9
$\begingroup$

I am looking at $H^1(\mathbb{R}^n,G)$ where $G$ is a finite 2-group. I'm wondering if such things have been calculated. I'm afraid I can't say I know anything here, past the result that this calculates group extensions of $\mathbb{R}^n$ by $G$.

Any related literature would also be appreciated.

$\endgroup$

1 Answer 1

10
$\begingroup$

I assume that you are looking at group cohomology. Then the action of $\Bbb{R}^n$ on $G$ is necessarily trivial, so $H^1(\Bbb{R}^n,G)$ is just $\mathrm{Hom}(\Bbb{R}^n,G)$ mod. conjugacy by $G$. But this is of course trivial - $\Bbb{R}^n$ has no finite quotient.

$\endgroup$
1
  • $\begingroup$ Ah, yes. And this would be true for any 2-group $G$, not just finite, as $\mathbb{R}^n$ is divisible. But, I think I asked the wrong question :-( This is still helpful, though $\endgroup$
    – David Roberts
    Commented Dec 24, 2013 at 9:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .