The surface area is
$$
\operatorname{Area}(\Sigma)
=\int_{\mathbb R^2}\frac{a^2}{(1+|z|^2)^2}\,dx\,dy
=2\pi a^2\int_0^\infty\frac{r\,dr}{(1+r^2)^2}
=\pi a^2.
$$
Integrating the <Taubes equation> over the compact surface eliminates the Laplacian and gives
$$
4\pi N=\int_\Sigma\Omega(1-e^u)d^2x
<\operatorname{Area}(\Sigma)
$$
for a nontrivial solution. For $N=2$, the <Bradlow bound> therefore requires
$$
\boxed{\pi a^2>8\pi,
\qquad a^2>8}.
$$
Equality is the dissolved-vortex limit with identically vanishing Higgs field and does not give the stipulated ordinary two-vortex solution.

Solved by gpt-5.6-sol high.

Solution

Paper 313 / 1 / b / iii / Solution

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Integrating the Taubes equation on a compact surface of area $A$ gives $A\geq4\pi N$. A nontrivial vortex with nonzero Higgs field requires the strict inequality $A>4\pi N$.


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 Incoming links: Bradlow bound