Let $X$ be an (naive) $O(n)$-spectrum (I'm choosing to work with orthogonal spectra). I've recently come across the following results,

$$(S^{n-1} \wedge X)_{hO(n)} \simeq X_{hO(n-1)}$$


$$\Omega^\infty (X_{hG)}) \simeq (\Omega^\infty X)_{hG} $$

where $G$ is a compact Lie group.

I've spent the last few days trying to find sources for these results - and trying to prove them myself to little avail.

Any help/suggestions or references would be greatly appreciated.

Added The answer by Tyler has shown that the second is false. This raises the question about how close can we get, in the following sense: If $X$ is $n$-connected, then how connected is the map $$ (\Omega^\infty X)_{hG} \to \Omega^\infty(X_{hG})\,? $$


Both of these are false.

The first is close to true: if $S(n-1)$ is the unit sphere in $\Bbb R^{n}$ with its standard $O(n)$-action, then we can identify $S(n-1)$ with $O(n) / O(n-1)$ and so get the identification $$ (S(n-1)_+ \wedge X)_{hO(n)} = EO(n)_+ \wedge_{O(n)} O(n)/O(n-1)_+ \wedge X = EO(n)_+ \wedge_{O(n-1)} X. $$

To show that the first one is false, let's take $X = \Sigma^\infty O(n)_+$. Then the first equation would give an equivalence between $\Sigma^\infty S^{n-1}$ and $\Sigma^\infty O(n)/O(n-1)_+ \cong \Sigma^\infty S(n-1)_+$. These have nonisomorphic homology for all $n$.

For the second one, we can take $X$ to be the desuspension of the Eilenberg-MacLane spectrum $H\Bbb Z$ with trivial $O(n)$-action. Then $\Omega^\infty X$ is contractible and so its homotopy orbit space is $BG$ (or contractible, if you use the based orbit space), while $$ \begin{align*} \pi_n \Omega^\infty (\Sigma^{-1} H\Bbb Z)_{hG} &= \pi_n (\Sigma^{-1} H\Bbb Z_{hG})\\ &= \pi_{n+1} (H\Bbb Z_{hG})\\ &= \pi_{n+1} (H\Bbb Z \wedge \Sigma^\infty BG_+)\\ &= H_{n+1}(BG, \Bbb Z) \end{align*} $$ which is typically very different from either $\pi_n BG$ or $\pi_n (*)$.

  • $\begingroup$ Thanks for your answer. For the second one, can we make some kind of connectivity argument, say for instance the spectrum in $n$-connected. Can we say anything about the connectivity of either side and get a result about the connectivity of the canonical map between them? (The canonical map here will go from right to left) - This argument won't give an equivalence but it's not a bad partial result $\endgroup$ – Niall Taggart Aug 16 '18 at 12:51
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    $\begingroup$ @274072 Yes. I believe that if X is n-connected, the map is (2n+1)-connected. (It is the same as the connectivity estimate for the map $\Sigma^\infty \Omega^\infty X \to X$.) For this, you need to use based homotopy orbits. $\endgroup$ – Tyler Lawson Aug 16 '18 at 19:30
  • $\begingroup$ On second thought, that should probably be 2n+2. $\endgroup$ – Tyler Lawson Aug 16 '18 at 19:39
  • $\begingroup$ I think (2n+1)-connected would be correct if we took X to be a suspension spectrum of a based space... albeit I’m not convinced - I’m obtuse when it comes to connect arguments $\endgroup$ – Niall Taggart Aug 16 '18 at 20:05

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