TITLE: Group cohomology of $\mathbf{R}^\ast$ acting on $\mathbf{R}$
QUESTION [5 upvotes]: I am interested in computing the first group cohomology $H^1(\mathbf{R}^\ast, \mathbf{R})$, where $\mathbf{R}^\ast$ is acting on $\mathbf{R}$ by multiplication (here $\mathbf{R}$ denotes the real numbers). One should probably take into account continuous cochains or something like that, but I would already be very happy to know if it is $0$ or not.
Of course this is equivalent to find a nontrivial crossed homomorphism $\mathbf{R}^\ast \to \mathbf{R}$, and I think that $\log$ or $\exp$ can help here, but I am stuck.
Any help is appreciated!

REPLY [6 votes]: There is no cohomology. No continuity condition is needed.
The cocycle condition is
$$a \phi(b) + \phi(a) = \phi(ab)$$
Since $\mathbb R^\times$ is commutative,
$$ a\phi(b) + \phi(a) = \phi(ab)=\phi(ba) = b\phi(a) + \phi(b)$$ so $$ (a-1) \phi(b) = (b-1)\phi(a)$$ and thus $$ \frac{ \phi(b)}{b-1} =\frac{\phi(a)}{a-1}.$$
So the only cocycles are scalar multiples of $\phi(x)=x-1$, i.e. the one JLA wrote down. These scalar multiples are indeed coboundaries, so there is no cohomology.
An alternate proof uses the fact that cocycles classify extensions of the trivial representation of $\mathbb R^\times$ by the standard representation, and all such extensions split, as we can see by diagonalizing the action of any nontrivial element of $\mathbb R^\times$.