Group cohomology of $\mathbf{R}^\ast$ acting on $\mathbf{R}$ 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!
 A: 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 $\mathbb R^\times$.
A: There are van Est theorems that can help with the computation of group cohomology, and in degree one it's pretty simply. One such cocycle seems to be $f(x)=x-1\,,$ and I believe all others are just scalar multiples of this (I mean as long as you are working with continuous cochains)$^1$.
This cocycle appears to be trivial in cohomology though, trivialized by applying the group differential to the constant 1, thought of as a function whose domain is only the identity element. So I believe that with continuous cochains, the cohomology is zero.
$^1$ To arrive at this I just decided to work with the identity component of $\mathbb{R}^*\,,$ which is isomorphic to $\mathbb{R}\,.$ The corresponding cocycle equation is $e^af(b)+f(a)-f(a+b)=0\,,$ from which you can differentiate with respect to $b$ to deduce that $e^af'(b)-f'(a+b)=0\,,$ which doesn't leave a whole lot of options for $f'\,,$ and of course $f$ just differs from $f'$ by a constant. I then checked, using the cocycle condition, that $f(-1)=-2\,,$ which forces $f(x)=c(x-1)$ for all $x\in\mathbb{R}^*\,.$
