Mùi invariants and Milnor operations

107 140 107 arxiv version: fonts, pagination and layout may vary from GTM published version Mùi invariants and Milnor operations MASAKI KAMEKO MAMORU MIMURA We describe Mùi invariants in terms of Milnor operations and give a simple proof for Mùi s theorem on rings of invariants of polynomial tensor exterior algebras with respect to the action of finite general linear groups. Moreover, we compute some rings of invariants of Weyl groups of maximal non-toral elementary abelian p subgroups of exceptional Lie groups. 55R40; 55S10 1 Introduction Let p be a fixed odd prime, q the power of p and F q the finite field of q elements. Let P n = F q [x 1,..., x n ] be the polynomial algebra in n variables x 1,..., x n over the finite field F q. Let E r n = Λ r (dx 1,..., dx n ) be the r th component of the exterior algebra of dx 1,..., dx n over the finite field F q and let n E n = r=0 be the exterior algebra of dx 1,..., dx n over F q. Let P n E n be the polynomial tensor exterior algebra in n variables x 1,..., x n over the finite field F q. The general linear group GL n (F q ) and the special linear group SL n (F q ) act on both the polynomial algebra P n and the polynomial tensor exterior algebra P n E n. In [4], the ring of invariants of the polynomial algebra is determined by Dickson. In [7], Mùi determined the ring of invariants of the polynomial tensor exterior algebra and described the invariants in terms of determinants. E r n Published: 14 November 2007 DOI: 10.2140/gtm.2007.11.107

108 Masaki Kameko and Mamoru Mimura In the first half of this paper, we describe the invariants in terms of Milnor operations and give a simpler proof for Mùi s theorem. With the notation in Section 2, we may state Mùi s theorems. Theorem 1.1 (Mùi) The ring of invariants of the polynomial tensor exterior algebra P n E n with respect to the action of the special linear group SL n (F q ) is a free P n SL n(f q) module with the basis {1, Q I dx 1... dx n }, where I ranges over A n. Theorem 1.2 (Mùi) The ring of invariants of the polynomial tensor exterior algebra P n E n with respect to the action of the general linear group GL n (F q ) is a free P n GL n(f q) module with the basis {1, e q 2 n Q I dx 1... dx n }, where I ranges over A n. The invariant Q i1... Q in r dx 1... dx n in Theorem 1.1 and Theorem 1.2 above is, up to sign, the same as the Mùi invariant [r : i 1,..., i n r ] in [7]. The first half of this paper has some overlap with M C Crabb s work [3]. However, our point of view on Mùi invariants seems to be different from his. The second half of this paper is a sequel to the authors work in [6] on the invariant theory of Weyl groups of maximal non-toral elementary abelian p subgroup A of simply connected exceptional Lie groups. For p odd prime, up to conjugation, there are only 6 of them, for p = 3, A = EF 3 4, E3E 4 6, E2E 4 7, EE 5a 8, EE 5b 8 and for p = 5, A = EE 3 8. They and their Weyl groups are described by Andersen et al [1]. We computed the polynomial part of the invariants of Weyl groups except for the case p = 3, A = EE 5b 8 as described by the authors [6]. In this paper, we compute rings of invariants of polynomial tensor exterior algebras with respect to the action of Weyl groups except for the case p = 3, A = EE 5b 8. In Section 2, we set up the notation used in the above theorems. In Section 3, we prove Theorem 1.1 and Theorem 1.2. In Section 4, we state Theorem 4.2 and using this theorem, we compute rings of invariants of Weyl groups of maximal non-toral elementary abelian p subgroups of simply connected exceptional Lie groups. In Section 5, we prepare for the proof of Theorem 4.1 and Theorem 4.2. In Section 6, we prove Theorem 4.1. In Section 7, we prove Theorem 4.2. In Appendix A, we prove that the invariant Q i1... Q in r dx 1... dx n in Theorem 1.1 and Theorem 1.2 above is, up to sign, equal to the Mùi invariant [r : i 1,..., i n r ]. We thank N Yagita for informing us that a similar description of Mùi invariants to the above form is also known to him.

Mùi invariants and Milnor operations 109 2 Preliminaries Let K n be the field of fractions of P n. For a finite set {y 1,..., y r }, we denote by F q {y 1,..., y r } the F q vector space spanned by {y 1,..., y r }. Let GL n (F q ) be the set of n n invertible matrices with coefficients in F q. We denote by M m,n (F q ) the set of m n matrices with coefficients in F q. In this paper, we consider the contragredient action of the finite general linear group, that is, for g GL n (F q ), we define the action of g on P n E n by gx i = n a i,j (g 1 )x j, gdx i = j=1 n a i,j (g 1 )dx j, j=1 for i = 1,..., n and g(x y) = g(x) g(y), for x, y in P n E n, where a i,j (g 1 ) is the entry (i, j) in the matrix g 1. For x i, dx i in the polynomial tensor exterior algebra P n E n, we define cohomological degrees of x i, dx i by deg x i = 2, deg dx i = 1 for i = 1,..., n and we consider P n E n as a graded F q algebra. Now, we recall Milnor operations Q j for j = 0, 1,.... The exterior algebra Λ(Q 0, Q 1, Q 2,...) over F q, generated by Milnor operations, acts on the polynomial tensor exterior algebra P n E n as follows; the Milnor operation Q j is a P n linear derivation defined by the Cartan formula Q j : P n E r n P n E r 1 n Q j (xy) = (Q j x) y + ( 1) deg x x (Q j y) for x, y in P n E n and the unstable conditions Q j dx i = x qj i, Q j x i = 0, for i = 1,..., n, j 0. Thus, the action of Milnor operations Q j commutes with the action of the finite general linear group GL n (F q ).

110 Masaki Kameko and Mamoru Mimura It is also clear that the action of Q j on P n trivially extends to the quotient field K n and we may regard Q j as a K n linear homomorphism Q j : K n E r n K n E r 1 n. We set up additional notations for handling Milnor operations. Definition 2.1 For a positive integer n, we denote by S n the set {0, 1,..., n 1}. Let A n be the set of subsets of S n. We denote by A n,r the subset of A n consisting of I = {i 1,..., i r } such that We write Q I for 0 i 1 < < i r < n. Q i1... Q ir. We consider A n,0 as the set of empty set { } and define Q to be 1. It is also convenient for us to define A n to be the union of A n,r, where r ranges from 0 to n 1. Definition 2.2 Let I, J be elements of A n. We define sign(i, J) as follows. If I J, then sign(i, J) = 0. If I J = and I J = {k 1,..., k r+s }, then sign(i, J) is the sign of the permutation ( i 1,..., i r, j 1,..., j s k 1,..., k r, k r+1,..., k r+s where I = {i 1,..., i r }, J = {j 1,..., j s }, i 1 < < i r, j 1 < < j s and k 1 < < k r+s. ), The following proposition is immediate from the definition above. Proposition 2.3 For I, J in A n, we have where K = I J. Q I Q J = sign(i, J)Q K = ( 1) rs sign(j, I)Q K,

Mùi invariants and Milnor operations 111 Finally, using Milnor operations in place of determinants, we describe Dickson invariants. We follow the notation of Wilkerson s paper [9]. Let n (X), f n (X) be polynomials of X over P n of homogeneous degree q n defined respectively by n (X) = ( 1) n Q 0... Q n dx 1... dx n dx = n ( 1) n i (Q 0... Q i... Q n dx 1... dx n )X qi, f n (X) = i=0 x F q{x 1,...,x n} (X + x), where the cohomological degrees of dx, X are 1, 2, and Q i dx = X qi, Q i X = 0, respectively. Proposition 2.4 The polynomial n (X) is divisible by the polynomial f n (X) and e n (x 1,..., x n )f n (X) = n (X), where e n (x 1,..., x n ) = Q 0... Q n 1 dx 1... dx n 0. Proof On the one hand, both n (X) and f n (X) have all x F q {x 1,..., x n } as roots. On the other hand, the coefficient of X qn in n (X) is e n (x 1,..., x n ) = Q 0... Q n 1 dx 1... dx n and f n (X) is monic. Since both n (X) and f n (X) have the same homogeneous degree q n as polynomials of X, we have the required equality. Thus, we have the following proposition. Proposition 2.5 We may express f n (X) as follows: n f n (X) = ( 1) n i c n,i (x 1,..., x n )X qi, where and c n,n (x 1,..., x n ) = 1. i=0 Q 0... Q i... Q n dx 1... dx n = e n (x 1,..., x n )c n,i (x 1,..., x n ) The above proposition defines Dickson invariants c n,i (x 1,..., x n ) for i = 0,..., n 1. When it is convenient and if there is no risk of confusion, we write e n, c n,i for e n (x 1,..., x n ), c n,i (x 1,..., x n ), respectively.

