# Why group elements associated with gauge transformations of finite action field configurations in QCD don't depend in $r$?

I am reading the chapter on instantons in Coleman's Aspects of Symmetry. I am puzzled by an argument i don't quite follow. In section 3.2, Coleman considers configurations of the gauge field with finite action. Using some arguments that (I think) are not relevant in what follows he concludes that for the action to be finite $F_{\mu\nu}$ must fall off at infinity like $O(1/r^3)$. He then says that this doesn't imply that $A_{\mu}$ falls off like $O(1/r^2)$ but rather that in general it can be a gauge transform of a configuration that does fall like $O(1/r^2)$. So far so good.

He then explicitly writes the gauge transformed field

$$A_{\mu}\to{}gO(1/r^2)g^{-1}+g\partial_{\mu}g^{-1}$$

$$A_{\mu}\to{}g\partial_{\mu}g^{-1}+O(1/r^2)$$

After writing this equation he says that $g$ denotes an element of the gauge group that does not depend on $r$. I don't see why this is legit. I mean, in principle we should be able to consider a general $g$, shouldn't we?

"...where $g$ is a function from four-space to $G$ of order one, that is to say, a function of angular variables only..."
This is the key to the answer. So first that you should note is that why we are interested only in the configurations for which $g = 1$. This is the key to the answer.
Let's look to amplitude of transition between gauge vacua: $$M \equiv \langle \bar{\text{vac}}|e^{iHT}|\text{vac}\rangle, \qquad |\text{vac}\rangle = \sum_{g \in \text{ equivalence class}}|g(\mathbf x)\rangle$$ Without loss of generality we may choose initial state being in the equivalence class (EQ) of $g(\mathbf x) = 1$. The reason of that is that we can simultaneously perform gauge transformation of initial an final state without changing $M$. All the states of this EQ has the property that $$\lim_{x \to \infty } g(\mathbf x ) = g_{0} \Rightarrow \lim_{x \to \infty}A(\mathbf x) = 0$$ Next, semiclassical approximation of an amplitude is indeed zero for final states, for which $$\lim_{x \to \infty}\bar{g}(\mathbf x ) \neq g_{0} \Rightarrow \lim_{x\to \infty}A(\mathbf {x}) \neq 0$$ Really, this condition means that we have nonzero $\dot{A}(\mathbf x)$. Being multiplied on infinite volume, it gives infinity, so thus energy of such state is infinity, and semiclassical amplitude is exact zero. So the only vacua for which semiclassical amplitude is nonzero are those which belongs to the EQ $\lim_{x\to \infty}\bar{g}(\mathbf x) = 1$.
We have then the restriction on all interesting group elements: $$\lim_{x \to \infty}g(\mathbf x) = 1$$ As immediate consequence we have the statement that $g(\mathbf x)$ doesn't depend on $\mathbf x$ on spatial infinity. This compactifies three-dimensional space, making it topologically equivalent to 3-sphere.
What does this mean in context of your question? In fact, coordinates of 3-sphere are given by directions on its surface, i.e., by angles. That's why $g(\mathbf x )$ in $$A(\mathbf x) \to g(\mathbf x)\partial g^{-1}(\mathbf x) \ \ \ \text{ at } \ \ x \to \infty$$ depends on angles only.