I've seen a couple of similar questions asking to verify computations of Bredon cohomology here and here, so I will ask one such question myself.

Let $\mathbb{Z}/2$ act on $S^3\subset \mathbb{C}^2$ by restriction of a permutation action on $\mathbb{C}^2.$ I wanted to compute Bredon cohomology $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}}).$

I have a cell decomposition based on a decomposition of complex $1$-dimensional disk into $3$ cells: $\mathbb{D}=D\sqcup T\sqcup *.$ Here $T\sqcup *=S^1=\partial \mathbb{D}$ and $D$ is the interior of $\mathbb{D}.$ Then we have a decomposition of $S^3=\mathbb{D}\times S^1 \cup S^1\times \mathbb{D}$ into cells compatible with the $\mathbb{Z}/2$ action.

The fixed point set of an action is a circle given by $\{z_1=z_2\}\cap S^3\subset \mathbb{C}^2.$ Since the orbit category of $\mathbb{Z}/2$ consists of $*$ and $\mathbb{Z}/2$ there are the following equivariant chains: \begin{array}{|c|c|c|c|} \hline \operatorname{dim} &*& \mathbb{Z}/2 & \operatorname{cells corresponding to} \underline{C}_n(S^3)(\mathbb{Z}/2)\\ \hline 0 & \mathbb{Z} & \mathbb{Z} & * \times *\\ 1 & 0 & \mathbb{Z}\oplus\mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & T\times *,*\times T\\ 2 & 0 & \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\\0\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 1\\0\end{pmatrix};\;\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 0\\-1\end{pmatrix} & D\times *, *\times D, T\times T\\ 3 & 0 & \mathbb{Z}\oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & D\times T, T\times D\\ \hline \end{array}

So it seems that the cochains valued in $\underline{\mathbb{Z}}$ are:

\begin{array}{|c|c|} \hline \operatorname{dim} & \\ \hline 0 & \mathbb{Z}\\ 1 & \mathbb{Z}\\ 2 & \mathbb{Z}\\ 3 & \mathbb{Z}\\ \hline \end{array} Since $(T\times T)^*=0$ in cochains, we have $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}.$ Differential $d_1$ is an isomorphism since $\partial(D\times *)=T\times *.$ It seems that $\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=H^*(S^3;\mathbb{Z}).$

It is a bit odd to me that the quotient is a homological sphere. Sure, the group $\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}$ since orientation is preserved, but maybe I've missed some $2$-torsion in lower degrees?

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    $\begingroup$ I think you're correct. $\underline{\Bbb Z}$ coefficients have the property that they compute the integral cohomology of the orbit space $S^3 / (\Bbb Z/2)$, and by a slightly different cell decomposition I think this space is homotopy equivalent to $S^3$. $\endgroup$ – Tyler Lawson Aug 16 at 15:26

Your final answer is correct, but the cell structure you're using isn't a $G$-CW structure: $T\times T$ can't be used as a cell in this way.

I would approach it like this: The action of $G = {\mathbb Z}/2$ on $\mathbb{C}\times\mathbb{C}$ can be written as the representation $\mathbb{C}\oplus\mathbb{C}^\sigma$, where $G$ acts trivially on $\mathbb{C}$ and by negation on $\mathbb{C}^\sigma$. The sphere $S(\mathbb{C}\oplus\mathbb{C}^\sigma)$ is also the one-point compactification $S^{1+2\lambda}$, where $\lambda$ denotes the real line with $G$ acting by negation. This has a $G$-CW structure with

  1. one $G$-fixed 0-cell,
  2. one $G$-fixed 1-cell,
  3. one $G$-free 2-cell, and
  4. one $G$-free 3-cell,

so that the skeleta are $*$, $S^1$, $S^{1+\lambda}$, and $S^{1+2\lambda}$. From here you can work out that the $\underline{\mathbb{Z}}$-cochain complex is $$ \mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{0} \mathbb{Z}. $$

A way to check that the answer is correct is to write $$ H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0) \oplus \tilde H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0)\oplus \tilde H_G^{n-1-2\lambda}(S^0) $$ and then use the known calculation of the $RO(G)$-graded cohomology of a point (originally due to Stong (unpublished), since published in various places).

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  • $\begingroup$ It is not a G-CW structure because fixed points are not a G-CW subspace? $\endgroup$ – Gregory G Aug 17 at 8:15
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    $\begingroup$ All the cells in a $G$-CW complex have to have the form $G/H\times D^n$ where $G$ acts trivially on $D^n$. $T\times T$ doesn't have that form, it looks like the disc of a nontrivial representation of $G$. $\endgroup$ – Steve Costenoble Aug 17 at 10:23
  • $\begingroup$ Is there a way to work with "cells" with nontrivial action of the stabilizer? I'm now aware that any cell with nontrivial action can be subdivided into "good" G-cells, but is there a way to avoid this step? $\endgroup$ – Gregory G Aug 17 at 12:28
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    $\begingroup$ Not easily, at least not if you want to calculate the integer-graded part of the cohomology. Any filtration gives rise to a spectral sequence that could, theoretically, be used for computation, but that's not going to be straightforward. There is a notion of $G$-CW($V$) complexes, using cells of the form $G/H\times D(V)$, but that calculates the cohomology in grading $V$. I think Stefan Waner first noticed this, Gaunce Lewis published an exposition, and then Stefan and I published a book using a very generalized version. $\endgroup$ – Steve Costenoble Aug 17 at 14:47
  • $\begingroup$ One other situation where you can use arbitrary cells: At least for $G = \mathbb{Z}/p$, if you have a space built out of cells of the form $G/H\times D(V)$ where $V$ is even-dimensional, but can vary, then, with some additional assumptions, you can conclude that the $RO(G)$-graded cohomology is a free module over the cohomology of a point, though the generators may not be where you expect them to be. I think this was first proved by Ferland and Lewis. $\endgroup$ – Steve Costenoble Aug 17 at 17:21

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