# Chern-Simons term for a non-abelian gauge multiplet

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.

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.