A question about the book "the geometry and dynamics of magnetic monopoles"

Question

In chapter 2 of the book "The geometry and dynamics of magnetic monopoles", by M.F. Atiyah and N.J. Hitchin (the chapter is called "Geometry of the monopole spaces"), it is written:

"If we identify gauge-equivalent monopoles (of charge $k$) we obtain a moduli or parameter space which we shall denote by $N_k$. In fact for most purposes it is best to enlarge this space by a circle or phase factor, so getting a space $M_k$. The simplest way to define this is to fix a direction in $\mathbb{R}^3$, (say the $x_1$-direction), use the gauge $A_1 = 0$ and allow only gauge transformations which tend to the identity as $x_1 \to \infty$."

And then they write a bit later

"It follows that $M_k$ is fibered over $N_k$ with fibre $S^1$."

I am not sure why this is the case.

Here are some thoughts to let you know about my confused state. I get that one can, using a gauge transformation, reduce to the case $A_1 = 0$, by essentially solving some first-order matrix PDE. I think that the residual group of gauge transformations consist of just constant maps into $SU(2)$ (please inform me if I am wrong).

I think that $SO(3)$ acts on $N_k$, by acting as "spatial" rotations in $\mathbb{R}^3$. Somehow, the subgroup of $SO(3)$ which fixes $x_1$ is a copy of $SO(2)$. Is this somehow the circle fiber they are talking about?

Please help clear my confusion if you can. Thank you!

Edit: I think things would go fine if the residual group consisted of constant maps from $\mathbb{R}^3$ to $U(1)$ (rather than $SU(2)$)... As of now, I don't see why this would be the case though.

Answer 1

There is a $U(1)$ in that story and there is an $SO(2)$ consisting of spatial rotations fixing $\{x_1 > 0\}$. It turns out that the relevant group is the first one, which has to do with rigidifying the monopole.

More specifically, it is always possible to reduce to the gauge $A_1 = 0$ by solving for

$$ -dg g^{-1} + g A_1 g^{-1} = 0, $$

i.e. for

$$ g^{-1} dg = A_1 $$

where $g: \mathbb{R}^3 \to SU(2)$. So from now on, we can assume WLOG that $A_1 = 0$. Note that a gauge transformation $g$ preserving $\{A_1 = 0\}$ must satisfy: $$-dgg^{-1} = 0$$ so that $g: \mathbb{R}^3 \to SU(2)$ is constant. Moreoever, the group of gauge transformations preserving both the set $\{A_1 = 0\}$ and $\Phi(*)$ is the set of constant maps from $\mathbb{R}^3 \to U(1)$, where $U(1)$ denotes the subgroup of $SU(2)$ which fixes $\Phi(*)$.

But then when we rigidify a monopole by further requiring that the gauge transformation goes to the identity as $x_1 \to \infty$, we see that the moduli space $M_k$ of rigidified monopoles (in the sense above) is a circle bundle over the moduli space $N_k$ of monopoles up to gauge equivalence.

This was the explanation. The subgroup $SO(2)$ of the group of Euclidean rotations of $\mathbb{R}^3$ which fix $\{ x_1 > 0 \}$ was actually not the "right" group to look at, in this case!

Edit: I think one can choose for example the (more complete) gauge given by: $$A_1 = 0$$ and $$\Phi(*) = \begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix},$$ then the residual group of gauge transformations would just be the group of constant maps from $\mathbb{R}^3 \to U(1)$, where $$ U(1) = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda^{-1} \end{pmatrix} $$ and $\lambda \in \mathbb{C}$ with $|\lambda| = 1$.