4
$\begingroup$

I am reading through "Morse Homology and Floer Homology" by Audin and Damian and I am confused about the definition of the differential in integral Morse homology.

Let $V$ be a compact manifold, $f: V \to \mathbb{R}$ a Morse function, and $X$ a pseudo-gradient vector field on $V$ adapted to $f$ such that given any two critical points $a,b$ of $f$, the stable manifold of $W^s(a)$ intersects the unstable manifold of $W^u(b)$ transversely. For the case of Morse homology with $\mathbb{Z}/2\mathbb{Z}$ coefficients we then take the $k$-chain groups to be formal $\mathbb{Z}/2\mathbb{Z}$ combinations of critical points of $f$ of index $k$. We take the boundary maps $\partial_X$ to be defined on critical points $a$ of index $k$ by $$ \partial_X(a) = \sum_{b \text{ critical point of index } k-1} n_X(a,b) b $$ where $n_X(a,b)$ is the number of trajectories going from $a$ to $b$ along the flow of $X$.

My question is: How do we define these numbers $N_X(a,b)$ when we are using integer coefficients? The authors touch on this briefly - they say to choose an orientation for each stable submanifold of each critical point (the stable and unstable submanifolds are all diffeomorphic to open discs) and then define $N_X(a,b)$ as the number of trajectories form $a$ to $b$ "counted with sign". This counting with sign is what I am confused about.

Thanks!

$\endgroup$

1 Answer 1

3
$\begingroup$

I am used to orienting the unstable manifolds. You can also do this for the stable ones. Basically, at a point of intersection $p\in W^u(x)\cap W^s(y)=:W(x,y)$ you will have a short exact sequence of vector spaces

$$0\rightarrow T_p(W(x,y))\rightarrow T_pW^u(x)\rightarrow NW^s(y)\rightarrow 0$$

Here $N_pW^s(y)$ is the normal bundle to the stable manifold of $y$. The point is now that the normal bundle $NW^s(y)$ is trivial, and moreover at $y$ we can identify $N_p\cong N_yW^s(y)\cong T_yW^u(y)$, the tangent space of the unstable manifold of $y$. But we have oriented this already! That means that the space $T_p(M(x,y))$ is oriented by the choice of orientation of $T_pW^u(x)$ and $NW^s(y)$. Now $M(x,y)$ is not the thing we count. If $\mathrm{ind} x=\mathrm{ind} y+1$, this is a one dimensional space. But we can quotient out the $\mathbb{R}$ action. We get another sequence $$ 0\rightarrow \mathbb{R}\rightarrow T_p(W(x,y))\rightarrow TM(x,y)\rightarrow 0 $$

Here $M(x,y)=W(x,y)/\mathbb{R}$ and $\mathbb{R}$ corresponds to the negative gradient flow that generates the free action. This is also oriented. But now $M(x,y)$ is oriented! and this is the thing we count. Working out all the signs in the moduli space is very instructive but likely cause a little headache.

If you want a concrete problem to do this counting you should construct a Morse function on $\mathbb{R}P^2$ with three critical points, and do the oriented counting there. This will give you insight in what is happening.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .