0
$\begingroup$

Let $f:\mathbb{R}^n\to\mathbb{R}$ be a Morse function whose all critical points are of index 0. Is it true that $f$ has a unique critical point if it has at least one critical point?

Since $\mathbb{R}^n$ is not compact, the relation between the number of critical points of different indices and the Euler characteristic does not apply here. So I have no idea about this question currently. Thank you.

$\endgroup$
1
$\begingroup$

This is false for compactness reasons. You can not do this in $\mathbb{R}$, but there exists a function on $\mathbb{R}^2$ with two local minima and no other critical points.

It is a bit hard to write an explicit formula, but consider the graph of the function $f(x,y)=(x-1)^2(x+1)^2+y^2$. This has two minima at $(\pm 1,0)$ and one saddle at $(0,0)$. Imagine dragging the saddle point to $(0,\infty)$.

Another way of seeing that you can do this is to Look at the function on $X=R^2\setminus [-1/2,1/2]\times [1/2,\infty)$, where it has only the two minimia. Convince yourself that $X$ is diffeomorphic to $R^2$.

There are Morse relations for Morse functions on non-compact manifolds. Keywords to look for are exhaustive Morse functions, proper functions and the Palais-Smale condition.

| cite | improve this answer | |
$\endgroup$

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.