DICKSON INVARIANTS HIT BY THE STEENROD SQUARE

K. F. TAN AND KAI XU

Abstract. Let D_{3} be the Dickson invariant ring of 𝔽_{2}[X_{1}, X_{2}, X_{3}] by GL(3, 𝔽_{2}). In this paper, we provide an elementary proof of Theorem 3.2 of [2]: each element in D_{3} is hit by the Steenrod square in 𝔽_{2}[X_{1}, X_{2}, X_{3}].

1. Introduction

Throughout this paper, we work on the polynomials over 𝔽_{2}. Let GL(n, 𝔽_{2}) be the n × n-general linear group over 𝔽_{2}. Then the standard group action:

GL(n, 𝔽_{2}) × 𝔽_{2}[X_{1}, X_{2}, …, X_{n}] ⟶ 𝔽_{2}[X_{1}, X_{2}, …, X_{n}]

gives rise to the ring of Dickson invariants

𝔽_{2}[X_{1}, X_{2}, …, X_{n}]^{GL(n,𝔽_{2})},

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 = 0, ..., n − 1) is a certain polynomial over 𝔽_{2}.

Definition 1.1. 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 1.2. Is each Dickson invariant hit?

The motivation of presenting this question was illustrated in detail in [2] and in an excellent expository paper [9], p501. When n = 1 and 2, the answer to the above question is negative. For example, when n = 1, the Dickson invariant X_{1} is not hit; when n = 2, X_{1}^{2} +X_{2}^{2} +X_{1}X_{2} is a Dickson invariant but is not hit.

Conjecture 1.3 (Hung, [1, 2, 3, 4, 8]). When n ≥ 3, all Dickson invariants are hit.

In general, this question is hard to answer. When n = 3, Hung has proved that the conjecture is true [2] by using many advanced results. We were told that Hung and his co-author have recently established the proof of the conjecture for any n > 3. The main goal of this paper is to give an elementary proof of the above conjecture when n = 3.

Write V_{1} = X_{1}, V_{2} = X_{2}(X_{2} + X_{1}) and V_{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_{2,0}^{2}+V_{3}Q_{2,1} = (V_{2}V_{1})^{2}+V_{3}(V_{1}^{2}+V_{2}) and Q_{3,2} = Q_{2,1}^{2}+V_{3} = (V_{1}^{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, Theorem 1.4 (Hung [1]). The Steenrod operation acts trivially on Q_{3,0}, Q_{3,1} and Q_{3,2} except for the following cases,

1991 Mathematics Subject Classification. 55S10, 55Q45, 55S10, 55T15.

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_{3,0}^{2}, Sq^{1}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_{3,1}^{2}, Sq^{2}Q_{3,2} = Q_{3,1}
Sq^{3}Q_{3,2} = Q_{3,0}, Sq^{4}Q_{3,2} = Q_{3,2}^{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

min{m ∈ ℕ | n = ∑_{i=1}^{m} γ(k_{i}) for some k_{i} ∈ ℕ}.

Let E and F be homogeneous polynomials of degrees e and f, respectively. Then we have the following,

Theorem 1.5 (Silverman [5]). Suppose that e < (2^{k+1} − 1)μ(f) for some k ≥ 0. Then the polynomial EF^{2^{k+1}} is hit.

.

For an integer t, let α(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 1.6 (Wood [8]). If α(deg(U) + n) > n, then U is hit.

.

John Milnor showed that there is an anti-automorphism χ from the Steenrod algebra to itself, satisfying the following condition,

χ(Sq^{k}) = ∑_{i=1}^{k} Sq^{i}χ(Sq^{k−i}) for k ≥ 1.

We will frequently use the so-called χ-trick in the following sections, which is based on the following,

Proposition 1.7 (see for example, [9]). 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 1.8. 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 K.G. Monks for providing his Maple program on the Internet. His source code benefits us a great deal.

2. Proofs

Clearly, to prove our result, we only need show that Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} is hit for any non-negative 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 2.1. The proof of Cases(1)-(4) is extracted from the first author’s MSc thesis [7].

Notation 2.2. 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.

.

2.1. 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,

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}^{2l_{0}+1}Q_{3,1}^{2l_{1}}Q_{3,2}^{2l_{2}}
= [Sq^{1}Q_{3,1}](Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}
= Sq^{1}[Q_{3,1}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}].

