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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 $G$?

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    $\begingroup$ 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
    Commented Jul 14, 2020 at 5:07
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    $\begingroup$ I believe this would not be geodesically complete--not all geodesic segments would be extendible to infinity $\endgroup$
    – Yellow Pig
    Commented Jul 14, 2020 at 5:17
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    $\begingroup$ Good point, thanks, I assumed I was missing something. $\endgroup$
    – Ian Agol
    Commented Jul 14, 2020 at 8:42
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    $\begingroup$ 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
    Commented Aug 12, 2020 at 9:45

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