I know very little about the fancy equivariant stable homotopy category, so I apologize if this question is silly for one reason or another, but:

I think that stable homotopy, in the non-equivariant case, corresponds to the homology theory associated to the sphere spectrum. And indeed, in the equivariant context we have $RO(G)$-graded homology and cohomology theories for every spectrum. On the other hand, equivariant stable homotopy groups are defined by looking at maps, not from representation spheres, but from spheres smashed with homogeneous spaces. The resulting object has a sort of quirky bigrading, it seems to me, over the integers and subgroups of $G$.

Is there some reason that this choice is more natural than defining homotopy groups as $\pi_{\nu}(E) = [S^{\nu}, E]_G$ for $\nu \in RO(G)$? Is this object somehow trivial or something?

If the answer to this question is that I'm missing something easy or that the objects I gave above are trivial for some reason- then let me know but don't tell me the answer! It'll probably be good for me to figure that out on my own...


If $G$ acts linearly and isometrically on $V$ then the unit disk $D(V)$ and its boundary $S(V)$ are an appealing kind of equivariant disk and sphere. You can try taking the pairs $(D(V),S(V))$ as the building blocks for equivariant complexes, but you won't get all the equivariant homotopy types that you want, not even the manifolds with smooth $G$-action.

You will get them all if you consider also representations of subgroups. If $H$ acts on $V$ then look at $(G\times_HD(V),G\times_HS(V))$, where $G\times_HX$ means the quotient of $G\times X$ with $(g,hx)$ identified with $(gh,x)$. (This converts an $H$-action into a $G$-action.)

On the other hand, you also get everything you want if you only use trivial representations of arbitrary subgroups rather than arbitrary representations of arbitrary subgroups.

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    $\begingroup$ This explanation definitely convinced me. I want G-manifolds! (And I don't want to pay more for them.) $\endgroup$ – Dylan Wilson May 17 '11 at 16:10
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    $\begingroup$ The question was about the stable theory ($G$-spectra). The answer was about $G$-spaces. It applies to $G$-spectra, too, once the right definitions have made. Of course, representation spheres occur in the stable theory in another way, too. Nonequivariantly spheres play two different central roles in homotopy, don't they? On the one hand, they lead to cell complexes, and on the other hand smashing with spheres leads to spectra. Equivariantly the $S^n\wedge G/H_+)$ play the first role while the representation spheres play the second role. $\endgroup$ – Tom Goodwillie May 17 '11 at 16:58
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    $\begingroup$ Right, this is sort of what I expected the answer to be related to; I guess I just saw "Hey that 'n' got generalized to a 'V' since integers should mean dimensions of vector spaces... so why didn't that other 'n' get generalized to a 'V'?" and I suppose you're answer is "Because that 'n' means a different thing than the other 'n' because it comes from using spheres in a different way." $\endgroup$ – Dylan Wilson May 17 '11 at 22:24
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    $\begingroup$ I'm glad you asked the question. Answering it clarified things for me a little. $\endgroup$ – Tom Goodwillie May 17 '11 at 23:21

Actually, this is a subtle and interesting question, and it is to some extent model dependent. When one defines the integer graded homotopy groups, one does so using colimits over representation spheres, so that ``RO(G)-graded information'' is encoded in the definition of homotopy groups. This is why one can get away with defining weak equivalence in terms of the Z-graded homotopy groups of all fixed point spectra.

As Justin says, $RO(G)$-graded homotopy groups are of definite interest. Calculating the integer part of $RO(G)$ graded cohomology theories obscures structure that is only visible in the $RO(G)$ graded world. But then the coefficients of the theory are the $RO(G)$-graded homotopy groups of the representing spectrum.

The cells that build up $G$-spectra in the model theoretic sense are model dependent. Using symmetric or orthogonal $G$-spectra, one must use cells of the form $G\times_H D^V$, and one can restrict to G-representations. In Lewis-May or EKMM $G$-spectra, where all objects are fibrant, one can use just cells $G/H_+\wedge D^n$. Here again, the $RO(G)$-graded information is built into these $G$-spectra, which is why integer grading suffices.

In Tom Goodwillie's space level answer the notion of ``get'' needs interpretation. One can get every weak homotopy type of $G$-space using $G$-CW complexes defined using only cells of the form $G/H \times D^n$, no representations in sight. Intuitively, as Justin says, this works because of the classical theorem that representation spheres are themselves finite $G$-CW complexes in this sense, but that is irrelevant to the proof. If one wants only finitely many cells to get a smooth $G$-manifold, then one must use representations of all subgroups, as Tom says (and even then I don't know a good reference). EDIT added: This last sentence is nonsense. I'll quote from a Math Review:

Illman, S\"oren The equivariant triangulation theorem for actions of compact Lie groups. Math. Ann. 262 (1983), no. 4, 487–501.

``Let $G$ be a compact Lie group. The main results of the paper are the following. Theorem 7.1: Let $M$ be a smooth $G$-manifold with or without boundary. Then there exists an equivariant triangulation of $M$. Corollary 7.2: Let $M$ be a smooth $G$-manifold with or without boundary. Then $M$ can be given an equivariant CW complex structure. The other results of the paper are the technical preparation for the above results. The paper is written very carefully and has a detailed history of results leading to the above ones.''

No representation spheres are needed. Where one needs representation spheres is to construct dual triangulations of smooth compact manifolds and for work in $RO(G)$-graded homology. See Chapter X, $G-CW(V)$ complexes and $RO(G)$-graded cohomology, by Stefan Waner, in Equivariant homotopy and cohomology theory (CBMS Conf. Series 91, AMS).

In any construction of the stable category of $G$-spectra, one wants to invert representation spheres; for one of many reasons, as Justin says, Poincare' duality requires that. However, when one inverts representation spheres, one actually inverts all homotopy spheres as well, whether or not one wants to. See (Fausk, Lewis, May. The Picard group of the equivariant stable homotopy category. Advances in Math 163(2001), 17--33). Philosophically, one might want to define homotopy groups using the entire Picard group of invertible objects, but that is calculationally unwise, to put it mildly. One can and should feel free to use either $Z$-graded or $RO(G)$-graded homotopy groups, adapting the choice to the application at hand.


Typically, you want something like Whitehead's theorem, which should say something like:

  • If $f: X\to Y$ is a map of "cell complexes" which induces an isomorphism on "homotopy groups", then it is a homotopy equivalence.

In the category of $G$-equivariant spaces, a "cell complex" is something built by iteratively attaching a $G$-space of the form $G/H\times D^n$ along the subspace $G/H \times S^{n-1}$, where $n\geq0$; in the stable world, an analogous statement is true, except you use smash product instead of products, and now you can allow $n$ to be any integer. This choice of "cell" determines the notion of "homotopy group" which should appear in the analogue of Whitehead's theorem.

Of course, you can still define a notion of $RO(G)$-graded homotopy groups. In which case, you might ask if it's possible to use $RO(G)$-graded homotopy groups in place of "cell"-homotopy groups in the statement of Whitehead's theorem? I don't know off the top of my head.

  • $\begingroup$ Are these cells somehow more natural than some analog with representation spheres? Although I guess it's not obvious what the analog of the disk would be? $\endgroup$ – Dylan Wilson May 16 '11 at 22:44
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    $\begingroup$ They are a natural thing to turn to because when you start looking at some obvious classes of $G$-spaces such as those admitting an ordinary CW structure in which the $G$-action permutes the cells, or smooth manifolds with a smooth $G$-action, they tend to be things you can build up in the way Charles describes. $\endgroup$ – Tom Goodwillie May 17 '11 at 0:11
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    $\begingroup$ Note that the basic example $G_+$ (i.e. $(G/H\times D^n)/(G/H\times S^{n-1})$ in the case $H=1$, $n=0$) admits no nontrivial maps from any representation sphere if $G$ is nontrivial. In the stable setting it's not quite that easy to see that representation spheres are inadequate for detecting weak equivalences. They are adequate when $G$ has order $2$ because the homotopy cofiber of $G_+\to S^0$ is a representation sphere. But things go wrong already when $G$ has order $3$. $\endgroup$ – Tom Goodwillie May 17 '11 at 0:17

I would like to add a few points:

  1. You can define $RO(G)$-graded homotopy groups of $G$-spectra, see for example Stefan Schwede's course notes on equivariant homotopy theory.

  2. These groups are interesting. For example the inclusion of the 0-skeleton $S^0\rightarrow S^{\sigma}$ of the sign representation stabilizes to a non-trivial (and non-nilpotent!) element in $\pi_0^{C_2} S^{\sigma}\cong \pi_{-\sigma}^{C_2} S^0.$ This can be proved by showing its Hurewicz image in Bredon homology is non-trivial.

  3. At least for finite groups $G$-representation spheres can be triangulated and constructed out of the standard $G$-cells ($G/H_+\wedge S^n,$ $G/H_+\wedge D^n$). So ordinary equivariant weak equivalences between cell complexes induce $RO(G)$-graded equivalences.

  4. One reason to put representation spheres into your theory is so that you have a reasonable form of equivariant Spanier-Whitehead duality. At the very least we want finite G-sets to be dualizable, so we want them to equivariantly embed into a G-sphere, which should be one of the spaces in the sphere spectrum. This will not work if we only allow trivial spheres.


A different perspective is that of Elmendorf's theorem: the homotopy theory of G-equivariant spaces (in the sense usually meant) is equivalent to the homotopy theory of diagrams on the orbit category $\mathcal{O}_G$ of G. Here we take the homotopy theory of diagrams to be in the obvious sense that we have a homotopy theory of diagrams on any category: the weak equivalences are the objectwise ones. If X is a G-space, the corresponding diagram on $\mathcal{O}_G$ sends an orbit G/H to the fixed-point space $X^H$, which is morally why this equivalence holds, since weak equivalences of G-spaces are also defined fixed-point-space-wise.

However, in the homotopy theory of diagrams on any category C, there is a standard notion of cell complex, in which cells are attached "at objects of C". More precisely, the generating cofibrations of diagrams are obtained from the standard ones $S^{n-1} \to D^n$ by applying the left adjoint of "evaluate at x", for various $x\in C$. Passing back across the equivalence to G-spaces, these cells correspond to those of the form $(G/H)_+ \wedge D^n$.

From this perspective, it is more mysterious why representation spheres appear at all!

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    $\begingroup$ I don't think that there is any reason why representation spheres have to appear in the foundations of a homotopy theory of $G$-spaces. They occur naturally when you look at $G$-manifolds. For $G$-spectra they are usually built into the definition, and I imagine there is more than good viewpoint as to why they should. $\endgroup$ – Tom Goodwillie May 18 '11 at 13:09
  • $\begingroup$ The point should also be made that the obvious analogue of Elmendorf's theorem for spectra does not hold for the "fancy" theory. $\endgroup$ – Tom Goodwillie May 18 '11 at 13:09

Weak equivalences in equivariant homotopy theory are equivariant maps $f\colon X\rightarrow Y$ such that for any subgroup $H\subset G$ the map between subspaces of fixed points $f^H\colon X^H\rightarrow Y^H$ is a weak equivalence in the classical sense. Equivariant homotopy groups detect these weak equivalences, this is why they are used.

  • $\begingroup$ While this is a good answer it sounds somewhat circular in my head- what's so bad about using the weak equivalences that come from the $RO(G)$-graded version? $\endgroup$ – Dylan Wilson May 16 '11 at 22:42
  • $\begingroup$ @Dylan: it need not be bad, maybe they are also the weak equivalences of another model category structure, I don't know, do you have any particular interest in it? If so, it could be worth to develop. $\endgroup$ – Fernando Muro May 17 '11 at 10:14

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