# Does the Morse homology depend on the orientation?

Before asking my question I need to define some objects. I will follow the book "M. Audin, M.Damian - Morse theory and Floer homology", but the terminology is quite standard:

Let $$M$$ be a smooth compact manifold and consider a Morse-Smale pair $$(X,f)$$ on $$M$$ ($$X$$ is a gradient-like vector field and $$f$$ is an adapted Morse function). If $$a,b$$ are two critical points of $$f$$, we indicate with $$\mathcal L(a,b)$$ the manifold such that every point is a trajectory of $$X$$ ''starting'' from $$a$$ and ''ending'' in $$b$$. One can show that if $$\text{ind}(a)=\text{ind}(b)+1$$ then $$\mathcal{L}(a,b)$$ is a finite set. Moreover if we orient the stable manifold $$W^s(a)$$, remember that it is a disk, we induce an orientation on $$\mathcal L(a,b)$$, namely at each point we associate $$\pm 1$$ if $$\text{ind}(a)=\text{ind}(b)+1$$.

At this point one can define the Morse-Smale complex: $$C_k:=\sum_{a\in\text{Crit}_k(f)}\mathbb Za$$ where clearly $$\text{Crit}_k(f)$$ is the set of critical points of index $$k$$. The map $$d_k:C_k\longrightarrow C_{k-1}$$ acts on the generators of $$C_k$$ in the following way: $$d_k(a)=\sum_{a\in\text{Crit}_{k-1}(f)}N(a,b)b$$ where $$N(a,b)\in\mathbb Z$$ is the sum of the $$\pm 1$$ (the orientations) attached to the points of $$\mathcal L(a,b)$$.

Question: From the above construction it is evident that the Morse-Smale complex (in particular the number $$N(a,b)$$) depends on the orientation that we fix on the stable manifolds $$W^s(a)$$. This sounds very strange to me, indeed I'd expect a complex independent from the orientation. Maybe by passing to the homology group one can recover the independence but I can't see it.

Many thanks.

• Yes, it turns out to be independent, because it turns out to be the same as singular homology! – user98602 Dec 29 '15 at 15:10
• Ok, this is true but there is another strange thing: we can pass to the $\mathbb Z/2\mathbb Z$-complex by the canonical homomorphism $\mathbb Z\rightarrow\mathbb Z/2\mathbb Z$ (it is unique). Now, if we have many $\mathbb Z$-complexes, depending on the orientations, then we have many $\mathbb Z/2\mathbb Z$-complexes. On the other hand if we construct directly the $\mathbb Z/2\mathbb Z$-complex, as in the book cited in my question, you can see that it is unique. – Dubious Dec 29 '15 at 15:19
• I don't think this is very strange - the ambiguity of the orientation on the differential is a $\pm 1$ on each term. When you work mod 2 there is no such ambiguity - in fact, this is the entire reason it is sometimes desirable to work mod 2. – user98602 Dec 29 '15 at 15:20
• Oh yes! changing a single $\pm 1$ doesn't modify the number mod $2$. – Dubious Dec 29 '15 at 15:22

## 2 Answers

Yes, it's independent of these choices - if only because it turns out to be isomorphic to singular homology. The need for orientations is a necessity of working with $\Bbb Z$ coefficients instead of $\Bbb Z/2$ - when writing down the differential, we need to know when a flow line is "positively oriented" or "negatively oriented" to properly count them with signs. This leads us to the unseemly need for this choice.

One reason many books first work mod 2 (and sometimes Floer homology books will do the same, at least at first) is precisely so that one doesn't have to worry about these orientation issues. The ambiguity from the orientations is a bunch of $\pm 1$s induced in the differential - but in $\Bbb Z/2$, there is no ambiguity at all, since $1=-1$. Indeed, when working mod 2, we don't need to assign signs to each flow line - we just need to count them.

If you change the orientation on $W^s(a)$ then an isomorphism from the old complex to the new is given by sending the generator $a$ of $C_k$ to $-a$ (and acting as the identity on all other critical points).