Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 313 / 1

With $D_i\varphi =({\partial }_i-i{A}_i)\varphi$ and $B=F_{12}$, the critically coupled Abelian Higgs model energy in the conventions of the question isFinite energy requires $|\varphi |\to 1$, $D_i\varphi \to 0$, and $B\to 0$ at spatial infinity.

For the Derrick theorem test, preserve gauge covariance by definingThe magnetic, covariant-gradient, and potential energies scale asStationarity at $\lambda =1$ requires $E_B=E_V$, which is possible for a nonconstant finite-energy configuration. The oppositely scaling magnetic and potential terms therefore evade the usual Derrick obstruction and allow vortex solitons.

At infinity write $\varphi \to e^{i\chi }$. The condition $D\varphi \to 0$ gives $A\to d\chi$, and the phase winding defines the Abelian Higgs vortex numberBy Stokes theorem,

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Away from zeros of $\varphi$, writeThe first Bogomolny vortex equation, $(D_x+i{D}_y)\varphi =0$, givesIf the zeros $z_r$ have multiplicities $N_r$, the phase curl and the logarithmic singularities give, distributionally,Equating this with $\Omega (1-e^u)/2$ yields the Taubes equation

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For a zero of order $N$ at the origin and no other zeros,Away from the zero, the flat Taubes equation isIf $u$ had a positive interior maximum, then the second-derivative test would give $\Delta u\le 0$ there, while $e^u-1>0$, a contradiction. The boundary values are $-\infty$ at the zero and $0$ at infinity, so the maximum principle gives $u\le 0$. Hence

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The surface area isIntegrating the Taubes equation over the compact surface eliminates the Laplacian and givesfor a nontrivial solution. For $N=2$, the Bradlow bound therefore requiresEquality is the dissolved-vortex limit with identically vanishing Higgs field and does not give the stipulated ordinary two-vortex solution.

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