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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 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
 \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
\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 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. :)

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