DICKSON INVARIANTS HIT BY THE STEENROD SQUARES

Transcription

1 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES K F TAN AND KAI XU Abstract Let D 3 be the Dickson invariant ring of F 2 [X 1,X 2,X 3 ]bygl(3, F 2 ) In this paper, we prove each element in D 3 is hit by the Steenrod square in F 2 [X 1,X 2,X 3 ] Our method provides a clue in attacking the question in the general case 1 Introduction Throughout this paper, we work on the polynomials over F 2 Let GL(n, F 2 )bethen n-general linear group over F 2 Then the standard group action: gives rise to the ring of Dickson invariants GL(n, F 2 ) F 2 [X 1,X 2,,X n ] F 2 [X 1,X 2,,X n ] F 2 [X 1,X 2,,X n ] GL(n,F2), which is isomorphic to the polynomial ring F 2 [Q n,0,q n,1,, Q n,n 1 ], where each generator Q n,i (i =1,, n 1) is a certain polynomial over F 2 Definition 11 A polynomial f(x 1,X 2,,X n ) is hit if it can be written as f(x 1,X 2,,,X n )= i 1 Sq i f i (X 1,X 2,,X n ), for some polynomials f i (X 1,X 2,,X n ) Question 12 Is each Dickson invariant hit? The motivation of presenting this question was illustrated in detail in an excellent expository paper [9], p501 In general, this question is hard to answer But when n = 1 and 2, the answer is negative For example, when n = 1, the Dickson invariant X 1 is not hit; when n =2,X 2 1 +X 2 2 +X 1 X 2 is a Dickson invariant but is not hit Conjecture 13 (Hung, [1, 2, 3, 4, 8]) When n 3, all Dickson invariants are hit The main result of this paper is to prove the above conjecture when n = 3 The method we use is elementary though a bit tedious We hope that it provides a clue in attacking the conjecture in the general case Write V 1 = X 1, V 2 = X 2 (X 2 + X 1 )andv 3 = X 3 (X 3 + X 2 )(X 3 + X 1 )(X 3 + X 2 + X 1 ) Then Q 3,0 = V 3 V 2 V 1, Q 3,1 = Q 2 2,0 +V 3Q 2,1 =(V 2 V 1 ) 2 +V 3 (V1 2+V 2)andQ 3,2 = Q 2 2,1 +V 3 =(V1 2+V 2) 2 +V 3 Clearly, deg(q 3,0 ) = 7, deg(q 3,1 ) = 6 and deg(q 3,2 ) = 4 The Steenrod operation acts on the Dickson invariants in the following way, 1991 Mathematics Subject Classification Primary 55S10, 55Q45, 55S10, 55T15 1

2 2 K F TAN AND KAI XU Theorem 14 (Hung [1]) The Steenrod operation acts trivially on Q 3,0,Q 3,1 and Q 3,2 except for the following cases, Sq 4 Q 3,0 = Q 3,0 Q 3,2, Sq 6 Q 3,0 = Q 3,0 Q 3,1, Sq 7 Q 3,0 = Q 2 3,0, Sq1 Q 3,1 = Q 3,0, Sq 4 Q 3,1 = Q 3,1 Q 3,2, Sq 5 Q 3,1 = Q 3,0 Q 3,2, Sq 6 Q 3,1 = Q 2 3,1, Sq2 Q 3,2 = Q 3,1 Sq 3 Q 3,2 = Q 3,0, Sq 4 Q 3,2 = Q 2 3,2 Now we define a function γ : N {0} N as follows: γ(0) = 0 and for k 1, γ(k) =2 k 1 Write µ(n) for m min{m N n = γ(k i ) for some k i N} Let E and F be homogeneous polynomials of degrees e and f, respectively following, i=1 Then we have the Theorem 15 (Silverman [5]) Suppose that e<(2 k+1 1)µ(f) for some k 0 Then the polynomial EF 2k+1 is hit For an integer t, letα(t) be the number of digits 1 in the binary expansion of t and let U be an n-variable homogeneous polynomial Then we have the following theorem, known as the Peterson Conjecture, Theorem 16 (Wood [8]) If α(deg(u)+n) >n,thenu is hit We will frequently use the above theorem in the next section John Milnor showed that there is an anti-automorphism χ from the Steenrod algebra to itself, satisfying the following condition, χ(sq k )= k Sq i χ(sq k i ) for k 1 i=1 We will frequently use the so-called χ-trick in the following sections, which is based on the following, Proposition 17 For any polynomials E 1 and E 2, E 1 [Sq m (E 2 )] [χ(sq m )(E 1 )]E 2 is hit, where m is any non-negative integer Acknowledgment 18 We would like to thank W M Singer for a correspondence, which provides us a brief introduction on the subject related to the Steenrod algebra We also would like to thank KG Monks for providing his Maple program on the Internet His source code benefits us a great deal

3 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES 3 2 Proofs Clearly, to prove our result, we only need show that 3,0 Qn1 3,1 Qn2 3,2 integers n 0,n 1 and n 2 The proof will split into the following cases, (1) At least two of n 0,n 1 and n 2 are even; (2) Exactly one of n 0,n 1 and n 2 is even; (3) n 0, n 2 are odd and n 1 1 mod 4; (4) n 0 is odd, n 1 3 mod 4 and n 2 1 mod 4; (5) n 0 1 mod 4, n 1 3 mod 4 and n 2 3 mod 4; (6) n 0 3 mod 4, n 1 3 mod 4 and n 2 3 mod 4 Now we will prove each of the above six cases Remark 21 The proof of Cases(1)-(4) is a part of the first author s MSc thesis[7] is hit for any non-negative Notation 22 For simplicity, we use the following notation: for any polynomials E 1 and E 2, E 1 E 2 means that E 1 E 2 is hit 21 Proof of Case(1) If n 0,n 1 and n 2 are all even, then the result is a direct consequence of the definition of the Steenrod square Suppose that n 1,n 2 are even and n 0 is odd Then putting n 0 =2l 0 +1,n 1 =2l 1 and n 2 =2l 2, we have the following, 3,0 Qn1 3,1 Qn2 3,2 = Q 2l0+1 3,0 Q 2l1 3,1 Q2l2 3,2 = Q 3,0 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 =[Sq 1 Q 3,1 ](Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 =Sq 1 [Q 3,1 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] 0 (see Notation 22) Suppose that n 0,n 2 are even and n 1 is odd Then by letting n 0 =2l 0, n 1 =2l 1 +1 and n 2 =2l 2, we have the following, Notice that Then 3,0 Qn1 3,1 Qn2 3,2 = Q 2l0 3,0 Q2l1+1 3,1 Q 2l2 3,2 = Q 3,1 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 =[Sq 2 Q 3,2 ](Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 =Sq 2 [Q 3,2 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ]+Q 3,2 Sq 2 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 Q 3,2 Sq 2 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 Q 3,2 = Q 2 2,1 + V 3 and V 3 =Sq 1 (Q 2,1 X 3 + X 3 3) (211) Q 3,2 Sq 2 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 =(Q 2 2,1 + V 3)Sq 2 [(Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] =[Q 2,1 Sq 1 (Q l0 +Sq 1 [(Q 2,1 X 3 + X3 3 )(Sq 1 (Q l0 3,0 Ql1 3,1 Ql2 3,2 ))2 ] 0

