The following was posted to math.stackexchange to no avail: <https://math.stackexchange.com/questions/908756/an-exercise-in-homology-computation-what-is-the-geometric-fixed-points-of-an-e>


The question I want to ask has a reasonably elementary formulation and I think there is a good chance it can be answered in this form (by someone more computationally skilled than me, or perhaps by someone recognising the construction). It is, however, motivated by equivariant stable homotopy theory and certainly insights can be drawn from this picture as well. I'm going to state the elementary form first, and then explain background and motivation.


For a pointed simplicial set $X\_\bullet$ I shall write $\tilde{\mathbb{Z}}[X]$ for the chain complex with $\tilde{\mathbb{Z}}[X]\_n = \mathbb{Z}\{X\_n\}/\*$ by which I mean the free abelian group on $X\_n,$ except that we treat the base point as zero. Now let $G$ be a finite group. I write $X^{\wedge G}$ for the simplicial $G$-set $X \wedge X \wedge \dots \wedge X$ ($|G|$ copies of $X$) with $G$ acting by permuting factors. Note that if $X\_\bullet$ is a pointed simplicial $G$-set then $\tilde{\mathbb{Z}}[X]$ has a natural $G$-action. Finally let $S^n = \Delta^n/\partial \Delta^n$ be the $n$-sphere (naturally pointed!).


**Question:** What is $H\_{n,k} := H\_{n+k}\left( \left[\tilde{\mathbb{Z}}[(S^n)^{\wedge G} \right]^G \right)$? Here the outer-most $[]^G$ means $G$-invariants in the chain complex, and $H\_\*$ is just homology groups.


One may show that there is a stabilisation map $H\_{n,k} \to H\_{n+1,k}$. Write $H\_{\infty, k}$ for the colimit. This is the group I'm really after.


**Some background:** Recall that there is a tensor triangulated category $SH(G)$, the $G$-equivariant stable homotopy category. This has two structures I am interested in here: firstly there is a tensor triangulated functor $\Phi^G: SH(G) \to SH$, the *geometric fixed points* functor. Secondly, the category has a natural $t$-structure whose heart is equivalent to the category of mackey functors. If we write $\underline{\mathbb{Z}}$ for the constant mackey functor with value $\mathbb{Z}$ and $H\underline{\mathbb{Z}}$ for the corresponding Eilenberg-MacLane spectrum, then what I would really like to understand is $\Phi^G(H\underline{\mathbb{Z}}).$ If I interpret correctly example 2.13 and paragraph 7.3 in [1], then $H\_{\infty, k} = \pi\_k \Phi^G(H\underline{\mathbb{Z}})$, hence my question.


**Some results:** Let $H$ be a proper subgroup of $G.$ Then $\Phi^G (G/H\_+ \wedge S\_G) = 0$. The transfer $H\underline{\mathbb{Z}} \to G/H\_+ \wedge H\underline{\mathbb{Z}} \to H\underline{\mathbb{Z}}$ is multiplication by $|G:H|.$ Hence multiplication by $|G:H|$ is zero on $\Phi^G(H\underline{\mathbb{Z}})$ and so also on $H\_{\infty, k}.$ In particular $H\_{\infty, k} = 0$ unless $G$ is a $p$-group, in which case $H\_{\infty, k}$ is $p$-torsion.


One may work out $\tilde{\mathbb{Z}}[(S^n)^{\wedge G}]$ fairly explicitly, but the combinatorics of the resulting chain complex is a big mess. Using some standard results about lifting homotopies, one may prove that in the definition of $H\_{n,k}$ we can replace $\left[\tilde{\mathbb{Z}}[(S^n)^{\wedge G} \right]^G$ by $(C\_\bullet^{\otimes G})^G$ for any simplicial free abelian group $C\_\bullet$ with $H\_\* C = H\_\* S^n,$ for example the Dold-Kan inverse to the chain complex $\mathbb{Z}[n]$, i.e. $C\_k = 0$ for $k < n$, $C\_n = \mathbb{Z}$ and the higher degree groups are just freely added degeneracies. Hence in particular $H\_{n,k} = 0$ for $k < 0.$ But the resulting complex still does not seem particularly amenable to computation (it is related to the orbits of $G$ on $G$-sets of the form $X^{\times G}$ where $X$ has trivial $G$ action). I believe I can use this description to show that $H\_{n,0} = \mathbb{Z}/p.$


Some wishful thinking led me to ask in the stackexchange post if in fact $H\_{\infty, k} = 0$ for all $k > 0.$ I'm thinking now that this may be wrong. In fact using paper [2] on "derived mackey functors" I seem to have convinced myself that for $G = C\_2$ (the unique group of order two), $H\_{\infty,\*} = \mathbb{Z}/2[t]$ where $t$ is placed in degre $2$ (i.e. the stabilised homology is two-periodic). Better take this computation with a grain of salt, though.


