CHERN CLASSES AND EXTRASPECIAL GROUPS OF
ORDER p5
DAVID J. GREEN
Abstract
A presentation is obtained for the Chern subring modulo nilrad-
ical of both extraspecial p-groups of order p5, for p an odd prime.
Moreover, it is proved that, for every extraspecial p-group of expo-
nent p, the top Chern classes of the irreducible representations do not
generate the Chern subring modulo nilradical. Finally, a related ques-
tion about symplectic invariants is discussed, and solved for Sp4(Fp).
The main innovation in this work is to consider extraspecial groups
as central products, and to partition the maximal elementary abelian
subgroups of the central product into those which lift to abelian sub-
groups of the corresponding direct product, and those which do not.
Introduction
The author and I. J. Leary argued in [5 ] that, when trying to determine the
mod-p cohomology ring of a p-group, the subquotient obtained by factoring
out the nilradical and then taking the subring generated by Chern classes
is a natural and worthy object of study. In contrast to the partial results
of a general nature obtained in that paper, the current article presents a
complete determination of this subquotient for the extraspecial p-group p1+4+
of order p5 and exponent p, for p an odd prime.
Once more, the theorem of Quillen that characterizes nilpotent elements
by their restrictions to elementary abelian subgroups will play a fundamental
role. The key insight of the current work is that, by employing a suitable
partition of the set of maximal elementary abelian subgroups, the search for
_________________________________________________________________________
Submitted 7 June 1995.
1991 Mathematics Subject Classification 20J06.
The author gratefully acknowledges the support of the Max Kade Foundation.
1
relations in the Chern subring may be pursued in a methodical fashion. This
partition is related to the fact that the extraspecial p-group in question is
the central product of two copies of the extraspecial group p1+2+of order p3
and exponent p.
This technique lends itself to a number of generalisations that could be
used to launch an attack by induction on the Chern subrings of all extraspe-
cial groups. It is to be hoped that further research will identify the correct
generalisation for this task.
Much of the paper is in fact taken up in solving a related problem in pure
algebra.
Problem. For an odd prime p and a positive integer n, let En be a
2n-dimensional Fp-vector space, carrying a nondegenerate symplectic bilinear
form. Let K0, : :,:Kn-1 be indeterminates, and let Fn denote the polynomial
algebra S(E*n) Fp Fp[K0; : :;:Kn-1].
For each maximal totally isotropic subspace I of En, there is then a unique
algebra homomorphism qI: Fn ! S(I*) which behaves on E*nas the restric-
tion map E*n! I*, and sends Kr to Dr(I*), the Dickson invariant in the
(pn - pr)th symmetric power of I*.
The intersection of the kernels of all the qI is an ideal in Fn; define Qn
to be the corresponding quotient algebra. Give a presentation for Qn.
A presentation for Q2 is achieved in Theorem 17. After that, we shall
prove in Theorem 23 that the Chern subring modulo nilradical for p1+4+is
isomorphic to Q2 Fp Fp[Z] for an indeterminate Z in degree 2n. Together,
these constitute the main result of the paper.
Afterwards, we will proceed with some further applications of this infla-
tion technique. Tezuka and Yagita studied the ring generated by the top
Chern classes of the irreducible representations of p1+2n+. This is contained
in the Chern subring modulo nilradical, and it is shown in Theorem 20 that
this containment is strict. After this, we discuss a problem about invariants
of the symplectic group which is related to the cohomology of extraspecial
groups. Last of all, the Chern subring modulo nilradical is obtained for the
extraspecial p-group p1+4-of order p5 and exponent p2.
It is obtaining the presentation for Q2 which requires the most work.
Define T2 to be the image in Q2 of S(E*2). The structure of T2 is known from
the work of Tezuka and Yagita. We need to understand the ring generated
by T2 and the images 0, 1 of K0, K1 respectively. Consider E2 as the or-
thogonal direct sum E0 ? E00of two nondegenerate 2-dimensional symplectic
2
spaces. Partition the set of maximal totally isotropic subspaces I of E into ,
the set of those I which decompose as I = (I \ E0) (I \ E00); and , the
set of those I which do not. Associated to this partition, an "inflation" map
Q2 ! Q1 Q1 is constructed in Lemma 3. In Proposition 4 it is shown how
to reduce questions about Q2 to questions about Q1 Q1. A key step in
the proof of this result is Lemma 6, which is in some sense the core of the
paper. It describes how T2 would look if Q2 had been defined using only
the maximal totally isotropic subspaces from , or only the maximal totally
isotropic subspaces from . After this, obtaining a presentation for Q2 is
merely a matter of hammering the relations out, although some of these are
rather complicated.
The author wishes to thank J. L. Alperin and I. J. Leary for encourage-
ment and helpful conversations.
1. A regular sequence
Let b denote the symplectic form En Fp En ! Fp, and denote by q the
quotient map Fn ! Qn. Then, for each maximal totally isotropic subspace
I of En, there is a unique map ^qI:Qn ! S(I*) such that ^qIq = qI.
Write Tn for the image of S(E*n) under q. Pick a symplectic basis A1, : :,:
An, B1, : :,:Bn for En: so Ai ? Aj, Bi ? Bj and b(Ai; Bj) = ffiij. Take the
corresponding dual basis A*1, : :,:A*n, B*1, : :,:B*nfor E*n. Define elements
of Tn by ffi = q(A*i) and fii = q(B*i). Then ff1; : :;:ffn,rfi1;r: :;:fin gener*
*ate Tn
as an Fp-algebra.rrDenote by Rr(En) the element A*1B*1p- A*1pB*1+ . .+.
A*nB*np- A*npB*nof S(E*n). If the vector space En is clearly determined by
the context, this notation will be shortened to Rr. Note that Rr equals the
b(v; F r(v)) of [3 ].
Theorem 1. (Tezuka-Yagita) The sequence R1; : :;:Rn in S(E*n) is a
regular sequence. The ideal generated by these elements contains Rr for all
r > 1, and is the kernel of the surjection S(E*n) ! Tn.
Proof. See Proposition 8.2 of [3 ]. The h of that paper is defined to be the
codimension in En of a maximal totally isotropic subspace, and so takes the
value n here. The first two parts are explicitly stated. The last part follows,
using the Nullstellensatz, from the fact that the ideal is radical and from the
description of the variety in terms of isotropic subspaces.
3
2. Dickson invariants
We now recall the salient facts about the Dickson invariants. See Chapter 8
of Benson's book [2 ] for proofs.
Let V be an m-dimensional Fp-vector space. For each 0 6 r 6 m - 1,
there is a Dickson invariant Dr(V ) in the (pm - pr)th symmetric power of V
such that
Y m m-1X r
(X - v) = Xp + (-1)m-r Dr(V )Xp in S(V )[X]. (1)
v2V r=0
The natural action of GL (V ) on S(V ) has as ring of invariants the polynomial
algebra Fp[D0; : :;:Dm-1 ].
In several papers, Dr(V ) is denoted cm;r. There are two reasons for using
nonstandard notation: to avoid a clash with the standard notation for Chern
classes, and to specify V explicitly.
Each Dickson invariant is a polynomial in the elements of any basis for V .
These polynomials are independent of the choice of basis, and in fact depend
only on p, r and m, since the Dr are invariant under GL (V ). If w1; : :;:wm
are elements of an Fp-vector space W , and r < m, then Dr(w1; : :;:wm ) shall
denote the evaluation at (w1; : :;:wm ) of the polynomial for Dr(Fmp). The
following lemma describes how this will be related to the Dickson invariants
of the space spanned by the wi. It takes a particularily elegant form when
we work with dual spaces.
Lemma 2. Let V be an m-dimensional Fp-vector space, and U a subspace
of codimension `. Then, for every 0 6 r 6 m - 1,
ae * p`
ResU (Dr(V *))= Dr-`(U ) if ` 6 r, and
0 otherwise.
Proof. Obvious from the definition of Dr.
This paper is particularily concerned with small cases. The first three
Dickson invariants are easily calculated: D0(w) = wp-1 and
2-1 p2-1
wp1 - w2
D0(w1; w2) = (w1w2 p- w1pw2)p-1 D1(w1; w2) = ______________p-1p-1:(2)
w1 - w2
4
Note in particular that if we take for any m-dimensional V a non-zero ele-
ment of each 1-dimensional subspace, and multiply these together, we get an
element of S(V ) which is well-defined up to a scalar, and has (p - 1)th power
D0(V ). If m = 2 and V has basis v1, v2, then this element is v1v2p- v1pv2.
In Qn, define n;r= q(Kr) for 0 6 r 6 n - 1. Then the ffi, fij and n;r
together generate the Fp-algebra Qn. Note that the algebraic independence
of the Dickson invariants, together with the definition of q, ensures that
the n;rnot only are algebraically independent over Fp in Qn, but also, no
polynomial in them over Fp is a zero divisor in Qn. For small values of n we
will abbreviate n;r, denoting 1;0, 2;0, 2;1by , 0, 1 respectively.
3. Partition
The object of this section is to prove a result (Proposition 4) which allows
the search for relations in Q2 to be carried out in Q1 Q1. This is achieved
by partitioning the set of maximal totally isotropic subspaces of E2 into two
families, and for each family, determining which elements of Q2 it fails to
detect. This partition can in fact be performed in En, and so we will only
restrict ourselves to E2 when this becomes necessary.
Suppose that n = ` + m. Then En is the orthogonal direct sum En =
E` ? Em of nondegenerate symplectic spaces E`, Em`. Partition the set of
maximal totally isotropic subspaces I of En as , where I 2 if and
only if I is the direct sum of (necessarily maximal) totally isotropic subspaces
of E` and Em .
Lemma 3. Let n = ` + m. Then the isomorphism S(E*n) ~= S(E*`)
S(E*m) induces an inflation homomorphism ss*: Qn ! Q` Qm such that,
for any x 2 Qn, we have ss*(x) = 0 if and only if ^qI(x) = 0 for all I 2 .
Proof. As Fnp~=F`pFmpand GL `(Fp)xGL m (Fp) 6 GL n(Fp), each Dr(Fnp)
is a polynomial in the Ds(F`p) and the Dt(Fmp). Define ss*(n;r) to be the
corresponding polynomial in the `;s1 and the 1m;t. The rest is obvious.
Remark. The above partition is related to the fact that the extraspecial
group Pn is the central product P` * Pm . The inflation map corresponds to
the cohomology inflation from Pn to P`x Pm .
5
Definition. For a polynomial f 2 T2[x0; x1] with coefficients in T2,
define e (f) 2 Q2 by
e (f) = f(0; 1) - f(D0(ff1; fi1); D1(ff1; fi1)) ;
and define fl2 to be the element D1(ff1; fi1) - D1(ff2; fi2) of T2.
Proposition 4. Let f 2 T2[x0; x1] be a polynomial with coefficients
in T2. Then the element f(0; 1) of Q2 belongs to T2 if and only if there ex-
ists t 2 T2 such that the equation e (f) = tfl2 holds after inflation to Q1Q1.
If such t does exist, then e (f) = tfl2 holds in Q2.
Before proving this proposition, we shall establish two auxilliary results.
We shall work with the partition associated to the orthogonal direct sum
decomposition En = E1 ? En-1, where n > 2 and E1 has basis A1, B1.
Lemma 5. For n > 2, let I be a maximal totally isotropic subspace of En.
Then I 2 if and only if the restrictions of ff1 and fi1 are linearly independ-
ent in I*.
Proof. Obvious.
By the definition of Qn, the maximal totally isotropic subspaces in and
combined detect every non-zero element of Qn. We shall now determine
the ideals of elements which and individually fail to detect. Recall that
Tn ~=S(E*n)=(R1; : :;:Rn).
Lemma 6. 1. In S(E*n)=(R1; : :;:Rn-1), there is a unique fln such that
(ff1fi1p- ff1pfi1)fln = Rn. In particular, fl2 = D1(ff1; fi1) - D1(ff2; fi*
*2),
as in the definition preceding Proposition 4.
2. Consider the ideal in Tn of classes whose image under ^qIis zero for
every I 2 . It is the principal ideal generated by ff1fi1p- ff1pfi1.
3. The corresponding ideal for is also principal; it is generated by fln
(considered as an element of Tn).
Proof. Write bTfor S(E*n)=(R1; : :;:Rn-1). For part 1, we have Rr(En) =
Rr(E1) + Rr(En-1). By the Tezuka-Yagita theorem for En-1, it follows
that Rn(En) lies in the ideal in S(E*n) generated by the Rr(En) for r < n, and
6
the Rr(E1) for r 6 n. Now apply the Tezuka-Yagita theorem again, this time
for E1. Hence (the image of) Rn(En) lies in the principal ideal of bTgenerated
by ff1fi1p- ff1pfi1. Therefore fln exists; it is unique since Rn(En) is a non-z*
*ero
2 p2
divisor in bT. Observing that (ff1fi1p- ff1pfi1)D1(ff1; fi1) = ff1fi1p- ff1 fi*
*1,
we can verify the equation for fl2.
For part 2, observe first that ^qI(ff1fi1p- ff1pfi1) is zero in S(I*) for ev*
*ery
I 2 , and non-zero for every I 2 . Since S(I*) is an integral domain, it
follows that ^qI(fln) = 0 in S(I*) for every I 2 . Therefore, if t 2 Tn satisfi*
*es
^qI(t) = 0 for every I 2 , then tfln = 0 in Tn.
Now pick ^t2 bT lying above t. Then for some s 2 bT, ^tfln = sRn(En).
Multiplying both sides by ff1fi1p- ff1pfi1 and rearranging then yields (^t-
s(ff1fi1p- ff1pfi1))Rn = 0. Since Rn is a non-zero divisor, ^tlies in the ideal
of bT generated by ff1fi1p- ff1pfi1, proving part 2. The same method works
for part 3.
Remark. It would appear that the main obstacle to obtaining a present-
ation for Qn is the current lack of a version of Lemma 6 for general n.
We can now proceed with the proof of Proposition 4.
Proof of Proposition 4.For r = 0 or 1, and I 2 , the images under ^qI
of r and Dr(ff1; fi1) both equal Dr(I*). So for every I 2 , the images
under ^qIof f(0; 1) and f(D0(ff1; fi1); D1(ff1; fi1)) must be equal. Therefore
if f(0; 1) 2 T2 then there exists t 2 T2 such that
e (f) = tfl2 : (3)
Conversely, suppose that there is a t 2 T2 such that Eqn. (3) holds after ^qI
for every I 2 . Then Eqn. (3) holds in Q2, since each side of the equation
is in ker^qIfor every I 2 . Hence f(0; 1) 2 T2.
4. Technical results
In light of Proposition 4, we want a presentation for Q1. This has generat-
ors ff1, fi1 and 1;0. We shall drop the subscripts from these generators.
Proposition 7. The Fp-algebra Q1 is generated by ff, fi and ; a suffi-
cient set of relations is 2 = ff2(p-1)- ffp-1fip-1 + fi2(p-1), ff - ffp = 0, and
fi - fip = 0.
7
Proof. These relations are easily verified after every ^qI. Note that they
imply the relation fffip - ffpfi. Therefore by Theorem 1, all relations in T1
are present. It only remains to prove that does not lie in T1. To see this,
2 p-1 p(p-1)
note that p = ffp(p-1)- ff(p-1)fi + fi . This is not the p-th power of
any polynomial in ff and fi.
