Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to learn a bit about equivariant homotopy theory. Let G be a compact Lie group. I guess there is a cofibrantly generated model category whose objects are (compactly generated weak Hausdorff or whatever) topological spaces with G-action and whose morphisms are G-maps, in which the generating cofibrations are maps of the form G/H x Sn-1 → G/H x Dn (n ≥ 0, H a closed subset of G) and the generating acyclic cofibrations are the obvious analogous thing. Apparently the weak equivalences in this category are those maps which induce weak equivalence on H-fixed points for every closed subgroup H of G. I assume the corresponding (∞,1)-category is presentable. (My preliminary question is, does anyone know a good source for this paragraph?)

My real question is: Can you give an (∞,1)-categorical description of this category, say via a universal property, or built somehow from the category of spaces? For instance, what is an explicit presentation as a localization of a category of presheaves of spaces? (An example of the kind of answer I am looking for is "functors from BG to Spaces", but that describes a model category of G-spaces whose weak equivalences are simply weak equivalences of the underlying spaces.)

(My next question would be asking for an analogous description of the equivariant stable homotopy category. I imagine this would be easy if I knew how to answer the first question, but if something special happens in the stable situation, I would like to know about it.)

share|improve this question
    
This reminds me that I had the problem tagging the posts about infinity categories. I think your tag is actually better than higher-category-theory, but perhaps there could be other suggestions as well? –  Ilya Nikokoshev Oct 28 '09 at 23:33
1  
I liked both these answers a lot, so I flipped a coin to pick one to accept. –  Reid Barton Oct 28 '09 at 23:59
    
@Ilya: I wanted to tag this (∞,1)-categories but of course that wouldn't work (the system changed it to 1-categories :P) so I adopted Jacob Lurie's abuse of terminology. I think of higher-category-theory as an related but distinct area, having to do with 2-categories and such. –  Reid Barton Oct 29 '09 at 0:07
2  
I don't think the question, has yet been anwered, below: what is the really natural (oo,1)-categorical way to understand genuine G-spectra? While it is true that one answer is: stabilization of oo-presheaves on the orbit category at spehers with a G-action, this doesn't look like the general insightful way of looking at it that one might hope for. I am guessing the answer must involve constructions as in motivic cohomology, where we start with a big oo-topos of oo-sheaves and then stabilize it both with respect to categorical as well as geometric spheres. I'd love to see such kind of answer –  Urs Schreiber Jan 28 '10 at 15:06
    
@Urs: Agreed. My question here was primarily about the unstable case, and I got a sufficient description of the genuine stable category for my purposes through off-line conversations. But I am also interested in the question you raise; I suggest you ask it as a separate question. I have heard that in both the equivariant and motivic situations, the "extra" stabilizations provide a better theory of duality, but I do not know whether this can be taken as a characterization of the resulting category, or merely a post hoc justification for its study. –  Reid Barton Jan 28 '10 at 15:29

4 Answers 4

up vote 12 down vote accepted

I think a good reference for the first paragraph is "Equivariant Homotopy and Cohomology Theory" by Peter May and a bunch of other people. Chapter 5 includes "Elmendorf's theorem" that this homotopy theory of G-spaces is equivalent to the homotopy theory of diagrams of spaces on the orbit category O(G) of G. In the latter homotopy theory, the weak equivalences are "levelwise" as is usual in the homotopy theory of diagrams.

I'm less sure about the (∞,1)-categorical versions, but I would expect that the (∞,1)-category associated to a levelwise model structure on O(G)-diagrams will be essentially the (∞,1)-category of functors from O(G) to the (∞,1)-category of spaces. That ought to imply that it is locally presentable as well.

One might guess that the equivariant stable homotopy category would be the "stabilization" of this (∞,1)-category, but that's not entirely obvious to me. The point at issue is that there are two kinds of G-spectra: "naive" G-spectra, which are indexed on integers, and "true" G-spectra, which are indexed on G-representations. It seems possible to me that the standard "stabilization" process of an (∞,1)-category will only stabilize with respect to integers.

share|improve this answer
1  
From the description of G-spaces as functors from O(G) to Spaces, it follows by generalities about presentable (∞,1)-categories that its stabilization is the category of functors from O(G) to Spectra. This seems unlikely to be the same as the category of naive G-spectra (which I believe is functors from BG to Spectra) so I guess it's probably the category of true G-spectra...? –  Reid Barton Oct 29 '09 at 0:02
5  
This always confuses me, since I know three kinds of G-spectra, I can never remember which one is supposed to be naive: (1) Ordinary equipped with a G-action. These are presheaves of spectra over BG. (2) G-spectra indexed on the universe built from the trivial representation: these are spectrum valued presheaves on O(G). (3) G-spectra indexed on a complete universe: these are something else ... –  Charles Rezk Oct 29 '09 at 0:16
    
(1) is supposed to read "Ordinary spectra equipped ...". –  Charles Rezk Oct 29 '09 at 0:26
    
Oh, hmm. I have been (naively?) using "naive G-spectrum" to refer to (1), but it seems that May uses it to mean (2). And now I would like a (∞,1)-categorical description of (3)... Maybe another question is in order... –  Reid Barton Oct 29 '09 at 0:31
3  
To clarify for other non-experts like myself: the confusion between (1) and (2) probably arises from the fact that they are the homotopy theories of two model categories with the same underlying category, but different weak equivalences (the weak equivalences between Ω-spectra are in (1) levelwise equivalences of underlying spaces but in (2) levelwise equivariant equivalences). –  Reid Barton Oct 29 '09 at 3:40

I have been trying to hold myself back from answering this, because I am not entirely sure my view on this is accurate. To me, it seems like a G-spectrum should be a spectrum with an action of G on it, full stop. Obviously you have to specify your notion of spectrum, and obviously if you want to include topological groups G it had better be a symmetric monoidal category of spectra that is enriched over topological spaces. So you could take S-modules of EKMM or orthogonal spectra (my personal favorite) or symmetric spectra based on topological spaces. With any of these categories, there is a notion of a G-spectrum, by which I mean a spectrum with an action of G.

I can hear you objecting--you must be being too naive--what about complete G-universes? I take the point of view that picking a universe corresponds to picking a model structure on the one God-given category of G-spectra. Picking a smaller universe just means localizing the model structure. So the complete universe is the "initial" one, and every other universe is a localization of the complete universe. The naive universe is the "terminal" one, in the sense that it is a localization of every other universe. There are lots of universes corresponding to model structures in between these.

I cannot now remember how these model structures are supposed to go, but I believe that both Neil Strickland and Tony Elmendorf have separately written something about this approach. Tony's might be part of a joint paper, I can't remember. I think it is just a different way to look at things, but it goes so much against the prevailing viewpoint that it has not gotten so much traction.

Again, I have to confess that I am working from memory from something I probably did not completely understand. Possibly Mike Shulman or someone else will be able to convince me I am completely wrong.

share|improve this answer
    
Perhaps I'm not aware of the issues, but how could a localization provide a sequence of deloopings for representation spheres? –  Tyler Lawson Nov 23 '09 at 5:42
1  
I was afraid someone would ask this. So the answer is that I don't know and I might be wrong. However, S^V is a G-space, right, so can't I declare the map X --> Omega^V S^V X a weak equivalence? Here I am just using the enrichment of G-spectra over G-spaces. Of course, I would have to do this carefully to be sure I got a model structure. Again, I am not sure this is right. –  Mark Hovey Nov 23 '09 at 14:19
1  
I ... don't think so. If you did this to spaces, you'd be decreeing that the map S^0 -> N_+ is a weak equivalence, and this annihilates S^0. You want to invert the suspension operator instead, but if you invert maps X -> Y that become equivalences after suspension you are only picking out suspension spectra. –  Tyler Lawson Nov 23 '09 at 16:02
3  
Mark, the paper of mine you're remembering is the one joint with Peter May, "Algebras over equivariant sphere spectra", JPAA 116 (1997) 139-149. It shows that, indeed, you can consider just one underlying category of G-spectra with a number of (cofibrantly generated) model structures on it. The "change of universe" functors usually considered from the Lewis-May-Steinberger point of view are then simply the derived functors of the identity functor. The category itself is the EKMM category with G acting in the obvious way on the objects. -- Tony Elmendorf –  Tony Elmendorf May 18 '10 at 16:31

