# A heart for stable equivariant homotopy theory

Let $G$ be a finite group. I wonder whether the following statement is true, known and written down:

There is a t-structure on the stable $G$-equivariant homotopy category such that the associated heart is isomorphic to the category of Mackey functors (on $B_G$).

I feel like someone has told me so, but I can't find a reference. Thanks for your help!

• Could you define "t-structure" and "heart"? – Dev Sinha Aug 23 '10 at 18:06
• @Dev: en.wikipedia.org/wiki/Triangulated_category#t-structures I am pretty sure that the answer to your question is "yes", with truncation functors given by connective covers, but my references aren't handy. – Tyler Lawson Aug 23 '10 at 19:16
• The answer is certainly "yes," but the only proof I know off the top of m head is that you first show that the homotopy category of G-equivariant spectra is equivalent to the homotopy category of "spectral Mackey functors," where the t-structure is easy to write down. (In fact this is an equivalence of infty-categories; I'm still writing up these details.) Presumably there's a more direct proof; I seem to remember some appendix of Gaunce Lewis ... Sorry I can't be of more help! – Clark Barwick Aug 26 '10 at 16:16

Since G is finite, there is no problem with just repeating the proof in the case $G=e$, using $Z$-graded homotopy group functors on the orbit category. Take $D^{\leq n}$ to be the spectra whose homotopy groups $\pi_q(X^H)$ are zero for $q>n$, and dually for $D^{\geq n}$. The intersection for $n\leq 0$ and $n\geq 0$ consists of the Eilenberg-MacLane $G$-spectra $K(M,0)$ for Mackey functors $M$.
• Sorry, can you explan why the couple $(D^{\le 0}, D^{\ge 0})$ yields a t-structure? Surely, abstract nonsense yields the existence of a t-structure whose non-negative part is precisely $D^{\ge 0}$; should one apply some "standard topological argument" to compute $D^{\le 0}$ then? – Mikhail Bondarko Jul 29 '16 at 19:29
• The proof is identical as for $G=e$, once you understand $G$-CW spectra, for which the source is math.uchicago.edu/~may/BOOKS/equi.pdf. The point is that they satisfy the same formal properties used in a standard proof for $G=e$. I can add full details, but I hope that is not necessary. – Peter May Jul 30 '16 at 18:00
• Ok, since I thought I had already answered this question, I've gone back and answered your other question mathoverflow.net/questions/245006/…. The negative homotopy groups of any suspension $G$-spectrum are zero. My long answer to mathoverflow.net/questions/65180/… is also relevant. – Peter May Jul 30 '16 at 23:34