4 4 K F TAN AND KAI XU Suppose that n 0,n 1 are even and n 2 is odd Putting n 0 =2l 0, n 1 =2l 1 and n 2 =2l 2 +1, we have the following, 3,0 Qn1 3,1 Qn2 3,2 = Q 2l0 3,0 Q2l1 3,1 Q2l2+1 3,2 So Case (1) has been done = Q 3,2 Q 2l0 3,0 Q2l1 3,1 Q2l2 3,2 = (Q 2 2,1 +Sq1 (Q 2,1 X 3 + X3 3))(Ql0 3,0 Ql1 3,1 Ql2 3,2 )2 using (211) = (Q 2,1 Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 +Sq 1 [(Q 2,1 X 3 + X3 3 ))(Q l0 3,0 Ql1 3,1 Ql Proof of Case(2) Suppose that n 0 =2l 0, n 1 =2l 1 +1 and n 2 =2l We will use the χ-trick (17) to prove the result Notice that 3,0 Qn1 3,1 Qn2 3,2 =[Sq4 (Q 3,1 )](Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 Q 3,1 χ(sq 4 )(Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 (221) Using the formulae χ(sq 4 )=Sq 4 +Sq 2 Sq 2 and Q 3,1 =Sq 2 (Q 3,2 ), we can easily see that the last term of (221) can be reduced to Sq 2 (Q 3,2 )[Sq 2 (Q l0 + Q 3,1 [Sq 1 (Sq 1 (Q l0 3,0 Ql1 3,1 Ql2 3,2 ))]2 Noting that Sq 1 Sq 1 = 0, so the last term is zero Again we use the χ-trick, the first term is hit if and only if Q 3,2 χ(sq 2 )[Sq 2 (Q l0 is hit Now Q 3,2 χ(sq 2 )[Sq 2 (Q l0 = Q 3,2 Sq 2 [Sq 2 (Q l0 =(Q 2 2,1 + V 3)[Sq 1 Sq 2 (Q l0 =[Q 2,1 Sq 1 Sq 2 (Q l0 + V 3 [Sq 1 Sq 2 (Q l0 V 3 [Sq 1 Sq 2 (Q l0 =Sq 1 (Q 2,1 X 3 + X 3 3 )[Sq1 Sq 2 (Q l0 =Sq 1 {(Q 2,1 X 3 + X3 3 )[Sq 1 Sq 2 (Q l0 3,0 Ql1 3,1 Ql2 0 3,2 )]2 } Suppose that n 0 =2l 0 +1,n 1 =2l 1 and n 2 =2l 2 +1 Then 3,0 Qn1 3,1 Qn2 3,2 = Q 2l0+1 3,0 Q 2l1 3,1 Q2l2+1 3,2 =Sq 1 (Q 3,1 )[Q 3,2 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] =Sq 1 [Q 3,1 Q 3,2 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] (since Sq 1 Q 3,2 =0) 0 Suppose that n 0 =2l 0 +1,n 1 =2l 1 +1andn 2 =2l 2 Then 3,0 Qn1 3,1 Qn2 3,2 = Q2l0+1 3,0 Q 2l1+1 3,1 Q 2l2 3,2 =Sq2 (Q 3,2 )[Q 3,0 (Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] Since Sq i (Q 3,0 )=0fori =1, 2 and Sq 1 Q 3,2 = 0, it is sufficient to show that Q 3,2 Q 3,0 Sq 2 [(Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] is hit Indeed, Q 3,2 Q 3,0 Sq 2 [(Q l0 3,0 Ql1 3,1 Ql2 3,2 )2 ] =(Q 2 2,1 + V 3)Q 3,0 [Sq 1 (Q l0 = V 3 V 2 V 1 Q 2 2,1 [Sq1 (Q l0 + V 3 Q 3,0 [Sq 1 (Q l0 3,0 Ql1 3,1 Ql2 =Sq 1 (Q 2,1 X 3 + X3)V 3 2 V 1 Q 2 2,1[Sq 1 (Q l0 +Sq 1 (Q 2,1 X 3 + X3 3)Q 3,0[Sq 1 (Q l0 3,2 )2 ] 3,2 )]2 (since Q 3,0 = V 3 V 2 V 1 )

