NILPOTENCE IN THE STEENROD ALGEBRA

Kenneth G. Monks

University of Scranton

August 1991

I. Introduction and Notation

While all of the relations in the Steenrod algebra, 𝒜, can be deduced in principle from the Adem relations, in practice, it is extremely difficult to determine whether a given polynomial of elements in 𝒜 is zero for all but the most elementary cases. In his original paper [Mi] Milnor states “It would be interesting to discover a complete set of relations between the given generators of 𝒜”. In particular Milnor shows that every positive dimensional element of 𝒜 is nilpotent. Thus it would be desirable to find a simple closed form for nilpotence relations in 𝒜.

Let x ∈ 𝒜. We say that x has nilpotence k, if x^{k} = 0 and x^{k−1} ≠ 0 (take x^{0} = 1). In this case we write Nil(x) = k. In this paper we investigate Nil(x) for several infinite families of Milnor basis elements of 𝒜 at the prime 2.

The paper is organized as follows. First, an infinite family of subalgebras and isomorphisms between them are constructed. The isomorphisms are used to produce infinite families of elements having the same nilpotence. Next, we compute strong upper and lower bounds for the nilpotence of Milnor basis elements in these subalgebras. Comparing these bounds and extending to the families produced via the isomorphisms shows that Sq(2^{m}(2^{k} − 1) + 1) has nilpotence k + 2. Finally a strong lower bound for the nilpotence of P_{t}^{s} is computed for all s, t ∈ ℕ. The main results are stated and discussed in Sections II and III. Detailed proofs are presented in Section IV.

II. Nilpotence in an Odd Subalgebra of 𝒜

There is a doubling isomorphism (see Section IV) which implies that

Nil(Sq(2r_{1}, …, 2r_{m})) ≥ Nil(Sq(r_{1}, …, r_{m}))

for every Milnor basis element in 𝒜. Thus it is natural to begin by asking what the nilpotence of Sq(r_{1}, …, r_{m}) is when some or all of the r_{i} are odd.

We begin by describing a family of isomorphic subalgebras 𝒪_{k} ⊂ 𝒜 and a family of isomorphisms between them.

Typeset by AMS-TEX 2 KENNETH G. MONKS

1980 Mathematics Subject Classification (1985 Revision). Primary 55S10, 55S05; Secondary 57T05.

Definition 2.1. Let k ∈ ℕ. Let 𝒪_{k} be the ℤ_{2}-subspace of 𝒜 whose basis is the set of Milnor elements

B_{k} = { Sq(r_{1}, …, r_{m}) | r_{i} ≡ −1 (mod 2^{k+1}) for i < m, and r_{m} ≡ 1 (mod 2^{k+1}) } .

We will write 𝒪 = 𝒪_{0}. Thus we have the vector subspace inclusions

𝒜 ⊃ 𝒪 = 𝒪_{0} ⊃ 𝒪_{1} ⊃ 𝒪_{2} ⊃ ….

Notice 𝒪 is just the subspace of 𝒜 generated by the Milnor basis elements Sq(r_{1}, …, r_{m}) with r_{i} odd for all 1 ≤ i ≤ m . Theorem 2.2. 𝒪_{k} is a sub-algebra of 𝒜 for all k ∈ ℕ.

𝒪_{k} is not a Hopf subalgebra, but we do not require this for our purposes. Definition 2.3. Let λ : 𝒪 → 𝒪 be the ℤ_{2}-linear map given by

λ(Sq(r_{1}, …, r_{m})) = Sq(2r_{1} + 1, 2r_{2} + 1, …, 2r_{m−1} + 1, 2r_{m} − 1)

on elements of the basis.

For example, λ (Sq(5) + Sq(3, 1, 3)) = Sq(9) + Sq(7, 3, 5). Theorem 2.4. λ is an algebra monomorphism.

If we let λ^{(0)} be the identity map on 𝒪 , and λ^{(k)} = λ ∘ λ^{(k−1)} for k > 1 then λ^{(k)} is also an monomorphism for every k. It is a routine calculation to check that

λ^{(k)}(Sq(r_{1}, …, r_{m})) = Sq (2^{k}r_{1} + (2^{k} − 1), …, 2^{k}r_{m−1} + (2^{k} − 1), 2^{k}r_{m} − (2^{k} − 1))

(2.5)

Using (2.5) it is elementary to see that λ(𝒪_{k}) = 𝒪_{k+1} and thus that the restriction of λ to 𝒪_{k} yields an isomorphism λ_{k} between 𝒪_{k} and 𝒪_{k+1}. Hence for any x ∈ 𝒪 we have Nil(x) = Nil(λ^{(k)}(x)) for all k ∈ ℕ. Thus Corollary 2.6. Let Sq(r_{1}, …, r_{m}) ∈ 𝒪. Then

Nil (Sq(r_{1}, …, r_{m})) = Nil (Sq (2^{k}r_{1} + (2^{k} − 1), …, 2^{k}r_{m−1} + (2^{k} − 1), 2^{k}r_{m} − (2^{k} − 1)))

for all k ∈ ℕ.

