Partial and Covariant derivatives in invarint actions

[AlphaNumeric]

It's physics based but actually a maths question so I'm asking it here rather than the physics forums.

$$I = \int \mathcal{L}\; d^{4}x$$

I is invariant under some transformation $$\delta_{\epsilon}$$ if $$\delta_{\epsilon}\mathcal{L} = \partial_{\mu}X^{\mu}$$ for some function/tensor/field thingy $$X^{\mu}$$. This I've no problem with.

However, is the same true for a covariant derivative? If $$\delta_{\epsilon}\mathcal{L} = D_{\mu}X^{\mu}$$ where $$D_{\mu}\varphi = \partial_{\mu}\varphi + g[A_{\mu},\varphi]$$, as you get in nonabelian field theory. Is the action still invariant? Obviously the $$\partial_{\mu}$$ part of $$D_{\mu}$$ represents no problem but I don't know if the $$g[A_{\mu},\varphi]$$ term vanishes or not within the integral.

I've been doing some supersymmetry and a number of times I've got the answer the question has asked to find plus a covariant derivative of something. If I just got a mess of terms I'd know I'm way off, but the fact everything collects nicely into a covariant derivative makes me feel I'm at least on the right track.

Thanks for any help :)

 

[fresh_42]

If $[A_\mu,X^\mu]=0$ we are done, otherwise we just do not have enough information to draw a conclusion. The reason is, that $\partial_\mu$ is a local property, and $D_\mu$ connects this local property with another one $[A_\mu,X^\mu]$ at another location. So $\mathcal{L}$ could still be invariant or equally not.