5 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES 5 The last two terms are hit, since Sq 1 acts on V 2 V 1 and Q 3,0 trivially 23 Proof of Case(3) Suppose that n 0 =4l 0 +1,n 1 =4l 1 +1andn 2 =2l 2 +1 Then 3,0 Qn1 3,1 Qn2 3,2 = Q 4l0+1 3,0 Q 4l1+1 3,1 Q 2l2+1 3,2 = Q 3,0 Q 3,1 Q 3,2 (Q 2l0 3,0 Q2l1 3,1 Ql2 3,2 )2 Write B for Q 2l0 3,0 Q2l1 3,1 Ql2 3,2 Then Q 3,0 Q 3,1 Q 3,2 (Q 2l0 3,0 Q2l1 3,1 Ql2 3,2 )2 = Q 3,0 Q 3,1 Q 3,2 B 2 = V 3 V2 3V 1 7B2 + V 3 V2 5V 1 3B2 + V3 2V 2V1 7B2 + V3 3V 2 1V 1 3B2 + V3 2V 2 2V 1 5B2 +V3 2 V2 4 V1 1 B 2 + V3 3 V2 2 V1 1 B 2 Now we will show that each term after the last equality is hit Using the following lemma, we will know that the first four terms are hit Lemma 23 For any positive odd integers k 0,k 1 and k 2, the polynomial V k0 3 V k1 2 V k2 1 B2 is hit Proof Put k 0 =2a +1,k 1 =2b +1andk 2 =2c +1 Thenwehave V k0 3 V k1 2 V k2 1 B2 = V3 2a+1 V2 2b+1 V1 2c+1 B 2 =[Sq 1 (Q 2,1 X 3 + X3 3 2a )]V3 V 2 2b+1 V1 2c+1 B 2 =Sq 1 [(Q 2,1 X 3 + X3 3 2a )(V3 V 2 2b+1 V1 2c+1 B 2 )] (since Sq 1 (V 2 V 1 )=0) Since V3 2 V2 2 V1 5 B 2 = V 1 (V 3 V 2 V1 2 B) 2 and V3 2 V2 4 V1 1 B 2 = V 1 (V 3 V2 2 B) 2,weverifythatthesetwoterms satisfy the condition in Theorem 15 So they are hit It remains to show that V3 3V 2 2V 1 1B2 is hit In fact V 3 3 V 2 2 V 1 1 B2 = X 3 1 X2 2 X12 3 B2 + X 1 X 4 2 X12 3 B2 + X 7 1 X2 2 X8 3 B2 + X 1 X 8 2 X8 3 B2 +X 8 1 X6 2 X3 3 B2 + X 5 1 X2 2 X10 3 B2 + X 4 1 X3 2 X10 3 B2 + X 2 1 X5 2 X10 3 B2 +X 1 X 6 2 X 10 3 B 2 + X 9 1 X 2 2X 6 3 B 2 + X 8 1X 3 2 X 6 3B 2 + X 2 1 X 9 2X 6 3 B 2 +X 4 1 X10 2 X3 3 B2 + X 1 X 10 2 X6 3 B2 + X 9 1 X4 2 X4 3 B2 + X 5 1 X8 2 X4 3 B2 +X 8 1 X5 2 X4 3 B2 + X 4 1 X9 2 X4 3 B2 + X 7 1 X6 2 X4 3 B2 + X 3 1 X10 2 X4 3 B2 +X 5 1X 3 2X 9 3 B 2 + X 3 1X 5 2 X 9 3B 2 + X 8 1 X 4 2X 5 3 B 2 + X 2 1X 10 2 X 5 3B 2 +X 9 1 X5 2 X3 3 B2 + X 5 1 X9 2 X3 3 B2 Using Theorem 15, we can prove all but the following terms are hit: X 3 1 X 5 2X 9 3 B 2, X 5 1X 3 2 X 9 3 B 2, X 9 1 X3 2 X5 3 B2, X 9 1 X5 2 X3 3 B2, X 3 1 X9 2 X5 3 B2 and X 5 1 X9 2 X3 3 B2 Indeed we only need the following proposition to conclude the result Proposition 24 X 3 1 X5 2 X9 3 B2 +X 5 1 X3 2 X9 3 B2, X 9 1 X3 2 X5 3 B2 +X 9 1 X5 2 X3 3 B2 and X 3 1 X 9 2X 5 3 B 2 + X 5 1 X 9 2X 3 3 B 2 are hit Proof Careful observations reveal the following, X1X X3 9 + X1X X3 9 =Sq 2 (X1 9 X2 3 X3)+Sq 3 1 (X1 10 X2X )+X1X X3, 4 (231) X1 9 X2X X1 9 X2X =Sq 2 (X1 3 X2X )+Sq 1 (X1X X3 3 )+X1 4 X2X and X1 3 X9 2 X5 3 + X5 1 X9 2 X3 3 =Sq2 (X1 3 X3 2 X9 3 )+Sq1 (X1 3 X3 2 X10 3 )+X4 1 X4 2 X9 3 We shall show that X1 3 X2X B 2 +X1X X3 9 B 2 is hit; the remaining two cases can be proved similarly

6 6 K F TAN AND KAI XU By Theorem 15, X1 9X4 2 X4 3 B2 = X 1 (X1 4X2 2 X2 3 B)2 in (231) is hit So we are left to show that Sq 2 (X1 9X3 2 X3 3 )B2 +Sq 1 (X1 10X3 2 X3 3 )B2 is hit This can be done by using the χ-trick Infact, X1 3 X3 2 X9 3 [χ(sq2 )(B 2 )] + X1 3 X3 2 X10 3 [χ(sq1 (B 2 )] = 0, where we use the facts that Sq 2 B 2 =0andSq 1 B 2 =0 To finish the proof of Case(3), we are left to show that Q 4l0+3 3,0 Q 4l1+1 3,1 Q 2l2+1 3,2 is hit The proof can be done as follows, Q 4l0+3 3,0 Q 4l1+1 3,1 Q 2l2+1 3,2 = Q 3 3,0 Q 3,1 Q 3,2 Q4l0 3,0 Q4l1 3,1 Q2l2 3,2 =Sq 2 [Q 3,0 Q 3 3,1Q 3,2 ](Q 4l0 3,0 Q4l1 3,1 Q2l2 3,2 )+Q 3,0Q 4 3,1(Q 4l0 3,0 Q4l1 3,1 Q2l2 Q 3,0 Q 3 3,1 Q 3,2 χ(sq2 )(Q 4l0 3,0 Q4l1 3,1 Q2l2 3,2 )+Q4l0+1 3,0 Q 4l1+4 3,1 Q 2l2 3,2 It is easy to see that the first term is zero and that the second term is hit by using the result of Case(1) 24 Proof of Case(4) Put n 0 =2l 0 +1,n 1 =4l 1 +3andn 2 =4l It is easy to verify that 3,2 ) Q 3,0 Q 3 3,1 Q 3,2 =Sq4 [Q 3,0 Q 3,1 Q 3 3,2 ]+Q3 3,0 Q2 3,2 + Q 3,0 Q 3,1 Q4 3,2 Thus 3,0 Qn1 3,1 Qn2 3,2 = Q2l0+1 3,0 Q 4l1+3 3,1 Q 4l2+1 3,2 = Q 3 3,0Q 3,1 Q 3,2 (Q 2l0 3,0 Q4l1 3,1 Q4l2 3,2 ) =Sq 4 [Q 3,0 Q 3,1 Q 3 3,2](Q 2l0 3,0 Q4l1 3,1 Q4l2 3,2 )+Q2l0+3 3,0 Q 4l1 3,1 Q4l2+2 3,2 + Q 2l0+1 3,0 Q 4l1+1 3,1 Q 4l2+4 3,2 The last two terms are hit by the results of Cases(1) and (2) Using the χ-trick and the result of Case(3), it can be shown that the first term is also hit We will leave the detail to the reader to verify the conclusion 25 Proof of Case(5) When n 0 =4l 0 +1,n 1 =4l 1 +3andn 2 =4l 2 +3, 3,0 Qn1 3,1 Qn2 3,2 = Q 4l0+1 3,0 Q 4l1+3 3,1 Q 4l2+3 3,2 = P 1 Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2, where P 1 = Q 3,0 Q 3 3,1 Q3 3,2, which is a symmetric polynomial in X 1,X 2 and X 3 Let X1 axb 2 Xc 3 be a term in the expansion form of P 1 For dimmesion reason, a + b + c = 37 Suppose that a is odd, b and c are even Then setting k =0,E = X 1 and F = X a X b 2 2 X c 2 3 Q 2l0 3,0 Q2l1 3,1 Q2l2 3,2, (a 1) + b + c deg F = +(7 2l 0 )+(6 2l 1 )+(4 2l 2 )=18+2(7l 0 +6l 1 +4l 2 ) 2 So µ(deg(f )) 2 Then we can use Theorem 15 to conclude that X1 a X2X b 3Q c 4l0 3,0 Q4l1 3,1 Q4l2 3,2 is hit Suppose that a, b and c are all odd Then after expanding P 1, we know that up to the permutation, (a, b, c) can only have the following choices, (3, 13, 21), (3, 9, 25), (5, 11, 21), (5, 13, 19), (7, 9, 21), (7, 11, 19), (7, 13, 17), (9, 11, 17) (251) Referring to the above list, we wish to show that X1 3X13 2 X21 3 Q4l0 3,0 Q4l1 3,1 Q4l2 3,2 is hit Check the condition of Theorem 15 as follows Setting k =1,E = X1 3X and F = X3 2 X5 3 Ql0 3,0 Ql1 3,1 Ql2 3,2, then deg E =5and degf =8+7l 0 +6l 1 +4l 2

