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

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):

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!

• Do you mean closed/open balls instead of spheres ? – Nicolas Hemelsoet Aug 23 '18 at 19:14
• Yes, typically a CW-complex works I think. – Nicolas Hemelsoet Aug 23 '18 at 19:27
• 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! – user43208 Aug 27 '18 at 0:04
• 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). – user43208 Aug 27 '18 at 0:51
• It might be easier to discuss this offline. Google my name IRL + in nLab and you should find my email. – user43208 Aug 27 '18 at 1:09

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.
• @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. – Nicolas Hemelsoet Aug 24 '18 at 4:21
• @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 $n=1$. – Nicolas Hemelsoet Aug 24 '18 at 18:51