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Let $E$ be a spectrum acted upon a finite group $G$. Is there a general way of computing the homology of the homotopy fixed point spectrum $E^{hG}$ in terms of that of $E$? (I'm aware that there is a spectral sequence for computing $\pi_* E^{hG}$ in terms of $\pi_* E$, but smashing with some other spectrum probably doesn't preserve homotopy fixed points.)

Here's a specific example I have in mind. Take connective $K$-theory $bu$. This has an action of $\mathbb{Z}/2$, which comes from the $\mathbb{Z}/2$-action on $K$-theory (given on the level of cohomology theories by complex conjugation of vector bundles). Then $bu^{\mathbb{Z}/2} = bo$. Actually, this is only true before taking connective covers.

Let's say I know how to compute the mod 2 homology of $bu$ (it's $\mathbb{Z}/2[\zeta_1^2, \zeta_2^2, \zeta_3, \zeta_4, \dots]$ as a comodule over the dual Steenrod algebra). Does that give any information about $H_*(bo; \mathbb{Z}/2)$? Is there a good reference for this material?

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I think that, while $KU^{\mathbb Z/2}=KO$, it is NOT true that $ku^{\mathbb Z/2}=ko$. If you run the homotopy fixed point spectral sequence, it's ptretty obvious that you don't get $\pi_*(ko)$ from $H^*(\mathbb Z/2;\pi_*(ku))$. –  André Henriques Jun 30 '12 at 20:17
    
I thought that taking homotopy fixed points would commute with truncation $\tau_{\geq 0}$ (which is a right adjoint, right?). Is there something that goes wrong here? –  Akhil Mathew Jul 1 '12 at 2:36
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@Akhil: The truncation is a right adjoint, but only if the target is a different category (connective spectra). In that category, the homotopy fixed point object of $ku$ is indeed $ko$. –  Tyler Lawson Jul 1 '12 at 5:40
    
Thanks for the clarification. –  Akhil Mathew Jul 1 '12 at 12:36

4 Answers 4

up vote 8 down vote accepted

Consider $G$ of order $2$ acting trivially on the sphere spectrum $S^0$. In this key example, smashing with $H\mathbb Z/2$ drastically fails to commute with $^{hG}$: If it did commute, then the mod $2$ homology of the homotopy fixed point spectrum would be the homotopy of $(H\mathbb Z/2)^{hG}$, so $\mathbb Z/2$ in nonpositive dimensions and $0$ in positive dimensions. But the (solved) Segal conjecture shows that the homotopy fixed point spectrum is connective.

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Thanks! This example is very nice. –  Akhil Mathew Jul 1 '12 at 12:53

A good reference for the sort of spectral sequence that you're looking for is an article by Bruner-Rognes, "Differentials in the homological homotopy fixed point spectral sequence." It treats some of the convergence questions that you're worried about.

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This paper looks pretty interesting. Thanks. –  Akhil Mathew Jul 1 '12 at 12:55
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I was just about to suggest this. Their construction is pretty explicit. While they are mostly interested in the case where $G=S^1$ you can get the general case by using your favorite model for $EG$ and following their construction. –  Sean Tilson Jul 2 '12 at 4:47
    
Also, that spectral sequence works the same for any generalized cohomology theory. The only issue might be convergence if it is not connective (or bounded below rather). –  Sean Tilson Jul 2 '12 at 4:49

There is a subject of equivariant stable homotopy theory. You are referring to $bu$ as what is called a naive $G$-spectrum (a spectrum acted on by a group), and your notation $E^G$ is confusing the categorical fixed point spectrum with the homotopical fixed point spectrum. The difference between the two is, for example, the subject of the Segal conjecture. There is a homotopy fixed point spectral sequence for homotopical fixed point spectra. There is a huge literature on equivariant stable homotopy theory (which was a key input to solution of the Kervaire invariant problem by Hill, Hopkins, and Ravenel). One expository introduction is:

Greenlees and May. Equivariant stable homotopy theory. Handbook of Algebraic Topology, edited by I.M. James, pp. 279-325. 1995.

But a lot has been done since then. Naive $G$-spectra are not so interesting. Genuine $G$-spectra have $RO(G)$-graded homology and cohomology.

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I did mean the homotopy fixed point spectrum, though I realize my notation was off (I've changed it). I'm aware of the homotopy fixed point sequence for the homotopy groups; I guess my question is whether one can use it to compute homology, because smashing with a big spectrum like $H \mathbb{Z}/2$ probably doesn't preserve homotopy fixed points, right? –  Akhil Mathew Jul 1 '12 at 2:36
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In a sense "naive $G$-spectra" is the natural home for questions about homotopy fixed-point spectra and homotopy orbit spectra. But in order to answer a question you sometimes have to go to the larger world of "genuine $G$-spectra". –  Tom Goodwillie Jul 1 '12 at 11:23
    
Just a note for people who find this thread in the future: after this answer the OP edited his question to replace $E^G$'s by $E^{hG}$'s in response to the first part of this answer. –  David White Jul 4 '12 at 13:13

Maybe to expand on the relationship between naive and genuine G-spectra, Tom's answer is stated in terms of naive G-spectra, but the actual mathematics, the Segal conjecture, is all about genuine G-spectra. The genuine sphere $G$-spectrum $S_G$ has non-trivial $G$-action and nontrivial $G$-fixed point spectrum. Consider a finite $G$-CW complex. The $G$-map $EG\times X \to X$ induces a map of genuine function $G$-spectra \[ F(X_+, S_G) \to F((EG\times X)_+, S_G). \] If $G$ is a $p$-group, the Segal conjecture says that this map is an equivalence after $p$-adic completion. This remains true after passing to $G$-fixed point spectra. By definition, that passage means that you first take underlying naive $G$-spectra and then take actual fixed points. Tom is taking $X$ to be a point, and then the map is \[ (S_G)^G \to F(EG_+, S_G)^G, \] going from the fixed point spectrum to the homotopy fixed point spectrum of the genuine $G$-spectrum $S_G$. The target is equivalent to the naive homotopy fixed point spectrum $F(EG_+, S)^G$ of the naive sphere $G$-spectrum with trivial action by $G$, and that is equivalent to $F(BG_+,S)$, giving the cohomotopy groups of $BG$ on passage to homotopy groups. But all of the real mathematics in validating Tom's answer takes place on the level of genuine $G$-spectra.

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Sure. I've been a bit loose about this in the question. I'll take a look at the equivariant homotopy book so that I can understand these distinctions. –  Akhil Mathew Jul 6 '12 at 0:39
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Your question was about spectra with $G$-action. There was no need to mention genuine $G$-spectra in the question. The remarkable thing is they come into the answer in an essential way. –  Tom Goodwillie Jul 6 '12 at 12:24

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