The category U III

This is part of a series of posts intended to understand some of the basic structure of the category ${\mathcal{U}}$ of unstable modules over the (mod ${2}$ ) Steenrod algebra, to prepare for the proof of the Sullivan conjecture. Here’s what we’ve seen so far:

${\mathcal{U}}$ is a Grothendieck abelian category, with a set of compact, projective generators ${F(n)}$ (the free unstable module on a generator in degree ${n}$ ). (See this post.)
${F(n)}$ has a tautological class ${\iota_n}$ in degree ${n}$ , and has a basis given by ${\mathrm{Sq}^I \iota_n}$ for ${I}$ an admissible sequence of excess ${\leq n}$ . (This post explained the terminology and the proof.)
${F(1)}$ was the subspace ${\mathbb{F}_2\left\{t, t^2, t^4, \dots\right\} \subset \widetilde{H}^*(\mathbb{RP}^\infty; \mathbb{F}_2)}$ .

Our goal in this post is to describe some of the additional structure on the category ${\mathcal{U}}$ , which will eventually enable us to prove (and make sense of!) results such as ${F(n) \simeq (F(1)^{\otimes n})^{ \Sigma_n}}$ . We’ll start with the symmetric monoidal tensor product and the suspension functor, and then connect this to the Frobenius maps (which will be defined below).

1. The symmetric monoidal structure

Our first order of business is to describe the symmetric monoidal structure on ${\mathcal{U}}$ , which will be given by the ${\mathbb{F}_2}$ -linear tensor product. In fact, recall that the Steenrod algebra is a cocommutative Hopf algebra, under the diagonal map

$\displaystyle \mathrm{Sq}^n \mapsto \sum_{i+j = n} \mathrm{Sq}^i \otimes \mathrm{Sq}^j.$

The Hopf algebra structure is defined according to the following rule: we have that ${\theta}$ maps to ${\sum \theta' \otimes \theta''}$ if and only if for every two cohomology classes ${x,y }$ in the cohomology of a topological space, one has

$\displaystyle \theta(xy) = \sum \theta'(x) \theta''(y).$

The cocommutative Hopf algebra structure on ${\mathcal{A}}$ gives a tensor product on the category of (graded) ${\mathcal{A}}$ -modules, which is symmetric monoidal. It’s easy to check that if ${M, N}$ are ${\mathcal{A}}$ -modules satisfying the unstability condition, then so does ${M \otimes N}$ . This is precisely the symmetric monoidal structure on ${\mathcal{U}}$ .

Example 4 Given an unstable ${\mathcal{A}}$ -module ${M}$ , we can define the suspension ${\Sigma M}$ with

$\displaystyle (\Sigma M)_t = M_{t-1},$

and with the same ${\mathcal{A}}$ -module structure. A short check shows that ${\Sigma M}$ is also unstable, so we get an endofunctor

$\displaystyle \Sigma: \mathcal{U} \rightarrow \mathcal{U},$

which is fully faithful. Another way to describe the suspension functor is that it is the tensor product with the object ${\Sigma \mathbb{F}_2 \in \mathcal{U}}$ (which is ${\mathbb{F}_2}$ in degree one and zero everywhere else). Below, we’ll describe the left adjoint to ${\Sigma}$ and its derived functors in terms of the Frobenius.

2. The Frobenius

The category ${\mathcal{U}}$ also comes with another endofunctor

$\displaystyle \Phi: \mathcal{U} \rightarrow \mathcal{U}$

which we can think of as a Frobenius (or Verschiebung) of sorts. Namely, given a module ${M \in \mathcal{U}}$ , we define

and the ${\mathcal{A}}$ -module structure on ${\Phi(M)}$ is such that the action of ${\mathrm{Sq}^n}$ on ${\Phi(M)}$ corresponds to the action of ${\mathrm{Sq}^{n/2}}$ for ${n}$ even and zero if ${n}$ is odd.