I would say that

a) cohomology is in any case something defined in some (oo,1)-topos (maybe secretly so, but still) -- details and further links on this point of view are at nLab:cohomology

b) from that general point of view there is a very general definition of equivariant cohomology, as indicated at nLab:equivariant cohomology

More precisely, this is (the generalization of) Borel equivariant cohomology . See the remark at A Survey of Elliptic Cohomology: equivariant cohomology - Borel equivariant cohomology.

share|improve this answer
    
That's a shame, because for some purposes the more interesting type of equivariant cohomology is Bredon cohomology, which is naturally RO(G)-graded and Mackey-functor-valued. Perhaps a more expansive notion of cohomology is in order? –  Mike Shulman Oct 29 '09 at 3:50
    
I doubt that the notion of cohomology as being the connected components in a hom-space of an (oo,1)-topos or stable (oo,1)-category is not expansive enough. I haven't looked into equivariant constructions with the orbit category, though. Would be interesting to figure out how to say that more generally. –  Urs Schreiber Oct 29 '09 at 8:49
    
Comment months after: this has meanwhile been sorted out. Details at on the nLab entry on equivariant cohomology: ncatlab.org/nlab/show/equivariant+cohomology –  Urs Schreiber Jan 28 '10 at 15:01

Let C be the category of homogeneous G-manifolds; the hom sets have a natural topology so you can consider C as an infinity-category. The equivariant homotopy category is the category of contravariant functors from C to the infinity-category of spaces. You build such a functor out of an honest G-space by restricting Hom_G(-,X) to C.

I think this answer is sort of disappointing: it says that all that algebraic topology can see in a G-space are the fixed point sets with respect to subgroups. What are the theorems along these lines that justify this definition?

According to the discussion below, the answer is Whitehead's theorem: any weak G-homotopy equivalence between tame enough G-spaces--at least, all G-CW complexes (Whitehead) and all smooth G-manifolds (Illman)--is a strong G-homotopy equivalence. "Weak" means that the map induces an isomorphism on homotopy groups of all fixed-point sets, and "strong" means that there's an equivariant map backwards so that the compositions are equivariantly homotopic to the identity maps.

share|improve this answer
    
I don't know why you would say that this is "all that algebraic topology can see in a G-space". Any G-space X can be reconstructed from its fixed point diagram O(G)->Spaces, so that diagram must keep all the information about X. –  Charles Rezk Oct 29 '09 at 0:22
    
I think you are recovering the homotopy type of X, not X itself. In fact you can recover, tautologically, the equivariant homotopy type of G->Aut(X) out of the diagram O(G)->Spaces. Is there some other natural way of saying "equivariant homotopy type" so that this tautology becomes an interesting theorem? –  David Treumann Oct 29 '09 at 0:41
1  
I believe I can actually recover X itself from its fixed point diagram. Of course, when I do so, I'm thinking "1-categorically", and not "infty-categorically", so perhaps that's not germane here ... I'm also not sure why it's a tautology. I thought the point was the equivariant Whitehead theorem: a map f:X->Y between G-CW complexes is an equivariant homotopy equivalence (i.e., there is G-map g:Y->X such that gf and fg are each homotopic to identity through G-maps) iff X^H -> Y^H is a weak equiv. of spaces for each subgroup H. –  Charles Rezk Oct 29 '09 at 0:58
    
Yes, another way to say it is that the model category--in particular, the class of weak equivalences--is determined by the generating cofibrations and acyclic cofibrations I wrote down; it doesn't seem tautological to me that the result is a diagram category. –  Reid Barton Oct 29 '09 at 1:09
1  
To make it more concrete: let C be a small category, Psh(C)= presheaves of spaces on C. Put a model category structure on Psh(C), where the generating cofibrations are built using all quotients of representable functors. Will this be equivalent to some other model category of presheaves? (In your case, Reid, C was the group G, and the quotients of representables are the G/H.) –  Charles Rezk Oct 29 '09 at 1:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.