In particular, if n is odd then Nil (Sq(n)) = Nil (Sq(2^{k}n − (2^{k} − 1))) for all k ∈ ℕ. For example, since Nil (Sq(7)) = 4, every element of the family

Sq(7), Sq(13), Sq(25), Sq(49), Sq(97), Sq(193), …

also has nilpotence 4.

Theorem 2.4 reduces the problem of computing the nilpotence of elements of 𝒪 to that of finding the nilpotence of elements in 𝒪 − 𝒪_{1}. For the case m = 1 this says that the nilpotence of the Milnor elements Sq(n) with n ≡ 1 (mod 4) is completely determined by the nilpotence of the elements Sq(n) with n ≡ −1 (mod 4). We begin to attack this question by obtaining a strong upper bound.

Theorem 2.7. Let Sq(r_{1}, …, r_{m}) ∈ 𝒪. Then

Nil(Sq(r_{1}, …, r_{m})) ≤ min { k | r_{m} < 2^{(k−1)m+1} − 1 } .

Corollary 2.8. If n is odd then Nil(Sq(n)) ≤ min { k | n < 2^{k} − 1 }.

For example Sq(15, 31)^{4} = 0 since 31 < 2^{(4−1)2+1} − 1 = 127. As a possible application, notice that Sq(r_{1}, …, r_{m})^{2} = 0 whenever r_{m} < 2^{m+1}−1 and Sq(r_{1}, …, r_{m}) ∈ 𝒪. Elements whose square is zero have been useful in the past in developing P_{t}^{s} homology theory.

It should be noted that this upper bound appears to be quite good. Computer calculations show that we actually have equality in Corollary 2.8 for every n ≡ −1 (mod 4) less than 143 with the exception of n = 67 and n = 131 (note also that these exceptions eliminate the possibility that one might actually be able to prove equality in all cases).

We now obtain a lower bound on nilpotence for certain of these elements.

Theorem 2.9. Let n be odd. Then

Nil(Sq(n)) > max { k | n ≡ −1 (mod 2^{k}) } .

Combining all of the previous results yields

Theorem 2.10. Nil (Sq (2^{m}(2^{k} − 1) + 1)) = k + 2 for all m, k ≥ 1.

Notice for m = 1 this implies Nil(Sq(2^{k} − 1)) = k + 1.

As an illustration of the theorem consider that 524281 = 2^{3}(2^{16} − 1) + 1. Then by Theorem 2.10 we have immediately that Sq(524281)^{17} ≠ 0 and Sq(524281)^{18} = 0, which would be a truly monumental computation by usual means.

Table 2.11 gives a comparison between the nilpotence bounds obtained in this section and the actual values of Nil(Sq(n)) for odd n less than 64. In the table the values labeled Nil are the actual values of Nil(Sq(n)) obtained from computer calculations. The values labeled High are the upper bounds for Nil(Sq(n)) obtained from Corollary 2.6 and Corollary 2.8. Similarly, the values labeled Low are the lower bounds obtained from Corollary 2.6 and Theorem 2.9. Finally, the values labeled Gap are just the difference between the upper and lower bounds. Thus the nilpotence is completely determined whenever the gap is zero. This occurs at the values of n given in Theorem 2.10.

Table 2.11: Comparison of Nilpotence Bounds with Computed Values

n  Nil High Low Gap  n  Nil High Low Gap
1  2  2  2  0  33  3  3  3  0
3  3  3  3  0  35  6  6  3  3
5  3  3  3  0  37  5  5  3  2
7  4  4  4  0  39  6  6  4  2
9  3  3  3  0  41  4  4  3  1
11  4  4  3  1  43  6  6  3  3
13  4  4  4  0  45  5  5  4  1
15  5  5  5  0  47  6  6  5  1
17  3  3  3  0  49  4  4  4  0
19  5  5  3  2  51  6  6  3  3
21  4  4  3  1  53  5  5  3  2
23  5  5  4  1  55  6  6  4  2
25  4  4  4  0  57  5  5  5  0
27  5  5  3  2  59  6  6  3  3
29  5  5  5  0  61  6  6  6  0
31  6  6  6  0  63  7  7  7  0
III. Nilpotence of P_{t}^{s}

Let P_{t}^{s} = Sq(r_{1}, …, r_{t}) where r_{i} = 0 for all i < t and r_{t} = 2^{s}. There is an old conjecture which has been growing in notoriety ([Da], [Conf]) which says Nil (Sq(2^{s})) = 2s + 2 for all k (or equivalently, Nil(P_{1}^{s}) = 2s + 2). One naturally might ask what the corresponding conjecture would be for Nil(P_{t}^{s}) for any t. Some sample calculation leads one immediately to the following. Let ⌊r⌋ denote the greatest integer less than or equal to the rational number r.

Conjecture 3.1. Nil(P_{t}^{s}) = 2⌊s/t⌋ + 2 for all s ≥ 0, t ≥ 1.

Our main result regarding this conjecture is

Theorem 3.2. Nil(P_{t}^{s}) ≥ 2⌊s/t⌋ + 2 for all s ≥ 0, t ≥ 1.

This theorem generalizes an original result of Davis [Da], who first proved this theorem for the special case t = 1.

It is well known that the conjecture is true if ⌊s/t⌋ = 0, i.e. if s < t. We can also prove the conjecture for ⌊s/t⌋ = 1.