In other words, we use an algebra homomorphism

$\displaystyle \phi: \mathcal{A} \rightarrow \mathcal{A}$

such that

and pull back the ${\mathcal{A}}$ -module structure along ${\phi}$ , but modifying the grading appropriately since ${\phi}$ halves the grading.

In order for this to make any sense, we need to check that the map ${\phi}$ we’ve defined is in fact a homomorphism. One way to do this is to check that ${\phi}$ respects the Adem relations. A more conceptual way is to consider the dual Steenrod algebra (discussed in this post)

$\displaystyle \mathcal{A}^{\vee} \simeq \mathbb{F}_2[\xi_1, \xi_2, \dots ]$

which is a commutative Hopf algebra, and consider the Frobenius

$\displaystyle \mathcal{A}^{\vee} \rightarrow \mathcal{A}^{\vee}, \quad x \mapsto x^2,$

which is a map of Hopf algebras. Unwinding the pairing between ${\mathcal{A}^{\vee}}$ and ${\mathcal{A}}$ , the dual map can be checked to be ${\phi}$ . More explicitly, we need to check that for any ${x \in \mathcal{A}^{\vee}}$ and any ${n}$ , we have

$\displaystyle \left \langle x^2, \mathrm{Sq}^n\right\rangle = \left \langle x, \mathrm{Sq}^{n/2}\right\rangle.$

But ${\left \langle x^2, \mathrm{Sq}^n \right \rangle}$ can be identified with the pairing

$\displaystyle \left \langle x \otimes x , \sum_{i+j=n} \mathrm{Sq}^i \otimes \mathrm{Sq}^j \right\rangle,$

using the duality between multiplication and comultiplication. Now here it’s easy to see that everything cancels except for ${\left \langle x, \mathrm{Sq}^{n/2}\right\rangle^2 = \left \langle x, \mathrm{Sq}^{n/2}\right\rangle}$ . This exhibits ${\phi}$ as the dual to the Frobenius map and is therefore a homomorphism (of Hopf algebras). Finally, a short check shows that ${\Phi M}$ still satisfies the unstability condition, and that ${\Phi}$ commutes with the symmetric monoidal structure.

This proves:

Proposition 33 ${\Phi}$ as defined above is a symmetric monoidal endofunctor ${\mathcal{U} \rightarrow \mathcal{U}}$ .

3. The Frobenius map

The next goal of this post is to connect the construction ${\Phi}$ with the suspension functor defined two sections ago. In order to do this, we’ll need to define a Frobenius map, which will be a natural transformation:

$\displaystyle \Phi M \rightarrow M.$

The map sends an element ${x}$ in degree ${2n}$ of ${\Phi M}$ , identified with ${x \in M_n}$ , to ${\mathrm{Sq}^n x \in M_{2n}}$ . By the unstability condition, ${\mathrm{Sq}^n x}$ is the top square that can’t vanish. We need to check that this is actually a map in ${\mathcal{U}}$ :

Proposition 34 The map ${\Phi M \rightarrow M}$ defined is ${\mathcal{A}}$ -linear.

Proof: This is a manipulation with the Adem relations. Namely, we need to show that if ${ x \in M_n}$ and ${i \in \mathbb{Z}_{\geq 0}}$ , then

$\displaystyle \mathrm{Sq}^i \mathrm{Sq}^n x = \mathrm{Sq}^{n+i/2} \mathrm{Sq}^{i/2} x.$

When ${i > 2n}$ , there is nothing to prove: both sides are zero by unstability. When ${i = 2n}$ , both sides are obviously equal. When ${i < 2n}$ , we can use the Adem relations to get

$\displaystyle \mathrm{Sq}^i \mathrm{Sq}^n = \sum_{k \leq i/2} \binom{n-k-1}{i - 2k} \mathrm{Sq}^{n+i - k} \mathrm{Sq}^k.$

When applied to a class in degree ${n}$ , the instability condition forces everything on the right-hand-side to vanish except when ${i}$ is even and ${k =i/2}$ , in which case we get the desired result. $\Box$

It’s also straightforward (from the Cartan formula) that the map ${\Phi \rightarrow \mathrm{Id}}$ of functors is symmetric monoidal. One reason that ${\Phi}$ is relevant is that ${\Phi}$ measures the failure of a module to be a suspension. Observe that if ${M}$ is a suspension of an object in ${\mathcal{U}}$ , then ${\mathrm{Sq}^n}$ vanishes on all classes of degree ${n}$ , and the map

$\displaystyle \Phi M \rightarrow M$

is zero. Conversely, if the map ${\Phi M \rightarrow M}$ is zero, we can desuspend ${M}$ and the result will still satisfy the unstability condition.

We get:

Proposition 35 The object ${M \in \mathcal{U}}$ is a suspension if and only if ${\Phi M \rightarrow M}$ is zero.

In the next section, we’ll elaborate on the relationship between ${\Phi}$ and the suspension.

4. The loop space and its derived functors

Earlier in this post, we defined a suspension functor

$\displaystyle \Sigma: \mathcal{U} \rightarrow \mathcal{U} .$

This functor is exact and commutes with limits, and it is also fully faithful. It isn’t invertible because the desuspension of an unstable ${\mathcal{A}}$ -module doesn’t have to be unstable. The adjoint functor theorem implies that it has a left adjoint ${\Omega}$ (observe that we have dualized and are working in cohomology rather than homology). In this section, we’ll try to understand a little about ${\Omega}$ and its derived functors.

Let’s start by observing that the loop space of a suspension is easy to describe: we have

$\displaystyle \Omega \Sigma M \simeq M, \quad M \in \mathcal{U}$

which is a formal consequence of the fact that ${\Sigma}$ is fully faithful. As another example, we have

$\displaystyle \Omega F(n) = F(n-1)$

by appealing to universal properties of both ${\Omega }$ and the free objects. Moreover, ${\Omega}$ (as a left adjoint) is right exact. In general, however, it is not left exact.

Let’s now describe ${\Omega}$ in general. If ${M}$ is any module, we need to define ${\Omega M}$ such that

$\displaystyle \hom_{\mathcal{U}}(\Omega M, N) \simeq \hom_{\mathcal{U}}(M, \Sigma N).$

This shows that we can describe ${\Omega M}$ as the largest quotient of ${\Sigma^{-1} M \in \mathrm{Mod}(\mathcal{A})}$ which satisfies the unstability condition. This has a direct connection with ${\Phi}$ above. The map

$\displaystyle \Phi M \rightarrow M$

has a cokernel, which is precisely the quotient of ${M}$ by ${\left\{\mathrm{Sq}^n x, x \in M_n\right\}}$ , and that desuspends in ${\mathcal{U}}$ . In fact, that’s exactly the largest quotient that does desuspend, and we find:

Proposition 36 We can describe ${\Omega M}$ as the desuspension (which exists in ${\mathcal{U}}$ ) of the cokernel of the Frobenius map ${\Phi M \rightarrow M}$ .

It follows that for any ${M}$ , we have an exact sequence

$\displaystyle \Phi M \rightarrow M \rightarrow \Sigma \Omega M.$

We would like to prove:

Proposition 37 The desuspension of the kernel of ${\Phi M \rightarrow M}$ is the first derived functor of ${\Omega}$ . All the higher derived functors of ${\Omega}$ vanish.

Stated another way, the derived functor of ${\Omega}$ , in the derived category, is a desuspension of the map ${\Phi M \rightarrow M}$ (considered as a two-term chain complex). In order to prove this, it suffices by abstract nonsense to show effaceability of the candidate derived functors: i.e., we must show that ${\Phi M \rightarrow M}$ is injective if ${M}$ is projective. It suffices thus to prove:

Proposition 38 For each ${n}$ , we have an exact sequence

$\displaystyle 0 \rightarrow \Phi F(n) \rightarrow F(n) \rightarrow \Sigma F(n-1) \rightarrow 0;$

in other words the Frobenius ${\Phi F(n) \rightarrow F(n)}$ is injective.

Proof: This is a straightforward consequence of the basis provided for ${F(n)}$ . $\Box$

Climbing Mount Bourbaki