To appear:
Proc. Lond. Math. Soc.
A q {analogue of the Jantzen{Schaper theorem
Gordon James and Andrew Mathas
Abstract
In this paper we prove an analogue of Jantzen's sum formula for the q{Weyl modules of
the q{Schur algebra and, as a consequence, derive the analogue of Schaper's theorem for the
q{Specht modules of the Hecke algebras of type A. We apply these results to classify the
irreducible q-Weyl modules and the irreducible (e{regular) q{Specht modules, dened over any
eld. In turn, this allows us to identify all of the ordinary irreducible representations of the
nite general linear group GLn(q) which remain irreducible modulo a prime p not dividing q.
1 Introduction
One of the most important tools for calculating decomposition numbers for the symmetric group Sr
is the theorem of Schaper [22], which is derived from Jantzen's sum formula [16, 18]. For example,
Schaper's Theorem tells us precisely which simple modules are composition factors of the Specht
module S , provided that the composition factors (together with multiplicities), are known for all
Specht modules S with B (here \B" is the usual dominance order on partitions). A good
account of such matters appears in [2]. Here we derive theorems for q {Schur algebras and Hecke
algebras of type A which reduce to the results of Jantzen and Schaper when q is set equal to 1.
We base our results upon a formula for the determinant of the Gram matrix of the q {Weyl
module W for the q {Schur algebra. Now, W may also be regarded as a Weyl module for quantum
GLn (see, for example, [3, Theorem 3.3.2]). Consequently, it is possible to construct a proof of
the formula we need using [1, Prop. 10.5] together with the techniques of Schaper's thesis [22, pp.
42{55]. The proof of [1, Prop. 10.5] is geometric, and relies upon quantum analogues of Kempf's
vanishing theorem [1, Theorem 5.7] and Serre duality [1, Theorem 7.3]. Here, we take a more direct
and purely algebraic approach by introducing generalisations of the Murphy operators [19, 6] for
the q {Schur algebra and using them to construct an orthogonal basis for the q {Weyl modules W .
By combining the properties of these Murphy operators with the combinatorial results in section 2
we arrive at the formula we seek in Theorem 3.26.
As an application of our results, we derive, for the rst time, a necessary and sucient condition for an ordinary irreducible representation of the nite general linear group GLn (q ) to remain
irreducible modulo a prime number which does not divide q .
We have made this paper largely self contained, by quoting little beyond basic facts about
q{Weyl modules [8]. In particular, our results include those of Jantzen and Schaper as special
cases.
Supported
in part by SERC grant GR/J37690
1
2
Gordon James and Andrew Mathas
2 Combinatorial results involving tableaux
In this section we develop the combinatorial machinery necessary to describe the determinant of
the Gram matrix, dened with respect to the semistandard basis, of the q {Weyl module (these
objects will be dened in section 3).
Throughout this paper we x two natural numbers n and r with n r. We regard a composition
of r as a sequence = (1 ; 2; : : :) of non{negative integers whose sum is r, and i = 0 for all i > n.
We write j= r if is a composition of r. If, moreover, 1 2 n , then is a partition
of r and we write ` r. We often suppress all zeros which are at the end of a composition.
Suppose now that j= r and j= r. The diagram [] consists of all ordered pairs of integers
(i; j ) with 1 i n and 1 j i. We refer to the elements of [] as nodes and we illustrate
diagrams as follows.
If = (4; 2; 1) then [] =
:
A {tableau of type is obtained from [] by mapping each node to an integer between 1 and n in
such a way that, for all i, the number of nodes mapped to i is i . Let T (; ) denote the set of all
{tableaux of type .
Suppose that t 2 T (; ). Then t is row{standard if the entries in t are non{decreasing along
each row. If ` r, then t is semistandard if it is row{standard and the entries in t are increasing
down each column. Let T0 (; ) denote the set of semistandard {tableaux of type . In the case
where t 2 T0(; (1r)), we say t is a standard {tableau; if t is a standard tableau then all of the
entries in t are distinct.
Of the three (4; 2; 1){tableaux of type (2; 3; 1; 1) illustrated below, only the rst two are row
standard, and only the rst is semistandard.
1 1 2 2
1 1 2 4
1 2 3 2
2 3
2 3
4 1
4
2
2
Let tij denote the entry in row i and column j of t. We dene a relation D on T (; ) as
follows.
(2.1) Let s; t 2 T (; ). Then s D t if and only if for all u and all k
jf (i; j ) 2 []jj k and tij u gj jf (i; j ) 2 []jj k and sij u gj:
Write s B t if s D t and s 6= t.
When ` r, the restriction of D to T0 (; ) is easy to understand, as we explain next. Assume
that s and t belong to T0 (; ) and consider the tableau obtained from s by omitting all entries
from s which are greater than u; since s is semistandard, this tableau has the shape of a partition,
which we call sju . Then, s D t if and only if sju D tju for all u, where the second symbol \D"
denotes the usual dominance order on partitions. For example,
1 1 1 2 2
2 3 3
3
B
1 1 1 2 3
2 2 3
3
B
1 1 1 3 3
:
2 2 2
3
A q {analogue of the Jantzen{Schaper theorem
3
Let t be the {tableau of type (1r ) which has the numbers 1; 2; : : :; r inserted in order along
the rows. For example, if = (4; 2; 1) then
1 2 3 4
t = 5 6
7
:
The symmetric group Sr acts on T (; (1r)) from the right by letter permutations, in the usual
way. Let S denote the row stabilizer of t . Then S is a standard Young subgroup of Sr .
The basic transpositions (m; m + 1), with 1 m r ? 1, generate Sr . If w 2 Sr , we denote by
`(w) the shortest length of an expression for w as a product of basic transpositions. Each right coset
Sw of S in Sr contains a unique element of minimal length; this element is called a distinguished
right coset representative of S. The distinguished right coset representatives of S are precisely
those permutations d such that t d is row{standard. Let D denote the set of distinguished right
coset representatives of S in Sr , and let D = D \ D?1 . Then the elements of D form a
complete set of S {S double coset representatives in Sr . If d 2 D then d is the unique element
of minimal length in S dS. Indeed, the following holds (see, for example, [5, 1.6]).
(2.2) Each w 2 Sr can be written in a unique way as a product w = udv with u 2 S, d 2 D,
and v 2 D \ S where is the composition of r which corresponds to the standard Young subgroup
d?1 Sd \ S of Sr ; moreover, `(w) = `(u) + `(d) + `(v):
We next relate this result to the row{standard tableaux in a way which will be useful later.
(2.3) Denition Suppose that ; j= r and that t is a row{standard tableau in T (; (1r)). Let
(t) be the row{standard tableau in T (; ) which is obtained from t by replacing each entry i by
u if i belongs to row u of t .
So, t 7?! (t) gives a mapping from the set of row{standard tableaux in T (; (1r)) into T (; ).
1 2 3
1 2 5 8 9
1 1 2 3 4
and = (3; 3; 2; 2), then t = 47 58 6 , and (t) = 1 2 3
.
For example, if t = 3 4 7
6 10
2
4
9 10
(2.4) Denition Suppose that ; j= r and that s is a row{standard tableau in T (; ). Then
?1 (s) is dened to be the set of row{standard tableaux of type (1r ) which are obtained from s by
replacing the 1's in s by 1; 2; : : :; 1, then the 2's in s by 1 + 1; : : :; 1 + 2 , and so on. If, at each
stage, we make the replacements so that the successive numbers are as high as possible, then we
obtain a tableau which we call rst(s). On the other hand, if we place successive numbers as low as
possible, we get a tableau which we call last(s).
For example, if s =
1 2 3 6 8 9 11
.
last(s) = 4 5 7
10
1 1 1 2 3 3 5
1 2 3 4 7 8 11
, then rst(s) = 5 6 9
and
2 2 3
4
10
3
4
Gordon James and Andrew Mathas
Note that if s is semistandard, then every tableau t in ?1 (s) is standard, and satises rst(s) D
t D last(s).
We now relate the row{standard tableaux in T (; ) to the S{S double cosets.
(2.5) Suppose that s is a row{standard tableau in T (; ) and that t = rst(s). Then t = td
where d is the unique permutation of minimal length in the double coset S dS, and
[
SdS =
Sd0:
d0
t d0 2?1 (s)
Moreover, this gives a 1{to{1 correspondence between the set of row{standard tableaux in T (; )
and the set of S{S double cosets.
The Young subgroup S which appears in (2.2) is the intersection of the row stabilizer of
rst(s) with S . For example, if s = 12 13 13 2 2 3 4 , then rst(s) = 16 28 39 4 5 7 10 and
= (3; 2; 1; 1; 2; 1).
We shall use the following special case of (2.5).
(2.6) Suppose that i < j and that x is the last number in row i of t and y is the rst number in
row j of t . Then (x; y ) is the permutation of minimal length in the double coset S (x; y )S and
[
S(x; y)S = Sd0;
d0
where d0 runs over the elements of D with the property that the entries in t d0 agree with those in
t except that some pair of numbers have been swapped between row i and row j .
1 2 3
For example, if = (3; 2; 2) then t = 4 5 , and, to calculate S(3; 6)S, we apply (2.5)
6 7
1 1 3
S
to the case where s = 2 2 , so rst(s) = t (3; 6). We nd that S (3; 6)S = d0 S d0 where
1 3
0
d runs over
f(3; 6); (3; 6)(2; 3); (3; 6)(2; 3)(1; 2); (3; 6)(6; 7); (3; 6)(2; 3)(6; 7); (3; 6)(2; 3)(1; 2)(6; 7)g:
The next result is a straightforward computation in a similar vein.
(2.7) Assume that i < j and that x belongs to row i of t, and y belongs to row j of t. Suppose
that there are a numbers to the right of x in row i of t , and b numbers to the left of y in row j .
Then (x; y ) = wd where w 2 S, d 2 D and `(w) = a + b.
1 2 3 4
1 3 4 9
For example, if t = 5 6 7 then (2; 9) = (2; 3)(3; 4)(8; 9)d where t d = 5 6 7 .
8 9
2 8
A q {analogue of the Jantzen{Schaper theorem
5
We now wish to introduce various rational functions associated with tableaux. To this end, let
q be an indeterminate. For each non{negative integer m, let
[m]q =
mX
?1
a=0
qa
and
fmgq =
m
Y
a=1
[a]q :
Note that [0]q = 0 and f0gq = 1. Also, setting q = 1, [m]q and fmgq reduce to m and m!
respectively.
