DSpace at VNU: The hit problem for the modular invariants of linear groups

Journal of Algebra 246, 367–384 (2001) doi:10.1006/jabr.2001.8974, available online at http://www.idealibrary.com on The Hit Problem for the Modular Invariants of Linear Groups1 Nguye˜ˆ n H V Hu· ng and Traˆ` n Ngo.c Nam Department of Mathematics, Vietnam National University, ˜ˆ Tr˜ Hanoi, 334 Nguyen Street, Hanoi, Vietnam E-mail: nhvhung@hotmail.com, trngnam@hotmail.com Communicated by G D James Received March 5, 2001 Let the mod Steenrod algebra, , and the general linear group, GLk = xk with deg xi = in the usual manner We GL k , act on Pk = x1 prove that, for a family of some rather small subgroups G of GLk , every element in Pk In other words, of positive degree in the invariant algebra PkG is hit by PkG + ⊂ + · Pk , where PkG + and + denote respectively the submodules of PkG and consisting of all elements of positive degree This family contains most of the parabolic subgroups of GLk It should be noted that the smaller the group G is, the harder the problem turns out to be Remarkably, when G is the smallest group of the family, the invariant algebra PkG is a polynomial algebra in k variables, whose degrees are ≤ and fixed while k increases It has been shown by Hu· ng [Trans Amer Math Soc 349 (1997), 3893–3910] GL that, for G = GLk , the inclusion Pk k + ⊂ + · Pk is equivalent to a weak algebraic version of the long-standing conjecture stating that the only spherical classes in Q0 S are the elements of Hopf invariant and those of Kervaire invariant  2001 Elsevier Science Key Words: Steenrod algebra; invariant theory; Dickson invariant; M` ui invariant INTRODUCTION Let Pk = x1 xk be the polynomial algebra over the field of two xk , each of degree It is equipped elements, , in k variables x1 with the usual structure of a module over GLk = GL k by means of The work was supported in part by the National Research Program, Grant 1.4.2 367 0021-8693/01 $35.00  2001 Elsevier Science All rights reserved hu· ng and nam 368 substitutions of variables The mod Steenrod algebra, use of the formula  j = 0,  xi j Sq xi = xi j=1  otherwise, , acts upon Pk by and subject to the Cartan formula Sqn fg = n Sqj f Sqn−j g j=0 for f g ∈ x1 xk Let G be a subgroup of GLk Then Pk possesses the induced structure of G-module Denote by PkG the subalgebra of all G-invariants in Pk Since the action of GLk and that of on Pk commute with each other, PkG is also an -module In [3], the first named author is interested in the homomorphism jG 2⊗ PkG → ⊗Pk G induced by the identity map on Pk He also sets up the following conjecture for G = GLk and shows that it is equivalent to a weak algebraic version of the long-standing conjecture stating that the only spherical classes in Q0 S are the elements of Hopf invariant and those of Kervaire invariant Conjecture 1 jGLk = in positive degrees for k > This has been established for k = in [3] and then for arbitrary k > in [6] That the conjecture is no longer valid for k = and k = is respectively shown in [3] to be an exposition of the existence of the Hopf invariant-1 and the Kervaire invariant-1 classes In the present paper, we are interested in the following problem: Which subgroup G of GLk possesses jG = in positive degrees? It should be noted that, as observed in the introduction of [3], jG = in positive degrees ⇐⇒ PkG + ⊂ + · Pk where PkG + and + denote respectively the submodules of PkG and consisting of all elements of positive degree Therefore, the smaller the group G is, the harder the problem turns out to be For instance, we have understood that jG = for G = , G = GL1 , or G = GL2 Furthermore, let Tk be the Sylow 2-subgroup of GLk consisting of all upper triangular matrices with entries on the main diagonal Then jTk = 0; indeed V1 = x1 is a Tk -invariant; however, x1 ∈ + · Pk The problem we are interested in is closely related to the hit problem of determination of ⊗ Pk This problem was first studied by Peterson the hit problem 369 [11], Wood [16], Singer [14], and Priddy [12], who showed its relationships to several classical problems in cobordism theory, modular representation theory, Adams spectral sequence for the stable homotopy of spheres, and stable homotopy type of classifying spaces of finite groups The tensor product ⊗ Pk has explicitly been computed for k ≤ (see [9]) It seems unlikely that an explicit description of ⊗ Pk for general k will appear in the near future There is also another approach, the qualitative one, to the problem By this we mean giving conditions on elements of Pk to show that they go to zero in ⊗ Pk , i.e., belong to + · Pk Peterson’s conjecture [11], which has been established by Wood [16], claims that ⊗ Pk = in certain degrees Recently, Singer, Monks, and Silverman have refined Wood’s method to show that many more monomials in Pk are in + · Pk (See Silverman [13] and references therein.) In this paper, we prove that jG = in positive degrees, or equivalently, PkG + ⊂ + · Pk , for a family of some rather small groups G This family contains most of the parabolic subgroups of GLk Observing the obstructions of the Hopf invariant-1 and the Kervaire invariant-1 classes, it seems necessary to make the hypothesis that G ⊃ GL3 in order to get jG = in positive degrees Let us consider the subgroup A ∗ B G1 • G2 = A ∈ G1 B ∈ G2 ⊂ GLk where G1 is a subgroup of GLn and G2 is a subgroup of GLk−n for n ≤ k We are especially interested in the case G1 = GLn and G2 = 1k−n , the unit subgroup of GLk−n We suppose n > so that GL3 ⊂ GLn • 1k−n Here is an interpretation of this group, which does not depend on coordinates Let V be an -vector space of dimension k and let W be a vector subspace of dimension n Then, the subgroup GLn • 1k−n can be interpreted as the subgroup of GL V consisting of all isomorphisms ϕ V → V with ϕ W = W and ϕ¯ = idV/W , where ϕ¯ denotes the induced homomorphism of ϕ on V/W We compute the algebra of GLn • 1k−n -invariants by combining the works of Dickson [1] and M` ui [10] M` ui’s invariant of degree 2n−1 is defined as follows: Vn = λj ∈ λ1 x1 + · · · + λn−1 xn−1 + xn Dickson’s invariant of degree 2n − 2s is defined by the inductive formula Qn s = Qn−1 where, by convention, Qn in [1] that x1 n = 1, Qn xn GLn = s−1 s + Vn Qn−1 s = for s < Then, Dickson proves Qn Qn n−1 hu· ng and nam 370 while M` ui shows in [10] that x1 xk Tk = V1 Vk To generalize these works, we set Vn+1 xi = λ x1 + · · · + λ n xn + x i λj ∈ for n < i ≤ k Then, we get Theorem 1.2 For k ≥ n, x1 xk GLn •1k−n = Qn Qn n−1 Vn+1 xn+1 Vn+1 xk The purpose of this paper is to prove Theorem 1.3 (Main theorem) jGLn •1k−n = in positive degrees if and only if n > Obviously, GL3 • 1k−3 is the smallest group