# My first group cohomology - did I screw up?

Published: Sun 21 May 2006

I thought the seminar on tuesday would possibly benefit from something not very often seen - explicit examples. So I started working through one. I wanted to calculate $$H^*(C_8,\mathbb F_2)$$ and give explicitly in a series of ways the product structure - as Yoneda splices, as chain map compositions and as cup products.

Now, $$\mathbb F_2$$ has a very nice resolution as a $$\mathbb F_2C_8$$-module - all cyclic (finite) groups have canonically a really cute minimal resolution - given by

$$\to \mathbb F_2C_8\to\mathbb F_2C_8\to\mathbb F_2C_8\to\mathbb F_2\to 0$$

with the last map taking $$\epsilon\colon a_0+a_1g+a_2g^2+\dots+a_7g^7\mapsto a_0+a_1+\dots+a_7$$ and the other maps alternatingly being multiplication with $$\partial_1\colon 1+g+g^2+\dots+g^7$$ and with $$\partial_0\colon g-1$$.

So this gives as a nice projective (in fact: free) resolution to work with. We now can observe that $$\Hom_{\mathbb F_2C_8}(\mathbb F_2C_8,\mathbb F_2)=\Hom_{\mathbb F_2C_8}(\mathbb F_2,\mathbb F_2)=\mathbb F_2$$ since any map has to respect the group action, which is trivial on $$\mathbb F_2$$, and so any map is determined by its value on 1. Thus we get the sequence of dual modules

$$0\to\mathbb F_2\to\mathbb F_2\to\mathbb F_2\to\dots$$

with the codifferentials given by the identity in the first step, since if $$f:\mathbb F_2\to\mathbb F_2$$, then $$\epsilon^*f(1)=f\epsilon(1)=f(1)$$ and zero everywhere else, since $$\partial_0^*f(1)=f\partial_0(1)=f(g-1)=gf(1)-f(1)=f(1)-f(1)=0$$ and $$\partial_1^*f(1)=f\partial_1(1)=f(1+g+\dots+g^7)=8f(1)$$ since $$C_8$$ acts trivially on $$\mathbb F_2$$.

So we have that $$H^0=\ker\mathbb 1=0$$ and $$H^n=\ker 0/\operator{im} 0=k$$ for all other n.

Now for the product structure. We have $$H^*=\bigoplus_{i\geq 1} \mathbb F_2C_8x_i$$ as a module; now we need to define the product of two cochains. It is enough to look at what happens on the generators; thus taking $$x_ix_j$$. Using the cochain map model, we need to lift the map that takes $$g^i\mapsto 1$$ to a chain map, run it i-j steps up and then compose it with the map that takes 1 to 1.

There are precisely two possible cochains we could want to lift - either the 0-map, or the augmentation map $$\epsilon$$. Furthermore, there are precisely two different cases for these maps - either i is odd or i is even. For the 0-map, it doesn't matter what i is; the lift is the 0-map. For the augmentation map, we actually get different results.

Firstly, for even i
$$\begin{diagram} \mathbb F_2C_8 &\rTo^{1+g+\dots+g^7} & \mathbb F_2C_8 &\rTo^{g-1}&\mathbb F_2C_8 &\rTo^{1+g+\dots+g^7}&\dots\\ \dTo^{f_2} && \dTo^{f_1} && \dTo^{f_0} & \rdTo^\epsilon \\ \mathbb F_2C_8 &\rTo^{1+g+\dots+g^7}& \mathbb F_2C_8 &\rTo^{g-1}& \mathbb F_2C_8 &\rTo^\epsilon &\mathbb F_2&\rTo&0\end{diagram}$$
so we can take $$f_0=\mathbb 1$$ and $$f_1=\mathbb 1$$, and in fact lift to the identity chainmap.
For odd i, we instead get
$$\begin{diagram} \mathbb F_2C_8 &\rTo^{g-1}& \mathbb F_2C_8 &\rTo^{1+g+\dots+g^7} & \mathbb F_2C_8 &\rTo^{g-1}&\dots\\ \dTo^{f_2} && \dTo^{f_1} && \dTo^{f_0} & \rdTo^\epsilon \\ \mathbb F_2C_8 &\rTo^{1+g+\dots+g^7}& \mathbb F_2C_8 &\rTo^{g-1}& \mathbb F_2C_8 &\rTo^\epsilon &\mathbb F_2&\rTo&0\end{diagram}$$
we can set $$f_0=\mathbb 1$$ and thus find the condition on $$f_1$$ that $$(g-1)f_1(m)=(1+g+\dots+g^7)m$$. One map fulfilling this is multiplication by $$1+g^2+g^4+g^6$$ since then $$(g-1)(1+g^2+g^4+g^6)m=(g+g^3+g^5+g^7+1+g^2+g^4+g^6)m$$.

Thus for our basis map products $$x_ix_j$$ we get that if either $$x_i$$ or $$x_j$$ are the 0-cochain, then the product is 0 (no surprise there!) and if both are even, then $$x_ix_j=x_{i+j}$$. If $$x_i$$ is odd, then we compose the relevant cochain for $$x_j$$ with the chain map in the second expose above - so we have two different cases to distinguish. If $$x_j$$ has odd degree, then the product is the representative of the map $$m\mapsto\epsilon((1+g^2+g^4+g^6)m)=4\epsilon(m)=0\pmod 2$$. If, however, $$x_j$$ has even degree, then we get the augmentation map; and thus $$x_ix_j=x_{i+j}$$.

Thus the product structure on $$H^*(C_8,\mathbb F_2)$$ annihilates products of odd degree elements, and in all other cases sends $$x_ix_j=x_{i+j}$$.

This proves that $$H^*(C_8,\mathbb F_2)=\mathbb F_2[x^2,\xi]/\langle \xi^2\rangle$$, which is exactly what Carlsen et.al. give.