TITLE: Cohomology of Structure Sheaves: Algebraic, Constructible and more
QUESTION [19 upvotes]: I am not an algebraic geometer, but I am a topologist who uses sheaves. I have studied some algebraic geometry and am interested in what happens as I reduce the amount of rigidity in the structure sheaves on a space. Specifically I want to know what the cohomology of the following structure sheaves tell you. Please do things over characteristic zero.
If $X$ is a topological space then the natural structure sheaf of continuous functions has no interesting cohomology because of the existence of partitions of unity. Consequently, $C^k$, smooth and topological manifolds have structure sheaves with no interesting cohomology.
To contrast, schemes ($X,\mathcal{O}_X$) with their structure sheaves of regular functions have lots of interesting information. In particular, there is non-trivial higher cohomology. However, I am still unsure what these groups tell you (aside from what all sheaf cohomology means - obstructions to extending sections). For example Hartshorne exercise 4.3 tells you that $H^1(U,\mathcal{O}_U)$ is infinite dimensional (spanned by $x^iy^j|i,j<0$) where $U=\mathbb{A}^2_k-(0,0)$. For $X$ a curve then the dimension of $H^1(X,\mathcal{O}_X)$ tells you the genus. For affine pieces this cohomology is trivial, so the cohomology of the structure sheaf detects "non-triviality" of a space. Are there any other characterizations of the higher cohomology groups of the structure sheaf?
I am actually interested in the definable/o-minimal/constructible setting. So I want to consider a constructible space $X$ along with it's structure sheaf of ($\mathbb{R}$ or $\mathbb{Z}$-valued) constructible functions as a ringed space. Since one implementation of definable spaces is the semi-algebraic (or semi-analytic) setting, I would like to know that the cohomology of the structure sheaf here tells you. So if someone could address any of the following:

a real analytic space with the sheaf of analytic functions (no partitions of unity, so potential higher cohomology?) Question: Since regular functions in AG are defined as locally being the quotient of polynomials, would regular for analytic be locally the ratio of analytic? (EDIT: I am interested primarily in the real case since GAGA shows in some cases complex analytic spaces are "as rigid as" complex algebraic ones.)
a semi-analytic space with the above structure sheaf(ves)
a semi-algebraic space with its structure sheaf (Which is? Do Nash functions come into play here?)
for a cell complex, is there a natural way of considering it as a ringed space? If so what would that cohomology tell you?

I apologize for the wide spread of questions. Partial answers will be voted up.

REPLY [7 votes]: Dear Justin, let me address your first question. First of all, the local quotient of two analytic functions is called meromorphic. In dimension one you can see meromorphic functions as regular functions with codomain $\mathbb P^1$. However in dimension $\geq 2$ this won't work : for example at the origin of $\mathbb C^2$ the function $z_1/z_2$ approaches  any complex number according to the path you take (you have to blow-up $\mathbb C^2$ if you want a well-defined holomorphic  map to $\mathbb P^1$). 
There is a related interesting question, the Poincaré problem, for a complex manifold $X$: is a meromorphic map globally
the quotient of two holomorphic ones, i.e. do we have $\mathcal M(X)=Frac (\mathcal O(X))$ ? Here are answers in two extreme cases.
Compact manifolds Since holomorphic functions are constant, the answer to Poincaré's problem is "no", unless the manifold has only constants as meromorphic functions (this happens for some  tori for example)
Stein manifolds The answer to Poincaré's problem is "yes" if the Stein manifold $X$ satisfies the purely topological condition $H^2(X,\mathbb Z)=0$. In this case we have indeed  $\mathcal M(X)=Frac (\mathcal O(X))$ .
Interestingly the proof uses that $H^1(X, \mathcal O)=H^2(X, \mathcal O)=0$ for the Stein manifold $X$. 
The result $\mathcal M(X)=Frac (\mathcal O(X))$ is due to  Weierstrass in the particular case  $X=\mathbb C$
and to Poincaré for $X=\mathbb C^2$. You can read the proof of the general result in Grauert-Remmert's Theory of Stein Spaces.