# Is a $SU(2)$ supergauge theory really a $SU(2)$ gauge theory?

Consider $SU(2)$ supergauge theory with $A$, a doublet of two chiral superfields in the fundamental representation.

$$A= \begin{pmatrix} \Phi_1\\ \Phi_2 \end{pmatrix}$$ where $\Phi_1$ and $\Phi_2$ are chiral superfields. Since we have said it is in the fundamental representation it transforms

$$A\to{}A'=e^{i\sigma_j\phi_j}A$$

where $\sigma_j$ are Pauli matrices. If $e^{i\sigma_j\phi_j}$ were to be a $SU(2)$ matrix then the $\phi_j$ ought to be real. Nonetheless, this would make $A'$ no longer chiral unless $\phi_j$ are themselves chiral superfields, that is complex functions making $e^{i\sigma_j\phi_j}$ not a $SU(2)$ matrix.

Therefore this is no longer a proper $SU(2)$ theory. What is going on?

• Could you present your precise definition of "chiral superfield"? Nov 29, 2015 at 18:04
• @ACuriousMind Following section 4.4 of arxiv.org/abs/hep-ph/9709356 a chiral superfield is a superfield $\Phi$ satisfying $\bar{D}_{\dot{\alpha}}\Phi=0$ Nov 29, 2015 at 20:01
• I'm not sure what your question really is, then. As far as I can see, the linear combination of chiral superfields is a chiral superfield (the constraint they fulfill is linear), so for any matrix $M$, the components of $MA$ are chiral superfields if the components of $A$ were, no? Nov 30, 2015 at 10:11
• @ACuriousMind notice that the elements of the matrix $M$ need not be constants so $\bar{D}_{\dot{\alpha}}=-\frac{\partial}{\partial\bar{\theta}^{\dot{\alpha}}}+i(\theta\sigma^{\mu})_{\dot{\alpha}}\partial_{\mu}$ applies to $M$ as well Nov 30, 2015 at 11:57
• Ah, yes, indeed. Nov 30, 2015 at 12:01

## 1 Answer

You are correct in stating that the doublet $A'$ ceases to be chiral if your gauge parameters $\phi_j$ are real. In fact, in an $\mathcal{N} = 1$ (global) supersymmetric gauge theory, a gauge transformation is not given by what you wrote above. Working as you are in the superfield formalism, a supersymmetric gauge transformation of a chiral superfield $\Phi$ in a representation $R$ of the gauge group $G$ is given by $$\Phi \rightarrow \Phi' = e^{i \Lambda} \Phi,$$ where $\Lambda \equiv \Lambda_a T_R^a$, with $T_R^a$ the generators in the appropriate representation (the Pauli matrices, in your example).

These gauge parameters $\Lambda_a$ are, as you suspected, chiral superfields, whose 'lowest components' $\lambda_a$ (see the equation below) are complex fields.

The $\Lambda_a$ are indeed complex functions of superspace coordinates, namely spacetime $x^\mu$ and the Grassmann numbers $\theta_\alpha, \overline\theta_{\dot \alpha}$. Explicitly (different conventions may be at play): $$\Lambda_a \equiv \Lambda_a(x,\theta,\overline\theta) = \lambda_a(x) + \sqrt{2} \theta \psi^\lambda_a(x) + i \theta \sigma^\mu \overline \theta \partial_\mu \lambda_a(x)- \theta \theta F^\lambda_a(x)\\ \qquad- \frac{i}{\sqrt{2}}\theta\theta \partial_\mu \psi^\lambda_a(x) \sigma^\mu \overline \theta - \frac{1}{4} \theta\theta\overline\theta\overline\theta \square \lambda_a(x).$$ Similarly, $\Phi(x,\theta,\overline\theta) = \phi(x)+ \sqrt{2} \theta \psi(x) - \theta \theta F(x) + (\ldots)$.

If my gauge group is, say, $G = \textrm{SU}(N)$, you claim that the theory is no longer a proper $\textrm{SU}(N)$ theory, since $e^{i\Lambda}$ is not (taking $R$ to be the fundamental representation) an $\textrm{SU}(N)$ matrix.

However, the fact is that after integrating over Grassmann coordinates -- thus hiding the superfield construction -- you end up with the actual Lagrangian of the theory where gauge invariance is manifest in the usual way.

In particular, the matter kinetic part of the Lagrangian of a gauge-matter SUSY theory is given, using superfields, by: $$\mathcal{L}_\textrm{kin} = \int d\theta^2 d\overline\theta^2 \, \overline\Phi e^{V} \Phi,$$ where $V \equiv V_a T_R^a$, with $V_a$ real/vector superfields (general superfields constrained by $\overline V_ a = V_a$).

The gauge transformation acts on the vector superfields as follows: $$e^V \rightarrow e^{i\overline\Lambda} e^V e^{-i\Lambda},$$ ensuring that in superspace our theory is SUSY-gauge invariant.

Integrating now over the Grassmann coordinates $\theta, \overline\theta$, one finds (omitting gaugino interactions and terms with auxiliary fields): $$\mathcal{L}_\textrm{kin} \supset |D_\mu \phi(x)|^2 -i\overline\psi(x) \overline\sigma^\mu D_\mu \psi(x),$$ where (up to a factor of $2g$, where $g$ is the gauge coupling) the covariant derivative $D_\mu$ is the usual one. This is a perfectly normal gauge-invariant Lagrangian, i.e. invariant under usual gauge transformations. Here, it is $\phi(x)$ (the $\Phi$ superfield's lowest component) which transforms as you wrote, in the same representation $R$.

N.B.: I have systematically omitted gauge indices, $\Phi$ is $\Phi^i$ (and thus $\phi$ is $\phi^i$, $i=1,\ldots,\textrm{dim }R$), corresponding to $A$ in your example ($i=1,2$).

Recommended reading: chapter 4.3.1 of R. Argurio's Introduction to Supersymmetry'', available online.