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Suppose a CW complex $M$ is given by the union of $n$-spheres, namely $M=\bigcup_{\alpha\in A}S^n$, without knowledge of intersections. The only requirement is that the union covers $M$. Let $\Sigma=\{S^n,\dots,S^n\}$ be a finite collection of sets, with cardinality $|A|$. The nerve consists of all subcollections whose sets have a non-empty common intersection, $\text{Nrv}(\Sigma)=\left\{X\subseteq\Sigma\big|\bigcap X\ne\emptyset\right\}$, which is an abstract simplicial complex. The nerve should look something like this (e.g. a Čech complex): enter image description here

That is, we are allowed to arrange the spheres in a configuration of our choosing, so long as the configuration still covers $M$. (Indeed, we can "pull" the spheres apart as much as possible so that they still cover $M$-an optimal configuration-with the least amount of spheres used).

How do we compute homology groups $H_k(M;\mathbb{Z})$ from the nerve $\text{Nrv}(\Sigma)$ for $\Sigma=\{S^n,\dots,S^n\}$ covering the CW complex $M$?

Idea: Since we need some information on intersections, suppose we construct the following optimal configuration. Begin with two $n$-spheres attached at a base point, namely $S^n\vee S^n$. Then construct two other $n$-spheres that pass through the intersection point of $S^n\vee S^n$. Finally, we continue the process by constructing other spheres $S^n$ that pass through the intersection "points" of other spheres. We write "points" realizing that the intersection of two $n$-spheres is actually an $(n-1)$-sphere. The points we refer to are those corresponding to the two intersections in the $S^2$ orthographic projection onto a plane. We can of course vary the radius of the sphere under this construction.

As mentioned by Mike Miller, the condition we want is that the $k$-fold intersections, for $k$ sufficiently large, are all empty or contractible. Then the Čech complex of this cover (with constant sheaf $\mathbb{Z}$) recovers the homology of the manifold. The general case where $k$-fold intersections are not contractible instead takes the form of a spectral sequence involving cohomology of the various intersections. I am not sure, however, how to make this mathematically precise.

Any help would be much appreciated. Thanks in advance!

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  • $\begingroup$ Do you mean closed/open balls instead of spheres ? $\endgroup$ – Nicolas Hemelsoet Aug 23 '18 at 19:14
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    $\begingroup$ Yes, typically a CW-complex works I think. $\endgroup$ – Nicolas Hemelsoet Aug 23 '18 at 19:27
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    $\begingroup$ Honestly, something about this question sounds like barking up a wrong tree or making your life unnecessarily difficult. If you want to compute homology of a CW complex, a standard procedure is to compute cellular homology of a chosen CW presentation. Is there a good example one can sink one's teeth into, which would explain why this utterly standard procedure isn't a good idea, or why you want to do it this other way? Is there a reason you want to avoid using homology of the nerve of a good open cover, and instead get your hands dirty with a complicated spectral sequence? Context, motivation! $\endgroup$ – user43208 Aug 27 '18 at 0:04
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    $\begingroup$ I don't think we're communicating well. A CW complex is not given by a union of spheres, or at least I don't follow what you mean. It's given (or presented) inductively by a sequence of skeleta $\{X_n\}_{n=0, 1, 2, \ldots}$ together with, for each $n$, a family of attaching maps of the form $S^n \to X_n$ which prescribe how to form $X_{n+1}$ from $X_n$. I think what Mike is describing doesn't have much to do with CW complexes per se, but applies to any (paracompact) space $X$ with a good open cover ("good" meaning that all nonempty finite intersections of opens in the cover are contractible). $\endgroup$ – user43208 Aug 27 '18 at 0:51
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    $\begingroup$ It might be easier to discuss this offline. Google my name IRL + in nLab and you should find my email. $\endgroup$ – user43208 Aug 27 '18 at 1:09
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Very interesting question, unfortunately this is not enough to know the nerve. For an explicit counterexample consider $M_1$ and $M_2$ given as the union of two circles, where the circles in $M_1$ intersect twice and the circle in $M_2$ intersect four times. The nerves are isomorphic but the homology groups are not isomorphic.

When the covering is given by open balls, in order to be able to compute homology using the nerve you need to assume strong conditions on the intersections (they should be all contractible or empty) on your cover. You need probably strong conditions here too but I can't think of a good condition right now.

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    $\begingroup$ The condition you want is that the n-fold intersections are all empty or contractible. Then the Cech complex of this cover (with constant sheaf Z) recovers the homology of the manifold. The general case where n-fold intersections are not contractible instead takes the form of a spectral sequence involving cohomology of the various intersections: en.m.wikipedia.org/wiki/… If I remember the Cech spectral sequence is well explained in the book by Bott and Tu. $\endgroup$ – user98602 Aug 24 '18 at 3:12
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    $\begingroup$ @MikeMiller : Thanks for your comment ! So you don't require anything on $k$-fold intersections for $k<n$ ? Also when you say "Cech complex" what do you mean ? The Cech complex associated to the nerve don't compute the homology correctly here but I'm not sure what would be a correct replacement for the Cech complex here. $\endgroup$ – Nicolas Hemelsoet Aug 24 '18 at 4:21
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    $\begingroup$ @Multivariablecalculus I don't understand Mike Miller comment so far (and I have very limited internet connexion as well) but if I understand it sure ! $\endgroup$ – Nicolas Hemelsoet Aug 24 '18 at 18:46
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    $\begingroup$ @MikeMiller : thanks for the reply. I don't think I understand, let us take a single copy of $S^n$. The Cech complex associated to its nerve is $Z$ in degree 0 and the cohomology is not the cohomology of the sphere. So given the nerve associated to this family of spheres, there should be a more complicated complex we should associate which computes correctly the cohomology, however I'm not sure I can see how to construct one it looks tricky. Also, another trivial remark but your condition doesn't work if $n=1$. $\endgroup$ – Nicolas Hemelsoet Aug 24 '18 at 18:51
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    $\begingroup$ @MikeMiller : I agree with your first comment. However the question here was "given the abstract nerve associated to a cover of M by spheres, how to compute the homology of M only using the nerve". I don't understand how you (correct) first comment helps to the question by the OP. $\endgroup$ – Nicolas Hemelsoet Aug 25 '18 at 8:41

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