In Q1 Q1, we need to be able to distinguish the of the first factor
from that of the second. Write 0 , 00 for 1, 1 respectively. Let T1;1
denote the subring T1 T1 of Q1 Q1. Both T2 and T1;1have generators ff1,
ff2, fi1, fi2. In T1;1however, the relations are generated by ff1fi1p- ff1pfi1 *
*and
ff2fi2p- ff2pfi2.
Lemma 8. Let f 2 T1;1[y1; y2] be a polynomial without constant term.
Then f(0 ; 00) lies in the sub T1;1-module of Q1 Q1 generated by 0 and
00 if and only if the coefficients in f of both y1y2 and y21y22are without zero
degree term.
Proof. We have 0 r00s2 T1;1if and only if neither r nor s is one. If we
remove, for all such (r; s), both yr+11ys2and yr1ys+12from the set of monomials
in y1, y2, then we are left with 1, y1y2 and y21y22. If ffi is one of the gener*
*ators
of T1;1, then ffi0 00 and ffi0 2002 lie in the submodule in question. So it rem*
*ains
to show that 0 00 and its square do not. As 0 00 lying there would imply
that 0 2002did too, it is enough to show that 0 2002is not in the T1;1-module
generated by 0 and 00.
Suppose that 0 2002 = g0 + h00, with g, h 2 T1;1. Now 0 2002 involves
ffp-11ffp-12fip-11fip-12, and so g0 + h00 must too. But every term in g0 must
involve either ffp1or fip1, and every term in h00 must involve either ffp2or fi*
*p2.
Since the only relations in T1;1are ff1fip1= ffp1fi1 and ff2fip2= ffp2fi2, we h*
*ave
derived a contradiction.
Lemma 9. Suppose that u, v 2 T1;1satisfy u0 = v00 . Then u - v lies in
the ideal in T1;1generated by ff1ff2p- ff1pff2, ff1fi2p- ff1pfi2, fi1ff2p- fi1p*
*ff2
and fi1fi2p- fi1pfi2.
Proof. Each of the four elements u - v listed in the statement satisfies
u0 = v00 , and so lies in the ideal in question. Conversely, define a monomial
ffr11ffr22fis11fis22to be admissible if both s1 < p unless r1 = 0, and s2 < p
unless r2 = 0. Then the admissible monomials form a basis for the Fp-vector
space T1;1. In particular, u and v may be expressed in terms of this basis.
8
Since u0 = v00 with both u and v 2 T1;1, it follows that u0 and v00 lie
in T1;1. Hence every admissible monomial in u has either r1 or s1 positive,
and every admissible monomial in v has either r2 or s2 positive. Then u0
is obtained from u as follows: the coefficients remain the same, and each
monomial has r1 increased by p - 1, unless r1 = 0, in which case s1 is
increased by p - 1. Similarily, v00 is obtained from v by increasing r2 or
s2. Since u0 = v00 , there is an induced bijection between the admissible
monomials in u and in v, and the difference between any admissible monomial
in u and the corresponding monomial in v is divisible by one of the four
elements listed in the statement.
Lemma 10. Let f 2 T2[x0; x1]; then, in Q1 Q1,
0 00 0 00 p-1 00 0 00 p-1 0 p 0p
ss*(e (f)) = f ( - ) ; ( - ) + - f(0; ) :
Moreover, ss*(fl2) = (0 - 00)p.
Proof. The maps Q1 Q1 ! S(I*), for all I 2 , detect the elements
of Q1 Q1. Using Lemma 6 and Eqn. (2), it is straightforward to verify
that the equations ss*(fl2) = (0 - 00)p, ss*(0) = 0 00(0 - 00)p-1, ss*(1) =
(0 p+1- 00p+1)=(0 - 00), ss*(D0(ff1; fi1)) = 0 and ss*(D1(ff1; fi1)) = 0 phold
after mapping to any such S(I*).
5. Deriving the relations
We are now in a position to derive the relations in Q2.
Lemma 11. The generator 1 does not belong to T2, but its pth power
does:
Xp
p1= D1(ff1; fi1)p-iD1(ff2; fi2)i : (4)
i=0
Proof. Inflate to Q1 Q1. Now, s t belongs to T1;1if and only if
neither s nor t equals 1. Then ss*(1) equals p-1 + p-1 modulo T1;1,
by Lemma 10. So 1 62 T2.
If I 2 then ^qI(D1(ffi; fii)) = D1(I*) for i = 1; 2. Also, ss*(D1(ffi; fii)*
*) is
0p, 00p for i = 1; 2 respectively. This establishes Eqn. (4).
9
We now turn our attention to the T2-submodule of Q2 generated by 1
and 1.
Proposition 12. Let f 2 T2[x0; x1] be a polynomial with coefficients
in T2. The element f(0; 1) of Q2 belongs to T21 + T2 if and only if the
coefficients in f of x0 and, if p = 3, of x21have no degree zero part.