Theorem 3.3. If ⌊s/t⌋ = 1 then Nil(P_{t}^{s}) = 4.

The conjecture has been verified by computer calculation for all s, t such that s+t < 16 and s − 2t < 4 as well as several other cases. For the case t = 1 the conjecture was originally verified by Davis for s ≤ 5 [Da]. A summary of the calculation is given in Table 3.4. It is interesting to note that for many of the 56,627 Milnor basis elements, x, which are a summand of Sq(64)^{13}, the product Sq(64) · x is nonzero, and yet the sum of all such products is still zero.

The Theorems in this section were first presented in the author’s Ph.D. thesis [Mo].

Table 3.4: Powers of Sq(2^{s})

Number of Milnor Basis Elements Which are a Summand of Sq(2^{s})^{k}

k  s = 0 s = 1 s = 2 s = 3 s = 4 s = 5 s = 6
1  1  1  1  1  1  1  1
2  0  1  1  2  3  5  8
3  1  3  5  13  32  84
4  0  2  8  27  131  629
5  2  15  72  473  3,632
6  0  7  93  876 11,454
7  8  153  2,222 37,128
8  0  59  2,070 61,447
9  69  3,297 136,738
10  0  1,093 100,618
11  1,313 158,089
12  0 46,325
13  56,627
14  0

IV. Proof of Results

We begin by recalling some results from [Mi] to which we will need to refer in the proofs that follow. The mod 2 Steenrod algebra is a graded ℤ_{2}-vector space with basis all formal symbols Sq(r_{1}, r_{2}, …) where r_{i} ≥ 0 and r_{i} > 0 for finitely many i. As usual, it is convenient to write Sq(r_{1}, …, r_{m}) for Sq(r_{1}, …, r_{m}, 0, 0, …) when r_{m} ≠ 0. Let R = (r_{1}, …, r_{m}). It will also be convenient to write Sq(R) for the Milnor basis element Sq(r_{1}, …, r_{m}).

The product is given by

Sq(r_{1}, r_{2}, …) · Sq(s_{1}, s_{2}, …) = ∑_{X} Sq(t_{1}, t_{2}, …)

where the summation is taken over all matrices X = (x_{ij}) satisfying:

∑_{i} x_{ij} = s_{j}  (4.1)

∑_{j} 2^{j}x_{ij} = r_{i}  (4.2)

∏_{h}(x_{h0}, x_{h−1 1}, …, x_{0h}) ≡ 1 (mod 2)  (4.3)

where (n_{1}, …, n_{m}) is the multinomial coefficient (n_{1} + ⋯ + n_{m})!/n_{1}! ⋯ n_{m}!. We will say such a matrix X is Sq(r_{1}, r_{2}, …)Sq(s_{1}, s_{2}, …)–allowable. Each such allowable matrix yields a summand Sq(t_{1}, t_{2}, …) given by

t_{h} = ∑_{i+j=h} x_{ij}  (4.4)

In this case we will say that X is the matrix associated with Sq(t_{1}, t_{2}, …) (for the product Sq(r_{1}, r_{2}, …) · Sq(s_{1}, s_{2}, …)). The value of x_{00} is never used and may be assumed to be zero.

When evaluating the multinomial coefficients in (4.3) it is well known (e.g. [Ma]) that (n_{1}, …, n_{m}) is odd if and only if the n_{i} have disjoint binary expansions. More formally, let n = ∑_{j} α_{j}(n)2^{j} be the binary expansion of an integer n. Then

Lemma 4.5. (n_{1}, …, n_{m}) is odd if and only if for each k < ∞, α_{k}(n_{i}) = 1 for at most one i.

In particular, if (n_{1}, …, n_{m}) is odd then at most one of the n_{i} is odd. We will make frequent use of this fact.

Proof of Theorem 2.2. It suffices to show that x · y ∈ 𝒪 for all x, y ∈ B_{0}. We will prove a slightly stronger result which we will need later, namely

Lemma 4.6. Let Sq(r_{1}, …, r_{m}), Sq(s_{1}, …, s_{n}) ∈ 𝒪. If Sq(t_{1}, …, t_{p}) is a summand of Sq(r_{1}, …, r_{m})Sq(s_{1}, …, s_{n}) then Sq(t_{1}, …, t_{p}) ∈ 𝒪 and p = m + n.

Proof. Let X = (x_{ij}) be the matrix associated with Sq(t_{1}, …, t_{p}). Sq(r_{1}, …, r_{m}) ∈ 𝒪 implies that r_{i} is odd for each 1 ≤ i ≤ m . Thus x_{i0} is odd for each 1 ≤ i ≤ m by (4.2). Combining this with (4.3) shows x_{ij} is even whenever i + j ≤ m, and j > 0. Let d < n and assume that x_{mj} is odd for j ≤ d and x_{ij} is even whenever i + j ≤ m + d, j > 0, and i < m. Then Sq(s_{1}, …, s_{n}) ∈ 𝒪 implies s_{d+1} is odd and thus x_{m d+1} is odd by (4.1). Once again invoking (4.3) shows x_{ij} is even whenever i + j = m + d + 1, and j > d + 1. Thus by finite induction on d we have shown x_{ij} is odd if and only if j = 0, i ≤ m or i = m, j ≤ n. Applying (4.4) shows Sq(t_{1}, …, t_{p}) ∈ 𝒪. Further t_{p} = x_{mn} is odd, therefore p = m + n. □ □

Proof of Theorem 2.4. It is easy to see from the definition that λ is injective. Let R = (r_{1}, …, r_{m}), S = (s_{1}, …, s_{n}), and T = (t_{1}, …, t_{m+n}). To show that λ is a homomorphism we will prove that Sq(T) is a summand of the product Sq(R)Sq(S) if and only if λ(Sq(T)) is a summand of λ(Sq(R))λ(Sq(S)) for every Sq(R), Sq(S) ∈ 𝒪. Let X̂ = (x̂_{ij}) be a λ(Sq(R))λ(Sq(S))–allowable matrix. As shown in the proof of Lemma 4.6, x̂_{ij} is odd if and only if j = 0, i ≤ m or i = m, j ≤ n. Thus there exist nonnegative integers x_{ij} such that

x̂_{ij} = \{ \begin{matrix}2x_{ij} + 1 & if j = 0, i ≤ m or i = m, j < n \\ 2x_{mn} − 1 & if i = m and j = n \\ 2x_{ij} & otherwise. \end{matrix}  (4.7)

Given such an allowable matrix X̂ we can define the matrix X = (x_{ij}). On the other hand, if we are given a Sq(R)Sq(S)–allowable matrix, X = (x_{ij}), we can define a matrix X̂ = (x̂_{ij}) by (4.7). We now wish to show that X̂ is λ(Sq(R))λ(Sq(S))–allowable if and only if X is Sq(R)Sq(S)–allowable. We must verify that each of the conditions (4.1),(4.2), and (4.3) hold for X if and only if they hold for X̂.

Let 1 ≤ j ≤ n and define ϵ_{j} = \{ \begin{matrix}−1 & if j = n \\ 1 & otherwise \end{matrix} . Then λ(Sq(S)) = Sq(ŝ_{1}, …, ŝ_{n})

where ŝ_{j} = 2s_{j} + ϵ_{j}. Thus checking (4.1) we have

∑_{i=0}^{m} x̂_{ij} = ŝ_{j} ⟺ ( ∑_{i=0}^{m−1} 2x_{ij}) + (2x_{mj} + ϵ_{j}) = 2s_{j} + ϵ_{j}

⟺ ∑_{i=0}^{m} x_{ij} = s_{j}

Again letting λ(Sq(R)) = Sq(r̂_{1}, …, r̂_{m}) we have r̂_{i} = 2r_{i} + 1 for 1 ≤ i < m and r̂_{m} = 2r_{m} − 1. Verification for (4.2) breaks up into two cases. If 1 ≤ i < m then

∑_{j=0}^{n} 2^{j}x̂_{ij} = r̂_{i} ⟺ (∑_{j=1}^{n} 2^{j}2x_{ij}) + (2x_{i0} + 1) = 2r_{i} + 1

⟺ ∑_{j=0}^{n} 2^{j}x_{ij} = r_{i}

But if i = m we have

∑_{j=0}^{n} 2^{j}x̂_{mj} = r̂_{m} ⟺ (∑_{j=0}^{n−1} 2^{j}(2x_{mj} + 1)) + 2^{n} (2x_{mn} − 1) = 2r_{m} − 1

⟺ 2 ∑_{j=0}^{n} 2^{j}x_{mj} + ∑_{j=0}^{n−1} 2^{j} − 2^{n} = 2r_{m} − 1

⟺ 2 ∑_{j=0}^{n} 2^{j}x_{mj} + (2^{n} − 1) − 2^{n} = 2r_{m} − 1

⟺ ∑_{j=0}^{n} 2^{j}x_{mj} = r_{m}

Verification of (4.3) follows easily from the observation that for any multinomial coefficient (a_{1}, …, a_{h}) we have

(a_{1}, …, a_{h}) ≡ (2a_{1} + γ_{1}, …, 2a_{h} + γ_{h}) (mod 2)

where γ_{i} = 1 for at most one 1 ≤ i ≤ h and is zero otherwise. This follows immediately from Lemma 4.5. Thus

∏_{h}(x_{h0}, x_{h−1 1}, …, x_{0h}) ≡ ∏_{h}(x̂_{h0}, x̂_{h−1 1}, …, x̂_{0h}) (mod 2)

Finally let Sq(t_{1}, …, t_{m+n}) be the summand of Sq(R)Sq(S) associated with X and let Sq(T̂) = Sq(t̂_{1}, …, t̂_{m+n}) be the summand of λ(Sq(R))λ(Sq(S)) associated with X̂. Then by (4.4) for h < m + n

t̂_{h} = ∑_{i+j=h} x̂_{ij}

= 2 ( ∑_{i+j=h} x_{ij}) + 1

= 2t_{h} + 1

and t_{m+n} = x̂_{mn} = 2x_{mn} − 1 = 2t_{mn} − 1. Thus Sq(T̂) = λ(Sq(T)) which completes the proof. □

Proof of Theorem 2.7. Let R = (r_{1}, …, r_{m}). It suffices to show that Sq(R)^{k} = 0 if r_{m} < 2^{(k−1)m+1} − 1. Let Sq(T) = Sq(t_{1}, …, t_{p}) be any summand of Sq(R)^{k−1}. By Lemma 4.6 p = (k − 1)m. Let X = (x_{ij}) be any Sq(R)Sq(T)–allowable matrix. As shown in the proof of Lemma 4.6, x_{ij} is odd if i = m and j ≤ (k − 1)m. Combining this with (4.2) we have

r_{m} = ∑_{j=0}^{(k−1)m} 2^{j}x_{mj} ≥ ∑_{j=0}^{(k−1)m} 2^{j} = 2^{(k−1)m+1} − 1.