112 Masaki Kameko and Mamoru Mimura Proposition 2.6 There holds c n,0 (x 1,..., x n ) = e n (x 1,..., x n ) q 1. Proof It is clear that (Q i1... Q in dx 1... dx n ) q = Q i1 +1... Q in+1dx 1... dx n and so e n (x 1,..., x n ) q = e n (x 1,..., x n )c n,0 (x 1,..., x n ). From the above definitions of c n,i (x 1,..., x n ) and e n (x 1,..., x n ) and from the fact that, for g GL n (F q ), gdx 1... dx n = det(g 1 )dx 1... dx n, it follows that and that ge n (x 1,..., x n ) = det(g 1 )e n (x 1,..., x n ) gc n,i (x 1,..., x n ) = c n,i (x 1,..., x n ). Thus, it is clear that P n SL n(f q) contains e n and c n,1,..., c n,n 1 and that P n GL n(f q) contains c n,0,..., c n,n 1. Indeed, the following results are well-known. For proofs, we refer the reader to Benson [2], Smith [8] and Wilkerson [9]. Theorem 2.7 (Dickson) The ring of invariants P SL n(f q) n is a polynomial algebra generated by c n,1,..., c n,n 1 and e n. Theorem 2.8 (Dickson) The ring of invariants P GL n(f q) n is a polynomial algebra generated by c n,0,..., c n,n 1. In addition, we need the following proposition. Proposition 2.9 Let be the obvious projection. Then, we have and, for i = 1,..., n 1, π : F q [x 1,..., x n ] F q [x 1,..., x n 1 ] π(e n (x 1,..., x n )) = 0 π(c n,i (x 1,..., x n )) = c n 1,i 1 (x 1,..., x n 1 ) q.

Mùi invariants and Milnor operations 113 Proof It is clear that e n (x 1,..., x n ) is divisible by x n, so we have On the one hand, we have f n (X) = π(e n (x 1,..., x n )) = 0. α F q = x F q{x 1,...,x n 1 } α F q f n 1 (X + αx n ) = α F q (f n 1 (X) + αf n 1 (x n )) (X + αx n + x) = f n 1 (X) q f n 1 (X)f n 1 (x n ) q 1. On the other hand, since f n (x n ) is divisible by x n, we have π(f n (X)) = f n 1 (X) q. Comparing the coefficients of X qi, we have, for i = 1,..., n 1, π(c n,i (x 1,..., x n )) = c n 1,i 1 (x 1,..., x n 1 ) q. 3 Proof of Theorem 1.1 and Theorem 1.2 In the case n = 1, the invariants are obvious. In the case r = 0, the invariants are calculated by Dickson. So, throughout the rest of this section, we assume n 2 and r > 0. To prove Theorem 1.1 and Theorem 1.2, it suffices to prove the following theorems. Theorem 3.1 The submodule (P n En) r SLn(Fq) is a free P SL n(f q) n module with the basis {Q I dx 1... dx n }, where I ranges over A n,n r. Theorem 3.2 The submodule (P n E r n) GLn(Fq) is a free P n GL n(f q) module with the basis {e q 2 n Q I dx 1... dx n }, where I ranges over A n,n r. To begin with, we prove the following proposition. Proposition 3.3 The elements Q I dx 1... dx n form a basis for K n E r n, where I ranges over A n,n r.

114 Masaki Kameko and Mamoru Mimura Proof Firstly, we show the linear independence of Q I dx 1... dx n. Suppose that a = a I Q I dx 1... dx n, I A n,n r where a I K n. For each I A n,n r, let J = S n \I. It is clear that J I if I I in A n,n r. Hence, we have sign(j, I) 0 and sign(j, I ) = 0 for I I A n,n r. By Proposition 2.3, we have Q J a = sign(j, I)a I Q 0... Q n 1 dx 1... dx n = sign(j, I)a I e n. Thus, if a = 0, then a I = 0. Therefore, the terms Q I dx 1... dx n are linearly independent in K n E r n. On the other hand, since dim Kn K n E r n = is equal to the number of elements in A n,n r, we see, for dimensional reasons, that Q I dx 1... dx n s form a basis for K n E r n. Lemma 3.4 Let h I be polynomials over F q in (n 1) variables, where I ranges over A n,n 1. Suppose that a 0 = h I (c n,n 1,..., c n,1 )e 1 n Q I dx 1... dx n I A n,n 1 is in P n E 1 n. Then h I = 0 for each I A n,n 1. Proof of Theorem 3.1 Suppose that a is an element in P n En r and that a is SL n (F q ) invariant. By Proposition 3.3, the elements Q I dx 1... dx n form a basis for K n En. r Hence, there exist a I in K n such that a = a I Q I dx 1... dx n. I A n,n r For I A n,n r, let J = S n \I. Then, Q J a is in P n. As in the proof of Proposition 3.3, we have Q J a = sign(j, I)a I e n. Therefore, there are polynomials f I,k over F q in (n 1) variables such that a I = k 0 ( n r ) f I,k (c n,n 1,..., c n,1 )e k 1 n.

Mùi invariants and Milnor operations 115 Thus, we have a = I A n,n r k 0 f I,k (c n,n 1,..., c n,1 )e k 1 n Q I dx 1... dx n. It remains to show that f I,0 = 0 for each I A n,n r. Let a 0 = a I A n,n r k 1 f I,k (c n,n 1,..., c n,1 )e k 1 n Q I dx 1... dx n. Then, we have that a 0 = f I,0 (c n,n 1,..., c n,1 )e 1 n Q I dx 1... dx n I A n,n r and that a 0 is also in P n E r n. For J A n,r 1, the element Q J a 0 is in P n En 1. By Proposition 2.3, we have Q J a 0 = sign(j, I)f I,0 (c n,n 1,..., c n,1 )Q K dx 1... dx n. K A n,n 1,I=K\J,I A n,n r Hence, by Lemma 3.4, we have sign(j, I)f I,0 = 0. For each I in A n,n r, there exists J A n,r 1 such that sign(j, I) 0. Therefore, we have f I,0 = 0 for each I. This completes the proof. Proof of Theorem 3.2 As in the proof of Theorem 3.1, if a P n En r is GL n (F q ) invariant, the element a can be expressed in the form a = f I,k (c n,n 1,..., c n,1 )en k 1 Q I dx 1... dx n. I A n,n r k 1 For g GL n (F q ), we have ga = det(g 1 ) k f I,k (c n,n 1,..., c n,1 )en k 1 Q I dx 1... dx n. I A n,n r k 1 Therefore, a is GL n (F q ) invariant if and only if f I,k = 0 for k 0 modulo q 1. Hence, we have a = f I,m(q 1)+(q 1) (c n,n 1,..., c n,1 )e m(q 1) n e q 2 n Q I dx 1... dx n. Since e q 1 n I A n,n r m 0 = c n,0, we may write a = f I (c n,n 1,..., c n,1, c n,0 )e q 2 n Q I dx 1... dx n, I A n,n r