Proof. By Proposition 4, f(0; 1) belongs to T21+T2 if and only if there
exist a; b 2 T2 such that the equation
e (f) = afl2 + be (x1)
holds after inflation to Q1 Q1. Lemma 10 says, that this is the case if and
only if there exist a0; b02 T1;1such that, in Q1 Q1,
0 00 0 00 p-1 00 0 00 p-1 0 p 0p
f ( - ) ; ( - ) + - f(0; )
= a0(0 - 00)p + b000(0 - 00)p-1 :
Observe that both sides of the equation are divisible by (0 -00 )p-1. Perform-
ing this division on the right hand side yields a00 + (b0- a0)00. By Lemma 8,
this means that such a0, b0exist if and only if the coefficients of both y1y2 a*
*nd
y21y22in (f(y1y2(y1 - y2)p-1; y2(y1 - y2)p-1 + yp2) - f(0;=yp1))(y1-y2)p-1 have
no zero degree term. This happens exactly when the coefficients in f(x0; x1)
of x0 and, if p = 3, of x21, have no zero degree part.
Remark. There now remain so few elements outside T21+T2, that their
degrees distinguish them.
2
Lemma 13. Let ffi 2 q(E*2). Then ffi0 = ffip1 - ffip .
Proof. Apply q^Ifor any maximal totally isotropic subspace I. This
sends ffi to some element of I*, and r to Dr(I*). Observe that the left
hand side of Eqn. (1) vanishes whenever X is an element of V .
Proposition 14. Let ffi1; ffi2 2 E*. Then
(ffi1ffi2p- ffi1pffi2) (1 - D1(ffi1; ffi2))= 0 :
Proof. Let I be a maximal totally isotropic subspace of E2. If ^qI(ffi1),
^qI(ffi2) are linearly independent in I*, then 1 and D1(ffi1; ffi2) both map to
D1(I*) under ^qI. If they are linearly dependent, then ^qI(ffi1ffi2p- ffi1pffi2*
*) = 0.
10
We can now derive a presentation for the T2-module generated by 1
and 1.
Lemma 15. The following ideals in T2 are equal:
o The ideal J1 of all u 2 T2 such that u1 2 T2.
o The ideal J2 in T2 generated by ff1ff2p- ff1pff2, ff1fi2p- ff1pfi2, fi1ff2*
*p-
fi1pff2, fi1fi2p- fi1pfi2 and ff1fi1p- ff1pfi1. Note that the last generat*
*or
equals -(ff2fi2p- ff2pfi2).
Proof. By Proposition 14, J2 J1. We shall show that J1 J2. First
we reduce this to a problem in Q1 Q1. Let u 2 J1. Then u1 2 T2, so
by Proposition 4, u(1 - D1(ff1; fi1)) = vfl2 for some v 2 T2. Now inflate to
Q1 Q1. We get ss*(u)00(0 - 00)p-1 = ss*(v)(0 - 00)p. Since (0 - 00)p-1 is
a non-zero divisor in Q1 Q1, we cancel and rearrange to get ss*(u + v)00 =
ss*(v)0 . By Lemma 9, it follows that ss*(u) lies in the ideal in T1;1generated
by the images under ss* of the first four generators of J2. Since the kernel
in T2 of inflation is principal, generated by the fifth generator of J2, we are
done.
The following lemma will help us to describe some elements of T2 involved
in relations in Q2.
Lemma 16. Suppose that u 2 Q2 satisfies q^I(u) = 0 for all I 2 .
Then for each t 2 T1;1, there is a unique v 2 Q2 such that ^qI(v) = 0 for
all I 2 and ss*(v) = tss*(u). It therefore makes sense to refer to v as
tu. In particular, this result holds for u = fl2, and for u = e (f) for any
f 2 T2[x0; x1].
Proof. The inflation map T2 ! T1;1is surjective. Pick any ^t2 T2 such
that ss*(^t) = t. Then ^tu satisfies the requirements on v. The uniqueness part
follows from Lemma 3 and the definition of Qn.
We can now put the above results together to obtain a presentation for Q2.
Define polynomials f1, f2 2 Fp[y1; y2] to be y21(yp+11-yp+12)=(y21-y22), respec*
*t-
ively y21y22(yp-31- yp-32)=(y1 - y2) + 2yp1+ yp-21y22+ 2yp2.
11
Theorem 17. A presentation for the commutative Fp-algebra Q2 con-
sists of six generators ff1, ff2, fi1, fi2, 0, 1, together with relations as fo*
*l-
lows:
ff1fi1p- ff1pfi1 + ff2fi2p- ff2pfi2 = 0 (5)
2 p2 p2 p2
ff1fi1p- ff1 fi1 + ff2fi2 - ff2 fi2 = 0 (6)
(ffi1ffi2p- ffi1pffi2) (1 - D1(ffi1; ffi2))= 0 for ffi1; ffi2 2({ff1;7fi1*
*;)ff2; fi2}
2
ffi0 - ffip1 + ffip = 0 for ffi 2 {ff1; fi1; ff2; fi2}(8)
Xp
p1= D1(ff1; fi1)p-iD1(ff2; fi2)i (9)
i=0
e (x20) = 0 2002(0 - 00)p-2fl2 (10)
e (x0x1) = (f1(0 ; 00) + f1(00; 0))e (x1) + f1(00; 0)fl2 : (11)
If p > 3 then
e (x21) = f2(0 ; 00)e (x1) + (0 p-2002 + 00p)fl2 (12)
If p = 3, then Eqn. (12) is replaced by the relations
e ("1x21) = ("1003 + 2"31002+ 2"71)e (x1) + "31002fl2 for "1 = ff1;(fi113)
e ("2x21) = (2"20 3+ "3202+ 2"72)e (x1) + "72fl2 for "2 = ff2; fi2.(14)
In fact, Eqn. (9) is a consequence of the other relations if p > 3, though this
would be hard to verify directly. Note that Eqn. (6) is redundant too: this
can be seen from above.