among all the ones of the form GLn • 1k−n for n > Being applied to this group, the main theorem shows that Q3 Q3 Q3 V4 x4 V4 xk + ⊂ + · Pk where deg Q3 = 7, deg Q3 = 6, deg Q3 = 4, deg V4 xi = for < i ≤ k This gives a large family of elements, which are hit by in Pk Remarkably, the degrees of all the generators of this polynomial algebra are small and not depend on k Let us now study the parabolic subgroup of GLk :    ∗ A1       A2    A ∈ GL with k +···+k = k GLk1 km =  i k m i        Am It is easily seen that GLk1 • 1k−k1 is a subgroup of GLk1 we have Corollary 1.4 jGLk km km Therefore, = in positive degrees if and only if k1 > Let G be a subgroup of GLk and let ω ∈ GLk It is easily seen that = ωPkG As the action of GLk on Pk commutes with that of , Theorem 1.3 and Corollary 1.4 also claim that jG = for any subgroup G, which is conjugate either to GLn • 1k−n with n > or to GLk1 km with k1 > Note that GLk is a special case of the parabolic subgroup GLk1 km with k = k1 and m = Hence we obtain an alternative proof for Conjecture 1.1: −1 PkωGω the hit problem Corollary 1.5 [6] 371 jGLk = in positive degrees if and only if k > The readers are referred to [4] and [5] for some problems, which are related to the main theorem and Corollary 1.5 Additionally, the problem GL of determination of ⊗ Pk k and its applications have been studied by Hu· ng and Peterson [7 8] The paper contains five sections and is organized as follows We determine the algebra of GLn • 1k−n -invariants in Section and study the action of on this algebra for n = in Section The main theorem and its corollaries are proved in Section assuming the truth of Lemma 4.2 as a key tool Finally, we show this lemma in Section and then complete the proof of the main theorem THE INVARIANT ALGEBRA OF GLn • 1k−n The action of GLk on Pk = x1 xk is precisely described as follows For every ω = ωij k×k ∈ GLk and any f ∈ Pk , one defines ωf x1 where ωx1 xk = f ωx1 ωxk ωxk are given by ωxj = 1≤i≤k ωij xi 1≤j≤k Then, each subgroup G of GLk possesses the induced action on Pk Using the notations given in the introduction, we get the following theorem, which is also numbered as Theorem 1.2 Theorem 2.1 For k ≥ n, x1 xk GLn •1k−n = Qn Qn n−1 Vn+1 xn+1 Vn+1 xk We prove this theorem by three lemmata Lemma 2.2 The polynomials Qn are GLn • 1k−n -invariants Qn n−1 , Vn+1 xn+1 Vn+1 xk Proof The polynomials Qn Qn n−1 depend only on x1 xn but not on xn+1 xk On the other hand, the action of A0 E ∗ ∈ GLn • 1k−n k−n on x1 xn , where Ek−n denotes the unit k − n × k − n -matrix, is exactly the same as that of A ∈ GLn Therefore, according to Dickson [1], Qn n−1 are GLn • 1k−n -invariants Qn Note that Vn+1 xi can be rewritten as Vn+1 xi = x∈ x + xi n nh R As deg Q2 = is the smallest number of the degrees of Q0 , Q1 , Q2 , W4 Wk , the degree information shows that h R < h T ≤ h R + i/4 for every T in the sum Consider an arbitrary term S = Sqj1 R1 · · · Sqjh Rh in the sum As h S = h R , we can see that jp = for every p with Rp being one of the invariants Q0 , W4 Wk Suppose the contrary that i2 S ≥ i2 R (Then, we have actually i2 S = i2 R because of h S = h R ) By Proposition 3.1, jp = for every p with Rp = Q2 So, jp could be nonzero just only in the case Rp = Q1 Furthermore, as h S = h R and by Proposition 3.1, if jp = 0, then jp = Therefore, i = j1 + · · · + jh ≤ i1 R This