If G is a finite group, I understand that the category of RO(G)-graded spectra, when rationalized, becomes Quillen equivalent to the category of Mackey functors valued in chain complexes of rational vector spaces.

How does RO(G) act on the category of Mackey functors? For instance, if F = F(G/H) is a Mackey functor and V is a real representation of G, what is (S^V F)(G/H)?

  • $\begingroup$ In characteristic zero the category of Mackey functors is semisimple. The simple Mackey functors over any field have been classified by Thevenaz and Webb -- they are one-to-one with conjugacy classes of pairs (subgroup $H$, irrep $L$ of $N_G(H)/H$). I think a sphere $\,\, S^V$ acts on the simple Mackey functor corresponding to $(H,L)$ by a shift, but a shift that depends on $H$: the dimension of the fixed-point space $V^H$. $\endgroup$ Commented Nov 6, 2014 at 1:54

2 Answers 2


The term ``RO(G)-graded spectrum'' is a misnomer, not to be used. It is never used in the literature I'm familiar with, and it never should be. There are several Quillen equivalent models for the category of genuine $G$-spectra, and such animals represent $RO(G)$-graded cohomology theories, but it is meaningless to think of them as $RO(G)$-graded themselves. They are not.

The zeroth stable homotopy group system of the sphere $G$-spectrum (take $H$-fixed point spectra and then take $\pi_0$ of those) is the Burnside ring Mackey functor $\mathbf{A}(G)$. It is a Green functor (think of it as a ring in the category of Mackey functors), and any Mackey functor is a module over it. That is where the action takes place. The category of rational $G$-spectra, $G$-finite, is equivalent to the category of chain complexes in the category of rational Mackey functors.

Via permutation representations, we get a map of Green functors from $\mathbf{A}(G)$ to the representation ring Mackey functor $\mathbf{RO}(G)$ that sends $G/H$ to $RO(H)$. That is one way that $RO(G)$ enters the picture.

In equivariant cohomology theory, the use of $RO(G)$ is a great convenience, but it is not the thing most intrinsic to the mathematics. Ignore the multiplication on $RO(G)$, which is irrelevant to its use for grading theories, and think of it just as an abelian group. Send a representation $V$ to the isomorphism class of the suspension $G$-spectrum of the one-point compactification $S^V$. This induces a homomorphism from $RO(G)$ into the Picard group $Pic(Ho G\mathcal S)$ of the stable homotopy category of $G$-spectra, namely the abelian group of equivalence classes of $G$-spectra that are invertible under the smash product. That homomorphism is neither a monomorphism nor an epimorphism. See http://www.math.uchicago.edu/~may/PAPERS/FLMJan01.pdf for a discussion of that Picard group. Logically, equivariant cohomology theories really should be graded on $Pic(Ho G\mathcal S)$, but that group is much less convenient than $RO(G)$.

  • $\begingroup$ Thanks Peter, sorry for the malapropism. I think my question still makes sense: how does RO(G) act on the category of chain complexes of rational-vector-space-valued Mackey functors? How to give a formula for (S^V F)(G/H) in terms of F? $\endgroup$
    – KHBG
    Commented Jul 27, 2014 at 2:20
  • $\begingroup$ Delete the first use of ``how''. No meaning, no formula. $\endgroup$
    – Peter May
    Commented Jul 27, 2014 at 19:03
  • $\begingroup$ If E is a rational G-spectrum, is its smash product with a representation sphere not also a rational G-spectrum? If that's a meaningful operation, I would like to know a formula for it in terms of Mackey functors. $\endgroup$
    – KHBG
    Commented Jul 27, 2014 at 19:39
  • $\begingroup$ Not an operation of RO(G). You've got to learn the math. $\endgroup$
    – Peter May
    Commented Jul 28, 2014 at 0:26
  • 1
    $\begingroup$ I think you are still missing the point: many inequivalent representations have homotopy equivalent S^V: that is, as I said in the first place, the map from RO(G) to Pic is neither injective or surjective. This is not something you expect to calculate naively. $\endgroup$
    – Peter May
    Commented Jul 28, 2014 at 1:49

Although I agree with Peter's comments, I believe I can add a few helpful comments of my own. First for $G$ finite, every rational Mackey functor is both injective and projective, so chain complexes are weakly equivalent to their homology. One can find this statement in appendix A of Greenlees-May 'Generalized Tate Cohomology.'

Given a representation sphere $S^V$ and a graded Mackey functor $M$ with associated Eilenberg-MacLane spectrum $HM$ we have an associated $G$-spectrum $S^V\wedge HM$ which represents the integer graded Bredon cohomology theory $$ X\mapsto H^{*-V}_{(-)}(X;M ).$$ Because rational equivariant cohomology theories are ordinary there is an equivalence $$S^V\wedge HM\simeq H(\pi_*^{(-)}(S^V\wedge HM)).$$ Hence the representation sphere takes a graded Mackey functor to another graded Mackey functor, namely the Bredon homology of $S^V$ with coefficients in $M$.

A similar argument shows that $H^{*+V}_{(-)}(X;M)$ is represented by $F(S^V,HM)\simeq D(S^V)\wedge HM$ which is again ordinary. Here $DX=F(X,S)$ is the equivariant Spanier-Whitehead dual of $X$.

If you like, after fixing $M$ you can extend this into a functor from the subcategory of the homotopy category spanned by the representation spheres, their Spanier-Whitehead duals, and their smash products to the category of graded Mackey functors. One can try to take a skeleton of this category to obtain an '$RO(G)$-graded' functor, but this involves keeping track of the automorphisms and solving a coherence problem. My understanding is that this coherence problem can be solved, but that it involves choices (which never seem to be specified).

As Peter pointed out the representation spheres are only distinguished by the fact that they are invertible, well understood, and play a pivotal role in the duality theory for $G$-manifolds. The canonical choice from a homotopical perspective would be to think of the Picard group of invertible objects instead.


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