116 Masaki Kameko and Mamoru Mimura where f I (c n,n 1,..., c n,0 ) = f I,m(q 1)+(q 1) (c n,n 1,..., c n,1 )c m n,0. m 0 This completes the proof. Proof of Lemma 3.4 For the sake of notational simplicity, let I i = S n \{i} and we write h i for h Ii. Since a is in P n E 1 n, there are ϕ 1,..., ϕ n in P n such that The coefficient ϕ n of dx n is given by a 0 = ϕ 1 dx 1 + + ϕ n dx n. = n 1 h i (c n,n 1,..., c n,1 )e 1 n Q Ii dx 1... dx n 1 i=0 { n 1 } h i (c n,n 1,..., c n,1 )c n 1,i i=0 e 1 n e n 1. Hence, we have { n 1 } h i (c n,n 1,..., c n,1 )c n 1,i e n 1 = e n ϕ n. i=0 By Proposition 2.9, the obvious projection π : F q [x 1,..., x n ] F q [x 1,..., x n 1 ] maps e n, c n,i to 0, c q n 1,i 1, respectively. So, we have n 1 h i (c q n 1,n 2,..., cq n 1,0 )c n 1,i = 0. i=0 Since c n 1,i (i = 0,..., n 2) are algebraically independent in F q [x 1,..., x n 1 ] and since c n 1,n 1 = 1, writing y i for c n 1,i, we have the following equation: n 2 (1) h n 1 (y q n 2,..., yq 0 ) + h i (y q n 2,..., yq 0 )y i = 0. i=0 Applying partial derivatives / y i, we have (2) h i (y q n 2,..., yq 0 ) = 0

Mùi invariants and Milnor operations 117 for i = 0,..., n 2. Hence, h i (y q n 2,..., yq 0 ) = 0 for i = 0,..., n 2. Substituting these to the previous equation (1), we also have h n 1 (y q n 2,..., yq 0 ) = 0. Since we deal with polynomials over the finite field F q, we have h i (y q n 2,..., yq 0 ) = h i(y n 2,..., y 0 ) q for i = 0,..., n 1. Therefore, we have h i (y n 2,..., y 0 ) = 0 for i = 0,..., n 1. Since y 0,..., y n 2 are algebraically independent, we have h i = 0 as polynomials over F q in (n 1) variables for i = 0,..., n 1. 4 Invariants of some Weyl groups In this section, we consider the invariant theory of polynomial tensor exterior algebras. In what follows, we assume that n 2. To state Theorem 4.2, which is our main theorem on the invariant theory, we need some notation. Let P n 1 = F q [x 2,..., x n ] be the subalgebra of P n generated by x 2,..., x n and let E n 1 be the subalgebra of E n generated by dx 2,..., dx n. Let G 1 be a subgroup of SL n 1 (F q ) which acts on P n 1 and P n 1 E n 1 both. Let G be a subgroup of SL n (F q ) consisting of the following matrices: ( 1 m 0 g 1 ), where g 1 G 1 and m M 1,n 1 (F q ). Obviously the group G acts on P n and P n E n. Finally, let O n 1 (x i ) = (x i + x). x F q{x 2,...,x n} Theorem 4.1 Suppose that the ring of invariants P G 1 n 1 is a polynomial algebra generated by homogeneous polynomials f 2,..., f n in (n 1) variables x 2,..., x n. Then, the ring of invariants P G n is also a polynomial algebra generated by O n 1 (x 1 ), f 2,..., f n.

118 Masaki Kameko and Mamoru Mimura This theorem is a particular case of a theorem of Kameko and Mimura [6, Theorem 2.5]. We use this theorem to compute the polynomial part of invariants P G n which appear in our main theorem, Theorem 4.2. So, Theorem 4.2 below works effectively together with Theorem 4.1. Theorem 4.2 Suppose that the ring of invariants P G 1 n 1 is a polynomial algebra generated by homogeneous polynomials f 2,..., f n in (n 1) variables x 2,..., x n and suppose that the ring of invariants (P n 1 E n 1 ) G 1 is a free P G 1 n 1 module with a basis {v i}, where i = 1,..., 2 n 1. Then, the ring of invariants (P n E n ) G is a free P G n module with the basis {v i, Q I dx 1... dx n }, where i = 1,..., 2 n 1 and I ranges over A n 1. We prove Theorem 4.1 and Theorem 4.2 in Section 5, Section 6 and Section 7. As an application of Theorem 4.1 and Theorem 4.2, we compute rings of invariants of the mod p cohomology of the classifying spaces of maximal non-toral elementary abelian p subgroups of simply connected exceptional Lie groups with respect to the Weyl group action. It is well-known that for an odd prime p, a simply connected exceptional Lie group G does not have non-toral elementary abelian p subgroups except for the cases p = 5, G = E 8, and p = 3, G = F 4, E 6, E 7, E 8 (see [1] and [5]). Andersen, Grodal, Møller and Viruel [1] described the Weyl groups of maximal non-toral elementary abelian p subgroups as well as their action on the underlying elementary abelian p subgroup explicitly for p = 3, G = E 6, E 7, E 8. Up to conjugate, there are only 6 maximal non-toral elementary abelian p subgroups of simply connected exceptional Lie groups. For p = 5, G = E 8 and for p = 3, G = F 4, E 6, E 7, there is one maximal non-toral elementary abelian p subgroup for each G. We call them EE 3 8, EF 3 4, E3E 4 6, E2E 4 7, following the notation in [1]. For p = 3, G = E 8, there are two maximal non-toral elementary abelian p subgroups, say EE 5a 8 and EE 5b 8, where the superscript indicates the rank of elementary abelian p subgroup. For a detailed account on non-toral elementary abelian p subgroups, we refer the reader to Andersen et al [1, Section 8], and its references. Let A be an elementary abelian p subgroup of a compact Lie group G. Suppose that A is of rank n. We denote by W(A) the Weyl group of A. Choosing a basis, say {a i }, for A, we consider the Weyl group W(A) as a subgroup of the finite general linear group GL n (F p ). We write H BA for the mod p cohomology of the classifying space BA.

Mùi invariants and Milnor operations 119 The Hurewicz homomorphism h : A = π 1 (BA) H 1 (BA; F p ) is an isomorphism. We denote by {dt i } the dual basis of {h(a i )}, so that dt i is the dual of h(a i ) with respect to the basis {h(a i )} of H 1 (BA; F p ) for i = 1,..., n. Let β : H 1 BA H 2 BA be the Bockstein homomorphism. Then, the mod p cohomology of BA is a polynomial tensor exterior algebra H BA = F p [t 1,..., t n ] Λ(dt 1,..., dt n ), where deg t i = 2, deg dt i = 1 and t i = β(dt i ) for i = 1,..., n. We denote by ΓH BA the polynomial part of H BA, that is, ΓH BA = F p [t 1,..., t n ]. The action of the Weyl group W(A) on A = π 1 (BA), given by ga i = j a j,i (g)a j, where {a i } is the fixed basis of A, induces the action of W(A) on H BA, which is given by gt i = a i,j (g 1 )t j, gdt i = a i,j (g 1 )dt j, j j for i = 1,..., n. Now, we compute (H BA) W(A) for A = E 3 E 8, E 3 F 4, E 4 3E 6, E 4 2E 7, E 5a E 8 using Theorem 1.1, Theorem 4.1 and Theorem 4.2. Proposition 4.3 For the above elementary abelian p subgroup A, the ring of invariants (H BA) W(A) is given as follows: (1) For p = 5, G = E 8, A = E 3 E 8, (H BA) W(A) is given by F 5 [x 62, x 200, x 240 ] F 5 {1, Q I u 3 }, where x 62 = e 3 (t 1, t 2, t 3 ), x 200 = c 3,2 (t 1, t 2, t 3 ), x 240 = c 3,1 (t 1, t 2, t 3 ), u 3 = dt 1 dt 2 dt 3 and I ranges over A 3 ; (2) For p = 3, G = F 4, A = E 3 F 4, (H BA) W(A) is given by F 3 [x 26, x 36, x 48 ] F 3 {1, Q I u 3 }, where x 26 = e 3 (t 1, t 2, t 3 ), x 36 = c 3,2 (t 1, t 2, t 3 ), x 48 = c 3,1 (t 1, t 2, t 3 ), u 3 = dt 1 dt 2 dt 3 and I ranges over A 3 ;

