I would like to show the following: Let $X$ be a complete metric space that is uniquely geodesic (i.e. each two distinct points are connected by a unique geodesic segment) and $\phi\colon X\to X$ an isometry of finite order $k$. Then $\phi$ has a fixed point.

My ideas so far:

  1. Pretty elementary: The statement is obvious if $k\leq 2$. So take an orbit $x_1,...,x_k$. It's fairly easy that the whole orbit can't lie on one geodesic (otherwise it would contradict uniquely geodesic). Take the midpoints of the geodesic segments $[x_i,x_{i+1}]$ (mod $k$) and call them $x_i^{(1)}$. Do this inductively. Again by unique geodesic we have $d_j:=d(x_1^j,x_2^{j})>d_{j+1}$. From that follows easily by triangle inequality that $d(x_1^j, x_1^{j+1})$ is strictly decreasing. If this sequence converges to $0$, the $x_1^j$'s converge to a point which is $\phi$ invariant. But I have no idea why it should.

  2. $X$ is contractible. If $\phi$ acts fixed point free, then $X\to X/\phi$ is a covering (is that even true?) that induces a metric on the quotient, which is a $K(\mathbb{Z}_k,1)$. Is it a contradiction that a $K(G,1)$ of a group having torsion is metrizable (e.g. because it's too complicated as a CW complex?

  3. Give up and assume $X$ is moreover a manifold and find a compact convex $\phi$ invariant set. Show that such a set is homeomorphic to a closed ball (is that always true?) and use Brouwer.

  • $\begingroup$ If $k=p$ is a prime, you can certainly combine 2 and 3: $K(\Bbb Z_p,1)$ cannot be finite dimensional. $\endgroup$ Dec 2 '14 at 20:19
  • $\begingroup$ Regarding 2, my understanding is that the quotient map is not necessarily a covering map. If the order of the isometry is not prime, then a priori orbits can have different cardinalities. This can happen for diffeomorphisms, see mathoverflow.net/questions/222567/… $\endgroup$ Sep 23 '16 at 14:04

If $X$ is a CAT(0) space, this is true (the standard reference is Bridson-Haefliger, corr. 2.8, though the result precedes them by several decades). I believe the argument also works for $p$-uniformly convex spaces (as in Naor-Silberman), for any $p$ (take the $p$-barycenter of the counting measure on the orbit of your cyclic group).


This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .