> To appear in Advances in Mathamtics
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# Vortex lattices and the Bogoliubov-de Gennes Equations

Ilias (Li) Chenn and I. M. Sigal

(Date: April 8, 2020)

###### Abstract.

We consider the Bogoliubov-de Gennes equations giving an equivalent
formulation of the BCS theory of superconductivity. We are interested in
static solutions with the magnetic field present. We carefully formulate the
equations in the basis independent form, discuss their general features and
isolate key physical classes of solutions (normal and vortex lattice states)
which are the candidates for the ground state. We prove existence of the
normal and vortex lattice states and stability of the normal states for large
temperature or magnetic fields and their instability for small temperature and
small magnetic fields.

Mathematics Subject Classification (MSC): 81Q80, 81V70, 81V74, 35Q40, 35Q56,
35Q82

##  1\. Introduction

###  1.1. Background

The Bogoliubov-de Gennes (BdG) equations describe the remarkable quantum
phenomenon of superconductivity.111For some physics background, see books [12,
27] and the review papers [11, 23]. They present an equivalent formulation of
the BCS theory of superconductivity and are among the latest additions to the
family of important effective equations of mathematical physics. Together with
the Hartree, Hartree-Fock (-Bogoliubov), Ginzburg-Landau, Gross-Pitaevskii and
Landau-Lifshitz equations, they are the quantum members of this illustrious
family consisting of such luminaries as the heat, Euler, Navier-Stokes,
Boltzmann and Vlasov equations.

The BdG equations describe the evolution of superconductors on nanoscopic and
macroscopic scales. There are still many fundamental questions about these
equations which are completely open, namely

  * •

Derivation;

  * •

Well-posedness;

  * •

Existence and stability of stationary magnetic solutions.

In this paper we address the third problem.222A formal derivation of the
(time-dependent) BdG equations is discussed in Appendix E. For a different but
related formal derivation (going back to Dirac and Frenkel) see [6, 26].
Presently, there are no rigorous derivations. For a recent book on rigorous
derivations of the simpler Hartree-Fock equation see [5], for some more recent
papers [29, 30] and for the related bosonic Hartree-Fock-Bogoliubov equations
[15, 16, 28]. For the literature before 2016, we refer only to the papers [21,
32], which led to the most of the recent developments and to the paper [14],
dealing with the relation between the mean-field limit and deformation
quantization. Following appearance of the first version of this paper on
arXiv, the existence problem in the absence of magnetic field was taken up in
[6] (cf. [3] for the global existence for the related bosonic Hartree-Fock-
Bogoliubov equations). By the magnetic solutions we mean solutions with non-
zero magnetic fields.

The key special solutions of the Bogoliubov-de Gennes (BdG) equations are
normal, superconducting and mixed or intermediate states. The superconducting
(or Meissner) states assume, by the definition, that the magnetic field is
zero, while mixed ones, to have non-vanishing magnetic fields. For type II
superconductors, according to experiments, the latter are (magnetic) vortex
lattices. In this paper, we prove the existence of the normal states for non-
vanishing magnetic fields and partial results on their stability and the
existence of the vortex lattices.

There is a considerable physics literature devoted to the BdG equations, but
despite the role played by magnetic phenomena in superconductivity it deals
mainly with the zero magnetic field case, with only few disjoint remarks about
the case when the magnetic fields are present, the main subject of this
work.333The Ginzburg-Landau equations give a good account of magnetic
phenomena in superconductors but only for temperatures sufficiently close to
the critical one (see e.g. [11, 12, 33]).

As for rigorous work, it also deals exclusively with the case of zero magnetic
field. The general (variational) set-up for the static BdG equations is given
in [4]. The next seminal works on the subject are [19], where the authors
prove the existence of superconducting states (the existence of the normal
states under the assumptions of [19] is trivial), to which our work is
closest, and [13], deriving the (macroscopic) Ginzburg-Landau equations. For
an excellent, recent review of the subject, with extensive references and
discussion see [20].