In the next section, we shall dene vectors fs , indexed by the semistandard tableaux s in
T0(; ), in a certain inner product space (namely, the {weight space of the q{Weyl module W ).
The square of the norm of fs turns out to be the rational function s 2 Z(q ), which we dene next.
(2.8) Denition Let ` r, j= r, and s 2 T0(; ). For each (i; j ) 2 [], let
?s (i; j ) = f (k; l) 2 []jl < j; skl < sij ; and sk0 l > sij for all k0 > k g;
and dene
s =
[j ? i + k ? l + 1]q :
[j ? i + k ? l]q
(i;j )2[] (k;l)2?s(i;j )
Y
Y
Thus ?s (i; j ) is the set of nodes (k; l) 2 [] such that the l{th column of s lies to the left of the
j {th column of s and contains no node labelled sij and, in addition, skl must be the largest entry in
the l{th column which is less that sij . Moreover, the factor that a node (k; l) 2 ?s (i; j ) contributes
q
to s is precisely [[+1]
]q where is the so{called axial distance from (k; l) to (i; j ) (see [13, p. 123]).
In order to help understand these denitions we illustrate them with several examples, many of
which will be needed later in the paper.
(2.9) Example (i) If s = 12 13 2 then s = [2]q[2]q.
[4]q [3]q :
(ii) If s = 12 12 3 then s = [3]
q [2]q
1
Y
(2.10) Example If t = t then t = figq .
i=1
1 2 3 4
For example, if t = 5 6 7 then the contributions to t from the (1; j ){nodes are
8
[4]q [3]q [2]q ;
[3]q [2]q ;
and
[2]q [1]q
[3]q [2]q [1]q
for 1 j 4 respectively, and the product of these terms is f4gq .
[1]q ;
[2]q ;
5
6
Gordon James and Andrew Mathas
(2.11) Example (cf. [19, 4.1], [6, 4.10]) Suppose that t and t0 are in T0(; (1r)) and that t0 =
t(m; m + 1), for some 1 m r ? 1. Assume that tij = m + 1, tkl = m where i < k; then j > l,
since t and t0 are standard tableaux. Then,
t = [ + 1][q []2 ? 1]q t0 ;
q
where = j ? i + k ? l.
Proof: To see this, rst note that ?t0 (a; b) = ?t (a; b) for (a; b) 6= (i; j ); in particular, ?t (k; l) =
?t0 (k; l). Also, the only dierence between the two sets ?t (i; j ) and ?t0 (i; j ) is that (k; l) 2 ?t (i; j )
and (k ? 1; l) 2 ?t0 (i; j ); these nodes contribute factors of
[]q
[ + 1]q
and
[]q
[ ? 1]q
to t and t0 respectively.
Recall the map t 7?! (t) from Denition 2.3. For any standard tableau t, we want to relate
the two functions t and (t); for this we need the next denition.
(2.12) Denition Assume that ` r, j= r, and t 2 T0(; (1r)). For each (i; j ) 2 [], let
Pt (i; j ) = f (k; l) 2 []jl < j; tkl < tij ; and (t)kl = (t)ij g;
and let
t =
[j ? i + k ? l + 1]q :
[j ? i + k ? l]q
(i;j )2[] (k;l)2P (i;j )
Y
Y
t
(2.13) Example Assume that = (2; 2).
(i) If t = 14 2 3 , then t = [2]q .
(ii) If t = 13 2 4 , then t = [4]q [2]q .
[3]q [1]q
(2.14) Example Suppose that t; t0 2 T0(; (1r)) with (t) = (t0) and t0 = t(m; m + 1) for some
m, with 1 m r ? 1. Suppose that tij = m + 1, tkl = m, and i < k. Then, by an argument
similar to that used in Example 2.11, we have
t = [ [+]1]q t0 ;
where = j ? i + k ? l.
q
A q {analogue of the Jantzen{Schaper theorem
7
(2.15) Example Suppose that s 2 T0(; ) and let t = rst(s). Let d 2 D be given by t = td
and let be the composition such that S = d?1S d \ S (see (2.2)). Then,
Proof: Notice that if (i; j ) 2 [], then
jPt(i; j )j, then
1
Y
t = figq :
i=1
Pt (i; j ) = f (k; l) 2 []jl < j
and skl = sij g. Let xij =
[j ? i + k ? l + 1]q = [xij + 1]q [xij ]q : : : [2]q = [x + 1] :
ij
q
[j ? i + k ? l]q
[xij ]q [xij ? 1]q [1]q
(k;l)2Pt(i;j )
Y
Therefore,
t =
Y
[xij + 1]q =
(i;j )2[]
1
1 Y
Y
i=1 u=1
fciugq ;
where ciu is the number of entries in the i{th row of s which are equal to u. However, the parts of
are precisely the numbers ciu , as i and u vary, by the remark following (2.5).
(2.16) Example Suppose that s 2 T0(; ) and let t = last(s). Comparing the denitions of t
and s we see that t = t s .
Next, we give a name to a product of the rational functions s . In the next section we will
show that this product is (essentially) the Gram determinant of the {weight space of the q {Weyl
module W .
(2.17) Denition Let ` r and j= r. Dene () =
Y
s2T0(;)
s .
The remainder of this section is devoted to establishing an explicit formula for ().
Let be a partition of r and let 0 = (10 ; 20 ; : : :) denote the partition conjugate to . That is,
0 is the partition whose diagram is obtained by interchanging the rows and columns in [ ]. Recall
that n is an integer with n r. If i = i0 + n ? i for 1 i n then 1; 2; : : :; n is a sequence of
{numbers for 0 (see [13, p. 77]). Note that 1 + + n = r + n(n ? 1)=2. Hereafter, whenever
we deal with integers 1; : : :; n, we shall implicitly assume that 1 + + n = r + n(n ? 1)=2.
(2.18) Denition Given integers 1; 2; : : :; n and a composition j= r, dene an integer d(1; 2; : : :; n)
as follows.
(i) If 1 > 2 > > n 0, then let d (1; 2; : : :; n ) = jT0(; )j where 0 is the partition
whose {numbers are 1 ; 2; : : :; n .
(ii) If i = j for some i 6= j , or i < 0 for some i, then let d (1; 2; : : :; n) = 0.
(iii) If w 2 Sn , then let d (1w ; 2w ; : : :; nw ) = sign(w)d(1 ; 2; : : :; n):
7
8
Gordon James and Andrew Mathas
Now we x our notation so that ` r and j= r. Let 1 > 2 > > n 0 be {numbers
for 0, and suppose that z is the largest integer such that z 6= 0.
Every s in T0 (; ) has z entries equal to z , and these entries are at the feet of distinct
columns. If we delete these entries from s, then we obtain a tableau in T0 (; ), say, where is
a partition of r ? z and = (1 ; : : :; z?1 ). The {numbers for 0 are 1 > 2 > > n 0
where for precisely z subscripts i = i1; : : :; iz we have i = i ? 1, and for the remaining subscripts
j 6= i1; : : :; iz we have j = j . Since d(1; : : :; n) = jT0(; )j and d (1 ; : : :; n ) = jT0(; )j,
it follows from our conventions in Denition 2.18 that
X
d (1 ; : : :; i ? 1; : : :; iz ? 1; : : :; n):
d(1 ; : : :; n) =
1
i1 <<iz
And now, further thought shows that in fact we have the following.
(2.19) For all integers 1; 2; : : :; n
X
d(1; : : :; i ? 1; : : :; iz ? 1; : : :; n):
d(1; : : :; n) =
1
i1 <<iz
Next, we extend Denition 2.17 of (), replacing by a sequence 1; : : :; n of integers.
(2.20) Denition Suppose that 1 2 n are integers. If 1 > 2 > > n 0
then let (1; : : :; n) = ( ) where 1 ; : : :; n are a sequence of {numbers for 0 ; otherwise,
let (1; : : :; n) = 1.
We now embark upon the task of evaluating () in earnest. Recall the denition of s from
Denition 2.8. We have,
Y
Y
Y
Y
[j ? i + k ? l + 1]q :
s =
(2:21)
() =
[j ? i + k ? l]q
s2T (;) (i;j )2[] (k;l)2?s(i;j )
s2T (;)
In order to relate () to (), where and are as above, notice that
Y
Y
Y
[j ? i + k ? l + 1]q
[j ? i + k ? l]q
s2T (;) i;j 2 (k;l)2?s(i;j )
(2:22)
sij <zY
(1; : : :; i ? 1; : : :; iz ? 1; : : :; n );
=
0
0
(
0
) [ ]
i1 <<iz
1
since this calculation involves only those semistandard tableaux obtained by removing all entries
labelled z from some semistandard tableau in T0(; ). The remaining contributions to ()
come from those s 2 T0 (; ) and (i; j ) 2 [] for which sij = z . For such s and (i; j ), the set ?s (i; j )
consists of those nodes (k; l) which lie at the foot of a column in [] with l < j and skl 6= z ; so we
have j ? i + k ? l = l ? j . Therefore,
Y
Y
Y
[j ? i + k ? l + 1]q
[j ? i + k ? l]q
2 z (k;l)2?s(i;j )
s2T (;) i;j
sij =
Y
Y (2:23)
[l ? j + 1]q d ( ;:::;i ?1;:::;iz ?1;:::;n )
=
[l ? j ]q :
l<j
i <<iz
0
(
) [ ]
1
1
l=2fi ;:::;iz g
j2fi11 ;:::;iz g
1
9
A q {analogue of the Jantzen{Schaper theorem
Let X1 be the numerator on the right hand side of this equation. Then
X1 =
(2:24)
=
1
Y
Y
Y Y
h=1 i1 <<iz
1
Y
Y
j
l
[h]dq ( ;:::;i ?1;:::;iz ?1;:::;n )
1
1
l=2fi1 ;:::;iz g
j2fi1 ;:::;iz g
l ?j +1=h
Y Y
h=1 i1 <<iz?1
[h]fq ;
1
j
l
fl;jg\fi1 ;:::;iz?1 g=?
l ?j +1=h
where f1 = d (1 ; : : :; i ? 1; : : :; j ? 1; : : :; iz? ? 1; : : :; n): Interchanging the entries l and
j ? 1 in f1, so that l goes in position j and j ? 1 goes in position l, we obtain
1
1
f1 = ?d (1 ; : : :; i ? 1; : : :; l; : : :; j ? 1; : : :; iz? ? 1; : : :; n)
= ?d (1 ; : : :; i ? 1; : : :; j ? 1 + h; : : :; l ? h; : : :; iz? ? 1; : : :; n )
since l ? j + 1 = h.
