Let $X$ be a $G$-space. There are a number of places in the literature where one can find the claim that under certain conditions rationalization and taking homotopy fixed points with respect to a finite group action commute, that is, the map $X^{hG} \to (X_{\mathbb{Q}})^{hG}$ is a rationalization or equivalently $(X^{hG})_{\mathbb{Q}} \to (X_{\mathbb{Q}})^{hG}$ is a weak equivalence.

I know how to prove a number of special cases of this (e.g. when $X$ is based with $G$-fixed base point and nilpotent with finitely many non-zero homotopy groups), but a quite general result is claimed in the 1989 thesis of Goyo (he uses $(-)^{(G)}$ as notation for homotopy fixed points):

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However, I don't understand his proof (it seems to use the claim that an infinite limit of rationalizations is a rationalization, which is not true). So I am looking for a reference which gives an answer to the following question:

When exactly does rationalization commute with homotopy fixed points?

  • $\begingroup$ What's $X$ here? A space, a spectrum, a complex... $\endgroup$ Jun 13 '19 at 7:59
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    $\begingroup$ @DenisNardin I think X is a space, since proposition 5.13 is about G-spaces. $\endgroup$
    – cellular
    Jun 13 '19 at 9:29
  • $\begingroup$ There is a spectral sequence $E^{p,q}_{2}=H^{-p}(G,\pi_{q}(X))$ converging to $\pi_{p+q}(X^{hG})$. I think that $E^{p,q}_{2}=H^{-p}(G,\pi_{q}(X)\otimes \mathbb{Q} )$ converges to $\pi_{p+q}(X^{hG})\otimes \mathbb{Q}$. Which should imply that $X_{\mathbb{Q}}^{hG}\simeq (X^{hG})_{\mathbb{Q}}$ under simply connected hypothesis. $\endgroup$
    – cellular
    Jun 13 '19 at 13:58

The statement appears to me to be false. The difficulty, from some point of view, is that in the spectral sequence going from $H^{-i}(G;\pi_j(M))$ to $\pi_{i+j}(M^{hG})$ an infinite number of torsion groups can conspire to make something rationally non-trivial.

As I understand it, the statement you are asking about says that if a finite group $G$ acts on a space $M$, and if both $M$ and the homotopy fixed point space $M^{hG}$ are simply connected, and if $M\to M_{\mathbb Q}$ is a rationalization of $M$ that is also a $G$-map for some action of $G$ on $M_{\mathbb Q}$, then the induced map $M^{hG}\to (M_{\mathbb Q})^{hG}$ is also a rationalization.

Let's make a counterexample in which $G$ acts trivially on $M$ (and on $M_{\mathbb Q}$). In this case $M^{hG}$ is the function space $Map(BG,M)$. I claim that it is enough if we can find a simply connected space $M$ and a finite group $G$ such that the based function space $Map_\ast(BG,M)$ is simply connected but not rationally trivial. If so, then the fibration sequence $$ Map_\ast(BG,M)\to Map(BG,M)\to M $$ shows that $Map(BG,M)$ is simply connected, and also in the fibration sequence $$ Map_\ast(BG,M)_{\mathbb Q}\to Map(BG,M)_{\mathbb Q}\to M_{\mathbb Q} $$ the right hand map is not a weak equivalence. On the other hand, in the fibration sequence $$ Map_\ast(BG,M_{\mathbb Q})\to Map(BG,M_{\mathbb Q})\to M_{\mathbb Q} $$ the right hand map is a weak equivalence because $H^j(BG;V)=0$ for $j>0$ and $V$ a rational vector space. It follows that $$ Map(BG,M)_{\mathbb Q}\to Map(BG,M_{\mathbb Q}) $$ is not a weak equivalence.

To come up with such an $M$ and $G$ we can use the Atiyah-Segal completion theorem. The basic idea is to take $M$ to be $BU$, but I have to modify this a little to make $Map_\ast (BG,M)$ simply connected.

Start with $BU$, whose homotopy groups are $\pi_{2k}\cong \mathbb Z$ with complex conjugation acting by $+1$ when $k$ is even and by $-1$ when $k$ is odd. Localize it by inverting $2$, and split the result as a product of two factors according to that conjugation action. $M$ will be the factor corresponding to $-1$, so its homotopy groups are $\pi_j\cong \mathbb Z[1/2]$ if $j$ congruent to $2$ mod $4$.

Let $G$ be the dihedral group of order $6$. Then $H^i(BG;\mathbb Z[1/2])$ is trivial when $i$ is not a multiple of $4$ and is of order $3$ if $i>0$ is a multiple of $4$. It follows that $Map_\ast(BG,M)$ is simply connected and that $\pi_2Map_\ast(BG,M)$ is the inverse limit of larger and larger finite $3$-groups. By Atiyah-Segal (which describes the homotopy groups of the related space $Map_\ast(BG,BU)$) there is actually a copy of the $3$-adic integers here, so the group is rationally nontrivial.

  • $\begingroup$ Yes, I expected it to be false, and this is a great counterexample. I will wait to see whether someone can find a simpler one (e.g. M a finite CW complex), but otherwise I will expect it. $\endgroup$
    – skupers
    Jun 13 '19 at 20:11
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    $\begingroup$ If M is finite then the Sullivan conjecture suggests that the statement should be true -- certainly for trivial action. $\endgroup$ Jun 13 '19 at 20:45
  • $\begingroup$ I wonder if Goyo's statement had an implicit finiteness assumption. I have not looked at the supposed proof. $\endgroup$ Jun 14 '19 at 12:01
  • $\begingroup$ @skupers here an article (just after theorem 2) using the same result "Rational homotopy of the (homotopy) fixed point sets of circle actions" Urtzi Buijs , Yves Félix , Aniceto Murillo It seems that the condition is that the $G$-action extends to $X_{\mathbb{Q}}$ whtever it means... $\endgroup$
    – cellular
    Jun 14 '19 at 15:08

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