A toy analog of the Sullivan conjecture

In this post, I’d like to describe a toy analog of the Sullivan conjecture. Recall that the Sullivan conjecture considers (pointed) maps from ${BG}$ into a finite complex, and states that the space of such is contractible if $G$ is finite. The stable version replaces ${BG}$ with the Eilenberg-MacLane spectrum:

Theorem 13 Let ${H \mathbb{F}_p}$ be the Eilenberg-MacLane spectrum. Then the mapping spectrum

$\displaystyle (S^0)^{H \mathbb{F}_p}$

is contractible. In particular, for any finite spectrum ${F}$ , the graded group of maps ${[H \mathbb{F}_p, F] = 0}$ .

In the previous post, I sketched a proof (from Ravenel’s “Localization” paper) of this result based on a little chromatic technology. The spectrum ${H \mathbb{F}_p}$ is dissonant: that is, the Morava ${K}$ -theories don’t see it. However, any finite spectrum is harmonic: that is, local with respect to the wedge of Morava ${K}$ -theories. It follows formally that the spectrum of maps ${H \mathbb{F}_p \rightarrow S^0}$ is contractible (and thus the same with ${S^0}$ replaced by any finite spectrum). The non-formal input was the fact that ${S^0}$ is in fact harmonic, which requires a little work.

In this post, I’d like to sketch an earlier proof of the above theorem. This proof is based on the Adams spectral sequence. In fact, the proof runs parallel to Miller’s proof of the Sullivan conjecture. There is a classical Adams spectral sequence for computing ${[H \mathbb{F}_p, S^0]}$ , with ${E_2}$ page given by

$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}}(\mathbb{F}_p, \mathcal{A}) \implies [ H \mathbb{F}_p, S^0]_{t-s} ,$

with ${\mathcal{A}}$ the (mod ${p}$ ) Steenrod algebra.

It turns out, however, for purely algebraic reasons, that the ${E_2}$ term is trivial. Miller’s proof of the Sullivan conjecture relies on more complicated algebra to show that the unstable version of all this has the same vanishing property at ${E_2}$ . Most of this material is from Margolis’s Spectra and the Steenrod algebra.

1. The Adams spectral sequence

Let’s start by describing a method for calculating ${[H \mathbb{F}_p, X]}$ when ${X}$ is any connective spectrum with finitely generated homology. The Adams spectral sequence is based on a cosimplicial resolution of ${X}$

$\displaystyle H \mathbb{F}_p \otimes X \rightrightarrows H \mathbb{F}_p \otimes H \mathbb{F}_p \otimes X \dots$

whose totalization (under these hypotheses) is the ${p}$ -adic completion ${\widehat{X}_p}$ . In particular, we have

$\displaystyle [H \mathbb{F}_p, \widehat{X}_p] = \mathrm{Tot}( H \mathbb{F}_p, H \mathbb{F}_p^{\otimes (s+1)} \otimes X],$

which leads to a ${\mathrm{Tot}}$ spectral sequence whose ${E_2}$ -page can be identified as

$\displaystyle E_2^{s,t} = \mathrm{Ext}_{\mathcal{A}}( H^*(X),H^*(H \mathbb{F}_p)) \implies [ H \mathbb{F}_p, \widehat{X}_p]_{t-s} .$

The ${\mathrm{Ext}}$ groups are computed in the category of modules over the Steenrod algebra ${\mathcal{A}}$ .

Of course, we wanted ${[H \mathbb{F}_p, X]}$ to begin with, while we have ${\widehat{X}_p}$ instead of ${X}$ . The good news is that we have an arithmetic square

Observe that the spectrum of maps from ${H \mathbb{F}_p}$ into a rational or ${q}$ -complete (for ${q \neq p}$ ) spectrum is trivial. The spectrum of maps from ${H \mathbb{F}_p}$ into ${X}$ is thus seen to be the same as the spectrum of maps from ${H \mathbb{F}_p}$ into ${\widehat{X}_p}$ . So we can restate the Adams spectral sequence:

$\displaystyle E_2^{s,t} = \mathrm{Ext}_{\mathcal{A}}( H^*(X) ,H^*(H \mathbb{F}_p)) \implies [ H \mathbb{F}_p, X]_{t-s} \ \ \ \ \ (1)$

2. Injectivity of the Steenrod algebra

The surprising result that leads to this computation is the fact that the Steenrod algebra is injective as a module over itself. This has an analog in the Sullivan conjecture: the cohomology of ${B \mathbb{Z}/p}$ is injective in the category of unstable modules over the Steenrod algebra. This implies that the Adams spectral sequence for ${[H \mathbb{F}_p, X]}$ has its ${E_2}$ term concentrated on the bottom role with ${s = 0}$ , and we get a complete description of maps out of ${H \mathbb{F}_p}$ :

Theorem 14 Let ${X}$ be a connective spectrum with finitely generated homology. Then

$\displaystyle [H \mathbb{F}_p, X] = \hom_{\mathcal{A}}( H^*(X), \mathcal{A}).$

There is no nonzero map of ${\mathcal{A}}$ -modules ${ \mathbb{F}_p \rightarrow \mathcal{A}}$ . Thus this also implies the above result on finite spectra.

In the unstable case, there is a purely algebraic way to describe the cohomology of the mapping space ${Z^{B \mathbb{Z}/p}}$ , for ${Z}$ a space.

