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I am trying to work out $2$-dimensional representation of $\operatorname{SL}_2(\mathbb{Z})$. I know that $\operatorname{SL}_2(\mathbb{Z})$ is generated by $S = \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, R = \begin{pmatrix} 0 & -1\\ 1 & 1\end{pmatrix}$ and $T= \begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}$.

My lecture notes say that for $\rho\colon\operatorname{SL}_2(\mathbb{Z}) \rightarrow \operatorname{GL}_2(\mathbb{C})$, I should diagonalize $\rho(S)$ and see what choices I have for $\rho(R)$. But I am not sure how to work these calculations out. Can someone help? Thank you in advance!

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  • $\begingroup$ Is $\rho$ arbitrary? $\endgroup$ Dec 9, 2020 at 18:09
  • $\begingroup$ @Julian Quast yes! $\endgroup$ Dec 9, 2020 at 20:33

1 Answer 1

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$\mathrm{SL}_2(\mathbb Z)$ is generated by $S$ and $R$ and we have $S^4 = \mathrm{id}$ and $S^2 = R^3 = -\mathrm{id}$. By diagonalizing $\rho$, we can assume, that $\rho(S) = \begin{pmatrix} \alpha & 0 \\ 0 & \beta\end{pmatrix}$ for $\alpha, \beta \in \{\pm 1, \pm i\}$. We have $R^6 =\mathrm{id}$, so $\rho(R)$ is diagonalizable with eigenvalues being $\gamma,\delta \in \{1,\zeta, \dots, \zeta^5\}$, where $\zeta = e^{\frac{2\pi i}{6}}$. Hence $\rho(R) = A\begin{pmatrix} \gamma & 0 \\ 0 & \delta\end{pmatrix}A^{-1}$ for some $A \in \mathrm{GL}_2(\mathbb C)$.

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    $\begingroup$ $PSL_2(\Bbb{Z})$ is the free product of $C_2=\langle S \rangle/\pm I$ with $C_3=\langle R\rangle/\pm I$ so $SL_2(\Bbb{Z})$ is defined by $S^4=1, S^2= R^3, S^2 R=RS^2$ thus by $S^4=1,S^2=R^3$. Any $\rho(S),\rho(R)$ satisfying those relations will generate an homomorphism $SL_2(\Bbb{Z})\to GL_2(\Bbb{C})$. $\endgroup$
    – reuns
    Dec 10, 2020 at 8:57

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