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I think it's pretty intuitive how singular/simplicial cohomology detects "holes" in a space.

How can we directly visualize how and in what sense the Cech cohomology of a cover does this?

In case it's of any interest, here are two examples I've looked at with the constant sheaf $\mathbb{Z}$:

(1) The disk, covered "Venn diagram style" with three open patches $U_1, U_2, U_3$ overlapping near the center (like this, but with overlaps), and

(2) The restriction of this cover to the boundary circle of the disk: three opens $U_1, U_2, U_3$ with 3 double intersections $U_{12}, U_{13}, U_{23}$ and no triple intersection.

If you look at the Cech complex in (2), the $H^1=\mathbb{Z}$ "comes from" the fact that you can write down a triple of elements $(1,0,0)$ on $U_{12}, U_{13}$,$U_{23}$ which "would" disagree on the triple overlap in (1), but since it's "missing", $(1,0,0)$ gets counted as a cocycle, which is not a coboundary. Even better, the presentation of this $H^1$ you get from the Cech complex is $\mathbb{Z}^3/\{(a,b,c)=(b,c,a)\}$, which is isomorphic to $\mathbb{Z}$ because you can "rotate" all the coordinates "around the missing intersection" into the first component.

I think a like minded analysis of higher dimensional analogues provides similar intuition. Are there any formulations of the Cech complex to really make precise how this intuition should work? What's going on here?


Follow up: Following Mariano's answer below, I started reading about Abstract simplicial complexes and their cohomology, which seem like just what I was looking for. What helped me most were the ideas that

1) The (constant sheaf) Cech cohomology of a cover $\cal{U}$ of $X$ "is" the simplicial cohomology of its nerve
$N(\cal{U})$, an abstract simplicial complex,

2) The simplicial cohomology of an abstract simplicial complex "is" the singular cohomology of its geometric realization, and

3) The geometric realization of the nerve of a covering of $X$ is a "simple approximation" of $X$,

So in this sense, we can say precisely that

Cech (constant sheaf) cohomology on a cover detects holes in a "simple approximation" to $X$ defined by that cover.

In particular, seeing the faces of a simplicial complex encoded as formal wedge products of its vertices totally made my day :)

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I appreciate very much your Follow-Ups. Way to be a good MO citizen! –  Aaron Mazel-Gee Nov 20 '09 at 10:37

4 Answers 4

up vote 15 down vote accepted

The complex which computes Cech cohomology for a covering is the "same" one as the one that computes the cohomology of the nerve of the covering. It is not hard to see that the geometric realization of that nerve is, in some sense, an aproximation to the original space. Since you apparently find it intuitive that simplicial cohomology detects the geometry, then this should convince you that Cech cohomology also does :P

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This is fantastic! Can anyone recommend some reading on this? For example, where can I read about nerves and their cohomology? I'll have lots of questions like "When/how is the cohomology of a nerve the same as the cohomology of its geometric realization?", and it would be nice to have a book or article to walk me through them. –  Andrew Critch Nov 12 '09 at 22:11
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Bott and Tu's book on Differential Forms in Algebraic Topology is a good start. –  Mariano Suárez-Alvarez Nov 12 '09 at 22:18

Here's an idea; I'm not sure if all of the steps are possible, but I think it should work.

Take a simplicial complex of dimension $n$ (all simplices are of dimension $\leq n$). "Thicken" the simplicial complex to get a new topological space containing the original simplicial complex as a deformation retract, such that there is an open neighborhood $U_\sigma$, homeomorphic to an $n$-ball, of each simplex $\sigma$ which deformation retracts to that simplex. Do this so that the intersections $U_\sigma \cap U_{\sigma'}$ are equal to $U_{\sigma \cap \sigma'}$. Now take the open cover $\{ U_{\sigma} \}$; then the Cech complex associated to this open cover should be the "same" as the complex that computes the simplicial (co)homology of the simplicial complex. "Same" here probably means "canonically homotopy equivalent".

For example, if you think of a circle as being the simplicial complex given by a triangle with 3 edges and 3 vertices, then this essentially matches up with your example (2).

This is essentially the same as Mariano's explanation, but sort of in the converse direction.

This sort of thing might be covered in some older books, like Munkres' "Elements of Algebraic Topology" or Spanier's "Algebraic Topology". Cech cohomology doesn't seem to really be covered in most contemporary algebraic topology books; I guess it's just not very useful for what people are interested in these days.

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I do not know why you think Cech cohomology stopped being interesting! For one thing, every time you construct something locally you are in fact doing it à la Cech. Also, Cech cohomology (or the Cech `idea') is one of the reasons cohomologies are computable... –  Mariano Suárez-Alvarez Nov 12 '09 at 22:13
    
I didn't say and I don't think that it's uninteresting. I'm aware that the Cech 'idea' is instructive and important. But Cech cohomology itself just doesn't seem to me to be used very much these days in the context of algebraic topology. Though, now that you've mentioned it, Bott-Tu is a pretty good counter-example to my claim. –  Kevin H. Lin Nov 12 '09 at 22:32

This is explained in "Principles of Algebraic Geometry," by Griffiths & Harris, p42.

Given a simplicial decomposition of your space, associate each vertex vα with its star Uα, where the star means the union of interiors of the simplices that contain it. This gives an open cover {Uα} where the intersection of p open sets is nonempty precisely when the corresponding vertices span a p-simplex. So a p-cochain maps (Uα1, ..., Uαp) to a nonzero section of the coefficient sheaf only if {vα1, ..., vαp} span a simplex.

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This was what I was shooting for in my idea in my answer. –  Kevin H. Lin Nov 20 '09 at 15:46

You're probably beyond this stage but I always thought nice intuition came from Penrose stairs. You can draw them because the "triple overlap" is missing from the middle of the staircase and a cunning artist can draw staircases based on cocycles that aren't coboundaries, even though a correct 3D representation is impossible.

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