In order to prove the above theorem, we are reduced by the Adams spectral sequence to showing the following purely algebraic result:

Proposition 15 (Adams, Margolis) The Steenrod algebra ${\mathcal{A}}$ is injective in the category of (graded) modules over ${\mathcal{A}}$ .

3. Injectivity for finite-dimensional Hopf algebras

Let ${A}$ be a finite-dimensional, graded, cocommutative, connected Hopf algebra over the ground field ${k}$ . “Connectedness” is the condition that ${A_0 = k}$ . Let ${\mathrm{Mod}(A)}$ be the category of graded ${A}$ -modules.

The main goal of this section is to prove:

Proposition 16 ${A}$ is injective in the category ${\mathrm{Mod}(A)}$ .

To start with, note that ${\mathrm{Mod}(A)}$ (with the ${k}$ -linear tensor product) is a symmetric monoidal category with monoidal product ${\wedge}$ , thanks to the Hopf algebra structure on ${A}$ . The unit object ${\mathbf{1}}$ is the field ${k}$ concentrated in degree zero. Another interesting object is ${A}$ itself, of course. A useful property of ${A}$ is the “shearing” isomorphism

$\displaystyle A \wedge M \simeq A \otimes M, \quad M \in\mathrm{Mod}(A) \ \ \ \ \ (2)$

with the following meaning: the first object is the internal tensor product (i.e., ${A}$ acts via the codiagonal) and the second object is an external tensor product (i.e., ${A}$ acts only on the first factor ${A}$ ).

Proof: The shearing map is determined by specifying a map ${M \rightarrow A \wedge M}$ of ${k}$ -vector spaces, and it is the evident map ${m \rightarrow 1 \otimes m}$ . In particular, this defines the natural transformation. We observe that the functors ${M \mapsto A \wedge M, \quad M \mapsto A \otimes M}$ preserve colimits, so it suffices to check the result for ${M}$ connective and finite-dimensional. Any connective, finite-dimensional module has a filtration whose quotients are shifts of ${k}$ itself, so it suffices to check that the map is an isomorphism when ${M = k}$ . Here, though, it is obvious. $\Box$

The category ${\mathrm{Mod}(A)}$ has dual objects as well (at least for finite-dimensional objects). Given a finite-dimensional ${A}$ -module ${M}$ , one has a dual object ${DM \stackrel{\mathrm{def}}{=} \hom_k(M, k)}$ where the ${A}$ -linear structure comes from the antipode on ${A}$ . To show that ${A}$ is injective, we can play with this duality and give a formal argument.

In fact, let ${\mathrm{Mod}^f(A) \subset \mathrm{Mod}(A)}$ be the full subcategory of finite-dimensional (hence dualizable) objects. The existence of duals shows that ${\mathrm{Mod}^f(A)}$ is anti-equivalent to itself. It suffices to show that ${A}$ is injective in ${\mathrm{Mod}^f(A)}$ to see that it is injective in ${\mathrm{Mod}(A)}$ by a straightforward argument.

Therefore, to show that ${A}$ is injective in ${\mathrm{Mod}^f(A)}$ , we can use duality and show that ${DA}$ is projective in ${\mathrm{Mod}^f(A)}$ . Since, up to a grading shift, ${\mathbf{1}}$ is a retract of ${A}$ , it follows that ${DA}$ is a retract of ${A \wedge DA}$ , and it suffices to show that ${A \wedge DA}$ is projective. But ${A \wedge DA \simeq A \otimes DA}$ is clearly projective. This completes the proof.

4. Proof of injectivity

Now let’s consider the case of the Steenrod algebra ${\mathcal{A}}$ . Our goal is to show that ${\mathcal{A}}$ is injective in the category of graded ${\mathcal{A}}$ -modules. In other words, for any graded ${\mathcal{A}}$ -module ${M}$ , we want to show that

$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}}(M, \mathcal{A}) = 0$

for all ${s > 0}$ and all ${t}$ . It suffices to assume that ${M}$ is finite-dimensional in each dimension. In fact, we may assume that ${M}$ is finitely presented as a module over ${\mathcal{A}}$ .

The Steenrod algebra is not finite-dimensional, but it is a cocommutative Hopf algebra. Moreover, it is the union

$\displaystyle \mathcal{A} = \bigcup_n \mathcal{A}^{(n)}$

where each ${\mathcal{A}^{(n)}}$ is a finite-dimensional, graded, connected Hopf algebra and therefore self-injective by the previous section. Moreover, $\mathcal{A}$ is free over $\mathcal{A}^{(n)}$ by Milnor-Moore structure theory. It turns out that this is enough to prove that ${\mathcal{A}}$ is self-injective with only a little extra work.

In order to do this, observe that there exists ${n}$ and an ${\mathcal{A}_n}$ -module ${M_n}$ such that

$\displaystyle M \simeq \mathcal{A} \otimes_{\mathcal{A}_n} M_n,$

so it suffices to show that

$\displaystyle \mathrm{Ext}^{s,t}_{\mathcal{A}_n}(M_n, \mathcal{A}) =0, \quad s> 0.$

Again, this follows because ${\mathcal{A}}$ is a direct sum of copies of ${\mathcal{A}^{(n)}}$ (by Milnor-Moore structure theory), so that ${\mathcal{A}}$ is injective over ${\mathcal{A}^{(n)}}$ . (Because ${\mathcal{A}^{(n)}}$ is finite-dimensional, the infinite direct sum of injectives is again injective.) This implies that the ${\mathrm{Ext}}$ -groups vanish and proves the claim.

Climbing Mount Bourbaki