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Write $\Omega =e^{2\rho }$, $F={\partial }_x{A}_2-{\partial }_y{A}_1$, and $s=|\Phi {|}^2$, so the physical magnetic field is $B=F/\Omega$. Use finite-energy Abelian Higgs vortex boundary data: the field tends to unit magnitude at the ideal boundary, its phase winds $N$ times, and the boundary limits in the decomposition exist. This vacuum asymptotic behavior is the setting in which the stated degree gives an energy bound. We retain the current boundary term; degree alone must not be silently equated with magnetic flux on an open disk.

Define $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$. The gauge-covariant derivative of a charged scalar field satisfies $[D_x,D_y]\Phi =-iF\Phi$. Differentiating the current givesOn the other hand,Combining the two identities and completing the square in the magnetic and potential terms yields the conformal-surface vortex square completionAll integrations on the noncompact metric disk can first be made over $|z|<R<1$ and then passed to the limit. By the Stokes theorem, the two boundary contributions combine to $\frac{1}{2}\oint (A+j)$. At the vacuum boundary write $\Phi =e^{i\chi }$. Then $j=d\chi -A$, so this combined integral is $\frac{1}{2}\oint d\chi =\pi N$, even if the current term does not vanish separately. The usual additional condition of tangential covariant decay makes that current term vanish and yields magnetic flux quantization of an Abelian Higgs vortex, but is not needed if the combined boundary term has this phase limit. Thus for $N\ge 0$, $V\ge \pi N$. Reversing both signs in the square completion gives $V\ge -\pi N$ for $N\le 0$, henceFor positive degree, equality is attained precisely when the Bogomolny vortex equations areDistinguishing the physical field $B$ from the coordinate curvature coefficient $F$ prevents an erroneous conformal factor.

For the proposed field, set $q=x^2+y^2$ and choose the smooth connection one-formIndeed, with ${\partial }_{\bar{z}}=\frac{1}{2}({\partial }_x+i{\partial }_y)$ and $A_{\bar{z}}=\frac{1}{2}(A_1+i{A}_2)$,The second Bogomolny vortex equation follows fromThus both squares vanish, including at the simple zero $z=0$ by smoothness. At $r\to 1$, $\Phi \to e^{i\vartheta }$ and $A=2r^{2}d\vartheta /(1+r^2)\to d\vartheta$, proving degree one. Moreover, $j_{\vartheta }=4r^2(1-r^2)/(1+r^2{)}^3\to 0$, so the current boundary term really vanishes for this solution. The magnetic flux is explicitlyConsequentlyThis is the hyperbolic one-vortex at curvature minus one half; its finite energy follows directly from the vanishing squares and the finite flux.