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Let $S_{g,n}^b$ denote a surface of genus $g$ with $n$ punctures and $b$ boundary components. Let us assume $\max\{b,n\}\geq 1$. It is then obvious that $S_{g,n}^b$ deformation retracts to a bouquet of $m:=2g+n+b-1$ circles and $\pi_1(S_{g,n}^b)$ is free on $m$ generators.

Let $m\geq 2$. If $S_{g,n}^b$ is endowed with a complete hyperbolic metric with geodesic boundary, then it is known that there is a Fuchsian group of the second kind $\Gamma$, acting on the disk, such that $S_{g,n}^b$ may be reconstructed as $$\mathbb{D}\cup(S^1-\Lambda(\Gamma))/\Gamma$$ where $\Lambda(\Gamma)$ is the limit set.

My question is as follows: given a discrete faithful representation $F_{2g+n+b-1}\to PSL_2(\mathbb{R})$, how can one find the topological type of the corresponding surface? Notice, for example, $F_4$ could describe a surface with genus $2$ and $1$ puncture or a surface of genus $1$ with $2$ punctures and $1$ boundary component.

Ideally, there is a source or paper where this kind of question has been studied. (Otherwise, I'll have to work it out myself).

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    $\begingroup$ If the representation is faithful but not discrete then it has a dense image and I'm not sure what you mean by "corresponding surface". $\endgroup$
    – YCor
    Commented Apr 30, 2019 at 21:19
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    $\begingroup$ $PSL_2(\mathbb R) \cup (S^1 - \Lambda(\Gamma))/\Gamma$ looks wrong. Perhaps you mean $\mathbb H^2 \cup (S^1 - \Lambda(\Gamma)) / \Gamma$? $\endgroup$
    – Lee Mosher
    Commented Apr 30, 2019 at 21:23
  • $\begingroup$ My bad, brain fart. thanks. And yes, should be discrete rep, lazy me. $\endgroup$
    – user470881
    Commented Apr 30, 2019 at 22:32
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    $\begingroup$ @onepotatotwopotato proper Zariski-closed subgroups in (P)SL$_2$ are virtually solvable. Hence any faithful rep of a group with a free subgroup has Zariski-dense image. So if $H$ is the closure of the image $L$, $H^0$ is normal, hence either $H^0$ is finite central (so $L$ is discrete), or $L$ is dense. $\endgroup$
    – YCor
    Commented Aug 26 at 11:01
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    $\begingroup$ @onepotatotwopotato I wrote "has a free subgroup", not "is free". The surface group has a free subgroup. $\endgroup$
    – YCor
    Commented Aug 27 at 8:44

2 Answers 2

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It suffices to count the punctures and boundary components. For each puncture or boundary component, the loop around it gives a conjugacy class in the fundamental group, well-defined up to inversion. It suffices to characterize which conjugacy classes arise from this construction applied to both punctures and boundary components, as by counting we recover $n$ and $b$, and then using the Euler characteristic we recover $n$.

For a puncture, a necessary and sufficient condition is that the conjugacy class be sent to a unipotent element in $PSL_2(\mathbb R)$ and not be a nontrivial power of another conjugacy class in the free group. The necessity is straightforward. For sufficiency, we need to observe that as soon as an element of $\pi_1$ is sent to a unipotent element of $PSL_2(\mathbb R)$, we can find closed loops representing the element whose lengths converge to zero, which must in the Riemann surface lie in a small neighborhood of a puncture, hence must be a power of the standard loop around that puncture or its inverse.

For a boundary condition, a necessary and sufficient condition is that the the conjugacy class be sent to a diagonalizable element of $PSL_2(\mathbb R)$, that one of the two components of $S^1$ minus the fixed points of that element contain no limit points of the group, and that the conjugacy class is not a nontrivial power of another conjugacy class. We can rephrase the last two conditions as saying that , for one of the two components of $S^1$ minus the two fixed points, every element of the group that sends a point in that component to another point in that component is an integer power of the chosen element. The proofs of necessity of both of these characterizations are fairly straightforward from drawing the fundamental domains, and the proof of sufficiency is clear from the definition you gave with boundary components.

You might find the boundary component characterization unsatisfactory as it is not clear how to compute with it. On the other hand I suspect the puncture characterization can't be improved much - this is essentially what one uses to count cusps of modular curves, for example.

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  • $\begingroup$ Thanks! Very helpful $\endgroup$
    – user470881
    Commented May 1, 2019 at 5:25
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I am essentially just repeating Will's answer, but giving a slightly different point-of-view (and a relevant reference).

Let $F_r$ be a free group of rank $r>1$. Given a discrete faithful representation $\rho:F_r\to \mathrm{PSL}(2,\mathbb{R})$ it is torsion-free and so $\mathbb{H}^2/\Gamma$ is a complete hyperbolic surface $\Sigma_\Gamma$, where $\Gamma:=\rho(F_r)$. You asked how to determine the homeomorphism type of $\Sigma_\Gamma$.

In general, you need to know the gluing data for the fundamental domain. For this, it suffices to know simple closed curves around each boundary and puncture (which tells you $n+b$). For then, using $r=2g+n+b-1$ you can determine $g$.

Around each collar neighborhood of a boundary, the holonomy $\rho$ must correspond to a translation (which determines a geodesic length) and so its absolute trace will be $>2$. Around each puncture the holonomy of the corresponding loop must be a horolation (rotation at infinity) and so its absolute trace must be 2. So from the traces you can determine which loops are at punctures and which are at boundaries.

I recommend reading concrete examples in Trace Coordinates on Fricke spaces of some simple hyperbolic surfaces by William M. Goldman. See Section 4, in particular.

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    $\begingroup$ Thanks for the reference! $\endgroup$
    – user470881
    Commented May 1, 2019 at 16:58

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