# Trivial holomorphic first sheaf cohomology

1. Jul 27, 2008

### 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

Last edited: Jul 27, 2008
2. Jul 27, 2008

### 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.

3. Jul 27, 2008

### 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.

4. Jul 28, 2008

### 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.
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

Last edited: Jul 28, 2008
5. Jul 28, 2008

### lark

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