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My first group cohomology – did I screw up?

May 21st, 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.

One Person had this to say...

Karlsen Said...

After error checking it several times, I couldn’t find anything wrong with it.
Congrats on your first group cohomology.

May 21st, 2006 at 14:54

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about: Michi is a recent PhD working in homological algebra and applied algebraic topology. This blog is his outlet for texts with some manner of thought put into them. Over at his LiveJournal intimate details and streams of consciousness might be found.
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