Let $f:M\to\mathbb{R}$ a Morse function, and $\pi:\tilde{M}\to M$ a covering map, show that $f\circ \pi:\tilde{M}\to \mathbb{R}$ is also a Morse function.

Let $\tilde{f}=f\circ \pi$, then $$d\tilde{f}_{\tilde{p}}=d(f\circ \pi)_{\tilde{p}}=df_{\pi(\tilde{p})}\circ d\pi_{\tilde{p}}=df_p\circ d\pi_{\tilde{p} } .$$ Hence $$\begin{array}{lll} \tilde{p}\in Cr(\tilde{f})&\iff & d\tilde{f}_{\tilde{p}}=0 \\ &\iff &df_p\circ d\pi_{\tilde{p} }=0\\ \end{array} $$

We know that $\pi$ is a local diffeomorphism, so $d\pi_{\tilde{p}}=0$ then, i will get $df_p=0$, i.e. $p\in Cr(f)$. Therefore. $$\boxed{Cr(\tilde{f})=\{\tilde{p}\in\tilde{M}|p\in Cr(f)\}.}$$

It just remains to show that if $\tilde{p}\in Cr(\tilde{f})$ then it is non-degenerated. How can I do this?

Thank you.


If $\tilde{p}\in Cr(\tilde{f})$, then $p\in Cr(f)$ and we have

$$\begin{array}{lll} d^2\tilde{f}_{\tilde{p}}(v_1,v_2)&= &d(df_p\circ d\pi_{\tilde{p}})\\ \\ &= &d^2f_{p}(d\pi_{\tilde{p}}(v_1),d\pi_{\tilde{p}}(v_2))+df_{p}(d^2\pi_{\tilde{p}}(v_1,v_2))\\ \\ &= &d^2f_{p}(d\pi_{\tilde{p}}(v_1),d\pi_{\tilde{p}}(v_2))\quad\mbox{ Using that $df_p=0$.} \end{array}$$

Knowing that $f$ is a Morse function, then $p$ is non-degenerated, then , $\tilde{p}$ is also non-degenerate. Therefore, $\tilde{f}=f\circ \pi$ is a Morse function.

I know that the second detivative is only defined at critical points, but I also used the chain rule, is that ok? Thank you.

  • $\begingroup$ Do you mean $d\pi$ is invertible? $\endgroup$ – Keshav Nov 30 '19 at 20:31
  • $\begingroup$ Since $\pi$ is a local diffeomorphism, you only need to show that the Hessian is well-defined at critical points. You can do this in coordinates with the chain rule, or appeal to bilinear form definition for a slicker proof. $\endgroup$ – Elliot G Nov 30 '19 at 20:37
  • $\begingroup$ @Elliot I will try that, thank you. $\endgroup$ – Framate Nov 30 '19 at 20:44
  • $\begingroup$ @ElliotG, I added what I tried, would you mind to take a look? Thank you. $\endgroup$ – Framate Dec 8 '19 at 5:49

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