It is cleaner to use the action of the charge than to expand every double sum. The commutators from part a give, for $v^\rho=x^\rho,p^\rho$, or any $\alpha_n^\rho$,
$$
[M^{\mu\nu},v^\rho]
=i\left(\eta^{\mu\rho}v^\nu-\eta^{\nu\rho}v^\mu\right).
$$
Thus $M^{\mu\nu}$ acts on every mode in the <vector representation> of the <Lorentz algebra>. Apply the <Jacobi identity> to the action of $[M^{\mu\nu},M^{\kappa\lambda}]$ on every mode. The result agrees with the action of
$$
i\left(\eta^{\mu\kappa}M^{\nu\lambda}-\eta^{\nu\kappa}M^{\mu\lambda}
+\eta^{\nu\lambda}M^{\mu\kappa}-\eta^{\mu\lambda}M^{\nu\kappa}\right).
$$
There is no extra scalar term: the orbital and oscillator pieces commute with one another, and direct use of their <canonical commutation relations> gives no central contribution. Hence the displayed operators obey precisely the standard <Lorentz algebra>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 1 / c / Solution

The vector representation is the defining action of a matrix group on the vector space on which its matrices act. For an orthogonal or Lorentz group, it preserves the corresponding symmetric bilinear form.


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