Therefore if r_{m} < 2^{(k−1)m+1} − 1 there are no Sq(R)Sq(T)–allowable matrices, and hence Sq(R)^{k} = 0. □

Before continuing we would like to outline an alternate proof of Theorem 3 that lends some insight into what is going on at the cost of being much more tedious.

Let Q_{t−1} = P_{t}^{0}. It is quite easy to see from the product formula that

Q_{i}Q_{j} = Q_{j}Q_{i}  for all i, j ∈ ℕ  (4.8)
Q_{i}^{2} = 0  for all i ∈ ℕ  (4.9)

and that for any Sq(r_{1}, …, r_{m}) with r_{i} even for all 1 ≤ i ≤ m

Sq(s_{1}, …, s_{m})Q_{i} = ∑_{j=0}^{m} Q_{j+i}Sq(s_{1}, …, s_{j} − 2^{i+1}, …, s_{m})  (4.10)

where we define Sq(t_{1}, …, t_{m}) to be zero if t_{i} < 0 for any i. Notice that (4.10) gives us a way to shift Q_{i}’s from the right side of a Milnor basis element with even entries to the left side. Also notice that the largest Q_{j} obtainable on the left by shifting a Q_{i} via (4.10) is Q_{m+i} and that this can only occur if s_{m} ≥ 2^{i+1}. For any Sq(r_{1}, …, r_{m}) ∈ 𝒪 we can write

Sq(r_{1}, …, r_{m}) = Q_{0}Q_{1} ⋯ Q_{m−1}Sq(r_{1} − 1, …, r_{m} − 1)

and hence

(Sq(r_{1}, …, r_{m}))^{k} = (Q_{0}Q_{1} ⋯ Q_{m−1}Sq(r_{1} − 1, …, r_{m} − 1))^{k}  (4.11)

Applying (4.8), (4.9), and (4.10) repeatedly to the right hand side of (4.11) in order to collect all of the Q_{i} on the left and computing the effect on the m^{th} position in the resulting Milnor elements yields the desired result. We leave this verification to the interested reader.

Proof of Theorem 2.9. Let k be the largest integer such that n ≡ −1 (mod 2^{k}). We can write n uniquely in the form n = 2^{k}a − 1 for some odd integer a ≥ 1. For each 1 ≤ h ≤ k define an h–tuple R_{n,h} = (r_{n,h,1}, r_{n,h,2}, …, r_{n,h,h}) by

r_{n,h,i} = \{ \begin{matrix}2^{k−i}a + 1 & if 1 ≤ i < h \\ 2^{k−i+1}a − 1 & if i=h \end{matrix}

For example, for n = 47 we have

R_{47,1} = (47)
R_{47,2} = (25, 23)
R_{47,3} = (25, 13, 11)
R_{47,4} = (25, 13, 7, 5)

We now wish to show that Sq(R_{n,h}) is a summand of Sq(n)^{h} for 1 ≤ h ≤ k , and thus that Sq(n)^{k} ≠ 0.

We proceed by finite induction on h. If h = 1 then Sq(R_{n,1}) = Sq(n), which is clearly a summand of Sq(n)^{1}. Assume as the induction hypothesis that Sq(R_{n,h}) is a summand of Sq(n)^{h} where h < k. Suppose Sq(R_{n,h+1}) is a summand of Sq(n)Sq(T) for some summand Sq(T) = Sq(t_{1}, …, t_{h}) of Sq(n)^{h}. Let X = (x_{ij}) be the associated matrix. Then by (4.4)

x_{1h} = r_{n,h+1,h+1} = 2^{k−h}a − 1

and from (4.2)

n = ∑_{j=0}^{h} 2^{j}x_{1j}

= (∑_{j=0}^{h−1} 2^{j}x_{1j}) + 2^{h}x_{1h}

= (∑_{j=0}^{h−1} 2^{j}x_{1j}) + 2^{h} (2^{k−h}a − 1)

= (∑_{j=0}^{h−1} 2^{j}x_{1j}) + n − 2^{h} + 1.

From which we obtain

∑_{j=0}^{h−1} 2^{j}x_{1j} = 2^{h} − 1

But once again using the fact from the proof of Lemma 4.6 that x_{1j} is odd for 1 ≤ j ≤ h we conclude that (4.2) is satisfied if and only if x_{1j} = 1 for 1 ≤ j < h (assuming x_{1h} = 2^{k−h}a − 1). But from (4.1) and (4.4) with 1 ≤ j < h we have

t_{j} = x_{0j} + x_{1j}
= (r_{n,h+1,j} − 1) + 1
= 2^{k−j}a + 1
= r_{n,h,j}

and

t_{h} = x_{0h} + x_{1h}
= (r_{n,h+1,h} − 1) + r_{n,h+1,h+1}
= ((2^{k−h}a + 1) − 1) + (2^{k−(h+1)+1}a − 1)
= 2^{k−h+1}a − 1
= r_{n,h,h}

Thus we have shown that if Sq(R_{n,h+1}) is a summand of Sq(n)Sq(T) for some summand Sq(T) = Sq(t_{1}, …, t_{h}) of Sq(n)^{h} then Sq(T) = Sq(R_{n,h}). But by our very construction the matrix X satisfies (4.1) and (4.2) for the product Sq(n)Sq(R_{n,h}). It also satisfies (4.3) as r_{n,h,j} is always odd and therefore the multinomial coefficient (1, r_{n,h,j} − 1) is odd also. Thus X is Sq(n)Sq(R_{n,h})–allowable and Sq(R_{n,h+1}) is a summand of Sq(n)^{h+1}, completing the induction and the proof. □

Proof of Theorem 2.10. By Theorem 2.7 we have Nil(Sq(2^{k} − 1)) ≤ k + 1 and by Theorem 2.9 Nil(Sq(2^{k} − 1)) > k. Therefore Nil(Sq(2^{k} − 1)) = k + 1. By (2.5)

λ^{(m−1)} (Sq(2^{k+1} − 1)) = Sq (2^{m−1}(2^{k+1} − 1) − (2^{m−1} − 1))
= Sq (2^{m}(2^{k} − 1) + 1))

for every m ≥ 1, k ≥ 0. Thus

Nil (Sq (2^{m}(2^{k} − 1) + 1)) = Nil (λ^{(m−1)} (Sq (2^{k+1} − 1)))
= Nil (Sq (2^{k+1} − 1))
= k + 2

□

In order to prove Theorem 3.2 we must first recall the following information from [Mi]. Let 𝒜_{∗} be the Hopf dual of 𝒜. 𝒜_{∗} is isomorphic to the polynomial algebra ℤ_{2}[ξ_{1}, ξ_{2}, …] on generators ξ_{i} in dimension 2^{i} − 1. If R = (r_{1}, …, r_{m}) we will write ξ^{R} to mean the monomial ξ_{1}^{r_{1}} ⋯ ξ_{m}^{r_{m}}. The basis of monomials ξ^{R} in 𝒜_{∗} is dual to the Milnor basis for 𝒜 . As is common we will write ⟨x, y⟩ for the evaluation of y ∈ 𝒜_{∗} on x ∈ 𝒜. Thus

⟨Sq(R), ξ^{S}⟩ = \{ \begin{matrix}1 & if R = S \\ 0 & otherwise \end{matrix}

The algebra homomorphism ϕ : 𝒜_{∗} → 𝒜_{∗} ⊗ 𝒜_{∗} by

ϕ(ξ_{k}) = ∑_{i+j=k} ξ_{i}^{2^{j}} ⊗ ξ_{j}

is the dual of the product map in 𝒜.

Let E be the exterior subalgebra of 𝒜 generated by { Q_{i} | i ∈ ℕ }. There is a doubling isomorphism D : 𝒜 → 𝒜//E given by

D (Sq(s_{1}, s_{2}, …)) = [Sq(2s_{1}, 2s_{2}, …)]

where [x] denotes the equivalence class in 𝒜//E of x ∈ 𝒜.

Finally, let 𝒜_{n} be the subalgebra of 𝒜 generated by { Sq (2^{i}) | i ≤ n }.

Proof of Theorem 3.2. Let n, t ∈ ℕ, t ≠ 0. For each i ∈ ℕ let j_{i} and ϵ_{i} be the unique integers satisfying i = 2j_{i} + ϵ_{i} where ϵ_{i} ∈ {0, 1}. Define an integer sequence

R_{n,t}(i) = (r_{i,1}, r_{i,2}, r_{i,3}, …)

recursively on i so that it satisfies the three conditions

R_{n,t}(1) = (2^{nt}, 0, 0, …)  (4.12)
r_{i,k} = \{ \begin{matrix}2^{−t}r_{i−1,k−1} & if i is even \\ r_{i−1,k} & if i is odd \end{matrix} for k > 1 and i > 1  (4.13)

∑_{k=1}^{j_{i}+1} r_{i,k} = 2^{nt+ϵ_{i}}.  (4.14)

Notice that (4.14) is used to compute r_{i,1} after obtaining r_{i,k} for k > 1 from (4.13). For example, for n = 3 and t = 2 (dropping trailing zeros)

R_{3,2}(1) = (64)
R_{3,2}(2) = (48, 16)
R_{3,2}(3) = (112, 16)
R_{3,2}(4) = (32, 28, 4)
R_{3,2}(5) = (96, 28, 4)
R_{3,2}(6) = (32, 24, 7, 1)
R_{3,2}(7) = (96, 24, 7, 1)

12 KENNETH G. MONKS

We will require the following implication of (4.14) for odd i.