contradicts the hypothesis that i > i1 R The lemma is proved Lemma 3.4 Suppose R is an H-monomial in PkH and n is a nonnegative R = Then integer such that i2 R ≡ 2n − (mod 2n ) and h2n−1 R = Sq2 i R where R = R/Q22 s S < n Proof n+1 i R −2n−1 RQ22 + S and each term S in the sum is an H-monomial with We have i R h R = h RQ22 = h R + i2 R ≡ h R + 2n − (mod 2n R Hence h R + 2n−1 − ≡ h R − 2n−1 (mod 2n ) As h2n−1 = 0, the term n−1 n−1 occurs in the 2-adic expansion of h R − Thus n−1 h RQ22 2n−1 −1 = h R + 2n−1 − 2n−1 = h R − 2n−1 2n−1 =1 hu· ng and nam 376 Applying Lemma 3.3(i) to RQ22 Sq2 n+1 RQ22 n−1 n−1 −1 −1 and i = 2n−1 , we get = RQ22 n −1 + S where each S is an H-monomial in PkH satisfying i2 S < i2 RQ22 n−1 −1 + 2n−1 = 2n − This inequality implies s S < n Put a = i2 R − 2n − ≡ (mod 2n ) By the Cartan formula and Proposition 3.1, we have Sq2 n+1 i R −2n−1 RQ22 = Sq2 n+1 RQ22 n−1 −1 = Sq2 n+1 RQ22 n−1 −1 = RQ22 n −1 + Q2a Q2a + RQ22 S Q2a + RQ22 S Q2a + RQ22 =R+ n−1 n−1 −1 Sq2 −1 n−1 n+1 Sq2 −1 n+1 Sq2 Q2a n+1 Q2a Q2a where each term S Q2a in the sum satisfies s S Q2a < n, because s S < n and a ≡ (mod 2n ) On the other hand, from Proposition 3.1, if n+1 Sq2 Q2a = 0, then it is not divisible by Q2 Therefore s RQ22 n−1 −1 Sq2 n+1 Q2a n+1 RQ22 = s Q22 n−1 −1 =n−1 To sum up, we can write R = Sq2 i R −2n−1 + S where each term S satisfies s S < n The lemma is proved Lemma 3.5 Suppose R is an H-monomial in PkH , which is not divisible by Q2 , while n and i are positive integers satisfying h R ≡ mod 2n i1 R ≤ 2n − 2n ≤ i ≤ 2n+1 Then Sqi RQ22 n −1 = S+ T where each term S is an H-monomial in PkH with s S < n, while each term T T = is an H-monomial in PkH with i2 T ≡ 2n − (mod 2n ) and h2n−1 the hit problem 377 Proof Note that i ≥ 2n > 2n − ≥ i1 R = i1 RQ22 n Lemma 3.3(ii) to RQ22 −1 and i, we get Sqi RQ22 n −1 = S+ n −1 < h T ≤ h RQ22 −1 Applying T where each S is an H-monomial with i2 S < i2 RQ22 each T is an H-monomial with h RQ22 n n −1 n −1 = 2n − 1, while + i/4 n For each S in the sum, as i2 S < − 1, it implies s S < n For each T in the sum, we have h RQ22 n −1 = h R + 2n − < h T ≤ h R + 2n − + i/4 ≤ h R + 2n + 2n−1 − Hence h R + 2n ≤ h T ≤ h R + 2n + 2n−1 − Combining these T inequalities with the hypothesis h R ≡ (mod 2n ), we obtain h2n−1 = n Finally, suppose i2 T = − + b, where b is an integer (that can be positive, negative, or zero) If b ≡ (mod 2n ), then i2 T ≡ 2n − (mod 2n ) Otherwise, if b ≡ (mod 2n ), then s T < n and such a T can be considered as a term in the sum S The lemma is proved PROOFS OF THE MAIN THEOREM AND ITS COROLLARIES The following two lemmata will play a key role in the proof of the main theorem Lemma 4.1 Let R = be a product of some distinct elements in the set Q0 Q1 Q2 W4 Wk Then R ∈ Sq1 Pk + Sq2 Pk Proof We write R = RS with R Q1 Q2 and S Q0 W4 · · · Wk If Q1 R, then from Proposition 3.1, Sq1 R = Sq1 RS = Hence, by [6, Lemma 2.5], R ∈ Sq1 Pk If R = Q1 , then by Proposition 3.1, R = Q1 S = Sq2 Q2 S = Sq2 Q2 S ∈ Sq2 Pk Finally, if R = Q1 Q2 , then by [6, Lemma B], we have Q1 Q2 = Sq1 u1 + Sq2 u2 for some elements u1 u2 ∈ Pk Then R = Q1 Q2 S = Sq1 u1 + Sq2 u2 S = Sq1 u1 S + Sq2 u2 S ∈ Sq1 Pk + Sq2 Pk The lemma follows We postpone the proof of the next lemma until the last section hu· ng and nam 378 Lemma 4.2 Suppose R is an H-monomial in PkH , u = is an arbitrary element in Pk , and n is a positive integer n (i) If s R < n, then Ru2 ∈ (ii) n + · Pk n If i2 R ≡ − (mod ) and n h R 2n−1 n = 0, then Ru2 ∈ n + · Pk n (iii) If i2 R = − ≥ i1 R , h R ≡ − (mod ), and u ∈ n Sq1 Pk + Sq2 Pk , then Ru2 ∈ + · Pk Proof of the Main Theorem It suffices to show the theorem for the group H = GL3 • 1k−3 , as this is the smallest one of the groups GLn • 1k−n for n > Moreover, using Theorem 1.2, we need only to prove that PkH + = Q0 Q1 Q2 W4 Wk + ⊂ + · Pk for every k > Suppose R is an H-monomial of positive degree in PkH We need to show that R ∈ + · Pk Set n = s S Then, by definition, i2 R ≡ 2n − (mod 2n+1 ) Let us consider the following four cases n Case Q22 divides R Combining this with the hypothesis i2 R ≡ 2n − (mod 2n+1 ), it implies n+1 dividing R Denoting R = R/Q22 , we have i2 R = i2 R − 2n+1 ≡ − (mod 2n+1 ) Thus s R = n < n + Applying Lemma 4.2(i) to the n+1 triple R Q2 n + , we get R = RQ22 ∈ + · Pk n+1 Q22 n Case There exists u ∈ Q0 Q1 W4 Wk such that u2 n+1 divides R n+1 Setting R = R/u2 , we have s R = s R = n < n + Applying n+1 Lemma 4.2(i) to the triple R u n + , we get R = Ru2 ∈ + · Pk Case i0 R , i1 R , i2 R , i4 R ik R all are ≤ 2n+1 − and there n Wk such that u2 divides R exists u ∈ Q0 Q1 Q2 W4 By Case 1, u = Q2 Furthermore, since i2 R ≤ 2n+1 − and i2 R ≡ − (mod 2n+1 ), it implies i2 R = 2n − We investigate the following three subcases n Case 3a n = Then, by Lemma 4.1, R ∈ Sq1 Pk + Sq2 Pk R Case 3b n ≥ and there exists m with < m ≤ n and h2m−1 = Obvim R ously i2 R ≡ 2m − (mod 2m ) Put R = R/u2 and we have h2m−1 = m h R −2 h R = 2m−1 = Applying Lemma 4.2(ii) to the triple R u m , we get 2m−1 m R = Ru2 ∈ + · Pk the hit problem 379 R = for every m with < m ≤ n It implies Case 3c n ≥ and h2m−1 n n n h R ≡ − (mod ) We write uniquely R in the form R = Rv2 , where v = is a certain product of distinct elements in the set Q0 Q1 W4 Wk (consequently, i2 v = 0), and R is a certain H-monomial with i0 R , i1 R , i2 R , i4 R ik R all ≤ 2n − Observe that i R = i R − n i v = n − ≥ i1 R h R = h R − 2n h v ≡ 2n − (mod 2n By Lemma 4.1, v ∈ Sq1 Pk + Sq2 Pk Applying Lemma 4.2(iii) to the triple n R v n , we get R = Rv2 ∈ + · Pk Case i0 R , i1 R , i2 R , i4 R ik R all are ≤ 2n − In particular, i2 R = 2n − 1, since i2 R ≡ 2n − (mod 2n+1 ) It should be noted that n > 0; otherwise, R = with degree We also examine the following three subcases Case 4a n = Then, by Lemma 4.1, R ∈ Sq1 Pk + Sq2 Pk R = It is obCase 4b n ≥ and there exists m with < m < n and h2m−1 m R m m vious that i2 R ≡ − (mod ) Put R = R/Q2 and we have h2m−1 = m h R −2 h R m , we = = Applying Lemma 4.2(ii) to the triple R Q 2m−1 2m−1 m get R = RQ22 ∈ + · Pk R = for every m with < m < n It implies Case 4c n ≥ and h2m−1 n−1 h R ≡ 2n−1 − (mod 2n−1 ) We write uniquely R in the form R = Ru2 , where u = is a certain product of distinct elements in the set Q0 Q1 Q2 W4 Wk with i2 u = 1, and R is a certain H-monomial with i0 R , i1 R , i2 R , i4 R ik R all ≤ 2n−1 − Note that i2 R = i2 R − 2n−1 i2 u = 2n − − 2n−1 = 2n−1 − ≥ i1 R h R = h R − 2n−1 h u ≡ 2n−1 − (mod 2n−1 ) By Lemma 4.1, we have u ∈ Sq1 Pk + Sq2 Pk Applying Lemma 4.2(iii) to n−1 the triple R u n − , we obtain R = Ru2 ∈ + · Pk The main theorem is completely proved Proof of Corollary 1.4 Note that GLk1 • 1k−k1 is a subgroup of GLk1 km So, by the main theorem, we have GLk1 Pk for k1 > km + GLk1 •1k−k1 + ⊂ Pk ⊂ + · Pk hu· ng and nam 380 If k1 = 1, then it is easily seen that Q1 ∈ GL1 x1 + GLk1 km ⊂ Pk + However, Q1 = x1 ∈ + · Pk Finally, if k1 = 2, then we observe that Q2 ∈ x1 x2 while Q2 = x21 + x22 + x1 x2 ∈ The corollary is proved + GL2 + GLk1 ⊂ Pk km + · Pk Since the general linear group GLk is a special case of the parabolic subgroup GLk1 km with k = k1 and m = 1, Corollary 1.5 follows PROOF OF LEMMA 4.2 The lemma is proved by induction Its starting case is handled by the following lemma Lemma 5.1 Suppose R is an H-monomial in PkH with s R = 0, and u = is an arbitrary element in Pk Then Ru2 ∈ Sq1 Pk + Sq2 Pk Proof We consider the following two cases Case i1 R ≡ (mod 2) By Proposition 3.1, we have Sq1 Ru2 = Sq1 R u2 = So, using [6, Lemma 2.5], we get Ru2 ∈ Sq1 Pk Case i1 R ≡ (mod 2) i Put S = R/Q1 Q22 with i2 = i2 R Since s R = 0, the number i2 is even Then we have i Ru2 = SQ22 u2 Q1 i = SQ22 u2 Sq2 Q2 i i = Sq2 SQ22 u2 Q2 + Sq2 SQ22 u2 Q2 i +1 = Sq2 SQ22 u i +1 + Sq2 Su2 Q22 the hit problem 381 It is easy to see that Sq2 Su2 = Sq2 S u2 + S Sq1 u Combining this with the fact i1 S = i1 Sq2 S ≡ (mod 2), we obtain Sq1 Sq2 Su2 = Then, by [6, Lemma 2.5], this gives Sq2 Su2 = Sq1 v for some v ∈ Pk Therefore i +1 i +1 Sq2 Su2 Q22 i +1 = Sq1 vQ22 = Sq1 vQ22 So, in any case, we have Ru2 ∈ Sq1 Pk + Sq2 Pk The lemma is proved Proof of Lemma The proof is divided into three steps Step If Lemma 4.2(i) and Lemma 4.2(ii) are valid for every n ≤ N, then so is Lemma 4.2(iii) for every n ≤ N Suppose u = Sq1 v1 + Sq2 v2 for some v1 v2 ∈ Pk We have n Ru2 = R Sq1 v1 + Sq2 v2 = R Sq1 v1 2n n = Sq2 Rv12 + Sq2 n n+1 n = Sq2 Rv12 n 2n 2n + R Sq2 v2 n + Sq2 R v12 Rv12 n n + Sq2 R Sq1 v2 + Sq2 n+1 Rv22 n n n + Sq2 R v12 + Sq2 n n+1 2n + Sq2 n+1 n + Sq2 R Sq1 v2 R v22 R v22 n 2n n Note that n Sq2 R Sq1 v2 2n n 2n n 2n = Sq2 R Sq1 v2 = Sq2 R Sq1 v2 + R Sq1 Sq1 v2 2n as Sq1 Sq1 = Thus n n n Ru2 + Sq2 R v12 + Sq2 n+1 n R v22 ∈ + · Pk n Set R = R/Q22 −1 Obviously, R is an H-monomial in PkH that is not divisible by Q2 with h R = h R − 2n − ≡ mod 2n ) and i1 R = i1 R ≤ 2n − Using Lemma 3.5, we get n n Sq2 R = Sq2 RQ22 Sq2 n+1 R = Sq2 n+1 n −1 RQ22 n = −1 S1 + = T1 S2 + T2 where each term S1 or S2 is an H-monomial with s S1 < n and s S2 < n, while each term T1 or T2 is an H-monomial with i2 T1 ≡ i2 T2 ≡ 2n − T1 T2 = h2n−1 = Hence (mod 2n ) and h2n−1 n Ru2 + n S1 v12 + n S2 v22 + n T1 v12 + n T2 v22 ∈ + · Pk hu· ng and nam 382 From the hypothesis, Lemma 4.2(i) is valid for the triples S1 v1 n and n n S2 v2 n ; that means that S1 v12 and S2 v22 are in + · Pk for every S1 , S2 Also by the hypothesis, Lemma 4.2(ii) holds for the triples T1 v1 n and n n T2 v2 n , so T1 v12 and T2 v22 both belong to + · Pk for every T1 , T2 n Therefore Ru2 ∈ + · Pk Step is proved Step If Lemma 4.2(i) holds for every n ≤ N, then so does Lemma 4.2(ii) for every n ≤ N Applying Lemma 3.4, we obtain R = Sq2 where R = s S < n So i R R/Q22 i R −2n−1 n+1 RQ22 + S and each term S in the sum is an H-monomial with n Ru2 = Sq2 n+1 i R −2n−1 RQ22 n u2 + Su2 n Since s S < n, by the hypothesis, Lemma 4.2(i) holds for the triple n S u n ; that means Su2 ∈ + · Pk for every S in the sum n−1 i R −2 By the Cartan formula, we get Set R = RQ22 Sq2 n+1 R u2 = Sq2 n n+1 Ru2 n + Sq2 R Sq1 u = Sq2 n+1 Ru2 n + Sq2 R Sq1 u + R Sq2 u = Sq2 n+1 Ru2 n n 2n + R