B,J(\Psi,A)\rangle_{\Omega^{1}_{\mathfrak{g}}}$. For the second term on the
r.h.s. of (B.10), integrating by parts yields $II=\int_{M}\langle
B,d_{A}^{*}F_{A}\rangle_{\Omega^{1}_{\mathfrak{g}}}$. Collecting the last two
equations gives

$\displaystyle\delta_{A}\mathcal{S}(\Psi,A)B=2\int_{M}\langle
B,-J(\Psi,A)+d_{A}^{*}F_{A}\rangle_{\Omega^{1}_{\mathfrak{g}}}.$ (B.14)

For this derivative to be zero for every variation $B$, (B.5) must hold. ∎

#### Conserved energy.

Again, the Gâteaux derivative calculations in the following subsection are
formal. Recall that $M:=\Omega\times[0,T]\subset\mathbb{R}^{d+1}$.

To find the expression for the energy, we use, as in classical mechanics, the
(partial, i.e. without passing to the momentum fields) Legendre transform of
(B.2) is given by

$\displaystyle E(\Psi,A)$
$\displaystyle=\partial_{\nabla_{0}\Psi}\mathcal{L}(\Psi,A)\nabla_{0}\Psi+\partial_{\overline{\nabla_{0}\Psi}}\mathcal{L}(\Psi,A)\overline{\nabla_{0}\Psi}$
$\displaystyle+\sum\limits_{i=1}^{d}\partial_{F_{0i}}\mathcal{L}(\Psi,A)F_{0i}-\mathcal{L}(\Psi,A).$
(B.15)

######  Proposition B.1.

The (partial) Legendre transform (B.15) of Lagrangian (B.2) yields the
conserved energy

$\displaystyle
E(\Psi,A):=\int_{\Omega}\big{(}\|\nabla_{A}\Psi\|_{\Omega^{1}_{V}}^{2}+U(\Psi)+\|F_{A}\|_{\Omega^{2}_{\mathfrak{g}}}^{2}\big{)},$
(B.16)

where the norms are taken using the Euclidean metric on $\mathbb{R}^{d+1}$
(rather than the Minkowski metric).

Note that for static (time-indepent) fields, $E(\Psi,A)=-\mathcal{L}(\Psi,A)$.

###### Proof.

Let $\partial_{\\#}$ denote the partial derivative with respect to the symbol
$\\#$, and recall that $\delta_{\\#}$ denotes the partial (real) Gâteaux
derivative with respect to $\\#$. We calculate

$\displaystyle\partial_{\nabla_{0}\Psi}\mathcal{L}(\Psi,A)\nabla_{0}\Psi=\int_{\Omega}\|\nabla_{0}\Psi\|_{V}^{2}=\partial_{\overline{\nabla_{0}\Psi}}\mathcal{L}(\Psi,A)\overline{\nabla_{0}\Psi}$
(B.17)

and

$\displaystyle\sum\limits_{i=1}^{d}\partial_{F_{0i}}\mathcal{L}(\Psi,A)F_{0i}=\int_{\Omega}2\sum\limits_{i=1}^{d}|F_{0i}|^{2}.$
(B.18)

(B.16) results.

It remains to show that (B.16) is conserved by the YMH equations (B.4) -
(B.5). This can be done by using the (partial) Legendre transform (B.15) as in
classical mechanics, or by a direct computation. We proceed in the second way.
Applying the chain rule gives

$\displaystyle\frac{d}{dt}E(\Psi,A)=\delta_{\Psi}E(\Psi,A)\partial_{0}\Psi+\delta_{\overline{\Psi}}E(\Psi,A)\partial_{0}\overline{\Psi}+\delta_{A}E(\Psi,A)\partial_{0}A,$
(B.19)

where, recall, $\partial_{i}\equiv\partial_{x^{i}}$. We now calculate the
first term using (B.4).

$\displaystyle\delta_{\Psi}E(\Psi,A)\partial_{0}\Psi=\int_{\Omega}\big{(}$
$\displaystyle\langle\nabla_{0}\Psi,\nabla_{0}\partial_{0}\Psi\rangle_{V}+\sum_{k=1}^{d}\langle\nabla_{k}\Psi,\nabla_{k}\partial_{0}\Psi\rangle_{V}$
$\displaystyle+$ $\displaystyle\langle
U^{\prime}(\Psi),\partial_{0}\Psi\rangle_{V}\big{)}.$ (B.20)

Integrating the second term by parts gives

$\displaystyle\delta_{\Psi}E(\Psi,A)\partial_{0}\Psi=\int_{\Omega}\big{(}$
$\displaystyle\langle\nabla_{0}\Psi,\nabla_{0}\partial_{0}\Psi\rangle_{V}+\sum_{k=1}^{d}\langle\nabla_{k}^{*}\nabla_{k}\Psi,\partial_{0}\Psi\rangle_{V}$
$\displaystyle+$ $\displaystyle\langle
U^{\prime}(\Psi),\partial_{0}\Psi\rangle_{V}\big{)}.$ (B.21)

By (B.4), we have

$\displaystyle\nabla_{0}^{*}\nabla_{0}\Psi-\sum_{k=1}^{d}\nabla_{k}^{*}\nabla_{k}\Psi=U^{\prime}(\Psi),$
(B.22)

so (B) becomes

$\displaystyle\delta_{\Psi}E(\Psi,A)\partial_{0}\Psi=\int_{\Omega}\big{(}\langle\nabla_{0}\Psi,\nabla_{0}\partial_{0}\Psi\rangle_{V}+\langle\nabla_{0}^{*}\nabla_{0}\Psi,\partial_{0}\Psi\rangle_{V}\big{)}.$
(B.23)

Here $\nabla_{0}^{*}=-\partial_{0}+A_{0}^{\dagger}=-\partial_{0}-A_{0}$, where
the second equality follows because the representation of $\mathfrak{g}$ is
unitary. Therefore,

$\displaystyle\delta_{\Psi}E(\Psi,A)\partial_{0}\Psi$
$\displaystyle=\int_{\Omega}\big{(}\langle(\partial_{0}+A_{0})\Psi,(\partial_{0}+A_{0})\partial_{0}\Psi\rangle_{V}$
$\displaystyle\quad+\langle(-\partial_{0}-A_{0})(\partial_{0}+A_{0})\Psi,\partial_{0}\Psi\rangle_{V}\big{)}$
$\displaystyle=\int_{\Omega}\partial_{0}\langle\Psi,A_{0}\partial_{0}\Psi\rangle_{V}.$
(B.24)

Similarly,

$\displaystyle\delta_{\overline{\Psi}}E(\Psi,A)\partial_{0}\overline{\Psi}=\int_{\Omega}\partial_{0}\langle
A_{0}\partial_{0}\Psi,\Psi\rangle_{V},$ (B.25)

and so

$\displaystyle\delta_{\Psi}E(\Psi,A)\partial_{0}\Psi+\delta_{\overline{\Psi}}E(\Psi,A)\partial_{0}\overline{\Psi}=\int_{\Omega}\partial_{0}J_{0}(\Psi,A),$
(B.26)

where $J_{0}(\Psi,A)$ is the time component of the YMH current (B.6).

One may show using (B.5) that

$\displaystyle\delta_{A}E(\Psi,A)\partial_{0}A=-\int_{\Omega}\partial_{0}J_{0}(\Psi,A).$
(B.27)

Hence, by (B.19) we have $\frac{d}{dt}E(\Psi,A)=0$, as required. ∎

#### Gauge symmetries.

We define the local action, $\rho_{g}A$171717Compared with the notation of
Appendix A, to simplify the notation we omit the tilde over $\rho_{g}$ in
action of the Lie algebra $\mathfrak{g}$ on $V$. , of the group $G$ on $A$, by
the equation $d_{\rho_{g}A}=gd_{A}g^{-1}$, for all $g\in C^{1}(N,G)$, where
$N$ is either $M$ or $\Omega$. We compute

$\displaystyle\rho_{g}A=gAg^{-1}+gdg^{-1}.$ (B.28)

######  Proposition B.2.

The Lagrangian (B.2) is invariant under the Poincaré group and the gauge
transformations

$\displaystyle T^{gauge}_{g}$
$\displaystyle:(\Psi,A)\mapsto(g\Psi,\rho_{g}A),\qquad\forall g\in
C^{1}(M,G).$ (B.29)

###### Proof.

The invariance under the Poincaré group follows from the definition of this
group and the choice of the Minkowski metric on $M\subset\mathbb{R}^{d+1}$.

For the gauge invariance, recall that $U(\Psi)$ is $\mathfrak{g}$-invariant,
and that the representations $g\mapsto\rho_{g}$ (on $V$) and the adjoint
representation $g\mapsto\text{ad}_{g}$ (on $\mathfrak{g}$) are unitary.
Therefore, to prove invariance under the gauge transformation (B.29), it
suffices to show that

$\displaystyle\nabla_{\rho_{g}A}g\Psi$ $\displaystyle=g\nabla_{A}\Psi,$ (B.30)
$\displaystyle F_{\rho_{g}A}$ $\displaystyle=gF_{A}g^{-1}.$ (B.31)

We shall use the equation

$hdh^{-1}=-dhh^{-1},\quad\forall h\in G$ (B.32)

which follows from $d(hh^{-1})=0$. For (B.30) we compute

$\displaystyle\nabla_{\rho_{g}A}g\Psi$
$\displaystyle=d(g\Psi)+({gAg^{-1}+gdg^{-1}})(g\Psi)$ (B.33)
$\displaystyle=(dg)\Psi+gd\Psi+gA\Psi+g{dg^{-1}g}\Psi.$ (B.34)

Since $g{dg^{-1}g}=-g{g^{-1}dg}=-dg$, this gives
$\nabla_{\rho_{g}A}g\Psi=g\nabla_{A}\Psi$.

For (B.31), computing in coordinates $\\{x^{i}\\}$ and writting
$F_{\rho_{g}A}:=(F_{\rho_{g}A})_{ij}dx^{i}\wedge dx^{j}$ and
$F_{A}:=(F_{A})_{ij}dx^{i}\wedge dx^{j}$, we find

$\displaystyle(F_{\rho_{g}A})_{ij}$
$\displaystyle=\frac{1}{2}[\partial_{i}(gA_{j}g^{-1}+g\partial_{j}g^{-1})-\partial_{j}(gA_{i}g^{-1}+g\partial_{i}g^{-1})]$
$\displaystyle+\frac{1}{2}[gA_{i}g^{-1}+g\partial_{i}g^{-1},gA_{j}g^{-1}+g\partial_{j}g^{-1}],$
(B.35)

where, recall, $\partial_{i}\equiv\partial_{x^{i}}$. Expanding the partial
derivative and commutators gives

$\displaystyle(F_{\rho_{g}A})_{ij}=$
$\displaystyle\frac{1}{2}[\partial_{i}gA_{j}g^{-1}+g\partial_{i}Ag^{-1}+gA_{j}\partial_{i}g^{-1}+\partial_{i}g\partial_{j}g^{-1}+g\partial_{i}\partial_{j}g^{-1}$
$\displaystyle+(gA_{i}g^{-1}+\partial_{i}gg^{-1})(gA_{j}g^{-1}+g\partial_{j}g^{-1})$
$\displaystyle-(i\leftrightarrow j)].$ (B.36)

Expanding the product on the second line gives

$\displaystyle(F_{\rho_{g}A})_{ij}=$
$\displaystyle\frac{1}{2}[\partial_{i}gA_{j}g^{-1}+g\partial_{i}Ag^{-1}+gA_{j}\partial_{i}g^{-1}+\partial_{i}g\partial_{j}g^{-1}+g\partial_{i}\partial_{j}g^{-1}$
$\displaystyle+gA_{i}A_{j}g^{-1}+\partial_{i}gA_{j}g^{-1}+gA_{i}\partial_{j}g^{-1}+\partial_{i}g\partial_{j}g^{-1}$
$\displaystyle-(i\leftrightarrow j)].$ (B.37)

Cancelling terms symmetrical in $i$ and $j$ and simplifying gives

$\displaystyle(F_{\rho_{g}A})_{ij}$
$\displaystyle=g(\frac{1}{2}[\partial_{i}A_{j}-\partial_{j}A_{i}]+\frac{1}{2}[A_{i}A_{j}-A_{j}A_{i}])g^{-1}$
(B.38) $\displaystyle=g(F_{A})_{ij}g^{-1},$ (B.39)

as required. ∎

Specifying (B.16) to the WS model gives (2.6).

#### The YMH equations in coordinate form.

In coordinate form, the differential form (gauge field) entering the YMH
Lagrangian (B.2) is written as $A=A_{i}dx^{i}$. The local coordinate
expression for the curvature is $F_{A}=F_{ij}dx^{i}\wedge dx^{j}$, where
$F_{ij}:=\frac{1}{2}(\partial_{i}A_{j}-\partial_{j}A_{i})+\frac{1}{2}[A_{i},A_{j}]$.
Furthermore, for the covariant derivatives $\nabla_{A}$ and $d_{A}$, we have
$\nabla_{A}\Psi=\nabla_{i}\Psi dx^{i}$ and
$d_{A}^{*}F_{A}=-\nabla^{i}F_{ij}dx^{j}$, where
$\nabla_{i}\Psi:=(\partial_{i}+A_{i})\Psi$ and
$\nabla^{i}F_{ij}:=\partial^{i}F_{ij}+[A^{i},F_{ij}]$.

For an arbitrary $\mathfrak{g}$-valued one-form $B=B_{i}dx^{i}$, we have
$d_{A}B=\nabla_{i}B_{j}dx^{i}\wedge dx^{j}$ and $d_{A}^{*}B=-\nabla^{i}B_{i}$,
where

$\nabla^{i}B_{j}:=\partial^{i}B_{j}+[A^{i},B_{j}].$ (B.40)

We write $F_{ij}=F_{ij}^{a}\gamma_{a}$ for an orthonormal basis $\gamma_{a}$
of $\mathfrak{g}$ and the lower case roman indices run over the spatial
components $1,2,\dots,d$. Note that $F_{ij}=[\nabla_{i},\nabla_{j}]$, but
$F_{ij}\neq\frac{1}{2}(\nabla_{i}A_{j}-\nabla_{j}A_{i})$.

Let $\Omega$ be either a bounded domain in $\mathbb{R}^{d}$ or
$\mathbb{R}^{d+1}$. In the former case, we assume either periodic or Dirichlet
boundary conditions.

######  Proposition B.3.

The Lagrangian and energy for the YMH model are given in coordinates by

$\displaystyle\mathcal{L}(\Psi,A)=\int_{\Omega}\langle\nabla_{k}\Psi,\nabla^{k}\Psi\rangle_{V}-U(\Psi)+\frac{1}{2}F^{a}_{ij}F^{a,ij},$
(B.41) $\displaystyle
E_{\Omega}(\Psi,A)=\int_{\Omega}\langle\nabla_{k}\Psi,\nabla_{k}\Psi\rangle_{V}+U(\Psi)+\frac{1}{2}F^{a}_{ij}F^{a}_{ij}$
(B.42)

(with different ranges of indices as mentioned above). The YMH equations are
given in coordinates by

$\displaystyle-\nabla^{i}\nabla_{i}\Psi=U^{\prime}(\Psi),$ (B.43)
$\displaystyle-\nabla^{i}F_{ij}=\operatorname{Re}\langle{\gamma_{a}}\Psi,\nabla_{j}\Psi\rangle_{V}\gamma_{a}.\
$ (B.44)

###### Proof.

Equations (B.41) and (B.42) follow from the coordinate expressions
$d_{A}\Psi=\nabla_{k}\Psi dx^{k}$ and $F_{A}=F_{ij}^{a}\gamma_{a}\otimes
dx^{i}\wedge dx^{j}$, together with the fact that $dx^{k}$ and
$\gamma_{a}\otimes dx^{i}\wedge dx^{j}$ form orthonormal bases for
$\Omega^{1}$ and $\Omega^{2}_{\mathfrak{g}}$, respectively.

Equations (B.43) - (B.44) follow from equations (B.4) - (B.6) and the
coordinate expressions for $d_{A}$ and $d_{A}^{*}$ above. ∎

##  Appendix C The WS equations in coordinate form

For the gauge group $G=U(2)=SU(2)\times U(1)$, we choose the standard inner
product

$\langle\gamma,\delta\rangle_{\mathfrak{u}(2)}:=2\operatorname{Tr}\gamma^{*}\delta=-2\operatorname{Tr}\gamma\delta$
(C.1)

on $\mathfrak{u}(2)$, for which $-\frac{i}{2}\tau_{a}$, $a=0,1,2,3$, (where
$\tau_{a}$, $a=1,2,3$, are the the Pauli matrices together with
$\tau_{0}:=\mathbf{1}$) form an orthonormal basis. It is customary to factor
out the coefficient of $-\frac{i}{2}$. In coordinates, we write

$\nabla_{Q}\Phi=\nabla_{i}\Phi dx^{i},\quad Q=-\frac{i}{2}Q_{i}dx^{i}\ \text{
and }\ F_{Q}=-\frac{i}{2}Q_{ij}dx^{i}\wedge dx^{j},$ (C.2)

with $Q_{i}(x)$, $Q_{ij}(x)\in i\mathfrak{u}(2)$. Using equation (2.3), we
compute
$Q_{ij}=\frac{1}{2}(\partial_{i}Q_{j}-\partial_{j}Q_{i})-\frac{i}{4}[Q_{i},Q_{j}]$.
Furthermore, we write $Q=V+X$ and

$V=-\frac{i}{2}V_{i}dx^{i}\ \text{ and }\ X=-\frac{i}{2}X_{i}dx^{i},$ (C.3)

with $V_{i}(x)\in i\mathfrak{su}(2)$ and $X_{i}(x)\in i\mathfrak{u}(1)$. Then
$Q_{ij}=V_{ij}+X_{ij}$ and

$\displaystyle\nabla_{i}\Phi$
$\displaystyle:=(\partial_{i}-\frac{ig}{2}V_{i}-\frac{ig^{\prime}}{2}X_{i})\Phi,$
(C.4) $\displaystyle V_{ij}$
$\displaystyle:=\frac{1}{2}(\partial_{i}V_{j}-\partial_{j}V_{i})-\frac{ig}{4}[V_{i},V_{j}],$
(C.5) $\displaystyle X_{ij}$
$\displaystyle:=\frac{1}{2}(\partial_{i}X_{j}-\partial_{j}X_{i}).$ (C.6)

We specify equation (B.41) - (B.44) for to the Weinberg-Salam (WS) model,
which has the gauge group $G=U(2)=SU(2)\times U(1)$. As was mentioned in
Appendix A, in this case, there is a slight discrepancy in the definition of
the covariant derivative due to the fact that $U(2)$ is not simple, but a
(semi-)direct product of the simple group $SU(2)$ and $U(1)$, with each
component having a coupling constant, see (C.2)-(C.6).

