I tried to compute Bredon cohomology of $\mathbb{S}^\sigma$, where $\sigma$ is a sign representation of $\mathbb{Z}/2$, following first chapter and first construction of cohomology from Bredon's "Equivariant cohomology theories". Could somebody please verify it, at least the result?

Throughout $G$ means $\mathbb{Z}/2$.

So I assume the following: $G$-CW structure is obvious, given by two points with trivial action as 0-cells and two arcs with swapping action as 1-cells. I am using simple coefficient system $\mathcal{L}$ on it, that is my functor from "cellular category" to abelian groups factors through some coefficient system $M$. $M$ consists of two groups $M(*)$ - trivial $G$-module and $M(G)$ - $G$-module, and an equivariant map $\epsilon:M(*)\rightarrow M(G)$.

$C^0(\mathbb{S}^\sigma;\mathcal{L})$ consists of the functions $f:\{e^0_1,e^0_2\}\rightarrow M(*)$. Since action of G is trivial on 0-cells, induced action on 0-chains is also trivial, therefore $C^0(\mathbb{S}^\sigma;\mathcal{L})=C^0_G(\mathbb{S}^\sigma;\mathcal{L})=M(*)^2$. $C^1(\mathbb{S}^\sigma;\mathcal{L})$ consists of the functions $f:\{e^1_1,e^1_2\}\rightarrow M(G)$. Action on 1-cells is non-trivial (even free), so $C^1_G(\mathbb{S}^\sigma;\mathcal{L})$ consists of equivariant $f$'s. Thus $C^1_G(\mathbb{S}^\sigma;\mathcal{L})=M(1)$.

The only non-trivial differential is $\delta :C^0\rightarrow C^1$ and is given by $(\delta f)(\tau)=\pm\epsilon(f(e^0_1))\mp\epsilon(f(e^0_2))$. Here $\tau$ of course means any of two 1-dimensional cells.

So $H^0_G(\mathbb{S}^\sigma;\mathcal{L})=M(*)$ and $H^1_G(\mathbb{S}^\sigma;\mathcal{L})=M(G)/M(G)^G$ - but for this I have to assume that $\epsilon$ is an iso on $M(G)^G$.

If this is not "mathoverflow" question, I can ask it also on MathStack.

up vote 5 down vote accepted

In general, you're going to get $H_G^0(\mathbb S^\sigma;\mathcal L) = M(*)\oplus\ker\epsilon$ and $H_G^1(\mathbb S^\sigma;\mathcal L) = M(G)/\operatorname{im}\epsilon = \operatorname{coker}\epsilon$. This is probably easier to see if you compute the reduced cohomology, where the cochain complex becomes simply $\epsilon\colon M(*)\to M(G)$.

  • Ok, which agrees with my statement assuming that $\epsilon$ is an iso on $M(G)^G$. Thanks! – Igor Sikora Apr 19 at 12:53

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.