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.


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.

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