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I'm trying to write ${\cal N}=4$ SYM in terms of ${\cal N}=1$ superfields. I have the Lagrangian

$$\mathcal{L}=\frac{1}{16 k} \int d^2 \sigma \text{Tr} \big[W^a W_a\big]+c.c+\int d^4\theta \text{Tr}\big[\bar{\Phi}^i e^V \Phi^i e^{-V}\big]+\frac{\sqrt{2}}{3}\int d^2\theta \text{Tr}\big[\phi^i [\phi^j,\phi^k]\big]\epsilon_{ijk}+c.c $$

Where the $\Phi^i$ are chiral superfields and V is a vector superfield. In components, this is all fine except for the Yukawa terms $$\mathcal{L} \supset i\sqrt{2} f^{ABC} Z^{i\dagger}_A \psi^i_B \lambda_C - \sqrt{2}\epsilon_{ijk} Z^i_A \psi^j_B \psi^k_C+c.c $$

Where $A,B,C$, are $SU(N)$ gauge group indices, $i,j,k$ number my 3 chiral superfields, which have an explicit $SU(3)$ symmetry, the $Z^i_A$ are the complex scalars from my chiral superfields, $\psi^i_A$ are the fermions from my chiral superfields, and the $\lambda_A$ is the fermion from my vector superfield.

The fermions combine into an fundamental $SU(4)$ multiplet $\chi^I=(\psi^i, \lambda)$, and I decompose my complex scalars into real ones in a fundamental $SO(6)$ (isomorphic to $SU(4)$) multiplet, $Z^i=X^a+iX^{a+3}$. I should be able to write the Yukawa terms as

$$\mathcal{L} \supset f^{ABC} X^a_A C^a_{IJ}\chi^I \chi^J +c.c $$

essentially putting the scalars into the antisymmetric matrix representation of $SU(4)$, $X_{IJ}=X_{[IJ]}=X^a_{IJ}X^a$.

So I need to show that the $C^a_{IJ}$ that I have are an invariant symbol of $SO(6)=SU(4)$, and thus my Lagrangian has that R symmetry. Not sure how to do that... a reference I found says they should be related to the $SO(6)$ gamma matrices (http://arxiv.org/abs/hep-th/0201253, below equation 3.1), but that hasn't been very helpful.

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this question is 2 years old, but I thought it's never too late. I'm not sure about the definite answer, but here are my thoughts.

Take the SO(6) algebra viewpoint. The $\mathbf{6}$ is the fundamental (vector) representation, and the $\mathbf{4}$ is the spinor representation. So we are looking for symbols $\Sigma_{AB}^I$ that combine two spinors into a vector.

This looks analogous to the usual SO(1,3) case, where $\sigma^\mu_{\alpha\dot\beta}$ convert two Weyl spinors into one vector. Write the gamma matrices in the Weyl basis $$ \gamma^\mu=\begin{pmatrix}0 & \sigma^\mu \\ \bar\sigma^\mu & 0\end{pmatrix} $$ and you can read off the sigmas.

It should be similar for the SO(6). Write its $8\times 8$ gamma matrices $\hat\gamma^I$ in Weyl basis $$ \hat\gamma^\mu=\begin{pmatrix}0 & \Sigma^\mu \\ \bar\Sigma^\mu & 0\end{pmatrix} $$ and one should be able to read off the sigma symbols.

One can find an explicit construction here (however mind the conventions and the signature!), or you can try to construct the gamma and sigma matrices from scratch.

What I'm not sure about is the fact that in the case of SO(1,3) the $\sigma^\mu_{\alpha\dot\beta}$ have one left-handed and one right-handed spinor index, this should probably also hold for SO(6), so I would be tempted to write $\Sigma^i_{A\dot B}$, but in the contraction in your example $\Sigma^I$ is contracted with two spinors of same chirality.

Edit

An addition about what I said in the last paragraph: in SO(4) (so I assume also in SO(1,3)) the left and right handed spinors are the $\mathbf{2}=[1,0]$ and the $\mathbf{\bar{2}}=[0,1]$, their tensor product gives the vector: $\mathbf{2}\otimes\mathbf{\bar{2}}=\mathbf{4}$, as expected.

In SO(6) it seems to be a bit different: the spinor is the $\mathbf{4}=[0,1,0]$, and there is also the $\mathbf{\bar{4}}=[0,0,1]$, which is presumable the other chirality spinor. But the difference is that unlike for $SO(4)$, here $\mathbf{4}\otimes\mathbf{\bar{4}}=\mathbf{1}\oplus\mathbf{15}$, so the $\mathbf{4}$ and the $\mathbf{\bar{4}}$ cannot be combined into a vector. Instead, $\mathbf{4}\otimes\mathbf{4}=\mathbf{6}\oplus\mathbf{10}$ and $\mathbf{\bar{4}}\otimes\mathbf{\bar{4}}=\mathbf{6}\oplus\mathbf{\bar{10}}$, so to get the vector one has really to combine two spinors of the same chirality.

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