Using Eqs (C.2)-(C.6), we express the Lagrangian and the energy in coordinates
as

$\displaystyle\mathcal{L}(\Phi,Q):=\int_{\Omega}\langle\nabla_{i}\Phi,\nabla^{i}\Phi\rangle_{\mathbb{C}^{2}}-U(\Phi)+\frac{1}{2}\operatorname{Tr}Q_{ij}Q^{ij},$
(C.7) $\displaystyle
E(\Phi,Q):=\int_{\Omega}\langle\nabla_{i}\Phi,\nabla_{i}\Phi\rangle_{\mathbb{C}^{2}}+U(\Phi)+\frac{1}{2}\operatorname{Tr}Q_{ij}Q_{ij},$
(C.8)

(with indices ranging from $0$ to $d$ and $1$ to $d$, respectively, as
mentioned above), and the Euler-Lagrange equations are written in coordinates
as

$\displaystyle-$ $\displaystyle\nabla^{i}\nabla_{i}\Phi=U^{\prime}(\Phi),$
(C.9)
$\displaystyle\nabla^{i}Q_{ij}=\frac{1}{2}g\operatorname{Im}\langle\tau_{a}\Phi,\nabla_{j}\Phi\rangle_{\mathbb{C}^{2}}\tau_{a}+\frac{1}{2}g^{\prime}\operatorname{Im}\langle\tau_{0}\Phi,\nabla_{j}\Phi\rangle_{\mathbb{C}^{2}}\tau_{0}.$
(C.10)

Eqs. (C.8) - (C.10) can be expressed in terms of the $W$, $Z$, Higgs and
electromagnetic fields resulting in 2D equations (3) - (3.15), see Appendix
D.2.

##  Appendix D The Weinberg-Salam energy in terms of the fields $W$, $A$, $Z$
and $\varphi$

###  D.1 Dimension $3$

We work in a fixed coordinate system, $\\{x^{i}\\}_{i=1}^{3}$ and write the
fields as $W=W_{i}dx^{i},$ $Z=-\frac{i}{2}Z_{i}dx^{i}$ and
$A=-\frac{i}{2}A_{i}dx^{i}$. We show

######  Proposition D.1.

Energy (2.6), written in terms of the fields $W,A,Z$ and $\varphi$ and
coordinates $\\{x^{i}\\}_{i=1}^{3}$, is given by (see also [40]):

$\displaystyle E_{\Omega}^{WS}$
$\displaystyle(W,A,Z,\varphi):=\int_{\Omega}\big{[}\sum_{ij}(\frac{1}{2}|W_{ij}|^{2}+\frac{1}{4}|Z_{ij}|^{2}+\frac{1}{4}|A_{ij}|^{2})$
$\displaystyle\qquad+\frac{1}{2}g^{2}\varphi^{2}|W|^{2}+\frac{1}{4\cos^{2}\theta}g^{2}\varphi^{2}|Z|^{2}+T(W,A,Z)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad+|\nabla\varphi|^{2}+\frac{1}{2}\lambda(\varphi^{2}-\varphi_{0}^{2})^{2}\big{]},$
(D.1)

where $W_{ij}:=\nabla_{i}W_{j}-\nabla_{j}W_{i}$, with
$\nabla_{k}:=\partial_{k}-igV^{3}_{k},\partial_{k}\equiv\partial_{x^{k}}$,
$Z_{ij}:=\partial_{i}Z_{j}-\partial_{j}Z_{i}$,
$A_{ij}:=\partial_{i}A_{j}-\partial_{j}A_{i}$ and $T(W,A,Z)$ is the sum of
super-quadratic terms,

$\displaystyle T(W,A,Z):=$
$\displaystyle\frac{g^{2}}{2}\sum_{ij}(|W_{i}W_{j}|^{2}-W_{i}^{2}\overline{W}_{j}^{2})-ig\sum_{ij}V_{ij}^{3}W_{i}\overline{W}_{j},$
(D.2)

where $V^{3}:=Z\cos\theta+A\sin\theta$ and
$V^{3}_{ij}:=\partial_{i}V_{j}-\partial_{j}V_{i}$, with the important property
that $T(W,A,Z)$ is invariant under the gauge transformation (3.7).

###### Proof of (D.1).

We proceed by rewriting the terms in the coordinate expression of the WS
energy (C.8), in terms of the fields $W=W_{i}dx^{i}$,
$Z=-\frac{i}{2}Z_{i}dx^{i}$, $A=-\frac{i}{2}A_{i}dx^{i}$ and $\varphi$.

For the first term, first we calculate $\nabla_{i}\Phi$. Recall the definition
$\nabla_{i}\Phi:=(\partial_{i}-\frac{ig}{2}V_{i}-\frac{ig^{\prime}}{2}X_{i})\Phi$.
We simplify the matrix representing the connection’s action on $\Phi$:

$\displaystyle-\frac{ig}{2}V_{i}$
$\displaystyle-\frac{ig^{\prime}}{2}X_{i}=-\frac{ig}{2}V_{i}^{a}\tau_{a}-\frac{ig^{\prime}}{2}X_{i}\tau_{0}$
$\displaystyle=-\frac{ig}{2}\left(\begin{array}[]{cc}0&V^{1}_{i}\\\
V^{1}_{i}&0\end{array}\right)-\frac{ig}{2}\left(\begin{array}[]{cc}0&-iV^{2}_{i}\\\
iV^{2}_{i}&0\end{array}\right)$ (D.7)
$\displaystyle-\frac{ig}{2}\left(\begin{array}[]{cc}V^{3}_{i}&0\\\
0&-V^{3}_{i}\end{array}\right)-\frac{ig}{2}\tan\theta\left(\begin{array}[]{cc}X_{i}&0\\\
0&X_{i}\end{array}\right)$ (D.12)
$\displaystyle=-\frac{ig}{2\cos\theta}\left(\begin{array}[]{cc}V^{3}_{i}\cos\theta+X_{i}\sin\theta&V^{1}_{i}\cos\theta-
iV^{2}_{i}\cos\theta\\\
V^{1}_{i}\cos\theta+iV^{2}_{i}\cos\theta&-V^{3}_{i}\cos\theta+X_{i}\sin\theta\end{array}\right).$
(D.15)

In terms of the fields $Z$, $A$ and $W$ (see equations (3.5) - (3.6) for the
definitions of these fields), (D.15) becomes

$\displaystyle-\frac{ig}{2}V_{i}-\frac{ig^{\prime}}{2}X_{i}$
$\displaystyle=-\frac{ig}{2\cos\theta}\left(\begin{array}[]{cc}Z_{i}\cos
2\theta+A_{i}\sin 2\theta&\sqrt{2}\ W_{i}\cos\theta\\\ \sqrt{2}\
\overline{W}_{i}\cos\theta&-Z_{i}\end{array}\right).$ (D.18)

Hence, for $\Phi=(0,\varphi)$,

$\displaystyle\nabla_{i}\Phi=\left(\begin{array}[]{cc}-\frac{ig}{\sqrt{2}}W_{i}\varphi\\\
\partial_{i}\varphi+\frac{ig}{2\cos\theta}Z_{i}\varphi\end{array}\right).$
(D.21)

Therefore, the first term of (C.8), written in terms of the fields $W,A,Z$ and
$\varphi$, becomes

$\displaystyle\langle\nabla_{i}\Phi,\nabla^{i}\Phi\rangle_{\mathbb{C}^{2}}$
$\displaystyle=\overline{\frac{ig}{\sqrt{2}}W_{i}}\frac{ig}{\sqrt{2}}W^{i}$
$\displaystyle+\overline{(\partial_{i}\varphi+\frac{ig}{2\cos\theta}Z_{i}\varphi)}(\partial^{i}\varphi+\frac{ig}{2\cos\theta}Z^{i}\varphi)$
$\displaystyle=\frac{g^{2}}{2}\varphi^{2}|W|^{2}+|\nabla\varphi|^{2}+\frac{g^{2}}{4\cos^{2}\theta}\varphi^{2}|Z|^{2}.$
(D.22)

The second term of (C.8) becomes

$\displaystyle U(\Phi)$
$\displaystyle=\frac{1}{2}\lambda(\|\Phi\|^{2}-\varphi_{0}^{2})^{2}=\frac{1}{2}\lambda(\varphi^{2}-\varphi_{0}^{2})^{2}.$
(D.23)

For the third term of (C.8), we will use the fact that
$\operatorname{Tr}Q_{ij}Q^{ij}=\operatorname{Tr}V_{ij}V^{ij}+\operatorname{Tr}X_{ij}X^{ij}$,
where $V_{ij}$ and $X_{ij}$ are defined in (C.5) and (C.6). Furthermore, we
have

$\displaystyle
V_{i}:=V_{i}^{a}\tau_{a}=\left(\begin{array}[]{cc}V_{i}^{3}&\sqrt{2}\ W_{i}\\\
\sqrt{2}\ \overline{W}_{i}&-V_{i}^{3}\end{array}\right).$ (D.26)

We recall $V_{ij}^{3}=\partial_{i}V_{j}^{3}-\partial_{j}V_{i}^{3}$ and
$W_{ij}^{0}=\partial_{i}W_{j}-\partial_{j}W_{i}$ and calculate

$\displaystyle\frac{1}{2}(\partial_{i}V_{j}-\partial_{j}V_{i})=\frac{1}{2}\left(\begin{array}[]{cc}V_{ij}^{3}&\sqrt{2}\
W_{ij}\\\ \sqrt{2}\ \overline{W}_{ji}^{0}&-V_{ij}^{3}\end{array}\right),$
(D.29)

and, with $K_{ij}:=V^{3}_{i}W_{i}-V^{3}_{j}W_{i}$,

$\displaystyle-\frac{ig}{4}[V_{i},V_{j}]$
$\displaystyle=-\frac{ig}{4}\left(\begin{array}[]{cc}V_{i}^{3}&\sqrt{2}\
W_{i}\\\ \sqrt{2}\
\overline{W}_{i}&-V^{3}_{i}\end{array}\right)\left(\begin{array}[]{cc}V_{j}^{3}&\sqrt{2}\
W_{j}\\\ \sqrt{2}\
\overline{W}_{j}&-V^{3}_{j}\end{array}\right)-(i\leftrightarrow j)$ (D.34)
$\displaystyle=-\frac{ig}{4}\left(\begin{array}[]{cc}V_{i}^{3}V_{j}^{3}+2W_{i}\overline{W}_{j}&\sqrt{2}\
K_{ij}\\\ \sqrt{2}\
\overline{K}_{ij}&-V_{i}^{3}V_{j}^{3}-2W_{i}\overline{W}_{j}\end{array}\right)-(i\leftrightarrow
j)$ (D.37)
$\displaystyle=-\frac{ig}{2}\left(\begin{array}[]{cc}W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j}&\sqrt{2}\
K_{ij}\\\ \sqrt{2}\
\overline{K}_{ji}&-W_{i}\overline{W}_{j}+\overline{W}_{i}W_{j}\end{array}\right).$
(D.40)