7 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES 7 Hence (2 k+1 1)µ(deg F )=3µ(8 + 7l 0 +6l 1 +4l 2 ) Suppose that l 0 is even Then the condition of Theorem 15 is satisfied Hence we are done Suppose that l 0 is odd, if the condition of the Peterson Conjecture(Wood s Theorem) is satisfied, namely, 3 <α( l l l 2 +3), then we have proved the result If the condition is not satisfied, then the Minimum Spike exists [6], p578 So there exist non-negative integers m r s, such that l l l 2 = 2 m 1+2 r 1+2 s 1 This implies that 7l 0 +6l 1 +4l 2 =2 m 2 +2 r 2 9, s =2andr 3 The property for Minimum Spike implies that we can assume m>rif r>ssoµ(8+7l 0 +6l 1 +4l 2 )=µ(8+2 m r 2 9) 2 Hence the condition for Theorem 15 is satisfied Therefore X1 3 X2 13 X3 21 Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 is hit Using the same method, we can show that all the polynomials corresponding to the list (251) are hit except for the case: (a, b, c) =(7, 11, 19) and its permutations Lemma 25 For any non-negative integers l 0,l 1 and l 2, the following statements are true (1) Sq i [Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 ]=0for i not divisible by 4; (2) For any non-negative integer t, Sq 4t [Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 ]= Q 4s0 3,0 Q4s1 3,1 Q4s2 3,2 for some integers s 0,s 1 and s 2 ; (3) Sq 4 Sq 4 [Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 ]=0 Proof The proposition is a basic consequence of the Cartan formula and Theorem 14 We leave the detail to the reader In the following, we will frequently use the above lemma without mentioning each time Denote by Y the product Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 Lemma 26 (X1 7 X11 2 X X11 1 X7 2 X19 3 )Y [X7 1 X11 2 X7 3 + X11 1 X7 2 X7 3 ]Sq8 Sq 4 Y We will provide the proof in the appendix, since it involves very elementary and tedious computation Let P be the set consisting of the index (1, 2, 3) and all its permutations Then we have the following, X1X X3 19 Y =(X1 7 X2 11 X X1 11 X2 7 X3 19 )Y +(X1X X X1 11 X2 19 X3 7 )Y P +(X 19 1 X11 2 X7 3 + X19 1 X7 2 X Proof of Case(6) When n 0 =4l 0 +3,n 1 =4l 1 +3andn 2 =4l 2 +3, 3,0 Qn1 3,1 Qn2 3,2 = Q 4l0+3 3,0 Q 4l1+3 3,1 Q 4l2+3 3,2 = P 2 Y, )Y 0 (using Lemma 26) where Y = Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 and P 2 = Q 3 3,0 Q3 3,1 Q3 3,2 After expanding P 2, by applying Theorem 15 and Peterson s Conjecture(Wood s Theorem), we can conclude that all the corresponding terms are hit, except the following terms, up to the permutations, X1 7 X2 19 X3 25 Y,X1X X3 33 Y and X1 11 X2 19 X3 21 Y (up to the permutation) (261) For example, X 5 1 X21 2 X25 3 Y appears in the expansion of P 2Y we will show that it is hit Set k =1, IfdegF is even, then the condition of Theorem 15 3 Y )ishit IfdegF is odd, then we can assume that α(3 + EF 4 ) < 3 E = X 1 X and F = X 1 X 5 2 X6 3 Ql0 3,0 Ql1 3,1 Ql2 3,2 satisfied So EF 4 (= X 5 1X 21 2 X 25

8 8 K F TAN AND KAI XU (otherwise the result is proved using the Peterson Conjecture (Wood s Theorem)) So the Minimum Spike exists [6], p578 Hence deg(ef 4 )=2 m 1+2 r 1+2 s 1 for some integers m r s Therefore we have the following, (7l 0 +6l 1 +4l 2 )=deg(ef 4 )=2 m 1+2 r 1+2 s 1 This implies that 7l 0 +6l 1 +4l 2 =2 m 2 +2 r 2 13, s =1andr 2 Using the property of the Minimum spike, we may assume that m>rso µ(deg F )=µ(12 + 7l 0 +6l 1 +4l 2 )=µ(2 m 2 +2 r 2 1) 2 Therefore the inequality: deg E<(2 2 1)µ(F ) is true Hence the condition of Theorem 15 is satisfied Using the same method we can show that all the terms of P 2 Y corresponding to the expansion of P 2 are hit, except the terms listed in (261) Recall that Y denotes Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 Since X1 7 X11 2 X X7 1 X19 2 X25 3 = X1 7 Sq8 [X2 11 X25 3 ], we have (X1 7 X2 11 X X1X X3 25 )Y = X1 7Sq8 [X2 11X25 3 ]Y X2 11 X3 25 χ(sq 8 )[X1Y 7 ] = X2 11X25 3 (Sq7 Sq 1 +Sq 5 Sq 2 Sq 1 +Sq 6 Sq 2 +Sq 8 )[X1 7Y ] = X1 11 X2 11 X3 25 Sq 4 Y + X1 7 X2 11 X3 25 Sq 8 Y Yχ(Sq 4 )[X1 11X11 2 X25 3 ]+X7 1 X11 2 X25 3 Sq8 Y =[X 11 1 X 12 2 X X 12 1 X 11 2 X X 12 1 X 13 2 X X 13 1 X X 13 1 X 13 2 X 25 3 ]Y + X1 7X11 2 X25 3 Sq8 Y Using Theorem 15, we obtain the following, (X1X X X1 7 X2 19 X3 25 )Y X1X X3 25 Sq 8 Y (262) It is easy to show that (262) implies that P (X 7 1 X11 2 X X7 1 X19 2 X25 3 )Y P X 7 1 X11 2 X21 3 Sq4 Sq 8 Y, (263) where P is the set consisting of the index (1, 2, 3) and all its permutations In fact Lemma 27 Sq 4 [X1 7X11 2 X21 3 Sq8 Y ] = Sq 4 [X1 7X11 2 X21 3 ]Sq8 Y + X1 7X11 2 X21 3 Sq4 Sq 8 Y = [X 7 1 X 11 2 X X 7 1 X X 8 1 X 13 2 X X 8 1 X 14 2 X X 9 1 X X 9 1 X 13 2 X X 10 1 X 11 2 X X 10 1 X X 11 1 X 11 2 X 21 3 ]Sq 8 Y + X1 7 X2 11 X3 21 Sq 4 Sq 8 Y [X 7 1 X11 2 X X11 1 X11 2 X21 +X 7 1X 11 2 X 21 3 ]Sq8 Y 3 Sq 4 Sq 8 Y (using Theorem 15) X 7 1 X11 2 X21 3 Sq4 Sq 8 Y (X 11 1 X21 2 X X19 1 X13 2 X19 3 )Y Again we will leave the proof to the appendix

9 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES 9 Combining the above lemma with (263), we have the following, P (X7 1 X11 2 X X7 1 X19 2 X X11 1 X21 2 X19 3 )Y P [X7 1 X2 11 X3 21 Sq 4 Sq 8 Y + X1 11 X2 21 X3 19 Y ] P [(X11 1 X21 2 X X19 1 X13 2 X19 3 )Y + X X21 2 X19 3 Y ] 27 Appendix 271 Proof of Lemma 26 Recall that Y := Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 Then (X 7 1 X11 2 X X11 1 X7 2 X19 3 )Y = X19 3 Sq4 [X1 7 X7 2 ]Y + R where R =(X1 8 X10 2 X X9 1 X9 2 X X10 1 X8 2 X19 3 )Y It is easy to see that R is hit by using Theorem 15 Then (X1 7X11 2 X X11 1 X7 2 X19 3 )Y X7 1 X7 2 χ(sq4 )[X3 19Y ] X1 7 X2X Sq 4 Y = Sq 15 [X1 3X3 2 X9 3 ]X 1 X Sq4 Y = X1 3X3 2 X9 3 χ(sq15 )[X 1 X Sq 4 Y ] Using Theorem 15, we can show that the last term is hit So X1 3X3 2 X9 3 Sq8 Sq 4 Sq 2 Sq 1 [X 1 X ]Sq 4 Y +X1 7X7 2 X11 3 Sq8 Sq 4 Y +(X1 5X7 2 X13 3 +X1X X X1X X X1 4 X2 11 X3 10 +X1 11X4 2 X10 3 )Sq8 Sq 4 Y (X1X X X1 11 X2X )Y X1 3X3 2 X9 3 Sq8 Sq 4 Sq 2 Sq 1 [X 1 X ]Sq 4 Y + X1 7X7 2 X11 3 Sq8 Sq 4 Y X1 11X11 2 X11 3 Sq4 Y +(X1 5X11 2 X X11 1 X5 2 X X4 1 X4 2 X X4 1 X19 2 X10 3 +X1 19X4 2 X10 3 )Sq4 Y + X1 7X7 2 X11 3 Sq8 Sq 4 Y Again using Theorem 15, we can prove that the middle term is hit Hence (X1X X X1 11 X2X )Q 4l0 3,0 Q4l1 3,1 Q4l2 3,2 X1 11 X2 11 X3 11 Sq 4 Y + X1 7 X2 7 X3 11 Sq 8 Sq 4 Y =Sq 15 [X1 5X5 2 X5 3 ]X 1 X Sq4 Y + X1 7X7 2 X11 3 Sq8 Sq 4 Y X1 5 X2X χ(sq 15 )[X 1 X Sq 4 Y ]+X1 7 X2X Sq 8 Sq 4 Y = X1 5X5 2 X5 3 (Sq8 Sq 4 Sq 2 Sq 1 )[X 1 X Sq 4 Y ]+X1 7X7 2 X11 3 Sq8 Sq 4 Y = X1 5 X2X (Sq 8 Sq 4 Sq 2 Sq 1 )[X 1 X ]Sq 4 Y +X1 5X5 2 X5 3 (Sq4 Sq 2 Sq 1 )[X 1 X ]Sq 8 Sq 4 Y + X1 7X7 2 X11 3 Sq8 Sq 4 Y =[X1 7X13 2 X X13 1 X7 2 X X13 1 X13 2 X7 3 + X6 1 X6 2 X X6 1 X21 2 X6 3 +X1 21 X2X ]Sq 4 Y + X1 7 X2X Sq 8 Sq 4 Y +[X1X X3 9 + X1X X3 9 +X1 9X9 2 X7 3 + X6 1 X6 2 X X6 1 X13 2 X6 3 + X13 1 X6 2 X6 3 ]Sq8 Sq 4 Y +X 7 1X 7 2X 11 3 Sq 8 Sq 4 Y Expanding it term by term, using Theorem 15, we can conclude that (X 7 1 X 11 2 X X 11 1 X 7 2 X 19 3 )Y X 7 1 X 7 2X 11 3 Sq 8 Sq 4 Y (271)

