6
$\begingroup$

Let $X$ be a projective variety (lets say normal and irreducible) with the topology coming from being a subspace of $\mathbb{P}^N$ (and not the Zariski topology). Surely one can then define the singular cohomology groups. My question is whether one can also make sense of the Cech cohomology groups $H^*(X,\mathbb{C})$ for the sheaf of locally constant $\mathbb{C}$-valued functions, and if the two cohomology groups agree, like would be the case if $X$ were a complex manifold. Is there also a notion of De Rahm cohomology where we look at forms in some suitable Sobolev space?

I apologize for the rather vague/soft question, but I was not able to find any references online. So it would also be great if someone could point out references for this sort of thing.

Improve this question
$\endgroup$
6
  • 7
    $\begingroup$ Surely a subset of $\mathbb CP^N$ defined by polynomial equations is known to be triangulable (and surely someone will provide a reference) and therefore singular cohomology of such a topological space coincides with Cech cohomology, or sheaf cohomology of the constant sheaf. $\endgroup$
    – Tom Goodwillie
    Jun 11, 2015 at 0:31
  • $\begingroup$ For triangulating a variety, I am guessing one should be able to proceed by induction. The regular set being a manifold should be triangulable, and the singular set is a lower dimensional variety. Is it easy to see that triangulable should imply that both the co-homologies are the same? $\endgroup$
    – Poincare-Lelong
    Jun 11, 2015 at 0:59
  • $\begingroup$ For finite $CW$-complexes, all (co-)homology theories coincide, which is an immediate consequence of the axioms. (The computation of via the cellular complex uses nothing but the axioms.) $\endgroup$
    – Alex Degtyarev
    Jun 11, 2015 at 6:56
  • 1
    $\begingroup$ Accidentally, the triangulizability of (semi-)algebraic sets is usually attributed to Hironaka. $\endgroup$
    – Alex Degtyarev
    Jun 11, 2015 at 6:57
  • 1
    $\begingroup$ As I recall, Hironaka credited the key ideas to Lojasiewicz, perhaps his article in Ann.Scu.Sup.Norm.Pisa(3) 18, 1964, on triangulations of semi analytic sets. see refs in Bierstone-Milman, Pub.Math IHES 67, 1988, p.5 $\endgroup$
    – roy smith
    Jun 11, 2015 at 15:44

1 Answer 1

Reset to default
8
$\begingroup$

Regarding your question on De Rham cohomology there are several approches to realize a De Rham complex that computes singular cohomology.

A. In Algebraic geometry.

You should look at R. Hartshorne "Algebraic De Rham cohomology" manuscripta math. 7, 125-140 (1972). It is a research announcement and survey on the cohomology of algebraic De Rham forms on algebraic varieties, details are published in the Publications of IHES (1975).

In particular for a scheme $Y$ of finite type over a characteristic zero field $k$, he defines its algebraic De Rham cohomology.

1) You embed $Y$ as a closed subscheme of a smooth scheme $X$.

2) You consider $\Omega^*$ the complex of sheaves of regular differential forms on $X$ over $k$.

3) You take $\hat{X}$ the formal completion of $X$ along $Y$: $$\hat{Y}=\bigcup_n Y(n)$$ where $Y(n)$ is the infinitesimal neighbourhood of order $n$ of $Y$ in $X$ and consider $\hat{\Omega}^*$ the completion of $\Omega^*$.

4) You define $H^*_{DR}(Y)$ as the hypercohomology of the complex $\hat{\Omega}^*$ on the formal scheme $\hat{X}$.

Then (theorem 1.6 of this paper) when $Y$ is a scheme of finite type over $k=\mathbb{C}$ we have a natural isomorphism $$H^i_{DR}(Y)\cong H^i(Y^{an},\mathbb{C})$$ where $Y^{an}$ is the corresponding complex analytic space and $H^i(-,\mathbb{C})$ is the singular cohomology.

...............................

B. As stratified spaces

You can use the fact that a complex algebraic variety is stratified, for example it is a stratifold in the sense of M. Kreck, then you have a notion of De Rham complex that computes singular cohomology with real coefficients:

Or you can use Whitney functions

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .