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I am new to this subject; so please correct me if I will say something wrong or if you don't like my notation. In particular, I don't know whether it is reasonable to consider an infinite group $G$ (should it actually be a compact Lie group?) in my question. Also, for a group $G$ the corresponding stable homotopy category $SH_G$ (should I use another notation?) also depends on the choice of a universe; does one get a canonical category taking a complete universe?

As far as I understand equivariant stable homotopy theory, in $SH_G$ one should take spectra $\Sigma_G (G/H)_+$ corresponding to all homogenious spaces $G/H$ to obtain a family of compact generators. Now, I want to know something on morphisms between these generators. In particular, is the group $Mor_{SH_G} (\Sigma_G (G/H_1)_+, \Sigma_G (G/H_2)_+[i])$ zero for any $i>0$ and all subgroups of $G$? This question seems to be related to A heart for stable equivariant homotopy theory Is there any text where I can find results of this sort? Possibly, one can say something about these morphism groups (say) using duality and apply some well-known results after that?

Any hints would be very welcome!

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    $\begingroup$ One of the basic results is the tom Dieck splitting that allows you to reduce the computation of equivariant homotopy groups of suspension spectra to the ordinary homotopy groups of different spectra. I think it is proven in Lewis-May-Steinberg but there are probably several references $\endgroup$ Jul 25, 2016 at 7:49

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Since Denis gave the right reference, namely http://www.math.uchicago.edu/~may/BOOKS/equi.pdf, I did not follow up and answer this question. We can work with any compact Lie group and any complete universe. Working in the equivariant stable category always, with $S = S^0$, $S^{-n}$ a negative sphere $G$-spectrum and $S^0$ and $S^n$ sphere spaces with trivial $G$-action, $$ \pi_{-n}(S) = [S^{-n},S]_G = [S,\Sigma^{\infty} S^n]_G = [S^0, QS^n]_G = 0 $$ where $Q^G$ is the functor $(colim_V \Omega^V \Sigma^V(-))^G$, with $V$ running through representations of $G$; $Q^G(X)$ is an $(n-1)$-connected $G$-space if $X$ is so. Replacing $S$ by $\Sigma^{\infty} G/H_+$, the same argument works, using change of groups so that one is now looking at $[-,-]_H$. More generally, the negative homotopy groups of the suspension $G$-spectrum of any $G$-space are zero, and the non-negative groups are calculated by the tom Dieck splitting theorem.

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  • $\begingroup$ Thank you very much! So, everything "works in the way I want it to"; I only have to read you book thoroughfully. $\endgroup$ Jul 31, 2016 at 14:41
  • $\begingroup$ Yeah, I'm sorry it is so lengthy, and pre tex. $\endgroup$
    – Peter May
    Aug 1, 2016 at 1:06
  • $\begingroup$ So, I have had a look at you book, and I have found references that give the vanishing in question. However, I believe that for $i=0$ the corresponding groups should be related to the Burnside ring of $G$; still I wasn't able to find any reference that gives this statement. I would be deeply grateful for any advice here! $\endgroup$ Jun 20, 2018 at 8:32
  • $\begingroup$ @Mikhail Bondarko. This is discussed in Section V.2 of the reference in my answer. Just click! For finite $G$, Segal first noticed that $\pi_0(S)$ is the Burnside ring of $G$. For compact Lie groups, tom Dieck defined the Burnside ring of $G$ in such a way that the same conclusion holds. For finite $G$, much more is true. Running through $\pi_0$ of the suspension $G$-spectra of orbits $G/H$ (with disjoint basepoints), one obtains the Burnside ring Mackey functor, and more recent work shows that there are multiplicative norms so that one actually obtains the Burnside ring Tambara functor. $\endgroup$
    – Peter May
    Jul 4, 2018 at 3:06
  • $\begingroup$ Thank you very much indeed! It turns out that the the Burnside rings is not quite what I want. I am rather interested in the direct sum of all $Mor_{SH_G} (\Sigma_G (G/H_1)_+, \Sigma_G (G/H_2)_+)$ for $H_i$ running through all subgroups of $G$. This is a ring (with a unit if $G$ is finite) with multiplication given by compositions of morphisms, and I wonder whether anybody has considered it previously. $\endgroup$ Jul 31, 2018 at 10:11

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