Winding number in the topology of magnetic monopoles

Question

I am reading on magnetic monopoles from a variety of sources, eg. the Jeff Harvey lectures.. It talks about something called the winding $N$, which is used to calculate the magnetic flux. I searched the internet but am not being able to understand the calculation done in this particular case.

$g=-\frac{1}{8}\int_{S^2_\infty} Tr([d\hat{\Phi},d\hat{\Phi}],\hat{\Phi})$

Then the author says that

Now $\Phi$ restricts to a map $\Phi : S_\infty^2 → S^2$ , where the target is the unit sphere in $su(2)$. This map has some degree $N$ , and it is easy to verify that the right-hand side of the above equation is $−2\pi$ times this. Therefore $g = −2\pi N$ .

What is $N$, the winding number also called as the degree on the map? By what i have learnt, it is the number of times you wind an object unto the another, then shouldn't the integral be $N*4\pi$, as $4\pi$ is the surface area of $S^2$.

Answer 1

I didn't find the equation and the argument you quoted in that paper. But, yes, it is the Brouwer degree, deg$(\hat{\Phi})$, which equals the monopole number $$ N\equiv\frac{1}{4\pi}\int_{\mathbb{R}^3} \mathrm{Tr}(F_A\wedge D_A(\Phi))=\frac{1}{4\pi}\int_{\mathbb{R}^3} d(\mathrm{Tr}(\Phi)F_A) =\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr}(\Phi F_A)$$ $$ =\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr}(\hat{\Phi} F_A) $$

where the one has used Bianchi identity, Stokes' theorem, to obtain the first two equalities, and Jaffe & Taubes show in their book that one can replace $\Phi$ by $\hat{\Phi}$. Now this coincides with the brower degree, for which there is an explicit formula:

$$N=\mathrm{deg}(\hat{\Phi})=-\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr(\hat{\Phi} d\hat{\Phi}\wedge d\hat{\Phi}})\in \mathbb{Z}=\pi_2(S^2)=[S^2,S^2]$$ (what you wrote.) This is physically understood as an infinite wall potential, separating the monopole sectors corresponding to different integers. Now, to actually answer your question, you can compute this integral for the t'Hooft-Polyakov monopole solution, for which $$ \hat{\phi}=(\sin(\theta)\cos\phi,\sin\theta \sin\phi,\cos\theta)_i\cdot \sigma^i, $$ and you will find $$N=-\frac{1}{4\pi}\int_{S^2_\infty} \mathrm{Tr(\hat{\Phi} d\hat{\Phi}\wedge d\hat{\Phi}})=+\frac{1}{4\pi}\int_{[0,2\pi]} \int_{[0,\pi]}\sin\theta d\theta\wedge d\phi=1.$$

Answer 2

N is equal to the number of points in the $S_2$ sphere at infinity mapped to the same point of the $S_2$ Higgs vacuum manifold.The integral is a topological invariant depending only on this number and not on the details of the map. In the following, I'll describe to you a family of these maps:

One way to perform the integral is to use the stereographic projection coordinate:

$ z = tan(\frac{\theta}{2}) e^{i\phi}$

In this coordinate system,a map of winding number $N$ will look as:

$ z \rightarrow Z= z^N$

The surface element of the sphere in these coordinates is:

$ d \mu = \frac{dzd\bar{z}}{1+z \bar{z}}$

Remark: In these coordinates, the Higgs components are given by:

$\Phi_x = 2\frac{Re(Z)}{1+Z\bar{Z}}$

$\Phi_y = 2\frac{Im(Z)}{1+Z\bar{Z}}$

$\Phi_z = \frac{1-Z\bar{Z}}{1+Z\bar{Z}}$

Using these coordinates,it is not difficult to see that the integral (1.43) value is N.