for $g^2 \simeq 1/2$ [@Giedt:2001zw]. Thus the perturbation $\epsilon$ could be the result of non-renormalizable operators in the superpotential of relative low-degree – perhaps involving only one or two powers of such a field vev, depending on the size of the dimensionless Yukawa couplings involved.

Considering the underlying $SU(2)_A \times SU(2)_B$ theory, fields bifundamental under both $SU(2)$ factors will decompose into a triplet and a singlet under the breaking to the diagonal subgroup. Let us denote this singlet representation as $\psi$. Then terms at dimension four in the superpotential that can populate the vanishing entries in (\[lambdaT1\]) include $$\begin{aligned}
%
\Delta W &=& \frac{\lambda_{11}}{M_{\PL}} L_1 (\mathbf{2}, 1) T
(\mathbf{2}, \mathbf{2}) \psi (\mathbf{2}, \mathbf{2}) L_1
(\mathbf{2}, 1) \nonumber \\
%
 & & +\frac{\lambda_{ij}}{M_{\PL}}L'_i (1, \mathbf{2})
 T(\mathbf{2},\mathbf{2}) \psi (\mathbf{2},\mathbf{2}) L'_j
 (1,\mathbf{2}) ,
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\label{psi} \end{aligned}$$ where $i, j = 2, 3$ and we denote the representations under $SU(2)_A \times SU(2)_B$ for convenience. The singlet field $\psi$ must have vanishing hypercharge, so it cannot be the singlet component of the same bifundamental that led to $T$ and $\oline{T}$, though it may be the singlet component of some bifundamental representation that served to generate the breaking to the diagonal subgroup in the first place, 