Group cohomology with coefficients in regular representation

Question

I am quite new to group cohomology and want to get an intuition for it. It seemed natural to me to calculate the groups $H^p(G,\mathbb C(G))$, where $G$ is a group and $\mathbb C(G)$ is the vector space of functions on $G$.

Is there something we can say about this in general, i.e. vanishing results or expressing the cohomology in terms of $G$ itself?

Edit: I didn't accept the answer so far because i was hoping for a more hands-on approach. Maybe somebody can jump in?

Answer 1

I can give you a proof for the case $p=1$:

Let $z\in Z^1(G,G(\mathbb C))$. Define $f\in C^0(G,\mathbb C(G))=\mathbb C(G)$ by $f(g)=[z(g)] (1)$. Then $$[df(x)] (g) = (x\cdot f-f)(g) = f(gx)-f(x) \\ [z(gx)](1)-[z(x)](1)=[z(x)](g),$$ where for the last equaility we used that $z$ is a cocycle.

You can probably generalize this approach for an arbitrary $p$.

Answer 2

$\Bbb C(G) = \operatorname{Hom}_{\mathbf{Set}}(G, \Bbb C) = \operatorname{Hom}_{\Bbb Z}(\Bbb Z[G],\Bbb C)$ since $\Bbb Z[G]$ is a free $\Bbb Z$-module (and the action on $G$ corresponds to the action on the basis)

Thus $\Bbb C(G)$ is a coinduced module and we get $H^p(G,\Bbb C(G)) = 0$ for all $p>0$ (compare Thm. 1.5.7, p.15 in these notes: http://math.ucla.edu/~sharifi/groupcoh.pdf)

Note that we can replace $\Bbb C$ with any abelian group.