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Let $\phi\in PSL(2,R)$ be hyperbolic and $\varphi\in PSL(2,R)$ be elliptic. Is it possible to find a local homeomorphism $f:H^2\rightarrow H^2$ such that $f(\phi(x))=\varphi(f(x))$ for all $x\in H^2$ ? I do not think so, but am unable to prove it.

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I assume that $\varphi$ has finite order; a similar construction works in the infinite order case, just it is a bit harder. Consider $D$, the complement to the fixed point of $\varphi$. Then $S'=D/<\varphi>$ is homeomorphic to the annulus $S=H^2/<\phi>$. Now, take a homeomorphism $g: S\to S'$ and let $f: H^2\to D\subset H^2$ be its lift.

Edit: Here is a more general theorem dealing with questions of this type.

Suppose that $M$ is a smooth noncompact connected oriented manifold of dimension $n\le 3$. Let $X$ be an open subset of $R^n$ and $G$ a Lie group acting smoothly on $X$.

Theorem. Every representation $\rho: \pi_1(M)\to G$ is the holonomy of an $(X,G)$-structure on $M$. In other words, there exists a $\rho$-equivariant local diffeomorphism $$ f: \tilde{M}\to X, $$ where $\tilde{M}$ is the universal cover of $M$.

Similar results hold in higher dimensions, but the conditions are more complex.

Conjecture. Let us omit the assumption $n\le 3$ in the above theorem, but assume instead that the universal cover of $M$ is parallelizable. Then, again, every representation $\rho: \pi_1(M)\to G$ is the holonomy of an $(X,G)$-structure on $M$.

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  • $\begingroup$ Thanks Misha. I should have added in my question that I wanted $\endgroup$
    – Dan Gallo
    May 28, 2014 at 1:04
  • $\begingroup$ Thanks Misha. I should have added in my question that I wanted the map f:H^2\rightarrow H^2 to be surjective. Regards. $\endgroup$
    – Dan Gallo
    May 28, 2014 at 1:09
  • $\begingroup$ @Dan Gallo: Surjective maps exist as well, just are harder to describe. I will add a description when I have more time. $\endgroup$
    – Misha
    May 28, 2014 at 1:33

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