# Bredon cohomology of a permutation action on $S^3$

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?

• 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 $S^3$. – Tyler Lawson Aug 16 '20 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).

• It is not a G-CW structure because fixed points are not a G-CW subspace? – Grisha Taroyan Aug 17 '20 at 8:15
• 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 $G$. – Steve Costenoble Aug 17 '20 at 10:23
• 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? – Grisha Taroyan Aug 17 '20 at 12:28
• 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. – Steve Costenoble Aug 17 '20 at 14:47
• 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. – Steve Costenoble Aug 17 '20 at 17:21