Periodic orbit property

Question

A topological space $X$ satisfies the "Periodic orbit property", briefly POP, if for every continuous map $f:X \to X$, there exist a natural number $n$ and a point $x_{0}\in X$ such that $f^{n}(x_{0})=x_{0}$.

Obviously fixed point property (FPP) implies POP.

For a natural number $n$, a topological space $X$ is called $n$-POP if for every continuous map $f$ on $X$, $f^{n}$ has a fixed point. (Ex: $\mathbb{S}^{2n}$ is a 2-POP manifold, because the degree of a fixed-point-free map on $\mathbb{S}^{2n}$ must be $-1$.)

The question:

Is there an example of a manifold $M$ which satisfies POP but for every $n\in \mathbb{N}$, there is a continuous map $f$ on $M$ such that $f^{n}$ has no fixed point?

Namely: we search for a manifold for which every self-map has a periodic orbit, but there is no any control on periods.

Equivalently:

Is there a manifold $M$ which is POP but not $n$-POP for all $n\in \mathbb{N}$?

In particular, can we say:

"every compact POP manifold is necessarily a $n$-POP manifold, for some $n$"?

Motivated by the Lefschetz fixed-point theorem, we ask that:

What algebraic topological criterion, can be introduced for consideration of this property(POP)?

Edit: According to the very interesting answer of Qiaochu Yuan, in the orientable case, the question is equivalent to the following:

Let M be a closed orientable manifold. Is it true that $M$ is not POP if and only if $\chi(M)=0$?

Note1 For a related question see this post and it is natural to ask that "Does $S^{2}\vee S^{2}$ satisfy the periodic orbit property?"

Note2 I think the continuation of the argument of Qiaochu Yuan for his first statement is not easy, for arbitrary manifold. Because for the simplest case $S^{3}$ we had the famous conjecture of "existence of a vector field on $S^{3}$ without periodic orbit. In fact consideration of non vanishing vector fields is necessary but not sufficient. Periodic orbits of vector fields are important, too. Moreover, perhaps an approach which is not based on "vector fields" could be useful, for example consideration of orientation-reversing diffeomorphisms.

Note 3: "pointwise periodic homeomorphism' is a concept which is indirectly similar to the subject of this post.

Answer 1

Nice question! Here's what I can show.

Let $X$ be a smooth closed manifold. Then:

(1) If $\chi(X) = 0$, then $X$ is not $n$-POP for any $n$.

(2) If $\chi(X) \neq 0$ and $X$ is orientable, then $X$ is $\text{lcm}(1, 2, ... n)$-POP with respect to maps $f : X \to X$ of nonzero degree, where $n = \text{max}(b_0 + b_2 + ..., b_1 + b_3 + ...)$ (where $b_i$ is the $i^{th}$ Betti number of $X$).

Proof of 1. We will use the converse of the Poincaré-Hopf theorem: if $\chi(X) = 0$, then $X$ admits a nonvanishing vector field. Let $\varphi(t)$ denote the flow of this vector field. Let $t_{0}>0$ be small enough so that $\varphi(t_0)$ has no fixed points. Such $t_{0}$ exists, because there is a positive uniform lower bound for the period of all periodic orbits.(As a consequence of the flow box theorem, around regular points of a vector field). For a given $n \in \mathbb{N}$, let $f = \varphi \left( \frac{t_0}{n} \right)$. Then $f^n$ has no fixed points, hence $X$ is not $n$-POP. $\Box$

(I strongly suspect that in this case $X$ is not POP either; it seems like we should be able to consider a small flow of a sufficiently generic nonvanishing vector field. But I don't know how to finish this argument.)

Proof of 2. We will need the following two observations.

Lemma 1: Let $f_0, f_1$ be linear operators acting on two finite-dimensional vector spaces $V_0, V_1$. If $\text{tr}(f_0^k) = \text{tr}(f_1^k)$ for $k$ between $1$ and $\text{max}(\dim V_0, \dim V_1)$, then $f_0$ and $f_1$ have the same nonzero eigenvalues with the same multiplicities.

Proof. The above condition implies, using the Newton-Girard identities, that $f_0$ and $f_1$ have the same characteristic polynomial up to factors of $t$. $\Box$

Lemma 2: Let $X$ be an $n$-dimensional smooth closed oriented manifold and let $f : X \to X$ be a map of nonzero degree. Then every eigenvalue of $f$ acting on cohomology (with complex coefficients) is nonzero.

Proof. Let $e_1, ..., e_d$ be a basis of generalized eigenvectors for the action of $f$ on $H^k(X, \mathbb{C})$. By Poincaré duality the cup product $H^k \otimes H^{n-k} \to H^n$ is nondegenerate, so we can find a dual basis $e_1^{\ast}, ..., e_d^{\ast}$ of $H^{n-k}(X, \mathbb{C})$. Since $f$ acts by a nonzero scalar, namely $\deg f$, on $e_i \smile e_i^{\ast}$ for all $i$, the generalized eigenvalue of $e_i$ must also be nonzero. $\Box$

Now back to the proof of 2. With hypotheses as above, let $f_0$ denote the map induced by $f$ on the direct sum $V_0$ of the even-dimensional complex cohomology of $X$ and let $f_1$ denote the map induced by $f$ on the direct sum $V_1$ of the odd-dimensional complex cohomology of $X$, so that the Lefschetz trace of $f^k$ can be written

$$L(f^k) = \text{tr}(f_0^k) - \text{tr}(f_1^k).$$

By Lemma 2, the eigenvalues of $f_0$ and $f_1$ are all nonzero, so if $f_0$ and $f_1$ have the same nonzero eigenvalues then in particular $\dim V_0 = \dim V_1$. By the contrapositive of Lemma 1, if $\chi(X) = \dim V_0 - \dim V_1 \neq 0$, then there exists some $k$ between $1$ and $n = \text{max}(\dim V_0, \dim V_1)$ such that $L(f^k) \neq 0$, hence, by the Lefschetz fixed point theorem, such that $f^k$ has a fixed point. In particular, $f^{\text{lcm}(1, 2, ... n)}$ has a fixed point. $\Box$

Answer 2

The following is proved by F. Brock Fuller in "The Existence of Periodic Points," Annals of Mathematics, Vol. 57, 1953, pp. 229-230:

Theorem. Let X be a compact simplicial complex with Betti numbers $B_i$, and nonzero Euler characteristic. If a continuous map $f: X\rightarrow X$ induces isomorphisms of homology groups, then $f$ has a point whose period is $\le \max_i \big (∑_i B_{2i},∑_i B_{2i-1} \big)$. Thus every homeomorphism of $X$ has a periodic orbit.