Elliptic subgroup of PSL(2,C)

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Consider the group $PSL(2,mathbb C)$ acting on $mathbbCP^1$ via Möbius transformations. Recall that an element of $PSL(2,mathbb C)$ different from the identity is called parabolic if it is conjugated to $z mapsto z+1$, loxodromic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| neq 0,1$ and elliptic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| =1$.



There is a natural embedding $PSU(2) subset PSL(2,mathbb C)$.



Question: Let $G$ be a finitelly generated subgroup of $PSL(2,mathbb C)$, not necessarily discrete. Is it true that if $G$ contains only elliptic elements (and the identity) then $G$ is conjugated to a subgroup of $PSU(2)$?



If $G$ is discrete it can be shown that it is finite, so by averaging the standard inner product on $mathbb C^2$ we get a $G$-invariant inner product, so $G$ is conjugated to a subgroup of $PSU(2)$. What about the case when $G$ is not discrete?




complex-analysis group-theory riemann-surfaces geometric-group-theory




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    Consider the group $PSL(2,mathbb C)$ acting on $mathbbCP^1$ via Möbius transformations. Recall that an element of $PSL(2,mathbb C)$ different from the identity is called parabolic if it is conjugated to $z mapsto z+1$, loxodromic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| neq 0,1$ and elliptic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| =1$.



    There is a natural embedding $PSU(2) subset PSL(2,mathbb C)$.



    Question: Let $G$ be a finitelly generated subgroup of $PSL(2,mathbb C)$, not necessarily discrete. Is it true that if $G$ contains only elliptic elements (and the identity) then $G$ is conjugated to a subgroup of $PSU(2)$?



    If $G$ is discrete it can be shown that it is finite, so by averaging the standard inner product on $mathbb C^2$ we get a $G$-invariant inner product, so $G$ is conjugated to a subgroup of $PSU(2)$. What about the case when $G$ is not discrete?




    complex-analysis group-theory riemann-surfaces geometric-group-theory




    share|cite|improve this question





















      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Consider the group $PSL(2,mathbb C)$ acting on $mathbbCP^1$ via Möbius transformations. Recall that an element of $PSL(2,mathbb C)$ different from the identity is called parabolic if it is conjugated to $z mapsto z+1$, loxodromic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| neq 0,1$ and elliptic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| =1$.



      There is a natural embedding $PSU(2) subset PSL(2,mathbb C)$.



      Question: Let $G$ be a finitelly generated subgroup of $PSL(2,mathbb C)$, not necessarily discrete. Is it true that if $G$ contains only elliptic elements (and the identity) then $G$ is conjugated to a subgroup of $PSU(2)$?



      If $G$ is discrete it can be shown that it is finite, so by averaging the standard inner product on $mathbb C^2$ we get a $G$-invariant inner product, so $G$ is conjugated to a subgroup of $PSU(2)$. What about the case when $G$ is not discrete?




      complex-analysis group-theory riemann-surfaces geometric-group-theory




      share|cite|improve this question











      Consider the group $PSL(2,mathbb C)$ acting on $mathbbCP^1$ via Möbius transformations. Recall that an element of $PSL(2,mathbb C)$ different from the identity is called parabolic if it is conjugated to $z mapsto z+1$, loxodromic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| neq 0,1$ and elliptic if it is conjugated to $z mapsto lambda z$ for some $lambda in mathbb C^* $ with $|lambda| =1$.



      There is a natural embedding $PSU(2) subset PSL(2,mathbb C)$.



      Question: Let $G$ be a finitelly generated subgroup of $PSL(2,mathbb C)$, not necessarily discrete. Is it true that if $G$ contains only elliptic elements (and the identity) then $G$ is conjugated to a subgroup of $PSU(2)$?



      If $G$ is discrete it can be shown that it is finite, so by averaging the standard inner product on $mathbb C^2$ we get a $G$-invariant inner product, so $G$ is conjugated to a subgroup of $PSU(2)$. What about the case when $G$ is not discrete?





      complex-analysis group-theory riemann-surfaces geometric-group-theory





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      asked Jul 27 at 4:15









      Lucas Kaufmann

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          After reading this question I found out that the result is indeed true. It can be found in the book Complex Functions by G. Jones and D. Singerman (Theorem 2.13.1).






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            1 Answer
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            up vote
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            down vote



            accepted










            After reading this question I found out that the result is indeed true. It can be found in the book Complex Functions by G. Jones and D. Singerman (Theorem 2.13.1).






            share|cite|improve this answer

























              up vote
              0
              down vote



              accepted










              After reading this question I found out that the result is indeed true. It can be found in the book Complex Functions by G. Jones and D. Singerman (Theorem 2.13.1).






              share|cite|improve this answer























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                After reading this question I found out that the result is indeed true. It can be found in the book Complex Functions by G. Jones and D. Singerman (Theorem 2.13.1).






                share|cite|improve this answer













                After reading this question I found out that the result is indeed true. It can be found in the book Complex Functions by G. Jones and D. Singerman (Theorem 2.13.1).







                share|cite|improve this answer













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 27 at 8:30









                Lucas Kaufmann

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