10
$\begingroup$

A geodesic in a proper CAT(0) space is said to be rank 1 if it does not bound a flat half-plane and zero-width if it does not bound a flat strip of any width. Let $X$ be a geodesically complete CAT(0) space that contains a rank 1 geodesic. Assume it admits a properly discontinuous cocompact action by a group $G$.

Under these conditions we know:

  1. $X$ contains a rank 1 geodesic which is an axis of an isometry in $G$ ([Link, Lemma 4.2] https://arxiv.org/pdf/1706.00402.pdf).

  2. $X$ contains zero-width geodesic ([Ricks, Theorem 2] https://arxiv.org/abs/1410.3921).

My question: Must $X$ necessarily contain a zero-width geodesic which is the axis of some isometry in $G$?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Maybe this is built in to your definition, but do you want to assume that $X$ is irreducible? Otherwise I think you can just take a product with an interval to get a counterexample (and to Ricks' theorem). $\endgroup$
    – Ian Agol
    Jul 14, 2020 at 5:07
  • 1
    $\begingroup$ I believe this would not be geodesically complete--not all geodesic segments would be extendible to infinity $\endgroup$
    – Yellow Pig
    Jul 14, 2020 at 5:17
  • 1
    $\begingroup$ Good point, thanks, I assumed I was missing something. $\endgroup$
    – Ian Agol
    Jul 14, 2020 at 8:42
  • 3
    $\begingroup$ The question has a positive answer at least for CAT(0) cube complexes. Caprace and Sageev constructed isometries skewering pairs of strongly separated hyperplanes, and the axis of such an isometry must have zero width. $\endgroup$
    – AGenevois
    Aug 12, 2020 at 9:45

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.