Let $f:M \to \mathbb{R}$ be a Morse function of a compact manifold $M$. Assume $\sigma:M \to M$ is a diffeomorphism such that $f$ is invariant under $\sigma$, i.e. $f(\sigma x)=f(x)$ for all $x \in M$.

I am trying to understand how $\sigma$ induces homomorphisms of the Morse homology groups $H_k(M)$ and how to describe them explicitly. If we denote the boundary maps by $\partial_k$, the elements of $H_k(M)$ are linear combinations of cosets of the form $x+\text{im }\partial_{k+1}$ where $x$ is a critical point of index $k$ with $x \in \ker \partial_k$.

Question: Does $\sigma$ act on $H_k(M)$ by application to representatives of the cosets? That is: Is the map $$x+\text{im }\partial_{k+1} \mapsto \sigma x+\text{im }\partial_{k+1}$$ well defined and does it extend to an automorphism of $H_k(M)$?

I think I am able to see that $\sigma$ permutes the critical points of a fixed index. This would show that $\sigma$ is an automorphism of the Morse chain groups $C_k(M)$. However, I don't know how to proceed. The standard approach would be to show that $\sigma$ commutes with the boundary maps $\partial_{k+1}$ but I have no ideas how to do this. I tried to analyze the action of $\sigma$ on the trajectories between critical points but up to now, this didn't result in anything useful for me. Not sure if this is a good strategy here...


1 Answer 1


You have to be careful, Morse chain complexes are undefined unless you pick a Riemannian metric on the manifold, however, it will be true that $\sigma$ induces a degree preserving isomorphism on the Morse homology of $M$ (and it does not depend on the choice of this Riemannian metric).

Indeed, recall the following result which is standard in Morse theory:

Let $M$ and $N$ be closed manifolds, let $(f,g)$ be a Morse-Smale pair on $N$ and let $\sigma\colon M\to N$ be a diffeomorphism, then $(f\circ\sigma,\sigma_*g)$ is a Morse-Smale pair on $M$ and $\sigma$ induces a degree preserving isomorphism on the Morse chains complexes of $(f,g)$ and $(f\circ\sigma,\sigma_*g)$ and it descends to a degree preserving isomorphism between the Morse homologies of $M$ and $N$.

Sketch of a proof. Recall that Morse homology is defined in a way that

  • Chains. The chains are spanned by the critical points of the Morse function.
  • Grading. The grading is given by the index of the critical points of this Morse function.
  • Differential. The differential is given by a count of gradient trajectories of the Morse function.

Notice that the chain rule implies that $\sigma$ is a bijection between the critical points of index $k$ of $f$ and the critical points of index $k$ of $f\circ\sigma$, it is thus a bijection between $MC_k(f,g)$ and $MC_k(f\circ\sigma,\sigma_*g)$.

Now, what about the differential? The chain rule implies that $\nabla_{\sigma_*g}(f\circ\sigma)=\sigma^*\nabla_gf$ and yet another use of the chain rule shows that the flows of $\nabla_{\sigma_*g}(f\circ\sigma)$ and $\nabla_gf$ are conjugated by $\sigma$, meaning that if $\gamma$ is a $g$-gradient trajectory of $f$ joining $\sigma(q)$ to $\sigma(p)$, then $\sigma^{-1}\circ\gamma$ is the unique $\sigma_*g$-gradient trajectory of $f\circ\sigma$ joining $q$ to $p$. In short, $\sigma$ commutes with the Morse differential. $\Box$

In your case, $M=N$ and $f\circ\sigma=f$, therefore the previous result shows that $\sigma$ induces an isomorphism between $MC_k(M,f,g)$ and $MC_k(M,f,\sigma_*\varphi)$ and if $\sigma$ is furthermore an isometry of $g$, then $\sigma$ induces an automorphism of $MC_k(M,f,g)$.

The map you defined in your question is indeed a degree preserving automorphism in homology.


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