The only relations in T2 are the first two relations above; the first three
relations above carry all information about T21 + T2.
Proof. It only remains to establish the last five relations. All are proved
using the method of the proof of Proposition 12. We give one example, the
case of e (x0x1). We have
ss*(e (x0x1)) = 0 00(0 - 00)p-1(0 p+1- 00p+1)=(0 - 00) ;
and require a0; b02 T1;1such that
a00 + (b0- a0)00= (0 p+200 - 0 00p+2)=(0 - 00)
= (0 p+300 - 0 200p+2- 0 00p+3+ 0 p+200)=(0 2- 002)
= f1(00; 0)0 + f1(0 ; 00)00 :
As both f1(00; 0) and f1(0 ; 00) lie in T1;1, we are done.
12
6. Extraspecial p-groups
The 2n-dimensional Fp-vector space En may be viewed as an elementary
abelian p-group of rank 2n. Let N be a cyclic group of order p. The nonde-
generate symplectic form b on E may be viewed as a map ExE ! N. Denote
by Pn the extraspecial p-group p1+2n+of order p2n+1 and exponent p. There
is a central extension 1 ! N ! Pn ! E ! 1, such that, for g1; g2 2 Pn, the
commutator [g1; g2] equals b ( (g1); (g2)). The maximal elementary abelian
subgroups of Pn have p-rank n+1, and are exactly the inverse images under
of the maximal totally isotropic subspaces of E.
To determine the irreducible characters of Pn, pick an embedding of the
additive group of Fp in Cx . There are p2n linear characters of Pn, all of
which factor through . These may be identified with the elements of the
dual space E*n. The p - 1 remaining irreducible characters all have degree pn
and are induced from any maximal elementary abelian subgroup of Pn. Let ^O
be a nontrivial linear character of N. Then for each 1 6 i 6 p - 1 there is an
irreducible character Oi of Pn whose restriction to any maximal elementary
abelian subgroup M is the sum of all linear characters of M whose restriction
to N is ^Oi.
For each OE 2 E*n, pick a representation aeOEof Pn whose character is linear,
corresponding to OE. Let ae1 be a representation of Pn affording the charac-
ter O1.
Definition. For any finite group G, define h*(G) to be the quotient of
the graded commutative ring H*(G; Fp) by its nilradical. Define ch(G) to
be the subring of h*(G) generated by the images under the homomorphism
H*(G; Z) ! H*(G; Fp) ! h*(G) of the Chern classes of the representations
of G.
The reader is referred to the appendix of Atiyah's paper [1 ] for a con-
cise introduction to Chern classes of group representations. A proof of the
following theorem may be found in Chapter 8 of Evens' book [4 ].
Theorem 18. (Quillen) Let G be a finite group, and let be a class
in h *(G). Then is zero if and only if Res E = 0 in h *(E) for every
elementary abelian p-subgroup E of G.
Recall that Qn is defined in terms of the polynomial algebra S(E*n) Fp
Fp[K0; : :;:Kn-1], denoted Fn.
13
Theorem 19. Let Z be an indeterminate. There is a unique Fp-algebra
homomorphism f :Fn Fp Fp[Z] ! H*(p1+2n+; Z) Z Fp which sends OE 2 E*n
to c1(aeOE), sends Kr to (-1)n-rcpn-pr(ae1), and sends Z to cpn(ae1). This ho-
momorphism induces an isomorphism f :Qn Fp Fp[Z] ! ch(p1+2n+).
Proof. Let ae, oe be degree one representations of Pn. Since c1(ae oe) =
c1(ae)+c1(oe), the algebra homomorphism f is well-defined; clearly it is unique.
Any elementary abelian p-group A may be viewed as an Fp-vector space.
There is an isomorphism ch(A) ! S(A*) which sends the first Chern class of
any degree one representation to the corresponding element of A*. Moreover,
ch(A) = h*(A).
For 1 6 j 6 p - 1, let aej be a representation of Pn which affords the
character Oj. Let I be any maximal totally isotropic subspace of En, and
let M be the corresponding maximal elementary abelian subgroup of Pn.
Then I* is the subspace of M* which annihilates N. Pick some fl 2 h2(M)
such that Res N(fl) = c1(O^). If we restrict the total Chern class of aej to M
and apply the Whitney sum formula, we have
Y
Res M c(aej)= (1 + v + jfl) ;
v2I*
n-1X n-1X !
n n-r * pr
= 1 + (-1)n-rDr(I*) + j flp + (-1) Dr(I )fl :
r=0 r=0
Quillen's Theorem then says that cpn-pr(aej) = cpn-pr(ae1) and cpn(aej) =
jcpn(ae1) in ch(Pn). Moreover, these are the only non-zero Chern classes
of the induced representations. Hence the map from H*(Pn; Z) Fp down
to h*(G) maps Im (f) onto ch(Pn). Observe that Z is the only generator of
Fn Fp[Z] whose image under Res M Of involves fl, and that Res M f(Z) is
transcendental over S(I*). Therefore, we only have to show that the induced
map Qn ! ch(Pn) is both injective and well-defined. But, for every y 2 Fn
and for every I, the elements qI(y) and Res M f(y) of S(I*) are equal. The
result then follows by Quillen's Theorem and the definition of Qn.
Remark. Observe that the maps f and f double the degree of homo-
geneous elements.
14
7. A general inequality
It is the business of this section to prove that Qn always strictly contains Tn.
Specifically, we prove the following theorem.
Theorem 20. For every n > 1, we have n;062 Tn. Therefore cpn-1(ae1)
lies outside the subring of ch (p1+2n+) generated by the first Chern classes
and cpn(ae1).
Proof. Proposition 7 gives us the case n = 1. We shall prove the rest
of the result by considering the inflation map ss*: Qn+1 ! Q1 Fp Qn, and
showing that ss*(n+1;0) lies outside T1 Qn for all n > 1.
The inflation map is associated to the orthogonal direct sum decomposi-
tion En+1 = E1 ? En. For each maximal elementary abelian subgroup In+1
of En+1 this induces the decomposition In+1 = I1 In. Using this decompos-
ition and Eqn. (1), we can express the Dickson invariants of In+1 in terms of
the Dickson invariants for I1 and In. In particular, if we define Dn(In) = 1,
then
n ! p-1
X pn-j-1_
D0(I*n+1) = (D0(I*1) D0(I*n)) (-1)jD0(I*1) p-1 Dn-j(I*n) ;
j=0
whence, also defining n;n= 1, we have
n ! p-1
X pn-j-1_
ss*(n+1;0) = ( n;0) (-1)j p-1 n;n-j : (15)
j=0
Since r 2 T1 if and only if r 6= 1, the right hand side of this equation equals
pn;0modulo T1 Qn.
8. Symplectic invariants
Closely related to the work of this paper is a question about symplectic
invariants. The symplectic group Sp2n(Fp) is by definition the group of those
linear transformations of En which preserve the nondegenerate symplectic
form b. The invariants of the corresponding action of Sp 2n(Fp) on S(E*n)
15
were determined by Carlisle and Kropholler, and are described in Section 8.3
of Benson's book [2 ].
The ring of invariants in S(E*n) is generated by R1(E*n), : :,:R2n-1(E*n),
Dn(E*n), : :,:D2n-1(E*n). Recall from Theorem 1 that the quotient of S(E*n)
by the ideal generated by the regular sequence R1(E*n), : :,:Rn(E*n) is Tn.
There is therefore an induced action of Sp2n(Fp) on Tn. It is natural to ask
what is the ring of invariants of this action.
By the Tezuka-Yagita Theorem, every Rr(E*n) is zero in Tn. Certainly
every Dr(E*n) is still invariant. But now there are other invariants as well.
Proposition 21. The natural action of Sp 2n(Fp) on Tn has as ring of
invariants the intersection of Tn with Fp[n;0; : :;:n;n-1].
Proof. The symplectic group permutes the maximal totally isotropic sub-
spaces I of En transitively. In addition, for any I, every automorphism of I
may be extended to a symplectic transformation on En. Hence, for every I
and for every symplectic invariant x 2 Tn, the element ^qI(x) of S(I*) is
invariant under GL (I), and this invariant is independent of I. Since the
Dickson invariants in S(I*) generate the invariants under GL (I), it follows
that x equals some polynomial over Fp in n;0, : :,:n;n-1. Conversely, any
such polynomial is invariant under the action of Sp2n(Fp) on Qn.
Theorem 22. The ring of invariants under the natural action of Sp4(Fp)
on T2 is the subring of the polynomial algebra Fp[0; 1] of polynomials whose
support contains neither any 0r1with r > 0 nor any r1with p - r. Over
the polynomial algebra Fp[20; p1], the ring of invariants is the free module
generated by 1, 20s1for 1 6 s 6 p - 1, and 30s1for 0 6 s 6 p - 1.
Proof. Let f(x0; x1) be any polynomial in Fp[x0; x1]. By Proposition 4,
f(0; 1) belongs to T2 if and only if there exists a 2 T2 such that
f(0; 1) - f (D0(ff1; fi1); D1(ff1; fi1))= afl2
holds after inflation to Q1 Q1. That is, if and only if there exists a02 T1;1
such that, in Q1 Q1,
0 00 0 00 p-1 00 0 00 p-1 0 p 0p 0 0 00 p
f ( - ) ; ( - ) + - f(0; ) = a ( - ) :(16)
Observe that both sides of Eqn. (16) are divisible by (0 - 00)p-1. Doing this
to the right hand side yields a0(0 - 00). Suppose f is the monomial xr0xs1. If
16
r > 2, then f(0; 1) 2 T2. If r = 1, then the left hand side of Eqn. (16) is
000 (0 -00 )p-1 (00(0 - 00)p-1 + 0 p)s, which is not divisible by (0 -00 )p. If
r = 0, then get (00(0 - 00)p-1 + 0 p)s- 0sp, which is divisible by (0 - 00)p
if and only if p | s. The monomials xr0xs1such that r0s162 T2 all have distinct
degrees when evaluated at (0; 1). Hence Fp[0; 1] \ T2 is the subring of
Fp[0; 1] described in the statement. By Proposition 21 it follows that the
ring of invariants is as claimed.