10 10 K F TAN AND KAI XU On the other hand, X1 7 X2X Sq 8 Sq 4 Y =Sq 9 [X1 2X2 2 X5 3 ]X3 1 X3 Sq8 Sq 4 Y X1X X3χ(Sq 5 9 )[X1 3 X2X 3 3 Sq 8 Sq 4 Y ] = 2X2 2 X5 3 (Sq8 Sq 1 +Sq 6 Sq 2 Sq 1 )[X1 3X3 Sq8 Sq 4 Y ] X1X X3(Sq 5 8 Sq 1 +Sq 6 Sq 2 Sq 1 )[X1 3 X2 3 X 3 ]Sq 8 Sq 4 Y +X1 7X7 2 X7 3 Sq4 Sq 8 Sq 4 Y (using Lemma 25 and Theorem 15) =[X 11 1 X 5 2 X X 5 1 X 11 2 X X 7 1 X X 10 1 X 6 2 X X 11 1 X 7 2 X X 6 1 X X 6 1 X 6 2 X X 7 1 X 5 2 X X 5 1 X 7 2 X 13 3 ]Sq 8 Sq 4 Y + X1 7X7 2 X7 3 Sq4 Sq 8 Sq 4 Y [X1X X3 7 + X1 11 X2X ]Sq 8 Sq 4 Y + X1 7 X2X Sq 4 Sq 8 Sq 4 Y So to prove the lemma, it suffices to show that X 7 1 X 7 2X 7 3 Sq 4 Sq 8 Sq 4 Y is hit This can be done as follows, X 7 1X 7 2 X 7 3Sq 4 Sq 8 Sq 4 Y =Sq 9 (X 3 1 X3 2 X3 3 )X 1 X Sq4 Sq 8 Sq 4 Y = X 3 1 X 3 2X 3 3 χ(sq 9 )[X 1 X Sq 4 Sq 8 Sq 4 Y ] = X 3 1 X3 2 X3 3 (Sq8 Sq 1 +Sq 6 Sq 2 Sq 1 )[X 1 X Sq 4 Sq 8 Sq 4 Y ] (X 4 1 X 5 2 X X 4 1X 4 2 X X 5 1X 4 2 X 4 3)Sq 8 Sq 4 Sq 8 Sq 4 Y (using Lemma 25 and Theorem 15) Proof of Lemma 27 First of all we claim the following, Proposition 28 X1 7 X2 11 X3 21 Sq 4 Sq 8 Y X1 19X9 2 X11 3 Sq4 Sq 8 Y +[X1 11X9 2 X X7 1 X13 2 X11 3 ]Sq8 Sq 4 Sq 8 Y +X1X X3 11 Sq 16 Sq 4 Sq 8 Y + X1 7 X2 21 X3 11 Sq 4 Sq 8 Y We will leave the proof to the next section Using the χ-trick and Theorem 15, we can easily verify that X 19 1 X 9 2 X 11 3 Sq 4 Sq 8 Y χ(sq 4 )[X 19 1 X9 2 X11 3 ]Sq8 Y =[X 19 1 X 12 2 X X 20 1 X 12 2 X X 20 1 X X 21 1 X 9 2 X X 21 1 X 10 2 X 12 3 ]Sq 8 Y 0 So X 7 1X 11 2 X 21 3 Sq 4 Sq 8 Y [X 11 1 X9 2 X X7 1 X13 2 X11 3 ]Sq8 Sq 4 Sq 8 Y +X1X X3 11 Sq 16 Sq 4 Sq 8 Y + X1 7 X2 21 X 11 3 Sq 4 Sq 8 Y

11 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES 11 Now we use the χ-trick and Theorem 15 again as follows, Therefore we have Hence we have [X1 11X9 2 X X7 1 X13 2 X11 3 ]Sq8 Sq 4 Sq 8 Y χ(sq 8 )[X1 11 X2 9 X X1X X 11 3 ]Sq 4 Sq 8 Y =[X 16 1 X 10 2 X X 13 1 X 10 2 X X 11 1 X X 12 1 X 10 2 X X 13 1 X 9 2 X X 16 1 X X 17 1 X 9 2 X X 17 1 X 10 2 X X 12 1 X X 19 1 X 9 2 X X 11 1 X 9 2 X X 8 1 X X 9 1 X 13 2 X X 7 1 X 16 2 X X 7 1 X X 7 1 X 21 2 X X 8 1 X 18 2 X X 9 1 X X 9 1 X 18 2 X X 9 1 X 14 2 X 16 3 ]Sq 4 Sq 8 Y [X 7 1 X13 2 X X7 1 X21 2 X11 3 ]Sq4 Sq 8 Y X1 5 X2X Sq 16 Sq 4 Sq 8 Y χ(sq 16 )[X1 5X7 2 X11 3 ]Sq4 Sq 8 Y =[X 8 1 X 19 2 X X 16 1 X 7 2 X X 17 1 X X 10 1 X 8 2 X X 10 1 X 9 2 X X 9 1 X X 9 1 X 11 2 X X 9 1 X 17 2 X X 16 1 X X 8 1 X 11 2 X X 17 1 X 11 2 X X 9 1 X X 18 1 X 8 2 X X 18 1 X 9 2 X X 5 1 X X 10 1 X 17 2 X X 10 1 X 16 2 X X 6 1 X X 6 1 X 16 2 X 17 3 ]Sq 4 Sq 8 Y X 17 1 X 11 2 X 11 3 Sq 4 Sq 8 Y [χ(sq 8 )χ(sq 4 )X1 17X11 2 X11 3 ]Y = X 18 1 X 21 2 X X 18 1 X 20 2 X X 18 1 X 13 2 X X 18 1 X X 17 1 X 21 2 X X 17 1 X 13 2 X X 17 1 X 16 2 X X 17 1 X X 20 1 X 13 2 X X 20 1 X 18 2 X X 24 1 X 11 2 X X 24 1 X X 24 1 X 14 2 X X 24 1 X 16 2 X X 20 1 X 11 2 X X 20 1 X X 20 1 X 19 2 X X 20 1 X 20 2 X X 17 1 X 17 2 X X 20 1 X X 20 1 X 17 2 X (usingTheorem15) X 7 1 X11 2 X21 3 Sq4 Sq 8 Y [X1 7 X2 13 X X1X X X1X X 11 = X1 7X13 2 X19 3 Sq4 Sq 8 Y χ(sq 8 )χ(sq 4 )[X1 7X13 2 X19 3 (X1 11 X2 21 X X1 19 X2 13 X 19 3 ]Sq 4 Sq 8 Y ]Y (using the χ-trick) 3 )Y (using Theorem 15) X 7 1 X 11 2 X 21 3 Sq 4 Sq 8 Y (X 11 1 X 21 2 X X 19 1 X 13 2 X 19 3 )Y (272) 273 Proof of Proposition 28 In this section, we will give a proof of Proposition 28 X1 7X11 2 X21 3 Sq4 Sq 8 Y = X1 3X3 Sq16 [X1 2X4 2 X10 3 ]Sq4 Sq 8 Y X1 2 X2X χ(sq 16 )[X1X X 3 Sq 4 Sq 8 Y ] = X1 2X4 2 X10 3 [Sq9 Sq 4 Sq 2 Sq 1 +Sq 12 Sq 3 Sq 1 +Sq 12 Sq 4 +Sq 13 Sq 2 Sq 1 +Sq 15 Sq 1 +Sq 16 +Sq 10 Sq 4 Sq 2 +Sq 14 Sq 2 +Sq 11 Sq 4 Sq 1 ][X1X X 3 Sq 4 Sq 8 Y ]

