DSpace at VNU: The hit problem for the Dickson algebra

Hung and Tran Ngọc Nam 5029-5040 Accessed: 13-01-2016 00:09 UTC REFERENCES Linked references are available on JSTOR for this article: http://www.jstor.org/stable/2693915?seq=1&cid=pdf-re

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The Hit Problem for the Dickson Algebra

Author(s): Nguyễn H V Hung and Tran Ngọc Nam

5029-5040

Accessed: 13-01-2016 00:09 UTC

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Volume 353, Number 12, Pages 5029-5040

S 0002-9947(01)02705-2

Article electronically published on May 22, 2001

THE HIT PROBLEM FOR THE DICKSON ALGEBRA

X NGUYEN H V HUNG AND TRAN NGOC NAM Dedicated to Professor Franklin P Peterson on the occasion of his 70th birthday

ABSTRACT Let the mod 2 Steenrod algebra, A, and the general linear group,

GL(k,F2), act on Pk := F2[xl, ,xk] with Ixil = 1 in the usual manner

We prove the conjecture of the first-named author in Spherical classes and

the algebraic transfer, (Trans Amer Math Soc 349 (1997), 3893-3910)

stating that every element of positive degree in the Dickson algebra Dk

shown to be equivalent to a weak algebraic version of the classical conjecture

on spherical classes, which states that the only spherical classes in QoS? are

the elements of Hopf invariant one and those of Kervaire invariant one

1 INTRODUCTION

Let Pk F2 [xl, ,Xk] be the polynomial algebra over (the field of two el- ements) F2 in k variables, each of degree 1 The general linear group GLk GL(k,F2) acts on Pk in the usual manner Dickson proves in [1] that the ring of invariants, Dk (Pk)GLk, is also a polynomial algebra Dk "- 2[Qk,k-1** Qk, where Qk,s denotes the Dickson invariant of degree 2k - 25 It can be defined by the inductive formula

Qk,s = Qk-l,s-1 + Vk * Qk-l,si

where, by convention, Qk,k = 1, Qk,s = 0 for s < 0 and

Vk= fJ (Alxl?+?*+Ak-lXk-1 + Xk)

Aj EF2

Let A be the mod 2 Steenrod algebra The usual action of A on Pk commutes with that of GLk So Dk is an A-module One of the authors has been interested

in the homomorphism

jk: F2 ?(Pk)GLk >' (F2 Pk)GLk

which is induced by the identity map on Pk (see [3]) Observing that ji is an isomorphism and i2 is a monomorphism, he sets up the following

Conjecture 1.1 (Nguyen H V Hu'ng [3]) ik = 0 in positive degrees for k > 2 Let Dk+ and A+ denote respectively the submodules of Dk and A consisting of all elements of positive degree Then Conjecture 1.1 is equivalent to Dk+ C A+ Pk

Received by the editors September 29, 1999 and, in revised form, February 22, 2000

2000 Mathematics Subject Classification Primary 55S10; Secondary 55P47, 55Q45, 55T15 Key words and phrases Steenrod algebra, invariant theory, Dickson algebra

This work was supported in part by the National Research Project, No 1.4.2

(@2001 American Mathematical Society

5029

for k > 2 (see [3]) In other words, it predicts that every GLk-invariant element of positive degree is hit by the Steenrod algebra acting on Pk for k > 2

Conjecture 1.1 is related to the hit problem of determination of F2 0Pk This

A problem has first been studied by F Peterson [9], R Wood [14], W Singer [12], and

S Priddy [10], who show its relationships to several classical problems in cobordism theory, modular representation theory, Adams spectral sequence for the stable ho- motopy of spheres, and stable homotopy type of classifying spaces of finite groups The tensor product F2 (Pk has explicitly been computed for k < 3 The cases

A

k = 1 and 2 are not difficult, while the case k = 3 is complicated and was solved

by M Kameko [8] It seems unlikely that a very explicit description of F2 0 Pk

A 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 F2 0Pk, i.e belong to A+ Pk Peterson's

A conjecture, which was established by Wood [14], claims that F2 0Pk = 0 in degree

A

d such that a(d + k) > k Here a(n) denotes the number of ones in the dyadic expansion of n Recently, W Singer, K Monks, and J Silverman have refined the method of R Wood to show that many more monomials in Pk are in A+ Pk (See Silverman [11] and its references.) Conjecture 1.1 presents a large family, whose elements are predicted to be in A+ Pk

In [3], one of the authors proves the equivalence of Conjecture 1.1 and a weak algebraic version of the conjecture on spherical classes stating that: There are no spherical classes in QoS? except the elements of Hopf invariant one and those of Kervaire invariant one He also gives two proofs of Conjecture 1.1 for the case

k = 3 In this paper, we establish this conjecture for every k > 2 That Conjecture 1.1 is no longer valid for k = 1 and 2 is respectively an exposition of the existence

of Hopf invariant one classes and Kervaire invariant one classes We have

Main Theorem D+ C A+ * Pk for k > 2

Recently, F Peterson and R Wood privately informed us that they had proved the theorem for k = 4 and probably for k 5 The readers are referred to [4] and [5] for some problems, which are closely related to the main theorem Additionally, the problem of determination of F2 0 Dk and its applications have been studied by

A Hu'ng and Peterson [6], [7]

The paper contains five sections Section 2 is a preparation on the action of the Steenrod squares on the Dickson algebra We prove the main theorem in Section

3 by means of two lemmata, which are later shown in Section 4 and Section 5 respectively

The action of the Steenrod squares on Dk is explicitly described as follows Theorem 2.1 ([2])

Sqi(Q-, ) Qk,rQk,t for i = 2k 2t + 2s - 2r r< s < t

k, s

s)for i 2 - - 2s,

0

0 otherwise

From now on, we denote Qk,s by Q, for brevity We get

Sq'(Qs) Qs-, if a = 2s81,

0 if 0a- <a281 or 2 < a < 2k for 0 < s < k Combining this with the Cartan formula, one obtains

Corollary 2.2 (a) Sqa(QsR) = QsSqa(R) if 0 < a < 2s-1,

(b) Sqa(QoR) = QoSqa(R) if O < a < 2k-1

for any polynomial R c Pk

Let In (n > 0) be the right ideal of A generated by the operations Sq2 for i=O, ,n

Definition 2.3 Suppose R1, R2 C Pk Then we write R1 -R2 (mod In) if R1 +R2 belongs to In Pk By convention, R1 R2 (mod In) means R, = R2 for n < 0 This is an equivalence relation

Lemma 2.4 (a) Sql (R1)R2 _ RiSql (R2) (mod Io),

(b) Sq2(Ri)R2- R,Sq2(R2) (mod I1)

for any polynomials R1, R2 c Pk*

Proof (a) From the Cartan formula Sql (R)R2 +RiSql(R2) Sql (RiR2), we get (a) by Definition 2.3

(b) We have

(by the Cartan formula)

- Sq2(Rf)R2 + R?Sq1Sq1(R2) + RfSq2(fR2) (modlo) (by Part (a))

Sq2(Rft)R2 + R? Sq2 (R2) (mod lo) (since Sq1Sql = 0)

Hence, Sq2 (R1)R2 + R1Sq2(R2) c h1 Pk and (b) follows L Lemma 2.5 Let R c Pk (k > 1) If Sql(R) = 0 and all the monomials of R are

of positive degree, then R 0 O (mod Io)

Proof The lemma is proved by induction on k For k = 1, it is easy to see that all the monomials of R are of even degree Since x2n = Sq1(x2l1) for n > 0, the lemma is proved Let k > 1 and suppose inductively that the lemma holds for polynomials in k - 1 variables Let us write

ft= S xiftR

O<i<2n

for some positive integer n and some polynomials Ri (O < i < 2n) in k -1 variables

x2, ,Xk We get

Sql((R) = 5 xiSql(Ri) + 7 x+1fRt

O<i<2n O<i<2n

= Sq'(Ro) + E x4Sql(Ri)

O<i<2n

i odd

+ S

xil+1[Sq1(Rf+j)+Rf]

O<i<2n

i odd

Since Sq'(R) = 0, we have Sq'(Ro) = 0 and Sql(Ri+,) = Ri for 0 < i < 2n, i odd Therefore,

