Michi’s blog » My first group cohomology – did I screw up?

## My first group cohomology – did I screw up?

• May 21st, 2006
• 1:42 pm

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 and give explicitly in a series of ways the product structure – as Yoneda splices, as chain map compositions and as cup products.

Now, has a very nice resolution as a -module – all cyclic (finite) groups have canonically a really cute minimal resolution – given by

with the last map taking and the other maps alternatingly being multiplication with and with .

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

with the codifferentials given by the identity in the first step, since if , then and zero everywhere else, since and since acts trivially on .

So we have that and for all other n.

Now for the product structure. We have 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 . Using the cochain map model, we need to lift the map that takes 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 . 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

so we can take and , and in fact lift to the identity chainmap.

For odd i, we instead get

we can set and thus find the condition on that . One map fulfilling this is multiplication by since then .

Thus for our basis map products we get that if either or are the 0-cochain, then the product is 0 (no surprise there!) and if both are even, then . If is odd, then we compose the relevant cochain for with the chain map in the second expose above – so we have two different cases to distinguish. If has odd degree, then the product is the representative of the map . If, however, has even degree, then we get the augmentation map; and thus .

Thus the product structure on annihilates products of odd degree elements, and in all other cases sends .

This proves that , which is exactly what Carlsen et.al. give.

### 1 Person had this to say...

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