12 12 K F TAN AND KAI XU After expanding, we have nine terms and we wish to look into these terms in detail: TERM 1 = X1 2X4 2 X10 3 [Sq9 Sq 4 Sq 2 Sq 1 3X3 Sq4 Sq 8 Y ] = X1 2 X2X [Sq 9 Sq 4 Sq 2 Sq 1 ][X1X X 3 ]Sq 4 Sq 8 Y +X1 2X4 2 X10 3 [Sq1 Sq 4 Sq 2 Sq 1 ][X1 3X3 ]Sq8 Sq 4 Sq 8 Y =[X 12 1 X 13 2 X X 11 1 X 14 2 X X 7 1 X 14 2 X X 7 1 X X 8 1 X 13 2 X X 8 1 X 20 2 X 11 3 X 11 1 X 10 2 X X 12 1 X X 18 1 X 9 2 X X 18 1 X 10 2 X X 5 1 X 8 2 X X 5 1 X X 6 1 X 7 2 X X 6 1 X 22 2 X X 20 1 X 7 2 X X 20 1 X 8 2 X 11 3 ]Sq 4 Sq 8 Y +[X 7 1 X 10 2 X X 7 1 X 12 2 X X 8 1 X 9 2 X X 8 1 X X 10 1 X 9 2 X X 10 1 X 10 2 X X 5 1 X 8 2 X X 5 1 X X 6 1 X 7 2 X X 6 1 X 14 2 X X 12 1 X 7 2 X X 12 1 X 8 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y 0 TERM 2 = X1 2X4 2 X10 3 Sq12 Sq 3 Sq 1 [X1 3X3 Sq4 Sq 8 Y ] = X1 2 X2X Sq 12 Sq 3 Sq 1 [X1 3 X2 3 X 3 ]Sq 4 Sq 8 Y 2X4 2 X10 3 Sq4 Sq 3 Sq 1 [X1 3X3 ]Sq8 Sq 4 Sq 8 Y +X1X 2 2X Sq 3 Sq 1 [X1 3 X2X 3 3 ]Sq 12 Sq 4 Sq 8 Y =0+[X 7 1 X 10 2 X X 7 1 X 12 2 X X 8 1 X 9 2 X X 8 1 X X 10 1 X 9 2 X X 10 1 X 10 2 X X 5 1 X 8 2 X X 5 1 X X 6 1 X 7 2 X X 6 1 X 14 2 X X 12 1 X X 12 1 X 8 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y +[X 5 1 X 8 2 X X 5 1 X 10 2 X X 6 1 X 7 2 X X 6 1 X X 8 1 X 7 2 X X 8 1 X 8 2 X 11 3 ]Sq 12 Sq 4 Sq 8 Y 0 TERM 3 = X1 2X4 2 X10 3 Sq12 Sq 4 [X1 3X3 Sq4 Sq 8 Y ] = X1 2X4 2 X10 3 Sq12 Sq 4 [X1 3X3 ]Sq4 Sq 8 Y +X1X X3 10 Sq 4 Sq 4 [X1X X 3 ]Sq 8 Sq 4 Sq 8 Y +X1 2X4 2 X10 3 Sq4 [X1 3X3 ]Sq12 Sq 4 Sq 8 Y =0+[X 5 1 X 12 2 X X 5 1 X 14 2 X X 6 1 X 13 2 X X 6 1 X X 7 1 X 10 2 X X 7 1 X 13 2 X X 8 1 X 9 2 X X 8 1 X X 10 1 X 7 2 X X 10 1 X 10 2 X X 11 1 X 8 2 X X 11 1 X X 12 1 X 7 2 X X 12 1 X 8 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y +[X 5 1 X 10 2 X X 6 1 X 9 2 X X 6 1 X 10 2 X X 7 1 X X 7 1 X 9 2 X X 8 1 X 7 2 X X 8 1 X 8 2 X 11 3 ]Sq 12 Sq 4 Sq 8 Y (X1X X X1 11 X2X )Sq 8 Sq 4 Sq 8 Y + X1X X3 11 Sq 12 Sq 4 Sq 8 Y

