Critical Points of a smooth map on a compact manifold

Show that a smooth map $f$ from a compact manifold $N$ to $mathbb R^m$ has a critical point. (Hint: Let $pi$: $mathbb R^m rightarrow mathbb R$ be the projection to the first factor. Consider the composite map $pi $ $circ$ $f$: $N rightarrow mathbb R$.

Assumptions given:

$N$ is a compact manifold

$pi $ $circ$ $f$: $N rightarrow mathbb R$.

$pi$: $mathbb R^m rightarrow mathbb R$

$f$ is a smooth map.

Attempt at proof:
Let $p in N$. Smoothness of $f: N rightarrow mathbb R^m$ means that there are charts $(U,phi)$ about $p in N$ and $(V,psi)$ about $f(p)$ $in mathbb R^m$ such that $psi circ f circ phi^-1$ is smooth.

I was pondering if I should use a chart $(V,pi)$ in $mathbb R^m $ and possibly $(U, pi circ f)$ in $N$. I'm also wondering how to use the compactness of N. I'm a bit lost, so without giving the full solution, I'm looking for a push in the right direction via some extra advise on how to approach this problem.

BTW I'm using Introduction to Manifolds by Loring Tu.

1 Answer
1

The hint given to you is related to the fact that every smooth real function (i.e., $f:M to mathbbR$) on a compact manifold has a critical point. This is, in turn, due to the fact that every continuous real function on a compact set attains a maximum at some point $p$. Then, if the function is differentiable on this point $p$, we must have that the derivative is zero (if you don't know how to prove this, you can pick curves passing through $p$ and use the one-variable result of "local maximum$implies f'=0$". The next step depends on your explicit definition of derivative on manifolds).

Given that, one now uses the chain rule in your case. Indeed, letting $p$ be a critical point of $pi circ f$, we know that $(pi circ f)'_p=pi'_f(p) circ f'_p=0,$ which implies that $f'_p$ cannot be surjective, since composition of surjective linear maps $(pi'_f(p)$ is obviously surjective) is surjective.

Just to be extremely explicit, since this may cause confusion: a critical point is not a point where the derivative is zero, in general. It is a point where the derivative is not surjective. These concepts coincide in the case of codomain with dimension $1$ because of the fact that linear maps with $1$-dimensional codomain are either surjective or zero.

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.