). r
r=l Proof~
First note that, by Lemma 8.7,
(9.12)
All elements of G* cODDDUte with n-s
E
when 1 S r
r
S
s
S
n.
Now assume that 1 s s < n and that the following results are true: s
(9.13)
l r=l
ME
n-r+1
G* s-l
= MA(2) n-1
n-2 s
L
r=l
G~_l
s
'
and
8-1
(11
i=r
These results are certainly true when s Since
A (2)
A (2)
(qi - 1) dim (ME
n-r+
I» •
= 1.
CH: C G;, we see that the right coset representatives for - 44 -
G*
s-l
where h E H* and in G* can be taken in the form h g j i, j s s {gi
I
1 S i
:$
s -l}
q
is a set of right coset representatives for H* in G*. s 8 { -1 A(2) gi gi 8
(9.14)
I
1
~
i
~
q
s -l}
= {A(i) s
2
Lemma 8.7 shows that
s i S
qS}.
Therefore,
= .(2)
A(2)
n-1
=•
n-2
A~2) (l gi)' by
(2) A (2)
n-l
n-2
Lemma
A (2) A (i)
s+l However, ME
= ~~~~ A~~~
and E n _ s
G*
n-s
s
•••
8.7,
i
= MEn-s ,
n
using (9.14).
'
since the elements of G* commute with E , s n-s
A(+2)1 A(l). 8 s
orthogonal idempotents in KG
s
I
Now, {A(i) 8
1 SiS q8} is a set of
Whose sum is 1, so
8+1
l
r=l
G*
ME
n-r+1
ME
s
n-r+1
= .(2)
G*
s
A (2)
n-1
n-2
This proves that the first part of (9.13) is true when s is replaced by s + 1 and hence (by putting s + 1
=n
in the last equation above) we have proved
conclusion (i) of the Theorem. We also obtain: s+l
dim
(L
ME
r=l
n-r+1
G*)
s
s
= dim
(ME
+ (qS - 1) dim ( n- s) - 45 -
I r=l
ME
n-r+l
G*)
s-l
s-l n (qi - 1) i=r
s
= dim (ME
=
s+l
l
n-s
s
n
I
) +
r=l i
(q
-
d~
1)
r=l i=r
d~ (MEn-r+ I)}'
n- r +1).
(ME
Therefore, the second part of (9.13) is true when s + 1 replaces s. (il) of the Theorem is the result of takinq s :: n in (9.13).
EXAMPLE.
If n
= 4, +
Conc Ius ic
•
then
U}
M = ME Gj 1
(ii)
dim M = dim (MEl) + (q3 _ 1) dim (ME
+
by (9.
ME Gj
2
+ ME3Gj + ME4 G) , and
(q3 _ 1) (q2 _ 1) dim (ME )
3
- 46 -
2)
+ (q3 _1)(q2 - 1) (q - 1) dim (ME
4)
Let A be a composition of n, and let
p~
be the corresponding parabolic
subgroup of G , described in Section 6. n
10.1
DEFINITIONS. Let p
x=
I 9(P
9
e KGn,
and let
M)"
A
Thus, II, is a right ideal of KG , and so it is a right KG -module. A n n Every element of II
is a K-linear combination of terms of the form
A
by Theorem 7.5, but M may be regarded equally well as the permutation A on the set of ~-flaqs.
representation of KG
n
10.2
THEOREM_ f
r
l 2
= {(i, = {(i,
Let
r
j)
r
C
be a closed subset of
row row
j) € r
T
1T
A
T ~1T
~
,
(i) > row T
A1T
Proof:
P~1T9
Sn-
Define
(j)}.
.!!
The cosets
1r f:
(j)} (i) :$ row T ~1r
SUppose that c and Xc are as in Definition 9.1.
=
and
Then
{2, 3, ••• , q } c
-1
~
r2
0, otherwise.
(9 £ G(r 2 » are all distinct.
If {2, 3, ••• , q}c
-1 ,J.
~ f
2
, then (k, t)c > 1 for some (k, t) E f l
Then
47 -
Usinq these results, we have
= L
(X(k, 1) c (n) P~ 1T~, 1 (0:»
clEFq •
=
(
2
a~F
X.{l~,~.c (0:»
-P~'Il'
q
=0
.
:Qut
P~1T
L
xc(g)g
gEG(f)
=P~1f (
2
gE:~,R.
x (g)g).(an element of KG ), c n
by Theorem 9.2,
= o. If{2, 3, ••• , q}c
-1
S;
f 2' then
But
by Theorem 9.2, so
Since r ~ ~-, Theorem 7.5 shows that the cosets PAwg (g E G(r distinct, and
ther~fore
2
»
the above element is non-zero.
Next, we investigate the consequences of Theorem 9.11 when Recall that fer)
= {(i,
are all
j)
Ii> j
S
r}, and
- 48 -
M
•
= M~.
= A(2)
A (2)
where (i, j)c
1 only if i - 1
~
r
Now, G
=j <
n-z'
r.
normalizes GCf(r» and G(rCr)}G
~r
A (1)
n-r+l
n-l
~r
is the group of matrices
of the form
(10.3) I
.",0
<::)
"*
"*
Using (9.12), we see that all elements of G(r(r»G (and, in particular, n-r all elements of U ) commute with E •
x
M is spanned by elements of the form P 1TU A A P
1TuE
A
-P)..'Il'E
= P X1TE r U .
r
r
~
(1T c!
W, u E: U-), and
Furthermore, an application of Theorem 10.2 shows that
0 only if
I
{ (j + 1, j)
< r} (; {( i,
1 S j
j)
j row
T A1T
(i) > row
(j) }. T
A1T
That is,
row T
1T
(1)
A
< rowT 1T (2) < ••• < rowT 1T (r). A A
For this to be possible, TA must have at least r rows.
10.5
THEOREM.
~-AEr =
0 for r > h.
Suppose that A has precisely h non-zero parts. Thus h
(i)
M, = M,E
/\
This proves:
/\ 1 •
(
L M,E
G* 1)
r=2 /\ r n-
- 49 -
Then
(11)
= ~L
dim M)"
(q
n-l
n-2 n-r+l 1) (q - 1) ••• (q - l)dim (M~Er)'
-
r=l For the remainder of this Chapter, let)" each~
>0
i
=
(AI' ),,2' ••• , )"h) with
Let the integer r
Clearly, this assumption involves no loss.
.
satisfy 1
10.6
O,T E ~
If
DEFINITION •
i 'd.ith 1 s i Sr.
10.7
LEMMA.
n
write a
,
"II
T
if row
T ),,0
(1)
=
If 9: E Gn-r and 11" E W, then there exist 11"' E: W and U f U
-
SUCh
'"
G
~
A = P x1I"'U
P 1I"g ~f __••
and
There exists a (r + 1) (11"0)
-1
~
11"
W
I'l
11"'.
"II
G
n-r
> (r + 2)
(11"0)
such that -1
> ••• > n(no) -1 •
UBin
9 the Bruhat Decomposition for G
n-r
-1
o
But p
9
>.. '!fOUl
ThElt
= hu l '[u 2
= P)" 11"0,
Elfor e P),,1Tg
, we may write
'
by the definition of a, and 11"0'[
= P),,(1TO'[)U2
= {R I
since o r E W "G _ • n r
11",
gives the expression we seek.
Assume 1 S r
DEFINITIONS.
"II
•
S h.
IR I = r}
(i)
Let ~
(ii)
For each R C '\., let 1I"R be any permutation such that row
T ),,1T
and
R
R .!i {I, 2, ••• , h} and
(1)
< row
R
= {row T
),,1T R
T ),,1T
(2)
< ••• < row
R
(il l I S i S r} • - 50 -
T ),,1T
•
(r), R
n
(iii)
For each R ERr' define the composition }"R of (n - r) by
By construction, no two elements 1n {'Il'R
the seWJe of ·10.6.
L
(10.9)
Ric1\-
Hence, by
L~
I
R
e
~} are equivalent, in
7.3 and. 10.7,
p\'Il'RCKG _ ) is a direct sum. n r
1\
NOw, Theorem 10.2 shows that (10.10)
P 'Il' E
AR r
r.1
where
=
tu.
=--J-rp'IT q
Ir 21
AR
\
GL(r) qr 2
X (q)q
c
'
j)
But we cannot have P},,'Il'Rg = P},,'Il'Rq ' with g E G(r ) 2 g = 1, by Lemma 10.7 and Theorem 7. 5.
~
u ,
g' € G unless n-r,
Therefore,
Bence, from (10.9), and. the fact that the elements of G n-r
COlllDUte
wi.th E , we have r
r
THEOREM.
10.12
P\'Il'RE (KG ) is a direct sum. r n-r
Rfi"l.r "
The KG
n-r
-modules
p\ 'Il' E (KG ) and P ,\ lT,R-(KG ) are isomorphic. -" R r - n-r n - r --';;~-"----""'" THEOREM.
10.13
For 1 :s; r :;; h,
e
K,E · = r 1\ RE~
Proof: P),,'lTUE
r
P, 'Il'RE (KG r
1\
Let 'Il' £ W, U € U. ;II!
0 only if
'IT '\I
lT
R
n-r
i .
Since PI..'Il'UE r for some R
= PI.. 'Il'ErU,
E~.
- 51 -
(10.4) shows that
But then P}"lT
= PlTRO'
for some
~L
(ii)
(q
n-l
n-2
-
1) (q
-
1)
•••
n-r+l
(q
1)d1m (M"E ) .
-
r
r=l For the remainder of this Chapter, let A :: (A A ••• , l, 2, each Ai > O. satisfy 1
Clearly, this assumption involves no loss. r
~
~
h; we shall examine MAE
~)
with
Let the inteqer r
r• ,
10.6
If a,T
DEFINITION.
i with 1
~
i
e
~
r.
~
n
, write a
~
t if row (i) T Aa
al~.,
=
I
1 .)
10.7
If 9 € G and --------n-r
LEMMA.
W, then there exist
E
71'
71' I
€
Wand u e U-
such that
1'\
G
4!
and. j I
,~
Proof:
There exists a
£
W~ G such that n-r
Using the Bruhat Decomposition for
where h e H, u
But P
A7I'OUl
Therefore P
10.8
l
~
U+ , r E. W, u
= PA~a, A7I'g
~
n-r
u
we may write
,
and h, u ' T, u £ G • 2 l n-r
by the definition of a, and 7I'at
= P A(7I'at)u2
Then
since at e W ~Gn_r.·
~ 71',
qives the expression we seek.
•
Assume 1 s r s h.
DEFINITIONS.
= {R I
(i)
Let~
(ii)
For each R row and
2
G
R
R
c<;~,
TA7I'R
={l,
2, ••• , h} and IRI
let 7I'R be any permutation such that
(l) < row
= {row
= r}.
TA7I'R
TA1f
(2)
< ••• < row
R
( i) i 1 s i s r} • - 50 -
TA7I'R
(r),
For each R ERr' define the composition A of (n ~ r) by R
(iii)
construction, no two elements in h'R
By
the sense of 10.6.
Hence, by
L~
I
R
e
~} are equivalent, in
7.3 and 10.7,
(10.9)
Now, Theorem 10.2 shows that (10.10)
P'~REr 1\
r1
where
1 =~ I r 21
r«.
I
j)
But we cannot have P
= 1,
A R
q
G
qC'
L
tr 2 )
Xc(q)q,
= {( 1, j)
r2 =
g
~
P ~
i
1T g == A R
PA~Rq'
with g E G(r
by Lemma 10.7 and Theorem 7.5.
2)
~
U , g' £ G _ , unless
n r
Therefore,
Bence, from (10.9), and the fact that the elements of G commute n-r wi.th E , we have r
10.12
P, 1l'RE r (KG ) -n-r
-1\
10.13
1\
Proof: 'll'UE A r
r
Let ~
=
'l1'
REl.r
P, ~REr (Ken-r ) 1\
P, 1l'R (KG ) -n-r
and
THEOREM. II, E .
-P
I
THEOREM.
1\
For 1
e
R.~
r
r
1\
1T
-v
-
U. 1T
R
The KG -modules n-r
are isomorphic. -
S h,
P, ~RE (KG
E W, u €
0 only if
~
is a direct sum.
n-r
i .
-
Since PA'll'UE for some R E
r
= P A~ErU,
6\..
- 51 -
(1O.4) shows that
But then P
1T
A
= P1l'RO
for some
(glE
r
= KEr
ctrren ,
because gl €
by a linear K-character.) M E
;1. r
=
L
RE~
r
is the group sum for G(r(r», weighted
Therefore,
P,7fRE (KG r
1\
and E
n-r
),
r
•
and (10.12) gives the stated result.
-
Proof: -We may regard P
(KG
A
R
n-r
-
) as the vector space over K spanned by the
o ,
<e r+l' e r+2' •.. , e n > (1
~
i
~
h).
For each AR-flag of this form, define a A-flag
by putting V~ = V 1
i
+ <e r , e r- l' ••• , e m> ,
where m is the least integer such that row
T ;1.7I"R
(m) > i.
-
Then P;1. R
(KG ) is n-r
isomorphic to the KG -module consisting of the vector space over K spanned n-r by the A-flags so constructed. to
P, 71"R 1\
(KG
n-r
But the latter module is, in turn, isomorphic
).
Combining our last three Theorems, we have:
10.15
COROLLARY.
For 1
~
r
~
h,- there is a KGn - r -isomorphism, ~=.::.;:;.;:;::._..-:.;;=:;;. - 52 -
•
• Ret"\.
M A
•
R
Applying Theorem 10.5, we obtain:
10.16
COROLLARY.
If A =
(Al~2'
••• , A all the parts being non-zero, then h), as
(i)
l-modules.
(qn-l _ 1) (qn-2 _ 1) ••• (qn-r+l _ 1)
(ii)
EXAMPLE.
KG
-n-
If A = (A A has two non-zero parts, then l, 2)
+ (qn-l - 1)
d~ M{A -1 A _!). 1
•
' 2
We remark upon the similarity between Corollary 10.16 and the correspondinq results for G: • n
In the case of symmetric qroups, M is replaced by A
the permutation module of Gt on the Young subqroup for the composition A. n
It is theI\ easy to see (using the same notation for KS -modules), that n
as a
n- l-module.
~
Thus, there is no module corresponding to the second term in 10.16(i).
But,
on the other hand, the second term has dimension divisible by q - 1, so when we "put q To
= 1,
the second term disappears"l
conclude this chapter, we give an alternative formula for
and derive a numericai result
10.17
DEFINITION.
d~
M A,
concerning this dimension.
(cf. Definition 2.5)
- 53 -
Let k be an integer, and define
He::}
=
if k
[k][k - 1} ••• (1)
~
1
{OJ = 1
=0
1/{k}
if k < O.
I.
Since M is the permutation module of G on p;\, dim M;\ ::: IGn:P A n A Bence (10.18)
we note that it is not obvious that this agrees with 10.16(ii).
10.19
DEFINITION.
~
If
is a partition of n, let
dim (ll)
EXAMPLE.
If II
=
(4, 3, 1) , then
1
1
1
1
rrr
1
1
rrr rrr dim (u)
:::
{s} •
TIT TIT 0
1
trr 1
«IT TiT I ; i
The reason for the notation is that dim (u) turns out to be the
I
KGn -module 10.20
S
II
is analogous to the Specht module
DBFIN7TZON.
(cf. 10.8(i».
i.artition of n, with l1
h
> 0,
Let
of {l, 2, ••• , h} which have the properties that l1 b
= ~+1'
then b + 1 C
= (J.l 1' l12, ••• , 4l.; be the set of [a] = r and if b
Assume that l1
and 1 S r S h.
l
j '. for (JI . 1 n
dimension of the module SJ.l which we shall construct in the next chapter.
J.l
h)
The
is a
subsets
R
~ a and
a.
Thus, an element a C £r lies in~; if and only if l1
n - r.
- 54 -
R
is a partition of i
10.21
THEOREM.
Assume that
=
dim (ll)
h
I
r=l
L
P is
a partition of n.
Then
(qn-l - 1) (qn-2 - 1) ••• (qn-r+1 - 1) dim (llR).
R€~*
r
All the terms in the expansion of the determinant for dim (u) have
Proof:
the form
where
21.
1
+ ••• +
~
= n.
Bence, by corollary 10.16 and (lO.18), h
=
L L
case.
= 0,
••• (q
n-r+1
-
1) dim (llR).
%'=1 Rt'£ r
But if b £ R and ~
and dim (llR)
n-1 (q - 1)
= 'b+1
and b
+ 1
4 R,
then
since rows b, b + 1 of the determinant are equal in this
•
The stated results follows.
EXAMPLE.
If II
= (4,
2, 2), then
x
[lll
X X X X • X X
=X
Dim Mll
=
dim M(3,2,2) + dim M(4,1,2) + dim M(4,2,1)
+ (q7 - 1) (dim M(3,1,2) + dim M(3,2,l) + dim M(4,1,1) )
+ (q7 _ 1) (q6 - 1) dim M(3,1,1). dim (u)
= dim
(3, 2, 2) + dim (4, 2, 1) + (q7 - 1) (dim (3, 2, 1) + dim (4, 1, 1)
+ (q' - l)(qb - 1) dim (3,1,1).
- 55 -
11
THE SUBl«>OOLE THEOREM
we
are now in a position to construct an irreducible KG -module for n
each composition A of n.
11... 1
DEFINITION.
