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I am trying to understand group cohomology, and I have a very basic question. So as I understand it, let $\Gamma$ be a group, and $V$ be a $\Gamma$-module (which is essentially another abelian group acted on by $\Gamma$). We shall construct the group cohomology of $\Gamma$ with coefficients in $V$ by constructing a chain complex and taking its homology groups.

1) We could consider the "homogenous" version as follows: let $$C^n(\Gamma,V)=\{f:\Gamma^{n+1} \to V\}$$ with a coboundary operator $$d^n:C^n(\Gamma,V) \to C^{n+1}(\Gamma,V)$$ $$d^nf(g_0,g_1,\dots,g_{n+1})=\sum \limits_{i=0}^{n+1} (-1)^i f(g_0,\dots,\hat{g}_i,\dots,g_{n+1})$$ Now the chain $$0 \to V \to C^1(\Gamma,V) \to C^2(\Gamma,V) \to \dots$$ is exact at every index, but we can consider the subspace of invariants under the action of $\Gamma$ on a function $f \in C^n(\Gamma,V)$ as follows $$g\cdot f (g_0,\dots,g_n)=g\cdot f(g^{-1}g_0,\dots,g^{-1}g_n)$$ to get a complex $$0 \to V^{\Gamma} \to C^1(\Gamma,V)^{\Gamma} \to C^2(\Gamma,V)^{\Gamma} \to \dots$$ and define $H^n(\Gamma,V)$ as the cohomology groups of this complex.

2) We can also define an "inhomogenous" complex to define the cohomology as follows: let $$\tilde{C}^n(\Gamma,V)=\{f:\Gamma^{n} \to V\}$$ with a coboundary operator $$\tilde{d}^n:C^n(\Gamma,V) \to C^{n+1}(\Gamma,V)$$ $$\tilde{d}^n f(g_1,\dots,g_{n+1})=g_1 \cdot f(g_2,\dots,g_{n+1}) + \sum \limits_{i=1}^{n} (-1)^i f(g_1,\dots,g_ig_{i+1},\dots,g_{n+1}) + (-1)^{n+1} f(g_1,\dots,g_n)$$ The complex obtained $$ V \to \tilde{C}^1(\Gamma,V) \to \tilde{C}^2(\Gamma,V) \to \dots$$ can be used directly to define the cohomology groups of $\Gamma$ in $V$ as the cohomology of this complex.

The equivalence of these two constructions can be seen by constructing isomorphisms between $C^n(\Gamma,V)^{\Gamma}$ and $\tilde{C}^{n+1}(\Gamma,V)$.

Note that in the first (homogenous) construction we explicitly specified an action of $\Gamma$ on functions in $f \in C^n(\Gamma,V)$ as follows $$g\cdot f (g_0,\dots,g_n)=f(g^{-1}g_0,\dots,g^{-1}g_n)$$ and used invariants of that action to define the complex and cohomologies. This is in addition to the given action of $\Gamma$ on $V$ which is used to define $V^{\Gamma}$.

On the other hand, in the inhomogenous construction, we do not use any action of $\Gamma$ on functions, but only the given action of $\Gamma$ on $V$, which is just applied pointwise on function values.

So my question is:

What happens if we use a different action of $\Gamma$ on functions in the homogenous construction? Or if we define our own action of $\Gamma$ on functions in the inhomogenous construction (not just using the action on $V$ pointwise)? Will the cohomologies change?

I really don't understand how the action of $\Gamma$ on functions fits together with the given action of $\Gamma$ on functions of $\Gamma^n$ into $V$. Apologies if this is elementary, but I'd be grateful for some explanation.

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  • $\begingroup$ There's a variety of cohomology theories that can be defined, but I think the answer you're looking for is that "reasonably similar" constructions should give the same cohomology by abstract nonsense: namely, that a derived functor is independent of the choice of injective resolution (modulo some techincal requirements). $\endgroup$ – anomaly Feb 11 at 19:09
  • $\begingroup$ In the homogeneous cochain, the action of $\Gamma$ is not the one you described. This is rather $(g.f)(x_0,...,x_n)=gf(g^{-1}x_0,...,g^{-1}x_n)$. With this action of $\Gamma$, the group $C^n(\Gamma,V)^\Gamma$ is actually the group of equivariant maps $Hom_{\mathbb{Z}[\Gamma]}(\mathbb{Z}[\Gamma^{n+1}],V)$. So it does use the action of $\Gamma$ on $V$. $\endgroup$ – Roland Feb 11 at 20:12
  • $\begingroup$ @Roland You're right, my mistake. I have corrected it now, thanks for pointing it out. But the broader question still stands. If we use a different action of $\Gamma$ on functions, say a conjugate action, then where would this break down? Is there any reason we define the action in that particular way? $\endgroup$ – BharatRam Feb 11 at 20:17
  • $\begingroup$ Well, if you use a different action, it is a bit as if you changed the action on $V$. Look at the particular degree $n=0$. Then $H^0(\Gamma,V)$ is supposed to be $V^\Gamma$ for the action of $\Gamma$ on $V$. Now with the homogeneous cochain, $H^0(\Gamma,V)$ is the set of maps $\Gamma\to V$ such that $(g.f)=f$ and for all $g,h\in\Gamma, f(g)-f(h)=0$. So $f$ is constant and invariant, thus $H^0(\Gamma,V)=V^\Gamma$. If you use another action, then you won't get the expected result it may look as if you had chosen another action on $V$. $\endgroup$ – Roland Feb 11 at 20:28
  • $\begingroup$ By the way, there is another mistake in the definition with the homogeneous cochain. You have a complex $0\to V\to C^0(\Gamma,V)\to C^1(\Gamma,V)\to...$ (including $n=0$) and you take the cohomology of $0\to C^0(\Gamma,V)^\Gamma\to C^1(\Gamma,V)^\Gamma\to ...$ (starting with $C^0$ and removing the $V^\Gamma$ part). $\endgroup$ – Roland Feb 11 at 20:31

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