Is there any noncompact manifold with the fixed point property?

Solution 1:

I have a feeling that this question was already asked at MSE some time ago, but it's easier to write an answer than to find a duplicate. So, here it is.

Let me assume that the noncompact manifold $M$ contains a noncompact connected component $X$. (I will leave it to you to find a fixed-point-free self-map of a manifold which has infinitely many connected components.)

The basic observation is that every noncompact connected manifold $X$ contains a ray, i.e. a closed subset $A$ homeomorphic to $[0,\infty)$, equivalently, that there is an injective proper continuous map $c: [0,\infty)\to X$. See, for instance, Kajelad's answer here.

In fact, the same is true for a much larger class of topological spaces:

Let $X$ be a metrizable, 2nd countable, locally compact, connected, locally path-connected, noncompact space. Then $X$ contains a ray.

Now, back to our manifold. Given a ray $A\subset X$, and a homeomorphism $$ c: [0,\infty)\to A, $$ define the continuous map $f: A\to [1,\infty)$, $f(a)=c^{-1}(a)+1$. Now, use the Tietze extension theorem to extend $f$ to a continuous map $$ g: M\to (0,\infty). $$ Lastly, take the composition $$ h: M\to A\subset M, h= c\circ g. $$ Clearly, $h$ has no fixed points (since its restriction to $A$ has no fixed points).

You can see from the proof that the existence of fixed-point free maps holds in much greater generality, say, for connected, locally path-connected, metrizable, 2nd countable, locally compact, noncompact spaces.

Edit. From the comments, it's clear that you are actually interested in constructing a diffeomorphism without fixed points on a noncompact connected manifold $M$. Here are the key steps in the construction:

i. Every smooth connected noncompact manifold admits a nonvanishing vector field. This was discussed on MSE and MO many times (my list is admittedly incomplete):

January 2011, November 2011, November 2013, November 2018, June 2021,

The most satisfactory answer (to my taste) is the one from December 2020: It avoids the obstruction theory and the details are easy to fill-in.

ii. Given a nonvanishing vector field $X$ on a manifold $M$, one can multiply $X$ by a positive function $\varphi\in C^\infty(M)$ such that the new vector field $Y=\varphi X$ is complete. My favorite way to do so is by using a complete Riemannian metric $g$ on $M$ and normalizing $X$ to a unit vector field on $M$ as it is done in this answer.

iii. Now, one can take the time-one flow $h=F(\cdot, 1)$ of the vector field $Y$. The only way $h$ can have a fixed-point $x$ in $M$ is when $x$ lies on some periodic trajectory $T$ of $Y$, whose length (with respect to the metric $g$ as above) is of the form $1/k$, $k\in {\mathbb N}$. Since $Y$ is nonvanishing, for each compact $K\subset M$ the infimum of lengths of periodic trajectories of $Y$ is bounded from below by some $\epsilon_K>0$. There exists a positive function $\psi\in C^\infty(M)$ such that for each compact $K\subset M$, $$ \inf \psi|_K < \epsilon_K. $$ Then the product $Z=\psi Y$ is still a complete vector field (since it is uniformly bounded from above) and, at the same time, the time-1 flow of $Z$ has no fixed points in $M$.

Lastly: it is unclear to me how to prove a similar results (existence of fixed-point free maps) in the PL and topological category, although, I have no doubt that they still hold.