Generators for tensor products over the Steenrod

algebra

Jérôme Scherer ^{∗}

Abstract

This short paper is for people who like to count generators and add binomial coefficients. Let F (1) be the free unstable module on one generator in degree 1. We prove that the n-fold tensor product of F (1) is finitely generated over the Steenrod algebra by exhibiting a small family of generators.

Introduction

During the winter semester 2000-2001, the author gave a course on unstable modules and Lannes’ T -functor, based on Lionel Schwartz’s excellent book [1]. A question arrose during the proof of the fact that the tensor product of two finitely generated unstable modules is again finitly generated (this is Corollary 1.8.5 in [1]). This proof relies on the fact that the n-fold tensor product F (1)^{⊗n} is finitely generated, where F (1) is the free unstable module on one generator in degree 1. The question, posed by Ueli Suter, was to know whether there is an obvious reason why this n-fold tensor product should be finitely generated over the Steenrod algebra.

In this short paper we prove that the answer is yes by writing down an explicit set of generators. We then refine slightly the rough analysis done to get a smaller number of generators. To understand how the Steenrod algebra acts on this tensor product of free modules, we first study the unstable module K = H̃^{∗}(ℝP^{2}; 𝔽_{2}) ≅ 𝔽_{2}[u]/u^{3}, which is a quotient of F (1). This is a nice example to start with, in particular because we get a nice formula for a minimal set of generators.

∗The author was partially supported by the Swiss National Science Foundation.

Theorem The n-fold tensor product K^{⊗n} is generated, as a module over the Steenrod

algebra, by ( \begin{matrix}n \\ [n/2] \end{matrix} ) generators.

Theorem The n-fold tensor product F (1)^{⊗n} is generated, as a module over the Steenrod algebra, by less than 2(n − 1)! generators.

We work at the prime 2 and follow the notations used by L. Schwartz in his book [1]. We denote by A the Steenrod algebra. The cohomology of ℤ/2 is H = H^{∗}(ℝP^{∞}; 𝔽_{2}) ≅ 𝔽_{2}[u], a polynomial algebra on a generator u of degree 1. The unstable module F (1) can then be identified with the subspace of H generated by the elements u^{2^{n}}, for all n ≥ 0. If M is an unstable module, the degree of an element x ∈ M is denoted by |x|.

Acknowledgments: The author would like to thank all the participants of the course.

1 A finite set of generators

We show here that there is a finite set of generators for both F (1)^{⊗n} and K^{⊗n}. In the following sections we will then refine this result and find a smaller set of generators.

Let K = H̃^{∗}(ℝP^{2}; 𝔽_{2}). This unstable module can be seen as the quotient of F (1) by Au^{4}. It consists as a graded vector space in two copies of 𝔽_{2}, one in degree 1 generated by u, and one in degree 2 generated by u^{2}. The Steenrod algebra acts by Sq^{1}u = u^{2}.

Proposition 1.1 Let L be an unstable module generated by a set X of generators. Then K ⊗ L is generated by Y = u ⊗ X ∐ u^{2} ⊗ X.

Proof. Since Sq^{m}u^{2} = 0 for any m ≥ 1, we see that u^{2} ⊗ Sq^{n}x = Sq^{n}(u^{2} ⊗ x) and u ⊗ Sq^{n}x = Sq^{n}(u ⊗ x) + Sq^{n−1}(u^{2} ⊗ x), for any x ∈ L. □

This gives immediately a first rough bound on the number of elements one needs to generate K^{⊗n}.

Corollary 1.2 The unstable module K^{⊗n} is generated by 2^{n−1} generators. □

We now adapt the above idea to the unstable F (1)^{⊗n}. Of course F (1) is not a finite module, so we have to be more cautious.

Proposition 1.3 Let L be an unstable module generated by a set X of elements having all degree < 2^{n−1}. Then F (1)⊗L is generated by Y = u⊗X ∐ u^{2} ⊗X ∐ … ∐ u^{2^{n−1}} ⊗X, a set of elements having all degree < 2^{n}.

Proof. First we show that the element y = u^{2^{m}} ⊗ x, with x ∈ X, belongs to AY. If m < n, the element y is already in Y. If m ≥ n, then Sq^{2^{m−1}}(u^{2^{m−1}} ⊗ x) = y, since |x| < 2^{m−1}.

We consider next an element of the form u^{2^{m}} ⊗ Sq^{I}x, where x ∈ X and I is an admissible sequence of length n. We prove by induction on n that these elements also belong to AY. If n = 0, we are done. Assume thus n > 0, so that Sq^{I} = Sq^{i}Sq^{J} with J an admissible sequence of length n − 1. Then

x = Sq^{i}(u^{2^{m}} ⊗ Sq^{J}x) + δ_{i2^{m}}u^{2^{m+1}} ⊗ Sq^{J}x

which belongs to AY by induction assumption. □ Since F (1) is generated by the element u of degree 1, we directly infer the following.

Corollary 1.4 The unstable module F (1)^{⊗n} is generated by n! elements of degree < 2^{n}. In particular it is finitely generated. □

We have thus given a positive answer to the original question. It is however possible with a little more work to improve the number of generators.

2 Cohomology of ℝP^{2}

This section is devoted to the study of K^{⊗n}. We improve the number of elements needed to generate K^{⊗n} thank to the following lemma. It states that the action of the Steenrod algebra on the elements in K^{⊗2m+1} of degree ≤ 3m + 1 allows to obtain the ones of degree 3m+2. Let 1 ≤ i_{1} < … < i_{k} ≤ n and denote by x_{i_{1} ... i_{k}} the element x_{1} ⊗ … ⊗x_{n} ∈ K^{⊗n} where x_{j} = u^{2} if j ∈ {i_{1}, … i_{k}} and else x_{j} = u. For example, in K^{⊗3}, x_{∅} = u ⊗ u ⊗ u and x_{13} = u^{2} ⊗ u ⊗ u^{2}. Notice also that the element x_{i_{1} ... i_{k}} ∈ K^{⊗n} has degree n + k.

Lemma 2.1 Let m ≥ 1. Then we have in K^{⊗2m+1}

x_{1 ... m+1} = ∑_{i=1}^{m+1} Sq^{i}( ∑_{m+2≤j_{1}< ... <j_{m+1−i}≤2m+1} x_{j_{1} ... j_{m+1−i}})

Proof. It suffices to check that any element x_{i_{1} ... i_{m+1}} which is different from x_{1 ... m+1} appears an even number of times in this sum. Assume i_{1}, …, i_{k} ≤ m + 1 and i_{k+1} ≥ m+2. As x_{i_{1} ... i_{m+1}} appears in Sq^{i}x_{j_{1} ... j_{m+1−i}} for all {j_{1}, …, j_{m+1−i}} ⊂ {i_{k+1}, …, i_{m+1}}, it

contributes, for a given i, to ( \begin{matrix}m − k + 1 \\ m + 1 − i \end{matrix} ) terms. The sum of these binomial coefficients,

for k ≤ i ≤ m + 1, is always an even number, unless k = m + 1, in which case we get 1. □

We need a few combinatorial lemmas.

Lemma 2.2 Let m ≥ 1 and n = 2m. Then ( \begin{matrix}n \\ [n/2] \end{matrix} ) = 2 · ( \begin{matrix}n − 1 \\ [(n − 1)/2] \end{matrix} ) . □

Lemma 2.3 Let m ≥ 1 and n = 2m+1. Then ( \begin{matrix}n \\ [n/2] \end{matrix} ) = (1+ \frac{m}{m+1})·( \begin{matrix}n − 1 \\ [(n − 1)/2] \end{matrix} ) .

□

The strategy to obtain a better bound on the number of generators is now to use Corollary 1.2, but to get rid of the superfluous generators in K^{⊗2m+1} of degree 3m + 2.

Proposition 2.4 Let m ≥ 1. The unstable module K^{⊗2m} is generated by a set X_{2m} of

generators with \frac{1+2i}{m+i+1} ( \begin{matrix}2m \\ m + i \end{matrix} ) generators in degree 3m − i.

