The antisymmetric $C_{\mu\nu}$ transforms as a <Lorentz tensor>, so its complete contraction
$$
\widetilde C_2=C_{\mu\nu}C^{\mu\nu}
$$
is a <Lorentz scalar>. Hence
$$
[M^{\rho\sigma},\widetilde C_2]=0.
$$
Parts ii and v give $[P_\rho,\widetilde C_2]=[Q_\alpha,\widetilde C_2]=[\bar Q_{\dot\alpha},\widetilde C_2]=0$. It therefore commutes with every generator of the <Super-Poincaré algebra> and is a <Casimir element>, called the <superspin Casimir>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 307 / 3 / vi / Solution

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A Lorentz tensor transforms by one Lorentz matrix for each contravariant index and one inverse matrix for each covariant index. Complete contraction of all indices produces a Lorentz scalar.


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