Let X2?1 be the denominator on the right hand side of (2.23). Thus,
(2:25)
1
1
1
X2 =
1
Y
Y
h=1 i1 <<iz
1
Y Y
j
l
[h]?q:d( ;:::;i ?1;:::;iz ?1;:::;n )
1
1
l=2fi1 ;:::;iz g
j2fi1 ;:::;iz g
l ?j =h
An argument similar to that which we just used for X1 shows that
(2:26)
X2 =
1
Y
Y
Y Y
h=1 i1 <<iz?1
l
j
[h]fq ;
2
fl;jg\fi1 ;:::;iz?1 g=?
l?j =h
where f2 = d (1; : : :; i ? 1; : : :; j + h; : : :; l ? h ? 1; : : :; iz? ? 1; : : :; n ).
Gathering together the formulae from (2.21) through to (2.26), we obtain a \branching rule"
for ().
1
(2:27)
() = X1 X2
1
Y
i1 <<iz
(1; : : :; i ? 1; : : :; iz ? 1; : : :; n );
1
where X1 and X2 are as in (2.24) and (2.26).
We may now derive our formula for ().
(2.28) Theorem Suppose that ` r and j= r. Let 1 > 2 > > n 0 be {numbers for
0. Then
() =
1 Y Y
Y
j
h=1 l
j >l ?h
[h]q?d ( ;:::;j +h;:::;l?h;:::;n ) :
1
9
10
Gordon James and Andrew Mathas
Note that in this formula, we do not require that j < l, merely that j > l ? h. Thus, in the
exponent, we add h to j and subtract h from l , whether or not j < l.
Proof: First, observe that if 1 ; : : :; n are integers with 1 2
n and i = j , for
some i 6= j , or n = ?1, then both () and the product given in the statement of the theorem
are equal to 1. This remark allows us to apply induction on r.
As before, assume that z is the last non{zero part of , and let = (1 ; : : :; z?1 ). By (2.27)
and induction on r, we have () = X1 X2Y1 Y2 Y3 Y4 ; where X1 and X2 are as above and Y1 , Y2 ,
Y3 , and Y4 are given below and correspond to the four ways in which fl; j g and fi1; i1; : : :; iz g can
intersect.
1 Y
Y Y
Y
[h]gq ;
Y1 =
1
h=1 i1 <<iz
j
l
j >l ?h
fl;jg\fi1 ;:::;iz g=?
where g1 = ?d (1; : : :; i ? 1; : : :; j + h; : : :; l ? h; : : :; iz ? 1; : : :; n );
1
Y2 =
1
Y
Y
h=1 i1 <<iz?1
Y Y
j
l
[h]gq ;
2
fl;jg\fi1 ;:::;iz?1 g=?
j ?1>l ?h
where g2 = ?d (1; : : :; i ? 1; : : :; j ? 1 + h; : : :; l ? h; : : :; iz? ? 1; : : :; n);
1
1
Y3 =
1
Y
Y
h=1 i1 <<iz?1
Y Y
j
l
[h]gq ;
3
fl;jg\fi1 ;:::;iz?1 g=?
j >l ?1?h
where g3 = ?d (1; : : :; i ? 1; : : :; j + h; : : :; l ? 1 ? h; : : :; iz? ? 1; : : :; n); and
1
1
Y4 =
1
Y
Y
h=1 i1 <<iz?2
Y Y
j
l
[h]gq ;
4
fl;jg\fi1 ;:::;iz?2 g=?
j >l ?h
where g4 = ?d (1; : : :; i ? 1; : : :; j ? 1 + h; : : :; l ? 1 ? h; : : :; iz? ? 1; : : :; n ).
Now, g2 is equal to f1 by (2.25); so by (2.24) and the denition of Y2 ,
1
2
X1Y2 =
1
Y
Y
h=1 i1 <<iz?1
Y Y
l
j
[h]gq :
2
fl;jg\fi1 ;:::;iz?1 g=?
j >l ?h
Also, g3 = ?f2 . Hence,
X2Y3 =
1
Y
Y
h=1 i1 <<iz?1
Y Y
l
j
fl;jg\fi1 ;:::;iz?1 g=?
j >l ?h
[h]gq :
3
A q {analogue of the Jantzen{Schaper theorem
11
Now, applying (2.19), we obtain
() = Y1 (X1 Y2 )(X2Y3 )Y4
1 Y Y
Y
=
[h]q?d ( ;:::;j +h;:::;l ?h;:::;n );
1
j
h=1 l
j >l ?h
as required.
(2.29) Example Suppose that = (3; 2). Then (6; 5; 3; 1; 0) are {numbers for 0, and
() = [2]?q d (4;7;3;1;0)[3]?q d (9;2;3;1;0)[4]?q d(2;5;7;1;0)?d(2;9;3;1;0)
= [2]qjT ((3;1 );)j[3]qjT ((1 );)j[4]q:jT ((2 ;1);)j?jT ((1 );)j
0
2
0
5
2
0
0
5
Now let = (22; 1). Then jT0 ((3; 12); )j = 1, jT0((15); )j = 0, and jT0((22; 1); )j = 1; so
() = [2]q [4]q . This agrees with Example 2.9 where we calculated s for the two elements s of
T0(; ).
There is another useful way to describe (); in order to state it we let hij = i + 0j ? i ? j +1
denote the length of the (i; j ){th hook in [] (see [10, x18]).
(2.30) Corollary (cf. [15, pp. 224, 233]) Suppose that ` r and j= r. Let 1 > 2 > >
n 0 be {numbers for 0 . Then
Y
() =
a;b;c
a;b)2[]
a;c)2[]
b<c
(
(
[hab ]q d ( ;:::;b+hac ;:::;c?hac ;:::;n ) :
[hac ]q
1
Proof: If we subtract h from l in (1 ; 2; : : :; n ) then we usually obtain an n{tuple where
some integer is repeated or some integer is negative. The only exceptions occur when h = hal for
some (a; l) 2 []; see [13, 2.7.13]. Therefore, by Theorem 2.28,
() =
=
Y
n
Y
[hal]q?d ( ;:::;j +hal ;:::;l?hal ;:::;n )
1
j=1
(a;l)2[]
j >l ?hal
Y Y
?d (1;:::;j +hal ;:::;l ?hal ;:::;n)
[hal ]q
(a;l)2[] j<l
Y
(a;l)2[]
Y
[hal]q?d ( ;:::;l ?hal ;:::;j +hal ;:::;n ) :
1
j>l
j >l ?hal
11
12
Gordon James and Andrew Mathas
In the product where j > l we have
?d(1; : : :; l ? hal; : : :; j + hal; : : :; n)
= d (1; : : :; j + hal ; : : :; l ? hal ; : : :; n )
d(1; : : :; l + haj ; : : :; j ? haj ; : : :; n );
=
0;
if (a; j ) 2 [];
if (a; j ) 2= [];
since l ? j = hal ? haj if (a; j ) 2 []. However, if (a; l) 2 [] and j < l then (a; j ) 2 [], so
() =
Y
[hal ]q?d ( ;:::;j +hal ;:::;l ?hal ;:::;n)
Y
1
a;l)2[]
a;j)2[]
j<l
(
(
[hal]qd ( ;:::;l+haj ;:::;j ?haj ;:::;n ) ;
1
a;l)2[]
a;j)2[]
j>l
(
(
which is the same as the expression in the statement of the corollary.
The exponent which appears in Corollary 2.30 has the following interpretation (cf. [15, p. 223]).
Given (a; b) and (a; c) 2 [], with b < c, unwrap the skew (a; c){hook from [] and wrap it back
onto the resulting diagram so that the hand of the new skew hook is in column b. Suppose that by
doing this we get a partition . Then
d (1; : : :; b + hac; : : :; c ? hac ; : : :; n ) = jT0(; )j;
where the sign is positive if and only if the sum of the leg lengths of the unwrapped and wrapped
skew hooks is even.
(2.31) Example Consider, once again, the partition = (3; 2). The pairs of nodes in the same
row which contribute to () are as shown (the nodes (a; b) and (a; c) in [] are circled and we
have labelled the nodes in the (a; c){hook by c).
c c ?! c
c
c
c
c ?! c
?! c
c
giving
giving
giving
[4]q ?jT ((1 );)j
[3]q
[4]q jT ((2 ;1);)j
[1]q
[2]q jT ((3;1 );)j
[1]q
5
0
0
0
2
2
The product of these three quotients agrees with the answer for () which was obtained in
Example 2.29.
A q {analogue of the Jantzen{Schaper theorem
13
3 The Hecke algebra and the q{Schur algebra
Let R be an integral domain and let q be a unit in R. The Hecke algebra H = HR;q (Sr ) is the
R{free R{algebra with basis f Tw jw 2 Sr g, where multiplication is dened by the rule that
Twv ;
if `(wv ) = `(w) + 1;
qTwv + (q ? 1)Tw; otherwise,
where v is a basic transposition in Sr and w 2 Sr .
If j= r, then let
Tw Tv =
(3:1)
x =
X
w2S
Tw
y =
and
X
w2S
(?q )?`(w) Tw :
We are interested in homomorphisms dened on the right ideals x H of H .
For ; j= r and d 2 D , the map 'd is dened by
'd(xh) =
X
w2S dS
Tw h;
(h 2 H ):
Then 'd 2 HomH (x H ; xH ), and it is easy to see that f 'djd 2 D g is a basis for this
R{module.
Recall from (2.5) that the elements of D are indexed by the row{standard {tableaux of type
. If d 2 D corresponds to a row{standard tableau s in T (; ) we dene 1s = d?1 2 D . Then
'1s is an element of HomH (xH ; xH ) and
f '1s js 2 T (; ) is row{standard g
is a basis for this R{module.
(3.2) Example Hereafter, ! denotes the partition (1r). Then x! H = H . For each h0 2 H , the
map which sends h 2 H to h0 h is an H {endomorphism of H considered as a right H {module.
Moreover, f 'x!! jx 2 Sr g is a basis of HomH (H ; H ); so we identify H with HomH (H ; H ) via
the map Tx 7?! 'x!! (x 2 Sr ).
Let (n; r) denote the set of compositions of r such that i = 0 for all i > n. The algebra
SR (n; r) = EndH (
M
2(n;r)
x H )
is the q {Schur algebra [7, 8]. The element '1s of HomH (x H ; x H ) is regarded as an element of
SR(n; r) by dening its action on x H to be zero for any composition 6= . Then
f '1s j; 2 (n; r) and s 2 T (; ) is row{standard g
is a basis of SR (n; r).
13
14
Gordon James and Andrew Mathas
In order to dene the q {Weyl modules, recall from [8] that q {tensor space is the left ideal
SR(n; r) '1!! of the q{Schur algebra. An R{basis of q{tensor space is
f '1!t j 2 (n; r) and t 2 T (; (1r)) g:
Given ` r, let w be unique permutation such that t w has the numbers 1; 2; : : :; r entered in
order down its columns. So, for example, if t = 14 25 3 then t w = 12 34 5 ; so w = (2; 3; 5; 4).
Let z = '1! Tw y0 , where, as usual, 0 is the partition conjugate to and y0 is dened in (3.1).
Then z is an element of q {tensor space and the right H {module generated by z is isomorphic
to the right ideal of H generated by xTw y0 , which is called the q {Specht module S . The left
H {module generated by '1!z is isomorphic to the left ideal of0 H generated by xTw y0 and
is the dual q {Specht module S~ (in [6] this module is denoted S~ ). The left ideal of the q {Schur
algebra generated by z is the q {Weyl module W .
For each s 2 T0 (; ), let es = '1s z . In [8, 8.1] it is proved that W is a free R{module with
basis
f esj 2 (n; r) and s 2 T0(; ) g:
This basis is called the semistandard basis of W .
Now, f '1 j 2 (n; r) g is a set of orthogonal idempotents in SR (n; r) whose sum is the identity element of SR(n; r). If U is a left SR (n; r){module, then '1 U is the {weight space of U
corresponding to the q {weight . We are interested in the {weight spaces W = '1 W of the
q{Weyl module W ; these spaces also have a semistandard basis.
(3:4)
f esjs 2 T0(; ) g is a basis of W:
(3:3)
It is the main aim of this section to calculate the determinant of the Gram matrix, with respect to
the semistandard basis, of W (we dene an inner product on W below).
Before we begin this calculation we introduce two results which will be needed in the next
section. The rst of these is taken from [8, 8.8,8.9].
(3.5) Suppose that R is a eld. Then every q-Weyl module W has a unique top composition factor
F , and f F j ` r g is a complete set of non{isomorphic irreducible SR (n; r){modules. Moreover,
F is a composition factor of W with multiplicity 1, and every composition factor of W is of the
form F with D .
Next, we generalize a well known property of the ordinary Weyl modules. By way of notation,
write
X
U ! a W (a 2 Z)
to mean that
`r
U
M
`r
a <0
(?a )W have the same composition factors.
and
M
`r
a >0
a W A q {analogue of the Jantzen{Schaper theorem
15
(3.6) Lemma Assume that R is a eld and that U is a left SR(n; r){module. Suppose that for all
` r,
X
dimR '1 U =
Then U !
P
`r aW .
Proof: From (3.5) we deduce that U
0=
X
`r
`r
a dimR W:
! P`r cW for some integers c; so, for all ` r,
(a ? c) dimR W =
X
`r
(a ? c)jT0(; )j:
Now, jT0(; )j = 1 and jT0(; )j = 0 if 4 , so the matrix of coecients jT0(; )j, as and run over all partitions of r, is unitriangular. Therefore, a = c for every ` r.
Now we begin our investigation of the weight spaces of the q {Weyl module W . Of these, the
!-weight space, where ! = (1r ), is the most easy to understand so we study maps between W and
W! for all j= r. In order to do this we rst consider the maps '1! and '1! . The reader may nd
it helpful to refer back to the results and denitions of (2.2){(2.5).
(3.7) Lemma Suppose that s 2 T (; ) is row{standard. Then
X
'1! '1s =
Proof: For all h 2 H ,
'1! '1s (xh) =
t2?1 (s)
X
w2S1sS
Tw h =
'1!t :
X
X
t2?1 (s) w21t S
Tw h;
P
since S 1s S = [t2? (s) 1t S by (2.5). Therefore, '1! '1s = t2? (s) '1!t as claimed.
1
1
For the next result we recall from [8, 1.11], that there is an antiautomorphism of SR(n; r)
which is given by
(3:8)
d? ;
('d ) = '
1
(; 2 (n; r); and d 2 D ):
Note that Tw = Tw? ; consequently, x = x for all j= r.
1
(3.9) Lemma Suppose that t 2 T (; (1r)) is row{standard and let s = (t). Write t = tdw,
where t d = rst(s) and w 2 S and let = (1; 2 ; : : :) be the composition of r given by S =
d?1 Sd \ S. Then
!
1
Y
'1! '1!t = q `(w) figq '1s :
i=1
15
16
Gordon James and Andrew Mathas
Proof: We have that `(dw) = `(d)+ `(w) by (2.2), and Tw x = q `(w) x since w 2 S . Therefore,
x Tdw x = xTdTw x = q `(w) xTd x :
P
If I = fi + 1; i + 2; : : :; i + j g is a subset of a row in t , then x w2S(I ) Tw = fj gq x . Hence, by
(2.2),
xTd x =
Therefore,
xT1t x = (xTdw x ) = q `(w)
1
Y
X
i=1
w2S dS
figq
1
Y
X
i=1
w2S1tS
figq
Tw :
Tw = q `(w)
1
Y
i=1
figq
X
w2S1sS
Tw :
This shows that x has the same image under the two homomorphisms which appear in the statement of the lemma, so the homomorphisms are equal.
We use Lemma 3.9 to show that '1! acts as a projection operator, mapping W! onto W .
(3.10) Lemma Suppose that t 2 T0(; (1r)), and let s = (t). Write t = t dw, where td =
rst(s) and w 2 S , and let = (1 ; 2 ; : : :) be the composition of r which is given by S =
d?1 Sd \ S.
!
1
Y
1
`
(
w
)
figq es:
(i) If s 2 T0 (; ) then '! et = q
i=1
(ii) If s 2= T0 (; ) then '1! et is a linear combination of terms es0 where s0 2 T0 (; ) and s0 B s.
Proof: By denition, '1! et = '1! '1!t z, so
'1! et = q `(w)
1
Y
i=1
figq '1s z
by Lemma 3.9. If s 2 T0 (; ), then '1s z = es . If s 2= T0(; ), then the construction of the
semistandard basis of W , given in [8], shows that '1s z is a linear combination of the terms es0
where s0 2 T0 (; ) and s0 B s.
Next, we introduce some elements of the q {Schur algebra which are generalizations of the
Murphy operators [19, 6].
(3.11) Denition Let = (1; 2; : : :) be a composition of r and suppose that m > 0. For each
i with i > 0, let xi be the last entry in the i{th row of t . Let y be the rst entry in the m{th row
of t , and note that (xi; y ) 2 D by (2.6). Dene L;m by
L;m =
m
?1
X
i=1
i >0
q?(1+i
+1
+i+2 ++m?1 )'(xi ;y) + ([1]
q + [2]q + + [m ? 1]q )'1 :
17
A q {analogue of the Jantzen{Schaper theorem
1 2 3 4
(3.12) Example Let = (4; 4; 3). Then t = 5 6 7 8 and
9 10 11
L;2 = q ?1 '(4;5) + ([1]q + [2]q + [3]q)'1 ;
L;3 = q ?5 '(4;9) + q ?1 '(8;9) + ([1]q + [2]q)'1:
Note that L!;m is the q {Murphy operator Lm of H [6]. That is,
L!;m = Lm = q ?(m?1) T(1;m) + q ?(m?2) T(2;m) + + q ?1 T(m?1;m):
Our next result relates the operators L;m and Lm ; the rationale being that we understand the
action of the q {Murphy operators on the (1r ){weight space of W , so we can understand how L;m
acts on W by pulling the calculation back to W! .
(3.13) Lemma Suppose that j= r and that m > 0. Let y be the rst entry in the m{th row of
t and let k = m . Then
'1! (Ly + Ly+1 + + Ly+k?1 ) = L;m'1! :
Proof: For this proof we adopt the notation that if w
2 Sr then w is the unique element of
minimal length in the coset S w.
Suppose that 0 u k ? 1. The denition of Ly+u gives
xLy+u =
y+X
u?1
j =1
q ?(y+u?j) xT(j;y+u) :
Dene (j ) to be the row index of j in t and let f (j ) be the number of entries to the right of j in
this row.
Assume rst that (j ) < m. Then, by (2.7), x T(j;y+u) = q u+f (j ) x T(j;y+u) , so
q ?(y+u?j) xT(j;y+u) = q j+f (j)?y x T(j;y+u) = q ?(1+ j
( )+1
++m?1 ) x
T(j;y+u) ;
since y = 1 + + m?1 + 1 and j + f (j ) = 1 + + (j ) .
Next, suppose that (j ) = m and j y + u ? 1. Then q ?(y+u?j ) x T(j;y+u) = q y+u?j ?1 x since
`(j; y + u) = 2y + 2u ? 2j ? 1. Therefore,
y+X
u?1
j =y
q ?(y+u?j) xT(j;y+u) = [u]q x:
17
18
Gordon James and Andrew Mathas
So the action of '1! (Ly + + Ly+k?1 ) on an element h of H is given by
h 7?!
(
m
?1 X
X
X
i=1 a2row i b2row m
q ?(1+i
+1
++m?1 ) x
T(a;b) + ([1]q + [2]q + + [k ? 1]q )x
)
h;
where a 2 row i means that (a) = i and b 2 row m means that (b) = m. By (2.6) and the
(xi;y)
denitions, '
is the map sending
x h 7?!
X
X
a2row i b2row m
xT(a;b)h;
so we deduce that '1! (Ly + Ly+1 + + Ly+k?1 ) = L;m '1! , as required.
(3.14) Corollary Let be a composition of r. Then, for all a and b for which L;a and L;b are
dened, L;a and L;b commute.
Proof: By [6, 2.2] the q {Murphy operators Lm commute. Therefore, by Lemma 3.13, L;a L;b '1! =
L;bL;a '1! , and hence L;a and L;b commute.
