Trivial holomorphic first sheaf cohomology

[lark]

What's an example of a set with trivial holomorphic first sheaf cohomology? I was thinking of subsets of C, and trying to think what would satisfy this.
For example, suppose you covered C by U_1=re^{i\theta}:r < 2 and U_2=re^{i\theta}:r> 1/2. Then if f(z)=1/z, \oint_{|r|=1} f(z)\neq 0, and he says the contour integral of a function in U_1\cap U_2 should be 0. So I don't think subdividing C this way is a good limiting open cover.
But what would be a good limiting open cover, other than just C by itself? One should be able to refine an arbitrary open cover of C into one which has trivial holomorphic first sheaf cohomology, since C by itself does.
Supposing you covered C by U_1=z: Re(z) > \pi/2 and U_2=z: Re(z) < \pi. I don't see how that open cover would have trivial first sheaf cohomology either since \displaystyle e^{1/ sin(z)} would be analytic in U_1\cap U_2 and I don't think e^{1/ sin(z)} could be expressed as the sum of a function that's analytic in U_1 and a function that's analytic in U_2. So would such an open cover need more refinement? Into what?
What's a good book on it?
Laura

 

[n_bourbaki]

I'm not sure I follow precisely what is going on (cohomology of which sheaf in particular?), but I'd suggest that simply connectedness is something to think about - if you have C\{0} then there's that standard logarithm thing isn't there.

 

[mathwonk]

I guess you mean H^1(O) = {0}. This is true of all connected open sets in C, but not in C^2, as I recall.

the point is whether there exists a function holomorphic in the set which does not extend to a larger connected open set. this is true of all regions in C.

another related problem is whether there is a divisor in that set which is not the zero set of a meromorphic function, but mittag leffler says there is not.

 

[lark]

I guess you mean H^1(O) = {0}. This is true of all connected open sets in C, but not in C^2, as I recall.

I guess the Mittag-Leffler theorem does say you could express e^{1/sin(z)} as the sum of two functions, one analytic in one half plane, one analytic in the other. It's pretty surprising.
I guess what Penrose says about trivial cohomology meaning that the contour integral is 0 for a function that's a coboundary, doesn't apply to the complex plane.
The idea of the sheaf cohomology group is that you have an open cover U_\alpha of the manifold, and holomorphic functions f_{\alpha \beta} defined on each intersection U_\alpha\cap U_\beta,, such that on triple intersections f_{\alpha \beta}+f_{\beta \gamma}+f_{\gamma \alpha}=0 (i.e. the f's are a cocycle), and a set of f's is considered equivalent to 0 iff f_{\alpha\beta}=g_\alpha-g_\beta, with g_\alpha analytic on U_\alpha, g_\beta analytic on U_\beta.
Laura

 

[lark]

I guess the Mittag-Leffler theorem does say you could express e^{1/sin(z)} as the sum of two functions, one analytic in one half plane, one analytic in the other. It's pretty surprising.

Surprising because e^{1/sin(z)} has essential singularities!
Laura