Let $(\Gamma_n)_n$ ($\subset PSL(2,\mathbb{R})$) be a sequence of discrete groups, if we say that $(\Gamma_n)_n$ converges to a group $\Gamma$ this means that there exist isomorphisms $\tau_n:\Gamma\rightarrow\Gamma_n$ such that for all $\sigma\in\Gamma,\tau_n(\sigma)$ converge to $\sigma$.

Do someone have an idea or a reference about the proof of the following Lemma:


Let $(\Gamma_n)_n$ be a sequence of Fuchsian groups converging to a group $\Gamma$ and $\tau_n$ the isomorphisms given in the definition above then:

There exist homeomorphisms $\phi_n:\mathbb{H}^2/\Gamma\longrightarrow\mathbb{H}^2/\Gamma_n$ whose lifts $\tilde{\phi}_n$ satisfies for any $\sigma\in\Gamma$


and such that $(\tilde{\phi}_n)_n$ converges to the identity map uniformly on compact sets i.e that $(\tilde{\phi}_n)$ satisfies, $$\forall x\in\mathbb{H}^2, \tilde{\phi}_n(x)\underset{n\rightarrow\infty}{\longrightarrow}x$$

($\mathbb{H}^2$ is the hyperbolic plane.)

Thanks for any answer

  • $\begingroup$ I think, this was asked earlier at MSE. The answer, as before, that without further assumptions, your lemma is false. $\endgroup$ – Moishe Kohan May 9 at 20:41
  • $\begingroup$ @MoisheKohan why it is false? For me the Lemma is true like that $\endgroup$ – Jongar Jongar May 9 at 20:47
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    $\begingroup$ You can find a sequence of free groups of infinite rank $\Gamma_n$ converging to a free group of infinite rank $\Gamma$ such that $H^2/Gamma$ is a planar surface, while each $H^2/\Gamma_n$ has infinitely many handles. Hence, homeomorphisms $\phi_n$ do not exist. $\endgroup$ – Moishe Kohan May 9 at 21:42
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    $\begingroup$ @SantanaAfton: Lemma holds for finitely generated groups, but I am not sure the result is stated anywhere in the literature. $\endgroup$ – Moishe Kohan May 10 at 14:25
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    $\begingroup$ In fairness to the OP, the phrase "Fuchsian groups" is used inconsistently in the literature. Often, it includes the hypothesis of finite generation. $\endgroup$ – HJRW May 11 at 10:38

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