Wess-Zumino Gauge in non-Abelian supersymmetric theory
I) The gauge transformation of the real gauge field $V$ reads
$$ e^{\widetilde{V}} ~=~e^Xe^Ve^Y, \qquad X~:=~i\Omega^{\dagger}, \qquad Y~:=~-i\Omega. \tag{1}$$
We next use the following BCH formulas
$$ e^Xe^V~\stackrel{\rm BCH}{=}~e^{V+B({\rm ad} V)X+{\cal O}(X^2)}, \qquad e^Ve^Y~\stackrel{\rm BCH}{=}~e^{V+B(-{\rm ad} V)Y+{\cal O}(Y^2)}.\tag{2} $$
Keeping only linear orders in $\Omega$, we get
$$\begin{align}\widetilde{V}~&\stackrel{(1)+(2)}{=}~B({\rm ad} V)X+V+B(-{\rm ad} V)Y\cr &~~~\stackrel{(4)}{=}~V+\frac{1}{2}[V,Y-X]+B_+({\rm ad} V)(X+Y),\end{align}\tag{3} $$
where
$$\begin{align} B(x)&~:=~\frac{x}{e^x-1}~=~\sum_{m=0}^{\infty}\frac{B_m}{m!}x^m~=~B_+(x)-\frac{x}{2}\cr &~=~1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\frac{x^6}{30240}+{\cal O}(x^8)\end{align} \tag{4} $$
and
$$\begin{align} B_+(x) &~:=~\frac{B(x)+B(-x)}{2}~=~\frac{x/2}{\tanh\frac{x}{2}} \cr &~=~1+\frac{x^2}{12}-\frac{x^4}{720}+\frac{x^6}{30240}+{\cal O}(x^8) \end{align} \tag{5} $$
are generating functions of Bernoulli numbers.
II) We would like $\widetilde{V}$ to be in WZ gauge
$$ \widetilde{V}~=~{\cal O}(\theta^2) .\tag{6} $$
For given $V$, $\widetilde{V}$, and $X-Y$, the eqs. (3+6) is an affine$^1$ equation in $X+Y=i\Omega^{\dagger}-i\Omega$. This has formally a solution if the operator
$$ B_+({\rm ad} V)~=~{\bf 1} + \ldots \tag{7} $$
is invertible, which is true, at least perturbatively. To finish the proof, one should write out the equation in its superfield components to check that the above affine shift mechanism really is realized at the component level. Recall e.g. that the gauge field $\widetilde{V}$ can not be gauged away completely (= put to zero), since $\Omega$ is a chiral superfield with not enough $\theta$'s to reach all components of $\widetilde{V}$, so to speak.
References:
- S.P. Martin, A Supersymmetry Primer, arXiv:hep-ph/9709356; p.43.
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$^1$ An affine equation is a linear equation with an inhomogeneous term/source term.