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This is probably a fairly basic question:

Poincaré Duality states the following: Given an $n$-manifold, the $k^{th}$ homology is isomorphic to the $(n-k)^{th}$ cohomology.

So I was curious is there some certain relation you get when dealing with de Rham cohomology? For example lets take the punctured plane, then the 1st cohomology group is simply $\mathbb{R}$. However, when looking at the first homotopy group of the punctured plane, we get $\mathbb{Z}$.

But, these two groups are not isomorphic. What am I missing?

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Dear Benjamin, The relationship between homotopy and homology with $\mathbb Z$-coefficients (to the extent that there is one) is given by the Hurewicz theorem. The relationship between homology with $\mathbb Z$ and $\mathbb R$ coefficients is given by universal coefficients. Also, Poincare duality holds for closed manifolds (and so does not apply, at least in the naive form that you have stated it, to a non-compact manifold such as the punctured plane). Regards, –  Matt E Nov 14 '11 at 1:32
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@Matt E: Why not make that an answer? –  Brandon Carter Nov 14 '11 at 1:39
    
@Brandon: Dear Brandon, I thought at first that the OP may have been looking for something more substantive. I will post a version as answer, though. Regards, –  Matt E Nov 14 '11 at 2:26
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When you allow the manifold to have boundary, frequently people call the appropriately-generalized version of Poincare Duality that applies to that situation Alexander Duality. It's also frequently phrased as $H_i(M,X) \simeq H^{m-i}(M,Y)$ where $X \sqcup Y = \partial M$. –  Ryan Budney Nov 14 '11 at 2:38
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There is a version Poincare duality between de Rham cohomology and compactly-supported de Rham cohomology. You can read about this in Section I of Bott & Tu's beautiful book, "Differential Forms in Algebraic Topology". –  Aaron Mazel-Gee Nov 14 '11 at 9:58

1 Answer 1

Here are some remarks related to the questions you raise:

First of all, you mention homotopy groups, rather than homology groups. The relationship between homotopy and homology with $\mathbb Z$-coefficients (to the extent that there is one) is given by the Hurewicz theorem. In any event, in discussing Poincaré duality, it is homology and cohomology groups that are directly relevant, rather than homotopy groups.

The relationship between singular homology with $\mathbb Z$ and $\mathbb R$ coefficients is given by the universal coefficient theorem. It states that $H_i(X,\mathbb R) = H_i(X,\mathbb Z)\otimes_{\mathbb Z} \mathbb R.$ (In general, there would be a Tor contribution as well, but that vanishes here, since $\mathbb R$ is torsion free.)

The de Rham Theorem states that the $k$th de Rham cohomology of a smooth manifold is isomorphic to the $k$th singular cohomology of the manifold with $\mathbb R$-coefficients, or, equivalently (by universal coefficients for cohomology), is dual to the $k$th singular homology with $\mathbb R$-coefficients.

Poincaré duality itself, when phrased in terms of de Rham cohomology, states that for a closed, connected, and orientable $n$-dimensional smooth manifold, the $n$th de Rham cohomology group $H^n$ is one-dimensional over $\mathbb R$, and the cup product from $H^k \times H^{n-k}$ to $H^n$ (which in de Rham cohomology is induced by wedge product on forms) is a perfect pairing.

Taking into account the relationship between de Rham and singular theory stated above, this can also be phrased as an isomorphism between $H^{n-k}$ and $H_k$ with $\mathbb R$-coefficients. But note that the homology under consideration has $\mathbb R$-coefficients, not $\mathbb Z$-coefficients! Note also that Poincaré duality is a statement for closed manifolds (and so does not apply, at least in the naive form that you have stated it, to a non-compact manifold such as the punctured plane).

There is a version of Poincaré duality with $\mathbb Z$-coefficients (as well as generalizations to non-closed and/or non-orientable manifolds); see the wikipedia page. Note though that if you want to consider homology with $\mathbb Z$-coefficients, then you will also need to consider cohomology with $\mathbb Z$-coefficients, and de Rham theory doesn't do this. (You will need to use some other form of cohomology, such as singular, Cech, or sheaf cohomology.)

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Not widely known about duality and Hodge star for closed oriented Riemannian, although you may have seen this already: mathoverflow.net/questions/3134/… –  Will Jagy Nov 14 '11 at 3:33
    
@Will: Dear Will, Thanks for the link, and also for implictly reminding me to add the adjective orientable to the above discussion. Regards, –  Matt E Nov 14 '11 at 4:02

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