From here until Theorem 3.26 we take our ground ring R to be F(q ), where F is an arbitrary
eld and q is an indeterminate. Thus we work with the Hecke algebra HF(q);q and the corresponding
q{Schur algebra. The case of Theorem 3.16 below, when R is any integral domain, follows from the
case where R = F(q ); however, in Denition 3.18 we will need to avoid certain rings.
As in [6, p. 71], dene the q -residue of the (i; j ){th node of a diagram to be [j ? i]q if j i and
j
?q ?i [i ? j ]q if i > j . The pattern of q{residues in a diagram is illustrated below.
0
?q?1
?q?1 ? q?2
..
.
1
0
?q?1
..
.
1+q
1
0
1 + q + q2 : : :
1+q
:::
..
.
If s 2 T (; ) and m is an entry in s, let rs (m) be the sum of the q {residues of the nodes which
m occupies in s.
For example, if
1 1 2 3
s= 2 3
;
4 4
then rs(1) = 1, rs (2) = ?q ?1 + 1 + q , rs (3) = 1 + q + q 2, and rs (4) = ?2q ?1 ? q ?2 .
It is an easy exercise to prove the following result.
(3.15) If s and s0 are distinct tableaux in T0(; ) then rs(m) 6= rs0 (m) for some m.
P
Given s 2 T0(; ), an expression such as uBs au eu , with au 2 (q ), denotes a linear combination of terms eu with u 2 T0(; ) and u B s.
F
A q {analogue of the Jantzen{Schaper theorem
19
(3.16) Theorem Let s 2 T0(; ) and suppose that m > 0. Then
L;mes = rs(m)es +
X
uBs
au eu ;
for some scalars au 2 F(q ).
Proof: The special case where = (1r ) is a version of the main property of the q {Murphy
operators. Although this property is normally stated in terms of the q {Specht modules (cf. [6, 3.15]
and [20, 4.6]), similar arguments prove the result for the dual q {Specht modules, and this is the
version which we use here.
Q
Let t = rst(s). Then, by Lemma 3.10(i), es = ?1'1! et ; where = 1
i=1 fi gq for some
composition of r. Let y and k be dened as in Lemma 3.13; then
L;m es = ?1 L;m '1! et = ?1 '1! (Ly + + Ly+k?1 )et :
Now, Ly+i et = rt(y + i)et+ a linear combination of terms ev where v 2 T0 (; (1r)) and v B t
because, as we remarked above, we know that the Theorem holds when t 2 T0 (; (1r)). But,
rt(y) + + rt(y + k ? 1) = rs(m), so
L;mes = ?1 '1!
rs(m)et +
X
vBt
bv ev
!
for some bv 2 F(q ). Now, if v B t then (v ) B (t); so, by applying Lemma 3.10, we obtain the
result.
There is an inner product < ; > dened on q {tensor space by
<'d! ; 'd!0 > =
q`(d) ;
0;
if = and d = d0 ;
otherwise.
Here ; 2 (n; r), d 2 D and d0 2 D . The inner product behaves nicely with respect to the
antiautomorphism dened in (3.8). Namely,
(3:17)
<su; v> = <u; sv>
<uh ; v> = <u; vh>;
and
for all s 2 SR(n; r) and h 2 H , where u and v are elements of q {tensor space (see [8, 4.1]).
We now use Theorem 3.16 to transform the semistandard basis of W into a basis of orthogonal
vectors (cf. [19]).
(3.18) Denition Assume that R = (q) and let s 2 T0(; ). Dene
Y
Y
L;m ? rs0 (m) ;
Es =
0
m with m 6=0 s0 2T ; rs (m) ? rs (m)
F
0(
)
rs0 (m)6=rs (m)
and let fs = Es es :
19
20
Gordon James and Andrew Mathas
Since L;a commutes with L;b by Corollary 3.14, there is no need to specify the order of the
products in the denition of Es .
(3.19) Lemma Assume that R = (q) and that s and s0 are distinct elements of T0(; ). Then
F
the following hold.
(i) L;m fs = rPs (m)fs ;
(ii) fs = es + uBs au eu for some au 2 F(q );
(iii) Es0 es = 0 if s B s0 ;
(iv) Es fs = fs and Es fs0 = 0.
Proof: The stated results follow from Theorem 3.16 by straightforward techniques (see [19,
(3.4){(3.11)] and [6, 4.6]). In order to prove that Es fs0 = 0, it is important that if s 6= s0 then
rs(m) 6= rs0 (m) for some m, as we noted in (3.15). We leave the reader to check the details.
Now, L;m = L;m where is the antiautomorphism of (3.8); so Es = Es for all s in T0 (; ).
Therefore, by (3.17) and Lemma 3.19(iv), if s and s0 are distinct semistandard tableaux then
<fs; fs0 > = <Esfs; fs0 > = <fs; Esfs0 > = 0:
Combining this observation with Lemma 3.19(ii), we deduce the following.
(3.20) f fsjs 2 T0(; ) g is an orthogonal basis of W.
Our next aim is to nd the inner products <fs ; fs > for all s 2 T0 (; ).
(3.21) Lemma Suppose that t and t0 2 T0(; (1r)) and that t0 = t(m; m + 1). Assume that
tij = m + 1, tkl = m and that i < k. Then
T(m;m+1) ft0 = [?1] ft0 + ft;
q
where = j ? i + k ? l.
Proof: The proof of this result is almost identical to that of [6, 4.9], so we omit it. The coecient
here is dierent from that in [6, 4.9] because the ordering of the tableaux has been reversed.
For the next lemma, recall the rational function t of Denition 2.12. In this lemma we study
the projection of the orthogonal basis of W! onto the orthogonal basis of W.
(3.22) Lemma Suppose that t 2 T0(; (1r)) and that s = (t) 2 T0(; ). Then '1! ft = tfs.
Proof: We show rst that '1! ft is a scalar multiple of fs . Using the notation in the statement
of Lemma 3.13, we have, for all m with m > 0,
L;m'1! ft = '1! (Ly + + Ly+k?1 )ft
= '1! (rt(y ) + + rt(y + k ? 1))ft ;
= rs (m)'1! ft :
by Lemma 3.13
by Lemma 3.19(i),
21
A q {analogue of the Jantzen{Schaper theorem
Hence, Es '1! ft = '1! ft . On the other hand, '1! ft is a linear combination of terms fu with
u 2 T0(; ) by (3.20); so Lemma 3.19(iv) now implies that '1! ft is a multiple of fs .
Write '1! ft = btfs . Then it remains to show that bt = t . To start with, consider the
case where t = rst(s). Then, by Lemma 3.19(ii), ft = et + a linear combination of terms ev
with (v ) B (t). Note also that '1! et = tes by Lemma 3.10(i) and Example 2.15. Therefore,
Lemma 3.19(ii) implies that
X
'1! ft = tes + au eu;
uBs
for some au 2 F(q ). However, since we already know that '1! ft is a multiple of fs , Lemma 3.19(ii)
shows that '1! ft = t fs in this case.
Now suppose that t 6= rst(s). Then there exists an integer m such that if t0 = t(m; m + 1)
then t0 B t and (t) = (t0 ). So, if tij = m + 1 and tkl = m then i < k and j > l. By induction,
'1! ft0 = t0 fs . Also, because m and m + 1 are in the same row of t, the map '1! T(m;m+1) sends
an element h in H to x T(m;m+1)h = qx h. Therefore, '1! T(m;m+1) = q'1! .
Let = j ? i + k ? l. Then, by Lemma 3.21, T(m;m+1) ft0 = [?]1q ft0 + ft : Writing '1! ft = btfs as
above, we have
qt0 fs = '1! T(m;m+1) ft0
= '1
!
?1 f 0 + f = ?t0 + b f :
[]q t t
[]q t s
q So, bt = [[+1]
]q t0 = t ; the last equality following from Example 2.14.
In contrast with Lemma 3.22, our next step is easy.
(3.23) Lemma Suppose that s 2 T0(; ) and let t = last(s). Then there exists scalars au 2 (q)
F
such that
'1! fs = ft +
X
uBt
a u fu :
Proof: Using the denitions and Lemma 3.7, we see that
'1! es = '1! '1s z =
X
u2?1 (s)
'1!u z =
X
u2?1 (s)
eu = et + terms eu with u B t;
since t = last(s). Therefore, there exist coecients bv , b0u , and au 2 F(q ) such that
'1! fs = '1! (es +
=
=
X
bv ev );
X vBs
et + b0u eu
uBt
X
ft + au fu ;
uBt
by Lemma 3.19(ii)
by Lemma 3.19(ii).
21
22
Gordon James and Andrew Mathas
Before we come to the main results of this section, we contract the inner product < ; > to the
left ideal of SR (n; r) which is generated by y0 . To do this we dene
uy0 ; vy0 = <u; vy0 >;
for all u; v 2 SR (n; r) (cf. [9, x 5.5], [8, 4.3]).
Now recall the rational function s from Denition 2.8.
(3:24)
(3.25) Theorem Suppose that s 2 T0(; ). Then fs ; fs = qas s for some as 2 .
Z
Proof: We consider rst the case where = (1r ); here the result is essentially the same as
in [6, 4.10], but we give the proof.
Let t = t 2 T0(; (1r)). Then ft = et = '1!t xTw y0 = xTw y0 , and
ft; ft = <xTw ; xTw y0 > = <xTw ; xTw >;
using (2.2) and the fact that w?1 S w \ S0 = f1g. Therefore,
1
Y
ft; ft = q`(w )<x; x> = q`(w ) figq = q`(w )t;
i=1
by Example 2.10.
Now assume that t 2 T0 (; (1r)) but t 6= t . Then there exist integers m and m + 1 such that if
tij = m+1 and tkl = m then i < k and j > l. Let t0 = t(m; m+1). By induction, ft0 ; ft0 = q at0 t0
for some at0 2 Z. By Lemma 3.21, T(m;m+1) ft0 = ?[1]q ft0 + ft , where = j ? i + k ? l. Therefore,
noting that ft0 ; ft = 0 by (3.20),
1 f 0 ; f 0 + f ; f = T
t t
(m;m+1)ft0 ; T(m;m+1)ft0 []2q t t
by (3.17),
= ft0 ; T(2m;m+1)ft0 = q ft0 ; ft0 + (q ? 1)T(m;m+1)ft0 ; ft0 = q ft0 ; ft0 ? 1 (q ? 1)ft0 ; ft0 :
[]q
Therefore,
2
ft; ft = q[]q ? (q[?]21)[]q ? 1 ft0 ; ft0 q
q
[
+
1]
[
?
1]q q at0 0
q
=
t
[]2q
= q at0 +1 t ;
the last equality coming from Example 2.11. This completes the proof of the theorem in the case
where = (1r ).
Assume now that 6= (1r ), let s 2 T0(; ), and set t = last(s). Then Lemma 3.22, (3.17), and
Lemma 3.23 give
ft; ft = '1!fs ; ft = fs ; '1!ft = tfs ; fs:
A q {analogue of the Jantzen{Schaper theorem
23
Since ft ; ft = q at t , Example 2.16 shows that fs ; fs = q at s as required.
We have now reached the main technical result of the paper. Choose a total order < on T0 (; )
such that s < s0 if s0 B s where s; s0 2 T0(; ). Let G be the matrix whose entries are the inner
products es ; es0 , where the rows and columns of G are ordered by <. That is to say, G is
the Gram matrix of W with respect to the semistandard basis f es js 2 T0(; ) g.
We show that det G is given by the rational function () which was introduced in Denition 2.17, and calculated in Theorem 2.28 and Corollary 2.30.
(3.26) Theorem Assume that R = (q) and that ` r and j= r. Then
det G = q a ()
F
for some integer a.
Proof: By Lemma 3.19(ii), the transition matrix from the semistandard basis fes g of W to the
orthogonal basis ffs g is unitriangular and hence has determinant 1. Therefore, because the vectors
fs are orthogonal,
Y
Y
s
det G =
fs ; fs = qa
s2T (;)
s2T (;)
for some a 2 Zby Theorem 3.25. The Theorem now follows from the Denition 2.17 of ().
0
0
(3.27) Corollary The rational function () belongs to [q]. If R is an arbitrary integral domain, and q is an invertible element of R, then the determinant of the Gram matrix G is equal
to q a (), for some a 2 , regarded as an element of R.
Z
Z
Proof: Assume rst that R = Q(q ), where q is transcendental. It is clear from the denitions
that es ; es0 2 Z[q; q ?1] for all s; s0 2 T0 (; ). Therefore () 2 Z[q; q ?1]. However, (),
as given in Theorem 2.28, can be calculated when q = 0 (where its value is 1), so () 2 Z[q ].
Now suppose that R is an arbitrary integral domain, and that q is an invertible element in R.
We work with HR;q (Sr ). For each f 2 Z[q; q ?1], let f denote the element of R obtained from f by
replacing q by q , and 1 by the identity of R. If gs;s0 is the (s; s0){th entry in the Gram matrix G
dened over Q(q ), where s; s0 2 T0 (; ), then gs;s0 is the (s; s0){th entry in the Gram matrix G
dened over R. Therefore, the determinant of the Gram matrix over R is q a (), as we wished
to show.
We do not know any direct way to prove that the rational function (), given by Theorem 2.28
or Corollary 2.30 is, in fact, a polynomial.
4 The main results
A technique, due to Jantzen, now enables us to derive information about the composition factors of
the q {Weyl modules and q -Specht modules from the formulae we have obtained for the determinant
of the Gram matrix G of W .
23
24
Gordon James and Andrew Mathas
Suppose that R is a principal ideal domain and that p is a prime in R. Note that R=pR is a
eld and let p : R nf0g?! Zbe the natural valuation on R associated with (R; p); thus, for x 2 R,
p (x) = i where i is the largest integer such that pi divides x.
Let M be a free R{module of nite rank, equipped with a non{singular bilinear form ; .
Denote by M (i) the R{submodule of M which consists of all elements m such that pi divides
m; m0 for all m0 2 M . If U is an R{submodule of M , let U denote the R=pR{submodule
(U + pU )=pU of M = M=pM . Then
M = M (0) M (1) M (2) gives a ltration of M with M (i) = 0 for all suciently large i.
Let e1 ; : : :; ed be an R-basis of M , and let G be the Gram matrix whose (i; j ){th entry is
ei ; ej . Then Jantzen's [17, Lemma 3] key observation is that
p (det G ) =
(4:1)
X
i>0
dimR=pR M (i):
The principal ideal domains R which are of particular interest for the q {Schur algebra are the
following.
(i) R = F[q; q ?1 ], where F is an arbitrary eld, q is transcendental over F, and p is a
cyclotomic polynomial.
(ii) R = F[q; q ?1 ], where F is an arbitrary eld, q is transcendental over F, and p = q ? q for some
non{zero element q of F.
(iii) R = Z[q ?1], where q is an integer, and p is a prime number which does not divide q .
In any case, we henceforth assume that R is a principal ideal domain in the denition of SR (n; r),
that p is a prime in R, and that q is an invertible element in R such that p (q ) = 0 and [a]q 6= 0 for
all 1 a r. The last condition on q ensures that ; is non{singular on W by Theorem 3.26.
(4.2)
To simplify the statements of our main results, we adopt conventions for the q {Weyl modules
like those in Denition 2.18. Namely, given a sequence 1; 2; : : :; n of integers we dene the
(virtual) SR=pR (n; r){module W (1 ; 2; : : :; n) as follows.
(i) W (1; 2; : : :; n ) = W if 1 > 2 > > n 0 are {numbers for the partition 0 ;
(ii) W (1; 2; : : :; n ) = 0 if i = j , for some i 6= j , or i < 0 for some i;
(iii) W (1w ; 2w ; : : :; nw ) = sign(w)W (1; 2; : : :; n ) for all w 2 Sn .
We can now prove a q {analogue of Jantzen's sum formula concerning the Weyl modules of
general linear groups.
(4.3) Theorem Let ` r and let 1 > 2 > > n 0 be {numbers for 0. Then, as
SR=pR(n; r){modules,
X
i>0
W (i) !
X
a;b)2[]
a;c)2[]
b<c
(
(
(p ([hab]q ) ? p ([hac ]q ))W (1; : : :; b + hac ; : : :; c ? hac ; : : :; n ):
25
A q {analogue of the Jantzen{Schaper theorem
Proof: Let ` r and recall that G is the Gram matrix of W with respect to its semistandard
basis. By Corollary 3.27, det G = q a () for some integer a. Therefore, applying (4.1) in the
case where M = W ,
X
i>0
dimR=pR W(i) = p (det G ) = p (q a ()) = p ( ());
since p (q a ) = 0. However, by Corollary 2.30,
p (()) =
X
?
(p ([hab ]q ) ? p ([hac ]q ))dim '1 W (1 ; : : :; b + hac ; : : :; c ? hac ; : : :; n ) ;
where the sum is over all a, b, c as in the statement of the Theorem. The desired result now follows
by Lemma 3.6.
The importance of Theorem 4.3 lies in the fact that W (1) is the unique maximal submodule
of W (dened over R=pR) { see [8, 4.6]. Thus,
W ! F + W (1);
and Theorem 4.3 gives us partial information about the composition factors F of W for 6= .
(4.4) Example Suppose that = (22; 1). Then the hook lengths in [] are given in the diagram
4 2
3 1 :
1
Thus, Theorem 4.3 gives
X
i>0
W (i) = (p ([4]q) ? p ([2]q))W (1 ) + p([3]q)W (21 ):
5
3
Now, [4]q = (1 + q 2)[2]q , so W (2 ;1) is irreducible unless p divides (1 + q 2 ) or (1 + q + q 2 ).
Assume for example that R = Q[q; q ?1] and that p is a cyclotomic polynomial. Then
2
W (2 ;1) ! F (1 ) + F (2 ;1) ; if p = 1 + q 2 ;
2
5
2
since W (1r ) = F (1r ) for all r. Also,
W (2 ;1) ! F (2;1 ) + F (2 ;1) ; if p = 1 + q + q 2 ;
2
3
2
since Theorem 4.3 shows that W (2;1 ) = F (2;1 ) in this case.
3
3
(4.5) Theorem The q{Weyl module W , dened over R=pR, is irreducible if and only if p([hab]q) =
p ([hac]q ) for all nodes (a; b) and (a; c) in [].
25
26
Gordon James and Andrew Mathas
Proof: The Theorem follows at once from Theorem 4.3 since W = F if and only if W (1) = 0.
Next, we convert our theorems on the q {Weyl modules W into results about the dual q {Specht
modules S~. Note that '1!! is an idempotent in SR(n; r) and that
(4:6)
'1!! SR (n; r) '1!! = H
and
'1!! W = S~:
We may therefore apply the \eSe{argument" as in [9, x 6.2] to convert SR(n; r){modules into
H {modules. In particular, if we work over a eld, then
f '1!! F j'1!! F 6= 0 g
is a complete set of irreducible H {modules.
Now, by applying (4.6), we deduce from Theorem 4.3 the q {analogue of Schaper's Theorem [22,
Satz 4.2] for the dual q {Specht modules (we extend the denition of the relation !, and our
conventions on indexing modules by {numbers, to the dual q {Specht modules in the obvious
way).
(4.7) Theorem Let ` r and let 1 > 2 > > n 0 be {numbers for 0. Then, as
HR=pR{modules,
X
i>0
S~(i) !
X
a;b)2[]
a;c)2[]
b<c
(p ([hab]q ) ? p ([hac ]q ))S~(1 ; : : :; b + hac ; : : :; c ? hac ; : : :; n ):
(
(
In order to be able to say something about which dual q {Specht modules are irreducible, we rst
need to investigate the relationship between the simple modules of the q {Schur algebra SR=pR(n; r)
and the simple modules of the Hecke algebra HR=pR(Sr ). Many properties of both of these algebras
do not depend directly on q , but rather upon the related integer e (possibly innite), which is dened
next.
(4.8) Denition Let e be the smallest positive integer such that p divides [e]q; let e = 1 if no
such integer exists.
A partition = (1; 2; : : :) is e{singular if and only if there is an integer i such that i+1 =
i+2 = = i+e > 0; otherwise is e{regular. By convention, all partitions are 1{regular.
The following is known (cf. [5, 7.6]).
(4.9) Suppose that the Hecke algebra H is dened over a eld. Then there exist H {modules D~ such that f D~ j0 is e{regular g is a complete set of irreducible H {modules. Moreover, if 0 is an
e-regular partition then D~ is a composition factor of S~ , appearing with multiplicity 1, and D~ is
a composition factor of S~ only if D .
A q {analogue of the Jantzen{Schaper theorem
27
Comparing (3.5), (4.6), and (4.9), we see that
~ ; if 0 is e{regular ;
1
(4:10)
'!! F = D
0;
otherwise.