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,

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}^{2l_{0}}Q_{3,1}^{2l_{1}+1}Q_{3,2}^{2l_{2}}
= Sq^{2}[Q_{3,2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}] + Q_{3,2}Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}
≡ Q_{3,2}Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}.

Notice that

Q_{3,2} = Q_{2,1}^{2} + V_{3} and V_{3} = Sq^{1}(Q_{2,1}X_{3} + X_{3}^{3}).  (2.1.1)

Then

Q_{3,2}Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2} = (Q_{2,1}^{2} + V_{3})Sq^{2}[(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}]
= [Q_{2,1}Sq^{1}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2} + Sq^{1}[(Q_{2,1}X_{3} + X_{3}^{3})(Sq^{1}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}}))^{2}]
≡ 0

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,

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}^{2l_{0}}Q_{3,1}^{2l_{1}}Q_{3,2}^{2l_{2}+1}
= (Q_{2,1}^{2} + Sq^{1}(Q_{2,1}X_{3} + X_{3}^{3}))(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2} (using (2.1.1))
= (Q_{2,1}Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2} + Sq^{1}[(Q_{2,1}X_{3} + X_{3}^{3}))(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}]
≡ 0

So Case (1) has been done.

2.2. Proof of Case(2). Suppose that n_{0} = 2l_{0}, n_{1} = 2l_{1} + 1 and n_{2} = 2l_{2} + 1. We will use the χ-trick (1.7) to prove the result. Notice that

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = [Sq^{4}(Q_{3,1})](Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2} ≡ Q_{3,1}χ(Sq^{4})(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}  (2.2.1)

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 (2.2.1) can be reduced to

Sq^{2}(Q_{3,2})[Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2} + Q_{3,1}[Sq^{1}(Sq^{1}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}}))]^{2}.

Noting that Sq^{1}Sq^{1} = 0, the last term is zero. Again using the χ-trick, the first term is hit if and only if Q_{3,2}χ(Sq^{2})[Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2} is hit. Now

Q_{3,2}χ(Sq^{2})[Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2}
= (Q_{2,1}^{2} + V_{3})[Sq^{1}Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2}
≡ V_{3}[Sq^{1}Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2}
= Sq^{1}{(Q_{2,1}X_{3} + X_{3}^{3})[Sq^{1}Sq^{2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2}}.

Suppose that n_{0} = 2l_{0} + 1, n_{1} = 2l_{1} and n_{2} = 2l_{2} + 1. Then

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}^{2l_{0}+1}Q_{3,1}^{2l_{1}}Q_{3,2}^{2l_{2}+1}
= Sq^{1}[Q_{3,1}Q_{3,2}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}] (since Sq^{1}Q_{3,2} = 0).

Suppose that n_{0} = 2l_{0} + 1, n_{1} = 2l_{1} + 1 and n_{2} = 2l_{2}. Then

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}^{2l_{0}+1}Q_{3,1}^{2l_{1}+1}Q_{3,2}^{2l_{2}} = Sq^{2}(Q_{3,2})[Q_{3,0}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}].  (2.2.2)

Since Sq^{i}(Q_{3,0}) = 0 for i = 1, 2, Sq^{1}Q_{3,2} = 0 and Q_{3,0} = V_{3}V_{2}V_{1}, the last term in (2.2.2) is reduced to

Q_{3,2}Q_{3,0}Sq^{2}[(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})^{2}] = Sq^{1}(Q_{2,1}X_{3} + X_{3}^{3})V_{2}V_{1}Q_{2,1}^{2}[Sq^{1}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2}

+ Sq^{1}(Q_{2,1}X_{3} + X_{3}^{3})Q_{3,0}[Sq^{1}(Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}})]^{2}.

The last two terms are hit, since Sq^{1} acts on V_{2}V_{1} and Q_{3,0} trivially.

2.3. Proof of Case(3). Suppose that n_{0} = 4l_{0} + 1, n_{1} = 4l_{1} + 1 and n_{2} = 2l_{2} + 1. Let B be the polynomial Q_{3,0}^{2l_{0}}Q_{3,1}^{2l_{1}}Q_{3,2}^{l_{2}}. Then Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}Q_{3,1}Q_{3,2}B^{2} = V_{3}V_{2}^{3}V_{1}^{7}B^{2} + V_{3}V_{2}^{5}V_{1}^{3}B^{2} + V_{3}^{2}V_{2}V_{1}^{7}B^{2} + V_{3}^{3}V_{2}^{1}V_{1}^{3}B^{2} + V_{3}^{2}V_{2}^{2}V_{1}^{5}B^{2} + V_{3}^{2}V_{2}^{4}V_{1}^{1}B^{2} + V_{3}^{3}V_{2}^{2}V_{1}^{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 2.3. For any positive odd integers k_{0}, k_{1} and k_{2}, the polynomial V_{3}^{k_{0}}V_{2}^{k_{1}}V_{1}^{k_{2}}B^{2} is hit.