Proof. We prove this by induction on m. The case m = 1 is verified by Corollary 1.2. Let us choose the set of generators X_{2m+2} as u ⊗ X_{2m+1} ∐ u^{2} ⊗ X_{2m+1}. Similarly, we choose Y_{2m+1} = u ⊗ X_{2m} ∐ u^{2} ⊗ X_{2m} and X_{2m+1} = Y_{2m+1} − Y_{2m+1}^{3m+2}, which we can do by Lemma 2.1. Assume now the formula is correct for X_{2m} and let us prove it for X_{2m+2}. Let 1 ≤ i ≤ m + 1. The generators X_{2(m+1)}^{3(m+1)−i} of degree 3(m + 1) − i are

u ⊗ u ⊗ X_{2m}^{3m−(i−1)} ∐ u ⊗ u^{2} ⊗ X_{2m}^{3m−i} ∐ u^{2} ⊗ u ⊗ X_{2m}^{3m−i} ∐ u^{2} ⊗ u^{2} ⊗ X_{2m}^{3m−(i+1)}

A straightforward computation shows that their number is exactly \frac{1+2i}{m+i+2} ( \begin{matrix}2(m + 1) \\ m + 1 + i \end{matrix} ) .

We are left with computing the cardinality of

X_{2(m+1)}^{3(m+1)} = u^{2} ⊗ u ⊗ X_{2m}^{3m} ∐ u^{2} ⊗ u^{2} ⊗ X_{2m}^{3m−1}

which is precisely \frac{1}{m+2} ( \begin{matrix}2(m + 1) \\ m + 1 \end{matrix} ) . □

Theorem 2.5 The unstable module K^{⊗n} is generated by ( \begin{matrix}n \\ [n/2] \end{matrix} ) generators.

Proof. We prove it by induction. The theorem is verified if n ≤ 2. Let us prove the formula for n > 2. If n is even, Lemma 2.2 together with Proposition 1.1 does the job. If n = 2m+1 is odd, Lemma 2.1 tells us that the set X_{2m+1} described in the preceding proof generates K^{⊗2m+1}. The cardinality of this set is twice that of X_{2m} minus the cardinality of Sq^{1} ⊗ X_{2m}^{3m}. That is

2 ( \begin{matrix}2m \\ m \end{matrix} ) − \frac{1}{m + 1} ( \begin{matrix}2m \\ m \end{matrix} ) = ( \begin{matrix}2m + 1 \\ m \end{matrix} )

by Lemma 2.3. □

Remark 2.6 Eventhough we do not prove it here, we think that the set of generators X_{n} we constructed above is minimal.

3 The free module F (1)

Recall that F (n) is the free unstable module on one generator in degree n. Thus F (1) is generated by an element u in degree 1 and contains all Steenrod operations on it, which are allowed by the unstability condition. So

F (1) = 𝔽_{2}{u, Sq^{1}u, Sq^{2}Sq^{1}u, …, Sq^{2^{n}} … Sq^{2}Sq^{1}u, … }

can be identified as a submodule of H^{∗}(ℝP^{∞}; 𝔽_{2}) ≅ 𝔽_{2}[u], where Sq^{2^{n}} … Sq^{2}Sq^{1}u = u^{2^{n+1}}. Notice that Sq^{k}u = 0 for any k ≥ 2.

Example 3.1 First F (1) is generated by u. Thus Corollary 1.4 tells us that u ⊗ u and u^{2} ⊗ u generate F (1) ⊗ F (1). It is easy to see that this is the smallest possible set of generators. Another application of Corollary 1.4 shows that the elements

u ⊗ u ⊗ u, u ⊗ u^{2} ⊗ u, u^{2} ⊗ u ⊗ u, u^{2} ⊗ u^{2} ⊗ u, u^{4} ⊗ u ⊗ u, u^{4} ⊗ u^{2} ⊗ u

generate F (1)^{⊗3}. But the elements u^{2} ⊗ u^{2} ⊗ u and u^{4} ⊗ u ⊗ u are superfluous so only 4 generators are necessary.

Here are two lemmas we will use to get rid of superfluous generators. The first one explains why u^{4} ⊗ u ⊗ u need not be included in our list of generators. We call X′_{n} the set of generators for F (1)^{⊗n} constructed inductively by Proposition 1.3. Notice that X′_{n} only contains elements with at least one component equal to u. Indeed, an element of the form u^{2} ⊗ x with Sq^{1}x = 0, can always be written as Sq^{1}(u ⊗ x).

Lemma 3.2 Let m ≥ 2 and x ∈ X′_{n−1} an element of degree ≤ 2^{m−1}. Then X′_{n}−{u^{2^{m}} ⊗x} still generates F (1)^{⊗n}.