< , >A be the unique bilinear form on H>.. such that
Let
This form is clearly non-singular, symmetric I and G -invariant. n
If
I is any subset of H we define I.,L by A, I"'" = {mi. H
A
I
<m, ml
>A =
for all m
0
€
I}.
Since the bilinear form is G -invariant, 1 4 is a KG -submodule of H, if I
n
n
I\.
is a KGn-submodule of H A• OUr plan is to apply the following general theorem:
11.2
THEOREM.
Let G be any group, K an arbitrary field and suppose that
H is any right KG-module I endowed with a G-invariant non-singular bilinear
form.
<
I
>.
Assume that there exists an idempotent E
=f
ai9i E KG (ai..!....!L
g1 C G) with the properties (a), (b), eel listed below: (a)
ME is a l-dimensional space, ME = lCmE, say em E: H).
(b)
If E
=f
(c)
<mE,
m>
Define S
= mE(KG).
-1 ai9i--L then mE(KG) ~
= mE(KG).
o.
Then each of the following results is true:
(i)
For every KG-submodule I of H, either I i2 S or I , s.L.
(ii)
~.
(iii)
If I is any ·KG-submodule of M and
S into H/I, then for some K (iv)
e K,
(mE)
e
= KmE
e
is a KG-homomorphism from
+ I.
S/(S n S.L) is a self-dual absolutely irreducible KG-module. - 56 -
The composition multiplicity of S/(S n SA) in H is one.
(v) Proof: and I
If, for some x
(i) ~
C
I, xE
=0
On the other hand, if xE
S.
0, then mE C I, by hypothesis (a),
;t
for all x £ I, then for all x E I,
g E G,
o = <xg-1E, since < ,
m> = <x, mEq>,
> is G-invariant. mE,
(ii)
-
mE E:
Then I S; S4, by hypothesis (b).
S, by hypothesis (b).
Also,
<mE, mE> = <mE 2 , m> = <mE, m> ~ 0, by hypothesis (c), so (iii) (mE)e
for some
K
If
=
s
4 s~.
e is a KG-homomorphism
(mE 2 ) e
=
(mE)eE
=
from S into MIl, then
~(mE) + I,
£ K, by hypothesis (a).
(iv) and (v):
s/(S
~ SA)
is an irreducible KG-module, in view of
conclusions (i) and (ii). (S (S + S.,
1'1
Is.J.,
S,L)"J.
=S
+ S~, so the dual of S/(S
which in turn is isomorphic to
Isomorphism Theorem.
Therefore,
sICs
1'1 s.L)
If I is a KG-submodule of H, and
e
(1
sl (S
S..&.) is isomorphic to 1'1 SoL) ,
by the second
is self-dual.
is a KG-homomorphism from
sl (S
f1
s~
into MIl, then by composing the maps S - -.....
canon. we get a KG-homomorphism
8:
mE + (S '" S~)
e 1+
from S into MIl. KmE
+ I
for some K € K, by conclusion (iii). Taking I
=S
f'I
Hence
Now conclusion (v) follows at once.
S~, we have also proved that every KG-endomorphism of
is just multiplication by an element of K.
57
Therefore, S/(S A s.L)
is
sl (S
f'I
S.J.)
(see~
absolutely irreducible
for example, Dornhoff [D), p. 20).
It is noteworthy that although the conclusions of this when G
= $n
Theo~em
•
bold
(taking M to be the permutation module in a Young subqroup -
see James [Jll), we do not know an appropriate idempotent E, except for certain fields K.
A similar theorem is true for weyl modules, see Theorem
8.3.2 in James and Kerber [JK).
In view of the results we have already
proved, there is no difficulty in devising a SUitable idempotent E to apply Theorem 11.2 in the case where G Recall that T
A
= Gn
and M = M • A
is the A-tableau which has 1, 2, ••• , n in order along
successive rows.
11.3
DEFINITION.
Let
1f
A
be the permutation which sends T
A
to the
A-tableau
which has 1, 2, ••• , n in order reading from top to bottom in successive columns.
EXAMPLE.
If A = (3, 2, 4), then 123
TA = 4 5
6 7 8 9
Inspired by the definition of the idempotent E which was used in the r last two sections, we give:
11.4
DEFINITION Let
C
(cf. 9.1 and 9.4).
be the function from t
to
{1, 2, ••• , q}, defined 'by
(i, j)c ... 2,
(i, j)c = 1,
otherwise.
Let
- 58 -
Aqain, the exact nature of e i. unilDportant.
For (9.6) tell. u. that
when c' is any other function from t- to {1, 2, ••• , q} such that (i, j)c'
>
1
(i, j)c'
~
1,
otherw1.e,
we have, for some h E H,
In particular,
EXAMPLES (i)
If ).
(ii)
If
A
,
1
a 1
0
• • 1
**81 **** the
sum beinq over all lower unitrianqular matrices.
(iii)
If A
E
11.7
1
x
=
1 (1, 1, 1), then TA'If). • 2 , and 3
=~ q
l
a,l:3,Yffl:
q
x2 (a ) X2 (B) [~ ~
gl .
y 8 1
The crucial properties of the idempotent
EX are qiven by:
THEOREM.
Then
Let)., 1.1 be compositions of n.
- 59 -
Proof:
P 'II'U lJ
Every element of M is a linear combination of terms of the fom ).l
('II' E W, u C U-).
I
(11.8) {(j + 1, j)
But
co1 (j + 1) = colT (j)} -TA'II'), A'II'),
E {(it j) Ii>
j and row
T \.I
'II'(i)
> rowT
(j)}. \.1'11'
This happens only if the numbers from each column of T), '11'), appear in different rows of TlJ'II'.
Hence P\.l'll'E), = 0 unless ),. lJ ( see Definition 4.1).
This proves
part (i) of the theorem. If ), • lJ I
(11'.8) holds only for those pe%1llUtations 'II' for whJ.ch
row (i) T ),'11'),
= row
T A7I'
(i) ,
for all i.
Applying Theorem 10.2 again I we qet
where
r = {(i, j)
i
>
j
•
and row (i) > row (j)}. T),'11'), Tx'11'),
Hence (noting thE: last sentence of Theorem 10.2): (11.10)
11.11
DEFINITIONS.
For each composition ), of n , let S), I D), b8 the following
KG -modules: n
- 60 -
Theorem 11.7(ii), (11.6) and (11.10) prove that the hypotheses of Theorem 11.2 hold when G
= Gn,
M
= MA,
S
= SA.
We now list the info::m&tion
we have proved about the modules SA' D : A ., (11.12)
(1)
u,
j) c
Let c be any function from ~- to {I, 2, .•• , q} such that
> 1
if i = j + 1 and colm (i) "').'lf A
(i, j)c = 1,
= col
(j)
T).'lf A
otherwise.
Then SA
= -PA'lf A = P). 'IfA
where
r
L_ gf:U
L
xc(g)g (KG) n
= {(i, j)
(i) > row (j) }. I i > j and row T A'If). T A'IfA
The cosets P A'IfAg (g f: G (r» (ii)
I
~
(KGn )
Xc (g) 9
g€G(r)
are all distinct.
THE SUBMODULE THEOREM.
For every submodule I of M , either A
.A-
S). or I .& SA'
(iii)
D
A
= SA/(SA 1\
S~) is a self-dual, absolutely irreducible
KG -module. n (iv)
D in M and in SA' is one • A, A
The composition multiplicity of
Theorem l1.7(i) gives information concerning two compositions ~
A and
•
of n:
11.13 Proof:
LEMMA.
.!!..p). is a composition factor of M
ll,
SUppose that I is a KG -submodule of M •
n
- -homomorphism Theorem 11. 2 (v), any non-zero KG n to a non-zero KG -homomorphism n
e
from SA into
- 61 -
lJ
then A ~
a-
'!'hen, as in the proof of
-e from
M/I.
D into A
But
Mil gives
rise
wMch is zero unless A ... lol, by Theorem 11.7 (i) •
11.14
e
•
COROLLARY
(i)
1!-Pl! is a composition factor of SA' then l! ~ A.
(il)
DA..!..Pll only if
(iii)
SA ~
!\
which map
'J Ker9,
].I
is obtained from A by rearranging the parts.
the intersection being over all KGn-homomorphisms
lnto some Mll with ).! I> A.
•
It is useful to give a qenerator of SA which consists of a linear combination of distinct terms PX'lfAU (u £ U-), each of which has coefficient
± 1.
Then the coefficients will belong to K, whereas the qenerator described
in 11.12(1) has coefficients which are roots of unity, which miqht belong to K \ K.
Let a, a + 1 be inteqers wMch lie in the same column of T
'IfA•
A
Applying:
Theorem 9.2, we know that
generates SA' where y.
aDd
r
II (i,j)er (i, j) ~ (80+1,80)
is as in 11.12(i). Now allow c to vary over the q - 1 functions for which (a + 1, a) c
~;I
{2, 3, ••• , q} but which aqree with the chosen function c on all other elements',
u,
j)€ ~-.
L
a~1F'
,
Then X(a+l,a) c (a) xa+l,a (a)
q
varies over a set of qenerators of the augmentation ideal of the group
- 62 -
x
a+1,a
•
Bence SA is generated by
Repeating this process, we deduce that SA is qenerated by (11.15)
where
~l
= {(j
+ 1, j)
and
We have now achieved our aim of producing a generator of SA which is a linear combination, with coefficients ± 1, of distinct terms
P)" 'If).u (u
~
u i.
If we define SA,K to be the right ideal of KG qenerated by the element n (11.15), then S)"
= SA,K e i.
Also, dim (S).,K
fI
St,K)
= dim
(S)" n
each is just the rank of the Gram matrix for the bilinear form respect to a basis of S).,K.
< ,
S~), since >11. with
Therefore,
is a self-dual absolutely irreducible KG -module. n
Although we proved this
result, with the qenerator (11.15) of S).,K in [J gl , the proof given here, using the extension field K, is superior, and the work in later chapters is greatly eased by the fact that our ground field is K.
11.16
THEOREM
.!A, ell
(The ordinary Irreducible Unipotent Representations of G ). n
is a se1f-dual, abs.o1utely irreducible flG -module. n
are distinct partitions of n, then SA,Q is not isomorphic to Proof:
When K
= ~, < , >A is
If). and 1.1
Sl.l,~.
an iMer product.
result now follows from 11.12(i1i)
~
Corollary 11.14(ii).
We now illustrate generators for the module SA.
- 63 -
The
•
11.17
EXAMPLES
(i1)
If
p).
~
=
(1, 1, 1), then
S~
L
[~~ ~].
x 2(a)x2(a)
a,~,Y~Fq
is generated by
Y~ 1
It is also generated by
P).
100] + [1 0O]} r { r ~ ~ ~] [~ ~ ~] [ Y1 1 Y1 1 Y~IF l~ -
0 1
q
(iii)
If).
=
010
110
Y 0 1
(3,2), then SA is generated by 1 0 0 0 0
o 0 100 o0 0 0 1 a 1 0 0 0 y 0 8 1 0 and by
1 000 0 .1 0 0 00001 01000 Y0 0 1 0
o0
(1v)
~
T
=4
). S).
=
If
-
~ ~ ~ ~ ~)j
0 0 001 11000 Y 0 0 1 0
-
10000 o 0 100 000 0 1 01000 Y 0 110
1 0 0 0 0
o 0 100
+
o 0 001 1 1 0 0 0 y 0 1 1 0
(3,2,4), then
123 5 6 7 8 9
is generated by
,
.
••
I
.. • ..•,. • I
~
,.. /It
where
I
, .. .
I
•
I
;-
•
jIl 1fr
I
• I It
I
is the sum over all matrices obtained by replacing each - 64 -
* by an
arbitrary element of
IF, q
and each. by 0 or 1, the sign being + if an even
number of .' s are replaced by 1 and -, otherwise. (v)
If
A=
(n - m, m) with n - m
~
m, then
S~
is generated by
I I
I
••
· . , • • ., *' ·· · I
oil
using the same conventions as in Example (iv). If we regard
M~
-
as the vector space over K spanned by the m-dimensional
subspaces of V, then the generator described above is
E
L
=0 or 1
ij (lijE Fq
± <Elle l + e 2 , + E e + e , (l2lel 4 23 3 el e3 + O,33 o,31
where the sign is +/- according to whether the number of E s equal to 1 is i j' even/odd. Thus, if m
=2
and n
~
4, as in Example (iii), we get
Several questions arise from the work in this section; later all the questions listed below will be answered in the affirmative: (11.18)
(i)
(ii)
SUppose that J.l is obtained from A by rearranging the parts.
Do
Is every composition factor of SA (and M of the form DlJ with A,
(iii)
Is dim SA independent of K?
(iv)
If K is the field of P elements (P a prime not dividinq q) , K K
is SA,K a IDDdule obtained by "t:educinq IDDdulo P the -module SA,,,,? K" (v)
Is the inclusion, S,
ene Kere,
described in Corollary 11.14(iii)
f\. -
actually equality? The last question is, in fact, the vital
ODe.
We shall call the
equality
"the Kernel Intersection Theorem", and the aim of the next few sections is to prove this result.
- 66 -
12
A LOWER BOUND FOR THE DIMENSION OF S
l.l
In this section we shall construct a certain number of linearly independent elements of S
II
when lJ. is a partition of n.
later be shown to be equal to the dimension of SA. ~
that SlJ. there is
This number will
Since we shall prove
SA when l.I is obtained from A be rearranging the order of the parts, DO
qreat loss in restricting l.I to be a partition, so we assume
throughout this chapter that
with
III ~ l.I
~
2
•••
~ lJ.
h
> O.
Recall from 10.20 that when 1 S r S h,
~*
r
is the set of those subsets
R of (i , 2, ••• , h} which have cardinality r, and for which l.lR is a partition of n - r.
12.1
DEFINITION •
Given R • -:' construct a }.I-tableau '1'R according to the
following recipe: (i)
u,
2,
~
For each b • R, place at the end of row b one of the numbers. in
.. ,
r} 1 do this in such a way that
row (1) T R (ii)
< rowT (2) < ••• < R
Complete T
R
row (r). T R
by inserting the numbers r + 1, r + 2, ••• , n in
order down successive columns in the remaining places.
These conditions determine T
R
EXAMPLE. T
and
~R
If lJ. = (3, 3, 2, 1) and R
II
= 123
'II"R •
4 5 6 7 8 9
T
~
II II
= 158 2 6 9 3 7
uniquely, and
= {l,
2, 4}, then
Tl.l'll"R
=4
7 1 5 8 2 6 9 3
4
(1 4 5 8 9 3) <'2 7 6). - 67 -
~R
,
satisfies 10.8(ii).
12.2
Assume that i > j ,
LEMMA.
i'll'~2-.i!llb..and REa;.
Then one of the
following holds-:
(iii)
< i'lf R
j'll'R
= j'll'~
column of T~, and i'll'~
Since i > j, row
prOof:
In this case, i
r.
S
T
+ 1 if and only if i'll'R ~
(i)
T
~
'II' ~
Since i'll', 1 ~
>
row
T
~
= j1l'R
+ 1.
(j).
~
Now, if j'll'R ~ r, then j
of row b.
and j belong to the same
lies at the end of a row of T , say at the end ll
j'lf , we cODClude that i1l' II
II
lies at the end of row c of
where
< c, ana.
b
= ll,+l = ••• .; ;:
~
lJ.
c
•
Therefore, j'll'R < i'll'R S r, and. the remainder of stateMnt (iii) follows. If r
< j1l'R' then we
case, then the construction of T Therefore, if r
<
< i'll'R < j'll'R; for if this were
do not have r R
would imply that
i1l'~
<
j'lf~,
a contradiction
j'lf , either conclusion (i) or conclusion (ii) holds •
R
COmpare the next result with Corollary 10.15 (cf. also 16.9).
12.3
For 1 fi r S h,
THEOREM.
dim S
Proof:
S
P
where
r
~
'IT
~R
is generated by
L
~ ~ qtG(r)
= {(i,
j)
Xc(g)g,
I i > j and row
and e is the function from
l) -
T
'IT
~ll
(i) > row T
to {I, 2,
... , q}
- 68 -
'If
(j)},
llll such that
the
•
(i, j)c (i, j)c
= 2, = 1,
if i
=j
+ 1 and COlT ~ (i) lA u
= COlT
~
(j)
u u
otherwise,
(see 11.12 (i» • SUppose that R e
L-1
giG(rn-
\.l
£;,
and apply
-1
~lA ~R
to this generator:
(q)q E S • 1J ~)
R
(Here we have used Theorem 9.2 and 5.10(i), note that 1T -1 f~1J ~R to
-1
R
~lAc
is a function from
{l, 2, ••• , q}).
NOw, i~
>
j~
lA
where
r 1_ f
f
2 3
= {(i~R' = {(i~R'
j1T
R
)
i1T
j~R)
i~
{(i'IT j1T ) R R,
i1T
by Lemma 12.2.
u
\.l
1l
> >
j~
j~
lA
\.l
lA
, i1T
and row (i) > row (j)} TlA TlA
R
:S
and row (i) > row (j)} r < j1T R, T T
1l
\.l
, r < J.~R < i1T , and row (i) > row (j)} T R T u lA
> j 1T , j1T
u
R
< i1TR
x , and row
:S
T
(i) > row lA
T
(j)},
lA
But f , f , f are disjoint closed subsets of-t,·so an 2 1 3
application of Theorem 9.2 shows that (12.4)
~
(q) q)
S • \.l
Next, recall that in Section 10 we defined row {~I
T C1 lA
I
(i)
= rowT
J.t
T
(i) for all i with 1
:S
~I '" 1T}, let
i
- 69 -
i
:S
r.