Proof. Put k_{0} = 2a + 1, k_{1} = 2b + 1 and k_{2} = 2c + 1. Then we have

V_{3}^{k_{0}}V_{2}^{k_{1}}V_{1}^{k_{2}}B^{2} = V_{3}^{2a+1}V_{2}^{2b+1}V_{1}^{2c+1}B^{2} = [Sq^{1}(Q_{2,1}X_{3} + X_{3}^{3})]V_{3}^{2a}V_{2}^{2b+1}V_{1}^{2c+1}B^{2}
= Sq^{1}[(Q_{2,1}X_{3} + X_{3}^{3})(V_{3}^{2a}V_{2}^{2b+1}V_{1}^{2c+1}B^{2})] (since Sq^{1}(V_{2}V_{1}) = 0).

Since V_{3}^{2}V_{2}^{2}V_{1}^{5}B^{2} = V_{1}(V_{3}V_{2}V_{1}^{2}B)^{2} and V_{3}^{2}V_{2}^{4}V_{1}^{1}B^{2} = V_{1}(V_{3}V_{2}^{2}B)^{2}, we can verify that these two terms satisfy the condition in Theorem 1.5. So they are hit.

It remains to show that V_{3}^{3}V_{2}^{2}V_{1}^{1}B^{2} is hit. In fact

V_{3}^{3}V_{2}^{2}V_{1}^{1}B^{2} = X_{1}^{3}X_{2}^{2}X_{3}^{12}B^{2} + X_{1}X_{2}^{4}X_{3}^{12}B^{2} + X_{1}^{7}X_{2}^{2}X_{3}^{8}B^{2} + X_{1}X_{2}^{8}X_{3}^{8}B^{2}
+X_{1}^{8}X_{2}^{6}X_{3}^{3}B^{2} + X_{1}^{5}X_{2}^{2}X_{3}^{10}B^{2} + X_{1}^{4}X_{2}^{3}X_{3}^{10}B^{2} + X_{1}^{2}X_{2}^{5}X_{3}^{10}B^{2}
+X_{1}X_{2}^{6}X_{3}^{10}B^{2} + X_{1}^{9}X_{2}^{2}X_{3}^{6}B^{2} + X_{1}^{8}X_{2}^{3}X_{3}^{6}B^{2} + X_{1}^{2}X_{2}^{9}X_{3}^{6}B^{2}
+X_{1}^{4}X_{2}^{10}X_{3}^{3}B^{2} + X_{1}X_{2}^{10}X_{3}^{6}B^{2} + X_{1}^{9}X_{2}^{4}X_{3}^{4}B^{2} + X_{1}^{5}X_{2}^{8}X_{3}^{4}B^{2}
+X_{1}^{8}X_{2}^{5}X_{3}^{4}B^{2} + X_{1}^{4}X_{2}^{9}X_{3}^{4}B^{2} + X_{1}^{7}X_{2}^{6}X_{3}^{4}B^{2} + X_{1}^{3}X_{2}^{10}X_{3}^{4}B^{2}
+X_{1}^{5}X_{2}^{3}X_{3}^{9}B^{2} + X_{1}^{3}X_{2}^{5}X_{3}^{9}B^{2} + X_{1}^{8}X_{2}^{4}X_{3}^{5}B^{2} + X_{1}^{2}X_{2}^{10}X_{3}^{5}B^{2}
+X_{1}^{9}X_{2}^{5}X_{3}^{3}B^{2} + X_{1}^{5}X_{2}^{9}X_{3}^{3}B^{2}.

Using Theorem 1.5, we can prove all but the following terms are hit: X_{1}^{3}X_{2}^{5}X_{3}^{9}B^{2}, X_{1}^{5}X_{2}^{3}X_{3}^{9}B^{2}, X_{1}^{9}X_{2}^{3}X_{3}^{5}B^{2}, X_{1}^{9}X_{2}^{5}X_{3}^{3}B^{2}, X_{1}^{3}X_{2}^{9}X_{3}^{5}B^{2} and X_{1}^{5}X_{2}^{9}X_{3}^{3}B^{2}. Indeed we only need the following proposition to conclude the result.

