Silly confusion about gauge invariance in supersymmetric Lagrangians - in particular, in the ${\cal N}=1$ superfield formulation of ${\cal N}=4$ SYM

Hoping to resolve a simple confusion I have about supersymmetric gauge theory, one which I ran into while trying to understand the $${\cal N}=1$$ superfield formulation of $${\cal N}=4$$ supersymmetric yang mills.

In various sources (e.g. this review: https://arxiv.org/abs/hep-th/9908171 see eq'n 2.9, or Erdmenger's book Gauge Gravity Duality, see eq'n 3.215), there are terms in the lagrangian of the form $$\mathcal{L}={\rm tr} \int d^4\theta ~\Phi^\dagger e^V \Phi e^{-V}$$ modulo different sign conventions in the textbook vs the review - this is from the book. Here $$\Phi$$ is a chiral superfield, and $$V$$ is a real superfield. Now as I understand it, the fields transform as follows under gauge transformations: $$\Phi\to e^{i\Lambda}\Phi$$ $$e^V \to e^{i\Lambda^\dagger} e^V e^{-i\Lambda}$$ Clearly, the term $$\mathcal{L}=\int d^4\theta \Phi^\dagger e^V \Phi$$ is invariant under such a transformation. But if I look at the lagrangian I wrote at the top of this question, it transforms as follows, as far as I can tell: $$\int d^4\theta \Phi^\dagger e^V \Phi e^{-V} \to \int d^4\theta \Phi^\dagger e^V \Phi e^{-i\Lambda^\dagger} e^{-V} e^{i\Lambda}$$ Doesn't seem very gauge invariant. What am I missing here?

• A possible answer is to consider decomposing under the basis. Mar 23 at 5:28

Presumably the $${\cal N}=1$$ chiral field $$\Phi\to e^{i\Lambda}\Phi e^{-i\Lambda}$$ and the $${\cal N}=1$$ vector field $$V\to e^{i\Lambda}V e^{-i\Lambda}$$ are assumed to transform in the adjoint representation of a unitary gauge group (so that $$\Lambda=\Lambda^{\dagger}$$ is Hermitian).

Also notice that the Lagrangian term has a gauge group trace in front unlike the standard construction.