# Gauge invariant supersymmetric transformations

Given the following action $$\mathcal{S}=\int{d^4x\;d^2\theta\;d^2\bar{\theta}\left(\bar{Q}_+e^{2V}Q_++\bar{Q}_-e^{-2V}Q_--2\xi V\right)}+\int{d^4x\;d^2\theta\left(mQ_-Q_++\frac{\tau}{16\pi i}W^\alpha W_\alpha\right)}+h.c.$$

Where $$\bar{Q}$$ is the hermitian conjugated $$Q$$, $$\tau = \frac{\theta}{2\pi}+i\frac{4\pi}{g^2}$$ the gauge coupling, $$\xi$$ the Fayet-Iliopoulos parameter, $$W_\alpha=-\frac{1}{4}\bar{D}^2D_\alpha V$$ is the gaugino superfield and $$V$$ the abelian vector superfield.

How can I show that the action is invariant under $$Q_+\rightarrow e^{i\Lambda}Q_+ \;\\ Q_-\rightarrow e^{-i\Lambda}Q_- \;\\ V\rightarrow V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda}$$

When I try the transformation on the first term I get the following: $$\bar{Q}_+e^{2V}Q_+\rightarrow \bar{Q}_+e^{2V}Q_+ e^{Im(\Lambda)}$$ I have a similar problem with the next term

Edit:

$$Q_+$$ is a chiral superfield with charge 1

$$Q_-$$ is a chiral superfield with charge -1

Let's check invariant of first term. Using transformation laws and fact, that transformation are abelian:

$$Q_+\rightarrow e^{i\Lambda}Q_+ \Rightarrow \bar{Q}_+\rightarrow \bar{Q}_+ e^{-i\bar{\Lambda}}$$

$$V\rightarrow V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda} \Rightarrow e^{2V}\rightarrow e^{i\bar{\Lambda}}e^{2V} e^{-i\Lambda}$$

We obtain:

$$\bar{Q}_+ e^{2V} Q_+ \to \bar{Q}_+ e^{-i\bar{\Lambda}} e^{i\bar{\Lambda}}e^{2V} e^{-i\Lambda} e^{i\Lambda}Q_+ = \bar{Q}_+ e^{2V} Q_+$$

For gauge multiplet term:

$$W_\alpha=-\frac{1}{4}\bar{D}^2D_\alpha V \to -\frac{1}{4}\bar{D}^2D_\alpha V -\frac{i}{8}\bar{D}^2D_\alpha\Lambda +\frac{i}{8}\bar{D}^2D_\alpha\bar{\Lambda}$$

Now using that $$\Lambda$$ is chiral, $$\bar{\Lambda}$$ antichiral and $$[\bar{D}^2,D_\alpha]=0$$: $$D_\alpha\bar{\Lambda} =0$$ $$\bar{D}^2D_\alpha\Lambda = D_\alpha\bar{D}^2\Lambda = 0$$

So $$W_\alpha$$ is gauge invariant quantity.

Fayet-Iliopoulos term:

$$\int d^4x \;d^2\theta d^2 \bar{\theta}\; \xi V\rightarrow \int d^4x \;d^2\theta d^2 \bar{\theta}\; \xi \left(V-\frac{i}{2}\Lambda +\frac{i}{2}\bar{\Lambda}\right)$$

Using that $$\int d\theta_\alpha = D_\alpha$$ and $$\int d\theta_{\dot{\alpha}} = \bar{D}_{\dot{\alpha}}$$ (up to the boundary term) and chirality properties, we immediately see, that FI term is gauge invariant.

The same idea works with other terms.

• Can you show this for the $W^\alpha W_\alpha$ term? Apr 25, 2020 at 23:21
• @fielder, I updated answer Apr 25, 2020 at 23:37
• thank you very much! Just a final part, how do you show that for the Fayet-Iliopoulos term ? Apr 26, 2020 at 0:22
• @fielder, I updated answer Apr 26, 2020 at 0:36
• @fielder, is my answer clear to you? Apr 26, 2020 at 18:55