(4.11) Lemma The irreducible SR=pR(n; r){module F is isomorphic to a submodule of SR=pR(n; r)
only if 0 is e{regular.
Proof: Let U be an irreducible submodule of SR=pR(n; r), and let u be a non{zero vector in U .
Then '1 u 6= 0 for some j= r. Hence, '1! '1 u 6= 0 (cf. [8, 2.5]), and so '1!! U 6= 0. If F = U,
0
it follows from (4.10) that is e{regular.
This allows us to prove the promised result concerning the irreducible dual q {Specht modules.
(4.12) Theorem Let be a partition of r. Then 0 is e-regular and the dual q{Specht module S~,
dened over R=pR, is irreducible if and only if p ([hab]q ) = p ([hac]q ) for all (a; b) and (a; c) in [].
Proof: If 0 is not e{regular then 0i+1 = = 0i+e > 0i+e+1 for some i; hence, for a = 0i+1 we
have p ([ha;i+1]q ) = p ([e]q ) > 0; but, [ha;i+e ]q = 1. Thus, p ([hab ]q ) = p ([hac ]q ) for all (a; b) and
(a; c) 2 [] if and only if 0 is e{regular and W is irreducible. These conditions in turn imply that
S~ is irreducible by (4.6).
Next, assume that 0 is e{regular and that S~ is irreducible. Let F be a bottom composition
factor of W . Then 0 is e{regular, by Lemma 4.11. Therefore, D~ is a composition factor of S~;
so = and W = F which is irreducible. Hence, the Theorem follows.
Now, S~ is mapped to S 0 (or, rather to a left ideal of H which corresponds to S 0 ), under the
outer automorphism # of H which is given by
T(#i;i+1) = (q ? 1)T1 ? T(i;i+1);
for 1 i < r; see [7]. Also, the bilinear form ; , of (3.24), on the one{sided ideal of SR(n; r)
generated by y0 converts into a bilinear form on the one{sided ideal generated by x . Therefore,
Theorem 4.7 gives the following theorem on q {Specht modules.
(4.13) Theorem Let be a partition of r and suppose that, as HR=pR{modules,
X
i>0
S~(i) !
X
`r
aS~;
where the coecients a are given by Theorem 4.7. Then
X
i>0
S 0 (i) !
27
X
`r
aS 0 :
28
Gordon James and Andrew Mathas
(4.14) Example As in Example 2.31, the calculations in Theorem 4.13 have a pictorial description
in terms of moving hooks in the diagram []; the dierence being that hooks hac and hbc in the
same column must be compared and moved. Let = (22; 1). Then, by taking conjugates in
Example 2.31, we see that
X
i>0
S (i) ! p ([2]q)S (3;1 ) + p ([4]q)S (3;2) ? (p ([4]q) ? p ([3]q))S (5):
2
We obtain from Theorem 4.12 a corresponding result about irreducible q {Specht modules.
(4.15) Theorem Let be a partition of r. Then is e{regular and the q{Specht module S ,
dened over R=pR, is irreducible if and only if p ([hac]q ) = p ([hbc ]q ) for all nodes (a; c) and (b; c)
in [].
It was shown in [6] that the determinant of the Gram matrix, with respect to the standard
basis of S , is given by q a (1r ) (0), for some a 2 Z. From this it is immediate that the condition
given in Theorem 4.15 is sucient to ensure that S is irreducible; until now, we knew of no way
to prove the converse (except in the case where F is a eld of characteristic zero and p = q + 1 in
(4.2); see [14]).
The e{regular partitions which satisfy the conditions of Theorem 4.15 have a particularly nice
form. In order to describe this, we attach a prime number p (or 1), to (R; p) in the denition
below.
If m 2 N, then m1R denotes the element 1R + 1R + + 1R (m times), of R.
(4.16) Denition (i) If char R = 0 and p (m1R) 6= 0 for some m 2 , then let p be the smallest
positive integer with p (p1R) =
6 0. In this case, p is a prime number and p = u(p1R) for some
N
unit u in R.
(ii) If char R = 0 and p (m1R) = 0 for all m 2 N, then let p = 1.
(iii) If char R 6= 0, then let p = char R.
If p is a prime number then p denotes the usual p{adic valuation on N. We adopt the convention
that if p = 1 then p (m) = 0 for all m 2 N.
The next lemma investigates the relationship between p, p, and e.
(4.17) Lemma Let a and b be integers with 1 a r and 1 b r. Then p ([a]q) = p ([b]q) if
and only if either
(i) e - a and e - b; or
(ii) eja and ejb and p (a) = p (b).
Proof: By denition, e is the smallest positive integer such that p divides [e]q . Therefore, if
m 2 N then p divides [m]q if and only if e divides m. It follows that p ([a]q) = p ([b]q) only if either
(i) e - a and e - b, or (ii) eja and ejb.
A q {analogue of the Jantzen{Schaper theorem
29
If e - a and e - b then p ([a]q ) = p ([b]q) = 0, so we may assume that eja and ejb.
First consider the case where p = 1. If m 2 N then [me]q = [e]q [m]qe and [m]qe m1R (mod p)
since p divides q e ? 1. Therefore, p ([me]q ) = p ([e]q ) since p - m1R. Hence, p ([a]q ) = p ([b]q) =
p ([e]q) and the lemma follows in this case.
Now consider the case where p is a prime number. Suppose rst that q = 1R . If char R 6= 0 then
e = char R and so [e]q = 0. However, in (4.2), we have assumed that [m]q 6= 0 for all 1 m r.
Since e r, it follows that char R = 0 if q = 1R . Thus, [a]q = a1R and the lemma holds in this
case also.
Finally, suppose that p is a prime number and that q 6= 1R . Write a = eupx and b = evpy where
p does not divide u or v. Then, since q 6= 1R,
p ([a]q ) = p ([b]q) () p(q a ?x 1R ) = p (q b ? 1R)
(4:18)
() p((qepx ? 1R)[u]qepx ) =y p ((qepy ? 1R)[v]qepy )
() p(qep ? 1R) = p (qep ? 1R);
the last line following because [u]qepx u1R (mod p), since p divides q e ? 1R, and u1R 6 0 (mod p)
(and similarly for [v ]qepy ). Now, if y > x then q epy ? 1R = (q epx ? 1R)[py?x ]qepx . However,
[py?x ]qepx py?x 1R 0 (mod p), so this forces p (q epy ? 1R) > p (q epx ? 1R ). Therefore,
p ([a]q) = p ([b]q) () p (a) = p (b);
completing the proof of the lemma.
Recall the denition of the e{core of a partition from [13, x 2.7].
(4.19) Example Suppose that = (14; 12; 7; 22; 12) and e = 3. We illustrate [], marking those
nodes in its e{core by , and circling those nodes which have hook lengths divisible by e.
We see then from Theorem 4.15 and Lemma 4.17 that S is irreducible for e = 3 if and only if
p 6= 2; 3.
Given partitions = (1 ; 2; : : :) and = (1 ; 2; : : :) and m 2 N, we dene + to be the
partition (1 + 1 ; 2 + 2 ; : : :) and m to be the partition (m1; m2; : : :).
We can now improve upon Theorem 4.15. The reader might nd it useful to refer to the previous
example during the proof,
(4.20) Theorem Let be a partition of r. Then is e-regular and the q{Specht module S ,
dened over R=pR, is irreducible if and only if there exist partitions (1); (2); : : : and (1); (2); : : :
such that
29
30
Gordon James and Andrew Mathas
(1)
(i) = (1) + e (1), (1) is an e{core, and i(1) = i(1)
?1 + e ? 1 for all i with i 6= 0; and
(ii) for all j 1, (j ) = (j +1) + p (j +1) , (j +1) is a p{core, and i(j +1) = i(?j +1)
1 + p ? 1 for all i
(j +1)
with i
6= 0.
Proof: For the duration of the proof, let hab denote the hook length of the (a; b){node in the
diagram [ ].
If e = 1, then S is always irreducible; in this case take (1) = and let the partitions
(2)
; : : :; (1); (2); : : : be partitions of 0. Thus, we may assume that e 6= 1.
By Theorem 4.15 and Lemma 4.17, is e{regular and S is irreducible if and only if the
following is true.
(4.21) For all nodes (a; c) and (b; c) in [] either e - hac and e - hbc , or ejhac and ejhbc and
p(hac) = p(hbc ).
We shall prove that (4.21) is equivalent to (i) and (ii) from the statement of the theorem.
Draw the diagram [] and circle those nodes which have hook lengths divisible by e, as in
Example 4.19. The condition that for all (a; c) and (b; c) 2 [] either e - hac and e - hbc , or ejhac
and ejhbc is readily seen to be equivalent to condition (i) in Theorem 4.20 (remove a skew e{hook
from the last row in [] which contains a circled node and apply induction). Assume therefore that
condition (i) holds.
Suppose that the (a; c){node of [] is circled (ie. e divides hac ), and that (a; c) lies in the k{th
column of circled nodes. Then hac = ehak . Therefore, p (hac ) = p (hbc ) for all b with (b; c) 2 []
if and only if p (hak ) = p (hbk ) for all b with (b; k) 2 [ (1)]. We now iterate this argument,
replacing by (1), and e by p, to deduce that (4.21) is equivalent to (i) and (ii) in the statement
of the theorem.
(1)
(1)
(1)
Now, the proof of [6, Theorem 2.14], which states that the centre of the Hecke algebra consists
of symmetric polynomials in the q {Murphy operators, contains a gap. Although we believe that
this result is correct, a consequence of the gap is that the proof of the Nakayama conjecture for
Hecke algebras [6, Theorem 4.13] is incomplete. We therefore provide a proof of the Nakayama
conjecture in Theorem 4.29 below.
First we classify the blocks of the q {Schur algebra. Now, by (3.5) the following holds.
(4.22) If P`r a W ! 0 then every integer a is zero.
Therefore, by applying a suitable sum of block idempotents, we deduce the next result.
(4.23) Suppose that U is a non{zero SR=pR(n; r){module with U ! P`r a W and suppose
that all of the composition factors of U lie in the same block. Then, W and U lie in the same
block if a 6= 0.
(4.24) Theorem Suppose that and are partitions of r. Then the SR=pR(n; r){modules W and W belong to the same block if and only if and have the same e{core.
