# Bredon cohomology of a sign representation for a cyclic group of order 4

Yet another question "I compute Bredon cohomology of something and I am not sure, whether it is correct".

So I am taking a sign representation $\sigma$ of cyclic group of order 4, $C_4$. Then I compactify $\sigma$ to get $\mathbb{S}^\sigma$ with two fixed 0-cells and one 1-cell of type $C_2$. Then I am taking $\underline{\mathbb{Z}}$, constant $\mathbb{Z}$-coefficients. It seems that $$C_{C_4}^0(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=\mathbb{Z}\oplus\mathbb{Z} \\ C_{C_4}^1(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=0,$$ therefore $H_{C_4}^0(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=\mathbb{Z}\oplus\mathbb{Z}$ and $H_{C_4}^1(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=0$.

Degree 1 looks ok for me, but I am worried about degree 0. Here I calculated similar thing for $C_2$: Bredon cohomology of $\mathbb{S}^\sigma$ and degree 0 answer is different. But it should not change, since sign action of $C_4$ factors through sign action of $C_2$.

Also, since I am using constant coefficient system, $H_{C_4}^*(\mathbb{S}^\sigma;\underline{\mathbb{Z}})=H^*(\mathbb{S}^{\sigma}/C_4;\mathbb{Z})$ (unless it is true only for cyclic groups of prime order) - so degree 0 cohomology should be only one $\mathbb{Z}$, as I am taking non-equivariant cohomology of a contractible space.

So is my answer correct?

No, I don't think so - I think there are more Bredon $1$-cochains than that.
The orbit category $\mathcal{O}C_4$ looks like $$C_4/e \to C_4/C_2 \to C_4/C_4$$ where the automorphism groups of the objects are $C_4$, $C_2$ and the trivial group, respectively. The Bredon cochains in degree $i$ are natural transformations from the functor $\underline{C}_i(\mathbb{S}^{\sigma}):\mathcal{O}C_4\to Ab$ given by $\underline{C}_i(\mathbb{S}^{\sigma})(C_4/H) = C_i((\mathbb{S}^\sigma)^H)$ to the constant functor $\underline{\mathbb{Z}}$. In degree $i=1$ we have $$\begin{array}{ccccc}\underline{C}_1(\mathbb{S}^{\sigma})(C_4/e) &\leftarrow &\underline{C}_1(\mathbb{S}^{\sigma})(C_4/C_2)& \leftarrow &\underline{C}_1(\mathbb{S}^{\sigma})(C_4/C_4)\newline \parallel & & \parallel & & \parallel \newline \mathbb{Z}\oplus\mathbb{Z} & \leftarrow & \mathbb{Z}\oplus\mathbb{Z} & \leftarrow & 0 \end{array}$$
with the action of $C_4$ on the bottom left and $C_2$ on the bottom centre both generated by $(a,b)\mapsto (-b,-a)$. So if I'm not mistaken there is a Bredon cochain given by mapping $(a,b)$ to $a-b$.
• Ok, I see where I made a mistake. Because of simplified notation, I got $0$ at $C_4/e$ level, which is obviously wrong. Thank you! – Igor Sikora May 20 '18 at 12:53
To expand on Mark Grant's answer, but looking at it slightly differently: You need to look at the chains as contravariant functors on the orbit category, and they will be projective functors. $\underline C_0(\mathbb S^\sigma)$ is a sum of two copies of $\mathbb Z\mathcal O C_4(-,C_4/C_4)$ while $\underline C_1(\mathbb S^\sigma)$ is one copy of $\mathbb Z\mathcal OC_4(-,C_4/C_2)$. Maps $\mathbb Z\mathcal OC_4(-,C_4/C_2) \to \mathbb Z\mathcal OC_4(-,C_4/C_4)$, by Yoneda, correspond to elements in $\mathbb Z\mathcal OC_4(C_4/C_2,C_4/C_4) = \mathbb Z$. The boundary map $\underline C_1(\mathbb S^\sigma)\to \underline C_0(\mathbb S^\sigma)$ is then specified by the element $(1,-1)\in \mathbb Z\mathcal OC_4(C_4/C_2,C_4/C_4)\oplus \mathbb Z\mathcal OC_4(C_4/C_2,C_4/C_4) = \mathbb Z\oplus\mathbb Z$.
Passing to cochains with $\underline{\mathbb Z}$ coefficients (and using Yoneda again) gives the cochain complex $$\mathbb Z \oplus \mathbb Z \to \mathbb Z \to 0 \to \cdots$$ where the coboundary is $(a,b) \mapsto a-b$. So $H_{C_4}^0(\mathbb S^\sigma;\underline{\mathbb Z}) = \mathbb Z$.