Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

5 Answers 5

You asked many questions, here are a few things related to some of them:

(1) What does $H^i(X,\mathcal O_X)$ mean?

In the examples you quoted it already shows that it depends a lot on $X$ what these groups mean.

(a1) if $X$ is smooth and projective, then there is the Hodge decomposition, which I am sure is something every topologists appreciate. It tells you that the singular cohomology groups $H^m(X,\mathbb C)$ may be decomposed as the direct sum of the Dolbeault cohmology groups $H^{p,q}(X)$ for $p+q=m$. One of these is isomorphic to $H^m(X,\mathcal O_X)$, so one possible answer to your question is that these cohomology groups give you a piece of the singular cohomology. A particular interesting case is $m=1$. Then $b_1=h^{0,1}+h^{1,0}=2\cdot h^{0,1}$. So, the vanishing or non-vanishing of $H^1(X,\mathcal O_X)$ is equivalent to the same for $H^1(X,\mathbb C)$.

(a2) still in the projective case, there is a duality, called Serre duality, between cohomology groups of $\mathcal O_X$ and those of $\omega_X$, the sheaf of top differential forms (i.e., the determinant of the cotangent bundle). So, $\dim H^i(X,\mathcal O_X)=\dim H^{n-i}(X,\omega_X)$ where $n=\dim X$.

(b) If $X$ is the complement of a closed subset in an affine variety, then higher cohomology of any coherent sheaf is isomorphic to a shifted local cohomology (with supports in the complement). In other words, $H^i(X,\mathcal O_X)\simeq H^{i+1}_Z(\bar X, \mathcal O_{\bar X})$ for $i>0$ where $X=\bar X\setminus Z$. Local cohomology tends to be big (if not zero), and that's the reason for that example you mention. Of course, now you can ask what local cohomology means, but I'll leave that for another answer/question.

(2) Replacing regular with analytic functions.

Quotients of analytic functions are analytic on their domains of definitions, so while you could define regular analytic functions as those that are locally quotients of analytic functions, you would not actually change the category. The point is that in AG, polynomials are the functions that we can originally define, but it makes sense that as long as the reciprocal of a function exists, then we should be able to use that reciprocal as a regular function. However, those are no longer polynomials, so we need this sort of extended definition.

share|improve this answer
    
This is an excellent exposition of the algebraic theory, which was a good percent of my question. I hope the other answers will be as clear! –  Justin Curry Jan 24 '11 at 0:18
add comment

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.

share|improve this answer
    
I was implicitly asking for real analytic spaces since we can use GAGA to go between complex analytic and algebraic. Nevertheless thank you for your answer. –  Justin Curry Jan 24 '11 at 15:03
1  
Dear Justin, there must be a misunderstanding here. Even in the complex case, GAGA can nowhere be used in my answer because I don't assume my manifolds to be algebraic in the compact case (the tori alluded to certainly aren't algebraic ) and GAGA definitely doesn't apply to Stein manifolds since they are never compact (if positive dimensional) and rarely algebraic (think of a ball or a polydisc). –  Georges Elencwajg Jan 24 '11 at 16:37
    
I was just trying to emphasize to others that in my view GAGA says that complex analytic functions are as rigid as polynomials. I wanted to step away from this rigidity and consider real analytic functions (among other things). Also, I didn't know that we use "meromorphic" in the real case, thus the confusion. –  Justin Curry Jan 24 '11 at 18:24
add comment

Your question is quite broad, so I will touch on a couple of sub questions, hopefully complementing Sándor's.

  • Sheaf cohomology gives accessible, often computable, invariants for spaces.
  • Under mild conditions, cohomology with coefficients in constant sheaves coincides with other theories such as singular cohomology.
  • Given a complex manifold, an analytic function can be defined as a $C^\infty$ function $f$ which lies in the kernel of the Cauchy-Riemann operator, given locally by $$\bar \partial = \sum \frac{\partial}{\partial {\bar z}_j} d \bar z_j$$ Alternatively, $f$ analytic means it can be expanded locally in a convergent power series (just like in one variable). The cohomology of the sheaf of analytic functions $\mathcal{O}_X$ is interesting. Here is one interpretation: the above operator extends to a complex, the $\bar \partial$-complex, which gives a fine resolution of $\mathcal{O}_X$. Thus $H^i(X,\mathcal{O}_X)$ is $\bar \partial$-cohomology.

Hope that helps.

share|improve this answer
    
Thank you for your answer. I am interested specifically in $\mathcal{O}_X$ and not more general sheaves such as $\mathcal{O}_X$-modules or quasi-coherent sheaves. –  Justin Curry Jan 24 '11 at 0:12
add comment

A couple little things to try to complement the previous answers:

(1) Partially addressing your first question: If you are interested in complex projective varieties, Serre's GAGA theorem says the cohomology groups of the structure sheaf of analytic functions (or other coherent analytic sheaf) are the same as those of the algebraic structure sheaf. So in this case, nothing changes when you move from regular algebraic functions to analytic ones.

(2) For real (semi)algebraic varieties, I think the natural sheaf would be that of semi-algebraic functions, though I know fairly little about it. The place to look would be the book by Bochnak, Coste, and Roy.

share|improve this answer
add comment

The higher cohomology groups of the structure sheaf (in any context) precisely capture the category of sheaves which are generated by the structure sheaf -- i.e. all sheaves which can be made by taking complexes built out of copies of the structure sheaf with arbitrary morphisms between them. Namely this (differential graded) category is equivalent to (dg) modules over the (dg) algebra made out of the cohomologies (or more precisely the self-Ext complex - ie the complex calculating the cohomology groups) of the structure sheaf.

In other words, if you want to make global sections into a (derived) equivalence you have to on the one hand restrict to sheaves generated by the structure sheaf and on the other consider the target of global sections not just vector spaces (or complexes) but rather modules over the Ext algebra of the structure sheaf (where the sheaf cohomology of any sheaf naturally lands).

share|improve this answer
    
thank you for your insights and for being general. do you have a paper/notes that elaborates on the proof of the above statements? –  Justin Curry Jan 24 '11 at 21:19
1  
There's a theorem (see [Keller, On differential graded categories] for the dg setting, [Schwede-Shipley, Stable model categories are categories of modules] for the model setting and Lurie, DAG II for the oo-categorical setting) that "enriched" versions of triangulated categories which are generated by a single object are equivalent to modules over the endomorphisms of the object (for abelian categories this is a standard result about endomorphisms of a projective generator, don't know what's a standard reference). –  David Ben-Zvi Jan 25 '11 at 2:39
1  
In any case if you look at the category generated by the structure sheaf, it will then be given by modules over derived endomorphisms of $O$, i.e., over derived global sections of $O$, i.e. over "derived global functions".. –  David Ben-Zvi Jan 25 '11 at 2:40
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.