A q {analogue of the Jantzen{Schaper theorem
31
Proof: If [ ] is a diagram which is obtained by unwrapping a skew hook of length divisible by
e from [], and wrapping the skew hook on somewhere else, then and have the same e{core.
Therefore, by Theorem 4.3, the composition factors of W , besides F , are composition factors
of q {Weyl modules W where and have the same e{core. (See the remark which follows
Corollary 2.30.) Thus, W and W belong to the same block only if and have the same e{core.
Now suppose that is an e{core of some partition of r, say ` r ? we. Let = (1; 2; : : :) be
the partition of r for which 1 = 1 + we and i = i for i > 1. Then has e{core .
Assume that ` r and that has e{core . To prove the theorem it suces to show that W and W are in the same block. Suppose that =
6 . Then, for some (a; b) the (a; b){skew hook
of [] has length e and has its foot node in a row later than the rst row. Remove this skew hook
from [] and add it back at the end of the rst P
row of [], to obtain the diagram [], say. Then the
expression which appears in Theorem 4.3 for i>0 W (i) includes W . Therefore, W and W belong to the same block by (4.23). By induction, W and W are in the same block, so W and
W are also in the same block.
To prove that the H -modules S and S belong to the same block if and only if and have
the same e{core is more tricky because we cannot replace q {Weyl modules by q {Specht modules in
(4.22). For example, it is known [12, 6.5], that
(4:25)
e?1
X
x=0
(?1)xS (e?x;1x ) ! 0:
However, (4.9) shows that the following is true.
(4.26) Suppose that P`r a S ! 0 where a = 0 for every e{singular . Then a = 0 for all
.
Therefore, as in (4.23), we obtain the following.
(4.27) Suppose that U is a non{zero HR=pR;q(Sr){module with U ! P`r a S and suppose
that all of the composition factors of U lie in the same block and that a = 0 for every e{singular
partition . Then S and U lie in the same block if a 6= 0.
The following generalization of (4.25) comes from applying the Littlewood{Richardson Rule
to combine the partition with the linear combination of partitions of e which appear in (4.25)
(cf. [10, Remark p. 84]).
(4.28) Let be a partition of r ? e. Then P S ! 0, where the sum is over all partitions where [ ] is obtained by wrapping a skew e{hook onto [], and the sign is positive if and only if the
leg length of the wrapped on skew hook is even.
We now prove the Nakayama conjecture for q {Specht modules.
(4.29) Theorem Suppose that and are partitions of r. Then the HR=pR(Sr ){modules S and
S belong to the same block if and only if and have the same e{core.
31
32
Gordon James and Andrew Mathas
Proof: It follows from (4.6) and Theorem 4.24 that S and S belong to the same block only if
and have the same e{core. Indeed, this implication is proved in [6, 4.4]. We concentrate upon
the converse.
Suppose that is an e{core of some partition of r, say ` r ? we. Let = (1; 2; : : :) be the
partition of r for which 1 = 1 + we and i = i for i > 1. Then has e{core .
Assume that ` r has e{core , where 6= . We prove that S and S are in the same block,
which will complete the proof of the theorem. Note that if is a partition of r with e{core then
D . Therefore, by induction, it suces to show that there exists a partition with e{core such that S and S are in the same block and B .
First, suppose that S is reducible. Then S shares a composition factor with some S where
is e{regular and B . Moreover, has e-core because S and S are in the same block; so
we may take = .
Next, if is e{singular and S is irreducible then S = D for some e{regular partition such
that B . Therefore, and have the same e{core and again we are done.
It remains to consider the case where is e{regular and S is irreducible. Such partitions are
described explicitly in Theorem 4.20; in particular, for all (a; b) and (b; c) 2 [], either e - hac and
e - hbc , or ejhac and ejhbc . (A typical partition of this kind is illustrated in Example 4.19; the reader
may care to keep this example in mind for the remainder of the proof.)
Let a be the largest integer such that ejhac for some c, and let b = a ? e + 1. Then hab = e
and ejhib for 1 i a; so
(4:30)
i i+1 ? 1 (mod e)
for 1 i a ? 1:
Let [] be the diagram obtained from [] by removing the (a; b){skew hook; so i = i for i 6= a
and a = a ? e. (In Example 4.19, a = 3 and = (14; 12; 4; 22; 12).)
As in (4.28), we now wrap a skew e{hook onto [] in all possible ways. Our rst aim is to show
that (except in one special case), the only one way to do this and obtain an e{singular diagram is
to add a vertical e{hook to the end of rst column of [].
First, recall that the only way to add a skew e{hook to an e{core to obtain an e{singular
diagram is to add a vertical e{hook to the end of the rst column (cf. [21]). A more general result
is the following, which can be proved by removing the rst part from [ ].
(4.31) Suppose that is an e{core and that is a partition such that 1 = 1 + ke, for some
non{negative integer k, and i = i for all i > 1. Suppose that is an e{singular partition whose
diagram is obtained by wrapping a skew e{hook onto [ ]. Then, either the skew e{hook was added
vertically to the end of the rst column of [ ], or e = 2, = (2), and = (2; 2).
We return now to the situation in hand. Assume that is an e{singular partition which is
obtained by wrapping a skew e{hook into [], and suppose that the hand of the wrapped on skew
e{hook lies in row x. We have that i i+1 ? 1 (mod e) for 1 i a ? 1 by (4.30), and
this implies that x a. Let = (a ; a+1 ; : : :). Then is a partition of the kind considered in
(4.31); also, a + e < a?1 . Therefore, either the skew e{hook which was added to [] to obtain [ ]
consisted of e nodes, placed vertically below the rst column of [], or e = 2, a = 4, a+1 = 0 and
= (1; : : :; a?1; 2; 2). So, we have proved the following.
(4.32) Unless e = 2, a = 4, and a+1 = 0, there is exactly one way to wrap a skew e{hook onto
[] to obtain an e-singular diagram.
33
A q {analogue of the Jantzen{Schaper theorem
Assume, for the moment, that we are not in the case where e = 2, P
a = 4, and a+1 = 0.
Then (4.28) and (4.32) give us a linear combination of q {Specht modules a S such that there
is a unique e{singular partition, say, with a 6= 0. Moreover, a = 1 and a = 1 where
= (1 + e; 2; : : :; a ? e; : : :): We can therefore apply (4.27), with U = S , to deduce that S and S are in the same block. Since B , this completes the proof in this case.
Finally, suppose that e = 2, a = 4, and a+1 = 0; we again show that and , where is
as above, are in the same block. Here, we can add a vertical 2{hook to [] in only
one way; we
P
2
obtain the diagram [ ] where = (1; : : :; a?1; 2; 1 ). Therefore, (4.28) gives a S ! 0,
where a = ?1, a = 1, a = 1, and a 0 for all other partitions . Hence, S and S have a
common composition factor, as do S and S . This again implies that S and S are in the same
block, and so completes the proof.
In conclusion, we classify the ordinary irreducible representations of the nite general linear
group GLn (q ) which remain irreducible modulo a prime number p which does not divide q . To
state and prove our theorem we adopt the notation of [4]. In particular, the ordinary irreducible
GLn(q){modules have the form S (I ) where I is an (n; 1){index [4, Denition 2.4].
(4.33) Theorem Let S (I ) be an ordinary irreducible GLn (q){module, where
I = ds 1 :: :: :: ds N k1(1) :: :: :: kN(N ) :
1
N
Suppose that p is a prime number which does not divide q . Then the reduction of S (I ) modulo p is
irreducible if and only if the following two conditions hold.
(i) If i 6= j and di = dj then the p{regular parts of si and sj are roots of distinct irreducible
polynomials over GF (q ).
(ii) For all i and for all nodes (a; c) and (b; c) in [(i)], p ([hac ]qdi ) = p ([hbc ]qdi ):
di ki . The module S (I )
Proof: First consider the GLdi ki (q ){modules S (Ii) where Ii =
i
si (i)
is irreducible modulo p if and only if the decomposition numbers d0 i 0 of the q di {Schur algebra are
zero for all 6= (i) by [8, Theorem 4.6]. Thus, Theorem 4.5, applied to the case where R = [q ?di ],
( )
Z
shows that condition (ii), in the statement of the theorem, is equivalent to the statement that all
of the modules S (Ii ), with 1 i N , are irreducible modulo p.
Let I(p) be the (n; p)-index which is given by
I(p) = ds01 :: :: :: ds0N k1(1) :: :: :: kN(N ) ;
1
N
where s0i is the p{regular part of si . Then S (I(p)) is the reduction modulo p of S (I ). The module
S (I(p)) is irreducible only if S (Ii ) is irreducible modulo p for all i; that is, condition (ii) holds.
Assume now that condition (ii) holds. Let e(di ) be the smallest positive integer such that p
divides [e(di)]qdi . Then the partition (i) is e(di ){regular, for all i, and S (I(p)) = D(I(p)) in the
notation of [4]. If condition (i) holds, then I(p) is a special foot index [4, 2.9]; so D(I(p)) is irreducible
by [4, 4.9]. If condition (i) does not hold, then we may use the Littlewood{Richardson Rule to
33
34
Gordon James and Andrew Mathas
express S (I(p)) as a linear combination of modules S (J ) where more than one (n; p){index J occurs
in the expression (cf. case 1 of the proof of [4, 6.2]); in particular, S (I(p)) is reducible.
An example where condition (i) of Theorem 4.33 fails is provided by GL6(2) with s1 a root of
1 + x + x3 , s2 a root of 1 + x2 + x3 , and
I=
3 3 1 1
s1 s2 (1) (1) :
If p = 7 then 1 is the p{regular part of s1 and s2 ; so S (I ) is an ordinary irreducible GL6(2){module
which is reducible modulo p.
We note that the special case of Theorem 4.33 where
I=
1 n
1 (so that S (I ) is a unipotent module), was conjectured in [11, 20.5]; the numerical condition which
appears there is equivalent to the condition that p ([hac ]q ) = p ([hbc ]q ); for all nodes (a; c) and (b; c)
in [], by Lemma 4.17.
Acknowledgements
We thank the referee and Jie Du for their comments on the manuscript. Also, our understanding
of the action of the operators L;m in the case where q = 1 was greatly enhanced by computations
made using the computer algebra package Gap [23].
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Department of Mathematics, Imperial College, Queen's Gate, London SW7 2BZ
e-mail: g.james@ic.ac.uk and a.mathas@ic.ac.uk
35