I'm trying to check by hand that the signs in the Morse complex, defined via choices of orientations on the unstable manifolds, lead to $\partial^2=0$. The books I've looked in seem to say either that this sign verification is easy (e.g. Audin-Damian), or prove $\partial^2=0$ by introducing more sophisticated notions like orientations of determinant line bundles of Fredholm operators (e.g. Schwarz).

Is there a good reference which explicitly works out that the broken trajectories at opposite ends of a one-dimensional moduli space count with opposite sign?

Here's what I've been attempting. Suppose we have critical points $a$, $b$, $b'$ and $c$, of indices $k+1$, $k$, $k$ and $k-1$ respectively, and broken trajectories $a \rightarrow b \rightarrow c$ and $a \rightarrow b' \rightarrow c$ which are joined by a path of genuine trajectories $a \rightarrow c$. We can view this as an embedding of $[0, 1]^2$ into our manifold $M$, with $a$ and $c$ at one pair of opposite corners and $b$ and $b'$ at the other, with the broken trajectories given by the edges.

We can trivialise $TM$ over this square, and after fixing orientations of the unstable manifolds $U_a$, $U_b$, etc we get induced orientations on the stable manifolds $S_a$, $S_b$, etc. The sign of the trajectory $a \rightarrow b$ comes from comparing the orientations of $U_a$ and $S_b$, whilst that of the trajectory $b \rightarrow c$ comes from comparing $U_b$ with $S_c$. The orientations of $S_b$ and $U_b$ are related by the orientation of $M$. What I can't see is how to piece this information together.

The phenomenon I'm worried about is the following. Suppose $V$ is a two-dimensional oriented real vector space with basis $e_1, e_2$. Then the vectors $a=e_2$ and $b=2e_1+e_2$ both have the same orientation relative to $e_1$, and similarly $c=e_1$ and $d=e_1+2e_2$ both have the same orientation relative to $e_2$, but $a, c$ and $b, d$ have opposite orientations to each other.

If I could relate the tangent space of $S_b$ to $S_c$ directly, say, rather than just in the quotient of $TB$ by $TU_a$, then I would be fine. Intuitively this seems plausible: it looks like near to the trajectory $a \rightarrow b$, the tangent space $TS_c$ is simply spanned by $TS_b$ along with the inward pointing normal vector to the edge of the square. But I don't know how to prove this.

Added later: The following example explains my confusion. Suppose we have trivialised $TM$ over the square as above, and let $e_1, e_2, f_1, f_2$ be a basis in this trivialisation. Let the square lie in the $e_1, e_2$ plane, with $a$ at $(0, 0)$, $b$ at $(1, 0)$, $b'$ at $(0, 1)$ and $c$ at $(1, 1)$. Now suppose

$TU_a=\langle e_1, e_2, f_2\rangle$

$TS_b=\langle e_1, f_2 \rangle$

$TU_b=\langle e_2, f_2\rangle$

$TS_{b'}=\langle e_2, f_1+2f_2\rangle$

$TU_{b'}=\langle e_1, 2f_1+f_2\rangle$

$TS_c=\langle e_1, e_2, f_1\rangle$

$TU_c=\langle f_2\rangle$

with the given bases for the tangent spaces to the unstable manifolds positively oriented. I calculate that all four edge trajectories count with positive signs.

It seems like the problem is related to the discrepancy between $TS_{b'}$ and $TS_c$, as mentioned in the final paragraph of the original version of the question.

This has now evolved into a separate question, here.


1 Answer 1


The thing that one uses is that if one is given an exact sequence of vector spaces

$$ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0 $$

orienting any two of them orients the third. This directly goes through for exact sequences of vector bundles. This one can use to prove that the transverse intersection of an oriented submanifold and a cooriented submanifold is oriented. With a coorientation I mean an orientation of the normal bundle. So lets $X$ be oriented and $Y$ be cooriented in $M$. Then

$$ 0\rightarrow T(X\cap Y)\rightarrow TX\rightarrow NY\rightarrow 0 $$

We use the fact here that the normal bundle of $Y$ is contained in the tangent bundle of $X$, because we assume that the intersection of $X$ and $Y$ is transverse. Now if $TX$ is oriented, and $NY$ is oriented, then $T(X\cap Y)$ is oriented. It is a very nice exercise to look at intersections of $\mathbb{R}P^k$ and $\mathbb{R}P^l$ in general position in $\mathbb{R}P^m$ for various choices of $k,m$ and $l$. The intersection will be an $\mathbb{R}P^n$ for some $n$, and depending if $n$ is odd or even it is orientable or not.

How does this apply to Morse theory? The unstable and stable manifolds are embedded discs, hence contractible and orientable. We start with a choice of orientation of the unstable manifolds $W^u(x)$. Not that at $x$ the normal space to $T_xW^s(x)$ is exactly $T_xW^u(x)$. So that means that the choice of orientation of $T_xW^u(x)$ coorients $T_xW^s(x)$. But as $W^s(x)$ is contractible, this coorients $W^s(x)$ everywhere. Let's look at the moduli space of parametrized orbits from $x$ to $y$

$$ 0\rightarrow T(W^u(x)\cap W^s(x))\rightarrow TW^u(x)\rightarrow N(W^s(y))\rightarrow 0. $$

Now the two bundles on the right hand side are oriented, so is the first one. But we want to look at the moduli space of unparametrized orbits. We have to quotient out the free $\mathbb{R}$ action given by the gradient flow. We have another exact sequence

$$ 0\rightarrow <-\nabla f>\rightarrow T(W^u(x)\cap W^s(y))\rightarrow T(W^u(x)\cap W^s(y)/\mathbb{R})\rightarrow 0 $$

Here the first bundle is the line bundle in the direction of the negative gradient flow, and is clearly oriented by the gradient. So we get an orientation of the moduli space of unparametrized orbits, which is what we wanted.

Proving that the the differential squares to zero, taking into account the orientations, is a bit technical but not incredibly hard. I remember this being done along these lines in a paper of Joa Weber in expositiones. But I also recall there being a small sign mistake there...

Anyway, the great thing about this description is that it never uses that the ambient manifold is orientable, and indeed this is not necessary. The Morse complex also recovers the integral homology of the ambient manifold, even if it is not orientable. Often one sees discussions about the Morse complex assuming that the ambient manifold is orientable, but this is not necessary and indeed confusing. The upshot of orienting the ambient manifold is that orientation and coorientation coincide, using an exact sequence

$$ 0\rightarrow TX\rightarrow TM\rightarrow NX\rightarrow 0. $$

  • $\begingroup$ Thanks for your answer. I understand how to orient the individual trajectories; it's specifically showing $\partial^2=0$ that's the difficulty. I found the paper of Weber you mentioned though (arxiv.org/abs/math/0411465), and in Theorem 3.11 he seems to do it. Now I can't see what was causing me a problem before! $\endgroup$
    – user149874
    Jun 2, 2016 at 15:39
  • $\begingroup$ I think choosing an orientation of $M$ over the square was a bad thing to do (although I assumed something like this would be necessary in order to take account of the path between my broken trajectories). Keeping track of the distinction between orientations and coorientations makes things much clearer. $\endgroup$
    – user149874
    Jun 2, 2016 at 15:40
  • $\begingroup$ Sorry to un-accept your answer after having accepted it, but I've now clarified my point of confusion. I'll add to my question. $\endgroup$
    – user149874
    Jun 3, 2016 at 8:25
  • $\begingroup$ I think the problem with Weber's proof is that the comparison of the orientations of $TW^u(x)$ with $T\mathcal{M}_{xy} \oplus TW^u(y)$ is meaningless unless they are the same space. This is the issue I raise in the final paragraph of the original version of the question. I think I'll ask this as a separate question, now I've distilled out the essence of my confusion. $\endgroup$
    – user149874
    Jun 3, 2016 at 8:55
  • $\begingroup$ @jez: No problem. I will look into it in the future, but unfortunately currently I don't have much time. $\endgroup$
    – Thomas Rot
    Jun 3, 2016 at 11:57

You must log in to answer this question.