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I am studying the de rham theorem and Poincaré Duality from http://www.few.vu.nl/~vdvorst/DeRham.pdf and I have a question about the Poincaré map

\begin{align*} \mathcal{PD} : \Omega^p(M) &\rightarrow\Omega^{n-p}_c(M) ^ { *} \\ \mathcal{PD} (\omega)(\eta) &= \int_{M}\omega \wedge \eta \end{align*}

$\Omega^p(M) $ is the space of p-forms and $\Omega^p_c(M) $ is the space of p-forms with compact support.

1) This conmutes up to a sign with the differential map, correct? The thesis says that it straight conmutes.

2) I am trying to understand why such a morphism from a CO-CHAIN complex \begin{align*} \cdots & \rightarrow \Omega^{n}(M) \overset{d}{\rightarrow} \Omega^{n+1}(M)\rightarrow \cdots \end{align*} to a CHAIN complex \begin{align*} \cdots & \rightarrow \Omega^{n-p}_c(M)^* \overset{d^*}{\rightarrow} \Omega^{n-p-1}_c(M)^*\rightarrow \cdots \end{align*}

induces a morphism from $\mathcal{PD} : H^p_{dr}(M) \rightarrow H^{n-p}_c(M) ^ { *}$

A priori, it would induce a morphism $\mathcal{PD} : H^p_{dr}(M) \rightarrow H_{n-p}({\Omega_c(M)^*})$ , where the later is the homology of the chain complex above.

Then I am trying to understand why

$H^{n-p}_c(M) ^ { *} \cong H_{n-p}({\Omega_c(M)^*})$

My guess: \begin{align*} H_{n-p}({\Omega_c(M)^*}) &\cong Hom_{\mathbb{R}}(H^{n-p}(\Omega_{c}(M)^*);\mathbb{R}) \end{align*}

via the universal coefficients theorem

\begin{align*} &= Hom_{\mathbb{R}}(H^{n-p}(\Omega_{c}(M));\mathbb{R}) \\ \end{align*}

because $V^{**}= V$, so the cohomology of the chain complex $\Omega_{c}(M)^*$ is by definition the cohomology of the co-chain complex $\Omega_{c}(M)^{**}= \Omega_{c}(M)$ (this is the handwaving part of the argument)

\begin{align*} &=Hom_{\mathbb{R}}(H^{n-p}_c(M);\mathbb{R}) \end{align*}

by definition

\begin{align*} =H^{n-p}_c(M)^* \end{align*}

again by definition.

Is this argument correct?

Werner Germán Busch
  • 803

1 Answers1

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Okay I think I got it, the problem was from a misunderstanding: Both are co-chain complexes because the variable is p and n is fixed.

$\{\Omega^{n-p}_c(M)\}_{p\in\mathbb{N}_0}$ is a chain complex because it depends on p, n is fixed:

\begin{align*} 0 & \leftarrow \overset{p=0}{\Omega^{n}_c(M)} \overset{d}{\leftarrow } \overset{p=1}{\Omega^{n-1}_c(M)} \leftarrow \cdots \leftarrow\overset{p=n}{\Omega^{0}_c(M)} \leftarrow 0 \end{align*}

Or, rewritting the thing:

\begin{align*} 0 & \rightarrow \overset{p=n}{\Omega^{0}_c(M)} \overset{d}{\rightarrow } \cdots \rightarrow \overset{p=1}{\Omega^{n-1}_c(M)} \overset{d}{\rightarrow }\overset{p=0}{\Omega^{n}_c(M)} \rightarrow 0 \end{align*}

So if you call $D_p :=\Omega^{n-p}_c(M)$

It is a chain complex

\begin{align*} 0 & \rightarrow {D_n} \overset{d}{\rightarrow } \cdots \rightarrow {D_1} \overset{d}{\rightarrow }{D_0} \rightarrow 0 \end{align*}

So $\{D_p^*\}_{p\in\mathbb{N_0}}=\{\Omega^{n-p}_c(M)^*\}_{p\in\mathbb{N}_0}$ is a co-chain complex.

Thus,

$H^{n-p}(Hom_\mathbb{R}(\Omega^{n-p}_c(M);\mathbb{R}))=H^{p}(Hom_\mathbb{R}(D_p;\mathbb{R}))=Hom_\mathbb{R}(H_p(D^\bullet);\mathbb{R})$

where the later equality is by the universal coefficients theorem for coefficients on a field (John Lee book: Introduction to topological manifolds)

But $H_p(D^\bullet)=\frac{Ker(D_p\overset{d}{\rightarrow}D_{p-1})}{Img(D^{p+1}\overset{d}{\rightarrow} D_p)}=\frac{Ker(\Omega^{n-p}_c(M)\overset{d}{\rightarrow}\Omega^{n-p+1}_c(M))}{Img(\Omega^{n-p-1}_c(M)\overset{d}{\rightarrow} \Omega^{n-p}_c(M))} =H^{n-p}_c(M)$

Thus

$H^{n-p}(\Omega^{n-p}_c(M)^*)=H^{n-p}_c(M)^*$

And when you got coefficients on a field, the universal coefficients isomorphism is natural, which is pretty important.

Werner Germán Busch
  • 803