# A Question about Wess-Zumino Gauge in Non-Abelian Gauge Theory

I have a question about the Wess-Zumino gauge in non-Abelian supersymmetric gauge theory. I am following BUSSTEPP Lectures on Supersymmetry.

The spinortial field strength is defined as

$$W_{\alpha}(x,\theta,\bar{\theta})=-\frac{1}{4}\bar{D}^{2}e^{-V}D_{\alpha}e^{V},$$

where $$V(x,\theta,\bar{\theta})$$ is a vector superfield. The vector superfield $$V$$ is $$\mathfrak{g}$$-valued, and can be written as

$$V(x,\theta,\bar{\theta})=iV^{i}(x,\theta,\bar{\theta})T_{i},$$

where $$\left\{T_{i}\right\}$$ is a fixed basis for the Lie algebra $$\mathfrak{g}$$.

The gauge transformation is given by

$$e^{V}\rightarrow e^{-\Lambda^{\dagger}}e^{V}e^{-\Lambda},$$

where $$\Lambda(x,\theta,\bar{\theta})$$ is a $$\mathfrak{g}$$-valued chiral superfield.

Using Baker-Campbell-Hausdorff-formula, one finds that the above gauge tranformation is equivalent to the following infinitesimal gauge transformation

$$V\rightarrow V-(\Lambda^{\dagger}+\Lambda)-\frac{1}{2}[V,\Lambda-\Lambda^{\dagger}]+\cdots.$$

On page 52, section 5.6, exercise V.11 says that from the above equation, one concludes that $$V$$ can be put into the Wess-Zumino gauge

$$V(x,\theta,\bar{\theta})\equiv\bar{\theta}\bar{\sigma}^{\mu}\theta v_{\mu}(x)+\bar{\theta}^{2}\theta\lambda(x)+\theta^{2}\bar{\theta}\bar{\lambda}(x)+\bar{\theta}^{2}\theta^{2}D(x),$$

where $$v_{\mu}(x)$$, $$\lambda(x)$$, and $$D(x)$$ are $$\mathfrak{g}$$-valued.

1. My first question is how to prove the above statement that Wess-Zumino gauge in non-Abelian gauge theory is always possible by a suitable choice of $$\Lambda$$?

Next, since in Wess-Zumino gauge $$V^{3}=0$$, the infinitesimal gauge transformation in Wess-Zumino gauge reduces to the following finite sum

$$V\rightarrow V-(\Lambda^{\dagger}+\Lambda)-\frac{1}{2}[V,\Lambda-\Lambda^{\dagger}]-\frac{1}{12}[V,[V,\Lambda+\Lambda^{\dagger}]]$$

The author concludes that the gauge transformation that preserves the Wess-Zumino gauge must take the form

$$\Lambda(x,\theta,\bar{\theta})=\omega(x)+i\bar{\theta}\bar{\sigma}^{\mu}\theta\partial_{\mu}\omega+\frac{1}{4}\bar{\theta}^{2}\theta^{2}\Box\omega(x),$$

where $$\omega(x)$$ is a $$\mathfrak{g}$$-valued scalar obeying $$\omega^{\dagger}=-\omega$$.

2. My last question is how to prove the above statement that the $$\Lambda$$ preserving the Wess-Zumino gauge must take the above form?

I just realized that the first question is answered here. My real question is the last one.