Proof. Let us compute Sq^{2^{m−1}}(u^{2^{m−1}}⊗x) = u^{2^{m}}⊗x+u^{2^{m−1}}⊗Sq^{2^{m−1}}x. Either |x| < 2^{m−1}, in which case the second summand is zero, or |x| = 2^{m−1}. Then x = u^{2^{m_{1}}} ⊗ … ⊗ u^{2^{m_{n−1}}} and 2^{m_{1}} + … + 2^{m_{n−1}} = 2^{m−1}. Therefore Sq^{2^{m−1}}x = u^{2^{m_{1}+1}} ⊗ … ⊗ u^{2^{m_{n−1}+1}}. Hence the second summand is an element of F (1)^{⊗n} which does not contain a single factor u. Such an element is not in X′_{n}. This implies that in this second case also u^{2^{m}} ⊗ x can be obtained from generators of smaller degree. □

Lemma 3.3 The element x_{n} = u^{2^{n−1}} ⊗ … ⊗ u^{2} ⊗ u is the only generator of maximal degree 2^{n} − 1 in X′_{n}. □

Let us define now a subset X_{n} ⊂ X′_{n}. For n ≤ 2, X_{n} = X′_{n} and for n ≥ 3, it is defined by the construction of Proposition 1.3 and then removing all elements x with 2^{n−1} < |x| < 2^{n} − 1.

Proposition 3.4 All elements x ∈ X′_{n} with 2^{n−1} < |x| < 2^{n} − 1 are superfluous, i.e. the set X_{n} still generates F (1)^{⊗n}.

Proof. The assertion is true for n = 3, see Example 3.1. We prove it now for n > 3 by induction. There are two kinds of elements in X′_{n} with the prescribed degree. The first type of elements are x = u^{2^{n−1}} ⊗ y for any y ∈ X_{n−1} with y ≠ x_{n−1}. By induction assumption, |y| ≤ 2^{n−2}. We conclude by Lemma 3.2 that such an element is superfluous.

The second type consists of elements x = u^{2^{m}} ⊗ x_{n−1} with 0 < m < n − 1. We define elements of smaller degree

a = u^{2^{m−1}} ⊗ u^{2^{n−1}} ⊗ … ⊗ u^{2^{m+1}} ⊗ u^{2^{m−1}} ⊗ u^{2^{m−1}} ⊗ … ⊗ u
b = u^{2^{m−1}} ⊗ u^{2^{n−1}} ⊗ … ⊗ u^{2^{m+1}} ⊗ u^{2^{m−1}} ⊗ u^{2^{m}} ⊗ u^{2^{m−2}} … ⊗ u

Then Sq^{2^{m}}a + Sq^{2^{m−1}}b = x. □

Let N_{n} denote the cardinality of the set X_{n}. The above proposition shows that X_{n} is obtained from X_{n−1} by taking all elements in u^{2^{m}} ⊗ X_{n−1} for 0 ≤ m ≤ n − 2, removing u^{2^{m}} ⊗ x_{n−1} if m > 0 and finally adding x_{n}. Thus

N_{n} = (n − 1)N_{n−1} − (n − 2) + 1 = (n − 1)(N_{n−1} − 1) + 2.

Theorem 3.5 The unstable module F (1)^{⊗n} is generated by a set of cardinality N_{n}. Moreover N_{n} ≤ 2(n − 1)!. Moreover there is a single generator in bottom degree n, a single one in top degree 2^{n} − 1, and n − 1 in degree 2^{n−1}. All other generators have degree in between n + 1 and 2^{n−1} − 1. □

Remark 3.6 The number of generators N_{n} found in Theorem 3.5 is not the best possible. Indeed Lemma 3.2 allows to get rid of even more elements in X′_{n} than what we have done in Proposition 3.4. For example in F (1)^{⊗5} the element u^{4} ⊗ u ⊗ u ⊗ u ⊗ u is superfluous. Similarly ten other elements are superfluous in X_{6}. In principle the inductive construction of the set of generators X_{n} could be improved by suppressing all elements indicated by Lemma 3.2. There just does not seem to exist a nice formula as in the case of K^{⊗n} that gives a better bound for the number of generators for F (1)^{⊗n}.

References

[1] Schwartz L., Unstable Modules over the Steenrod Algebra and Sullivan’s Fixed Point Set Conjecture, Chicago Lectures in Mathematics, University of Chicago Press, Chicago, IL, 1994.