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, 2014 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, 2016 at 14:04

1 Answer 1


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).


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