Fuchsian Groups of the First Kind and LatticesFirst, a Fuchsian group is a discrete subgroup of PSL 2 ( R ), which we can view as a group of transformations of the upper half-plane H that acts discontinuously. A lattice is a Fuchsian group with finite covolume. In other words, it is a Fuchsian group that has a fundamental domain in H with finite hyperbolic area (and then any two fundamental domains have the same hyperbolic area).My confusion arises with the notion of a Fuchsian group of the first kind. Every definition I have seen for this term is roughly the same. A Fuchsian group Γ is said to be of the first kind if every point in R ∪ { ∞ } (the boundary of H) is a limit point of the orbit Γz for some z ∈ H . Here, the notion of "limit point" is with respect to the topology on the Riemann sphere C ∞ .

Question

Fuchsian Groups of the First Kind and LatticesFirst, a Fuchsian group is a discrete subgroup of       PSL    2    (      R    ), which we can view as a group of transformations of the upper half-plane H that acts discontinuously. A lattice is a Fuchsian group with finite covolume. In other words, it is a Fuchsian group that has a fundamental domain in H with finite hyperbolic area (and then any two fundamental domains have the same hyperbolic area).My confusion arises with the notion of a Fuchsian group of the first kind. Every definition I have seen for this term is roughly the same. A Fuchsian group Γ is said to be of the first kind if every point in       R    ∪  {  ∞  } (the boundary of H) is a limit point of the orbit Γz for some   z  ∈      H  . Here, the notion of &quot;limit point&quot; is with respect to the topology on the Riemann sphere             C        ∞  .

Bordenauaa · Accepted Answer

Step 1I&#039;ll describe a Fuchsian group Γ which is not finitely generated such that   Λ  (  Γ  )  =      R    ∪  {  ∞  }.If Γ is a Fuchsian group and if   Λ  (  Γ  )  ≠      R    ∪  {  ∞  } then there is an interval   I  ⊂      R    ∪  {  ∞  } whose endpoints are in   Λ  (  Γ  ) but whose interior is disjoint from   Λ  (  Γ  ). Let   γ be the geodesic in H with the same endpoints as   Λ, let   P  ⊂      H   be the half-plane with finite boundary γ and infinite boundary I, and let       S    I    &amp;lt;  Γ be the infinite cyclic subgroup that stabilizes I, and   γ, and P. In this situation, the infinite quotient cylinder   P      /        S    I   embeds in the Riemann surface H/Γ and is a neighborhood of an end of the topological space H/Γ, so that space has an isolated end. So it suffices to describe a Riemann surface with no isolated end and whose fundamental group is not finitely generated.Step 2For this purpose, simply take       S    2    −  C where C is a Cantor set. Restricting the conformal structure on       S    2   to get a conformal structure on       S    2    −  C, apply the uniformization theorem to get a Fuchsian group Γ such that H/Γ is conformally equivalent to       S    2    −  C.

Skye Vazquez · Accepted Answer

Step 1Theorem. The following are equivalent for a Fuchsian group   Γ  &amp;lt;  P  S  L  (  2  ,  R  ) of the first kind:1. Γ is finitely generated.2. Γ is a lattice in PSL(2,R), i.e.                     H              2        /    Γ has finite area.3. One (equivalently every) fundamental polygon of Γ has only finitely many sides.Step 2Take a torsion-free Fuchsian subgroup       Γ    1    &amp;lt;  P  S  L  (  2  ,  R  ) such that                     H              2        /    Γ has finite area. Take a nontrivial normal subgroup       Γ    2    &amp;lt;      Γ    1   of infinite index. It is a general fact about general Fuchsian groups that the limit set of a nontrivial normal subgroup       Γ    2    ◃      Γ    1   equals the limit set of       Γ    1  . Since our group       Γ    1   is of the first kind, so is the normal subgroup       Γ    2  . On the other hand, area is multiplicative under isometric coverings between Riemannian surfaces: If       S    2    →      S    1   is an isometric covering of degree d of Riemannian surfaces, then  A  r  e  a  (      S    2    )  =  d  A  r  e  a  (      S    1    )  .(The same holds for manifolds in all dimensions, but the area would mean volume.) In our case, the degree of the covering map                    H              2        /        Γ    2    →                    H              2        /        Γ    1  equals the index       |        Γ    1    :      Γ    2        |    =  ∞. Hence,  A  r  e  a  (                    H              2        /        Γ    2    )  =  ∞  .

Answered question

Answer & Explanation

New Questions in College Statistics