Adding (D.29) and (D.40), using that $W_{ij}=W_{ij}^{0}+K_{ij}$ and denoting
$L_{ij}:=V^{3}_{ij}-ig(W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j})$ gives

$\displaystyle V_{ij}=\frac{1}{2}\left(\begin{array}[]{cc}L_{ij}&\sqrt{2}\
W_{ij}\\\ -\sqrt{2}\ \overline{W}_{ij}&-L_{ij}\end{array}\right).$ (D.43)

Since $V_{ij}$ and $X_{ij}$ are Hermitian, $\operatorname{Tr}V_{ij}V^{ij}$ and
$\operatorname{Tr}X_{ij}X^{ij}$ are the sum of the squared absolute values of
the matrix coefficients of $V_{ij}$ and $X_{ij}$, respectively. Thus

$\displaystyle\frac{1}{2}$
$\displaystyle\operatorname{Tr}Q_{ij}Q^{ij}=\frac{1}{2}\operatorname{Tr}V_{ij}V^{ij}+\frac{1}{2}\operatorname{Tr}X_{ij}X^{ij}$
$\displaystyle=\frac{1}{8}\sum\limits_{ij}2|L_{ij}|^{2}+4|W_{ij}|^{2}+2|X_{ij}|^{2}.$
(D.44)

Using $L_{ij}=V^{3}_{ij}-ig(W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j})$ and
expanding the first term gives

$\displaystyle\frac{1}{2}$
$\displaystyle\operatorname{Tr}Q_{ij}Q^{ij}=\sum\limits_{ij}\frac{1}{2}|W_{ij}|^{2}+\frac{1}{4}|V^{3}_{ij}|^{2}+\frac{1}{4}|X_{ij}|^{2}$
$\displaystyle+\frac{g^{2}}{4}\sum\limits_{ij}|W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j}|^{2}-\frac{ig}{4}\sum\limits_{ij}2V^{3}_{ij}(W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j}).$
(D.45)

Recall that $A_{ij}=V^{3}_{ij}\sin\theta+X_{ij}\cos\theta$ and
$Z_{ij}=V^{3}_{ij}\cos\theta-X_{ij}\sin\theta$. Writing the first line of
(D.45) in terms of these fields gives

$\displaystyle\frac{1}{2}$
$\displaystyle\operatorname{Tr}Q_{ij}Q^{ij}=\sum\limits_{ij}\frac{1}{2}|W_{ij}|^{2}+\frac{1}{4}|Z_{ij}|^{2}+\frac{1}{4}|A_{ij}|^{2}$
$\displaystyle+\frac{g^{2}}{4}\sum\limits_{ij}|W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j}|^{2}-\frac{ig}{2}\sum\limits_{ij}V^{3}_{ij}(W_{i}\overline{W}_{j}-\overline{W}_{i}W_{j}).$
(D.46)

Expanding the first term of the second line, and using
$V_{ij}^{3}=-V^{3}_{ij}$ in the second term, (D.46) becomes

$\displaystyle\frac{1}{2}\operatorname{Tr}Q_{ij}Q^{ij}$
$\displaystyle=\sum\limits_{ij}\frac{1}{2}|W_{ij}|^{2}+\frac{1}{4}|Z_{ij}|^{2}+\frac{1}{4}|A_{ij}|^{2}$
$\displaystyle+\frac{g^{2}}{4}\sum\limits_{ij}(|W_{i}|^{2}|\overline{W}_{j}|^{2}-W_{i}^{2}\overline{W}_{j}^{2}+(i\leftrightarrow
j))$
$\displaystyle-\frac{ig}{2}\sum\limits_{ij}(V^{3}_{ij}W_{i}\overline{W}_{j}+(i\leftrightarrow
j)).$ (D.47)

Recalling the definition (D.2) of $T(W,A,Z)$ gives

$\displaystyle\frac{1}{2}\operatorname{Tr}Q_{ij}Q^{ij}$
$\displaystyle=\sum\limits_{ij}\frac{1}{2}|W_{ij}|^{2}+\frac{1}{4}|A_{ij}|^{2}+\frac{1}{4}|Z_{ij}|^{2}+T(W,A,Z).$
(D.48)

Adding (D.22), (D.23) and (D.48) gives (D.1).∎

###  D.2 Dimension $2$: Proof of (3)

###### Proof of (3).

Now, we consider the Weinberg-Salam (WS) model in $\mathbb{R}^{2}$ with fields
independent of the third dimension $x_{3}$, and correspondingly choose the
gauge with $V_{3}=X_{3}=0$ (and hence $W_{3}=A_{3}=Z_{3}=0$). In this case the
summation in (D.1) contains only two terms, $(ij)=(12)$ and $(ij)=(21)$, and
we use this to simplify (D.1).

We proceed by simplifying the terms of (D.2) and the first line of (D.1); the
remaining terms are unchanged.

$\displaystyle\sum_{ij}(\frac{1}{2}|W_{ij}|^{2}+\frac{1}{4}|Z_{ij}|^{2}+$
$\displaystyle\frac{1}{4}|A_{ij}|^{2})=\sum_{i<j}(|W_{ij}|^{2}+\frac{1}{2}|Z_{ij}|^{2}+\frac{1}{2}|A_{ij}|^{2})$
$\displaystyle=|\operatorname{curl}_{gV^{3}}W|^{2}+\frac{1}{2}|\operatorname{curl}Z|^{2}+\frac{1}{2}|\operatorname{curl}A|^{2};$
(D.49) $\displaystyle\sum_{ij}(|W_{i}W_{j}|^{2}$ $\displaystyle-
W_{i}^{2}\overline{W}_{j}^{2})$
$\displaystyle=W_{1}W_{2}\overline{W}_{1}\overline{W}_{2}-W_{1}^{2}\overline{W}_{2}^{2}+W_{2}W_{1}\overline{W}_{2}\overline{W}_{1}-W_{2}^{2}\overline{W}_{1}^{2}$
$\displaystyle=\overline{(\overline{W}_{1}W_{2}-W_{1}\overline{W}_{2})}(\overline{W}_{1}W_{2}-W_{1}\overline{W}_{2})$
$\displaystyle=|\overline{W}\times W|^{2};$ (D.50)
$\displaystyle-\sum_{ij}V^{3}_{ij}W_{i}\overline{W}_{j}$
$\displaystyle=\sum_{i<j}V^{3}_{ij}(-W_{i}\overline{W}_{j}+W_{j}\overline{W}_{i})$
$\displaystyle=(\operatorname{curl}V^{3})\overline{W}\times W.$ (D.51)

Replacing corresponding terms in (D.1) - (D.2) with (D.49) - (D.51) proves
(3). ∎

###### Proof of (3.12) - (3.15).

We proceed by calculating the (complex) Gâteaux derivatives of (3).

Let $\delta_{\\#}$ denote the partial (real) Gâteaux derivative with respect
to $\\#$. Let $W_{z}=W+zW^{\prime}$, $z\in\mathbb{C}$, and define
$\partial_{\overline{z}}\equiv\frac{1}{2}(\partial_{\operatorname{Re}z}+i\partial_{\operatorname{Im}z})$
and
$\delta_{\overline{W}}\equiv\frac{1}{2}(\delta_{\operatorname{Re}W}+i\delta_{\operatorname{Im}W})$.
Then

$\displaystyle\delta_{\overline{W}}E_{\Omega}^{WS}(W,A,Z,\varphi)\overline{W^{\prime}}$
$\displaystyle=\partial_{\overline{z}}E_{\Omega}^{WS}(W_{z},A,Z,\varphi)|_{z=0}$
$\displaystyle=\int_{\Omega}$
$\displaystyle\operatorname{curl}_{gV^{3}}W\cdot\overline{\operatorname{curl}_{gV^{3}}W^{\prime}}+\frac{1}{2}g^{2}\varphi^{2}W\cdot\overline{W^{\prime}}$
$\displaystyle-
ig(\operatorname{curl}V^{3})JW\cdot\overline{W^{\prime}}+g^{2}(\overline{W}\times
W)JW\cdot\overline{W^{\prime}}.$ (D.52)

Integrating the first term by parts and factoring out $W$ and
$\overline{W}^{\prime}$ gives

$\displaystyle\delta_{\overline{W}}E_{\Omega}^{WS}(W,A,Z,\varphi)\overline{W^{\prime}}=\int_{\Omega}$
$\displaystyle[\operatorname{curl}_{gV^{3}}^{*}\operatorname{curl}_{gV^{3}}+\frac{g^{2}}{2}\varphi^{2}-ig(\operatorname{curl}V^{3})J$
$\displaystyle+g^{2}(\overline{W}\times W)J]W\cdot\overline{W^{\prime}}.$
(D.53)

For the derivative to be zero for every variation $W^{\prime}$, (3.12) must
hold.