120 Masaki Kameko and Mamoru Mimura (3) For p = 3, G = E 6, A = E 4 3E 6, (H BA) W(A) is given by F 3 [x 26, x 36, x 48, x 54 ] F 3 {1, Q I u 3, Q J u 4 }, where x 26 = e 3 (t 2, t 3, t 4 ), x 36 = c 3,2 (t 2, t 3, t 4 ), x 48 = c 3,1 (t 2, t 3, t 4 ), x 54 = (t 1 + t), t F 3 {t 2,t 3,t 4 } u 3 = dt 2 dt 3 dt 4, u 4 = dt 1 dt 2 dt 3 dt 4, I ranges over A 3 and J ranges over A 3; (4) For p = 3, G = E 7, A = E 4 2E 7, (H BA) W(A) is given by F 3 [x 26, x 36, x 48, x 108 ] F 3 {1, Q I u 3, x 54 Q J u 4 }, where x 26 = e 3 (t 2, t 3, t 4 ), x 36 = c 3,2 (t 2, t 3, t 4 ), x 48 = c 3,1 (t 2, t 3, t 4 ), x 108 = x54 2, x 54 = (t 1 + t), t F 3 {t 2,t 3,t 4 } u 3 = dt 2 dt 3 dt 4, u 4 = dt 1 dt 2 dt 3 dt 4, I ranges over A 3 and J ranges over A 3; (5) For p = 3, G = E 8, A = E 5a E 8, (H BA) W(A) is given by F 3 [x 4, x 26, x 36, x 48, x 324 ] F 3 {1, Q I u 3, x 2 u 1, (Q I u 3 )x 2 u 1, x 2 x 162 Q J u 5 }, where x 4 = x2 2, x 26 = e 3 (t 2, t 3, t 4 ), x 36 = c 3,2 (t 2, t 3, t 4 ), x 48 = c 3,1 (t 2, t 3, t 4 ), x 324 = x162 2, x 2 = t 5, x 162 = (t 1 + t), t F 3 {t 2,t 3,t 4,t 5 } u 1 = dt 5, u 3 = dt 2 dt 3 dt 4, u 5 = dt 1 dt 2 dt 3 dt 4 dt 5, I ranges over A 3 and J ranges over A 4, where the subscripts of u and x indicate their cohomological degrees. Proof (1), (2) In the case p = 5, G = E 8, A = E 3 E 8 and in the case p = 3, G = F 4, A = E 3 F 4, the Weyl group is SL 3 (F p ). Therefore, it is the case of Mùi invariants and it is immediate from Theorem 1.1. (3) In the case p = 3, G = E 6, A = EE 4 6, the Weyl group W(A) is the subgroup of SL 4 (F 3 ) consisting of the following matrices: ( ) 1 m, 0 g 1

Mùi invariants and Milnor operations 121 where m M 1,3 (F 3 ), g 1 SL 3 (F 3 ). The result is immediate from Theorem 1.1, Theorem 4.1 and Theorem 4.2. (4) In the case p = 3, G = E 7, the Weyl group W(A) is the subgroup of GL 4 (F 3 ) consisting of the following matrices: ( ) ε1 m, 0 g 1 where ε 1 F 3 = {1, 2}, m M 1,3(F 3 ), g 1 SL 3 (F 3 ). Firstly, we compute the ring of invariants of a subgroup W 0 of W(A). The subgroup W 0 is the subgroup of W(A) consisting of the matrices ( ) 1 m, 0 g 1 where m M 1,3 (F 3 ), g 1 SL 3 (F 3 ). By Theorem 1.1, Theorem 4.1 and Theorem 4.2, we have (H BA) W 0 = F 3 [x 26, x 36, x 48, x 54 ] F 3 {1, Q I u 3, Q J u 4 }, where I ranges over A 3 and J ranges over A 3. Let R = F 3 [x 26, x 36, x 48, x 108 ], and let M = F 3 {x δ 54, xδ 54 Q Iu 3, x δ 54 Q Ju 4 }, where x 108 = x 2 54, δ {0, 1}, I ranges over A 3 and J ranges over A 3. Then, we have (H BA) W 0 = R M. Next, we calculate the ring of invariants (H BA) W(A) as a subspace of (H BA) W 0. Put 2 0 0 0 0 1 0 0 α = 0 0 1 0. 0 0 0 1 Then, for x R, we have αx = x and we also have the following direct sum decomposition: M = M 1 M 2, where M i = {x M αx = ix} for i = 1, 2. In particular, we have M 1 = F 3 {1, Q I u 3, x 54 Q J u 4 }.

122 Masaki Kameko and Mamoru Mimura Since the Weyl group W(A) is generated by W 0 and α, an element x in R M is W(A) invariant if and only if αx = x. Hence, we have (H BA) W(A) = R M 1. (5) In the case p = 3, G = E 8, A = EE 5a 8, the Weyl group W(A) is the subgroup of GL 5 (F 3 ) consisting of the following matrices: ε 1 m 0 m 1 0 g 1 0 0 0 ε 2 where ε 1, ε 2 F 3 = {1, 2}, m 0 M 1,3 (F 3 ), m 1 M 1,1 (F 3 ), g 1 SL 3 (F 3 ). We consider the subgroup W 0 of W(A) consisting of the following matrices: 1 m 0 m 1 0 g 1 0, 0 0 1 where g 1 SL 3 (F 3 ), m 0 M 1,3 (F 3 ), m 1 M 1,1 (F 3 ). By Theorem 4.1 and Theorem 4.2, we have, (H BA) W 0 = F 3 [x 2, x 26, x 36, x 48, x 162 ] F 3 {1, Q I u 3, u 1, (Q I u 3 )u 1, Q J u 5 }, where I ranges over A 3 and J ranges over A 4. Let and let R = F 3 [x 4, x 26, x 36, x 48, x 324 ], M = F 3 {x δ 1 2 xδ 2 162, xδ 1 2 xδ 2 162 Q Iu 3, x δ 1 2 xδ 2 162 u 1, x δ 1 2 xδ 2 162 (Q Iu 3 )u 1, x δ 1 2 xδ 2 162 Q Ju 5 }, where δ 1, δ 2 {0, 1}, I ranges over A 3 and J ranges over A 4. Consider matrices 2 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 0 0 0 α = 0 0 1 0 0, β = 0 0 1 0 0. 0 0 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 2 Then, we have αx = x and βx = x for x R. Furthermore, it is also clear that we have the following direct sum decomposition: M = M 1,1 M 1,2 M 2,1 M 2,2,

Mùi invariants and Milnor operations 123 where M i,j = {x M αx = ix, βx = jx}. In particular, we have M 1,1 = F 3 {1, x 2 u 1, Q I u 3, x 2 (Q I u 3 )u 1, x 2 x 162 Q J u 5 }. Since W(A) is generated by W 0 and α, β in the above, x (H BA) W 0 is W(A) invariant if and only if αx = βx = x. Hence, we have (H BA) W(A) = R M 1,1. Remark 4.4 Our computation of the ring of invariants of polynomial tensor exterior algebra in Proposition 4.3 is based on the computation of the ring of invariants of polynomial algebra and the assumption that the ring of invariants of polynomial algebra is also a polynomial algebra. In the case A = E 5b E 8, however, the Weyl group does not satisfy the condition we assume in this section and the ring of invariants of polynomial algebra is no longer a polynomial algebra. Hence, both Theorem 4.1 and Theorem 4.2 do not apply in this case. 5 O n 1 (x i ) and D n 1 In this section, we collect some facts, which we need in the proof of Theorem 4.1 and Theorem 4.2. For i = 1,..., n, the element O n 1 (x i ) in F q [x 1,..., x n ] is defined to be O n 1 (x i ) = (x i + x). x F q{x 2,...,x n} We also define O n 2 (x i ) in F q [x 1,..., x n 1 ] by O n 2 (x i ) = for n 3 and by for n = 2. x F q{x 2,...,x n 1 } O 0 (x i ) = x i (x i + x) Using the same argument as in the proof of Proposition 2.5, we can easily obtain the following proposition.

124 Masaki Kameko and Mamoru Mimura Proposition 5.1 For i = 1,..., n, we may express O n 1 (x i ) and O n 2 (x i ) in terms of Dickson invariants as follows: n 1 O n 1 (x i ) = ( 1) n 1 j q c n 1,j (x 2,..., x n )x j i for n 2, j=0 n 2 O n 2 (x i ) = ( 1) n 2 j q c n 2,j (x 2,..., x n 1 )x j i for n > 2. j=0 We need Proposition 5.2 and Proposition 5.3 below in the proof of Proposition 7.5. Proposition 5.2 In F q [x 1,..., x n 1 ], we have the following equality: for n 3, and for n = 2. e n 1 (x 1,..., x n 1 ) = O n 2 (x 1 )e n 2 (x 2,..., x n 1 ) e 1 (x 1 ) = x 1 Proof For n = 2, the proposition is obvious. For n 3, by Proposition 2.4, Proposition 2.5 and Proposition 5.1, we have e n 1 (x 1,..., x n 1 ) = Q 0... Q n 2 dx 1... dx n 1 n 2 = ( 1) n 2 j (Q 0... Q j... Q n 2 dx 2... dx n 1 )x qj 1 = j=0 n 2 j=0 ( 1) n 2 j e n 2 (x 2,..., x n 1 )c n 2,j (x 2,..., x n 1 )x qj 1 = e n 2 (x 2,..., x n 1 )O n 2 (x 1 ). Proposition 5.3 The obvious projection π : F q [x 1,..., x n ] F q [x 1,..., x n 1 ] maps O n 1 (x 1 ), e n 1 (x 2,..., x n ) to O n 2 (x 1 ) q, 0, respectively. Proof Since e n 1 (x 2,..., x n ) is divisible by x n, we have π(e n 1 (x 2,..., x n )) = 0

Mùi invariants and Milnor operations 125 as in the proof of Proposition 2.9. For n = 2, the equality π(o 1 (x 1 )) = O 0 (x 1 ) q = x q 1 is obvious. For n 3, by Proposition 2.9, we have π(c n 1,j (x 2,..., x n )) = c n 2,j 1 (x 2,..., x n 1 ) q for j = 1,..., n 1. Hence, we have n 1 π(o n 1 (x 1 )) = π ( 1) n 1 j c n 1,j (x 2,..., x n )x qj 1 For a in P n E n, let = j=0 n 1 j=1 = O n 2 (x 1 ) q. ( 1) n 1 j c n 2,j 1 (x 2,..., x n 1 ) q x qj 1 n 1 D n 1 (a) = ( 1) n 1 j c n 1,j (x 2,..., x n )Q j a. j=0 Then, D n 1 induces a P n linear homomorphism which extends naturally to D n 1 : P n E r n P n E r 1 n, D n 1 : K n E r n K n E r 1 n. Proposition 5.4 For i = 1,..., n, we have D n 1 (dx i ) = O n 1 (x i ). In particular, for i = 2,..., n, we have D n 1 (dx i ) = 0. Proof By Proposition 5.1, we have D n 1 (dx i ) = = n 1 ( 1) n 1 j c n 1,j (x 2,..., x n )Q j dx i j=0 n 1 j=0 = O n 1 (x i ). ( 1) n 1 j c n 1,j (x 2,..., x n )x qj i

126 Masaki Kameko and Mamoru Mimura On the other hand, by the definition of O n 1 (x i ), we have O n 1 (x i ) = 0 for i = 2,..., n. Hence, we have D n 1 (dx i ) = 0 for i = 2,..., n. Let g 1 be an element in GL n 1 (F q ). We consider the following matrix ( ) 1 0 ḡ 1 =. 0 g 1 We need Proposition 5.5 and Proposition 5.6 below in the proof of Proposition 7.8. Proposition 5.5 For g 1 in GL n 1 (F q ), there holds D n 1 (ḡ 1 a) = ḡ 1 D n 1 (a). Proof Suppose that a = where a i K n for i = 1,..., n. Since n a i dx i, i=1 ḡ 1 dx 1 = dx 1, and since, for i = 2,..., n, ḡ 1 dx i is in E 1 n 1, we have ḡ1a = (ḡ1a1)dx1 + a 2dx 2 + + a ndx n for some a i K n for i = 2,..., n. Hence, by Proposition 5.4, we have D n 1 (ḡ 1 a) = (ḡ 1 a 1 )O n 1 (x 1 ). On the other hand, by Proposition 5.4, we have ḡ 1 D n 1 (a) = ḡ 1 (a 1 O n 1 (x 1 )) = (ḡ 1 a 1 )O n 1 (x 1 ). Let P n 2 be the projective space (F n 1 q \{(0,..., 0)})/, where l l if and only if there is α F q such that l = αl.

Mùi invariants and Milnor operations 127 Proposition 5.6 If a P n is divisible by α 2 x 2 + + α n x n for arbitrary (α 2,..., α n ) P n 2, then a is divisible by e n 1 (x 2,..., x n ). Proof Since the number of elements in P n 2 is equal to the homogeneous degree 1 + q + + q n 2 of e n 1 (x 2,..., x n ), it suffices to show that e n 1 (x 2,..., x n ) is a product of elements of the form α 2 x 2 + + α n x n, where (α 2,..., α n ) ranges over P n 2. It is clear that e n 1 (x 2,..., x n ) is divisible by x n. So, we have e n 1 (x 2,..., x n ) = bx n for some b P n. It is also clear that e n 1 (x 2,..., x n ) is invariant under the action of SL n 1 (F q ). There is g 1 SL n 1 (F q ) such that α 2 x 2 + + α n x n = ḡ 1 x n. Hence, on the one hand, we have and, on the other hand, we have ḡ 1 e n 1 (x 2,..., x n ) = (ḡ 1 b)(α 2 x 2 + + α n x n ) ḡ 1 e n 1 (x 2,..., x n ) = e n 1 (x 2,..., x n ). Therefore, e n 1 (x 2,..., x n ) is divisible by arbitrary α 2 x 2 + + α n x n. This completes the proof. 6 Proof of Theorem 4.1 In order to prove Theorem 4.1, we recall the strategy to compute rings of invariants given by Wilkerson in [9, Section 3]. It can be stated in the following form. Theorem 6.1 Suppose that G is a subgroup of GL n (F q ) and G acts on the polynomial algebra F q [x 1,..., x n ] in the obvious manner. Let f 1,..., f n be homogeneous G invariant polynomials in F q [x 1,..., x n ]. Let R be the subalgebra of F q [x 1,..., x n ] generated by f 1,..., f n. Then, R is a polynomial algebra F q [f 1,..., f n ] and the ring of invariants F q [x 1,..., x n ] G is equal to the subalgebra R if and only if F q [x 1,..., x n ] is integral over R and deg f 1... deg f n = G. In the statement of Theorem 6.1, deg f is the homogeneous degree of f, that is, we define the degree deg x i of indeterminate x i to be 1. For the proof of this theorem, we refer the reader to Smith s book [8, Corollaries 2.3.2 and 5.5.4, and Proposition 5.5.5] and Wilkerson s paper [9, Section 3].