Remark. Using Theorem 17, we could in principle give expressions in
terms of the ffi and fij for each generator of this ring of invariants: however,
these expressions would be very complicated. The current form of the result
is likely to be the more illuminating.
9. The other extraspecial group
In this section we shall study the extraspecial group p1+4-of order p5 and
exponent p2, and determine the Chern ring ch(p1+4-).
The groups p1+2n+and p1+2n-are very similar. There is a central extension
1+2n
1 ! N ! p1+2n-! En ! 1, such that, for g1; g2 2 p- , the commut-
ator [g1; g2] equals b ( (g1); (g2)). The element g of p1+2n-has exponent p if
A*1( (g)) = 0. If not, then g has exponent p2, and gp lies in N, the centre
of p1+2n-. The maximal elementary abelian subgroups of p1+2n-have p-rank
n + 1, and are exactly the inverse images under of those maximal totally
isotropic subspaces of E which contain B1. Once more, the irreducible rep-
resentations consist of a one-dimensional representation aeOEfor each OE 2 E*n,
together with representations ae1, : :,:aep-1 induced from a maximal abelian
subgroup, such that the restriction of aej to any maximal elementary abelian
subgroup M is the sum of all one-dimensional representations of M whose
characters restrict to N as ^Oj.
Define Q-nto be the quotient of Fn by the intersection of the kernels of the
maps qI for all maximal totally isotropic subspaces I of En containing B1.
These are precisely those I such that qI(A*1) = 0. Observe that Q-n is a
quotient of Qn.
Theorem 23. Let Z be an indeterminate. There is a unique Fp-algebra
homomorphism f :Fn Fp Fp[Z] ! H*(p1+2n-; Z) Z Fp which sends OE 2 E*n
to c1(aeOE), sends Kr to (-1)n-rcpn-pr(ae1), and sends Z to cpn(ae1). This ho-
momorphism induces an isomorphism f :Q-nFp Fp[Z] ! ch(p1+2n-).
17
Proof. The proof of Theorem 19 is easily adapted to this case.
Remark. Once more, the maps f and f double the degree of homogen-
eous elements.
Now we determine the structure of Q-2.
Theorem 24. The algebra Q-2has a presentation consisting of the gen-
erators ff1, ff2, fi1, fi2, 0, 1 of Q2 together with the relations
ff1 = 0 (17)
ff2fi2p- ff2pfi2 = 0 (18)
2-1
0 = fip-111 - fip1 (19)
"2(1 - D1(fi1; "2)) = 0 for "2 = ff2; fi2 (20)
(1 - D1(fi1; ff2))(1 - D1(fi1; fi2)) = 0 : (21)
In particular, (the image of) fi1 is a non-zero divisor.
Proof. Denote by q- the quotient map Fn ! Q-n. Then, for any x 2 Fn
we have q- (x) = 0 if and only if qI(x) = 0 for all I containing B1. Also,
q- factors through q: there is a unique algebra homomorphism ^q-:Qn ! Q-n
such that q- = ^q-q.
If I contains B1, then qI(B*1) is a non-zero divisor and qI(A*1) is zero
in S(I*). Hence ^q-(fi1) is a non-zero divisor in Q-n, and ^q-(ff1) = 0. Equa-
tion (18) is now an immediate consequence of Eqn. (5). Putting ffi = fi1 in
Eqn. (8) and then cancelling yields Eqn. (19). Equation (18) implies that,
for any I, the restrictions of either B*1, A*2or B*1, B*2must span I*; therefore
Eqn. (20) and Eqn. (21) hold. The necessity of the relations is established.
By the Tezuka-Yagita Theorem for p1+4-(see Proposition 8.2 of [3 ]), all
the relations in ^q-(T2) are generated by Eqn. (17) and Eqn. (18). It therefore
only remains to prove that ^q-(1) 62 ^q-(T2). Equation (4) implies that
Xp 2
2-p (p-1)2 p-1 p2-p i -
p1= fi(p-i)(p1-p)(ffp2 - ff2 fi2 + fi2 ) in Q2 .
i=0
But the right hand side is not the pth power of any element of ^q-(T2).
18
References
[1]M. F. Atiyah. Characters and cohomology of finite groups. Inst. Hautes
Etudes Sci. Publ. Math. 9 (1961), 23-64.
[2]D. J. Benson. Polynomial Invariants of Finite Groups. London Math.
Soc. Lecture Note Ser. no. 190 (Cambridge Univ. Press, 1993).
[3]D. J. Benson and J. F. Carlson. The cohomology of extraspecial groups.
Bull. London Math. Soc. 24 (1992), 209-235. Erratum: Bull. London
Math. Soc. 25 (1993), 498.
[4]L. Evens. The Cohomology of Groups. (Oxford Univ. Press, 1991).
[5]D. J. Green and I. J. Leary. Chern classes and extraspecial groups.
Preprint No. 16 (1995), Inst. f. Exp. Math., Univ. GH Essen.
Department of Mathematics
University of Chicago
5734 S. University
Chicago, IL 60637
U.S.A.
19