# When does rationalization commute with homotopy fixed points?

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):

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?

• What's $X$ here? A space, a spectrum, a complex... – Denis Nardin Jun 13 '19 at 7:59
• @DenisNardin I think X is a space, since proposition 5.13 is about G-spaces. – cellular Jun 13 '19 at 9:29
• 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. – 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.

• 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. – skupers Jun 13 '19 at 20:11
• If M is finite then the Sullivan conjecture suggests that the statement should be true -- certainly for trivial action. – Tom Goodwillie Jun 13 '19 at 20:45
• I wonder if Goyo's statement had an implicit finiteness assumption. I have not looked at the supposed proof. – Tom Goodwillie Jun 14 '19 at 12:01
• @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... – cellular Jun 14 '19 at 15:08