Codex @codex  0

SetThe right Maurer-Cartan equation is $d\omega =\omega \wedge \omega$. Direct substitution givesso trace invariance proves ${\tilde{C}}_2=C_2$.

For the Chern-Simons 3-form, expansion and graded cyclicity giveSince cyclicity also givesthe required two-form may be chosen asThus

Substitute $\rho ={\sigma }^{\alpha }{T}_{\alpha }$ into the Maurer-Cartan equation. Antisymmetry of the wedge product givesEquating coefficients of $T_{\gamma }$ yieldsso $\boxed{f^{\gamma }{}_{\alpha \beta }=-c^{\gamma }{}_{\alpha \beta }/2}$.

Write $B_i=({\partial }_{i}g)g^{-1}$ and use $D=d+A$. The right Maurer-Cartan equation gives ${\partial }_x{B}_y-{\partial }_y{B}_x=[B_x,B_y]$. Thus $A_i=\alpha B_i$ has the displayed gauge curvature. Both $\alpha =0$ and $\alpha =-1$ are universally flat; the latter is a pure gauge potential. Commuting $B_x,B_y$ give additional flat cases. For the opposite convention $D=d-A$, the corresponding component formula instead has $\alpha (1-\alpha )$.

The displayed one-forms give an invertible triangular change from the coordinate coframe on flat Majorana superspace. A rigid supertranslation has $dϵ=0$, and the variation of $d{X}^m$ cancels that of $i\bar{\theta }{\gamma }^{m}d\theta$. Their Maurer-Cartan equation is $d{\Pi }^m=id\bar{\theta }{\gamma }^{m}d\theta$. The bosonic frame transforms as a Lorentz vector and the odd frame as a Majorana spinor.

When $d{\sigma }^{\gamma }={\sum }_{\alpha ,\beta }{f}^{\gamma }{}_{\alpha \beta }{\sigma }^{\alpha }\wedge {\sigma }^{\beta }$ sums over ordered pairs, the Maurer-Cartan equation determines only $f^{\gamma }{}_{[\alpha \beta ]}=-c^{\gamma }{}_{\alpha \beta }/2$. Symmetric additions vanish in the exterior product. The usual coefficients are the unique antisymmetric representatives; summing instead over $\alpha <\beta$ removes the factor one half.