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In equation (20.9) of Freedmann and Van Proeyen's Supergravity, it is stated that for the following Chern-Simons term:

$$S_{\mathrm{CS}} = C_{IJK}\int A^I\wedge F^J \wedge F^K$$

to be invariant under a non-abelian gauge transformation, the symmetric tensor $C_{IJK}$ has to satisfy ${f_{I(J}}^MC_{KL)M} = 0$. Finding this condition is the subject of exercice 20.2, but I don't manage to do it. Varying this term, doing an integration by parts and using the symmetry of the tensor $C_{IJK}$, I find:

$$\delta S_{\mathrm{CS}} = 3C_{IJK}\int \delta A^I \wedge F^J \wedge F^K$$

requiring invariance under the non-abelian gauge transformation: $$\delta A^I = d\theta^I + \theta^MA^L{f_{LM}}^I$$

then yields the condition ${f_{LM}}^IC_{IJK} = 0$, which looks like the one that is mentioned, up to the symmetrization of some indices.

I wanted to know where this symmetrization comes from.

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While writing the question, I found the answer. The part of $\delta S_{\mathrm{CS}}$ that should be made to vanish is:

$$\delta S_{\mathrm{CS}} = 3\theta^M{f_{LM}}^IC_{IJK}\int A^L \wedge F^J \wedge F^K = 0$$

The wedge product on the right is symmetric in $J, K, L$ (since $F$ is a $4$ form), so the previous equation constrains only the part of ${f_{LM}}^IC_{IJK}$ which is symmetric in these indices, so we indeed find ${f_{I(J}}^MC_{KL)M} = 0$ after some renaming of the indices.

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