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## Main Question or Discussion Point

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

[tex]I = \int \mathcal{L}\; d^{4}x[/tex]

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

I've been doing some supersymmetry and a number of times I've got the answer the question has asked to find

Thanks for any help :)

[tex]I = \int \mathcal{L}\; d^{4}x[/tex]

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

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