R = > xR?Ri +>j x Sql(Ri+,)

O<i<2n O<i<2n ieven iodd

-

Ro+ E [xS + Ri+l + x'Sql(Ri+,)]

O<i<2n

% odd

- Ro+Sql( E X1Ri+)

O<i<2n

i odd

- Ro (mod lo)

- 0 (mod Io) (by the inductive hypothesis)

This lemma immediately implies that if all monomials of R E Pk are of positive degree, then R2 -0 (mod lo)

Corollary 2.6 Let k > 1 and suppose S is a non-empty subset of {O, , k - 1} such that 1 , S Then

QR2 0(mod lo), where Q = Hses Q8 and R is an arbitrary polynomial in Pk

Proof As k > 1 and 1 0 S, one gets Sq1(Q) = 0 This implies Sq1(QR2) = 0

Thus QR2 0 O (mod Io) by Lemma 2.5 The corollary is proved L

3 PROOF OF THE MAIN THEOREM

Let Q be a non-zero Dickson monomial If Q =A 1, it can be written as

Q= [I A 2

O<i<n

where n is some non-negative integer and Ai is some Dickson monomial dividing

Q for i = 0, , n with An =y 1

O<s<k

Indeed, suppose Q fi Q' Since Q :8 1, there exists at least one aes O

O<s<k

Consider the 2-adic expansions of all the non-zero a's:

Oi<n(s)

where asn(s) = 1 Now denoting

n := max nr(s),

O<s<k

Xsi := Oifrn(s) <i <n (O< s < k),

Ai : fl Qs (O < i < n),

O<s<k

one can easily check that Q = - A? and each Ai divides Qs Moreover,

there exists an integer r such that 0 < r < k, c 0r 7 0 and n = n(r) Then

O<s<k

Definition 3.1 (i) We call n the height of Q The monomial A?2 = Ai(Q)2' is called the ith cut of Q It is said to be full if Ai is divisible by f Qs The

O<s<k

monomial Q is called full if its cuts are all full

(ii) A Dickson monomial is called a based cut if it is the 0th cut of some Q 78 0 and :8 1

The main theorem is proved at the end of this section by means of the following two lemmata, whose proofs will be given in the last two sections

Lemma A Let k > 2 and suppose R is an arbitrary polynomial in Pk

(a) If Q = J7 A?i 7 1 and it is not full, then QR2n+' c A+ Pk

O<i<n

(b) If Q= AJ 4 is full, then QSq2m+n+l (R2+ ) ,A A PkfoO < m<k-1

O<i<n

Lemma B Suppose k > 2 If A is a full based cut, then A_ 0 (modI,)

Proof of the Main Theorem Suppose Q = A7 Ai is a Dickson monomial with

O<i<n

If Q is not full, then applying Lemma A(a) with R = 1, one gets Q c A+ * Pk

If Q is full and n = 0, then Q is the full based cut of itself So using Lemma B, one obtains Q- 0 (mod 11) In particular, Q cE A+ * Pk

If Q is full and n > 0, then An is the full based cut of itself By Lemma B, one has An = Sq1(R1) + Sq2(R2), with some R1, R2 E Pk Noting that Q' = A7 4i

O<i<n

is also full with the height n - 1, one can apply Lemma A(b) to it and get

Q'Sq (R)= 1 7E Ai Sq (R A+ Pk

O<i<n

QSq 2+(R2) J A i Sq (R2 A+ Pk

O<i<n

(It should be noted that 1 < k - 1.) Hence

Q ]I A2 A AnA2 = ]7 [Sq2n (R2n) + Sq2n+1 (R2n)] c P+P

O<i<n O<i<n

4 PROOF OF LEMMA A

In this section, we prove Lemma A by using Lemma 4.1 and Lemma 4.2 Lemma 4.1 Suppose k, m, j are integers satisfying k > 2, 0 < m < k - 1 and

0 < j < 2m Let Q be a full Dickson monomial of height n and B any Dickson monomial of Sq2 i(Q) Suppose B = Bi , with B?i the ith cut of B and

o<i<p

Bp =A1 We have

(a) p > n,

(b) If B'- rj Bi2 /- 1, then it is not full

O<i<n

Proof (a) Suppose to the contrary that p < n We get

degQ + 2n+1j = deg( fJ B 2)

?ip o<i<p

< ( 22)deg( fJ Qs)

< (2n- _1) deg( Jl Qs)

O<s<k

and

degQ+2n+1j > deg Q

- (2 - 1) deg( Hl Qs)

O<s<k

Therefore,

(2n _1)deg( Hl Qs) > (2n+1 - 1)deg( Hl Qs),

(2-n _1)degQo > 2ndeg( Hl Qs))

O<s<k

degQo > deg( Hl Qs)

O<s<k

The last inequality is false for every k > 2 This contradiction shows part (a) (b) Suppose to the contrary that fi B?" is full Then

O<i<n

degQ + 2n+lj = deg( Hl B 2)

0??

o<i<p

deg( Hl B 2 ) (mod2n+1),

O<i<n

degQ-deg( J Bi2) O(mod2n+'),

O<i<n

S 22(deg Ai-deg Bi) O(mod 2n+1)

O<i<n

It is easy to see that deg Ai - deg Bi = Ei deg Qo, with Ei c {O, 1, -1} Further- more, if Ei = 0, then Ai = Bi So E 22i degQo -0 (mod2n+1) It should be

O<i<n

noted that degQo 2 1 has no common divisor with 2n+1* So , 2'E 0

O<i<n

(mod 2n+1) This implies Ei = 0 for i = O, ,n In other words, Ai = Bi for

i=O, ., n and Q = f B2 We have

O<i<n

degQ + 2n+1j -deg( 17 BY2) +?deg( 11 BR )

- degQ +deg( ]7 B2 ),

n<i<p

n<i<p

Since j > 0, we get deg B = deg Q + 2n+lj > deg Q, so p > n Hence

degB2 > degBP2 > degQk1 - 12k-1

It implies j > 2k-1* Combining this and the fact 2k-1 > 2' > j, we obtain j > j This contradiction comes from the hypothesis that B' is full Therefore, the lemma

Lemma 4.2 Let A /- 1 be an unfull based cut Denote by s the smallest integer

s > 1 such that Q8 ,1IA If s > 1, then there exists for every R c Pk an expansion

AR2 = Sq2Sl (R1) + E BR2 where Rf c Pk, every R2 c Pk and every B is a Dickson mTonomTial with B

H Qr, B #- 1, Qs-, IB

O<r<k

Proof FRom the hypothesis we can write A = A fJ Qr with a certain Dickson

o<r<s

monomial A J QrQO By the Cartan formula

s<r<k

+ E Sq2 (AQ8 1 Qr)Sq2sl_2i (R2)

Denoting R1 AQ, := fJ QrR2, we get

O<r<s-1

Sq (AQs 17 Qr)R2 Sq2S(R)

o<r<s-1

? E Sq2 (AQ8 Qr)2q2Sl- 171 -2j (2)

0<j<2S92 O<r<s-1

We will prove that (a) A - Sq2S 1 (AQs 17 Qr) and that (b) every polyno-

O<r<s-1

mial Sq2j(AQs H Qr)Sq2S-l-2i(R2) for 0 < j < 2s-2 can be written in the

O<r<s-1

form ZBR2 , where B, R2 satisfy the conclusions of Lemma 4.2 Thus, the required expansion will be obtained

First we prove (a) By Corollary 2.2, we have

Sq 28(AQS H: Qr) - ASq2 (Qs Jl Qr)

So it suffices to show that Sq29 1 (Qs 1 Qr) 171 Qr By the Cartan formula

Sq28-1 (Qs f Qr) = QsSq2S( fi Qr)

+ E Sqa(Qs)Sq2s-la( f Qr)

- QsSq2s( fJ Qr) + Qs-i fJ Qr

(since Sqa(Qs) = 0 for 0 < a < 2s-1 and Sq (Qs) - Qs-,)

- QsSq2sl( H Qr) + Hl Qr

It is sufficient to prove Sq2sl ( 17 Qr) =0 Note that, by the Cartan formula,

O<r<s-1

Sq2S-1( H Qr) E S H SqarjQr),

O<r<s-1 O<r<s-1 where the sum is taken over all (ar)o<r<s-1 satisfying ar = 2s-1 and ar > 0-