Proposition 2.4. X_{1}^{3}X_{2}^{5}X_{3}^{9}B^{2}+X_{1}^{5}X_{2}^{3}X_{3}^{9}B^{2}, X_{1}^{9}X_{2}^{3}X_{3}^{5}B^{2}+X_{1}^{9}X_{2}^{5}X_{3}^{3}B^{2} and X_{1}^{3}X_{2}^{9}X_{3}^{5}B^{2}+ X_{1}^{5}X_{2}^{9}X_{3}^{3}B^{2} are hit.

Proof. Careful observations reveal the following,

X_{1}^{3}X_{2}^{5}X_{3}^{9} + X_{1}^{5}X_{2}^{3}X_{3}^{9} = Sq^{2}(X_{1}^{9}X_{2}^{3}X_{3}^{3}) + Sq^{1}(X_{1}^{10}X_{2}^{3}X_{3}^{3}) + X_{1}^{9}X_{2}^{4}X_{3}^{4},  (2.3.1)
X_{1}^{9}X_{2}^{3}X_{3}^{5} + X_{1}^{9}X_{2}^{5}X_{3}^{3} = Sq^{2}(X_{1}^{3}X_{2}^{9}X_{3}^{3}) + Sq^{1}(X_{1}^{3}X_{2}^{10}X_{3}^{3}) + X_{1}^{4}X_{2}^{9}X_{3}^{4}

and

X_{1}^{3}X_{2}^{9}X_{3}^{5} + X_{1}^{5}X_{2}^{9}X_{3}^{3} = Sq^{2}(X_{1}^{3}X_{2}^{3}X_{3}^{9}) + Sq^{1}(X_{1}^{3}X_{2}^{3}X_{3}^{10}) + X_{1}^{4}X_{2}^{4}X_{3}^{9}.

We shall show that X_{1}^{3}X_{2}^{5}X_{3}^{9}B^{2}+X_{1}^{5}X_{2}^{3}X_{3}^{9}B^{2} is hit; the remaining two cases can be proved similarly.

By Theorem 1.5, X_{1}^{9}X_{2}^{4}X_{3}^{4}B^{2} = X_{1}(X_{1}^{4}X_{2}^{2}X_{3}^{2}B)^{2} in (2.3.1) is hit. So we are left to show that Sq^{2}(X_{1}^{9}X_{2}^{3}X_{3}^{3})B^{2} + Sq^{1}(X_{1}^{10}X_{2}^{3}X_{3}^{3})B^{2} is hit. This can be done by using the χ-trick. In fact,

X_{1}^{3}X_{2}^{3}X_{3}^{9}[χ(Sq^{2})(B^{2})] + X_{1}^{3}X_{2}^{3}X_{3}^{10}[χ(Sq^{1}(B^{2})] = 0,

where we use the facts that Sq^{2}B^{2} = 0 and Sq^{1}B^{2} = 0.

To finish the proof of Case(3), we are left to show that Q_{3,0}^{4l_{0}+3}Q_{3,1}^{4l_{1}+1}Q_{3,2}^{2l_{2}+1} is hit. The proof can be done as follows,

Q_{3,0}^{4l_{0}+3}Q_{3,1}^{4l_{1}+1}Q_{3,2}^{2l_{2}+1}
= Q_{3,0}^{3}Q_{3,1}Q_{3,2}Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{2l_{2}}
= Sq^{2}[Q_{3,0}Q_{3,1}^{3}Q_{3,2}](Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{2l_{2}}) + Q_{3,0}Q_{3,1}^{4}(Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{2l_{2}})
≡ Q_{3,0}Q_{3,1}^{3}Q_{3,2}χ(Sq^{2})(Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{2l_{2}}) + Q_{3,0}^{4l_{0}+1}Q_{3,1}^{4l_{1}+4}Q_{3,2}^{2l_{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).

2.4. Proof of Case(4). Put n_{0} = 2l_{0} + 1, n_{1} = 4l_{1} + 3 and n_{2} = 4l_{2} + 1. It is easy to verify that

Q_{3,0}Q_{3,1}^{3}Q_{3,2} = Sq^{4}[Q_{3,0}Q_{3,1}Q_{3,2}^{3}] + Q_{3,0}^{3}Q_{3,2}^{2} + Q_{3,0}Q_{3,1}Q_{3,2}^{4}.