C1 '" '[
Cfor C1, r E (;; )
•
n
For each "'-class, [n-]
if
=
Now, U G is a group (it is the set of matrices described in (10.3», and n-r by Lemma 10.7 (12.5)
For each 11 € &'n' M[1I] is U Gn_r-invariant. Given 0, 't C Sn' we write to.}
row
T
< ['t) if for some i 1 , with
yei) for all i with i
u
1
1 S i
l
S x,
< i S r.
We now aim to prove: (12.6) ['tl:
<
If 1
;t
q c G(r ) , then p ...1I 9 1 R
= p ... 'tu
for some y € ~n' u € U , with
['lr ] .
R
A non-singular matrix m belong'S to p... if and only if m ij
row (i) T lJ
(12.7)
> rowT (j).
Therefore, qiven a,
Y ~
when
&'n'
...
mea
-1
P u r . . (amy
-1
) ij
=0
=0
SUppose that 1
11;~...'t,
Among' all such is maximal.
;t
T
when row a(i) T
T
l!
(j' l!
> rowT y(j). l!
9 E G(r ) and PlJ1I 9 = PlJYU (Y € c;'n' U C U). 1 R
and since 9
u,
(i) > row
when row
l!
gu-1 C
=0
;t
j) choose
1, there exists (i, j) with i (i
l,
j1) such that i
1
;t
j and 9
Then ij
;t
O.
is maximal, arid then j1
Since 9 6 G(r l "
(12.8)
The choice of (iI' jl) ensures that (gu-1'i j for i
>
i •. Bearin9 in mind that gu l
-1 E
row y(i) for i > i T 1" u
- 70 -
0 and (gu-l)ii
=1
-1 1 1 1I P lJY' these results yield, by an
application of (12.7) that
~
;t
R
= rowT~ttR(il).
Since i l S r, rowT tt (b) < c for all ]J R b < i by the definition of 'II'R. Thus, if row 11' (b) ~ s , then b T l, Let
€
u
b
~
row
T
with
2:
i
R
l
and
Conditions (12.9) now imply that
jl (using (12.8». {b
b
T(b) = o}
= {b
row (b) T]J'II'R
~
= 6}
if 0 > €,
and
In particular,
row T(b) T u
and
= rowT
~
tt (b) for i l < b
r.
S
R
Therefore, [T] < ['II'R l , and this completes the proof of (12.6). Since G( r2)'
err 3)
S' U-, when we apply (12.6) to the element of Sj.J
described in (12.4), and use (12.5), we obtain: (12.10)
Pj.J'lTR (
+
l
X -1
g.G(f 2 ) (an
(g)g) (
ttR ttj.JC
~
X -1
g£G(f ) 3
L
element of
M['II']) belongs to S •
[tt]<[wRl But r
3
(g)g)
'lTR tt]Jc ~
is a subset of f(r) (see Definition 9.4), and for (i, j) e f , 3
=1 by Lemma 12.2(iii).
if i
~
j + 1,
Thus, for all 9 E G(r ) , 3
Therefore, multiplying the element in (12.10) on the right by q obtain:
- 71 -
-Ir I
3 E , we r
L
+ (an element of
[11'1< [1TRl
M[ft]) belongs to 8lJEr•
Onder the isomorphisms of Theorems 10.12 and 10.14,
maps to a qenerator of S
IJR
, by condition (ii)
in the definition 12.1 of T
Therefore, multiplying (12.11) on the right by appropriate elements of we can obtain (dim S
~
)
R•
KGn-r
linearly independent elements in S E of the form lJr
(Here, we are using (12.5).
Note that G(r ) 2
~
G
~r
and the elements of G
~r"
commute withE ). r
As a vector space, M ll
= E9 M[1T)'
the sum being over distinct "-classes
(see Leuma 7.3), so we have proved that ~
dim (8 E ) IJ r
•
L
Rel2* r
Theorem 9.11(ii) now gives:
12.12
COROLLARY (cf. Theorem 10.21 and 16.9)
dim S
IJ
~
h
L L
(qn-l _ 1) (qn-2 _ 1) •••
(qn~r+l - 1) dim S
•
lla
r=l Rl!'ft?
In fact, it is possible to extract more information about 8 E
II r
the proof of Theorem 12.3.
To do this, for R, R' c~, define R
That is, the set
1*r
<
• from
R' if
is ordered by "last differences", in
other words, (12.13)
For distinct elements R, R' of
llL*, we write r
R
< R' if the largest
element of R , R' is less than the largest element of R 1
- 72 -
,
R.
"i !
Now suppose that ~ parts. )
=
{R
l
< R < ••• 2
<~}
(thus k
=
(~J
if
~
has distinct
Let s. be the element of S E described in (12.11), taking R Vr J
= Rj
,
and define KG
n-r
Then the part of the proof of Theorem 12.3 after (12.11) proves
12.14
THEOREM (cf. Theorem 16.9)
and for each 8 j6 j
j
with 1 ::; j
= S~R.
::!>
k, the exists a KG _ -homomorphism 6 such that n r j
and Sj_ 1 6 j
= O.
•
]
OUr linearly independent elements of S
).I
(which turn out to be a basis)
use, via the construction of T and induction on n, R
~-tableaux
which tend to
have the property that the numbers decrease along each row and increase down each column.
One might expect to get ).I-tableaux resembling standard ones by
redefining G* 1 to be a subgroup of the stabilizer of the subspace <e >,
n-
instead of <e >. l
n
However, closer examination reveals that this alternative
does not work smoothly, and it is likely that some reversal of the nUIQbers is intrinsic to the result. Since dim S
jJ
(as we shall see) is a polynomial in q, where the sum of
the coefficients equals the number of standard v-tableaux, it is interesting to see combinatorially that a standard jJ-tableau may be associated in a natural way with each power of q in the polynomial for dim 8 Can this association be used in constructing a basis of S? ).I not always.
~
(Thomas [T]). Sometimes, but
For example, dim 8(2,2) ~. q2 + q~, and there is a natural (but
difficult) way of constructing a basis where q2 and q~ elements are associated with the tableaux - 73 -
1 3 2 4
respectively. q3
1 2 3 4 '
Similar methods applied to S(3,3) give, however, qS
q5
q7 + q6 _ q5
q9
basis elements associated, respectively, with the tableaux 1 3 5 246
1 3 4 2 5 6
1 2 5 346
124 356
1 2 3
4 5 6 •
This gives some indication of the difficulties involved in devising a basis for S. l.l
In our 9pinion, Theorem 12.3 is the hardest result in the essay, this
is in marked contrast to the situation for CS , where a lower bound for the n
dimension of a Specht module is much easier to find than an upper bound.
- 74 -
13
THE KERNEL INTERSECTION TBEORBM FOR S (
we
shall prove that S
certain KG -homomorphisms 1/J
n
u
d,
n-m,m )
equals the intersection of the kernels of i' defined on M
(when \.I has h non-zero parts,
\.I
the subscripts d, i are bounded by 1 S d S h - 1,0 SiS \.ld+l - 1). special case where lJ
= (n
- m, m) with n - m
~
The
m will be presented first,
because there is an easy trick which works here, in general, the maps 'd,i involve only adjacent parts of \.I, so at least the full Kernel Intersection Another good reason
Theorem will look more plausible after this section I for dealing first with \.I
=
(n - m, m) is that the result for this case is
sufficient to show in the next section that the parts of an arbitrary \.I can be rearraDg'ed without loss. Assume, throughout this section, that n - m
~
m
> O.
From Corollary
12.12, we know that
(13.1) dim S( n-m,m )
~
dim S( n-m,m-l ) + dim S (n-m-l,m)
+ (q where the second term DIlst be omitted if n
n-1
-
1) dim S( n-m- 1 ,m- 1)'
= 2m.
We shall prove that we have equality here, and that
dim S(n-m,m) - [:] -
~~
J
By induction on n, we may assume that the number on the right-hand side of
(13.1) equals
rp [l!rn -- 3Jn _ ~In -- ~no) + (rnL m- ijJ _ l!arn --lJJ The second term is automatically zero if n
+ ( n-l _ 1)
= 2m.
q
(rn - 21 _ rn - 21) . l! - Jj
~- ~
Rearranging, and applying
Theorem 3.4, this number equals
~ =~ = [:] -
+
r~ ~ k~
+ (qm -
1)
r: 1- ~: fl - ~: ~ -
J. - 75 -
(qm - 1)
~: ~
Therefore, (13.2)
dim S{n-m,m)
we
reg-arc! M{
~
I:] -b~ J
n-m,m
)
as the vector space over K which has a basis consistil
of all. the m-c!1mensional subspaces of V, and c1efine the KG -homomorphism '1, i n from M( n-m,m ) into M( n _"1, i) as in 2.16, that is, for 0 SiS m, '1 , i senc!s
an m-dimensional space to the sum of all the i-dimensional spaces containec! in it.
13. 3
Note that
1/1
1 ,m
is the identity map.
THEOREM m-l S (n-m,m)
="
i=O
Ker 1/I l,i
-
and
dim S(n-m,m) ::
[:1 - ~~ J.
8-1 Furthermore, if we define
...
M
s
n
i ... O
Ker 1/Il,i (0 ~ s
Proof:
COnsider the generator t of S(
1l.17(v).
m+l
) (0 S sSm), describec! in Example'
Let V be the m-dimensional subspace of V spanned by l
The image of V
V 1.
n-s,s
m + 1), ~ .
= 0,
M ~ M
m
s
1
under 1/Il,s is the sum of all the s-dimensional subspaces of
But of these s-dimensional subspaces, precisely one, namely that. spanned
by
occurs in the expression for MctJI1,s
t
t.
Therefore,
S~_S,s)·
By the SUbmodule Theorem, (11.12 (ii) ) ,
- 76 -
;>
= 1, and so
M 1/1
(13.4)
o
~
1,s-
S
(n-s,s)
, f o r 0 s s s m,
At this stage, to avoid repeating the remainder of the proof later, we weaken our hypothesis as follows: Assume that M is a 8ubmodule of M( ) such that n-m,m O
(13.5)
for 0 s s s m,
Redefine M for 0 S s S m + 1 by s s-l
(n
M n O
M s
Ker Ij!l ,~.i .
i=O
We shall deduce from this hypothesis that M O
= M( n-m,m i '
and the
results in the statement of the theorem. The next step is to prove: If 0 S t S s
(13.6) to D
(n-s ,s)
:c:;;
m, then no composition factor of MO/Mt is isomorphic
•
This is certainly true if t D
(n-s ,s)
= 0, so we assume that t > o and that
is not a composition factor of MO/Mt_1·
But MO/{MO n Ker Ij!l,t-l) is
so it does not have D isomorphic to a KG -submodule of M (n-t+l,t-l) , (n-s,s) n Since M = M t t_ 1
as a composition factor, by Lemma 11.13.
D(n-s,s) is not a composition factor of MO/Mt.
n
Ker 1/It-l'
This proves (13.6).
From (13.6), we know that no composition factor of (MOlj!l is isomorphic to D(
n-s,s
)' for 0 :::; s ::;; m.
But
Therefore, M Ij!l ~ SL( )' since s ,s ~ n-s,s D
by
(n-s,s)
definition.
=
S
(n-s,s)
I(S
(n-s,s)
(l
S.J.
)
(n-s,s)'
Thus, - 77 -
,8
)/(M 1j!1 s
,s
)
for 0
by the Submodule Theorem.
s
~
~
m,
Furthermore, M ~ s( i ' by Corollary 11.14(iii) f: m n-m,m
so we have now proved the existence of submodules Mi
(0
s iSm) of
MO such
that
and
M'/M s s+l
S
S
for 0
(n-s,s)
sSm.
~
But m
L [:] = s=O
m
([:] - ~~J)
S
L
s=O
= dim
by (13.2),
dimS (n-s,s )'
m
m =
L
s=O
s~o
dim (M'/M 1) S s s+
M
O
dim(MsIMs+l)
dim M( n-m,m) = [:] •
'$
Therefore, we have equality in all possibl-e places. M
(n-m,m)
M'
s
and
In particular,
= Ms
for 0
dim S(n-s,s)
s sSm,
~~
= '[:] -
J
for 0
~
s S. m.
This concludes the proof of the Theorem.
13.7
COROLLARY
•
(see also Theorem 16.12)
The trivial KGn-module is a submodule of S( the characteristic of K (= P
K,
)
if and only if .',,
,
say) divides
... , Proof:
n-m,m
Clearly, the only trivial
,
m+ 1
KGn -submodule
the sum of all the m-dimensional subspaces of V.
- 78 -
il J.
of M(
)
n-m,m
is spanned by
But this element of M(
n-m,m
)
belongs toKer 1/Il,i if and only if for each i-dimensional subspace V of V l PK divides the number of m-dimensional subspace V
2
of V such that V
2
=-
V
l
That is, by Corollary 3.3(i),
PK divides
rn-if ~ _ ~. m-l
Since
S
(n-m,m)
=
n
i=O
Ker 111
1 ' i ' the result follows.
- 79 -
•
14
x
REORDERING THE PARTS OF
we
can now prove that when
p
is a composition of n obtained by re-
arranq1nq the parts of A, then Mp
~
M A•
trivial, we begin with
a4
To demonstrate that this is not
example to show that
M
u
need not be isomorphic to
M if we drop'the restriction that the characteristic of K does not divide A
q.
EXAMPLE.
= 3,
Let n
2, K
~
q
~
~,
A = (2,1), p
=
(1, 2).
, ~
the KGn-homomorphisms from M(1,2) to M(2,1) such that 60 sends a 2-dimensionaJ subspace of V to the sum of all the l-dimensional subspaces of V, and 6
1
sends a 2-dimensional subspace to the sum of all the I-dimensional subspaces which intersect it in zero.
By considering
orbits, it is clear that the
only non-zero KGn-homomorphisms from M(1,2) into M(2,1) are 60, 61 and 6
0
+ 8
1• Rank 8
0
= 1.'
so 8
The matrix of 8
1
0
is not an isomorphism.
(with respect to the natural bases) is:
(14.1) <e > l
<e > <e > 2
3
<e
l+e2
> <e
l+e 3
> <e
2+e3
>
(
0
0
1
0
1
1
0
1
0
1
0
1
<e , e >
1
0
0
1
1
0
<e + e 2, e 3> l <e + e , e >
1
1
0
0
1
1
1
0
1
1
0
1
<e + e , e > l 2 3 <e + e , e + e > 2 3 l 2
0
1
1
1
1
0
1
1
1
0
0
0
<e
l, 2
1
3
e > 3 3
2
I
Calculation reveals that the rank of this matrix (over F ) is 3, so rank 2
e1
= 3,
and hence rank (eO + e 1)
s 4.
an isomorphism, and M(1,2) , M(2,1)' Incidentally, suppose we were to define S(2,1)' in this case where the characteristic
of K divides q, to be the subset of M(2,1) - 80 -
consisting
of those vectors for which the sum of the coefficients equals zero (cf. Theorem 13.3). dim s(2,1)
Then 1m 8 1
= 6,
t
S(2,1) and Im 8 1
dim st2,1)
4 st2,1)'
= 3,
since dim DB 8 1
Thus, the SUbmodule Theorem fails when the
1.
•
characteristic of K divides q.
We now adopt once more the hypothesis that the characteristic of K does not divide q, and continue to regard
M(
n-s,s
K spanned by the s-dimensional subspaces of V.
that M( n-m,m )
14.2
aE
LEMMA.
OUr first aim is to prove
m,n-m i ' and a preliminary Lemma is needed:
Me
Assume that s
dimensional subspace of V.
:s;
min (m, n - m), and let V
described in Example 11.17 (v)
(n-s,s)
s-dimensional subspaces of <e
l,
e
2s
Thus,
,
..• , e
2,
precisely those where each £ij
.•. , e
=1
have
>.
ze~o
~
is a linear combination of Of the terms appearing in intersection with <e l
=
Proof:
2s
(_l)s qS(S-1)/2 ~
THEOREM.
=
Let , be the generator of
>' and hence have zero intersection with V • s and V2 " Vl
14.3
be an (n - m)-
Then
••• , e 2 s' e 2 s +1' e 2 s+ 2' ... , e n-m+s >.
"
l
By applying an appropriate element of G , we may assume that V n l
Proof:
s
as the vector space over
)
e ,
4
2,
Therefore,
= O}>( n-s,s )
•
o.
M and M -(n-m,m) (m,n-m) __ar~e-....;;;i;.;:s;.;:o..;;;mo;;.;;..;;r;"jjp~h;;:i;.;c;...;;~~Gn-modules. -
we may assume that m Let 8 be the
:s;
n - m.
KGn -homomorphism
each (n - m) -dimensional subspace of spaces which intersect it in zero.
from M( V
m,n-m
) to M(
n-m,m
) which maps
to the sum of the m-dimensional sub-
We shall prove that
a
is an isomorphism.
Assume that V is an (n - m)-dimensiona1 subspace of V. 1
- 81 -
For 0
:s;
sSm
let 1IJ ,s be the KGn -homomorphism from M( n-m,m ) to M( n-s,s ) defined in 2.16. l
we
claim that
Let V2 be an s-dimensional subspace of V. of m-dimensional subspaces V
3
q(n-m) (m-s) by Theorem 3.1.
