Gauge fixing and instanton calculation

Question

I am reading Cheng&Li's book "Gauge theory of elementary particle physics". In section 16.2, I am confused by some assumptions.
Suppose we have a $SU(2)$ gauge theory in $mathbb{R}^4$
$$
S=int d^4x Tr(F_{mu nu}F_{mu nu})qquad F_{mu nu}=partial_{mu}A_{nu}-partial_{nu}A_{mu}+[A_{mu},A_{nu}].
$$
For instanton solution, i.e., $S<infty$ we have following boundary condition
$$
F_{mu nu} rightarrow 0quad A_{mu}rightarrow U^{-1}partial_{mu}Uqquadmbox{for some }U in SU(2)tag{1}
$$
$U$ is a map from $S^3$ to $SU(2)$ and can be classified by winding number. An example of $U$ is $$U=frac{x_0+ivec{x}cdot vec{tau}}{r},qquad r=sqrt{x_0^2+vec{x}^2} $$ and corresponding $A$ is $$A_0=frac{-ivec{x}cdot vec{tau}}{r^2+lambda^2},qquadvec A=frac{-i(x_0vec{tau}+vec tau times vec x)}{r^2+lambda^2}. $$
Now the book chooses another gauge such that $A'_0=0$, i.e., for some $V$
$$ A'_0=V^{-1}A_0V+V^{-1}partial_0V=0.$$
In the next, the book claims we can set spatial component of $A rightarrow  0$, as $r rightarrow infty$, and hence $A_i rightarrow V^{-1}partial_i V$, $r rightarrow infty$.
Here is my question: why can we do this? In $(1)$, we have assumed $A_i$ goes to a pure gauge $U^{-1}partial_i U$. I think we must have
$$A'_i rightarrow V^{-1}U^{-1}(partial_i U) V+V^{-1}partial_iV=(UV)^{-1}partial_i (U V).$$
Please correct me if I am wrong.

ɪdɪət strəʊlə · Answer

$newcommand{rto}{overset{scriptscriptstyle rtoinfty}{longrightarrow}} newcommand{v}[1]{boldsymbol{#1}} newcommand{t}{tau} newcommand{pd}{partial} newcommand{demeqq}{overset{!}{=}}$
One should make a distinction between time-dependent and time-independent gauge transformations. I will denote $x=(t,v x)$. What is described in the book is the following. You have boundary condition:
$$A_mu(x) demeqq U^{-1}(x),pd_mu, U(x)tag{bc}label{bc}$$
and gauge-fixing condition
$$A_0(x) demeqq 0,, qquad text{for all} x. tag{gfc}label{gfc}$$
What they now say is that a time-independent gauge transformation doesn't change (ref{gfc}). Such a gauge transformation is one that $pd_0tilde{U}(v x)=0$
begin{align}
A_0(x)mapsto A_0'(x) &= tilde{U}^{-1}(v x) A_0(x) tilde{U}(v x) + tilde{U}^{-1}(v x)pd_0, tilde{U}(v x) 
&= tilde{U}^{-1}(v x); color{red}{0}; tilde{U}(v x) + tilde{U}^{-1}(v x);color{blue}{underset{0}{underbrace{pd_0, tilde{U}(v x)}}} 
&= color{red}{0}+color{blue}{0}=0.
end{align}
This means then that having gauge-fixed in (ref{gfc}) we have not completely gauge-fixed, as those time-independent gauge transformations are still allowed. The next claim in the book is that the only gauge field consistent with both (ref{bc}) and (ref{gfc}) is one that is always pure gauge, and moreover, time-independent gauge, so
$$A_mu(x) = left(begin{array}{cc} A_0(x)  A_i(x) end{array} right) = left(begin{array}{cc} 0  A_i(v x) end{array} right) = left(begin{array}{cc} 0  V^{-1}(v x) pd_i V(v x) end{array} right).tag{$star$}label{*} $$
You can see that because if you had any time-dependence in (ref{*}), the gauge transformations would necessarily generate some $A_0$, and this is forbidden by (ref{gfc}).
A comment: What you also have wrong in your understanding is in the phrase "chooses another gauge". In fact, this was the first time the authors chose a gauge; previously they merely said it is some pure gauge at $rtoinfty$.

Gauge fixing and instanton calculation

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