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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 308

Away from a zero of $\phi$, the first <Bogomolny vortex equation> gives
$$
a_{\bar z}=-i\partial_{\bar z}\log\phi,
\qquad
f_{z\bar z}=-2i\partial_z\partial_{\bar z}\log|\phi|.
$$
Substitution into the second equation, with $\nabla^2=4\partial_z\partial_{\bar z}$, gives
$$
\boxed{-\frac2\Omega\nabla^2\log|\phi|=1-|\phi|^2}.
$$
Now set $\widetilde\Omega=\Omega|\phi|^2$. The <Gaussian curvature> formula gives
$$
\widetilde K\widetilde\Omega=K\Omega-\nabla^2\log|\phi|.
$$
Therefore $(\widetilde K+1/2)\widetilde\Omega=(K+1/2)\Omega$ implies
$$
-\nabla^2\log|\phi|=\frac\Omega2(1-|\phi|^2),
$$
which is exactly the vortex equation.

The metric
$$
ds_n^2=\frac{8n^2|z|^{2n-2}}{(1-|z|^{2n})^2}dzd\bar z
$$
is the pullback of the $n=1$ <Poincare disc model> metric under $w=z^n$. It consequently has $K=-1/2$ away from the origin. Comparing it with $ds_1^2$ gives the <Witten hyperbolic vortex>
$$
\boxed{|\phi|=\frac{n|z|^{n-1}(1-|z|^2)}{1-|z|^{2n}}}.
$$
Putting $n=N+1$ gives winding number $N$. A gauge choice with positive radial factor is
$$
\phi=\frac{n z^N(1-|z|^2)}{1-|z|^{2n}},
$$
and then
$$
\boxed{
a_{\bar z}=i\left(\frac{z}{1-|z|^2}-\frac{nz|z|^{2n-2}}{1-|z|^{2n}}\right),
\qquad a_z=\overline{a_{\bar z}}.
}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 308 / 2 / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 308 / 2 / Solution

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 308 / 2 / Solution

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