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Let $\Gamma$ be a Fuchsian group. Choosing any point $p\in \Bbb H^2$ not fixed by any non-identity element of $\Gamma$, we can construct a hyperbolically convex connected fundamental region, called Dirichlet region, denoted by $D_p$. Also, $\partial D_p$ is contained in the geodesic $\{z\in \Bbb H^2:d(z,p)=d(z,\gamma p)\}$ for any $\gamma\in \Gamma\backslash \{\text{Id}\}$, i.e., hyperbolic area of $\partial D_p$ is zero.

Note that $D_p$ is closed in $\Bbb H^2$ but may not be in $\widetilde{\Bbb H^2}:=\Bbb H^2\cup \Bbb R\cup\{\infty\}$. Let $\widetilde{D_p}$ be the closure w.r.t. $\widetilde{\Bbb H^2}$. Then, $\widetilde{D_p}\backslash D_p$ may have uncountably many components. A component having positive $($Euclidean$)$length is called free side of $D_p$$($there are at most countably many free sides$)$. Any other component is a point: a point in $\widetilde{D_p}\backslash D_p$ is a proper vertex if it is the end-point of two sides$($in $\Bbb H^2)$ of $D_p$, and a point in $\widetilde{D_p}\backslash D_p$ is a improper vertex if it is the end-point of a side$($in $\Bbb H^2)$ and a free-side of $D_p$.

$\textbf{Question 1:}$ Is it possible that there is a point in $\widetilde{D_p}\backslash D_p$ that is neither proper nor improper vertex?

$\textbf{Question 2:}$ Are the number of free sides, proper or improper vertices independent of the chosen point $p$? In other words, is it possible to find some invariants for $\Gamma$ for counting free sides, proper or improper vertices $($after classifying under some equivalence relations$)$?

The above two questions are motivated by the following statements for any $p$ not fixed by any non-identity element of $\Gamma$:

$(1)$ $D_p$ has no free side $\iff$ $(2)$ $D_p$ has a finite hyperbolic area $\implies$ $(3)$ $D_p$ has finitely many sides$($in $\Bbb H^2)$ $\impliedby$ $(4)$ $\Gamma$ is finitely generated $\iff$ $(5)$ $\Bbb H^2/\Gamma$ is topologically finite.

Here, $\Bbb H^2/\Gamma$ topologically finite means $\Bbb H^2/\Gamma$ is an orbifold with finite genus, finite number of marked points$($the number of these points corresponds to the number of conjugacy classes of maximal elliptic cyclic subgroups$)$, a finite number of punctures$($the number of these punctures corresponds to the number of conjugacy classes of maximal parabolic subgroups$)$.

Someone
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    What is your definition of a Fuchsian group? – Moishe Kohan Jun 14 '21 at 21:02
  • Discrete subgroup of $\text{PSL}(2,\Bbb R)=\text{Iso}^+(\Bbb H^2)$. But I am not getting any idea from this hint. Could you elaborate more? – Someone Jun 15 '21 at 03:46
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    It was not a hint but a request for a clariufication. With your definition, the equivalence statements you are quoting are simply false. You need to assume finite generation. https://math.stackexchange.com/questions/2716033/fuchsian-groups-of-the-first-kind-and-lattices/3376910#3376910 – Moishe Kohan Jun 15 '21 at 09:26
  • Many thanks for pointing out my fault. Actually, $(2)\implies (3)$ by Siegel's Theorem, but $(3)\implies (2)$ if we assume $\Gamma$ is of the first kind, otherwise, the Dirichlet region may have a free side. $(1)\iff(2)$, $(4)\iff(5)$ and $(4)\implies (3)$ in any case. But, $(3)$ does not imply $(4)$ in general, and this can be interpreted(though I have no concrete counterexample) as there is an infinite-type surface having a geometrically finite Dirichlet region. Can you tell me something about $(2)\implies (4),(4)\implies(2)$ in the general case or some counterexample? – Someone Jun 15 '21 at 12:42
  • Also, while writing my question, I used this .pdf which is the second chapter of this book. Theorem 2.10. here is certainly not hold in general, as you already said. Similarly, here theorem 2.4. The authors of these two are the same. Thanks again. – Someone Jun 15 '21 at 12:53
  • Searching further, I got this: If a Fuchsian group has a fundamental region of a finite hyperbolic area then is of the first kind. So, by your answer, in this case, the Fuchsian group is finitely generated, i.e. we have $(2)\implies (4)$. So, I need a finite type surface with an infinite hyperbolic area(may be $\Bbb H^2$) and an infinite-type surface having a geometrically finite Dirichlet region. – Someone Jun 15 '21 at 13:24
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    I will add an answer later today. – Moishe Kohan Jun 15 '21 at 13:41
  • Sure, thanks, I will wait for it. – Someone Jun 15 '21 at 14:10

1 Answers1

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The correct implications are:

(0) The limit set $\Lambda$ of $\Gamma$ is the entire circle $S^1$ (i.e. $\Gamma$ is of the first kind) $\iff$ (1) $D_p$ has no free side $\Leftarrow$ (2) $D_p$ has a finite hyperbolic area $\implies$ (3) $D_p$ has finitely many sides $\iff$ (4) $\Gamma$ is finitely generated $\iff$ (5) the orbifold $\Bbb H^2/\Gamma$ is topologically finite.

The equivalence (0)$\iff$(1) comes from the fact that $D_p$ is a fundamental domain of $\Gamma$ in ${\mathbb H}^2$ and, moreover, the complement $\overline{D_p} \setminus \Lambda$ is a fundamental domain for the action of $\Gamma$ on $\overline{{\mathbb H}^2}\setminus \Lambda$.

There are many examples where $\Gamma$ is of the first kind but $D_p$ has infinite area (equivalently, $\Bbb H^2/\Gamma$ has infinite area), one is given in my answer here. The implication (2)$\Rightarrow$(1) is clear. The implications

(2)$\Rightarrow$(3)$\iff$(4)$\iff$(5)

are tricky and you can find proofs in

A.F.Beardon, "Geometry of Discrete Groups", Springer Verlag, 1983.

Moishe Kohan
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  • Is not it "equivalently $\Bbb H^2/\Gamma$ has infinite area"? in the first line of the third paragraph. Also, $(0)$ I can think of as an isolated one (as it is specific to the first kind), and the rest of (bi)implications do hold for any case(means irrespective of kind of $\Gamma$). Am I right? – Someone Jun 16 '21 at 04:54
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    @User: Right, corrected. (0) is included because it is a standard condition equivalent to (1) in your list. – Moishe Kohan Jun 16 '21 at 12:16