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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?

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  • $\begingroup$ A possible answer is to consider decomposing under the basis. $\endgroup$
    – Errorbar
    Mar 23 at 5:28

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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.

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