A misunderstanding about Sullivan's conjecture

Question

In this old blog post, Akhil Mathew describes the Sullivan conjecture and part of Miller's proof of a special case.

There's a point in the beginning which is not clear to me, about $p$-completions at different primes than $2$ (or, more generally, at primes $p$ with $p\land |G|=1$)

Indeed, his argument for "the homotopy fixed points of $\mathbb Z/2$ acting on $X(\mathbb C)$ completed at $p$ for $p$ odd is simply connected" is that $\pi_*((X(\mathbb C)^\wedge_p)^{h\mathbb Z/2}) \cong \pi_*(X(\mathbb C)^\wedge_p)^{\mathbb Z/2}$.

Now I understand the isomorphism, but I don't understand the connection with the Sullivan conjecture.

Indeed, as far as I understand, the Sullivan conjecture states (under some hypotheses):

After completion at $p$, $X^G\to X^{hG}$ is an equivalence.

So the statement would be about $(X^{hG})^\wedge_p$, not $(X^\wedge_p)^{hG}$.

So I don't understand how his argument relates to the Sullivan conjecture, and how it proves the point that "there's no hope for it to be true at odd primes".

There is a natural map $(X^{hG})^\wedge_p\to (X^\wedge_p)^{hG}$, but I don't expect it to be an equivalence.

Is this natural map an equivalence in the cases of concern ? If so, why ? If not, is tere an argument to compute $\pi_*((X^{hG})^\wedge_p)$ (specifically, it should work for $*=1$) ?

And if the answers to those are "no",

How does the argument about $\pi_*((X(\mathbb C)^\wedge_p)^{h\mathbb Z/2})$ relate to the Sullivan conjecture ?

Answer 1

I think I might have an answer, but assuming that $X$ is simply-connected (and it matters at more than one step).

Note that I'm in the case $|G|\land p =1$; I will also need to assume that $X$ is a pointed $G$-space (as I want to take a fiber as a $G$-space). So I will want to assume $X^{hG}\neq \emptyset$. For instance, in the case of $X(\mathbb C)$, we might want $X$ to have at least one real point.

$X^\wedge_p$ is $p$-complete, and those are stable under homotopy limits, so $(X^\wedge_p)^{hG}$ is also $p$-complete.

Moreover, the map $X\to X^\wedge_p$ induces a canonical map $X^{hG}\to (X^\wedge_p)^{hG}$ from $X^{hG}$ to a $p$-complete space. Therefore, to show that $(X^{hG})^\wedge_p \simeq (X^\wedge_p)^{hG}$, it suffices to prove that $X^{hG}\to (X^\wedge_p)^{hG}$ is an $\mathbb F_p$-equivalence.

First of all, note that $X$ being simply-connected implies that $X^\wedge_p$ is too, and so by the homotopy fixed points spectral sequence, $(X^\wedge_p)^{hG}$ is too (this spectral sequence has $E_2$-page $H^{-p}(G;\pi_q(X^\wedge_p))$, but $\pi_q(X^\wedge_p)$ is derived $p$-complete, so $|G|$ acts invertibly on them, which means that this $E_2$-page is $0$ for $p\neq 0$, which yields the desired result)

In particular, if we take a fiber sequence $F\to X\to X^\wedge_p$, then take homotopy fixed points to get a new fiber sequence $F^{hG}\to X^{hG}\to (X^\wedge_p)^{hG}$, we get in particular that it suffices to shows that $F^{hG}$ has homotopy groups that are $\mathbb Z[\frac 1 p ]$-modules (if they are, just look at the Serre spectral sequence of the fibration).

Note that since $X,X^\wedge_p$ are simply-connected, the converse to the above also holds for the first fiber sequence, in particular the homotopy groups of $F$ are $\mathbb Z[\frac 1 p]$-modules.

But now, $F$ is simple because $X$ is simply-connected, so we also get a homotopy fixed points spectral sequence with $H^{-p}(G,\pi_q(F)) \implies \pi_{q-p}(F^{hG})$. $p$ acts invertibly on $\pi_q(F)$, so it does so on $H^{-p}(G,\pi_q(F))$ too, and therefore, by convergence of the spectral sequence, it does so on $\pi_*(F^{hG})$.

Therefore $X^{hG}\to (X^\wedge_p)^{hG}$ is an $\mathbb F_p$-equivalence, which implies the desired result: $(X^{hG})^\wedge_p\simeq (X^\wedge_p)^{hG}$

In particular, in the case of the blog post, since $(X(\mathbb C)^\wedge_p)^{h\mathbb Z/2}$ is simply-connected, so is $(X(\mathbb C)^{h\mathbb Z/2})^\wedge_p$, and so $X(\mathbb R)= X(\mathbb C)^{\mathbb Z/2} \to X(\mathbb C)^{h\mathbb Z/2}$ can't be an equivalence after $p$-completion (for $p$ odd of course)