13 DICKSON INVARIANTS HIT BY THE STEENROD SQUARES 13 TERM 4 = X1 2X4 2 X10 3 Sq13 Sq 2 Sq 1 [X1 3X3 Sq4 Sq 8 Y ] = X1 2X4 2 X10 3 Sq13 Sq 2 Sq 1 [X1 3X3 ]Sq4 Sq 8 Y +X1X X3 10 Sq 5 Sq 2 Sq 1 [X1X X 3 ]Sq 8 Sq 4 Sq 8 Y +X1 2X4 2 X10 3 Sq1 Sq 2 Sq 1 [X1 3X3 ]Sq12 Sq 4 Sq 8 Y +0 + [X 7 1 X 10 2 X X 7 1 X 12 2 X X 8 1 X 9 2 X X 8 1 X X 10 1 X 9 2 X X 10 1 X 10 2 X X 5 1 X 8 2 X X 5 1 X X 6 1 X 7 2 X X 6 1 X 14 2 X X 12 1 X X 12 1 X 8 2 X 11 3 ]Sq 4 Sq 8 Y +[X 5 1 X 8 2 X X 5 1 X 10 2 X X 6 1 X 7 2 X X 6 1 X X 8 1 X 7 2 X X 8 1 X 8 2 X 11 3 ]Sq 12 Sq 4 Sq 8 Y 0 TERM 5 = X1 2X4 2 X10 3 Sq15 Sq 1 [X1 3X3 Sq4 Sq 8 Y ] = X1 2X4 2 X10 3 Sq15 Sq 1 [X1 3X3 ]Sq4 Sq 8 Y +X1X X3 10 Sq 7 Sq 1 [X1X X 3 ]Sq 8 Sq 4 Sq 8 Y +X1 2X4 2 X10 3 Sq3 Sq 1 [X1 3X3 ]Sq12 Sq 4 Sq 8 Y =0+[X 7 1 X 10 2 X X 7 1 X 12 2 X X 8 1 X X 8 1 X 12 2 X X 10 1 X 9 2 X X 10 1 X 10 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y +[X 5 1 X 8 2 X X 5 1 X 10 2 X X 6 1 X 7 2 X X 6 1 X X 8 1 X 7 2 X X 8 1 X 8 2 X 11 3 ]Sq 12 Sq 4 Sq 8 Y 0 TERM 6 = 2X4 2 X10 3 Sq16 3X3 Sq4 Sq 8 Y ] = X1X X3 10 Sq 16 [X1X X 3 ]Sq 4 Sq 8 Y + X1X X 10 +X 2 1 X4 2 X10 3 Sq4 [X 3 1 X3 ]Sq12 Sq 4 Sq 8 Y + X 2 1 X4 2 X10 3 Sq 8 [X1 3 X2 3 X 3 ]Sq 8 Sq 4 Sq 8 Y 3 X3 1 X3 Sq16 Sq 4 Sq 8 Y =0+0+[X 1 5 X 2 10 X X 1 6 X 2 9 X X 1 6 X 2 10 X X 1 7 X 2 8 X X 1 7 X 2 9 X X 1 8 X 2 7 X X 1 8 X 2 8 X 3 11 ]Sq 12 Sq 4 Sq 8 Y +[X 1 5 X 2 7 X 3 11 ]Sq 16 Sq 4 Sq 8 Y X 7 1X 9 2 X 11 3 Sq 12 Sq 4 Sq 8 Y + X 5 1 X 7 2X 11 3 Sq 16 Sq 4 Sq 8 Y TERM 7 = X1 2X4 2 X10 3 Sq10 Sq 4 Sq 2 3X3 Sq4 Sq 8 Y ] = X1 2 X2 4 X3 10 Sq 10 Sq 4 Sq 2 [X1X X 3 ]Sq 4 Sq 8 Y +X1 2X4 2 X10 3 Sq2 Sq 4 Sq 2 [X1 3X3 ]Sq8 Sq 4 Sq 8 Y =[X 5 1 X 20 2 X X 18 1 X 7 2 X X 7 1 X 20 2 X X 18 1 X X 5 1 X 22 2 X X 6 1 X 21 2 X X 6 1 X 22 2 X X 7 1 X X 19 1 X 8 2 X X 19 1 X 9 2 X X 20 1 X 7 2 X X 20 1 X 8 2 X 11 3 ]Sq 4 Sq 8 Y +[X 7 1 X 12 2 X X 10 1 X 9 2 X X 5 1 X 12 2 X X 5 1 X X 6 1 X 13 2 X X 6 1 X 14 2 X X 7 1 X 13 2 X X 10 1 X X 11 1 X 8 2 X X 11 1 X 9 2 X X 12 1 X 7 2 X X 12 1 X 8 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y (X1 7 X2 21 X X1 19 X2X )Sq 4 Sq 8 Y +(X1X X X1 11 X2X )Sq 8 Sq 4 Sq 8 Y

14 14 K F TAN AND KAI XU TERM 8 = X1 2X4 2 X10 3 Sq14 Sq 2 [X1 3X3 Sq4 Sq 8 Y ] = X1 2X4 2 X10 3 Sq14 Sq 2 [X1 3X3 ]Sq4 Sq 8 Y + X1 2X4 2 X10 3 Sq6 Sq 2 [X1 3X3 ]Sq8 Sq 4 Sq 8 Y +X1X X3 10 Sq 2 Sq 2 [X1X X 3 ]Sq 12 Sq 4 Sq 8 Y =0+[X 7 1 X 12 2 X X 10 1 X 9 2 X X 5 1 X X 5 1 X 14 2 X X 6 1 X 13 2 X X 6 1 X 14 2 X X 7 1 X X 10 1 X 7 2 X X 11 1 X 8 2 X X 11 1 X 9 2 X X 12 1 X X 12 1 X 8 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y +[X 5 1 X 8 2 X X 5 1 X X 6 1 X 7 2 X X 6 1 X 10 2 X X 8 1 X 7 2 X X 8 1 X 8 2 X 11 3 ]Sq 12 Sq 4 Sq 8 Y (X 11 1 X 9 2 X X 7 1 X 13 2 X 11 3 )Sq 8 Sq 4 Sq 8 Y TERM 9 = X1 2X4 2 X10 3 Sq11 Sq 4 Sq 1 [X1 3X3 Sq4 Sq 8 Y ] = X1 2X4 2 X10 3 Sq11 Sq 4 Sq 1 [X1 3X3 ]Sq4 Sq 8 Y +X1 2X4 2 X10 3 Sq3 Sq 4 Sq 1 [X1 3X3 ]Sq8 Sq 4 Sq 8 Y =[X 11 1 X 14 2 X X 12 1 X 13 2 X X 7 1 X 14 2 X X 7 1 X X 8 1 X 13 2 X X 8 1 X 20 2 X X 11 1 X 10 2 X X 12 1 X X 18 1 X 9 2 X X 18 1 X 10 2 X 11 3 ]Sq 4 Sq 8 Y +[X 7 1 X X 7 1 X 12 2 X X 8 1 X 9 2 X X 8 1 X 12 2 X X 10 1 X X 10 1 X 10 2 X 11 3 ]Sq 8 Sq 4 Sq 8 Y 0 Therefore 9 i=1 TERM i X19 1 X9 2 X11 3 Sq4 Sq 8 Y +[X1 11X9 2 X X7 1 X13 2 X11 3 ]Sq8 Sq 4 Sq 8 Y +X1X X3 11 Sq 16 Sq 4 Sq 8 Y + X1 7 X2 21 X3 11 Sq 4 Sq 8 Y This completes the proof of Proposition 28 References [1] Hung, HHV, The action of the Steenrod squares on the modular invariants of linear groups, Proc Amer Math Soc 113(1991), [2] Hung, HHV, Spherical classes and the algebraic transfer, Trans Amer Math Soc 10(1997), [3] Hung, HHV, Theweakconjectureonsphericalclasses,MathZ231(1999), [4] Hung, HHV and Minh, PA The action of the mod p Steenrod operations on the modular invariants of linear groups, Vietnam J Math 23(1995),39 56 [5] Silverman, J, Hit polynomials and conjugation in the dual Steenrod algebra, Math Proc Caambridge Philos Soc 123(1998), [6] Singer, WM, On the action of Steenrod squares on polynomial algebras, Proc Amer Math Soc 111(1991), [7] Tan, KF, Hit polynomials, Dickson invariants and related topics, MSc thesis (National University of Singapore)(1999) [8] Wood, RMW, Steenrod squares of polynomials and the Peterson conjecture, Math Proc Cambridge Philos Soc 105(1989), [9] Wood, RMW, Problems in the Steenrod algebra, Bull London Math Soc 30(1998), address, KFTan:tan khengfeung@pacificnetsg address, KaiXu:matxukai@leonisnusedusg Department of Mathematics, National University of Singapore, 2 Science Drive 2, Singapore