Let $A_{s}=A+sA^{\prime}$, $s\in\mathbb{R}$. Then

$\displaystyle\delta_{A}E_{\Omega}^{WS}(W,A,Z,\varphi)A^{\prime}$
$\displaystyle=\partial_{s}E_{\Omega}^{WS}(W,A_{s},Z,\varphi)|_{s=0}$
$\displaystyle=\int_{\Omega}$
$\displaystyle\operatorname{curl}_{gV^{3}}W\overline{(-ieA^{\prime}\times
W)}+\overline{\operatorname{curl}_{gV^{3}}W}(-ieA^{\prime}\times W)$
$\displaystyle+(\operatorname{curl}A)(\operatorname{curl}A^{\prime})+ie(\operatorname{curl}A^{\prime})\overline{W}\times
W.$ (D.54)

Using $A^{\prime}\times W=-JW\cdot A^{\prime}$ in the first two terms, and
integrating the last two terms by parts, gives

$\displaystyle\delta_{A}E_{\Omega}^{WS}(W,A,Z,\varphi)A^{\prime}$
$\displaystyle=\int_{\Omega}[-ie(\operatorname{curl}_{gV^{3}}W)J\overline{W}+ie\overline{(\operatorname{curl}_{gV^{3}}W)J\overline{W}}$
$\displaystyle+\operatorname{curl}^{*}\operatorname{curl}A+ie\operatorname{curl}^{*}(\overline{W}\times
W)]\cdot A^{\prime},$ (D.55)

which simplifies to

$\displaystyle\delta_{A}E_{\Omega}^{WS}(W,A,Z,\varphi)A^{\prime}=\int_{\Omega}$
$\displaystyle[\operatorname{curl}^{*}\operatorname{curl}A+2e\operatorname{Im}[(\operatorname{curl}_{gV^{3}}W)J\overline{W}$
$\displaystyle-\operatorname{curl}^{*}(\overline{W}_{1}W_{2})]]\cdot
A^{\prime}.$ (D.56)

For the derivative to be zero for every variation $A^{\prime}$, (3.13) must
hold.

The proof of (3.14) is essentially the same as the proof of (3.13), so we omit
it.

Let $\varphi_{s}=\varphi+s\varphi^{\prime}$, $s\in\mathbb{R}$. Then

$\displaystyle\delta_{\varphi}E_{\Omega}^{WS}(W,A,Z,\varphi)\varphi^{\prime}$
$\displaystyle=\partial_{s}E_{\Omega}^{WS}(W,A,Z,\varphi_{s})|_{s=0}$
$\displaystyle\int_{\Omega}$ $\displaystyle
g^{2}\varphi\varphi^{\prime}|W|^{2}+\frac{g^{2}}{2\cos^{2}\theta}\varphi\varphi^{\prime}|Z|^{2}$
$\displaystyle+2\nabla\varphi^{\prime}\cdot\nabla\varphi+2\lambda(\varphi^{2}-\varphi_{0}^{2})\varphi\varphi^{\prime}$

Integrating the third term by parts and factoring out $2\varphi^{\prime}$
gives

$\displaystyle=\int_{\Omega}$
$\displaystyle[\frac{g^{2}}{2}|W|^{2}+\frac{1}{2}\kappa|Z|^{2}$
$\displaystyle-\Delta+\lambda(\varphi^{2}-\varphi_{0}^{2})]\varphi\cdot
2\varphi^{\prime}.$ (D.58)

For the derivative to be zero for every variation $\varphi^{\prime}$, (3.15)
must hold. ∎

##  Appendix E Proof of (9.12)

In the proof below, we will use the following result:

######  Lemma E.1.

Let $L^{2}_{per}$ denote any of the spaces (5.8) - (5.10), and let
$\mathcal{H}^{2}_{per}$ denote the corresponding Sobolev space. Suppose that
$f_{s},g_{s}:\mathbb{R}\to\mathcal{H}^{2}_{per}$ satisfy
$||f_{s}||_{\mathcal{H}^{2}_{per}}=\mathcal{O}(|s|^{k})$ and
$||g_{s}||_{\mathcal{H}^{2}_{per}}=\mathcal{O}(|s|^{l})$ for some
$k,l\in\mathbb{Z}$. Then for $i,j=1,2$ and $p,q=0,1$,

$\displaystyle|\int_{\Omega^{\prime}}\partial_{i}^{p}f_{s}\partial_{j}^{q}g_{s}|=\mathcal{O}(|s|^{k+l}).$
(E.1)

Furthermore, if $f_{s}$ and $g_{s}$ have continuous derivatives of all orders
in $s$, then so does the above integral.

###### Proof.

Equation (E.1) follows from the following chain of inequalities:

$\displaystyle|\int_{\Omega^{\prime}}\partial_{i}^{p}f_{s}\partial_{j}^{q}g_{s}|$
$\displaystyle\lesssim||\partial_{i}^{p}f_{s}||_{\mathcal{L}^{2}_{per}}||\partial_{j}^{q}g_{s}||_{\mathcal{L}^{2}_{per}}$
$\displaystyle\lesssim||f_{s}||_{\mathcal{H}^{2}_{per}}||g_{s}||_{\mathcal{H}^{2}_{per}}=\mathcal{O}(|s|^{k+l}).$
(E.2)

If $f_{s}$ and $g_{s}$ have continuous derivatives of all orders in $s$, then
their $s$-derivatives of all orders are in $\mathcal{H}^{2}_{per}$. In
particular, this means that $\partial_{s}^{k}(f_{s}g_{s})$,
$k\in\mathbb{Z}_{\geq 0}$, remains integrable, so the $s-$derivatives of the
above integral (obtained by differentiation under the integral sign) are well-
defined. ∎

###### Proof of (9.12).

To prove (9.12), we use the $w$-field Equation (4.3), and
$\nu_{s}:=g(a_{s}\sin\theta+z_{s}\cos\theta)$, to get

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot[\operatorname{curl}_{\nu_{s}}^{*}\operatorname{curl}_{\nu_{s}}$
$\displaystyle+\frac{g^{2}}{2}(\psi_{s}+\xi_{s})^{2}$
$\displaystyle-i(\operatorname{curl}\nu_{s})J+g^{2}(\overline{w_{s}}\times
w_{s})J]w_{s}=0.$ (E.3)

We shall calculate each term of the integral (E.3) up to order $s^{3}$ using
Lemma E.1 and the Taylor expansions (9.8).

Integrating the first term of (E.3) by parts gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot\operatorname{curl}^{*}_{\nu_{s}}$
$\displaystyle\operatorname{curl}_{\nu_{s}}w_{s}=\int_{\Omega^{\prime}}\overline{\operatorname{curl}_{\nu_{s}}\chi}\cdot\operatorname{curl}_{\nu_{s}}w.$
(E.4)

Plugging in the Taylor expansions (9.8) gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot\operatorname{curl}^{*}_{\nu_{s}}\operatorname{curl}_{\nu_{s}}w_{s}=\int_{\Omega^{\prime}}$
$\displaystyle[\overline{\operatorname{curl}_{a^{n}}\chi}+\mathcal{O}(|s|^{2})]$
$\displaystyle\cdot$
$\displaystyle[s\operatorname{curl}_{a^{n}}\chi-s^{3}i\nu^{\prime}w^{\prime}+\mathcal{O}(|s|^{5})],$
(E.5)

where, recall, $\nu^{\prime}:=g(a^{\prime}\sin\theta+z^{\prime}\cos\theta)$.
Recall from Equation (5.22) that $\operatorname{curl}_{a^{n}}\chi=0$.
Therefore, applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot\operatorname{curl}^{*}_{\nu_{s}}$
$\displaystyle\operatorname{curl}_{\nu_{s}}w_{s}=\mathcal{O}(|s|^{5}).$ (E.6)

Plugging the Taylor expansions (9.8) into the second term of (E.3) gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot\frac{g^{2}}{2}(\psi_{s}+\xi_{s})^{2}w_{s}$
$\displaystyle=\int_{\Omega^{\prime}}\overline{\chi}\cdot\frac{g^{2}}{2}(\frac{\sqrt{2n}}{g}+s^{2}(\psi^{\prime}+\xi^{\prime})+\mathcal{O}(|s|^{4}))^{2}$
$\displaystyle\quad\quad\quad\quad\times(s\chi+\mathcal{O}(|s|^{5})).$ (E.7)

Expanding this product and applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot\frac{g^{2}}{2}(\psi_{s}+\xi_{s})^{2}w_{s}$
$\displaystyle=s\int_{\Omega^{\prime}}n|\chi|^{2}+s^{3}\int_{\Omega^{\prime}}g\sqrt{2n}(\psi^{\prime}+\xi^{\prime})|\chi|^{2}$
$\displaystyle\quad\quad+s^{3}\int_{\Omega^{\prime}}n\overline{\chi}\cdot
w^{\prime}+\mathcal{O}(|s|^{5}).$ (E.8)

Recall that $\chi\in\operatorname{Null}(H_{1}(n))$ and that $w^{\prime}$ is
orthogonal to $\text{Null}\ H_{1}(n)$. Therefore the third term vanishes:

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot\frac{g^{2}}{2}(\psi_{s}+\xi_{s})^{2}w_{s}$
$\displaystyle=s\int_{\Omega^{\prime}}n|\chi|^{2}+s^{3}\int_{\Omega^{\prime}}g\sqrt{2n}(\psi^{\prime}+\xi^{\prime})|\chi|^{2}$
$\displaystyle\quad\quad+\mathcal{O}(|s|^{5}).$ (E.9)

Plugging the Taylor expansions (9.8) into the third term of (E.3) gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(-i(\operatorname{curl}\nu_{s})Jw_{s})$
$\displaystyle=\int_{\Omega^{\prime}}\overline{\chi}\cdot(-in-s^{2}i(\operatorname{curl}\nu^{\prime})+\mathcal{O}(|s|^{4}))$
$\displaystyle\quad\quad\quad\quad\times(sJ\chi+s^{3}Jw^{\prime}+\mathcal{O}(|s|^{5})).$
(E.10)

