# homology with compact supports

In one of the exercises in McDuff and Salamon, they mention homology with compact supports. I know how to define *co*homology with compact supports, but I can't picture the homology version. How do I say that a chain has compact support? If I use singular chains, don't they all have compact support anyway?

Google isn't a big help here, so any references would really help me out.

Also, I've added some basic sub-questions that would also help me out tremendously. This must all be pretty simple, but my background in algebraic topology is weak and it completely baffles me!

1. McDuff-Salamon go on to state that for the open annulus $(1/2 < r < 1)$ in the plane, the first compact homology group is generated by the arc $\theta = 0$, $1/2 < r < 1$, which I can understand with hindsight: this is just the generator of the homology rel. the boundary and most likely there will be an isomorphism between the compact and relative homologies, just like there is one between the compact and relative co-homologies.

2. They also implicitly use an isomorphism between the compact homology and compact cohomology in certain dimensions. Should I just use this as the definition for the compact homology? I.e. $H_{k, c} = (H^k_c)^\ast$?

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My guess is that they are referring to what is usually called Borel-Moore homology (or homology with closed support). That fits at least with their given example. (A quick glance at Wikipedia indicates that the entry on Borel-Moore homology gives a correct description of it.) –  Torsten Ekedahl Feb 24 '10 at 16:15
Great!! You're right, this fits the example perfectly, but it also helped me to understand some constructions that I was thinking about. It turns out that BM homology is just the tool that I need. –  jvkersch Feb 24 '10 at 20:20

To elaborate on Ekedahl's comment:

It will be easiest to describe things for a triangulated space, so I can work with simplicial chains and cochains. (But my space could be infinitely triangulated; e.g. think of Escher's famous picture of the infinitely triangulated hyperbolic plane. I will also assume that my triangulation is locally finite.) As you observed, usual chains have compact support. This is why you can pair them with cochains (which have arbitrary support). Borel--Moore chains can have unbounded support. These can be paired with not with arbitrary cochains, but only with compactly supported ones.

So Borel--Moore homology is the "homology analogue" of compactly supported cohomology. (But the support conditions are reversed, since homology is dual to cohomology.)

One can often interpret Borel--Moore homology as relative homology. E.g. if $M$ is a compact manifold with boundary $\partial M$, then the Borel--Moore homology of $M\setminus \partial M$ (the interior of $M$) is the same as the homology of the pair $(M,\partial M)$.

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Thanks for the elaboration -- as I mentioned in my answer to Torsten Ekedahl's comment, Borel-Moore homology turns out to be precisely what I need. Your explanation made the remaining pieces fall into place. –  jvkersch Feb 24 '10 at 20:27

I think what McDuff-Salamon call homology with compact support is more commonly known as homology of infinite chains. The chains are formal infinite sums of singular simplices that are locally finite in the sense that each compact subset meets only finitely many singular simplices. The boundary is defined in the usual way.

Note that the usual singular homology are with compact support: the cycles have compact images. By contrast, the usual singular cohomology do not have compact support as a cocycle may take nonzero value on a sequence of cycles that run off to infinity. There is a book by Massey, "Homology and cohomology theory. An approach based on Alexander-Spanier cochains", where these matters are discussed in a very general setting.

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This is the same as Borel--Moore homology, as far as I understand. –  Emerton Feb 24 '10 at 19:51
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It seems they define homology with compact support as the direct limit of homology of compact subspaces. But this is just the ordinary homology (if we are considering singular homology), since every singular chain is contained in some compact subspace anyway (so the natural map between the direct limit and the ordinary homology is an isomorphism). –  Andrea Ferretti Feb 24 '10 at 15:39
but I think there are also homology theories, wehre this map is not an isomorphism. In this situation homology with compact support could make sense/be useful. –  HenrikRüping Feb 24 '10 at 16:44
Yes, but I don't see how this is relevant to the original poster question. –  Andrea Ferretti Feb 24 '10 at 17:30

For any manifold (compact or not), compactly supported cohomology is Poincare dual to (ordinary) homology, via capping with the fundamental class, which is an infinite chain (i.e the sum of all the top simplices in a triangulation). Likewise, (ordinary) cohomology is Poincare dual to homology with locally finite infinite chains. In notation, $H^{n-k}_{comp}(X)\cong H_{k}(X)$ and $H^{n-k}(X)\cong H_{k, inf}(X)$.

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I'm not sure that I understand the claims here. For example, one cannot pair an ordinary cocycle with a locally finite chain (since the resulting sum may still be infinite), but only with a finite chain. Also, it is true that $H^k_c$ and Borel--Moore homology in degree $k$ (which is what I think you mean by $H_{k,inf}(X)$) are canonically dual (with field coefficients). (This is not a statement of Poincare duality, but is easier, and is analogous to the canonical duality of cohomology and usual homology with field coefficients given by one of the universal coefficient-type formulas.) –  Emerton Feb 25 '10 at 2:16
$H^{n−k}_{comp}(X)=H_k(X)$ and $H^{n−k}(X)=H_{k,BM}(X)$ seems OK to me: If you think of a regular cell structure and its dual cell structure, the Poincare dual to a finite chain is a compactly supported cochain and the dual to a locally finite infinite chain is a regular cochain. But I agree your assertion contradicts what I said in the second paragraph. What if $X$ is an infinite genus 2-manifold? Then (say rational coeffs) $H^1_{comp}(X)=H_1(X)$ is a direct sum of infinitely many $Q$s, so $(H_1(X))^*$ is a infinite product of $Q$s. I'm confused. –  Paul Feb 25 '10 at 3:31
OK, I see my mistake. $H^1_c(X)$' is not dual to $H_1(X)$', but it is dual to `$H_{1,BM}(X)$. I'll get rid of the second paragraph of my answer, but then the first paragraph becomes irrelevant to the question. –  Paul Feb 25 '10 at 3:39