= Solution
Construct the <Lorentz-spinor generators from a Clifford algebra> as
$$
M^{\rho\sigma}=\frac14[\gamma^\rho,\gamma^\sigma]
=\frac12\gamma^\rho\gamma^\sigma-\frac12\eta^{\rho\sigma}I_4.
$$
They are antisymmetric in $\rho,\sigma$. Reordering a third <gamma matrix> with the <Clifford algebra> yields
$$
[\gamma^\rho\gamma^\sigma,\gamma^\tau]
=2\eta^{\sigma\tau}\gamma^\rho-2\eta^{\rho\tau}\gamma^\sigma,
\qquad
[M^{\rho\sigma},\gamma^\tau]
=\eta^{\sigma\tau}\gamma^\rho-\eta^{\rho\tau}\gamma^\sigma.
$$
Use the <commutator derivation identity> on the two factors in $M^{\tau\nu}=\frac14[\gamma^\tau,\gamma^\nu]$:
$$
\begin{aligned}
[M^{\rho\sigma},M^{\tau\nu}]
&=\frac14\left([[M^{\rho\sigma},\gamma^\tau],\gamma^\nu]
+[\gamma^\tau,[M^{\rho\sigma},\gamma^\nu]]\right)\\
&=\eta^{\sigma\tau}M^{\rho\nu}-\eta^{\rho\tau}M^{\sigma\nu}
+\eta^{\rho\nu}M^{\sigma\tau}-\eta^{\sigma\nu}M^{\rho\tau}.
\end{aligned}
$$
This derives every <Lorentz algebra> commutation relation, so \b[the matrices $M^{\rho\sigma}$ form its spinor representation]. Their finite exponentials describe the connected <Spinor representation of the Lorentz group>, more precisely the lift to its double cover.
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