Recall from Equation (5.22) that $\chi$ is of the form
$\chi=(\omega,i\omega)^{T}$, so
$\overline{\chi}\cdot J\chi=-i|\chi|^{2}$ and $\overline{\chi}\cdot
Jw^{\prime}=-i\overline{\chi}\cdot w^{\prime}$. Therefore (E.10) simplifies to

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(-i(\operatorname{curl}\nu_{s})Jw_{s})$
$\displaystyle=\int_{\Omega^{\prime}}(-in-s^{2}i(\operatorname{curl}\nu^{\prime})+\mathcal{O}(|s|^{4}))$
$\displaystyle\quad\quad\quad\quad\times(-si|\chi|^{2}-s^{3}i\overline{\chi}\cdot
w^{\prime}+\mathcal{O}(|s|^{5})).$ (E.11)

Expanding this product and applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(-i(\operatorname{curl}\nu_{s})Jw_{s})$
$\displaystyle=-s\int_{\Omega^{\prime}}n|\chi|^{2}-s^{3}\int_{\Omega^{\prime}}(\operatorname{curl}\nu^{\prime})|\chi|^{2}$
$\displaystyle\quad\quad-s^{3}\int_{\Omega^{\prime}}n\overline{\chi}\cdot
w^{\prime}+\mathcal{O}(|s|^{5}).$ (E.12)

Recall that $\chi\in\operatorname{Null}(H_{1}(n))$ and that $w^{\prime}$ is
orthogonal to $\text{Null}\ H_{1}(n)$. Therefore the third term vanishes:

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(-i(\operatorname{curl}\nu_{s})Jw_{s})$
$\displaystyle=-s\int_{\Omega^{\prime}}n|\chi|^{2}-s^{3}\int_{\Omega^{\prime}}(\operatorname{curl}\nu^{\prime})|\chi|^{2}$
$\displaystyle\quad\quad+\mathcal{O}(|s|^{5}).$ (E.13)

Using $\overline{\chi}\cdot Jw_{s}=-\overline{\chi}\times w_{s}$, the fourth
term of (E.3) becomes

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(g^{2}\overline{w}_{s}\times
w_{s})Jw_{s}$
$\displaystyle=\int_{\Omega^{\prime}}-g^{2}(\overline{\chi}\times
w_{s})\times(\overline{w}_{s}\times w_{s}).$ (E.14)

Plugging in the Taylor expansions (9.8) gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(g^{2}\overline{w}_{s}\times
w_{s})Jw_{s}$
$\displaystyle=\int_{\Omega^{\prime}}-g^{2}(s\overline{\chi}\times\chi+\mathcal{O}(|s|^{3}))$
$\displaystyle\quad\quad\quad\quad\times(s^{2}\overline{\chi}\times\chi+\mathcal{O}(|s|^{4})).$
(E.15)

Recall from Equation (5.22) that $\chi$ is of the form
$\chi=(\omega,i\omega)$, so $\overline{\chi}\times\chi=i|\chi|^{2}$. This fact
and Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\overline{\chi}\cdot(g^{2}\overline{w}_{s}\times
w_{s})Jw_{s}$
$\displaystyle=s^{3}\int_{\Omega^{\prime}}g^{2}|\chi|^{4}+\mathcal{O}(|s|^{5}).$
(E.16)

The $s^{3}$ terms of (E.6), (E.9), (E.13) and (E.16) must sum to $0$, and so
(9.12) results. ∎

##  Appendix F Proof of Lemma 9.4

###### Proof of Lemma 9.4.

Recall the definitions (9.5) and (9.16) of $\eta$ and $\eta^{\prime}$,
respectively. We proceed by relating the functions
$\alpha_{M_{z},M_{h}}(\tau)$ and $\alpha^{\prime}_{m_{z},m_{h}}(\tau)$
defining $\eta$ and $\eta^{\prime}$.

We define the unitary map $V_{r}:f(x)\rightarrow r^{-1}f(r^{-1}x)$ of
$L^{2}_{\mathcal{L}^{\prime}}$ to $L^{2}_{\mathcal{L}}$ and observe the
relation

$\displaystyle V_{r}G_{m,m^{\prime}}V_{r}^{-1}=r^{-2}G_{m/r,m^{\prime}/r},$
(F.1)

where the $U$-operators on the l.h.s. and r.h.s. act on the spaces
$L^{2}_{\mathcal{L}^{\prime}}$ and $L^{2}_{\mathcal{L}}$, respectively. Using
the relations (F.1) and
$|\chi|^{2}=|V_{r}^{-1}V_{r}\chi|^{2}=rV_{r}^{-1}|X_{r}|^{2}$, where
$X_{r}:=V_{r}\chi$, we find

$\displaystyle\langle|\chi|^{2},G_{m_{z},m_{h}}(|\chi|^{2})\rangle=\langle|X_{r}|^{2},G_{m_{z}/r,m_{h}/r}(|X_{r}|^{2})\rangle.$
(F.2)

Relation (F.2), together with $|\Omega|=r^{2}|\Omega^{\prime}|$ (recall that
$\Omega$ and $\Omega^{\prime}$ are fundamental domains of the original and
rescaled lattices, $\mathcal{L}$ and
$\mathcal{L}^{\prime}:=\frac{1}{r}\mathcal{L}$), gives

$\displaystyle\langle|\chi|^{2}G_{m_{z},m_{h}}(|\chi|^{2})\rangle=r^{2}\langle|X_{r}|^{2}G_{m_{z}/r,m_{h}/r}(|X_{r}|^{2})\rangle,$
(F.3)

where, recall, $X_{r}:=V_{r}\chi$.

We will now estimate the difference of the r.h.s. in (F.3) and the
corresponding unrescaled term:

$\displaystyle R_{G}(\omega)$
$\displaystyle:=\langle|X_{r}|^{2}G_{m_{z}/r,m_{h}/r}(|X_{r}|^{2})\rangle-\langle|X_{r}|^{2}G_{M_{Z},M_{H}}(|X_{r}|^{2})\rangle,$
(F.4)

where the dependence on $\omega=1-\frac{M_{W}^{2}}{eb}$ comes from the
parameter $r=\sqrt{\frac{n}{eb}}$ (which depends on $\omega$ through $b$). By
definition (9.4) and the first resolvent equation, we find

$\displaystyle R_{G}(\omega)$
$\displaystyle=\sum_{1}^{2}\big{(}\langle|X_{r}|^{2}G_{m_{i}/r}(|X_{r}|^{2})\rangle-\langle|X_{r}|^{2}G_{M_{i}}(|X_{r}|^{2})\rangle\big{)}$
$\displaystyle=\sum_{1}^{2}\langle|X_{r}|^{2}G_{m_{i}/r}\big{(}M_{i}^{2}-(m_{i}/r)^{2}\big{)}G_{M_{i}}(|X_{r}|^{2})\rangle,$
(F.5)

where $m_{1}=m_{z},m_{2}=m_{h}$ and $M_{1}=M_{Z},M_{2}=M_{H}$.

Now, we estimate the differences $M_{i}^{2}-(m_{i}/r)^{2},i=1,2,3$, where
$M_{3}=M_{W}$ and $m_{3}=m_{w}$. First, from $r=\xi_{s}/\varphi_{0}$ and the
Taylor expansions (9.8), we have

$\displaystyle m_{p}/r=M_{p}+R_{p}(s^{2}),$ (F.6) $\displaystyle
R_{p}(s^{2})=\mathcal{O}(|s|^{2}),\ p=1,2,3.$ (F.7)

Recall that $\omega=1-\frac{M_{W}^{2}}{eb}$, with
$M_{W}=\frac{1}{\sqrt{2}}g\varphi_{0}$. By (9.22),
$s^{2}=\mathcal{O}(|\omega|)$, so (F.6) - (F.7) become

$\displaystyle m_{p}/r=M_{p}+\tilde{R}_{p}(\omega),$ (F.8)
$\displaystyle\tilde{R}_{p}(\omega)=\mathcal{O}(|\omega|),\ p=1,2,3.$ (F.9)

Relation (F), together with (F.8)-(F.9) and the elementary estimate
$\|G_{m}\|\leq m^{-2}$, yields

$\displaystyle|R_{G}(\omega)|$ $\displaystyle\leq
C\langle|X_{r}|^{2}\rangle^{2}\big{(}|\tilde{R}_{Z}(\omega)|+|\tilde{R}_{H}(\omega)|\big{)}$
$\displaystyle=\mathcal{O}(|\omega|).$ (F.10)

Using estimate (F.10) in (F.4), recalling (F.3), and using
$|\Omega|/|\Omega^{\prime}|=r^{2}$, we obtain:

$\displaystyle\langle|\chi|^{2}G_{m_{z},m_{h}}(|\chi|^{2})\rangle$
$\displaystyle=r^{2}\langle|X_{r}|^{2}G_{M_{Z},M_{H}}(|X_{r}|^{2})\rangle+r^{2}R_{G}(\omega),$
(F.11) $\displaystyle R_{G}(\omega)$ $\displaystyle=\mathcal{O}(|\omega|).$
(F.12)

Recall the definitions (9.3) and (9.17) of $\alpha$ and $\alpha^{\prime}$,
respectively. The relations (F.11) - (F.12) and
$\langle|\chi|^{2}\rangle=r^{2}\langle|X_{r}|^{2}\rangle$ imply

$\displaystyle m_{w}^{2}\alpha^{\prime}_{m_{z},m_{h}}(\tau)$
$\displaystyle=\big{(}\frac{m_{w}}{r}\big{)}^{2}\alpha_{M_{Z},M_{H}}(\tau)+\frac{1}{r^{2}\langle|X_{r}|^{2}\rangle^{2}}R_{G}(\omega).$
(F.13)

Using Eqs (F.8) - (F.9), we obtain

$\displaystyle m_{w}^{2}\alpha^{\prime}_{m_{z},m_{h}}(\tau)$
$\displaystyle=M_{W}^{2}\alpha_{M_{Z},M_{H}}(\tau)+R_{\alpha}(\omega),$ (F.14)
$\displaystyle R_{\alpha}(\omega)$ $\displaystyle=\mathcal{O}(|\omega|).$
(F.15)

Plugging (F.14) - (F.15) into (9.16) gives (9.23) - (9.24).

It remains to show that $R_{\alpha}$ has continuous derivatives of all orders
in $\omega$. We proceed by first showing the remainder terms $R_{p}$,
$\tilde{R}_{p},\ p=1,2,3,$ and $R_{G}$ have continuous derivatives of all
orders. The continuous differentiability of $R_{p}$ follows from the
continuous differentiability of $\xi_{s}$ of all orders in $s$. The continuous
differentiability of $\tilde{R}_{p}$ follows from the continuous
differentiability of $R_{p}$ and the continuous differentiability of $s^{2}$
as a function of $\omega$ in all orders (see Eq. (9.22) and the surrounding
discussion). The continuous differentiability of $R_{G}$ follows from the
differentiability of the right-hand side of (F.12) in $r$ (see the comment
after (F.12)). The continuous differentiability of $R_{\alpha}$ follows from
(F.8) - (F.9), (F.13) and the continuous differentiability of $\tilde{R}_{p}$
and $R_{G}$. The continuous differentiability of $R_{\eta}$ follows from the
continuous differentiability of $R_{\alpha}$ and the definition (9.16) of
$\eta^{\prime}$ (since $x\mapsto[x+\sin^{2}\theta]^{-1}$ is analytic in $x$
for $x>0$). ∎

##  Appendix G Proof of (10)

###### Proof of (10).

We shall calculate each term in the integral (4) up to order $s^{6}$ using
Lemma E.1 and the Taylor expansions (9.8).

Plugging the Taylor expansions (9.8) into the first term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}|\operatorname{curl}_{\nu}w_{s}|^{2}$
$\displaystyle=\int_{\Omega^{\prime}}|s\operatorname{curl}_{a^{n}}\chi+\mathcal{O}(|s|^{3})|^{2}.$
(G.1)

Recall from Equation (5.22) that $\operatorname{curl}_{a^{n}}\chi=0$.
Therefore, applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}|\operatorname{curl}_{\nu}w_{s}|^{2}$
$\displaystyle=\mathcal{O}(|s|^{6}).$ (G.2)

Plugging the Taylor expansions (9.8) into the second term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}z_{s}|^{2}$
$\displaystyle=\int_{\Omega^{\prime}}\frac{1}{2}|s^{2}\operatorname{curl}z^{\prime}+\mathcal{O}(|s|^{4})|^{2}.$
(G.3)

Expanding the square and applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}z_{s}|^{2}$
$\displaystyle=s^{4}\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}z^{\prime}|^{2}+\mathcal{O}(|s|^{6}).$
(G.4)

Plugging the Taylor expansions (9.8) into the third term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}a_{s}|^{2}=\int_{\Omega^{\prime}}$
$\displaystyle\frac{1}{2}|\operatorname{curl}\frac{1}{e}a^{n}+s^{2}\operatorname{curl}a^{\prime}+s^{4}\operatorname{curl}a^{\prime\prime}+\mathcal{O}(|s|^{6})|^{2}.$
(G.5)

Recall that $\operatorname{curl}a^{n}=n$. Expanding the square gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}a_{s}|^{2}=\int_{\Omega^{\prime}}[\frac{1}{2}\frac{n^{2}}{e^{2}}+s^{2}\frac{n}{e}\operatorname{curl}a^{\prime}$
$\displaystyle+s^{4}\frac{n}{e}\operatorname{curl}a^{\prime\prime}$
$\displaystyle+s^{4}\frac{1}{2}|\operatorname{curl}a^{\prime}|^{2}+\mathcal{O}(|s|^{6})].$
(G.6)

The second and third terms vanish because $a^{\prime}$ and $a^{\prime\prime}$
are $\mathcal{L}^{\prime}$-periodic. Therefore, applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}a_{s}|^{2}$
$\displaystyle=\frac{1}{2}\frac{n^{2}}{e^{2}}|\Omega^{\prime}|+s^{4}\int_{\Omega^{\prime}}\frac{1}{2}|\operatorname{curl}a^{\prime}|^{2}+\mathcal{O}(|s|^{6}).$
(G.7)

Plugging the Taylor expansions (9.8) into the fourth term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}g^{2}\phi_{s}^{2}|w_{s}|^{2}=\int_{\Omega^{\prime}}$
$\displaystyle\frac{1}{2}g^{2}[\frac{\sqrt{2n}}{g}+s^{2}(\xi^{\prime}+\psi^{\prime})+\mathcal{O}(|s|^{4})]^{2}$
$\displaystyle\qquad\qquad\qquad\times|s\chi+s^{3}w^{\prime}+\mathcal{O}(|s|^{6})|^{2}.$
(G.8)

Expanding the square terms gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}g^{2}\phi_{s}^{2}|w_{s}|^{2}=\int_{\Omega^{\prime}}$
$\displaystyle\frac{1}{2}g^{2}[\frac{2n}{g^{2}}+s^{2}2\frac{\sqrt{2n}}{g}(\xi^{\prime}+\psi^{\prime})+\mathcal{O}(|s|^{4})]$
$\displaystyle\qquad\qquad\times[s^{2}|\chi|^{2}+s^{4}2\operatorname{Re}(\overline{\chi}\cdot
w^{\prime})+\mathcal{O}(|s|^{6})].$ (G.9)

Expanding this product and applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}$ $\displaystyle
g^{2}\phi_{s}^{2}|w_{s}|^{2}=s^{2}\int_{\Omega^{\prime}}n|\chi|^{2}$
$\displaystyle\qquad\qquad+s^{4}\int_{\Omega^{\prime}}[g\sqrt{2n}(\xi^{\prime}+\psi^{\prime})|\chi|^{2}+2n\operatorname{Re}(\overline{\chi}\cdot
w^{\prime})]+\mathcal{O}(|s|^{6}).$ (G.10)

Recall that $\chi\in\operatorname{Null}(H_{1}(n))$ and that $w^{\prime}$ is
orthogonal to $\operatorname{Null}(H_{1}(n))$. Therefore the third term
vanishes:

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{2}g^{2}\phi_{s}^{2}|w_{s}|^{2}=s^{2}$
$\displaystyle\int_{\Omega^{\prime}}n|\chi|^{2}+s^{4}\int_{\Omega^{\prime}}g\sqrt{2n}(\xi^{\prime}+\psi)|\chi|^{2}+\mathcal{O}(|s|^{6}).$
(G.11)

Plugging the Taylor expansions (9.8) into the fifth term of (4) and expanding
the square terms gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{4\cos^{2}\theta}$ $\displaystyle
g^{2}\phi_{s}^{2}|z_{s}|^{2}=\int_{\Omega^{\prime}}\frac{1}{4\cos^{2}\theta}g^{2}$
$\displaystyle\times[\frac{2n}{g^{2}}+s^{2}2\frac{\sqrt{2n}}{g}(\xi^{\prime}+\psi^{\prime})+\mathcal{O}(|s|^{4})][s^{4}|z^{\prime}|^{2}+\mathcal{O}(|s|^{6})].$
(G.12)

Expanding this product and applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}\frac{1}{4\cos^{2}\theta}g^{2}\phi_{s}^{2}|z_{s}|^{2}$
$\displaystyle=s^{4}\int_{\Omega^{\prime}}\frac{n}{2\cos^{2}\theta}|z^{\prime}|^{2}+\mathcal{O}(|s|^{6}).$
(G.13)

Plugging the Taylor expansions (9.8) into the sixth term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}$ $\displaystyle|\overline{w}_{s}\times
w_{s}|^{2}=\int_{\Omega^{\prime}}|s^{2}\overline{\chi}\times\chi+\mathcal{O}(|s|^{4})|^{2},$
(G.14)

Recall from Equation (5.22) that $\chi$ is of the form
$\chi=(\omega,i\omega)$, so $\overline{\chi}\times\chi=i|\chi|^{2}$.
Therefore, applying Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}$ $\displaystyle|\overline{w}_{s}\times
w_{s}|^{2}=s^{4}\int_{\Omega^{\prime}}|\chi|^{4}+\mathcal{O}(|s|^{6}).$ (G.15)

Plugging the Taylor expansions (9.8) into the seventh term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}i(\operatorname{curl}\nu_{s})\overline{w}_{s}\times$
$\displaystyle
w_{s}=\int_{\Omega^{\prime}}i[g\sin\theta\operatorname{curl}\frac{1}{e}a^{n}+s^{2}\operatorname{curl}\nu^{\prime}+\mathcal{O}(|s|^{4})]$
$\displaystyle\times[s\overline{\chi}+s^{3}\overline{w^{\prime}}+\mathcal{O}(|s|^{5})]\times[s\chi+s^{3}w^{\prime}+\mathcal{O}(|s|^{5})].$
(G.16)

where, recall, $\nu^{\prime}:=g(a^{\prime}\sin\theta+z^{\prime}\cos\theta)$.
Recall that $\operatorname{curl}a^{n}=n$ and $e=g\sin\theta$. Expanding the
wedge product of the second and third terms gives

$\displaystyle\int_{\Omega^{\prime}}i(\operatorname{curl}\nu_{s})\overline{w}_{s}\times
w_{s}$
$\displaystyle=\int_{\Omega^{\prime}}i[\frac{n}{g}^{2}+s^{2}\operatorname{curl}\nu^{\prime}+\mathcal{O}(|s|^{4})]$
$\displaystyle\times[s^{2}\overline{\chi}\times\chi+s^{4}(\overline{\chi}\times
w^{\prime}+\overline{w^{\prime}}\times\chi)+\mathcal{O}(|s|^{6})].$ (G.17)

