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Let $X$ be a rank $1$ symmetric space of non-compact type and $G$ its isometry group. $G$ is a semisimple linear algebraic Lie group of non-compact type with trivial center. Let $\rho$ be a representation of a discrete group $\Gamma$ into $G$. It is known that if the Zariski closure of $\rho$ is reductive then there is a finite subset $F\subset \Gamma$ and a constant $D\geq 0$ such that for any $\gamma \in \Gamma$ there is an $f\in F$ such that $$|\mu(\rho(\gamma f)) - \lambda(\rho(\gamma(f))|\leq D$$ Here, $\mu$ is the displacement function with respect to a fix point $p$, $$\mu(g) = d(p,g\cdot p)$$ and $\lambda$ is the translation length $$\lambda(g) = \inf_{x\in X} d(x,g\cdot x)$$

In the literature people say that $\gamma f$ is $\textit{proximal}$. This fact is stated on page 71 of this paper: https://arxiv.org/abs/1307.0250 and the ingredients of a proof are outlined on pages 7 and 8 of this document: https://arxiv.org/pdf/0704.3499.pdf

I am wondering about the following: suppose $X$ is instead a complete $CAT(-1)$ metric space and $\rho$ is reductive, which in this case means the number of fixed points of $\rho(\Gamma)$ on the ideal boundary $\partial_\infty X$ is different than $1$ (this is equivalent to Zariski closure being reductive in symmetric space setting), does the existence of $F$ and $D$ as above still hold? This seems to me like a general fact that should not rely on the fact that we have a matrix Lie group, although the proof heavily uses this machinery.

I must admit I have not spent too long trying to prove this; if the result is already out there I don't want to reinvent the wheel. I should also comment that all I really need for my purposes is for $X$ to be a $CAT(-1)$ Hadamard manifold, but I would expect the general result to be true.

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  • $\begingroup$ How do you define the Cartan projection $\mu$ in such a general setting (when $X$ is a complete CAT($-1$)-space, and even in the Hadamard context)? $\endgroup$
    – YCor
    Commented Sep 6, 2019 at 8:18
  • $\begingroup$ @YCor, what's wrong with the definition given above? I suppose one should at least assume that the isometry group of $X$ acts cocompactly on $X$, if one doesn't the choice of $p$ to matter too much. $\endgroup$
    – HJRW
    Commented Sep 6, 2019 at 8:57
  • $\begingroup$ Oops, sorry I didn't notice, and this meaning seems weird. This is a function from the group into real numbers, it sounds weird to call it a projection. $\endgroup$
    – YCor
    Commented Sep 6, 2019 at 18:57
  • $\begingroup$ My function $\mu$ reflects the geometry of the Cartan projection when $X$ is, say, $\mathbb{H}^n$. I'll just edit my question so there's no possible confusion. $\endgroup$
    – user470881
    Commented Sep 6, 2019 at 19:03
  • $\begingroup$ Actually, unless I misunderstood, it seems to me that it fails even in the context you're considering. Say, $G\subset PSL_2(R)$ is dense. Choose $g_i$ loxodromic with axis $a_i$, with $s_i=d(a_i,p)\to\infty$. Then for fixed $F$, $d(fa_i,p)$ also tends to infinity for each $f$. Then $d(p,f^{-1}g_ifp)$ is roughly (=up to bounded error) equal to $2s_i+\lambda(g_i)$, and also roughly equal to $\mu(g_if)$. Hence $\min_{f\in F}(\mu(g_if)-\lambda(g_if))$ tends to $+\infty$. $\endgroup$
    – YCor
    Commented Sep 6, 2019 at 19:30

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