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.


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.