Recall from Equation (5.22) that $\chi$ is of the form
$\chi=(\omega,i\omega)$, so $\overline{\chi}\times\chi=i|\chi|^{2}$ and
$\overline{\chi}\times w^{\prime}=i\overline{\chi}\cdot w^{\prime}$. Therefore

$\displaystyle\int_{\Omega^{\prime}}i(\operatorname{curl}\nu_{s})\overline{w}_{s}\times
w_{s}=\int_{\Omega^{\prime}}$
$\displaystyle[in+s^{2}i\operatorname{curl}\nu^{\prime}+\mathcal{O}(|s|^{4})]$
$\displaystyle\times[s^{2}i|\chi|^{2}+s^{4}2\operatorname{Re}(i\overline{\chi}\cdot
w^{\prime})+\mathcal{O}(|s|^{6})].$ (G.18)

Expanding this product and using Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}i(\operatorname{curl}\nu_{s})\overline{w}_{s}\times
w_{s}=-s^{2}$
$\displaystyle\int_{\Omega^{\prime}}n|\chi|^{2}-s^{4}\int_{\Omega^{\prime}}[2in\operatorname{Im}(\overline{\chi}\cdot
w^{\prime})$
$\displaystyle\qquad-s^{4}\int_{\Omega^{\prime}}(\operatorname{curl}\nu^{\prime})|\chi|^{2}+\mathcal{O}(|s|^{6}).$
(G.19)

Recall that $\chi\in\operatorname{Null}(H_{1}(n))$ and $w^{\prime}$ is
orthogonal to $\operatorname{Null}(H_{1}(n))$. Therefore the second term
vanishes:

$\displaystyle\int_{\Omega^{\prime}}i(\operatorname{curl}\nu_{s})\overline{w}_{s}\times
w_{s}=$
$\displaystyle-s^{2}\int_{\Omega^{\prime}}n|\chi|^{2}-s^{4}\int_{\Omega^{\prime}}(\operatorname{curl}\nu^{\prime})|\chi|^{2}$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad+\mathcal{O}(|s|^{6}).$
(G.20)

Plugging the Taylor expansions (9.8) into the eigth term of (4) gives

$\displaystyle\int_{\Omega^{\prime}}$
$\displaystyle|\nabla\phi_{s}|^{2}=\int_{\Omega^{\prime}}|s^{2}\nabla\psi^{\prime}+\mathcal{O}(|s|^{4})|^{2}.$
(G.21)

Expanding the square and using Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}|\nabla\phi_{s}|^{2}=s^{4}\int_{\Omega^{\prime}}|\nabla\psi^{\prime}|^{2}+\mathcal{O}(|s|^{6}).$
(G.22)

Plugging the Taylor expansions (9.8) into the ninth term of (4) and expanding
the inner squares gives

$\displaystyle\int_{\Omega^{\prime}}$
$\displaystyle\frac{1}{2}\lambda(\phi_{s}^{2}-\xi_{s}^{2})$
$\displaystyle=\int_{\Omega^{\prime}}\frac{1}{2}\lambda[\frac{2n}{g^{2}}+s^{2}2\frac{\sqrt{2n}}{g}(\xi^{\prime}+\psi^{\prime})-\frac{2n}{g^{2}}-s^{2}2\frac{\sqrt{2n}}{g}\xi^{\prime}+\mathcal{O}(|s|^{4})]^{2}$
$\displaystyle=\int_{\Omega^{\prime}}\frac{1}{2}\lambda[s^{2}2\frac{\sqrt{2n}}{g}\psi^{\prime}+\mathcal{O}(|s|^{4})]^{2}.$
(G.23)

Expanding the outer square gives and using Lemma E.1 gives

$\displaystyle\int_{\Omega^{\prime}}$
$\displaystyle\frac{1}{2}\lambda(\phi_{s}^{2}-\xi_{s}^{2})=s^{4}\int_{\Omega^{\prime}}\frac{4\lambda
n}{g^{2}}\psi^{\prime 2}+\mathcal{O}(|s|^{6}).$ (G.24)

Adding (G.2) - (G.24) and dividing by $|\Omega^{\prime}|$ gives (10), where
$R_{\varepsilon}$ collects the $\mathcal{O}(|s|^{6})$ remainder terms.
$R_{\varepsilon}$ has continuous derivatives of all orders because it is a sum
of integrals of the form (E.1) with $f_{s}$ and $g_{s}$ coming from the
continuously differentiable remainder terms $\mathcal{O}(|s|^{p})$ of (9.8). ∎

##  Appendix H Spectral analysis of the operator $-\Delta_{a^{n}}$

Recall from the main text, but in vector notation, that
$a^{n}:=\frac{n}{2}x^{\perp}$, where $(x^{1},x^{2})^{\perp}=(-x^{2},x^{1})$,
$\nabla_{q}:=\nabla-iq=(\nabla_{1},\nabla_{2})$,
$\nabla_{j}:=\partial_{j}-iq_{j},\partial_{j}\equiv\partial_{x^{j}}$, and
$\Delta_{q}:=\nabla_{q}^{2}=-\nabla_{q}^{*}\nabla_{q}$. The next proof follows
Section $5$ of [18].

###### Proof of Proposition 5.4.

The self-adjointness of the operator $-\Delta_{a^{n}}$ is well-known. To find
its spectrum, we introduce the complexified covariant derivatives (harmonic
oscillator annihilation and creation operators), $\bar{\partial}_{a^{n}}$ and
$\bar{\partial}_{a^{n}}^{*}=-\partial_{a^{n}}$, with

$\bar{\partial}_{a^{n}}:=(\nabla_{a^{n}})_{1}+i(\nabla_{a^{n}})_{2}=\partial_{x^{1}}+i\partial_{x^{2}}+\frac{1}{2}n(x^{1}+ix^{2}).$
(H.1)

One can redily verify that these operators satisfy the following relations:

$\displaystyle[\bar{\partial}_{a^{n}},(\bar{\partial}_{a^{n}})^{*}]$
$\displaystyle=\operatorname{curl}a^{n}=n;$ (H.2)
$\displaystyle-\Delta_{a^{n}}-n$
$\displaystyle=(\bar{\partial}_{a^{n}})^{*}\bar{\partial}_{a^{n}}.$ (H.3)

As for the harmonic oscillator (see e.g. [22]), this gives explicit
information about the spectrum of $-\Delta_{a^{n}}$, namely (5.24), with each
eigenvalue is of the same multiplicity. Furthermore, the above properties
imply (5.25).

We find $\operatorname{Null}\bar{\partial}_{a^{n}}$. A simple calculation
gives the following operator equation

$e^{-\frac{n}{2}(ix^{1}x^{2}-(x^{2})^{2})}\bar{\partial}_{a^{n}}e^{\frac{n}{2}(ix^{1}x^{2}-(x^{2})^{2})}=\partial_{x^{1}}+i\partial_{x^{2}}.$

(The transformation on the left-hand side is highly non-unique.) This
immediately proves that

$\bar{\partial}_{a^{n}}\psi=0,$ (H.4)

if and only if $\theta=e^{-\frac{n}{2}(ix^{1}x^{2}-(x^{2})^{2})}\psi$
satisfies $(\partial_{x^{1}}+i\partial_{x^{2}})\theta=0$. We now identify
$x\in\mathbb{R}^{2}$ with $z=x^{1}+ix^{2}\in\mathbb{C}$ and see that this
means that $\theta$ is analytic and

$\psi\left(x\right)=e^{-\frac{\pi
n}{2\operatorname{Im}\tau}(|z|^{2}-z^{2})}\theta(z,\tau),\
z=(x^{1}+ix^{2})/\sqrt{\frac{2\pi}{\operatorname{Im}\tau}}.$ (H.5)

where we display the dependence of $\theta$ on $\tau$. The quasiperiodicity of
$\psi$ transfers to $\theta$ as follows:

$\theta(z+1,\tau)=\theta(z,\tau),\qquad\theta(z+\tau,\tau)=e^{-2\pi
inz}e^{-in\pi\tau}\theta(z,\tau).$

The first relation ensures that $\theta$ have a absolutely convergent Fourier
expansion of the form $\theta(z,\tau)=\sum_{m=-\infty}^{\infty}c_{m}e^{2\pi
miz}.$ The second relation, on the other hand, leads to relation for the
coefficients of the expansion: $c_{m+n}=e^{-in\pi z}e^{i2m\pi\tau}c_{m}$,
which together with the previous statement implies (5.27). ∎

Next, we claim that the solution (H.5) satisfies

$\displaystyle\psi(x)=\psi(-x).$ (H.6)

By (H.5), it suffices to show that $\theta(z)=\theta(-z)$. We show this for
$n=1$. Denote the corresponding $\theta$ by $\theta(z,\tau)$. Iterating the
recursive relation for the coefficients in (5.27), we obtain the following
standard representation for the theta function

$\displaystyle\theta(z,\tau)=\sum_{m=-\infty}^{\infty}e^{2\pi
i(\frac{1}{2}m^{2}\tau+mz)}.$ (H.7)

We observe that $\theta(-z,\tau)=\theta(z,\tau)$ and therefore
$\psi_{0}(-x)=\psi_{0}(x)$. Indeed, using the expression (H.7), we find, after
changing $m$ to $-m^{\prime}$, we find

$\theta(-z,\tau)=\sum_{m=-\infty}^{\infty}e^{2\pi i(\frac{1}{2}m^{2}\tau-
mz)}=\sum_{m^{\prime}=-\infty}^{\infty}e^{2\pi i(\frac{1}{2}m^{\prime
2}\tau+m^{\prime}z)}=\theta(z,\tau).$ (H.8)

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