Closed Poincaré dual of the boundary of a closed manifold.

Question

Let $S$ be the boundary of a closed manifold $T$ embedded in $M$. I have to prove that the Poincaré dual of $S$ is $0$.

Assume $dim(M)=n,dim(T)=k$ with $k\le n$. Hence, $dim(S)=k-1$.

Let $[\eta_S]\in H_{DR}^{n-(k-1)}(M)$ be the (closed) Poincaré dual of $S$.

We have $i:S\to M$ embedding. We denote $\omega|_S:=i^*\omega$.

For every $\omega\in\Omega_c^{k-1}(M)$

$$ \int_S\omega|_S=\int_{\partial T}\omega|_S=\int_Td\omega|_S=0. $$

How can I conclude?

Nope (at least, not in de Rham cohomology or over $\mathbb{Z}$, etc.). I was wondering if there was an a priori reason to assume that $S$ (or $T$, etc.) was orientable. — anomaly, Oct 01 '15 at 14:25
I think the idea is to use Stokes theorem in order to prove that $\eta_S=d\eta_T$. — Andreas Cap, Oct 01 '15 at 15:11
But, I get $$\int_S\omega|S=(-1)^k\int_M\omega\wedge d\eta_T $$. — avati91, Oct 01 '15 at 16:58
What you did proves the claim. Your formula shows that $\eta_S=(-1)^kd\eta_T$. This is an exact form, so its cohomology class is zero. — Andreas Cap, Oct 01 '15 at 17:41
Right haha. I was a bit tired yesterday! Thanks a lot @AndreasCap — avati91, Oct 02 '15 at 09:01

score 1 · Answer 1 · answered Oct 04 '15 at 20:24

Here is a topological argument:

the inclusion $i:S\hookrightarrow M$ factors through $T$, hence by functoriality $i_*: H_*S \to H_*T \to H_*M$. The first map maps the fundamental class to zero. In particular the homology class which $S$ represents in $M$ is trivial.

Note that the boundary of a manifold is nullhomologous, by either stratifold theory, or the long exact sequence of the pair $H_k (T,\partial) \to H_{k-1}(\partial) \to H_{k-1} T$, where the first map maps fundamental class to fundamental class (which is then mapped to zero by exactness).

Closed Poincaré dual of the boundary of a closed manifold.

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