Wess-Zumino Gauge in non-Abelian supersymmetric theory I've got a question concerning  non-Abelian supersymmetric gauge theories.
Consider supersymmetric non-Abelian theory realized on chiral superfields $\Phi_i$ in a representation $R$ with matrix generators $T_{i}^{aj}$. Let us define supergauge transformation as
$$\Phi_i \rightarrow (e^{2\imath g_a \Omega^a T^a})_{i}{}^{j} \, \Phi_j.$$
The supergauge-invariant term in lagrangian is 
$$\mathcal{L} = \Bigl[\Phi^{*i}\,(e^V)_i{}^j \, \Phi_j\Bigr]_D.$$
For this to be gauge-invariant, the non-Abelian gauge transformation for the vector field must be
$$e^V \rightarrow e^{\imath \Omega^\dagger}\,e^V\,e^{-\imath \Omega}.$$
Using Baker-Hausdorff formula, we obtain
$$V^a \rightarrow V^a + \imath(\Omega^{a*}-\Omega^a)+g_a \, f^{abc}\,V^b(\Omega^{c*}+\Omega^c)+...$$
Usually at this moment they argue that since the second term on the right side does not depend on $V^a$, one can always do a supergauge transformation to Wess-Zumino gauge by choosing $\Omega^{a*}-\Omega^a$ appropriately.
This is the moment that I don't get. What does it mean?  Strictly speaking, the latter expression is complicated non-linear equation on components of $V^a$ superfield.
I guess they mean, that since the second term on r.h.s. doesn't depend on $V^a$, it's possible to solve it within the framework of perturbation theory in the coupling constant(s) $g_a$. Is it correct? If so, how to prove it strictly in all orders?
 A: 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.


--
$^1$ An affine equation is a linear equation with an inhomogeneous term/source term.
A: Wess-Zumino gauge is a particular choice of gauge where the vector superfield has a particular form and has less components than the generic vector super field. So if i'm free to make a gauge transformation i can choose the components of the chiral super field $\Omega$ in a manner that the sum of the $\theta$ (or any other "$\theta$ component" i want to eliminate to reach the WZ gauge) of the chiral and vector super field equal zero. 
A: I think, the gauge transformation has in its definition the superspace $y_\mu$, which is equal with $x_\mu + i \theta \sigma_\mu \bar \theta$. Thus in its Taylor expansion appears only 3 terms:  $Ω(y) = Ω(x) + i \theta \sigma_\mu \bar \theta \partial_\mu Ω(x) + \frac{1}{4} \theta \theta \bar \theta \bar \theta \Box{Ω(x)}$.