Sq2 u 2n n 2n + R Sq1 Sq1 u n 2n + R Sq2 u 2n 2n + Sq2 R Sq1 u 2n Thus Sq2 n+1 n R u2 + R Sq2 u i R −2 2n ∈ + · Pk n−1 = n − < n So, from the hypothIt is easy to see that s R = s Q22 esis, Lemma 4.2(i) holds for the triple R Sq2 u n − ; that means n n+1 n R Sq2 u ∈ + · Pk Hence Sq2 R u2 ∈ + · Pk Finally, we have n Ru2 = Sq2 n+1 n R u2 + n Su2 ∈ + · Pk Step is proved Step Lemma 4.2(i) is valid for every n This is proved by induction on n For n = 1, from the hypothesis s R < it yields s R = By Lemma 5.1, Ru2 ∈ Sq1 Pk + Sq2 Pk So Lemma 4.2(i) holds for n = Now let n > and suppose inductively that Lemma 4.2(i) has been proved for every smaller value of n By Steps and above, Lemma 4.2(ii) and Lemma 4.2(iii) are also valid for every smaller value of n We consider the following three cases the hit problem 383 Case s R = Then, by Lemma 5.1, n Ru2 = R u2 n−1 ∈ + · Pk Case There exists an integer m with ≤ m < s R and h2mR = Combining the facts m < s R < n and i2 R ≡ 2s R − (mod 2s R +1 ), we get m + < n and i2 R ≡ 2m+1 − (mod 2m+1 ) Since m + < n and by the inductive hypothesis, we can apply Lemma 4.2(ii) to the triple n−m−1 n n−m−1 2m+1 R u2 m + and have Ru2 = R u2 ∈ + · Pk Case s R > and h2mR = for every m with ≤ m < s R It implies h R ≡ 2s R − (mod 2s R ) Set p = s R > We write p uniquely R in the form R = RS , where R and S are certain H-monomials ik R all ≤ 2p − with i0 R i1 R i2 R i4 R Note that i2 R ≡ i2 R (mod 2p ) ≡ 2p − (mod 2p ), as p = s R Combining this with the fact i2 R ≤ 2p − 1, we obtain i2 R = 2p − Since p = s R , so 2p does not occur in the dyadic expansion of i2 R = i2 R + 2p i2 S = 2p − + 2p i2 S Hence, it implies s S = n−p−1 Applying Lemma 5.1 to the H-monomial S and v = u2 = 1, we get Sv2 ∈ Sq1 Pk + Sq2 Pk On the other hand, we observe that h R = h R − 2p h S ≡ h R (mod 2p ≡ 2p − (mod 2p i2 R = 2p − ≥ i1 R Using the inductive hypothesis together with the assumption p = s R < n, we can apply Lemma 4.2(iii) to the triple R Sv2 p and get n Ru2 = R Sv2 2p ∈ + · Pk Step is proved Therefore, Lemma 4.2 follows REFERENCES 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 N H V Hu· ng, The action of the Steenrod squares on the modular invariants of linear groups, Proc Amer Math Soc 113 (1991), 1097–1104 N H V Hu· ng, Spherical classes and the algebraic transfer, Trans Amer Math Soc 349 (1997), 3893–3910 N H V Hu· ng, The weak conjecture on spherical classes, Math Z 231 (1999), 727–743 N H V Hu· ng, Spherical classes and the lambda algebra, Trans Amer Math Soc., 353 (2001), 4447–4460 N H V Hu· ng and T N Nam, The hit problem for the Dickson algebra, Trans Amer Math 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operations, Ann Math Stud 50, 1962 16 R M W Wood, “Modular Representations of GL n p and Homotopy Theory,” Lecture Notes in Math., Vol 1172, 188–203, Springer-Verlag, Berlin, 1985  ... is proved Therefore, Lemma 4.2 follows REFERENCES 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... classical problems in cobordism theory, modular representation theory, Adams spectral sequence for the stable homotopy of spheres, and stable homotopy type of classifying spaces of finite groups The. .. + · Pk , for a family of some rather small groups G This family contains most of the parabolic subgroups of GLk Observing the obstructions of the Hopf invariant-1 and the Kervaire invariant-1