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

$$ F_{\mu \nu} \rightarrow 0\quad A_{\mu}\rightarrow U^{-1}\partial_{\mu}U\qquad\mbox{for some }U \in SU(2)\tag{1} $$

$U$ is a map from $S^3$ to $SU(2)$ and can be classified by winding number. An example of $U$ is $$U=\frac{x_0+i\vec{x}\cdot \vec{\tau}}{r},\qquad r=\sqrt{x_0^2+\vec{x}^2} $$ and corresponding $A$ is $$A_0=\frac{-i\vec{x}\cdot \vec{\tau}}{r^2+\lambda^2},\qquad\vec A=\frac{-i(x_0\vec{\tau}+\vec \tau \times \vec x)}{r^2+\lambda^2}. $$

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.

Answer 1

$\newcommand{\rto}{\overset{\scriptscriptstyle r\to\infty}{\longrightarrow}} \newcommand{\v}[1]{\boldsymbol{#1}} \newcommand{\t}{\tau} \newcommand{\pd}{\partial} \newcommand{\demeqq}{\overset{!}{=}}$ 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 $\pd_0\tilde{U}(\v x)=0$ \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 $r\to\infty$.