O<r<s-1

It is easy to show that there exists an r such that 0 < r < s - 1 and ar >

2r Since ar < 2s1 < 2k-1, we have 2r < ar < 2k1* So, by Theorem 2.1, Sqa-(Qr) = 0 Hence ]J Sqa-(Qr) = 0 This is true for every (ar)o<r<s_1i so

O<r<s-1

Sq 28-1 ( 7I Qr) = 0 Part (a) is shown

O<r<s-1

Next we prove (b) FRom Corollary 2.2 and since 2j < 2s-1 < 2k-1 we have

Sq2 (AQ, ]J Qr) - AQsSq2( ] Qr)

By the Cartan formula we get

Sq 2j( H Qr) = 5 Sq'r(Qr),

where the sum is taken over all sequences (jr)O<r<s-1 satisfying Z ir 2j and

O<r<s-1

ir > 0 From Theorem 2.1 and since jr < 2j < 2k-1 we have Sqji (Qr) = either 0 or

Qt with 0 < t < r So 17 Sqj,(Qr) is not divisible by Qs-1,Qs, Qk-l

O<r<s-1

Therefore, the 0th cut of every Dickson monomial in Sq2 ( nj Qr) is not divisi-

O<r<s-1

ble by Qs-i) Qs) Qk-1 Let us write Sq2J( 17 Qr) as the sum of its Dickson

O<r<s-1

monomials Sq2( ]j Qr) H = ]7j , where COi is an ith cut Then

Sq2 (AQs J Qr) = AQsSq2 ( J Qr)

O<i<p

We have shown that Co is not divisible by Qs-i, Qs, ) Qk-1 Note that deg Qo

is odd, while deg Qr is even for every r > 0 Thus, the 0th cut Co of every term in Sq2j( J7 Qr) is not divisible by Qo Recall that A is a divisor of f QrQ0-

So AQsCo is not divisible by Qs- Moreover, it is a Dickson monomial, which is

different from 1 and divides rI Qr

0<r<k

Putting B := AQsCo and R2 := ECi lSq2s 2(R) for each Co, we get

o<i<p

Sq2I(AQS 1 Qr)Sq2S2J(f2) Z EAQsCo 17 Ci sq2iS 2i(R2)

- EBR2

It has been shown that B = AQsCo satisfies the conclusions of Lemma 4.2 Hence, part (b) and therefore Lemma 4.2 is proved L Proof of Lemma A The proof is divided into 2 steps

Step 1 If Lemma A(a) is true for every n < N, then so is Lemma A(b) for every

n < N

Indeed, suppose Q = f A?2 (with n < N) is full and m satisfies 0 < m < k-1

O<i<n

One needs to prove QS 2m+n+l (R2n+l) E A+ Pk, where R c Pk Recall that

Sqa(R2n) f [Sq / (R)] if 2 la,

Sq (R 0 otherwise

Then, by the Cartan formula, one gets

O<j<2m

QSq 2m+n+l (R2n+l) ? Sq 2+i (Q)Rj 2+,

O<j<2m

where Rj := Sq i(R) for j = 1, 2

In order to prove that QS 2m+n+l (R2n+1) is A-decomposable, it suffices to show that each Sq2n+li(Q)R2n+1 is A-decomposable We do this by showing BR 2n 1

*A+ Pk for every Dickson monomial B of Sq2 + (Q) Let B = fJ B?2 , with B?i

o<i<p

the ith cut of B By Lemma 4.1(a), we have p > n If Bi ? = 1, then p > n, so

O<i<n

BR2+' = B? 0R (mod lo) If B?" /i4- 1, then it is not full by

Lemma 4.1(b) So we can choose an integer q such that Bq /- 1(0 < q < n < N) and B? ! is not full Applying Lemma A(a) to ]7 B?2 we obtain

2n+1 I 2' ( IIB2-q R2n- 2q+1 A

0<i<q q<i<p

Therefore, Step 1 is shown

Step 2 Lemma A(a) holds for every non-negative integer n