= 0,
If V2 n V1
of V such that V _ V 3
and V
2
On the other hand, if V
1'\
2
then the number 3
"
V 1
=0
is
,~
V ;e 0, there is no l
i,
This proves (14.4). Therefore, applying the
j
SulmM>dulel
Theorem, we deduce that "
M(m,n-m) OWl,s ~ S(n-s,s)
(14.5)
for 0 S s s; m.
1
Now, this shows that if we let M = M( m,n,..m ) a, then O
13.5, from which we concluded that MO dim Me
n-m,m
)' so
a
l
O
satisfies Hypothesis
But dim M(m,n-m) =
•
is an isomorphism, as we wished to show.
We have proved that the map subspace V
= M(n-m,m).
M
e which
of V to the element of M(
n-m,m
sends each (n - m)-dimensional ) consisting the sum of the
m-dimensional subspaces which intersect V1 in zero is an isomorphism when This shows that the elementary divisors of the matrix 14.1 (Viewed
2m s; n,
as an integer-valued matrix) are all powers of 2; they are, in fact, 1, 1, 1, 2, 2, 2, 4.
It seems remarkable that every matrix constructed in a'
similar way must have elementary divisors which are powers of the prime whic divides q. Since
a
is an isomorphism,
For each m-dimensional subspace V* of V (with 2m s n), there is a
(14.6)
K linear combination of (n - m) -dimensional subspaces of V, say ~e
~,
such that:'
= V*.
14.7
THEOREM.
If ~ is a composition of n obtained frOID A by permuting its'l ,
- 82 -
parts, then M and A Proof:
M~
are isomorphic KG -modules. n
It is sufficient to prove the theorem in the case where
~
is obtained
by interchanging two adjacent parts of A, so we assume that
The elements of M may be regarded as A
Define
a
to be the
of all the
~-fla9s
KGn -homomorphism
it
linear combinations of A-flaqs
which s~s the above A-flag to the sum
which have the form
with
In view of (14.6), for each
~-flag
like
there is a K linear combination of A-flaqs of the form
which is sent by
e
to this particular l1-flaq.
since dim M • dim M~ I we have proved that M A A It now looks very likely that SA knew that SAe
4 s~,
or if we knew that
~ S~
Thus,
e
maps M onto A
M~,
and
•
II M~.
(and hence also D !lil D~); if we A
the result would come out using the SUbmodule Theorem, D~
deduce what we want.
is not a composition factor of MA/S
A,
we could again
However, neither of these results is straightforward
yet, so. we postpone the proof that SA .$ S11 to Section 16, when it is easy.
- 83 -
15
THE KERNEL INTERSECTION '1'BEOREM
We have seen already that SA is contained in the intersection of the kernels of all
KG
-homomorphisms which map M, into so_ M with 11
n
~
A
I>
A.
In
this section we shall prove equality, and thereby brinq toqether several results from earlier chapters.
Actually, we shall prove that SA equals the
intersection of the kernels of several carefully selected KG -homomorphisms. n
DEFINrrION.
15.1
and 0
s i :s; Ad.
so that
=
(A A , ••• , A and assume that d l, 2 h),
if j
for PAn PII in P A•
~
d, d + 1.
Choose riqht coset representatives ql' q2'
Let W . be the KG -homomorphism from M, into M defined d ,1. n A II
by P,~
Wd,i= It
A
. . . pII (I:j
q.)~ J
ce
E
KGn i.
is easy to describe Wd,i in terms of A-flaqs1 it sends the flaq
to the sum of all the flags
which have the property that
In particular, the notation here is consistent with Definition 2.16. proved in Theorem 13.3 that when A := (A l,
71.
2
) WithAl
~
We
A we have 2,
A 2-1 SA:=
"
Define
= Aj
~j
SUppose that A
i=O
Ker Wl
'
s:
The general Kernel Intersection Theorem deals with arbitrary compositions - 84 -
=
but when A
(A
l,
A ••• , A is a partition, we shall have 2, h)
h
S
A
fl
...
d=2
Besides proving that SA is a kernel intersection, we shall also describe generators for many other kernel intersections which are contained in M (not, we may add, through any desire for generality, but because we A have found no other route to the result for SA.) own work on the symmetric groups (cf. James
[J
4
We are guided here by our
]) .
For a start, we define "a pair of compositions Ai, A for nil and a correspondinq
15.2 A* ...
KG
A of n-submodule S,# A ,
DEFINITION.
(A~' A~'
A~
a
M
A such that SAi , A 2 SA·
Let A ... (A A ••• , l, 2,
~)
be a composition of n and let
••• , A:> be a partition of some n' ~ n, such that
S Ai for all i
<
A~ < Ai for at most one i.
we shall call ~#, A a pair of compositions for n.
If Ai ~ A, we let d'" d(A
be the least non-negative integer such that A: < AdIn examples, we shall denote a pair of compositions for n by drawing the diagram for A with lines between the rows, then boxing in the part corresponding to Ai. If A ~ (3, 2, 3, 2) and A#
EXAMPLE.
=
(3, 2, 1, 0), then we draw:
[§i: x
X
XX XXX XX
If T is a A-tableau, then T # denotes the part of T which is boxed in during this process.
Thus, in the example above, then T#
3 B 4 6 7 2
- 85 -
15.3
DEFINITION.
Given a pair of compositions A", A for n (with d • d(A *»
r.ll. = {(i, AW,A
j)
vr«.
j)
let
row
Tx'11').
= rowT
(i)
.
A~ < COlT W
vtu,
X'I1'A
= d,
row (i) TAwA
j)
co1-
A'I1').
j
(j).
(j) <
a,
co~
~ (TX~A)
r
*
A ,X
(i)}
and
I
"
The definition of T X'I1'X ensures that
AWA
*
(j) S Ad < colT
' r X'I1'X
and
X'I1'X
.
(i) }.
~.-.
Let us explain why we take this complicated definition of r #
-
X
*
first set is natural enouqh; it equals. when X = A.
.x
•
The
In the proof of
(15. e) below, we need
to be a subset of
have to take
r
r"
* ' and once we are committed including this set, we
A ,X
in the form qiven, in order to ensure that
A ,X
r t A ,X
closed subset of .-.
EXAMPLE.
and
If A a (3, 2, 4) and A* • (3, 2, 1) then
r.ll. = {(i, X'II',).
j)
I
i , j
e
{l 2 3 4 5 7} and i
>
j}
V {(e 6), (9 6), (9 e)} V{(i, j)
15.4
Ii
e{6 8 9}, j C {l2 3}}.
DEFINITION (cf. 9.1 and 11.4).
Let c be the function from r.ll. to {l, 2, ••• , q} defined by X'II',X - 86 -
is a
(i, j)c .. 2,
otherwise.
(i, j)c '" 1,
Let
15.5
(KG n i ,
DEFINITION•
,
If A '" (0, 0, ••• , 0), then r
is empty, E
~
A",A
~
A'II',A
= 1,
and so S
If A' '" A, then A is a partition, and the definition of r
ensures A',A
that S, '" SA" A ,A By Theorem 10.2, PAWAE. A ,A
r '" t u ,
v
is a non-zero multiple of PAW A i, j
j)
~
•
(T, w,) , i > j, and row (i) > row (j)} TAWA TAWA ~ ~
colT j)
xc (q)q, where
,
s,
[
j
AWA
~ (TAWA) " rowT
(j)
AWA
~
A: < colT
, '"
' f ••• ,
(~1' ~2'
#
~h)'
by
- 87 -
AWA
(il}
(i) > rowT (j), and AWl
Now assume that d '" deAf) '" 1, and consider the above set. ~
.
rowT (i) '" d > rowT (j), j £ (TAWA) , and AWA AWA
{(i, j)
= {(i ,
r
gfG(r)
~
A'II',A
If we define
c
= 0,
then vi
d'
= d(V*),
where d'
since 0 < A* < A for at most one i. i
i
Hence
the set r above is the same as the set we get when A*,A is replaced by V*,A.
*
=S
Thus S A···,A (15.6)
v* , A
If d(A*)
when d(A
= .~,
*
=
~
1.
ThiE;l proves
=S
then S * A ,A
I
' where y* is obtained from A* by
v ,A
absorbing the firs'C part of A Therefore, when we are investigating S * ' we may assume that d(A*) > A ,A 1, in particular, there is little loss in writing this hypothesis into the next two theorems.
The conclusion of each of these theorems concerns an
inclusion between two KG n-modules, afterwards we prove, by very indirect means, that we have equality.
15.7 A*
~
THEOREM.
= d(A*)
A and d lld-l
*
Assume that A , A is .a pair of compositions for n, with
=
> 1.
Let
H = (¥1~2'
••• ) be defined by
Ad_i + Ad - A:
"tAd = A* d ~i
-Then
=
*
Ai
~
for i
s A ,A ~ d-l,A # d
a
d - 1, d.
*
s A,ll I
Proof:
First of all, note that A*, II is a pair of compositions for n; it is obtained
from AI, A by removing the nodes of row d which are not boxed in, and placing them back at the end of row (d - 1).
Also, MAljld-l,b # for notational convenience we write b = Ad"
~
M by definition, where ll,
I 1
I I
'I ~
.1
'1
where
- 88 -
j
The right coset representatives of ~d-l,b'
PA~P~
in P
A,
used in the definition of
may therefore be taken in the form
where
and for all i and k, iLk and
=i
~ € <X.. J.]
if row
T
( L)
;t:
d,
A
i > j and row
T
( i) A
= row T
(j) A
= d>.
Then
But d> i > j and row
T A'ITA
(i) = by the definition of 'ITA.
So
Now, by Theorem 10.2,
{(i, j)
i
- 89 -
#
{(i, j)
s tu., <=>
i, j
e
*A,
T
row (i) T A
= rowT
(j)
+ 1, and
CO~ (i)
A
= co~
A
(j
A
j)
row
{(i, j)
TA
(i)
= d,
row (j) TA
=d
- 1, and colT (i) A
= col
(j) S
TA
since T and TlJ agree on all rows except rows (d - 1) and d. A But Tk permutes only the numbers in row d of TA, so PlJTkWAEA*'A is non-zero if and only if T k
~ P A"
step 2
P lJ' and
k .. 1.
't
k
stabilizes row d of TlJ. We have now proved that
Let T be the lJ-tab1eau obtained from TAW
row d which lie outside (TATI
e.g.
while
A)
A
* to the end of row
then T
T W lJ lJ
However, in that case,
by moving those numbers in (d - 1).
10
10 •
we describe r
as a set of pairs of numbers in T: A*,A
r.
={(i,j)
A'II',A U{(i, j)
U{
u,
*
j)
E (TA'II'A) , and cO~(j)
Let
0
€ ~
n
be defined by To
= Tu TI. lJ - 90 -
~ A: s Ad_1 < co1 T (! ) }. Then
r
* a. tu ,
j)
s.,
tu ,
j)
rowT
A ,A
v
11 11
1r (1)
II II
*
A d_ l vt
u,
I
j)
>
j E (T 'If ) i and i
rowT'If (1)
II 11
= rowT < co~
1r
j}
(j)
=d
(j)
< colT
- 1, and
II II 1r
11 11
=d
comparing this with the definition of
1, j
-
(i)}
'If
11 11 €(T 1r ) ' , and
11 11
r. '
we see that
A ,11
(In general, the third subset of third set in the definition of
r
r.
a may be strictly contained in the A ,A
* .)
A ,11
Therefore,' by Theorem 9.2,
Then, applyinq the result of Step 1, we have:
Finally, it is easy to see that T permutation, so
P\.I 21\ a
1r
11 11
differs from T.
~
1r'\ a
only by a row
1\
P111rII •
Thus, we have constructed an element of S • A ,A whose image under ~d-l,b is a generator of S f ' and the theorem follows. A ,lI • =
1\
It is worth noting that the symmetric group result which correspon4s to this one (see James [J ] , Theorem 17.13) has equality in the conclusion. 6
Fortunately, the inclusion in Theorem 15.7 is the way round which enables us later to prove that it is an equality.
15.9
THEOREM.
Assume that A', A is a pair of compositions for n, with
A*;II: A and d · dO.*> defined
>
1, and that
A:~:_l.
bY - 91 -
~
Let ll*
= ()J~W~'
... ) be
Then
S It
A ,A
1"1
Ker
ljI
d-l,A
* Sp,A * ;2
d
First of all, note that p*, A 1.s a pair of composit1.ons for n, s1.nce
proof:
A: < A:_ l. Now,
r *
r *
~
A,A
' by CONJtruct1.on, so
Jl,A for some E;
and this proves that S *
~ S
A,A 1/1 d-l,b
maps
A into
M
n
* 1.I,A (b
Mp
E KG ,
*
= Ad)'
where
Jl
is the. composition def1.ned
in Theorem 15.7, and as 1.n Step 1 of the proof of that theorem, SJl*,A 1/Id-l,b 1.s non-zero only 1.f, for some k,
But for thi.s we need {(i, j)
I
row
T
(i)
= d,
A
row (j) T
=d
- 1, and
A
and this is impossible (the first set has cardinality
.
has cardinali.ty Ad).
Therefore, S ~
Jl'll',>..
1/Id-l,b ~
The last two theorems are illustrated by: EXAMPLE.
If
A*' A =~
,
~ X X X X
X
- 92 -
o.
A: + 1 and the
second'
•
o.nto a module containing' S , where then Wi 1 maps S # , A ,A A u
*,
=~ X X X X X
and
s
*
() Ker Wl,l contains S *
A ,A
*,
lJ
' where
lJ ,A
A =~
~
•
X X X X
X
By (15.6), s
*
=s
A ,lJ
v*, lJ =~X
# ' where v ,\.1
X X
xl.
X X X X X
*,
Now, transforming A
A to A*, u always moves some nodes up, and when
they reach row 1 they can be boxed in. to
\.I
# , A boxes in an extra node.
*
On the other hand, transforming A , A
Repeated applications of such manoeuvres
to ).*, A always lead to a situat1.on where all the nodes are boxed 1.n.
To formalise this process, we give:
15.10 A*
DEFINITION.
;t)..
*
SUppose that A , A is a pa1.r of composit1.ons for n, with
We define two new pairs of compositions (A*, A)I and (A*, A)K for n
as follows (I stands for "Dnage", K for "Kernel"):
for i > 1 (cf. 15.6). If deAf) > 1, let (A*, A)I .(ii)
= A*, lJ~ as defined in Theorem 15.7.
*
*
*
We define (A , A)K only 1.f d(A ) > 1 and Ad# < Ad-I.
case, let (A*, A)K
= \.1*,
A, as defined in Theorem 15.9.
- 93 -
In this
BXAMPLE
-x x
xxx
The only pairs of compositions to which we can apply neither I nor K
are those where A' = A (and so A is a part1.tion of nl ,
For every other pair
of composit1.ons, we can apply the maps I and K successively to reach a pair of compositions lli, ll, where ll* the paJ.r Ai
=
(0, 0), A
= (2,
DBFINITION.
In order to describe the partitions we
1DU1t1.p1icit~es),
reach (together with
The example above shows that from
3) we can get to p, ll, where pisoneof the
partit1.ons (5), (4,1), (3,2).
15.11
= p.
we g1.ve:
If A is a composition of n and 11 is a partition of n,
then a semistandard J.I-tab1eau of t)'R! A is an array of integers obtained by replacing the nodes in the diagram (11) by integers (allow1.ng repeats) such that
EXAMPLE
(i)
The numbers are strictly increasing down each column.
(i1.)
The numbers are non-decreasing along each row.
(1.i1.)
For each i
~
1, the number of entries equal to 1. is >"i.
1 1 2 2 3 2 2 3 3 3
is a semistandard (5, 3, 2)-tableau, of type (2,4,4).
*,
15.12
LEMMA.
Let A
A be a pa1.r of composit1.ons for n, and )J be a partition
of n.
Then the number of d1.stinct sequences
- 94 -
(each J
k
=I
or K)
such that
equals the number of semistandard ).I.-tableau T of type A which have the property that for each i, every number in row i of T# is 1. The result is certainly true when A*
Proof:
= A,
so assume that A*
(A~
that the result is true for (A#, A)I and also for The lemma is then trivial if d(A#) J
l
J
2
= 11
;II!
A and
A)K, if it is defined.
in this case, all sequences
••• start with I, and a semistandard tableau of type A has A 1 1 s in l
row 1.
Assume, therefore, that d
= deA') >
1.
The statement of the lemma may be rephrased by saying that the ).I., ).I.
which arise by applying sequences of I's and Kls to A*, A are the partitions
#
we get by first replacing the nodes in row i of A by i, for all i, then adding
(~'j
-
A~)
numbers j, for all j, to create a semistandard tableau.
In this process, all the (Ad - A:) numbers d which are to be added must 90 in row d or higher, by the definition of "semistandard".
If we put one of
these d' s in row d we end up with those semistandard tableaux which arise from (At, A)K.
If we do not put one of these d's in row d, then all of them
must 90 in row (d - 1) or higher.
Since all the (d - l)'s are already in
*
row (d - 1) (because A = A we get the same shaped tableaux by d_ l), d_ l renaming the d's, which are to be added, as d - 1.
#
tableaux which arise from (A , A) I. EXAMPLE.
*,
If A
A
Hence the result.
= ~ xr , X X X X
Thus, we get the
•
/
the various tableaux described in the Lemma, together with the sequences leading to them are:
- 9.5 - '
Bfl'2 3
3
mx
from IS from I~K2
.2 3 3
2 3 ~ 3
from 1 2KIK
fuH!I thEE
from KIKI 2
2 2 ~
from KI 3
fhPTI= futil
from K2I 3
from I 2KI 3
2 2 3
223
from K2IK
from K3 •
2 2 3 3
223 3
15.13
2 3 ~ ~ 3
from I 3KI 2
DEFINITION.