r_{i,1} = 2^{nt+1} − ∑_{k=2}^{j_{i}+1} r_{i,k}

= 2^{nt} + 2^{nt} − ∑_{k=2}^{j_{i}+1} r_{i−1,k}

= 2^{nt} + ∑_{k=1}^{j_{i}+1} r_{i−1,k} − ∑_{k=2}^{j_{i}+1} r_{i−1,k}

= 2^{nt} + r_{i−1,1}

Define the monomial ξ^{R_{n,t}(i)} ∈ 𝒜_{∗} by ξ^{R_{n,t}(i)} = ∏_{k=1}^{j_{i}+1} ξ_{kt}^{r_{i,k}} (notice this is not the same as the definition of ξ^{R} given before because of the kt subscript). Then for i > 1

ϕ (ξ^{R_{n,t}(i)}) = ϕ ( ∏_{k=1}^{j_{i}+1} ξ_{kt}^{r_{i,k}})

= ∏_{k=1}^{j_{i}+1} ϕ (ξ_{kt})^{r_{i,k}}

= ∏_{k=1}^{j_{i}+1} (ξ_{kt} ⊗ 1 + ξ_{(k−1)t}^{2^{t}} ⊗ ξ_{t} + S_{1})^{r_{i,k}}

= ∏_{k=1}^{j_{i}+1} (ξ_{kt} ⊗ 1 + ξ_{(k−1)t}^{2^{t}} ⊗ ξ_{t})^{r_{i,k}} + S_{2}

where S_{1} is a sum of terms of the form a ⊗ b with b ∉ {1, ξ_{t}} and S_{2} is a sum of terms of the form a ⊗ b with b ≠ ξ_{t}^{2^{nt}}.

Continuing this derivation with i even yields

ϕ (ξ^{R_{n,t}(i)}) = ( ∏_{k=1}^{j_{i}+1} ξ_{(k−1)t}^{2^{t}r_{i,k}}) ⊗ ξ_{t}^{2^{nt}} + S_{3}

= ( ∏_{k=1}^{j_{i}+1} ξ_{(k−1)t}^{r_{i−1,k−1}}) ⊗ ξ_{t}^{2^{nt}} + S_{3}

= (∏_{k=1}^{j_{i}} ξ_{kt}^{r_{i−1,k}}) ⊗ ξ_{t}^{2^{nt}} + S_{3}

= ξ^{R_{n,t}(i−1)} ⊗ ξ_{t}^{2^{nt}} + S_{3}

where S_{3} is a sum of terms of the form a ⊗ b with b ≠ ξ_{t}^{2^{nt}} because ∑_{k=1}^{j_{i}+1} r_{i,k} = 2^{nt}.

On the other hand, continuing the derivation with i odd yields

ϕ (ξ^{R_{n,t}(i)}) = (ξ_{t} ⊗ 1 + 1 ⊗ ξ_{t})^{r_{i,1}} ∏_{k=2}^{j_{i}+1} (ξ_{kt} ⊗ 1 + ξ_{(k−1)t}^{2^{t}} ⊗ ξ_{t})^{r_{i,k}} + S_{2}

= (ξ_{t} ⊗ 1 + 1 ⊗ ξ_{t})^{2^{nt}} (ξ_{t} ⊗ 1 + 1 ⊗ ξ_{t})^{r_{i−1,1}} ∏_{k=2}^{j_{i}+1} (ξ_{kt} ⊗ 1 + ξ_{(k−1)t}^{2^{t}} ⊗ ξ_{t})^{r_{i,k}} + S_{2}

= (ξ_{t}^{2^{nt}} ⊗ 1 + 1 ⊗ ξ_{t}^{2^{nt}}) (ξ_{t} ⊗ 1 + 1 ⊗ ξ_{t})^{r_{i−1,1}} ∏_{k=2}^{j_{i}+1} (ξ_{kt} ⊗ 1 + ξ_{(k−1)t}^{2^{t}} ⊗ ξ_{t})^{r_{i−1,k}} + S_{2}

= (ξ_{t}^{r_{i−1,1}} ∏_{k=2}^{j_{i}+1} ξ_{kt}^{r_{i−1,k}}) ⊗ ξ_{t}^{2^{nt}} + (ξ_{t}^{2^{nt}} ∏_{k=2}^{j_{i}+1} ξ_{(k−1)t}^{2^{t}r_{i−1,k}}) ⊗ ξ_{t}^{2^{nt}} + S_{4}

= ( ∏_{k=1}^{j_{i}+1} ξ_{kt}^{r_{i−1,k}}) ⊗ ξ_{t}^{2^{nt}} + (ξ_{t}^{2^{nt}+2^{t}r_{i−1,2}} ∏_{k=3}^{j_{i}+1} ξ_{(k−1)t}^{2^{t}r_{i−1,k}}) ⊗ ξ_{t}^{2^{nt}} + S_{4}

= ξ^{R_{n,t}(i−1)} ⊗ ξ_{t}^{2^{nt}} + ξ_{t}^{2^{nt+1}} (ξ_{t}^{r_{i−3,1}} ∏_{k=3}^{j_{i}+1} ξ_{(k−1)t}^{2^{t}r_{i−1,k}}) ⊗ ξ_{t}^{2^{nt}} + S_{4}

where S_{4} is a sum of terms of the form a ⊗ b with b ≠ ξ_{t}^{2^{nt}} because ∑_{k=1}^{j_{i}+1} r_{i−1,k} = 2^{nt} and in the last equality we have used the fact that 2^{t}r_{i−1,2} = r_{i−2,1} = 2^{nt} + r_{i−3,1} (taking r_{0,1} = 0).

Thus in both cases we have shown that

ϕ (ξ^{R_{n,t}(i)}) = (ξ^{R_{n,t}(i−1)} + ξ′) ⊗ ξ_{t}^{2^{nt}} + S_{5}

where S_{5} is a sum of terms of the form a⊗b with b ≠ ξ_{t}^{2^{nt}} and ξ′ is divisible by ξ_{t}^{2^{nt+1}} so that its evaluation on all elements of 𝒜_{t(n+1)−1} is zero. This shows that for any 1 ≤ i ≤ 2n + 1

⟨(P_{t}^{nt})^{i} , ξ^{R_{n,t}(i)}⟩ = ⟨(P_{t}^{nt})^{i−1} , ξ^{R_{n,t}(i−1)}⟩ · ⟨P_{t}^{nt}, ξ_{t}^{2^{nt}}⟩ = ⟨(P_{t}^{nt})^{i−1} , ξ^{R_{n,t}(i−1)}⟩ .

(4.15)

Noting that ⟨P_{t}^{nt}, ξ^{R_{n,t}(1)}⟩ = ⟨P_{t}^{nt}, ξ_{t}^{2^{nt}}⟩ = 1 we can use (4.15) and finite induction on i to see that

⟨(P_{t}^{nt})^{i} , ξ^{R_{n,t}(i)}⟩ for all 1 ≤ i ≤ 2n + 1 .

Thus (P_{t}^{nt})^{2n+1} ≠ 0 for all n, t ∈ ℕ, t ≠ 0.

Invoking the doubling isomorphism we notice that D (P_{t}^{s}) = [P_{t}^{s+1}]. Since D is an

algebra isomorphism we have D ((P_{t}^{s})^{i}) = [(P_{t}^{s+1})^{i}]. Thus (P_{t}^{s})^{i} ≠ 0 ⟹ D ((P_{t}^{s})^{i}) ≠

0 ⟹ [(P_{t}^{s+1})^{i}] ≠ 0 ⟹ (P_{t}^{s+1})^{i} ≠ 0. So by induction on w, (P_{t}^{s})^{i} ≠ 0 ⟹

(P_{t}^{s+w})^{i} ≠ 0 for all w ∈ ℕ. Since any s can be written uniquely as s = nt + w with n = ⌊s/t⌋ we see that (P_{t}^{nt})^{2n+1} ≠ 0 ⟹ (P_{t}^{nt+w})^{2n+1} ≠ 0 ⟹ (P_{t}^{s})^{2⌊s/t⌋+1} ≠ 0 for all s, t ∈ ℕ, t ≠ 0. □

Finally, we can prove Theorem 3.3 by the following Lemma.

Lemma 3. Let s, t ∈ ℕ with ⌊s/t⌋ = 1 and let w = s − t. Then

(1) (P_{t}^{s})^{2} = Sq(t_{1}, t_{2}, …) where

t_{i} = \{ \begin{matrix}2^{w}(2^{t} − 1) & if i = t \\ 2^{w} & if i = 2t \\ 0 & otherwise \end{matrix}

(2) (P_{t}^{s})^{3} = Sq(t_{1}, t_{2}, …) where

t_{i} = \{ \begin{matrix}2^{w}(2^{t+1} − 1) & if i = t \\ 2^{w} & if i = 2t \\ 0 & otherwise \end{matrix}

(3) (P_{t}^{s})^{4} = 0

The proof of this Lemma is an elementary, though tedious, exercise in using the product formula and we shall not present it here. Computer calculations indicate that an analogous method should work for the case ⌊s/t⌋ = 2 but that this method will not work for the case ⌊s/t⌋ = 3.

Acknowledgement

I would like to express my sincere gratitude to D. Davis. Since introducing me to this problem he has been a tireless source of encouragement and support. Our numerous mathematical discussions were quite beneficial. In addition, his proof of the case t = 1 of Theorem 3.2 provided the model from which I was able to construct the proof of Theorem 3.2. This paper surely would not exist without his help.

References

[Conf] N. Ray and Y. Rudyak, List of Problems, J. F. Adams Symposium Proceedings (to appear).

[Da] D. M. Davis, On the Height of Sq(2^{i}), unpublished (1985).

[Ma] H. Margolis, Spectra and the Steenrod Algebra, vol. 29, North-Holland Math Library, 1983.

[Mi] J. Milnor, The Steenrod Algebra and Its Dual, Ann. of Math. 67 (1958), 150–171.

[Mo] K. G. Monks, Nilpotence and Torsion in the Steenrod Algebra and Its Cohomology, Ph.D. thesis, Lehigh University (1989).

Kenneth G. Monks Department of Mathematics University of Scranton Scranton, PA 18510 E-mail: kgm303@jaguar.uofs.edu