Thus

Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} = Q_{3,0}^{2l_{0}+1}Q_{3,1}^{4l_{1}+3}Q_{3,2}^{4l_{2}+1} = Q_{3,0}^{3}Q_{3,1}Q_{3,2}(Q_{3,0}^{2l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}})
= Sq^{4}[Q_{3,0}Q_{3,1}Q_{3,2}^{3}](Q_{3,0}^{2l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}) + Q_{3,0}^{2l_{0}+3}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}+2} + Q_{3,0}^{2l_{0}+1}Q_{3,1}^{4l_{1}+1}Q_{3,2}^{4l_{2}+4}

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.

2.5. Proof of Case(5). When n_{0} = 4l_{0} + 1, n_{1} = 4l_{1} + 3 and n_{2} = 4l_{2} + 3,

\begin{aligned}Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} & = Q_{3,0}^{4l_{0}+1}Q_{3,1}^{4l_{1}+3}Q_{3,2}^{4l_{2}+3} \\ & = P_{1}Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}, \end{aligned}

where P_{1} = Q_{3,0}Q_{3,1}^{3}Q_{3,2}^{3}, which is a symmetric polynomial in X_{1}, X_{2} and X_{3}. Let X_{1}^{a}X_{2}^{b}X_{3}^{c} 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 = X1^{\frac{a−1}{2}} X2^{\frac{b}{2}} X3^{\frac{c}{2}} Q_{3,0}^{2l_{0}}Q_{3,1}^{2l_{1}}Q_{3,2}^{2l_{2}},

we have

deg F = \frac{(a − 1) + b + c}{2} + (7 · 2l_{0}) + (6 · 2l_{1}) + (4 · 2l_{2}) = 18 + 2(7l_{0} + 6l_{1} + 4l_{2}).

So μ(deg(F)) ≥ 2. Hence we can use Theorem 1.5 to conclude that X_{1}^{a}X_{2}^{b}X_{3}^{c}Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{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). (2.5.1)

Referring to the above list, we wish to show that X_{1}^{3}X_{2}^{13}X_{3}^{21}Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}} is hit. Use the condition of Theorem 1.5 as follows. Setting k = 1, E = X_{1}^{3}X_{2}X_{3} and F = X_{2}^{3}X_{3}^{5}Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}}, then

deg E = 5 and deg F = 8 + 7l_{0} + 6l_{1} + 4l_{2}.

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 1.5 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(37 + 4 · 7l_{0} + 4 · 6l_{1} + 4 · 4l_{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 37 + 4 · 7l_{0} + 4 · 6l_{1} + 4 · 4l_{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 = 2 and r ≥ 3. The property for Minimum spike implies that we can assume m > r if r > s. So μ(8+7l_{0}+6l_{1}+4l_{2}) = μ(8+2^{m−2}+ 2^{r−2} − 9) ≥ 2. Hence the condition for Theorem 1.5 is satisfied. Therefore X_{1}^{3}X_{2}^{13}X_{3}^{21}Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}} is hit. Using the same method, we can show that all the polynomials corresponding to the list (2.5.1) are hit except for the case: (a, b, c) = (7, 11, 19) and its permutations.

Lemma 2.5. For any non-negative integers l_{0}, l_{1} and l_{2}, the following statements are true.

(1) Sq^{i}[Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}] = 0 for i not divisible by 4;

(2) For any non-negative integer t, Sq^{4t}[Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}] = ∑ Q_{3,0}^{4s_{0}}Q_{3,1}^{4s_{1}}Q_{3,2}^{4s_{2}} for some integers s_{0}, s_{1} and s_{2};

(3) Sq^{4}Sq^{4}[Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}] = 0.

Proof. The proposition is a basic consequence of the Cartan formula and Theorem 1.4. 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_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}.

Lemma 2.6.

(X_{1}^{7}X_{2}^{11}X_{3}^{19} + X_{1}^{11}X_{2}^{7}X_{3}^{19})Y ≡ [X_{1}^{7}X_{2}^{11}X_{3}^{7} + X_{1}^{11}X_{2}^{7}X_{3}^{7}]Sq^{8}Sq^{4}Y.

Proof. This is an elementary exercise, using the χ-trick, Theorem 1.5 and Lemma 2.5.

Let P be the set consisting of the index (1, 2, 3) and all its permutations. Then we have the following,

∑_{P} X_{1}^{7}X_{2}^{11}X_{3}^{19}Y = (X_{1}^{7}X_{2}^{11}X_{3}^{19} + X_{1}^{11}X_{2}^{7}X_{3}^{19})Y + (X_{1}^{7}X_{2}^{19}X_{3}^{11} + X_{1}^{11}X_{2}^{19}X_{3}^{7})Y

+ (X_{1}^{19}X_{2}^{11}X_{3}^{7} + X_{1}^{19}X_{2}^{7}X_{3}^{11})Y ≡ 0 (using Lemma 2.6).

2.6. Proof of Case(6). When n_{0} = 4l_{0} + 3, n_{1} = 4l_{1} + 3 and n_{2} = 4l_{2} + 3,

\begin{aligned}Q_{3,0}^{n_{0}}Q_{3,1}^{n_{1}}Q_{3,2}^{n_{2}} & = Q_{3,0}^{4l_{0}+3}Q_{3,1}^{4l_{1}+3}Q_{3,2}^{4l_{2}+3} \\ & = P_{2}Y, \end{aligned}

where P_{2} = Q_{3,0}^{3}Q_{3,1}^{3}Q_{3,2}^{3} and Y = Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}. After expanding P_{2}, by applying Theorem 1.5 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,

X_{1}^{7}X_{2}^{19}X_{3}^{25}Y, X_{1}^{7}X_{2}^{11}X_{3}^{33}Y and X_{1}^{11}X_{2}^{19}X_{3}^{21}Y.  (2.6.1)

For example, X_{1}^{5}X_{2}^{21}X_{3}^{25}Y appears in the expansion of P_{2}Y. We will show that it is hit. Set k = 1, E = X_{1}X_{2}X_{3} and F = X_{1}X_{2}^{5}X_{3}^{6}Q_{3,0}^{l_{0}}Q_{3,1}^{l_{1}}Q_{3,2}^{l_{2}}. If deg F is even, then the condition of Theorem 1.5 satisfied. So EF^{4}(= X_{1}^{5}X_{2}^{21}X_{3}^{25}Y) is hit. If deg F is odd, then we can assume that α(3 + EF^{4}) < 3 (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,

3 + 48 + 4(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 = 1 and r ≥ 2. Using the property of the Minimum spike, we may assume that m > r. So

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 1.5 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 (2.6.1).

Recall that Y denotes Q_{3,0}^{4l_{0}}Q_{3,1}^{4l_{1}}Q_{3,2}^{4l_{2}}. Using the χ-trick, Theorem 1.5 and Lemma 2.5, we can show that

(X_{1}^{7}X_{2}^{11}X_{3}^{33} + X_{1}^{7}X_{2}^{19}X_{3}^{25})Y ≡ X_{1}^{7}X_{2}^{11}X_{3}^{25}Sq^{8}Y,  (2.6.2)

It is easy to show that (2.6.2) implies that

∑_{P}(X_{1}^{7}X_{2}^{11}X_{3}^{33} + X_{1}^{7}X_{2}^{19}X_{3}^{25})Y ≡ ∑_{P} X_{1}^{7}X_{2}^{11}X_{3}^{21}Sq^{4}Sq^{8}Y  (2.6.3)

where P is the set consisting of the index (1, 2, 3) and all its permutations. Again using the χ-trick, Theorem 1.5 and Lemma 2.5, we can show the following,

Lemma 2.7.

X_{1}^{7}X_{2}^{11}X_{3}^{21}Sq^{4}Sq^{8}Y ≡ (X_{1}^{11}X_{2}^{21}X_{3}^{19} + X_{1}^{19}X_{2}^{13}X_{3}^{19})Y.

.

Combining the above lemma with (2.6.3), we have

∑_{P} (X_{1}^{7}X_{2}^{11}X_{3}^{33} + X_{1}^{7}X_{2}^{19}X_{3}^{25} + X_{1}^{11}X_{2}^{21}X_{3}^{19})Y
≡ ∑_{P} [X_{1}^{7}X_{2}^{11}X_{3}^{21}Sq^{4}Sq^{8}Y + X_{1}^{11}X_{2}^{21}X_{3}^{19}Y]
≡ ∑_{P} [(X_{1}^{11}X_{2}^{21}X_{3}^{19} + X_{1}^{19}X_{2}^{13}X_{3}^{19})Y + X_{1}^{11}X_{2}^{21}X_{3}^{19}Y]
≡ 0.

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  E-mail address, Kai Xu: matxukai@leonis.nus.edu.sg

Department of Mathematics, National University of Singapore, 2 Science Drive 2, 117543 Singapore