I am reading the book "Morse theory and Floer Homology" by Michele Audin and Mihai Damian. Now I am reading the proof of the following theorem. Link to the statement of the theorem

Basically we want to prove that the Morse homology is independent of the pseudo-gradient field and the morse function. Since the proof is a little bit long and I do not want to just summarize it, here is a link to the proof presented in the book.


So, my confusion starts when the author proves that $(C_*(\tilde{F}|_{V \times A}),\partial _{\tilde{X}}) = (C_{*+1}(f_0), \partial_{X_0})$ (page 2 of pdf almost at the end of the page) and similarly for the other equality. After that he states that there are only two trajectories connecting critical points of $\tilde{F}$. So, I have three questions:

  • The first type of trajectories are the ones staying always in $A$, and I argued that because of $(C_*(\tilde{F}|_{V \times A}),\partial _{\tilde{X}}) = (C_{*+1}(f_0), \partial_{X_0})$. So, like analyzing the trajectories in such section is just as analyzing the trajectories of $X_0$, similarly for $X_1$. Is this the correct idea?
  • I do not get why there can only be trajectories from critical points of $f_0$ to those of $f_1$ (which are the second type of trajectories between critical points of $\tilde{F}$). My first thought about why we cannot have trajectories from critical points of $f_1$ to $f_0$ is because of how their indexes are related to the indexes of $\tilde{F}$ (almost at the beginning of page 2 of pdf) , but I am not sure why this would be true. Any suggestion?
  • My final question is in the definition of $\partial_{\tilde{X}}$. I do not know why in the first row, second column of its matrix representation, it is the zero matrix. Any ideas? (My hypothesis is that since we can only have trajectories as those described in the second bullet point, the number of trajectories from critical points of $f_1$ to those of $f_0$ and by definition of $\partial_{\tilde{X}}$, then that is why it is the zero matrix).

Again, I would really appreciate your help here. Maybe this is the easiest part of the proof and I am missing a basic fact, but I do not see it. Thanks in advance.


1 Answer 1


Oh, I happen to know the guy who wrote that PDF. As to your questions.

  1. Yes, that's the idea. In $V \times A$, $F = f_0$ so the critical points are in one-to-one correspondence with those of $f_0$. The shift in degree comes from this $g$ which has a maximum at $0$. Similarly in $V \times B$ but there is no shift in degree.

  2. Intuitively, the way $g$ and $\tilde{X}$ are chosen is so that the flow is going from a "high" position in the region of $V \times A$ to a "low" position in the region of $V \times B$. This is made precise by the fact that $g$ has very negative 1st derivative on $(0,1)$ and $\tilde{X}$ is a pseudo-gradient for $\tilde{F}$ so the trajectories cannot run the opposite way.

  3. The first row, second column entry has to be an operator $C_{k+1}(f_1) \to C_{k-1}(f_0)$ but it also has to use Morse trajectories since we're trying to describe $\partial_{\tilde{X}}$ which is defined in Morse terms. From (2), there are no trajectories running from the region associated to $f_1$ backwards to the region associated to $f_0$ so this operator must be zero.

  • $\begingroup$ Thank you this makes much more sense now. Just one more question, in bullet point 2, why the fact that $\tilde{X}$ is pseudo-gradient does not allow us to run in the opposite direction? $\endgroup$ May 14, 2021 at 19:34
  • 1
    $\begingroup$ I suppose this is true for any vector field; the trajectories are defined to run from $-\infty$ to $+\infty$, not the other way. So you can't run backwards. Moreover, for a pseudo-gradient adapted to $f$, the flow moves points from higher to lower values of $f$. So you have a "preferred direction" which is given by the real line and how $f$ decreases. $\endgroup$
    – inkievoyd
    May 14, 2021 at 19:48
  • $\begingroup$ Awesome, thanks :) $\endgroup$ May 14, 2021 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.