Magnetic field due to singular gauge transformation

Question

In $SU(2)$ gauge theory over $\mathbb R^3$, consider the following gauge transformation (in spherical polar coordinates) $$ \Omega=\begin{bmatrix}e^{i\phi}\cos(\theta/2)&\sin(\theta/2)\\-\sin(\theta/2)&e^{-i\phi}\cos(\theta/2)\end{bmatrix}.$$ It is multivalued (aka singular) on the half-infinite line $\theta=0$. Such gauge transformations are said to create magnetic monopoles (see https://www.sciencedirect.com/science/article/abs/pii/0550321378901530).

To test this I'm doing the following. I start with the vacuum configuration $\vec A=0$, where $\vec A\equiv \vec A_a\sigma_a$ is the gauge field 3-vector (each component being a $2\times 2$ matrix), and I apply the gauge transformation $\Omega$. The gauge field is then $\vec A=i\vec \partial \Omega~\Omega^{-1}$ (I've set the gauge coupling to unity). Doing the calculation, we find that the gauge field is finite everywhere except on the line $\theta=0$, where it is infinite (I can provide the explicit expression if required).

Now I want to compute the magnetic field $B_i= \epsilon_{ijk}F_{jk}$. At $\theta\neq 0$ we can easily see that $\vec B=0$ because we started with the vacuum configuration and applied a gauge transformation. On $\theta=0$ though, since the gauge field is infinite, the magnetic field might be non-zero -- in fact, I suspect it is some kind of delta-function smeared along $\theta=0$. How exactly can I compute the magnetic field on $\theta=0$ to verify this?

Edit: If the group was Abelian, I would compute $\oint \vec A.d\vec{l}=\int \vec B.d\vec S$ for a loop enclosing $\theta=0$. This would tell me if any magnetic flux is enclosed.

Answer 1

So I used Stokes law to write \begin{equation} \int_S \vec B.d\vec S=2\oint_l \vec A.d\vec l+\int_S i\epsilon_{ijk}[A_i,A_j]dS^k \end{equation} where $l$ is a loop around $\theta=0$, and $S$ is a surface bounded by $l$. The integrals are well defined despite the singularity. This shows that \begin{equation} \vec B=-4\pi \sigma_3 \hat z \end{equation} on $\theta=0$, and zero elsewhere.