= \sigma_{2} \sigma_{1} \sigma_{2}$ holds. We also have $(\sigma_{1} \sigma_{2} \sigma_{1})^{2} = \omega^{6} \cdot 1$, so if $\omega$ is a 6th root of unity the modular relation holds. One also has that $(\sigma_{1} \sigma_{2})^{3} = \omega^{6} \cdot 1$. This is the operator usually chosen to represent $ST$ in the modular group.