If >. is a couposition of n and p is a partition of n,
then K is defined to be the number of semistandard pA
}.I-tableaux of type A
(K is a Kostka number). pA For example, if we write 0 for the unique composition of zero, then Lemma 15.12 shows that the number of distinct sequences J 1 J 2 ••• (each J
k
=I
15.14
or K) such that
LEMMA.
(cf. Theorems 13.3 and 15.16)
KGn - submodu l es
There is a series of
of M A,
such that each M1/Ms+ 1 is isomo;phic to some Sp , where -
==-.o.....;;,;;,;;,~;",;o,;----s
n.
The number of factors MI/M
Proof:
s
s+
}.I
is a -partition of
1 which are isomorphic to S -
equals K ,.
p -
}.I1\
This follows from repeated applications of Theorems 15. 7 and 15.9,
and the remark preceding the statement of the lemma.
- 96 -
•
The following well-known (and rather startling) fact now enables us to tie everything toqether: (15.15)
Let Xl' x
for i < O.
2'
••• be commuting indeterminates, and define
For each partition
~
=
(Pi' P , ••• ,
2
(the determinant of an h x h matrix).
~h)
X
o
of n, define
Then for each composition A of n,
=IK),.Y. P
EXAMPLE.
If A
P
~
= (2, 3), then the semi standard tableaux of type A are
1 1 2 2 2
1 1 2 2
112 •
2
2 2
Hence
=
Ixsl + IX 4 xSI + Ix) x 4
Ixo Xli
15.16
IX I
x2
THEOREM
(i)
If P is a partition of n, thenCcf. Definition 10.17)
(ii)
If A is a composition of n, then there is a series of
KG -submodules of M"1\ --on
/
such that for each s (0
;:£
s ::iii t), Ms/M + is isomor,phic s l
to
some
S~
where
}J.
.;;;;i..;;,s-..;;;a;.. .a.p,;;;;a;;;.r.;:;t;;;.it=io,;;.;n;;,;;,.....o;;..f;;.....:n~. --..:Th::.;;;;.:e::;",. -;n;;,;I='m=b;.;e;.:r;;.. ;:o:..:;f;;.. ;:f;.:a;.:c;.:t:..:;o,;;;;r,;;;;s;.. .;;,M·s /Ms +1 which are isomorphic to
- 97 -
Proof.
Note that the claimed·dimension of S
in Definition
\.I
is the number dim (u) , given
We assume, inductively, that dim S
10~19.
.. dim (v) when v
v
is a composition of n l < n(this is certainly true when. n' .. 0). By Lemma 15.14, dim M, ~ /\
LK\.1/\,
lJ
dim S , II
the sum being over all partitions II of n.
Then,
applying Corollary 12.12,
Theorem 10.21, and our induction hyPOthesis, we get:
= {i} -1
But, if we put Xi
in (15.15), and mUltiply the las". m,{Uation
therein by {n}, we obtain
.. l
lJ
A dim
K
II
un ,
Therefore all the inequalities earlier in the proof are equalities and the
•
theorem follows.
15.17
COROLLARY.
Mopting the notation of Theorems 15.7 and 15.9, we have:
(i)
~
S
A#,A d-1,A
.d = SAIt
and this is an isomorphism if A# d
(ii)
Proof:
s
# {\ Ker A ,A
~
d-l,A
# d
, ,ll
#
= Ad_I·
..
J j
S .. II IA
if
A# < # Ad-I· d
"-/
.r .1
First note that any zero parts in a composition of n may be ignored. It is easy to see that there exists a composition v and a sequence
J
l
J
2
••. (each Ji .. I or K) such that (0, v)J J ••• 1 2
Corollary now follows by applying Theorem 15.16(i1)
- 98 -
=
to M •
v
.
(A , A).
The
•
*, A =
EXAMPLE
If A
"':':'''':::'''''':':''I'"-L- ,
\I
then we take
= XXXX ,
XXX X
xx
X X
we
deduce that M has a subquotient iscmorphic \I
Another immediate corollary of Theorem 15.16 (since S
•
* l1 ,IJ
= Sl1
when
l1) is:
15.18
'1'BB KERNEL INTERSPX:TION TBEORBM FOR PARTITIONS
If IJ = (¥lL-¥2' ••• , h
S
u
=
n
~)
is a partition of n, then
lJd- l
f\
d=2 i=O
Ker
ljI
e-i
i·
'
combininq this with corollary 11.14(iii), we have certainly Rroved the· next result in the case where A is a partition.
A little thouqht reveals that
it has been proved for arbitrary compositions A, but at the beqinn1nq of the next section we shall qiven an alternative proof.
15. 19
THE KERNEL IN"I'ERSEX::TION THEOREM
If A is any composition of n, then
the intersectionbei.ng over all
KGn ..homomo!'Phisms
with l1 r> A.
- 99 -
which map M
A
into
SOlIe
MlJ
16
CONSEQUENCES OF TIlE KERNEL IHTERSECl'ION TBEOREM
Several important consequences follow quickly from the results proved in the last chapter. Recall first that if 11 is the partition of n obtained from A by rearranging the parts in non-increasing order, then M and M are isomorphic A l1 KGn-modules; this was proved in Chapter 14.
16.1
THEOREM.
Let A be a Composition of n, and let \.I be the partition of
n obtained by rearranging the parts of A. M
to M'I.' then S
-;1-/\
e
= S'I.. \.I-II.
If
e
is any
KGn -isomorphiSill
from
In particular, SA and Sp are isomorphic KGn-modules,
and D'\ and Dare i801DOrphic KG -modules.
n----
11
-1\
At the same tilDe as proving this theorem, we shall show
16.2
THEOREM.
O"f MA/S with"
A, I>
Proof: and SA
Let A be a colllpOsition of n.
and every composition factor of SA..D-.§~, is isomorphic to some 0"
A.
Theorem 16.2 is certainly true when A J..
SA are both zero.
I'l
Then every composition factor
= (n),
for in that case
~/SA
We may therefore assume that Theorem. 16.2 is correc
for all compositions w such that w ~ A. We prove Theorem 16.2 first for the partition p which is obtained by rearranging the parts of A. All the homomorphisms involved in the Kernel Intersection Theorem for the partition p
into some M with w
map M
l!
II)
I>
11, and, by induction, every
(being either D
the form D" with v
f>
Therefore, every composition factor of M /5
isomorphic to some
D
w
"
w.
II)
or a factor of M
Is (.II or
composition factor of M
W
\I
t> 11.
W
11
S·) has W
is 11 is isomorphic to the dual of M Is \.I
with " ~ 11, But s.J. "11
and since every D" is self-dual, every composition factor of 0" with
S
5;
1.1
l!
,
has the form
This completes the proof of Theorem 16.2 for the partition 11. - 100 -
Now let factor of M
'1.1
a
be a KGn-isomorphism from M to M". Since no composition ll a is isomorphic to 0" by what we have just proved, while
a/s u
1\
S~)/s~ is isomorphic to 0)., it follows that sl.la
4
s~.
Therefore,
On the other hand, no composition factor of S; is isomorphic to either.
Therefore, S).
4 s~e,
so sA
;2
D ,
A
Slte, by the Submodule Theorem applied
to M e. 11
We have now proved that sl.la
= SA'
Thus, every composition factor of
.J..
MA/s)., and hence also of S)., must be isomorphic to D'V for some 'V e- L
Finally, since DA is the unique top composition factor of SA' it must
•
be KG -isomorphic to D • n It
Theorem 16.1 shows immediately that the Kernel Intersection Theorem 15.19 for an arbitrary composition A follows from the special case where A is a partition.
16.3 factor
COROLLARY.
Let A be a composition of n.
of SA (and of M with composition multiplicity one, and all other A)
cOmpc?s1t10n factors have the form D'V, where 'V Proof:
Then D is a composition A
I>
A.
Consider the following picture, and apply Theorem 16.2:
• - 101 -
16.4
COROLLARY •
Every irreducible unipotent representation of KG
n
..!!.
isomorphic to some DJ.I' where J.I is a partition of n. Proofz
By definition, the irreducible unipotent modules are composition on B+•
factors of the permutation representation of KG
n
representation is M(l ,1 ,
... ,1)' so
But this permutation ;
the Corollary follows from Theorem 16.1
and Corollary 16.3.
16.5
16.6 S).,Q.
~A
COROLLARY.
Prooft
• is independent of the field K.
Combine Theorems l5.l6(i) and 16.1.
THEOREM.
If the prime p does not eli vide
•
<see Section 11) , .when restricted to p-regular classes of G equals n,
the p-modular character of S). F. , P Proof:
We may regard P). as an element of ZG (ll- the rinq of inteqers) and n
each homomorphism 1/J J.I I> )..
d,
i' defined in 15.1, sends P,ZG 1\
n
into some
PJ.IZGn
with
The intersection of the kernels of several such maps may be viewed
as a sinqle kernel, by combininq the maps into one ZG -homomorphism n
The Kernel Intersection Theorem shows that s).,elequals ~.* Ker 1/1 Z for some such map 1/J z. pP,ZG 1\
n
But
-+p ~ PZG, J.II>)'
\.l
n
f.:
and hence we obtain a map 1P P, such that E
1/1 P:
(PlI.Gn>/(PPAI.Gn) .... ($ PJ.lZGn>/(P lJ~).
- 102 -
E9 lJ~).
PJ.lZGn>.
If a € ICer 1fJ I, then
a E Xer
1JJ
Fp
Therefore,
However, by Corollary 16.5 F. dim (Ker 1fJ p)
7L
Thus, Ker '"
= leer
1/J
= dim
F. (Q~ Ker 1fJ p)
= dim
(Ker
fp
1fJ~ •
•
stated result follows.
•
combined with Theorem 16.2, the last theorem proves:
16.7
For primes p which do not divide q, part of the p-modular
THEOREM.
decomposition matrix of G has rows and columns indexed by partitions A of n, n and has the form
(n)
(In)
(n-l,l)
1
(n)
(n-l,l)
0
1
. (In)
•
*
1
The entq in row A and column 11 is zero unless lJ
~
A.
Because we now know that, when 11 is a partition of n, dim Sl1
= dim
(it)
as in Definition 10.19, it follows from Theorem l2.} and Corollary 12.12 that (16.8)
=
dim (5 E ) J1 r h
=
I
r-=l
for 1 :S r '5 h
I
(qn-l _ 1) (qn-2 _ 1)
•.• (qn-r+l _ 1) dim S
•
IJ
R
RE~
From Theorem 12.14, we get
16.9
THEOREM.
Suppose that
}.l
is a partition of n.
- 103 -
Then there is a series
of KG -submodules of S E , -n-r 1.I.r
, .»
suCh that for each j with lei!> j S k, Sj/Sj_l is KGn_r-isomorphic for S l.IJ!tj
(Bere R , ••• , R are the elements of k l
~,
ordered as in 12.l3d
In particular, by Theorem 9.11,
16.10
THE BRANCHING 'l'BE0REM.
If 11 .. (1.I LJ1 2 ' ••• , 1
lb)
is a partition of n, then as KG~_l-modUles, ,
I
h
S
11
= S1.11 E
$
t L
r=2
S E G* l' 1.Irn-
and S E has a series of KG
-1.1 1
Q
n-
1 .,,
l-submodules,
1
.'j .~
j )
,
where
I I
s. J
and
if
=
1
S./5 is KGn-isomorphic to J j-1 ~~=~;,;;,;;;.-.;:;.;;;.
.l
.~
~
j
if
Compare the Branching Theorem with Corollary 10.16.
"
'j
Again, the ana1ogy~
with the symmetric group is clear1 the term
1 ·1 .
h
I
r=2
S E G* 1.1 r n-l
>
rj'
has dimension divisible by q - 1, so it "disappears when we put q the symmetric group, the restriction of M to II
form
- 104 -
~
n-
= 1".
1 is a direct sum
,
"
For
of the
'j
but the best we can do when restricting a Specht module to g 1 (except for nsome special fields, such as those of characteristic zero) is to construct a series of submodules like those in the description of SpE appearing in the l Branching Theorem. By considering the effect of E
l
on SJ.I' we obtain from the Branching
Theorem:
16.11
Let Xx denote the p-modular character of
THEOREM.
not divide
q) °
S)..,~
(where p does
suppose that the integers all have the property that
(swmned over the partitions }1 of n) is a p-modular character of G • n E a (E X ) is zero or a p-modular character of G 1 • J.I u Reali J.I R nEXAMPLE.
b alJXlJ Then
XC1,1,1) - X(2,1) is not a p-modular character of G3 , if p is coprime
to q, since X(l,l) - X(2) - X(l,l)
= -X(l,l)
is not a p-modu1ar character of
The Kernel Intersection Theorem is a very powerful tool when investigating the composition factors of S , for it gives a straightforward II
test to determine whether or not a particular element of
M)..
belongs to SJ.l0
We illustrate this by:
16.12
THEOREM
(see also Corollary 13.7 and Theorem 19.6) •
Suppose that II
KG
-n
=
(Pl~2'
••• , J.I h ) is a partition of n.
The trivial
-module is a submodule of S
if and only if the characteristic of K J . I - - - - -....- - - - - - - - - - - - - -
divides fd + Pd+1 -
L
J.I d +1 - i
~
J
for each d, i with 1 s d Proof:
Regard M
u
~
h - 1, 0 S i s J.I
d
+1 ...::......!..
as the vector space over K spanned by Il-flags. - lOS -
Then the
only trivial KG -submodule of M is that spanned by the sum of all the lI-flags n \.I If 1
~
d
~
h - 1 and 0 S i
s ""d+1 - 1, then the sum of all the \.I-flags belongs~"
to Ker ~d,i if and only if for each (lI l , ~2' ••• , lI d - l , lid + lI d +l - i, i,
lId +2, ••• ,
h ) - f 1ag ,
l.I
the characteristic of K divides the number of lI-flags of the form
such that Vd
But the number of such flags is
::::> U.
fd dl - ~ L J + lI +
l.I d +l - i
by Corollary 3.3(i) •
The result is now given by the Kernel Intersection
•
Theorem 15.18.
Using the Kernel Intersection Theorem, we have also been able to prove the following theorem on hook partitions:
16.13 \.I
=
THEOREM.
(n - m, 1
m).
Suppose that 1 ~ m ~ n - 1, and let ~
= (n
- m + 1, lm-l),
If the characteristic of K divides en], then there exists
a non-zero KGn:-homomorphism from S~ into SlI. ~1.I
is isomorphic to
1 i
Hence some composition factor
D;l..
•
For the symmetric group, there is a theorem of Peel [p] which states that, when the characteristic of K divides n, there is a non-zero KGbhomomorphism from the Specht module for It can be shown that for n
:S
;l.
into that for lI.
4, each SA is irreducible
characteristic of K divides [1][2] ••• en].
unle~s
the
Using Theorems 16.11, 12, 13, am
further calculations, we have found that the part of the decomposition matrix of Gn corresponding to unipotent representations is: - 106 -
(2)
(1, 1)
1
0
1
1
]
if char K divides [2]
( 3)
(2, 1)
(1, 1, 1)
1
0
0
0
1
0
1
0
1
( 3)
(2, 1)
(1; 1, 1)
1
0
0
1
1
0
0
1
1
=1
+
q
if char K divides [2]
=1
+ q
if char K divides [3J
=1
+ q + q2
(4)
(3, 1)
(2, 2)
(2, 1 2)
(1 4)
1
0
0
0
0
1
1
0
0
0
0
1
1
0
0
if char K divides
1
1
1
1
0
[2]
1
0
1
1
1
(4)
(3, 1)
(2, 2)
(2, 1 2)
(11+)
1
0
0
0
0
0
1
0
0
0
1
0
1
0
0
0
0
0
1
0
0
0
1
0
1
= 1 + q.
if char K divides [3]
- 107 -
=1
+ q + q2
(4)
(3, 1)
"(2, 2)
(2', l2)
(11J)
(4)
1
0
0
0
0
(3, 1)
1
1
0
0
0
(2,. 2)
0
0
1
0
Q
(2, 1 2 )
0
1
0
1
0
(lit)
0
0
0
1
1
if Char K divides 1 + q2, . but not 1 + q
:In each matrix,. the row labelled A corresponds to SA and the column labelled
11 corz:esponas to D ; thus the entry in row A.and column J.l is the
u
composition multiplicity of DJ.l in SA. Example 1.7.
Compare these matrices with those in
The order of the partitions has been reversed, but this could
be a matter of notation.
Each of the above matrices describes the
decomposition of ordinary. representations of GL (q) n characteristic coprime to q.
(n
s 4) over a field of
Replacinq [r] by r ("puttinq q
= 1")
we obtain
matrices concerned with representations of GL (F),. for all d, over the field d F of characteristic r.
We conjecture that this is part of a qeneral phenomeno
- a really astonishinq one, at that:
- 108 -
I J
,
17
RE*>VING THE FIRST COLUMN FROM
[~]
We are in a position to prove a theorem which relates the composition multiplicities of some of the modules
D~
in
S~
to composition multiplicities
in smaller general linear groups. The integer h partition
~
n will be fixed throughout this section.