[1] <http://www.math.uni-bonn.de/people/schwede/equivariant.pdf>


[2] <http://arxiv.org/abs/0812.2519>



What is the geometric fixed points of an (equivariant) Eilenberg Maclane Spectrum?
Your identification with the geometric fixed points and the calculation for $C\_2$ is correct. As Akhil remarked, one obtains the same answer for $C\_{2^n}$ and this is calculated in Hill-Hopkins-Ravenel. In fact, for $G=C\_{p^n}$, the same method shows $\Phi^G H\mathbb{Z}=\mathbb{Z}/p[t]$ where $|t|=2$. The key ingredients of this calculation are: 


1. All proper subgroups of $G$ are contained in the kernel of a non-trivial map $$\psi\colon C\_{p^n}\rightarrow C\_p.$$
2. The geometric fixed points $\pi\_\* \Phi^{C\_p}H\mathbb{Z}=\pi\_\*^GE\tilde{P}\wedge H\mathbb{Z}$ can be calculated as $H\_\*^{C\_p}(S^{\infty \rho};\mathbb{Z})$. Here $\rho$ is a non-trivial one dimensional complex representation of $C\_p$. This is because $S^{\infty\rho}\simeq E\tilde{P}\_{C\_p}$, (which follows from calculating the fixed points and that $E\tilde{P}\_{C\_p}$ is determined up to weak equivalence by its fixed points).
3. By pulling back along $\psi$ and applying 1. We see that $\psi^\*S^{\infty \rho}\simeq E\tilde{P}\_{C\_{p^n}}$ and because we are using *constant* Mackey functor coefficients we have $H\_\*^{C\_{p^n}}(\psi^\*S^{\infty\rho};\mathbb{Z})\cong H\_\*^{C\_p}(S^{\infty \rho};\mathbb{Z})$. This follows explicitly from the characterization of the homology you have given.
4. The unit sphere space $S(\infty\rho)$ is a model for $EC\_p$ (again look at the fixed points) and $H\_\*^{C\_p}(S(\infty\rho)\_+;\mathbb{Z})\cong H\_\*(BC\_p;\mathbb{Z})$. This follows from the definition of the Mackey functor homology as a coend (See May: Equivariant homotopy and cohomology theory).
5. Finally the cofiber sequence $$S(\infty\rho)\_+ \rightarrow S^0\rightarrow S^{\infty\rho}$$ induces a long exact sequence in homology. In degree 0 the left hand map induces a transfer map on chain complexes, because it is induced by a projection map $C\_p\rightarrow C\_p/C\_p$. The transfer induces multiplication by the index $$\mathbb{Z}\xrightarrow{|C\_p|}\mathbb{Z}$$ and we see that $\pi\_0 \Phi^G H\mathbb{Z}$ is $\mathbb{Z}/p$.


These arguments determine the graded group structure. To get the ring structure one analyzes what happens in homology when you smash copies of $S^\rho$ together: the orientation class $t$ in $H\_2^{C\_p}(S^\rho;\mathbb{Z})$ gives a polynomial generator. 


For more general groups, I can say the following. For non-trivial $p$-groups one can apply a similar argument but one needs to make some changes. First replace $\rho$ with $V$ where $V=\oplus\_{i\in I} V\_i$ is a direct sum of representations indexed over the set $I$ of non-trivial maps $G\rightarrow C\_p$. The representation $V\_i$ is obtained by pulling back the representation $\rho$ from above along the corresponding map. If we use all such maps then $S^{\infty V}$ is a model for $E\tilde{P}$. The calculation of the degree 0 term above can again be applied to show that $\pi\_0 \Phi^G H\mathbb{Z}=\mathbb{Z}/p$. 


More generally I believe one gets $\pi\_0 \Phi^G HR=R(G)/(\mathrm{Ind}\_H^G R(H))\_{H<G}$ which is consistent with your stated results. Here I mean quotient by the ideal generated by elements which are induced up from proper subgroups. You can see that this is true if $HR=S$ and $R(H):=\pi\_0^HS=A(H)$. It then follows for $HR=HA=P^0S$, the Burnside Mackey functor (which is the first Postnikov section of the sphere), by the Hurewicz theorem. Finally this implies it for general $HR$ by universal coefficients (which always works for $H\_0$). 


If $G$ is non-trivial and not a $p$-group, and $R$ is a constant Green functor (such as $\mathbb{Z}$) then 1 is in the ideal $(\mathrm{Ind}\_H^G R(H))\_{H<G}=(|G/H|)\_{H<G}$ so the bottom homology group is 0. Since this is a graded ring the geometric fixed points are trivial (as you already observed).


