# Gauge fixing and instanton calculation

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.

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 $$r\to\infty$$.
• So why is V determined by the equation $V^{-1}A_0V+V^{-1}\partial_0V=0$?. It seems in your (*), V hasn't been determined. Jul 22, 2020 at 12:31