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Suppose we have two different deck transformations $\tilde{S} \to \tilde{S}$ (where $S$ hyperbolic surface, which is closed, possibly punctured), say $\tau$ and $\kappa$ such that $\tau^i \ne \kappa^j$. We can extend them to be maps on the closure of $\tilde{S}$. By abuse of notation, we will denote these maps by $\tau$ and $\kappa$. Is there anyone proved that they don't share their fixed point(s) (in $\partial \tilde{S}$)?

Motivation: suppose $\tau$ is a deck transformation corresponding to an essential closed curve $\alpha$ in hyperbolic surface $S$ (the definition of $\tau$ can be seen here - in the answer of Riccardo). In the case of closed, possibly punctured hyperbolic surface $S$, $\tau$ is an isometry of $\mathbb{H}^2$ of hyperbolic type. And moreover, the two fixed points of $\tau$ are not lifts of a puncture of $S$.


It is sure that $\tau$ and $\kappa$ don't share their fixed point in $\tilde{S} = \mathbb{H}^2$. For an isometry of parabolic or hyperbolic type of $\mathbb{H}^2$, it will have one or two fixed point(s) in $\partial\mathbb{H}^2$ (then it is clear that an isometry of elliptic type, which has no fixed point in $\partial \mathbb{H}^2$, does not share its fixed point with two other types). I am wondering about the case of one isometry of hyperbolic type and one isometry of parabolic type (also for two hyperbolic types and two parabolic types).

I'm thinking about the following picture:

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Any isometry $f \in \text{Isom}^{+} (\mathbb{H}^2)$ can be extended uniquely to a map $\overline{f}: \overline{\mathbb{H}^2} \to \overline{\mathbb{H}^2}$. Since $\overline{f}$ is a self-homeomorphism of a closed disk, the Brouwer fixed point theorem gives that $\overline{f}$ has fixed point in $\overline{\mathbb{H}^2}$.

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This has nothing to do with hyperbolic geometry or the mapping class group. This is just algebraic topology.

The deck transformation group is a group. So if $\tau$ and $\kappa$ are deck transformations, it follows that $\kappa^{-1} \tau$ is a deck transformation.

You started by saying that $\tau$ and $\kappa$ are different deck transformations, so $\tau \ne \kappa$. It follows that $\kappa^{-1} \tau$ is not the identity element of the deck transformation group.

A nonidentity element of the deck transformation group has no fixed points, so $\kappa^{-1}\tau(x) \ne x$ for all $x \in \tilde S$.

It follows that $\tau(x) \ne \kappa(x)$ for all $x \in \tilde S$.

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  • $\begingroup$ Hi Lee Mosher, thank you for the answer. For an isometry of parabolic or hyperbolic type of $\mathbb{H}^2$, it will have one or two fixed point(s) in $\partial\mathbb{H}^2$ (then it's clear that an isometry of elliptic type, which has a fixed point in $\mathbb{H}^2$, does not share its fixed point with two other types). What you were writing is to show that an isometry of elliptic type does not share its fixed point with two other types. I am wondering about the case of one isometry of hyperbolic type and one isometry of parabolic type (also for two hyperbolic types and two parabolic types). $\endgroup$ – user578196 Aug 14 '18 at 9:10
  • $\begingroup$ Dear Lee Mosher, I understood now. The reason is the discreteness of the group. $\endgroup$ – user578196 Aug 16 '18 at 19:40

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