S
For every
of n which has precisely h non-zero parts, we define the partition
~ of n - h by A
~
= (~l
- 1, U - 1, ••• , 2 '" U
Thus, the diagram for column, and
17.1
and
MA, SA, ~
1..1
THEOREM
If
~h
- 1).
is obtained from that for DA
u
are KG
n-
h-modules.
and U are partitions of n, each haVing precisely h non-
~
in
D~
by removing the first
We shall prove:
zero parts, then the composition multiplicity of multiplicity of
~
D~~A
equals the composition
S~.
The symmetric group result corresponding to this one has only recently been proved (James [J ] ) , and it is interesting to note that our proof of the
a
theorem for g
n
used Weyl modules, we know of no proof which lies entirely within
the framework of the representation theory of symmetric groups. Recall the Definition 9.4 of so NEh is a KGn_h-module,
The elements of Gn-h commute with /
E... --h
for every KG
n-
h-modu1e M.
E
h
,
We have seen (Theorem
10.5) that (17.2)
If v is a composition of n with fewer than h non-zero parts, then
The main step in proving Theorem 17.1 is:
17.3
THEOREM.
If
~
is a partition of n with precisely h non-zero parts then
and
- 109 -
the iSOlllCrphisms being' KG Proof:
n_h
-isomorphisms.
By Theorem 15.16, there is a series of KG -submodu1es of :M)., n
some 1.1 e- )..
The number of factors M 1M
s
~
1 which are isomorphic to S
In a similar way, there is a series of KGn_h-Submodules of A
A
::> M z» M
u
such that ....
some 1.1
l>
Mu
.A.
II::
S1
u+1
equals
Mi'
= 0,
and for 0 :S s s u - 1,
The number of factors
1.1
Ms /Ms +1
Ms 1Ms+1
ll!
SA 1.1
(KG h-isomorphism) for '; n-
7
which are isomorphic to SA equals 1.1
, ,J
= K~i
The definition of Kostka numbers (15.13) shows that Kl.IA
both ). and 1.1 have precisely h non-zero parts.
if
l
Thus,
,i
(17.6)
If 1.1 has precisely h non-zero parts, then the number of factors
M/M
in (17.4) which are isomorphic to Sl.I equals the nUIllber of factors
..
S+ l
'" M /M in (17.5) which are isomorphic to s s+1
modules.
t>
A, then S
E. a; S" ICh 1.1
and D E
1.1 h
ill
D'" as KG n-h1.1
t> )..)
h
.. Mi (as a KG
series in (17.4) by
n_ h
-module), by Corollary 10.15.
Act on the
~:
This is a series of KGn_h-submodules of
M~,
- 110 -
:~
~
(There is no need to "start the induction", since we are assuming'
Now, MAE
i .~
nothing' in the case where there is no 1.1' with precisely h non-zero parts such that 1.1
~
,~
S~.
We next assume, inductively, that if 1.1 is a partition of n with
precisely h non-zero parts and 1.1
,
and for 0
~
s
~ t
-
1,
'; ,1
!)
~
M B.. 1M lB.. ;;; SA, if M. 1M 1:;;; S and 1J has h non-zero parts, sn s+ n 1I s s+ 1J
In particular,
Let 8 denote our
KGn _h - isomorphism .10.
Theorem, either S).~ 2 Sle or S).~ r; S18.
from
Ml
to M).~.
By the SUbmodule
But a composition factor of
•
(Mi8)/(Si8) is isomorphic to Di' so by (17.7)
S).~
~
.L
• Sie.
Bence, using (17.8)
we have
and this proves the first part of the theorem• ..I.
We next wish to show that S).E
oJ.
h
= S18.
'I"he proof of this is made
awkward by the fact that the bilinear form on M). (see Definition 11.1) is usually singular when
restricte~ to
M).E • h
Let us recall the definition of the
M).E -
KGn _ h - isomorphism
Ml
to
First of all, in Theorem. 10.14, we constructed an injection" from
h
This injection clearly has the property that. I
Then
8 from
e
is defined by
Now,
Let
- 111 ...
~
1 .. ;1~r(~h)-rl
and define the map
e from M1
into M l\ by A
Usinq (10.10), we find that
where
r
2
= {(i,
j)
1
> j
i
~ h and row
T ). 1T
(i) R
> rowT
A1T R
(j)},
and R is the unique element of ~.
e E. IIi
Now, for all t E S~, there exists M~e),
and for all
nc
with tl\ = ~e (because MAl\
- = nSh
Si there exists n'. S·A with n'E
s~e, by analogy with the result we have already proved).
which equals zero, since 1;
E:
s~ and n 'E 2 h
E
SA.
Hence
~
(because
SA~
Therefore,
J.
E S~, and we have
proved that
On the other hand, -: has the same
-
....
KGn
composition factors as MA/S A,
so SA~ has the same ICGn-h composition factors as IIAEh/SA~.
Therefore,
Since E is an idempotent element, h
_ ",
-
Thus,
Therefore,
=
Die
This is what we wished to prove. Now, assume that M is a KG -module with a composition series n
which has the property that, for each s, Ms/Ms+l is isomorphic to D (s) IJ
where IJ (s) is a partition of n into at most h non-zero parts.
is a series of KG _ - submodu1e s of n h
~,
Then
and by Theorem 17.3 and (17.2),
Ms--h E.. 1M E.. '" D s+l--h - '" (s) if JJ (s) has h non-zero parts, IJ
MsE
h
= Ms + 1Eh
If we take M =
s~,
if JJ(s) has fewer than h non-zero
parts.
then Theorem 17.1 follows at once, since
\
- 113 -
S~Eh ~ s~.
18
ISOTROPIC SPACES
1
There is a highly "symmetric" generator for SA when A has precisely two non-zero parts.
I
We could avoid usinq this generator later on, but the
construction of it is particularly elegant, and the power of the SUbmodule,
Theorem and the Kernel Intersection Theorem is well illustrated at a critical
i
point.
we
shall assume that A
as the vector space over subspaces of V.
=
K with
(n - m, m) with n - m
~
m
> 0, and regard MAoJ
basis elements indexed by the m-dimensional
The generator of SA described in the second equality of
11.12{i) involves
q
[r]
= q (m+l)m/2
subspaces of V.
In Example 11.17, we gave a generator using only
subspaces of V.
The generator we shall construct in this section involves ~
m-l i IT (q + 1)
= 2(q
+ l){q
2
m-l + 1) ••• (q + 1)
i=O subspaces of V, which is a smaller number still, and we feel confident that. no non-zero element of SA involves fewer subspaces.
for the correspondinq ~ -module involves 2 n
m
The standard generator
terms.
',I 18.1
(i)
f l, f 2 ,
:1
DEFINITIONS
... , f (ii)
"t~
l
Let V* be an arbitrary subspace of V of dimension 2m, and let .,::'.t,
m,
gl' g2' ••• , gm be a basis for
Define a bilinear form
, ) on
V*.)
V* by
.~
'il
lj
(olf l + ••• + °mfm + ~lgl + •.. + ~mgm' °ifl + •.. + o~fm + ~igl + ••
l
+ IS 'g ) ~ m mij :
- 114 -j
(iii)
Define the function ,. from V* to
(iv)
If W is a subspace of V*, let
w.j" = Iv
E V*
I
radW= {w ~ W (v)
IFq by
(v, w. = 0 for all w E w},
I wei> = 0
and (w, w·)
=0
for all w· E W}.
A subspace W of V* is said to be isotropic if W = rad W.
The Gram matrix for the bilinear form ( , ) with respect to the chosen
basis of V* is
one readily verifies that for vI' V2 € V, ex, 1$ € ~ I we have
The function • has been introduced only to cope with the case where q is a power of 2. Jo
If W is a subspace of V*, then (18.2) shows that W and rad W are subspaces of V* I and rad W c;, W " W~ with equality if q is odd. Althouqh the next four results are known, we supply the proofs; since they are short and
~ause
they are normally qiven only for the case where
\
q is odd.
(OUr source is !anq .lL).)
18.3
tEMMA.
(v, w)
;II!
If v and ware linearlY independent elements of V* with
0 and w.
=0
I
then there exists VI € V* such that
- IlS -
is a linear combination of v and
(i)
v
(ii)
J!l'
(iii)
v ... 0
1
but v
W,
l
,I
is independsnt
of w.
Proof:
::
1
1
i " ,1
-1
j
,j
W)"!
Replacinq v be a scalar multipl.e of itself, we may assume that (v, Since w,
.. 1.
w)
= 0,
we have (w, w) ... O.
Let
= -v,
(I
E:
Fq
and vI ...
+ v;
CJW
Then our results follow, because
18.4
LEMMA.
(aw +
and
(aw + v, w) '" 1
v)~
s
(i)
o.
(I
SUppose that W is a subspace of V* and let U be w
exist vI' ••• ,
.. v+ +
V
s
be a basis for rad W.
such that
For every non-zero element u of U, u, w ••• , wsL-Y l' l,
···f
V
s
are linearly independent.
=0
(ii)
(u, v i)
for all u E tJ, 1 $ iSs.
(iii)
(vi' v.) ]
= v., J.
(iv)
(wi' v j )
=1
if i .. j
=0
if i
£
.L
~
~
j
r
s
s,
j S s. .4
Proof: V
for 1 S i , j S s ,
.. 0
Then U C W, so W l .L
U ' W. l
Thus, (v, w l)
~
> • U.
Proof:
l
such that
WS>I
Then rad W = <w ' ••• , 2 l
and we may continue inductively to construct v2i ••• , v
COROLLARY.
.L
U and there exists l,
0, and Lemma 18.3 qives an element v
Now let W = <w l' ••• , w > •
18.5
C
s•
•
If W is an isotropic subspace of V*, then dim W :ii m.
This is an immediate consequence of conclusion (i) of the Lemma •
- 116 -
•
J J
I ~
1
1 j
,\,
COROLLARY.
18.6
SUppose that W is an m-dimensional isotropic subspace of
v*, and let wI' ••• , wm be a basis of W.
Then there exists a basis wI' .•. ,
..• , vm of v* such that
~m !.....::! l'
(w., w .) ~ J
(w., v ~
i
o
(v., v .) = w. ~ = v . ~ J
~
=
)
J
~
1
if i
=
j
=0
if i
~
j
~
for 1
~
i, j
m.
~
•
This says that we may extend any basis of an m-dimensional isotropic space to a basis of V* in such a way that the Gram matrix takes the form
OUr next main objective is to prove:
18.7
Suppose that WI L!! 2 ~ 3 are isotropic subspaces of V*, each
THEOREM.
of dimension m.
Then
is even. There is a neat proof of this theorem when q is odd.
For it can be
shown that if W and W are m-dimensional isotropic spaces, with dim {W 2
= r
l,
3
then every isometry 8
Thus, if dim (WI
~
W ) 3
=
1
sending W onto W has determinant (-1) 2 3
r , dim (WI n W ) 2
2
=
r , and 8 3
2
A W' 3
2
m-r l
.
is an isometry from
W to WI' then 8 is an isometry from W to WI' and 2 3 18 2
m-r (-1)
m-r 1
(-L)
2
m-r - det 8 det 8 = det 8 = (-I) 1 2 18 2
This proves Theorem 18.7 in the case where q is odd.
3
Apparently though, more
work must be done to prove the theorem in the general case; we will be rewarded, however, by other useful results on the way.
- 117 -
18.8
LEMMA.
SUppose that
Wi isotropic and dim (W
Wl~2
l.!!.lf2)
are m-dimensional subspaces of V*, with
= r.
wll.......: •• , wm be a basis of W l
Let
••• «
he a basis of V* having the properties described in Corollary 18.6.
v
Then
W is an isotropic subspace if and only if W has a basis of the form 2 2 ••• , x
m
where (a . j ~
IF, q
(r + 1
$
1.
r + 1 :S i
:S m) ,
with
Proof:
i, j
$
m) •
There is a basis for W of the form 2
where
(a
SUppose that W is isotropic. 2
ij
, ~ij
E
=0
for l S i S r < k S m.
Therefore, we may assume that
for r + 1 :S iSm.
- 118 -
~
m,
But wl ' ••• , wm' y r+l' ••• , y.m
are linearly independent, so the matrix
is non-singular.
:S m) •
Then
for r + 1 S i
since (Wi' Y.k)
~, r + 1 S i
m
(r + 1
0,
a ..
J.J.
~
i, j S m) •
Conversely, if these conditions hold, then W is isotropic. 2
18.9
COROLLARY.
Suppose that W is an m-dimensional isotropic subspace of l
V* and U is an r-dimensional subspace of W • l isotropic subspaces W such that W n W 2 2 l q
Proof:
(m-r) (m-r-l)/2
=U
Then the number of m-dimensional is
.
q(m-r) (m-r-l)/2 is the number of choices for the matrix
in the statement of the lemma.
18.10
•
COROLLARY. m-l IT
(q
i
•
The number of m-dimensional isotropic subspaces of V* is
+ 1).
i=O
... ,
Proof: For each r with 0
S
r
~
f > is an m-dimensional isotropic subspace of V*. m
m, the number of r-dimensional subspaces of W is l
[~J . Therefore, using the last corollary, the number of isotropic subspaces of V* is
I
fm1
r=O LrJ
(m-r) (m-r-1) /2 q
m-:l (qi
IT
the equality being obtained by putting x Proof of Theorem 18.7:
+ 1)
,
=1
in Theorem 3.5.
i=O
•
Suppose we have three isotropic subspaces W , W , W 2 1 3
of V*, each of dimension m.
Let
- 119 -
Choose a basis wl' w2 ' : .• , wm for Wl such that
W
l ' ••• ,
W
d
+C is a basis for Wl n W 2
l ' ••• , wd' wd+ c+ l' ••• , wd+ c+b is a basis for Wl n W3•
W
Construct a basis wl' ••• , wm' v l ' ••• , v of V* satisfying the m conclusion of Corollary 18.6.
Take a basis for W as in Lemma 18.8, 2,
form "i ' ..., wd +c ' Yd +c+ 1 ' · • ., y m
where (~ij E ~' d + c + 1
$
i
~
m).
Apply Lemma 18.8 again, to choose a basis of W 3: ' •.• , wd' zd+l' .•• , zd+c' wd+c+ l' ••• , wd+c+b' zd+c+b+l' ••
Wl
where
(Y Then
W
2
e IF i q,
ij
+
W
3
€
{d + 1, •.• , d + c} v {d + c + b + 1, ••• , m}).
is spanned by
m
v{
I j=d+c+b+l
~
.. w. + v.
J.)
)
J.
I
d + c + 1 .,; i
~
m}
m U {
I
j=d+c+b+l
YijW j + Vi I d + 1 :;; i - 120 -
~
d + c}
m
I
U{
I
y 1jWj + vi
d + c + b + l S i S m},
j=d+c+b+l Hence W + W is spanned by the first three of these sets together with 2 3 m
{
L
.
(~ j -
j=d+c+b+l
v• j) W. I
1
J
1
d + c+ b + l S i ~
m },
The number of linearly independent elements in this last set is even, since the matrix
is alternating. dim (W
2
Therefore, + W :: (d + c + b) + 3) =b+c+m
(m -
d - c) + c
(modulo 2).
Thus, dim (WI
fl
W 2)
+ dim (W
2
fl
W + dim (W 3 3)
1\
WI)
- (d + c) + (b + c + m) + (d + b) - m
(modulo 2),
as stated in Theorem 18.7.
18.11
•
Choose any m-dimensional isotropic subspace J of V*.
DEFINITION.
For
each m-dimensional isotropic subspac.e W of V*, define sign W E K by sign W = (-1 )dim(JIlW) K
Theorem 18.7 shows that for any two m-dimensional isotropic subspaces
(18.12)
sign WI • sign W = (-1) 2
m-dim(W 1"W2)
•
In particular, the ratio of sign WI to sign W is independent of the isotropic 2 space J chosen in the definition:
that is, if We alter J, then the signs of - 121 -
the isotropic spaces either all remain the same, or they all change. Consider the element
L (sign
W)W
W
of M( )' the sum being over all m-dimensional isotropic subspaces W of n-m,m V* •
We shall employ the Kernel Intersection Theorem and the Submodule
Theorem, respectively, to prove that this element belongs to, and indeed generates, S(
n-m,m
in the eaSe' where m
18 .13
Proof:
The pictures (2.7) and (2.8) illustrate this generator
).
= 2.
~ (sign W)W belongs to S(n-m,m)'
THEOREM.
Suppose that 0
~
i
~
m - 1, and let U be an i-dimensional subspace of
v* contained in some m-dimensiona1 isotropic subspace W of V*. 1
The number
of r-dimensional subspaces of W which contain U is 1
~ - i1 ~ -~' by Corollary 3.3 (i). isotropic.
Each of these r-dimensional spaces is automatically
Using (18.12) and Corollary 18.9, we have
E{sign w 1 w is an m-dimensiona1 isotropic subspace of v» and W 2 2 2
=
sign Wl
~
r~i
= sign w m-i I 1
(-1) m-r
(-l).j
rm l!- -
r:: LIm;
i
i1
~
(m-r) (m-r-l) /2 q
~ qj (j-l) /2 .
j=O
Since m - i > 0, we find that the last sum is zero, by putting x Theorem 3. 5. We have proved that
L (sign
W)W
E
Ker l/Jl,i
when 0
~
i
W
- 122 -
~
m - 1,
= -1
in
;:!
U
and the result now follows from the Kernel Intersection Theorem 13.3 •
18.14
THEOREM.
t (sign W)Wgenerates
W
The generator t of S(
Proof:
only subspaces of <e
l,
n-m,m
e , ••• , e 2
s(
n-m,m
•
) as a KG -module.
n
) described in Example 1l.17(v) involves
>, so we take this 2m-dimensional space
2m
to be V*, and let f
1
= e 2i-l
in Definition 18.1.
Every space which occurs in t has the form
+ with (lij E~.
The definition of the bilinear form on
space we have described is isotropic if and only if (lij
v* ensures that the
=0
for all i, j.
'rherefore,
<~,
L (s1qn
W
W)W>~
= sign
«91,92, ••• , gm» ~ 0,
which proves that I: (sign W) W E S ( W
n-m,m
) \
generated by our element must then equal S (
- 1.23
..&.
S(
n-m,m)
n-m,m
.
)' by
The KG -module n
the Submodule Theorem
•
19
THE PRIME DIVISORS OF GAUSSIAH POLJNOMIALS The form of the Kernel Intersection Theorem means that the problem
of deciding whether or not a particular
eleme~t
of M belongs to SA often A
reduces to one of determining which Gaussian polynomials are divisible by the characteristic of K - Theorem 16.12 illustrates this point.
Indeed,
in the next section we shall determine all the composition factors of SA' when A has just two parts, using techniques which rely heavily on the information we glean about Gaussian polynomials in the present chapter.
19.1
DEFINITIONS.
(i) such that p
Suppose that p is a prime number.
If b is a positive integer, let '" (b) be the largest integer i p
i
divides b. If b is a non-negative integer, let 1 (b) be the least non-
(ii)
p
i
negative integer i such that b < P . For the results we prove here, it is sufficient to assume that q is a positive integer coprime
to p, so let this be the case throughout the
chapter.
19.2
If P is an odd prime, q > 1, and s, t are positive inteSlers
LEMMA.
such that p divides (qS - 1), then
v [qst -
1]
=v
P
pqS- I
Proof:
If q
s
- 1
='
(t).
b
ap
= ( apb -
with a coprime
+ l)P - 1 b
ap _ p Suppose that t
=:
i1.
to
(p)i
p, then
b 1. 1 (ap)-
i=l .
(modulo p2), since p > 2.
= cpd ,
with c coprime to p. - 124 -
Then
qst _ 1
= qSC
qS _ 1
(
d
1
-
n
qS _ 1
i=l
qSC~ 1
l
1
1
-
1J
qSCP ~-1 _
But S S (c-1) q se - 1 :: C .... - - -... =l+q + ••• + q S q - 1
(modulo p) ,
so
[gst _11 Vp qS _ 1
J
I
v
i=l
= d
19.3
r q scpi
d
=
p
d
lq sop - 1J
2
=
i-1
1
i=l
= 'Vp (t).
Let p be a prime number.
LEMMA.
(i)
If P divides (q - 1), then
(ii)
If n
m
=:
(~)
::
modulo £.
k
nO + n 1 p +
= m0
[:]
+ ~p, ~ k +~
+ m p +
1
(0 s n i,
m
i
< p)
then
[:] = [:~] [:~] ...
[~]
mcdu.Io p.
n
In particular, p divides (m) if and only if n i Proof:
-i.Eli
for some i.
Since
(1 + x) (1 + xq)
•.. (1 + xq
n-1
)
= ~
L
in]
ii
i=O ..
q
i(i-1)/2 xi ,
we get n
I
i=O
[~J
x
i
=
(1 +
I
i=O
x)
.(~) l~
n
if P divides (q - 1) xi
- 125 -
_
= (1
+ x)
"o
+~)
(1
nl
k ~ ••• (1 + ~ )
modulo p.
m All the results follow by considering the coefficient of x •
19.4
•
DEFINITIONS e Let e be the least positive integer such that p divides (q - 1)
If b is a non-negative integer, define b* by
(ii) b
=:
o
b*e + b l
S b
< e.
l
Thus, e is the exponent of q modulo p, and b* is the integral part of
b/e.
19.5
Let a and b be positive integers.
THEOREM.
Then the prime p divides
each of
[a]
1 '
ra+b-il fa+i1 ra+21 L 2 J' L 3 J' ... , L b J
if and only if both the following conditions hold: (i)
e divides a, so a
(ii)
a*
=:
a*e •
.t (b*)
=0
modulo p p
•
We remark that condition (ii) holds, trivially, if b* Proof:
= O.
First note that p divides each of a *) (la* + ( l' 2
1]
(a* + b* - 1 'J l b*
' ••• ,
if and only if p divides each of (a*) (a1*) ' (a*) 2 ' ••• , lb* ' (consider Pascali s Triangle), and this happens if and only if 1 (b*)
a* - 0
mod p p
, - 126 -
by applying the last part of Lemma 19.3. If e .. 1, then p divides (q - 1), and the theorem follows immediately from Lemma 19.3 (i) •
Assume, therefore, that e
>
1.
Then p is odd, q
>
1,
and
[~J =
p divides
! _-
11 .. P divides (qa - 1) .. e divides a.
We may thus assume that a .. a*e.
COnsider
a
i - I f = (qa+i-l - 1) (qa+i-2 - 1)
J
i
Let i
= i*e
+ i
l
(qi _ 1)
• (qi-1 _ 1)
(q
(q -
-
1) 1)
Then {a + i - I , a + i - 2, ••• , a} contains i* integers
•
divisible by e if i' :: 0, and the set contains (i* + 1) integers divisible by e if i'
V
p
~
1.
r
+
Therefore,
~
by Lemma 19.2.
-
j
> 0,
if i'
~
1,
Thus, p divides each of
if and only if e divides a, and p divides each of
... ,
[
a * + b* b*
1) •
Again, the theorem follows from the first paragraph of the proof.
19.6
THEOREM.
Then the trivial (i)
•
Let).l = (Jl t...J;!2 ' ••• , Ph) };)e a partition of n, with P 2-..Q.:. l h
KGn-module
e divides
(~d
is a submodule of S + 1), and
(ii)
- 127 -
-
if and only if Jl-----.. ---
for each d with 1 proof:
~
~
d
h - 1.
This follows immediately from Theorems 16.12 and 19.5. ~ve
Condition (1) of the
•
theorem guarantees that the diagrams [lJ)
and [nl have the same e-core, in agreement with the Theorem 1.2 of Fong and
Srinivasan, although this is not obvious.
It is interesting to compare the
theorem with the corresponding situa.tion for $ , where the relevant n condition is
(1
~
d
~
h - 1).
This is equivalent to conditions (1) and (ii), when q
= 1,
for then p diVides
= 1.
q - 1, so e
The remaining results..in this section are purely numerical.
The
'j
integer q is not mentioned again, and we may take p to be any integer greater , than 1, and e to be an arbitrary positive integer, although when we apply the results, q will be a prime power, p a prime number not dividing q, and
we
e will be the exponent of q modulo p.
continue to denote the integral :4
"'j
part of (b/e) by b*.
19.7
DEFINITIONS.
Let p
~
2, e
~
I
1 be integers.
i
Given two non-negative integers, a and b, we say that a contains
(i)
b to base p if and only if a
.~ .~
> 0,
~ .~
~
.~
.!
+ ••• b
k
= b k +l = (ii) f
p,e
= 0,
(0
s b. < J.
and for each i, either b
i
~
p) ,
=0
or b
i
= a i.
Given two non-negative integers, n and m, let (n, m)
= 1,
if Cn + 1)* contains m* to base p, and either e divides m or e divides (n + 1 - m),
- 128 -
~
f f
p,e p,e
(n, m)
= 1,
(n, m)
= 0, otherwise.
if m
= 0,
A few remarks about these definitions are in order. Definition 19.7(i) we require that b then a
~
if n
2m.
~
2b + 1.
= 0;
k
hence if a contains b to base p,
It is easy to verify from this that f
The statement that f
p,e
Note that in
(n, m)
p,e
=1
1, if m = 0, in Definition 19.7(ii)
(n, m)
is not redundant; consider the case n + 1 < e, m = O.
The numbers m,
(n + 1 -
are
m)
only
appearing in the definition of f
(n , m)
p,e
the "first column
hook lengths" (see Definition 20.1) for the diagram [n - m, m], provided that
n - m
~
m > O.
We introduce the function f
next chapter that f
KGn -module
in
S(
(n, m) is the composition multiplicity of the trivial
Pre
n-m,m
since we shall prove in the
p,e
)'
when n - m ~
has the characteristic p, and e is
D1, K
the exponent of q modulo p.
we consider f p,e
(n -
1, m) + f
1, m - 1) in the next two
(n -
p,e
Lemmas, as this number enables us to apply, inductively, results for G 1" n-
_ _f p,e (n, -.::.:A;,;;s.;;s..;;u;..me;.;. . . ;..th_a.;,;t~m_~_l_and
19.8
LEMMA.
+ f
(n - 1, m - 1)
-p,e Proof: n
~
Then f
= 1.
m)
p,e
(n - 1,
~)
= 1.
Note first that n - 1 is positive since f
p,e
(n,
m)
=1
implies that
2m. Hereafter, we adopt the notation a
I b for
"a divides b" and a
:=>
b for
"a contains b to base p". Since m If e f
p,e
I
~
1, we have (n + 1) *
~
m* r and e
m
~ and e ) (n +.1 - m), then e ~ 1, so e
(n - 1, m - 1)
~
O.
or e
I
I
(n + 1 - m) •
(m - 1) ~ 1.
Also, (n + 1)* = n*, so n*' m*, and f
= 1. - 129 -
p,e
Therefore (n - 1, m)
i
If e f
p,e
and f
m and e 1 (n + 1 - m), then e
= o.
(n - 1, m)
;t
1, so e
(n - m) •
Therefore,
Also, (n + 1)* "'" n* and m* = (m - I) *, so n*
:>
(m - 1) *
1) :::; 1. p,e (n - 1, m If e
(n + 1)*.
= m*
(n + 1 - m), then (m - 1)*
m and e
- 1 and n* + 1 :::;
since (n + 1)*:> m*,
j+l + n* + 1 :::; a pj + a j+1P j m* where 0
~
a
b pj + b + P j+l + j 1 j
=
< p, 0 S b
i
i
< p, a j
... , 0, and either b
;t
j
:::; 0 or b
then n* ;) m* and. n* :;, m* - 1; also m > 1, since m* ~ 1. b
j
:::;
a
and hence n* :> m* - 1, n* ~ m*.
j
f
p,e
(n - 1, m) + f
If b
j
;t
If b
j
= 0,
0, then
Thus, in either case
(n - 1, m - I)
p,e
= aj •
j
= 1.
•
The proof of the next lemma· requires a great deal of delicate footwork!
19.9
Assume that m
LEMMA. f
p,e
1, m) + f
(n -
1, and
~
1, m - I) > f
(n -
p,e
p,e
(n, m).
Then condition (A), (B) or (C), below, is true: (A)
m :::; 1, e
i
n.
(B)
elm, e
I
(n + 1) , ~ m*
(C)
There exists an integer j satisfying all the following conditions:
proof:
(i)
1
s
j
(i1)
e
I
(n + 1 - m -
(iii)
(n + 1 - m -
(iv)
f
(v)
f
p,e p,e
= pi
for some i ~
o.
< m,
(n , j)
(n -
j)
j) ,
* ::
0
= f p,e (n,
1, j) + f
By the last lemma, f
p,e
p,e
R. (m*) modulo p P , m) :::;
0,
(n -
1, j - 1) :::; f
p,e
(n, m) :::; 0, so we have: - 130 -
(n -
1, m) + f
p,e
(n - 1
(19.10)
I
If elm or e
(n + 1 - m), then (n + 1)* ~ m*.
Assume that conclusions (A) and (B) of the lemma are false.
we
step 1
prove that m > 1.
If m
= 1,
then e n , since conclusion (A) is' false, and e Ther~fore,
conclusion (B) is false.
m*
=0
and (n + 1) *
;i
;t
1, since
m*, contraQictinq
(19.10). Either (19.11) or (19.12) is true:
Step 2 (19.11)
elm, e
(n + 1), and either n*
(19.12)
n* ~ m*, e
I
SUppose that f e
I
(n - m).
I
n, and either e p,e
(n - 1, m)
= 1..
If elm, then (n + 1)*
conclusion (19~11) holds.
;t
~
m* or n*
(m - 1) or e
~
I
(m - 1)*. (n - m).
Then n* ~ m* and either elm or n* by (19.10), so e
If elm, then e
I
I
(n + 1), and
(n - m), and conclusion (19.12)
follows. SUppose that fp,e (n - 1, m - 1) and either e
m*, by (19.10), while n* e e
I I
I
j (m - 1) or e ~
= 1.
Then, since m - 1;t 0, n*
(n + 1 - m).
If e
(m - 1)*, thus either e
1 and (m - 1)*
;z:
= m*,
(m - 1)*
(n + 1 - m) , then (n + 1) *
I
m or e
(n + 1 - m), we therefore get conc 1usion (19.11). (m - 1) and e ) n, so e
~
If e
I i
('n
+ 1).
~
If
(n + 1 - m) , then
we now have conclusion
(19.12) • step 3
I
Assume that elm, e
(n + 1), and either n*
We have here that (m - 1)*
(by 19.10), in particular, m* since n*
~
;t
O.
= m*
- 1, (n + 1)*
;>
m* or n* =- (m - 1) *.
= n* +
Because conclusion (B) is false, m*
:$ k
< m*
;t
m* or n* =- m* - 1, there exists a unique integer k such
that
o
1, and n* + 1 ~ m*
and
n*
::
m* + k - 1
But k ;t 0, for otherwise n* + 1 :a m*. - 131 -
1. (m*) modulo p p
1.
Let j ::;: ke. 1 s j
I
e
<
=k
Then j* ::;: k, (j - 1)*
- 1, and
m,
+ 1 - m - j), and
(n
~
(m*)
m* - k - 0 modulo p p
(n + 1 - m - j)* ::;: n* + 1
AlSo, n* + 1 ~ j*, since n* +-1 ~ m, so f
p,e
(n, j)
= O.
Using the fact that e divides j, m, (n + 1 - j)., and (n + 1 - m), and R. (m*) the congruence n* m* + j* - 1 modulo p p ,we have:
=
f
f
p,e
p,e
=1
(n - 1, j)
(n - 1
'
~
=1~
j - 1)
~
n*;> j*
n*
;;l
m* - 1 # f
n* :> J' * - 1
¢>'
p,e
n* :) m*
~
(n - 1, m - 1) :: l.
f
p,e
(n - 1
'
m) :: l.
This proves that (19.11) implies conclusion (e). Step 4
Assume that n*
;;l
m*, e ) n, and either e
Since n* ~ m*, either m*
=0
I
(m - 1) or e
I
(n - m).
(m - 1), and we let
(in which case, e )
k :: 0), or there exists a unique integer k such that
Now, e
R. (m*) n* - m* + k modulo p p
o
;S k
t
n (in particular, e ~ 1) , so we may write n :: n*e + t, with 1 ~ t
j
::
ke + t
if
e
j
::;:
ke + 1
if
e ) (m
~
j
< m and n*
< m*
and
<e
Let
Then 1
:>
j*
I
::
(m
(j
1) 1) ; in this case t
- 1)*
::;: k .
I
(m
if
e
n +
if
e ) (m -
m-
j
=
n - m - ke,
But if e
t
(m - 1), then e
e
I
(n + 1 - m - j)
Because e
rj
i ,
Also,
n + 1 - m - j = n*e - m*e - ke, 1 -
~
1) , 1).
(n - m), and therefore, in either case,
and
and e )
(n + 1 - m - j)* :: n* - m* - k
(n + 1 - j) , we have f
- 132 -
p,e
(n, j)
=0 o.
R. modulo p p
I
If e and e
I f f
(j p,e p,e
If e
(m -
1), then n* :>
I
e- e
1)
(n -
% (m
I
- 1), then e
e
I
(n -
j), e
% (n + 1 -
j)
- 1, m - 1) ; 1, and
=1~
1, j - 1)
(n -
= m*,
1) *
Thus,
m).
= f p,e (n
(n - 1 , j)
(m -
f
p,e
(n -
1, m)
I
(n - m), e
= l.
(j -
1), and none of j , (n - j)
(n + 1 - j), (n + 1 - m) is divisible bye, so f
f
p,e p,e
= f p,e (n
1, j)
(n -
-
= f p,e (n
(n - 1, j - 1)
= 0,
1, m - 1)
= 1.
- 1, m)
This proves that (19.12) implies conclusion (e), and completes the proof of
•
the lemma.
The converse of the next lemma is also true, but we do not need it.
19.13
LEMMA.
Assume that m
integers jOc.. j 1 ' m
= j0 >
... , jl
>
and for each i with 1
~
1, and f
p,e
= 1.
(n, m)
Then there exist
jS+l such that
... > js ~
i
~
> jS+1
= 0,
s + 1,
e divides (n + 1 - j. 1 - j,), and ~-
J?roof:
If m*
e i'm but e
I
= 0,
then i
(n + 1 - m).
therefore, that m* > kt~1
P
o.
(m*)
~
= 0,
and because f
Thus, we may take m
p,e
= jo
= k O'
>
jt
=1 = O.
and m
~
1,
Assume,
We begin by defining certain integers kO' k , ••• , 1
which depend upon m*. Let m*
(n, m)
and suppose that we have defined (r
such that - 133 -
~
0)
for 1 Then there exists a unique integer k O:iikr+ l
r+
Thus, there exist integers t
~
i
r.
S;
1 such that t
=k r
(n + 1)*
and
:£
0, and k
O'
+ k r+ 1. modulo p p k
••• , k
l,
t+1
(k )
r.
such that
and
(n
+ 1)*
=k i-I
+ k
Define x by n + 1 If elm, let ji
If e
I
= kie
i
=
(n + 1 - m), then m
=t
if jt+l
= 0,
s
for 1
(n + l)*e + x, so 0
+ x,
Further, let jt+2 = k t+ 2
Put s
1. (k - l) i
modulo P P
:£
~
x < e.
~
i
i
~
t + 1.
if i is odd,
= m*e
= o.
=t
+
XI
Then ji
let
=ki
(1
+ 1 if jt+l > O.
:£
t + 2), and
Then (1 s: i ::;; s + 1)
Therefore, e divides (n + 1 - ji-l - ji)' and
• Theorem 19.6 shows that, when n - m is a submodule of S{ n-m,m ) if and. only if
- 134 -
~
m > 0, the trivial .KGn -module
e
I
.t (m*) (n
+ 1 -
m)
and
(n + 1)* - m*
These conditions certainly imply that f last lemma by showinq that if f modules S(n_j,j)' starting at j
p,e
(n,
=0
p,e
m)
modulo p P
(n, m)
= 1,
:=
- 135
We shall apply the
then there is a sequence of
and ending at j
trivial composition factor.
1.
:=
m, each of which has a
20
THE COMPOSITION FACTORS OF S (
n-m,m )
We shall now determine the composition factors (together with composition multiplicities) of SA when A has just two non-zero parts.
In
:.i ~
"'l t
i
view of the form of the Kernel Intersection Theorem, where all the homomorphislD8!l 'i
involve only changes in two adjacent parts of a partition, it is likely that
.~ .~
the result for two-part partitions will shed some light on the g e n e r a l i .~
problem of determining the composition factors of SA. Remember that Sen) = D(n} = the trivial KGn-module, and S(n-m,m) ~ S( m,n-m )' so we may assume that n - m
~
m.
Since S( n-m,m ) is irreducible
when K has characteristic zero, we consider only the case where the characteristic of K is a prime p, and p .( q.
As
usual, we view
M(
n-m,m ) as
the vector space over K spanned by the m-dimensional subspaces of V. We first prove a necessary and sufficient criterion for S
(n-m,m)
to be
irreducible.
20.1
DEFINITION.
Let
~
= (~l'
~2'
... ) be a partition of n.
(i)
The hook lenqth h
(ii)
The hook graph for lJ. is obtained by replacing each (i, j) node
(iii)
Given q and p, let e be the least positive integer such that
ij
of the (i, j) mode in the diagram [lJ] is
e p divides (q - 1), as in Definition 19.4.
symbols
obtained by replacing each (i, j) node in [lJ] by
W,
20.2
Let [lJ]p,e be the array of
if e does not divide h
EXAMPLE.
If
J.l
=
ij
•
(4, 3, 1), then the hook graph for
- 136 -
~
is
643 1 421 1
If P
= 2,
3 and q
= 2,
then e
and
10""00
=0
0
m
00
If P
= 13
and q
= 3,
then e
= 3,
and
=
[]..I]l3,3
00
We emphasize that we revert to taking q to be the integer such that
= GL n (q),
G n
K is a field of prime characteristic p, with p
exponent of q modulo p.
S
is a
(n-m,m)
KGn -module
remarks following Corollary 11.14), and n - m
20.3
THEOREM.
[n - m, m] Proof:
S(
p,e
S(
- n-m,m
)
q, and e is the
(or a KG -module; see the n
m.
is irreducible if and only if no column of
contains two different symbols.
n-m,m
) is reducib1-e if and only if it has a composition factor
isomorphic to some D(n_j,j) with 0 by Theorem 17.1, S( S(
~
~
n-m,m
~ j
< m, by Corollary 16.3.
) is reducible if and only if m
Therefore,
1 and either
~
n-m-l,m- 1 ) is reducible or every composition factor of S( n-m,m )' besides
D(n-m,m)' is isomorphic to D(n)·
Therefore, S(
induction, if some column of [n - m, m] two different symbols. [n - m, m]
p,e
p,e
n-m,m
) is reducible, by
after the first column contains
Let us assume, therefore, that m
~
1 and no column of
, after the first, contains two different symbols.
Then, since
D( n-m,m ) occurs only as a top composition factor of S( n-m,m )' S (n-m,m) is \
reducible if and only if it contains a trivial KG -submodule. n equivalent to e divides (n + 1 - m), and R. (m*)
(n + 1 - m)* - 0
modulo p p - 137 -
But this is
1
by Theorem 19.6.
If m < e, then the second of these cOnditions holds, trivially, and the entry in the (2, 1) place of [n - m, m) place is not
CD
is
p,e
;
The entry in the (1, 1)
CD.
J
if and only if e divides n + 1 - m, and this proves the theorem
i:j
in the case where m < e.
1.' "
Assume henceforth that m
~
e.
Since no column of [n - m, m)
after
p,e
:,
the first column contains two different symbols, e divides m (20.4)
~
p
*e
divides (n + 1 - m), and
«m - 1)* - i)
= ~p «n
- m)* -
~)
for i
~
(m -
If both symbols in the first column of [n - m, m) are p,e so S{
n-m,m) is irreducible.
elm.
Let us therefore assume that e
then e .( (n + 1 -
I
(n + 1 - m) and
Then (n + 1 - m)*
Let k
00,
1)* - 1.
=
~
p
=
(n - m)* + 1
Then p
(m*).
k-l
and
- 1 S (m - 1)*
m*
=
k
(m - 1)* + 1.
- 1, and we deduce from
conditions (20.4) that + ....
+ ••• +
+ ••• + either 0 < ~-l
S
S;
b. < p), 1.
where
b k- l ,
or But
(0
and
'ie-l
=
o.
now =b
k-2
#
(n + 1 - m) *
=b
k-l
:: 0 modulo pk •
That is, the two numbers in the first column of [n - m, m) and only if S(
) is reducible. n - m, m
=p-l
p,e
are unequal if
•
It has been proved (James [J ] and James and Murphy [JMl) that in S - 138 -
~.
symmetric group theory, the specht module for the partition \.I (\.I p-regular) is irreducible if and only if no column of [\.I] numbers.
=1
Since e
p, 1 contains two different
when "q = 1", this lends credence to the following
conjecture, which is supported by the last theorem..
20.5
CONJECTURE.
4\.1
is irreducible if and only if no column of (\.Il p , e
contains two different SymI)ols. In view of Corollary 16.3, which tells us that every composition factor of S( n-m,m) has the form D( n- j , j) with 0
s
j :so m, the main theorem of
this chapter determines all the composition factors of S (
20.6
THEOREM.
~(n-m,m)
n-m,m
):
be defined over a field of prime characteristic Then the composition multiplicity of DC
p, where p does not divide q.
n- j ,Jl)
in S( ,for 0 :so j S m is f (n -2j, m - j), given by Definition 19.7. n-m,m, p,e Theorem 17.1 reduces this theorem to a matter of proving:
20.7
THEOREM.
Proof:
step 1.
Se
)
The Composition multiplicity of DC n ) in- n-m,m -is f , e (n, m). -p Assume that f
p,e
(n, m)
= 1.
We prove that the composition
multiplicity of D(n) in SCn-m,m) is at least 1. Suppose that 0 S j
S
n - k.
Let
e be
the KG -homomorphisDi from
n
M ,j) into MCn-k,k) Which sends each j-dimensional subspace V of V to the l Cn_ j sum of the k-dimensional subspaces Which intersect V ~ in zero. We claim that J..er
e~
SC j
.).
n- ,J
The simplest way to see this is in
terms of the generator of S(n_j ,j) described in section 18, so let V* be a 2j-dimensional sUbSPace of V, spanned by f , .... , f j, gl' .... , gj' say. 1
t
=I
Then
(sign W)W,
w summed over the j-dimensional isotropic subspaces W of V*, belongs to Scn-j,i). - 139 -
~J};
(See Definition 18.11 and Theorem 18.13). to V* in V.
~.
·Let be a complemen\l: l
IJ
Then the subspace
.~
.~
;t,
occurs in f;8 with coefficient equal to t{sign
wI
.~
.~
W is a j-dimensional isotropic subspace of V*
-: ~
~
'1j ',
But
••• , f
l,
j>
j
is isotropic, and there exist qj(j-l)/2 j-dimensianal
isotropic subspaces of V* which intersect this one in zero, by Corollary 18.9 all these qj(j-l)/2 subspaces have the same sign.
(n-k,k)
= ± qj(j-l)/2
~
Therefore,
)
.j ~
I
I
o.
'~
'.'"
In particular, ~e ~ 0, and Ker 8 ~ S(n_j,j)' We
now
n- j ,Jn ' by
know that Ker 8 ,s.A.(
the SUbmodule Theorem.
Since
J.
.';..
Me n-,J j .)/S( n-, j j) is isomorphic to the dual of Se n-J, . j)' (20.8)
Every composition factor of S(n_j ,j) is isomorphic to a composition
factor of Men_j,j)e. Next, we consider M(n_j ,j) 8. of V with dim V
l
=
j, dim V
3
= b,
Let 0
s b <
and V tl V l 3
number of k-dimeosional subspacell of
X, and V , V
= O.
l
j (k-b)
be subspaces "=I; •
Then, by Theorem 3.1, the-:;J:.
Vwhich both contain V3 an
,,
':i :1!.: 'tt,
in zero is
q
3
rn - j - bl
L
k - b
~,
J.
.~
Therefore, M(n_j ,j) 8
c Ker
lPl,b # p divides
rn-j-bl L k - b J.
,~
ii,
.~ ~,
:~
The Kernel Intersection Theorem and Theorem 19.5 now give:
,~
} '·1
- 140 -
M(n_j ,j) e 'S(n-k,k)
~
R. (k*)
e divides (n + 1 - j - k) and (n + 1 - j - k)* - 0
modulo p P
Combining this with (20.8), we have proved: (20.9)
Every composition factor of S(
j
is isomorphic to a composition
l)
n- ,)
factor of S(n-k,k) if conditions (i) - (iii) hold:
n - k
(i)
0
(ii)
e divides (n + 1 - k - j)
(iii)
(n + 1 - k - j)*
$
j
Now assume that f
S
p,e
(n,
R. (k*)
=0 m)
modulo p p
= 1.
Then, by
• Lemma
19.13, there exist
integers,
such that, for each i with 1
$
iSs + 1,
e divides (n + 1 - ji-1 - ji)' and modulo P since D(n) k
= ji-l
= S(n)
for i
=
R.p (j 1.~ 1) •
, we now deduce, by repeatedly applying (20.9) with j
= ji'
s + 1, s, ••• , 1 in turn, that D(n) is isomorphic to a
composition factor of S( n-m,m ). Step 2
The composition multiplicity of D( ) in S( n n-m,m) is at most f p,e (n, m) This is certainly true if m
n
~
1).
= 0,
so it is true when n
Therefore, we may assume that m
~
=
0 (and when
1 and that the result of Step 2
is true for all two-part partitions of n - 1. Let
Xi
denote the p-modu1ar character of SA,a"
If the compOsition
multiplicity of D(n) in S(n-m,m} is x, then X(n-m,m) - XX(n} is a p-modu1ar character of G , and Theorem 16.11 shows that n
- 141 -
X(n-m-l,m) + X(n-m,m-l) - XXCn - 1 ) is a p-modular character of
G
n-
n :: 2m, but we have seen that f
no odds.) f
1.
(Here, the first term must be omitted if
p,e
(n - 1, m) "" Oif n "" 2m, so this makes
Hence, by induction,
p,e
(n - 1, m) + f
p,e
~
(n - 1, m - 1)
x.
Therefore, the result of step 2 is immediate unless
',I j
f
p,e
(n - 1, m) + f
p,e
,1 'i
(n - 1, m - 1) > f (n, m) ,
I
p
.I ,j
"«I
'~
so we assume that this inequality holds.
"
1
We aim to prove:
,~
;~
(20.10)
There exists j with 1 S j S m such that some composition factor of
s (n-m,m)
is isomorphic to D(n- j ,J')' and when we reqard D(n- j ,Ju as a KGn- 1-
,
1
j
1 .~
.'
module, it has D(n_l) as a composition factor with multiplicity at least f
p,e
(n - 1, m) + f
p,e
(n - 1, m - 1).
At least one of conclusions (A), (B), (C) of Lemma 19.9 is true.
We
examine each possibility in turn. If m "" 1 and e
.'1
I
n, then the two symbols in the first column of
"
"
[n - m, m]
p,e
are
Therefore Sen
GO.
Here, we may take j
= m,
when (20.10) follows from the Branchinq Theorem.
Assume next that elm, e show that S ( e
I
n-m,m
) is irreducible, by Theorem 20.3. - m, m
I
(n + 1)
~
and m*
= pk
) is irreducible in this case, too.
(m - i) ~ e
I
(n + 1 - m -
where k
~ O.
1
We shall
Certainly,
for-O SiS m - 1.
i)
i,
'J
\i Also, (m - 1)*
= m*
- 1, n*
=
'~
(n + 1)* - 1, and n* contains either m* or m* - ~
to base p, and n* + 1 does not contain m* to base p (by Lemma 19.8). n* with 0 < b
= k
(p - 1) + ••• + (p - l)p
< p, 0 S a.
k-l
+ bkP
Thus, - 142 -
k
k+l + ap
~
Therefor~
.''/ ~
'!
.,.
,
= n*
(n + 1 - m)*
+ 1 - m*
= bk Pk
k+l , + ap
and hence, for 0 SiS m* - 1,
v (m* - i) = P
'j
p
.
Therefore, Theorem 20.3 proves that S ( take j
.,'
«n + 1 - m)* - i).
V
=m
n-m,m
i I I
:j
Finally, assume that there exists an integer j satisfying the conditi.ons Applying (20.9), we have:
(i)
1 S j < m,
(ii)
f
(iii)
f
(iv)
Every composition factor of S(n_j ,j) is isomorphic to a composition
·r
·1
Aqain, we may
in (20.10) •
listed in conclusion (e) of Lemma 19.9.
1
) is irreducible.
p,e p,e
(n, j) = f
p,e
(n, m)
(n - 1, j) + f
p,e
=0
(n - 1, j - 1)
=
f
p,e
(n - 1, m) + f
p,e
(n - 1, m -1)
factor of S( n-m,m) • Since 1 S j f
p,e
(n -
< m and
1, j) + f
p,e
(n -
1, j - 1) > f
p,e
(n, j),
we deduce, by induction in (20.10), that there exists k with 1 S k S j such that some composition factor of S(n_j ,j) is isomorphic to D(n-k,k)' and when we regard D(n-k,k) as a KGn_l-module, it has D(n-l) as a composition factor with multiplicity at least f
p,e
(n -
1, j)
+ f
p,e
(n -
1, j - 1)
= f p,e (n
-
1, m) + f
p,e
(n -
1, m - 1).
But some composition factor of S ( n-m,m ) is isomorphic to D( n-k , k)' and so the proof of (20.10) is complete. Let ~(n-j,j) be the p-modular character of D(n_j,j) in (20.10), and again denote by x the composition multiplicity of D i n S (n) (n-m,m)· X(noom,m) - ~(n-j,j) - XX(n) is a p-modular character of Gn, and
- 143 -
Then
X(n-m-l,m) + X(n-m,m-l) - (x + Y)X(n-l) is a p-modular character of
Y
~
f
Therefore, x
p,e
(n - 1, m) + f
= 0 = f p,e (n,
n- l' where,
G
p,e
by (20.10),
(n - 1, m - 1).
m), and this completes the proof of Step 2 •.
Puttinq toqether the results of steps 1 and 2, Theorem 20.7 is prove Theorem 20.6 in
s (n-m,m)
~ells
us that the composition multiplicity of D(n_j,j)
•
j
is 1 if
I
(n + 1 - 2j) * contains (m - j) * to base p, and either e
I
~
(m - j) Or e
I
Cn + 1 - m - j) ,
~,
I
and is 0, otmerwise.
ern
(James (J , J 2 3)
The correspondinq theorem for the symmetric qroup states that the p-modular irreducible representation
indexed by (n - j, j) is a composition factor of the Specht module for (n - m, m) with composition multiplicity 1 if (n + 1 - 2j) contains (m - j) to base p, and with composition multiplicity 0, otherwise.
But if q = 1 then e
so the symmetric group theorem is just that obtained by putting q
= 1,
=1
in the
theorem for the general linear group. We could hardly conclude with a more striking illustration of our mai.n tenet, namely that the representation theory of the symmetric group hal
a ltq-analogue" in the general linear group theory.
Everything we have
done~
easily reduces to a result for the symmetric group when q is replaced by 1, but it is far from clear how to reverse the process at each stages
conside~ !
~he
main theorem of this chapter, for example - it would be extremely hard '
to quess the correct q-analoque of the statement that (n
(m - j) to base p.
+ 1 - 2j)
contai~
We believe therefore, that the theory of unipotent
representation of GL (q) is a deep and siqnificant topic which will fUlly n
repay further research.
.....
~s
I am indebted to Jorn Olsson for his suggestion in February 1980 that I
should undertake an investigati.on of unipotent representations of the
;Jeneral li.near groups.
My
gratitude is also due to Hanafi Farahat for many
aseful conversations during a visit to the University of Calgary in the summer of 1982, when the final parts of the research for this essay were completed.
- 145 -
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CL
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n
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- 146 -
JK
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- 147
~trique"
