Fixed-flux cylinders and uniform end-collar energy

Decompose the integrand into the transverse density $e_{\perp }(x,z)$, containing $F_{12}$, $D_{1}q,D_{2}q$ and the potential, and the nonnegative longitudinal density $e_{\parallel }(x,z)$, containing $F_{31},F_{32},D_{3}q$. Since the total density is nonnegative and integrable, [F4] gives ${\mathcal{E}}_L={\int }_0^{L}E({\mathbf{A}}_{\perp }(\cdot ,z),q_z)dz+{\int }_{{\mathrm{\Omega }}_L}{e}_{\parallel }dxdz$, where ${\mathbf{A}}_{\perp }=A_{1}d{x}_1+A_{2}d{x}_2$ and $E({\mathbf{A}}_{\perp }(\cdot ,z),q_z)={\int }_{{\mathbb{R}}^2}{e}_{\perp }(x,z)dx$ for almost every $z$. The slice hypotheses and [F3] give $E({\mathbf{A}}_{\perp }(\cdot ,z),q_z)\ge T_n$ for each slice, while the longitudinal integral is nonnegative; hence ${\mathcal{E}}_L\ge T_{n}L$.

For finite $R$, apply [F5] to $D_R\times \{z\}$ to identify its magnetic flux with ${\int }_{\mathrm{\partial }{D}_R}{A}_{\perp }(\cdot ,z)$. Apply [F5] to the side $\mathrm{\partial }{D}_R\times [0,L]$: its flux is the difference of the two oriented end-circle integrals. The uniform angular one-form condition makes both integrals $2\pi n+o(1)$, so the side flux tends to zero. Thus the stated angular condition enforces the required no-leakage condition and is consistent with the fixed slice flux.

For an end trace with exterior phase $n\theta +\varphi$, choose a smooth radial cutoff $\eta$ which is zero on a disk containing the noncanonical region and one outside a larger disk. The real function $\chi =-\eta \varphi$ extends smoothly by zero to the whole plane; the allowed gauge transformation $(q,A)\mapsto (e^{i\chi }q,A+d\chi )$ therefore makes the exterior phase exactly $n\theta$. This constructs the asserted degree-zero exterior phase cleanup without attempting to extend a degree-$n$ angular map across the disk. For the planar minimizer in [F2], put $N=\mathrm{\mid }n\mathrm{\mid }$ and use its monic polynomial $P$ of degree $N$. Since $P(z)\mathrm{/}{z}^N=1+O(\mathrm{\mid }z{\mathrm{\mid }}^{-1})$, this quotient has a smooth logarithm tending to zero outside a sufficiently large disk; write $\psi =\mathrm{Im}\log (P\mathrm{/}{z}^N)$. With $\epsilon$ as in [F2], the signed reconstruction has exterior fields $q={\rho }_{\ast }{e}^{i(n\theta +\epsilon \psi )}$ and $A=nd\theta +\epsilon d\psi -\frac{\epsilon }{2}\ast dh$: for $n<0$ both connection terms change sign under conjugation. Applying the cutoff gauge $\chi =-\epsilon \eta \psi$ gives $q^{\ast }={\rho }_{\ast }{e}^{in\theta }$ and $A^{\ast }=nd\theta -\frac{\epsilon }{2}\ast dh$ outside the larger disk. Hence $A^{\ast }-nd\theta$ decays exponentially for either sign; for $n=0$, $P=1$ and $h=\psi =0$.

In canonical exterior gauge write $q=\rho e^{in\theta }$ and $\alpha =A-nd\theta$. Directly, $D_{A}q=e^{in\theta }(d\rho -i\rho \alpha )$, so the real and imaginary terms are orthogonal and $\mathrm{\mid }{D}_{A}q{\mathrm{\mid }}^2=\mathrm{\mid }d\rho {\mathrm{\mid }}^2+{\rho }^2\mathrm{\mid }\alpha {\mathrm{\mid }}^2$. Uniform convergence gives $\rho \ge \sqrt{\xi }\mathrm{/}2$ outside a disk; thus finite kinetic energy implies $\alpha \in L^2$ there. Also $\mathrm{\mid }\rho -\sqrt{\xi }\mathrm{\mid }\le \mathrm{\mid }{\rho }^2-\xi \mathrm{\mid }\mathrm{/}\sqrt{\xi }$, so finite potential energy gives $\rho -\sqrt{\xi }\in L^2$ there. On compact disks all fields are smooth and bounded. Consequently, for any two cleaned end traces and the cleaned minimizer, differences of connections and Higgs fields are globally in $L^2$; each Higgs field is bounded.

