When is the action of the gauge group on the space of connections free?

Question

Let $G$ be a compact Lie group. Let $\mathcal{A}$ be the space of connections on the principal trivial $G$-bundle $G\times \mathbb{R}^4$ possibly with some growth condition (to specify it is a part of the question).

ADD: Let us comment on the growth condition. Let $\nabla^0$ be the trivial connection on $G\times X$. Any connection $\nabla$ has the form $\nabla=\nabla^0+A$ where $A$ is the section of the bundle $\Omega^1\otimes Lie(G)$ of 1-forms on $\mathbb{R}^4$ with values in $Lie(G)$. The growth condition should be imposed on $A$.

The gauge group $\mathcal{G}:=Maps(\mathbb{R}^4\to G)$ acts on $\mathcal{A}$ in the usual ways.

Can the action of $\mathcal{G}$ on $\mathcal{A}$ be free? E.g. for $G=SU(2)$? If not, is it true that the set of connections with non-trivial stabilizers (or infinitesimal stabilizers) is 'very small' in some sense?

Remark. If $G=U(1)$ then the action of $\mathcal{G}$ on $\mathcal{A}$ is free provided we impose a growth condition on connections such that they should vanish at infinity at least along a given direction.

Answer 1

It suffices that there is a straight line (identified with $\Bbb R$ below) in $\Bbb R^4$ so that the connection matrix $A$ is $L^1$ (with respect to the trivial connection) when restricted to that line, and so that the gauge transformation $\sigma$ is $L^1_1$ (meaning that $\sigma: \Bbb R^4 \xrightarrow{a.e.} G \subset \Bbb R^N$ for some large $N$ has $\sigma - 1 \in L^1(\Bbb R^4, \Bbb R^N)$ and $\nabla \sigma \in L^1$). I have not made any effort to ensure that $\mathcal G_{decay}$ is a Banach Lie group acting smoothly on $\mathcal A_{decay}$.

Given any smooth connection $A$ on any principal bundle, the action of the gauge transformation $\sigma$ on $A$ is $\sigma(A) = A - (d_A \sigma) \sigma^{-1}$. So the stabilizer of $A$ inside the gauge group of all smooth gauge transformations is the set of $A$-parallel gauge transformations. So what you are asking for is a decay condition on $A$ that guarantees that any $A$-parallel gauge transformation cannot decay at infinity; this is reasonable, since the notion of being parallel means that your section is constant "with respect to $A$".

To do this, recall that the holonomy $H(t) = \text{Hol}_{0 \to t}(A) \in G$ of a connection along an interval $I = [s_1,s_2]$ with $-\infty \leq s_1 < 0 < s_2 \leq \infty$ defines a continuous map $L^1(I; \mathfrak g) \to C^0(I;G)$, where in the case that one of the $s_i = \pm \infty$, I am claiming that $H(\pm \infty) = \lim_{t \to \pm \infty} H(t)$. This follows from the claim that just the map $\text{Hol}_{0 \to s_2}: L^1(I;\mathfrak g) \to G$ is continuous in $L^1$, as for any $L^1$ function $f$, subintervals of $I$ contained in $(N,\infty)$ for $N$ sufficiently large will have $\|f\|_{L^1}$ arbitrarily small.

As a consequence, one sees that we have two continuous maps $\text{Hol}_{\pm\infty}: L^1(\Bbb R;\mathfrak g) \to G$.

Next recall that if $\sigma$ is an $A$-parallel gauge transformation on $\Bbb R$, this means that $\sigma(t) = H(t) \sigma(0) H(t)^{-1}$. Thus, if $\sigma(\infty) = 1$, we see that $\sigma(t) = 1$ for all $t$.

This implies that if you demand your gauge transformations decay to the identity at infinity on this line --- which is guaranteed by demanding that the gauge transformations are in $L^1_{1}$ --- then they must be constant on our line. But parallel gauge transformations are globally constant (so long as $A$ is locally $L^1$ on every line, which is guaranteed if it's bounded), and so a gauge transformation that decays to the identity at infinity is constant.

Answer 2

Let us assume that the growth condition at infinity is implemented by requiring that all fields (connections and gauge transformations) extend to a given compactification $M$ of $\mathbb R^4$. The classification of the gauge orbit types over compact manifolds can be found here for SU(N) and here for the other classical groups. The starting point for this classification is that a gauge transformation preserving a connection $A$ is constant on the holonomy subbundle of $A$. Thus, the classification boils down to classifying certain subbundles (so called Howe subbundles). In general, the classification involves certain characteristic classes and thus strongly depends on the topology of $M$ (i.e. on which growth condition you require).

If the gauge group has a center, then the action is never free: every element of the center defines a constant gauge transformation that is preserved by every connection. The center is the minimal orbit type in the sense that the space of connections whose stabilizer subgroup is the center is dense in the space of all connections (this is a consequence of the approximation theorem Theorem 2.7 and should answer your question "If not, is it true that the set of connections with non-trivial stabilizers (or infinitesimal stabilizers) is 'very small' in some sense?").

For $SU(2)$, the possible orbit types are $SU(2)$, $U(1)$ or $\mathbb Z_2$ (the latter being the center, and hence the generic orbit type). Which orbit types occur indeed in the final classification depends on the topology of $M$, see the discussion of examples in Section 8.