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Let $a,b$ be two loxodromic isometries of a CAT(0) space. Assume that, for every $n \geq 1$, $a^nb$ is also loxodromic. Is it possible for the translation length of $a^nb$ to be bounded independently of $n$?

First, I thought as obvious that the translation length of $a^nb$ has to tend to $+ \infty$ as $n \to + \infty$, but I may have been misled by the CAT(-1) case (where this is clearly true). Now, I go back and forth between a possible counterexample and an easy argument I am missing...

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  • $\begingroup$ Could you remind us of the definition of loxodromic? $\endgroup$
    – IJL
    Commented Mar 5, 2021 at 10:08
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    $\begingroup$ Loxodromic = there exists a bi-infinite geodesic on which the isometry acts as a translation. $\endgroup$
    – AGenevois
    Commented Mar 5, 2021 at 10:41
  • $\begingroup$ @PeterKosenko: You should write $\ell(a^nb)= \lim\limits_{k \to + \infty} d(x_0,(a^nb)^kx_0)/k$. $\endgroup$
    – AGenevois
    Commented Mar 5, 2021 at 11:13
  • $\begingroup$ Yes, you are correct -- and now I realize that it doesn't immediately imply your statement... $\endgroup$
    – Peter Kosenko
    Commented Mar 5, 2021 at 11:51

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Consider the following two transformations of $\mathbb{R}^3$: $a:(x,y,z)\mapsto (x+1,y,z)$ and $b:(x,y,z)\mapsto (-x,-y,z+1)$. The translation axis for $a^nb$ is the line $x=n/2$, $y=0$ and $a^nb$ translates this line by a distance of 1, independent of $n$.

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  • $\begingroup$ I've just realized that the $y$-coordinate isn't needed: $a$ and $b$ both preserve the plane $y=0$, so there is a simpler example in $\mathbb{R}^2$. $\endgroup$
    – IJL
    Commented Mar 5, 2021 at 16:59
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    $\begingroup$ This even works restricted to the $\mathbb{R}^2$ that is the xz-plane. $\endgroup$
    – Matt Zaremsky
    Commented Mar 5, 2021 at 16:59
  • $\begingroup$ (Sorry, simultaneous comment!) $\endgroup$
    – Matt Zaremsky
    Commented Mar 5, 2021 at 17:00
  • $\begingroup$ You were quicker than me Matt - should I delete my comment? $\endgroup$
    – IJL
    Commented Mar 5, 2021 at 17:00
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    $\begingroup$ I don't know the rules! I'm inclined to just leave everything. $\endgroup$
    – Matt Zaremsky
    Commented Mar 5, 2021 at 17:01

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