For any two such canonical profiles $p_a=(A^a,q^a)$ and $p_b=(A^b,q^b)$, put $A_t=(1-t)A^a+t{A}^b$, $q_t=(1-t)q^a+t{q}^b$, $0\le t\le 1$, and $\delta A=A^b-A^a$, $\delta q=q^b-q^a$. The curvature is $B_t=(1-t)B_a+t{B}_b\in L^2$, so linearity of the flux integral gives flux $2\pi n$ for every blend because both endpoint fluxes equal $2\pi n$. On the exterior, $q_t=e^{in\theta }{\rho }_t$ with ${\rho }_t=(1-t){\rho }_a+t{\rho }_b$; the uniform bounds and ${\rho }_t-\sqrt{\xi }\in L^2$ imply $\mathrm{\mid }{q}_t{\mathrm{\mid }}^2-\xi \in L^2$. Expanding the covariant derivative gives $D_{A_t}{q}_t=(1-t)D_{A^a}{q}^a+t{D}_{A^b}{q}^b+it(1-t)\delta A\delta q\mathrm{.}$ The first two terms are in $L^2$ by finite end energy, and the last is in $L^2$ because $\delta A\in L^2$ and $\delta q$ is bounded. Thus every transverse energy term of the blend is integrable, with a bound uniform for $t\in [0,1]$. Its exterior phase, vacuum limit and angular one-form limit are preserved by convexity and linearity.

Fix $\mathrm{\ell }>0$ and a smooth function $\sigma :[0,1]\to [0,1]$ which is zero near $0$ and one near $1$. On a collar of width $\mathrm{\ell }$, set $t(z)=\sigma (z\mathrm{/}\mathrm{\ell })$, $A_1,A_2=A_{t(z)}$, $q=q_{t(z)}$, and $A_3=0$. Then $D_{3}q=({\sigma }^{\mathrm{\prime }}(z\mathrm{/}\mathrm{\ell })\mathrm{/}\mathrm{\ell })\delta q$ and $F_{3i}=({\sigma }^{\mathrm{\prime }}(z\mathrm{/}\mathrm{\ell })\mathrm{/}\mathrm{\ell })\delta A_i$. Their squared integrals are finite by [F1] and step 1.4. Step 2.1 bounds the transverse density uniformly in $z$, so the complete collar energy is a finite constant depending only on its two profiles, $\mathrm{\ell }$, and $\sigma$. Since $\sigma$ is constant near each end, this collar joins smoothly to constant-in-$z$ profiles. The angular limit is the convex combination of the endpoint limits; its side flux tends to zero by step 1.2.

Apply step 3.1 first to $p_0$ and $p_{\ast }$, defining its finite energy to be $C_0$, and then to $p_{\ast }$ and $p_1$, defining its finite energy to be $C_1$. For $L\ge 2\mathrm{\ell }$, place the first collar on $[0,\mathrm{\ell }]$, the constant planar minimizer on $[\mathrm{\ell },L-\mathrm{\ell }]$, and the second collar on $[L-\mathrm{\ell },L]$. The profiles match smoothly because the collar cutoff is constant near its ends. The middle has no longitudinal energy and contributes exactly $T_n(L-2\mathrm{\ell })$ by [F2]–[F3]. At $L=2\mathrm{\ell }$ this middle interval has zero length; the two collars meet at the constant value $p_{\ast }$, so the same upper bound holds. This admissible concatenation proves $\inf {\mathcal{E}}_L\le C_0+T_n(L-2\mathrm{\ell })+C_1$.

Combining steps 1.1 and 4.1 gives $0\le \inf {\mathcal{E}}_L-T_{n}L\le C_0+C_1-2\mathrm{\ell }{T}_n$ for every $L\ge 2\mathrm{\ell }$; the right side is a finite constant independent of $L$, proving the bounded-remainder conclusion. If $p_0=p_1=p_{\ast }$, the constant extension $(A_1,A_2,q)=(A^{\ast },q^{\ast })$, $A_3=0$, has energy $T_{n}L$ and satisfies the angular and no-leakage conditions. The lower bound from step 1.1 then proves it is an exact translation-invariant minimum.