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There is a natural presentation $SL(2,\mathbb{Z})\hookrightarrow GL(2,\mathbb{R})$, are there other presentations in real dimension 2? Or there is a classification of all the presentation of $SL(2,\mathbb{Z})\to GL(2,\mathbb{R})$? Thanks in advance.

user50402
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  • Well, such a homomorphism would have to map into $SL(2, \mathbb{R})$ and there has been work on discrete subgroups of $PSL(2, \mathbb{R})$ (search for Fuchsian groups) and these groups correspond to discrete subgroups of $SL(2, \mathbb{R})$ that contain $\pm 1$. Hence, if the real representation is faithful, then the image of the homomorphism must contain $\pm 1$ and hence, a classification of Fuchsian groups would classify all homomorphisms that you are looking at as well. – Siddharth Venkatesh May 08 '14 at 11:58
  • Thanks, and could you list a reference for classification of Fuchsian groups? @SiddharthVenkatesh – user50402 May 08 '14 at 12:09
  • I actually don't know if there is a classification. My comment was more to refer you to a term that would help with your problem. We talked a little Fuchsian groups in a course I was taking so I don't exactly have a reference. I will try to look one up for you. – Siddharth Venkatesh May 08 '14 at 12:12
  • I don't think I will be able to provide you with a reference sorry. I am now not sure that a classification would entirely interest you because when we talked about the groups, we only cared about classification up to isomorphism, whereas you want to classify the subgroups that look like $SL(2, \mathbb{Z})$ but are different as subgroups of $SL(2, \mathbb{R}).$ Sorry I wasn't enough of a help here. – Siddharth Venkatesh May 08 '14 at 12:15
  • @user50402: I think you're confusing between "presentation" and "representation". Your question concerns representations, or is senseless. – YCor May 10 '14 at 17:04

3 Answers3

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J.P. Serre, Trees, Springer-Verlag (1980), p. 81, $$\mathrm{SL}_2(\mathbb{Z}) = \langle \,S, T \mid S^4 = 1, (ST)^3 = S^2 \,\rangle$$ with, \begin{align} S &= \begin{pmatrix} \phantom{-}0& 1 \\ -1 & 0 \end{pmatrix}, & T &= \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}. \end{align}

Samuel Vidal
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The classification is known. It is a bit easier to work with the projetive modulear group $PSL(2,\mathbb{Z})=SL(2,\mathbb{Z})/\pm I$. This group is isomorphic to the free product of $\mathbb{Z}/2\ast \mathbb{Z}/3$, and finite-dimensional representations of $PSL(2,\mathbb{Z})$ correspond bijectively to finite-dimensional modules of the group algebra $k[PSL(2,\mathbb{Z})]=k[\mathbb{Z}/2\ast \mathbb{Z}/3]\simeq k[\mathbb{Z}/2]\ast k[\mathbb{Z}/3]\simeq k\langle x,y,\rangle /( x^3-1,y^2-1)$. The $2$-dimensional modules are classified here, in section $1.3$.

Dietrich Burde
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  • What's the bijection between finite-dimensional representations and finite-dimensional modules? – user50402 May 13 '14 at 07:03
  • The natural bijection is the obvious one; for example see Prop. $9.7$ in http://math.berkeley.edu/~teleman/math/RepThry.pdf. – Dietrich Burde May 13 '14 at 08:40
  • I'm a little confused about this. We know $PSL(2,\mathbb{Z})={x,y|x^2=y^3=1}$, so does it tell us $x\mapsto ?$ and $y\mapsto ?$ in $SL(2,\mathbb{R})$ ? – user50402 May 14 '14 at 05:56
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To supplement the answer of @Dietrich Burde, the representations $PSL(2,\mathbb{Z}) = \mathbb{Z}/2 * \mathbb{Z}/3 \to PSL(2,\mathbb{R})$ correspond bijectively to ordered pairs of elements $X,Y \in PSL(2,\mathbb{R})$ such that $X$ has order $1$ or $2$ and $Y$ has order $1$ or $3$; equivalently, $X$ is the identity or has trace $0$, and $Y$ is the identity or has trace $\pm 1$.

It is also interesting that amongst all such representations, the ones which are discrete, faithful, and have the same parabolics as $PSL(2,\mathbb{Z})$ are precisely the ones which are conjugate to the inclusion $PSL(2,\mathbb{Z}) \to PSL(2,\mathbb{R})$.

Lee Mosher
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    In lemon terms: After conjugation you can assume that $X$ fixes the center of the Poincare disk. Assuming $X\ne 1, Y\ne 1$, $Y$ is determined (up to change of direction of rotation and conjugation) by the hyperbolic distance between its fixed point $p$ and the origin. Therefore, up to conjugation, such representations are in natural bijective correspondence with ${\mathbb R}_+\times {\pm 1}$, where the first parameter measures the distance between $p$ and the origin. The second parameter determines the direction of rotation of $Y$. – Moishe Kohan May 08 '14 at 14:49
  • @studiosus: I like this, but why "lemon"? – Lee Mosher May 08 '14 at 16:21
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    Layman's :) $\ $ – Matthew Towers Oct 02 '18 at 16:19