# 2 dimensional representation of $\operatorname{SL}_2(\mathbb{Z})$

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!

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

$$\mathrm{SL}_2(\mathbb Z)$$ 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 $$A \in \mathrm{GL}_2(\mathbb C)$$.
• $PSL_2(\Bbb{Z})$ 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 $SL_2(\Bbb{Z})\to GL_2(\Bbb{C})$. Dec 10, 2020 at 8:57