Is there any noncompact manifold with the fixed point property?

This question refers to manifolds without boundary. I was thinking about Brouwer´s fixed point theorem when this came to mind, and I found the name of the property ("fixed point property"), but the information I´ve seen about it treats just compact spaces.

At first I thought the answer would be negative and you could create some vector flow going towards infinity in the manifold, but after thinking a bit I see no reason why this should be possible. Maybe the Lefschetz fixed point theorem could be useful if the answer is positive.

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):

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.

• Oh, with your answer I realised that the fixed point property allows any continuous function from the manifold to itself not fixing points. I was originally thinking of homeomorphisms from the manifold to itself not fixing points and at some point the 2 concepts merged in my head. But this is an interesting answer so I´ll leave the question as it is. I would ask the other question but yesterday I found this math.stackexchange.com/questions/586862/… and it shouldn´t be that difficult using it, I´ll try today. Commented Dec 3, 2021 at 10:17
• @SaúlRodríguezMartín: Yes, the existence of such a vector field (you need to assume connectivity) is an exercise in Spivak's book, although Spivak's suggestion is better than what Ryan proposes. An alternative proof is via the obstruction theory. Once you have such a vector field, there are two more obstacles you have to overcome: Make sure that the flow-map is defined and that the time-one flow map does not have fixed points. Both are not hard, try this. Commented Dec 3, 2021 at 14:00
• Thanks for the advice! It seems that both obstacles can be overcome by multiplying the vector field by a small enough scalar function and then taking the time 1 flow. The second one is a bit more tricky but it works locally using the tubular flow theorem and then you can glue with partitions of unity. I had been thinking of using a flow but with a time dependent on the point but this is better, because the other way makes it harder to check the function will be a homeomorphism. Also I like that the functions with no fixed points can be as close to the identity as you want. Commented Dec 3, 2021 at 14:18
• You can overcome the 2nd obstruction the same way you overcome the first, by slowing down the flow via multiplication by a suitable function. Commented Dec 3, 2021 at 14:23