# Prove Poincare duality theorem with Morse theory.

First let us consider a smooth n-manifold. And find a Morse function $$f$$. Now let's consider $$-f$$. A singular point of $$f$$ with index $$k$$ is a singular point of -f with index n-k. Thus we have a canonical one-one correspondence between $$C_k(M)$$ and $$C^{n-k}(M)$$ where I'm considering the cellular chain and cochain groups. My question is can I deduce the Poincare duality theorem by analyzing carefully the behavior of boundary and co-boundary maps? But I don't see where is the condition orientable needed.

• So I think the issue might be that to get the dual homology data you need to orient the belt disks (which become the cores of the dual presentation) in a way determined by the original homology data. The original data include orientations for the cores of your handles. To orient the belt disks you need an orientation for the whole manifold. Then you can say "orient the belt so that the core-orientation followed by the belt-orientation is the chosen orientation for the ambient manifold." I may have a better answer in a few days but I think that's the right idea. Apr 9 '13 at 8:36
– lee
Apr 9 '13 at 15:13
• You can see the details in Schwarz' book. May 4 '14 at 2:03

The Morse complex on a manifold (without orientation assumptions) can be defined using the following data: A Morse function $f$, a metric $g$ and a choice of orientation $\mathfrak{o}$ of the unstable manifolds. Then the space $W(x,y)=W^u(x)\cap W^s(y)$ is given an orientation by the following exact sequence

$$0\rightarrow TW(x,y)\rightarrow TW^u(x)\rightarrow NW^s(y)\rightarrow 0.$$

The two spaces on the right are oriented by the choice $\mathfrak{o}$, as $NW^s(y)\cong TW^u(y)$. If two of the three vector bundles in an exact sequence are oriented, so is the third.

This is a general phenomenon (see for example Hirsch' book): the transverse intersection of a oriented and a cooriented submanifold is a oriented submanifold.

Now to define the dual complex one needs an orientation of the stable manifolds. Note that $\mathfrak{o}$ doesn't give this, it gives a coorientation. However, in an oriented manifold we have the exact sequence of oriented vector bundles

$$0\rightarrow TW^u(x)\rightarrow TM\rightarrow TW^s(x)\rightarrow 0.$$

Thus the stable manifolds are oriented, and we can do the count for $-f$. The orientation of $W(x,y,f)$ and $W(y,x,-f)$ need not agree, but if the manifold is oriented, the orientation difference depends only on the degree of $x,y$ and the dimension of the manifold. Thus we get a nice isomorphism.

It is good to work this out in the case where it fails, i.e. your favourite function on $\mathbb{R}\mathbb{P}^2$.

• Thanks for your answer. The orientation of unstable manifold can be chosen without any assumption?
– lee
Sep 28 '14 at 13:53
• Yes, the unstable manifolds are embedded disc, hence contractible, hence orientable Sep 29 '14 at 6:13