8

Assume that $M$ is a non compact smooth manifold. Is there a smooth map $f:M\to \mathbb{R}$ such that $f$ has no critical point?

The motivation comes from the conversations on this post.

Community
  • 1
Ali Taghavi
  • 925
  • 1
    I don't think so. Take the infinite genus surface with one end, for instance. Or I think a punctured torus should be a serviceable example. –  Sep 11 '15 at 16:08
  • @MikeMiller do you think that the infinite mobious strip is another example? – Ali Taghavi Sep 11 '15 at 16:24
  • I think any example is diffeomorphic to $M \times \Bbb R$ in such a way that the function $f$ is the projection. –  Sep 11 '15 at 16:28
  • 3
    @MikeMiller: Not necessarily. Take any open subset of $\mathbb R^n$, and let $f$ be any linear function. Such an open subset need not be diffeomorphic to a product. Your conclusion does hold if $f$ is proper, though. – Jack Lee Sep 11 '15 at 17:07
  • I agree. Thanks for pointing that out. I guess I don't really know anything about non-proper Morse functions. –  Sep 11 '15 at 17:08
  • @JackLee is there an smooth function on infinite mobius band without critical point? – Ali Taghavi Sep 11 '15 at 18:11
  • 1
    I think the method of this answer https://math.stackexchange.com/a/4324977/402997 applies here – Grisha Taroyan May 24 '23 at 08:48
  • 1
    @GrishaTaroyan : yes, it works if one starts with the gradient vector field of a Morse function and instead of pull back of vector field via diffeomorphisms one uses pull back of the function. – Moishe Kohan Mar 29 '25 at 14:41

1 Answers1

12

Interesting question. I did a little research, and found out that apparently the answer is yes: A 1961 paper by Morris Hirsch (Theorem 4.8) showed that every noncompact (connected) smooth manifold admits a smooth real-valued function with no critical points.

Jack Lee
  • 50,850