128 Masaki Kameko and Mamoru Mimura Proof of Theorem 4.1 As we mentioned, in order to prove Theorem 4.1, it suffices to show the following: (1) homogeneous polynomials O n 1 (x 1 ), f 2,..., f n are G invariant; (2) indeterminates x 1,..., x n are integral over R; (3) the product of homogeneous degrees of O n 1 (x 1 ), f 2,..., f n is equal to the order of G, that is, deg O n 1 (x 1 ) deg f 2... deg f n = G. By definition, f 2,..., f n are G 1 invariant, and so they are also G invariant. It follows from Theorem 6.1 that x 2,..., x n are integral over R 1 = F q [f 2,..., f n ], and so they are integral over R. It is also immediate from Theorem 6.1 that deg f 2... deg f n = G 1. It is clear from the definition of O n 1 (x 1 ) that deg O n 1 = 2 n 1. Hence, we have So, it remains to show the following: (1) O n 1 (x 1 ) is G invariant and (2) x 1 is integral over R. deg O n 1 (x 1 ) deg f 2... deg f n = 2 n 1 G 1 = G. First, we deal with (1). By the definition of O n 1 (x 1 ), we have that is a product of g(x 1 + x) = x 1 + go n 1 (x 1 ) n a 1,j (g 1 )x j + gx, where x ranges over F q {x 2,..., x n }. As x ranges over F q {x 2,..., x n }, the sum n a 1,j (g 1 )x j + gx j=2 j=2 also ranges over F q {x 2,..., x n }. Hence, we have go n 1 (x 1 ) = O n 1 (x 1 ). Next, we deal with (2). By Proposition 5.1, we have n 2 O n 1 (X) = X qn 1 + ( 1) n 1 j c n 1,j (x 2,..., x n )X qj. j=0

Mùi invariants and Milnor operations 129 Since Dickson invariants c n 1,j (x 2,..., x n ) are in R 1 = F 2 [x 2,..., x n ] G 1, the polynomial ϕ(x) = O n 1 (X) O n 1 (x 1 ) is a monic polynomial in R[X]. It is clear that ϕ(x 1 ) = 0. Hence, the indeterminate x 1 is integral over R. This completes the proof. 7 Proof of Theorem 4.2 Let G 0 be the subgroup of G consisting of the following matrices: ( ) 1 m, 0 1 n 1 where m M 1,n 1 (F q ), 1 n 1 is the identity matrix in GL n 1 (F q ). Let B n be the set of subsets of {2,..., n}. Let B n,r be the subset of B n such that J B n,r if and only if We write dx J for and we define dx to be 1. J = {j 1,..., j r } and 1 < j 1 < < j r n. dx j1... dx jr The following proposition is nothing but the particular case of Theorem 4.1 and Theorem 4.2. Proposition 7.1 The ring of invariants P G 0 n is given as follows: P G 0 n = F q [O n 1 (x 1 ), x 2,..., x n ]. The ring of invariants (P n E n ) G 0 is a free P G 0 n module with the basis {Q I dx 1... dx n, dx J }, where I ranges over A n 1 and J ranges over B n. Now, we consider a K n basis for K n E n.

130 Masaki Kameko and Mamoru Mimura Proposition 7.2 The elements form a K n basis for K n E 1 n. Q 0... Q n 2 dx 1... dx n, dx 2,..., dx n Proof For dimensional reasons, it suffices to show that the above elements are linearly independent in K n E 1 n. Suppose that a 1 Q 0... Q n 2 dx 1... dx n + a 2 dx 2 + + a n dx n = 0 in K n En 1, where a 1,..., a n are in K n. Then, since n Q 0... Q n 2 dx 1... dx n = ( 1) n i e n 1 (x 1,..., x i,..., x n )dx i, we have i=1 ( 1) n 1 a 1 e n 1 ( x 1,..., x n ) = 0, a 2 + ( 1) n 2 a 1 e n 1 (x 1, x 2,..., x n ) = 0, a n + ( 1) 0 a 1 e n 1 (x 1,..., x n 1, x n ) = 0. Thus, solving this linear system, we obtain a 1 = 0,..., a n = 0. Proposition 7.3 The elements Q I dx 1... dx n, dx J form a K n basis for K n E r n, where I A n 1,n r and J B n,r. Proof Again, for dimensional reasons, it suffices to show that the above elements are linearly independent in K n En. r Suppose that a I Q I dx 1... dx n + b J dx J = 0, I A n 1,n r J B n,r where a I, b J are in K n. The linear independence of the terms dx J is clear. Hence, it remains to show that a I = 0 for each I. Fix I A n 1,n r and let K = S n 1 \I. Then, applying Q K to the both sides of the above equality, we have sign(k, I)a I Q 0... Q n 2 dx 1... dx n + α = 0 in K n E 1 n, where α is a linear combination of dx 2,..., dx n over K n. Hence, by Proposition 7.2, we have a I = 0 for each I A n 1,n r..

Mùi invariants and Milnor operations 131 Suppose that a is in P n E 1 n. Then, on the one hand, we may express a as follows: a = ϕ 1 dx 1 + + ϕ n dx n, where ϕ 1,..., ϕ n are in P n. On the other hand, by Proposition 7.2, we may express a as follows: a = a 1 Q 0... Q n 2 dx 1... dx n + a 2 dx 2 + + a n dx n, where a 1,..., a n are in K n. Observe that terms a and ϕ are unique in the above expressions. We need to show that a 1,..., a n are in P n if a is G 0 invariant. Proposition 7.4 There are polynomials a i over F q in n variables such that for i = 1,..., n. a i e n 1 (x 2,..., x n ) = a i(x 1,..., x n ) Proof For i = 1, we apply D n 1 to a. Then, we have On the other hand, we have Hence, we obtain D n 1 (a) = ϕ 1 O n 1 (x 1 ). D n 1 (a) = ( 1) n 1 a 1 Q 0... Q n 1 dx 1... dx n = ( 1) n 1 a 1 O n 1 (x 1 )e n 1 (x 2,..., x n ). a 1 = (e n 1 (x 2,..., x n )) 1 ϕ 1 and a 1(x 1,..., x n ) = ϕ 1. For i = 2,..., n, applying Q 0,..., Q n 2 to a, we have a linear system Q 0 a = a 2 x 2 + + a n x n, Q 1 a = a 2 x q 2 + + a nx q n,. Q n 2 a = a 2 x qn 2 2 + + a n x qn 2 n.

