Global energy minima for critical Abelian Higgs vortices

Define the real current one-form $j=j_{1}d{x}_1+j_{2}d{x}_2$ by $j_i=\mathrm{Im}(\bar{q}{D}_{A,i}q)$. Expanding the squares and using $[D_{A,1},D_{A,2}]q=-iBq$ gives $\mathrm{\mid }{D}_{A}q{\mathrm{\mid }}^2=\mathrm{\mid }(D_{A,1}+i{D}_{A,2})q{\mathrm{\mid }}^2+{\mathrm{\partial }}_1{j}_2-{\mathrm{\partial }}_2{j}_1+B\mathrm{\mid }q{\mathrm{\mid }}^2\mathrm{.}$ Also, direct expansion gives $\frac{B^2}{2e^2}+\frac{e^2}{2}(\xi -\mathrm{\mid }q{\mathrm{\mid }}^2{)}^2=\frac{(B-e^2(\xi -\mathrm{\mid }q{\mathrm{\mid }}^2){)}^2}{2e^2}+B(\xi -\mathrm{\mid }q{\mathrm{\mid }}^2)\mathrm{.}$ Adding these identities yields the positive-winding disk identity with boundary term ${\int }_{D_R}({\mathrm{\partial }}_1{j}_2-{\mathrm{\partial }}_2{j}_1)$. Repeating the expansion with $D_{A,1}-i{D}_{A,2}$ and $B+e^2(\xi -\mathrm{\mid }q{\mathrm{\mid }}^2)$ gives the negative-winding identity with the opposite signs.

Enlarge the exterior radius in [F1] if necessary so $R_0\ge 1$. For $r>R_0$, let $Q(r)={\int }_{\mathrm{\partial }{D}_r}(\mathrm{\mid }q{\mathrm{\mid }}^2-\xi {)}^{2}ds$ and $K(r)={\int }_{\mathrm{\partial }{D}_r}\mathrm{\mid }{D}_{A,\tau }q{\mathrm{\mid }}^{2}ds$, where $D_{A,\tau }$ is the covariant derivative in the unit tangent direction and $ds=rd\theta =rd\sigma$ by [F7]. The polar formula [F7] and Fubini [F4] identify ${\int }_{R_0}^{\mathrm{\infty }}Q(r)dr$ with ${\int }_{\mathrm{\mid }x\mathrm{\mid }>R_0}(\mathrm{\mid }q{\mathrm{\mid }}^2-\xi {)}^{2}dx$ and ${\int }_{R_0}^{\mathrm{\infty }}K(r)dr$ with the integral of $\mathrm{\mid }{D}_{A,\tau }q{\mathrm{\mid }}^2$ there, which is bounded by the finite kinetic energy. In particular $Q,K\in L^1([R_0,\mathrm{\infty }))$ and are measurable. For each integer $m\ge 1$, the set $S_m=\{r\ge R_0:Q(r)>1\mathrm{/}m\}$ has finite measure. Its complement has divergent $\int dr\mathrm{/}r$, since removing a finite-measure set from a half-line cannot make the logarithmic integral finite. If $rK(r)\ge 1\mathrm{/}m$ for every sufficiently large $r\mathrm{\notin }{S}_m$, then $\int K(r)dr\ge m^{-1}{\int }_{[R,\mathrm{\infty })\setminus S_m}dr\mathrm{/}r=\mathrm{\infty }$, contradiction. Thus there is an unbounded set of radii with $Q(r)\le 1\mathrm{/}m$ and $rK(r)<1\mathrm{/}m$. By [A1] select an increasing sequence $r_m\to \mathrm{\infty }$ from these sets. Along it, [F6, F7] give ${\int }_{\mathrm{\partial }{D}_{r_m}}\mathrm{\mid }q{\mathrm{\mid }}^{2}ds\le 2\pi \xi r_m+\sqrt{2\pi r_{m}Q(r_m)}=O(r_m)$, and $\mid {\int }_{\mathrm{\partial }{D}_{r_m}}j\mid \le {\left({\int }_{\mathrm{\partial }{D}_{r_m}}\mathrm{\mid }q{\mathrm{\mid }}^{2}ds\right)}^{1\mathrm{/}2}K(r_m{)}^{1\mathrm{/}2}⟶0.$

Uniform convergence $\mathrm{\mid }q\mathrm{\mid }\to \sqrt{\xi }$ gives $\mathrm{\mid }q{\mathrm{\mid }}^2\ge \xi \mathrm{/}2$ for sufficiently large $r$. On $\mathrm{\partial }{D}_r$, $j=\mathrm{\mid }q{\mathrm{\mid }}^2(d(n\theta +\varphi )-A)$, so ${\int }_{\mathrm{\partial }{D}_r}A=2\pi n-{\int }_{\mathrm{\partial }{D}_r}j\mathrm{/}\mathrm{\mid }q{\mathrm{\mid }}^2$ because $\varphi$ is single-valued and periodic. Also $\mathrm{\mid }{j}_{\tau }\mathrm{\mid }=\mathrm{\mid }\mathrm{Im}(\bar{q}{D}_{A,\tau }q)\mathrm{\mid }\le \mathrm{\mid }q\mathrm{\mid }\mathrm{\mid }{D}_{A,\tau }q\mathrm{\mid }$, hence $\mathrm{\mid }{j}_{\tau }\mathrm{\mid }\mathrm{/}\mathrm{\mid }q{\mathrm{\mid }}^2\le \sqrt{2\mathrm{/}\xi }\mathrm{\mid }{D}_{A,\tau }q\mathrm{\mid }$. Along the sequence from step 1.2, Cauchy--Schwarz and [F7] give $\mid {\int }_{\mathrm{\partial }{D}_{r_m}}\frac{j}{\mathrm{\mid }q{\mathrm{\mid }}^2}\mid \le \sqrt{\frac{2}{\xi }}(2\pi r_{m}K(r_m){)}^{1\mathrm{/}2}⟶0.$ By [F3], ${\int }_{D_{r_m}}B\to 2\pi n$.

Integrate the positive-winding disk identity from step 1.1 over $D_{r_m}$ and use [F3] on the current term. The two square integrals are nonnegative, while the boundary-current integral tends to zero by step 1.2 and the flux integral tends to $2\pi n$ by step 2.1. By [F5], the integrals of the nonnegative energy and square densities over the increasing disks converge to their plane integrals, so taking the limit gives the positive-sign exact identity and hence $E(A,q)\ge 2\pi \xi n$ for $n\ge 0$. For $n\le 0$, apply the positive-sign identity from step 1.1 to $(\bar{q},-A)$, whose winding is $-n\ge 0$; its energy is unchanged and its curvature changes sign. The resulting identity is exactly the negative-sign identity for $(A,q)$ and gives $E(A,q)\ge -2\pi \xi n$. Thus the bound is $E(A,q)\ge 2\pi \xi \mathrm{\mid }n\mathrm{\mid }$, and both identities hold when $n=0$ as well.

For each fixed divisor, [F2] supplies a smooth pair satisfying the relevant first-order equations. Its square terms in step 1.1 vanish identically; the sequence estimates of steps 1.2–2.1 apply because [F2] gives finite energy and the same boundary sector. Hence its energy is exactly $2\pi \xi \mathrm{\mid }n\mathrm{\mid }$. The lower bound of step 3.1 applies to every admissible same-winding configuration, so this pair is an absolute transverse energy minimum, without radial or unit-winding assumptions.

Let $(a,\eta )$ be a smooth compactly supported variation of a minimizing pair, and set $A_t=A+ta$, $q_t=q+t\eta$; for any field expression $F$, define $\delta F:={\frac{d}{dt}\mid }_{t=0}F(A_t,q_t)$. For sufficiently small $t$, the fields remain in the same boundary sector; outside the compact support they equal the original fields. Thus the boundary-current limit in the disk identity is unchanged and the flux is unchanged by Stokes applied to $da$. The energy along this variation is the constant $2\pi \xi \mathrm{\mid }n\mathrm{\mid }$ plus the squared $L^2$ norms of the two first-order residuals in step 1.1. Both residuals vanish at $t=0$; differentiating the squared norms gives $E^{\mathrm{\prime }}(0)=0$ and $E^{\mathrm{\prime }\mathrm{\prime }}(0)=2{\int }_{{\mathbb{R}}^2}(\mathrm{\mid }\delta (D_{A,1}\pm i{D}_{A,2})q{\mathrm{\mid }}^2+\frac{\mathrm{\mid }\delta (B\mp e^2(\xi -\mathrm{\mid }q{\mathrm{\mid }}^2)){\mathrm{\mid }}^2}{2e^2})d{x}_{1}d{x}_2\ge 0,$ with the upper signs for $n\ge 0$ and lower signs for $n\le 0$ (either choice when $n=0$). Therefore the first variation vanishes and no smooth compactly supported same-sector perturbation lowers the energy. Distinct divisors give distinct possible minimizing fields, so no cross-divisor uniqueness follows.