132 Masaki Kameko and Mamoru Mimura Writing this linear system in terms of matrix, we have Q 0 a a 2 Q 1 a. = A a 3., Q n 2 a where A = a n x 2... x n x q 2... xn q..... x qn 2 2... xn qn 2 It is clear that det A = e n 1 (x 2,..., x n ) 0. It is also clear that each entry of A is in P n 1. Therefore, for some ϕ i,j in P n 1, we have n a i = e n 1 (x 2,..., x n ) 1 ϕ i,j Q j a. Since Q j a is in P n, by letting we obtain the required results. a i(x 1,..., x n ) = j=1. n ϕ i,j Q j a, Proposition 7.5 Suppose that a is G 0 invariant. Then a i(x 1,..., x n ) in Proposition 7.4 are also G 0 invariant for i = 1,..., n. j=1 Proof For g G 0, we have g(dx i ) = dx i for i = 2,..., n and, since G 0 SL n (F q ), we have g(dx 1... dx n ) = dx 1... dx n. Since the action of Milnor operations Q j commutes with the action of the general linear group GL n (F q ), we have g(q 0... Q n 2 dx 1... dx n ) = Q 0... Q n 2 dx 1... dx n.

Mùi invariants and Milnor operations 133 Hence, we have ga = (ga 1 )Q 0... Q n 2 dx 1... dx n + (ga 2 )dx 2 + + (ga n )dx n. Thus, if ga = a, we have ga 1 = a 1,..., ga n = a n. It is also clear that e n 1 (x 2,..., x n ) is G 0 invariant. Hence, a i(x 1,..., x n ) are also G 0 invariant for i = 1,..., n. Proposition 7.6 Suppose that a is G 0 invariant. Then, a 1 (x 1,....x n ) is divisible by x n. Proof Let us consider the coefficient ϕ n of dx n ; we have ϕ n e n 1 (x 2,..., x n ) = a n(x 1,..., x n ) + a 1(x 1,..., x n )e n 1 (x 1,..., x n 1 ). Since a 1 (x 1,..., x n ) and a n(x 1,..., x n ) are G 0 invariant, there are polynomials a 1, a n over F q in n variables such that a 1 (x 1,..., x n ) = a 1 (O n 1(x 1 ), x 2,..., x n ), a n(x 1,..., x n ) = a n(o n 1 (x 1 ), x 2,..., x n ). Since O n 1 (x 1 ), x 2,..., x n are algebraically independent, it suffices to show that a 1(y 1, x 2,..., x n 1, 0) = 0 for algebraically independent y 1, x 2,..., x n 1. Substituting x n = 0, we have the obvious projection π : F q [x 1,..., x n ] F q [x 1,..., x n 1 ]. It is clear from Proposition 5.2 and Proposition 5.3 that π(e n 1 (x 1,..., x n 1 )) = e n 1 (x 1,..., x n 1 ) = O n 2 (x 1 )e n 2 (x 2,..., x n 1 ) for n 3, π(e 1 (x 1 )) = O 0 (x 1 ) for n = 2, π(e n 1 (x 2,..., x n )) = 0, π(o n 1 (x 1 )) = O n 2 (x 1 ) q. Hence, for n 3, we have 0 = a n(y q, x 2,..., x n 1, 0) + ya 1(y q, x 2,..., x n 1, 0)e n 2 (x 2,..., x n 1 ), where y = O n 2 (x 1 ) and y, x 2,..., x n 1 are algebraically independent. Applying the partial derivative / y, we have a 1(y q, x 2,..., x n 1, 0)e n 2 (x 2,..., x n 1 ) = 0.

134 Masaki Kameko and Mamoru Mimura Hence, we have a 1(y q, x 2,..., x n 1, 0) = 0. Since y q, x 2,..., x n 1 are algebraically independent, we have the required result. For n = 2, we have 0 = a 2(y q, 0) + ya 1(y q, 0). Applying the partial derivative / y, we have and the required result. a 1(y q, 0) = 0 Lemma 7.7 Suppose that g 1 GL n 1 (F q ) and that a is G 0 invariant. Then, ḡ 1 a is also G 0 invariant. Proof Suppose that for each g, there is a g G 0 such that gḡ 1 = ḡ 1 g. If it is true, then for any G 0 invariant a, we have gḡ 1 a = ḡ 1 g a = g 1 a. Hence, g 1 induces a homomorphism from P G 0 n to P G 0 n. So, it suffices to show that for each g in G 0, there is a g G 0 such that gg 1 = g 1 g, which is immediate from the following equality: ( ) ( ) ( ) ( ) 1 m 1 0 1 0 1 mg 1 =, 0 1 n 1 0 g 1 0 g 1 0 1 n 1 where m M 1,n 1 (F q ) and 1 n 1 stands for the identity matrix in GL n 1 (F q ). Proposition 7.8 Suppose that a is G 0 invariant. Then, a 1,..., a n are in P n. Proof Firstly, we verify that a 1 is in P n. To this end, we prove that the element a 1 (x 1,..., x n ) is divisible by e n 1 (x 2,..., x n ). Let l = α 2 x 2 + + α n x n, where α 2,..., α n F q and l 0. By Proposition 5.6, it suffices to show that a 1 (x 1,..., x n ) is divisible by l. There is g 1 in GL n 1 (F q ) such that ḡ 1 (x n ) = l. Since, by Lemma 7.7, ḡ 1 1 a is also in (P n E 1 n) G 0, there is an element f in P n such that D n 1 (ḡ 1 1 a) = fx no n 1 (x 1 ).

Mùi invariants and Milnor operations 135 Here we have So we have O n 1 (x 1 )a 1(x 1,..., x n ) = D n 1 (a) = ḡ 1 D n 1 (ḡ 1 1 a) = O n 1(x 1 )ḡ 1 (f )l. a 1(x 1,..., x n ) = (ḡ 1 f )l. Secondly, we verify that a i are in P n for i = 2,..., n, which follows from the fact that ϕ i = a i + ( 1) n i a 1 e n 1 (x 1,..., x i,..., x n ). Proof of Proposition 7.1 Suppose that a is in P n En r and G 0 invariant. By Proposition 7.3, there are a I, b J K n such that a = a I Q I dx 1... dx n + b J dx J. I A n 1,n r J B n,r It suffices to show that a I, b J are in P n. Firstly, we verify that a I is in P n. Choose I and let K = S n 1 \I. Then, we have Q K a = sign(k, I)a I Q 0... Q n 2 dx 1... dx n + J B n,r b J Q K dx J. By Proposition 7.8, sign(k, I)a I is in P n and, by definition, sign(k, I) 0, hence a I is also in P n. Secondly, we prove that b J is in P n. Put a = a a I Q I dx 1... dx n = b J dx J. I A n 1,n r J B n,r It is clear that a is also in P n E r n. Hence, b J is in P n. This completes the proof. Now, we complete the proof of Theorem 4.2. Proof of Theorem 4.2 Suppose that a is an element in P n E n and that a is also G invariant. It suffices to show that a is a linear combination of {v i, Q I dx 1..., dx n } over P G n. It is clear that a is also G 0 invariant. Hence, by Proposition 7.1, there are a I, b J in P G 0 n = F q [O n 1 (x 1 ), x 2..., x n ] such that a = I a I Q I dx 1... dx n + J b J dx J.

136 Masaki Kameko and Mamoru Mimura Thus, we have a = I k 0 a I,k O n 1 (x 1 ) k Q I dx 1... dx n + J b J,k O n 1 (x 1 ) k dx J, where a I,k, b J,k are in P n 1 = F q [x 2,..., x n ]. Since, by Theorem 4.1, g G acts trivially on O n 1 (x 1 ), and since g G SL n (F q ) acts trivially on Q I dx 1... dx n, we have ga = (ga I,k )O n 1 (x 1 ) k Q I dx 1... dx n + (gb J,k )O n 1 (x 1 ) k (gdx J ). k 0 I k 0 J It is clear that gdx J is in E n 1. As a P n 1 module, (P n E r n) G 0 is a free P n 1 module with the basis k 0 {O n 1 (x 1 ) k dx J, O n 1 (x 1 ) k Q I dx 1... dx n }. Hence, we have Thus, a I,k is in P n 1 G 1 g(a I,k ) = a I,k. and so a I is in P G n. Put a = a I a I Q I dx 1... dx n. Then, a is also in the ring of invariants (P n E n ) G, and we have a = ( ) b J,k dx J O n 1 (x 1 ) k. k 0 J Hence, J b J,k dx J is in the ring of invariants (P n 1 E n 1 ) G 1. By the assumption on the ring of invariants (P n 1 E n 1 ) G 1, there are polynomials b 1,k,..., b 2 n 1,k in P G 1 n 1 such that 2 n 1 b J,k dx J = b i,k v i. Thus, writing b i for k 0 b i,k O n 1 (x 1 ) k, we have J i=1 a = b i v i + I 2 n 1 i=1 where b i, a I are in P G n. This completes the proof. a I Q I dx 1... dx n,

Mùi invariants and Milnor operations 137 A Appendix In [7], Mùi used the determinant of the k k matrix ( x qi l ) j whose (l, j) entry is x q i l j to describe the Dickson invariant [i 1,..., i k ]. Using these Dickson invariants, he defined the Mùi invariant [r : i 1,..., i n r ]. In this appendix, we verify in Proposition A.4 that the Mùi invariant Q i1... Q in r dx 1... dx n in this paper is indeed equal to the Mùi invariant [r : i 1,..., i n r ], up to sign. Firstly, we recall the definitions. See [7, Section 2] for the definition of [i 1,..., i k ] and and [7, Defintion 4.3] for the definition of [r : i 1,..., i n r ]. Definition A.1 defined by The Dickson invariant [i 1,..., i k ](x 1,..., x k ) F q [x 1,..., x k ] is σ sgn(σ)x qi 1 σ(1)... xqi k σ(k) = det(xqi l j ), where σ ranges over the set of permutations of {1,..., k} and sgn(σ) is the sign of the permutation σ. Definition A.2 The Mùi invariant [r : i 1,..., i n r ] P n E r n is defined by [r : i 1,..., i n r ] = J sgn(σ J )dx j1... dx jr [i 1,..., i n r ](x jr+1,..., x jn ), where σ J = ( 1,..., n j 1,..., j n ) ranges over the set of permutations of {1,..., n} such that j 1 < < j r and j r+1 < < j n. The above σ J corresponds to the subset J = {j 1,..., j r } of order r of {1,..., n}. Secondly, we prove the following proposition. Proposition A.3 There holds [i 1,..., i k ](x 1,..., x k ) = Q ik... Q i1 dx 1... dx k = ( 1) k(k 1)/2 Q i1... Q ik dx 1... dx k.

138 Masaki Kameko and Mamoru Mimura Proof We prove the first equality in this proposition by induction on k. Indeed, in the case k = 1, the proposition holds. Suppose that k 2 and that there holds the equality [i 2,..., i k ](x 1,..., x k 1 ) = Q ik... Q i2 dx 1... dx k 1. Using the cofactor expansion (or the Laplace development) of the k k matrix (x qi l j ) along the first row we have [i 1,..., i k ](x 1,..., x k ) = (x qi 1 1, xqi 1 2,..., xqi 1 k ), k s=1 ( 1) s+1 [i 2,..., i k ](x 1,..., x s,..., x k )x qi 1 s k = ( 1) s+1 (Q ik... Q i2 dx 1... dx s... dx k )x qi 1 s s=1 ( k ) = Q ik... Q i2 ( 1) s+1 x qi 1 s dx 1... dx s... dx k s=1 = Q ik... Q i1 dx 1... dx k. So, the first equality holds. The second equality is immediate from the fact that Q ik... Q i1 = ( 1) k(k 1)/2 Q i1... Q ik. Finally, we state and prove the following proposition. Proposition A.4 There holds [r : i 1,..., i n r ] = ( 1) (n r)r Q in r... Q i1 dx 1... dx n = ( 1) (n r)r+(n r)(n r 1)/2 Q i1... Q in r dx 1... dx n. Proof As in the definition of [r : i 1,..., i n r ], let σ J be a permutation of {1,..., n} with σ J (1) < < σ J (r), σ J (r + 1) < < σ J (n) and we denote by j k the value σ J (k) of σ J at k. Let I(J) be the ideal of P n E n generated by dx jr+1,..., dx jn. Let be the projection. It is clear that p J : P n E n P n E n /I(J) P n E r n/((p n E r n) I(J)) = P n

Mùi invariants and Milnor operations 139 and for f in P n E r n, we have f = J p J (f )dx j1... dx jr. So, by Proposition A.3, in order to prove the proposition, it suffices to show that p J (Q in r... Q i1 dx 1... dx n ) = ( 1) (n r)r sgn(σ J )Q in r... Q i1 dx jr+1... dx jn. Suppose that ψ(q in r... Q i1 ) = 1 Q in r... Q i1 + a a, where ψ is the coproduct of the Steenrod algebra. We may choose a, so that a = Q e1... Q el and l < n r. Thus, a (dx jr+1... dx jn ) belongs to I(J). Then, there holds Hence, we have as required. Q in r... Q i1 dx 1... dx n = sgn(σ J )Q in r... Q i1 dx j1... dx jn = ( 1) (n r)r sgn(σ J )dx j1... dx jr Q in r... Q i1 dx jr+1... dx jn + ( 1) r deg a sgn(σ J )(adx j1... dx jr )(a dx jr+1... dx jn ). p J (Q in r... Q i1 dx 1... dx n ) = ( 1) (n r)r sgn(σ J )Q in r... Q i1 dx jr+1... dx jn References [1] K Andersen, J Grodal, J Møller, A Viruel, The classification of p compact groups for p odd arxiv:math.at/0302346 [2] D J Benson, Polynomial invariants of finite groups, London Mathematical Society Lecture Note Series 190, Cambridge University Press, Cambridge (1993) MR1249931 [3] M C Crabb, Dickson Mùi invariants, Bull. London Math. Soc. 37 (2005) 846 856 MR2186717 [4] L E Dickson, A fundamental system of invariants of the general modular linear group with a solution of the form problem, Trans. Amer. Math. Soc. 12 (1911) 75 98 MR1500882 [5] R L Griess, Jr, Elementary abelian p subgroups of algebraic groups, Geom. Dedicata 39 (1991) 253 305 MR1123145

140 Masaki Kameko and Mamoru Mimura [6] M Kameko, M Mimura, On the Rothenberg-Steenrod spectral sequence for the mod 3 cohomology of the classifying space of the exceptional Lie group E 8, from: Proceedings of the Nishida Fest (Kinosaki 2003), Geom. Topol. Monogr. 10 209 222 [7] H Mùi, Modular invariant theory and cohomology algebras of symmetric groups, J. Fac. Sci. Univ. Tokyo Sect. IA Math. 22 (1975) 319 369 MR0422451 [8] L Smith, Polynomial invariants of finite groups, Research Notes in Mathematics 6, A K Peters Ltd., Wellesley, MA (1995) MR1328644 [9] C Wilkerson, A primer on the Dickson invariants, from: Proceedings of the Northwestern Homotopy Theory Conference (Evanston, Ill., 1982), Contemp. Math. 19, Amer. Math. Soc., Providence, RI (1983) 421 434 MR711066 A corrected version is available at the Hopf Topology Archive. Department of Mathematics, Faculty of Regional Science Toyama University of International Studies, 65-1 Higashikuromaki, Toyama, 930-1292, Japan Department of Mathematics, Faculty of Science, Okayama University 3-1-1 Tsushima-naka, Okayama, 700-8530, Japan kameko@tuins.ac.jp, mimura@math.okayama-u.ac.jp Received: 26 Feb 2005 Revised: 12 Sept 2005