In the rest of this section we introduce the BdG equations, describe their
properties and the main issues and present the main results of this paper. In
the remaining sections we prove these results, with technical derivations
delegated to appendices. In the last appendix, following [3], we discuss a
formal, but natural, derivation of the BdG equations.

###  1.2. Bogoliubov-de Gennes (BdG) equations

In the Bogoliubov-de Gennes approach states of superconductors are described
by the pair of bounded operators $\gamma$ and $\alpha$, acting on the one-
particle complex Hilbert space, $\mathfrak{h}$, with a complex conjugation (an
anti-linear involution), and satisfying

$0\leq\gamma=\gamma^{*}\leq 1,\ \quad\alpha^{*}=\overline{\alpha}\ \quad{\rm\
and\ }\ \quad\alpha\alpha^{*}\leq\gamma(1-\gamma)$ (1.1)

where $\overline{\gamma}:=\mathcal{C}\gamma\mathcal{C}$, with $\mathcal{C}$,
the operation of complex conjugation (see Appendix E for the origin of these
operators). $\gamma$ is a one-particle density operator and $\alpha$ is a two-
particle coherence operator, or diagonal and off-diagonal correlations.
$\gamma(x,x)$ is interpreted as the one-particle density.

The one-particle space $\mathfrak{h}$ is determined by the many-body quantum
problem. For zero density (or ‘finite’) systems, it is $L^{2}(\mathbb{R}^{d})$
and for positive density ones, $L^{2}(\Omega)$, where $\Omega$ is an arbitrary
fundamental cell in a lattice $\mathcal{L}\subset\mathbb{R}^{d}$, with
magnetic field dependent (twisted) boundary conditions (see Subsection 1.6 for
more details). To fix ideas, we take $\mathfrak{h}$ to be the latter,
specifically,

$\displaystyle\mathfrak{h}:=$ $\displaystyle\\{f\in
L^{2}_{loc}(\mathbb{R}^{2}):u_{s}^{\mathcal{L}}f=f\text{ for all
}s\in\mathcal{L}\\},$

where $u_{s}^{\mathcal{L}}$ are the magnetic translations given in (1.29)
below, with the scalar product given by $L^{2}(\Omega)$ for an arbitrary
fundamental cell $\Omega$ of $\mathcal{L}$. Furthermore, we understand $\int$
without specifying the domain of integration as taken over $\Omega$.

The BdG equations form a system of coupled, nonlinear equations for $\gamma$
and $\alpha$. It is convenient to organize the operators $\gamma$ and $\alpha$
into the self-adjoint operator-matrix

$\displaystyle\eta:=\left(\begin{array}[]{cc}\gamma&\alpha\\\
\alpha^{*}&\mathbf{1}-\bar{\gamma}\end{array}\right).$ (1.4)

The definition of $\gamma$ and $\alpha$ in terms of the many-body theory (see
(E.2) of Appendix E) implies that

$\displaystyle 0\leq\eta=\eta^{*}\leq 1\ {\rm\ and\ }\ J^{*}\eta
J=\mathbf{1}-\bar{\eta},\ J:=\left(\begin{array}[]{cc}0&\mathbf{1}\\\
-\mathbf{1}&0\end{array}\right).$ (1.7)

These relations imply relations (1.1).

Since the BdG equations describe the phenomenon of superconductivity, they are
naturally coupled to the electromagnetic field. We describe the latter by the
vector potential $a$ in the gauge in which the electrostatic potential is zero
so that the equations have a slightly simpler form). In what follows, for an
operator $A$, we denote by $A(x,y)$ its integral kernel. Then the time-
dependent BdG equations state (see e.g. [12, 11, 27])

$\displaystyle i\partial_{t}\eta=[\Lambda(\eta,a),\eta],$ (1.8)
$\displaystyle\Lambda(\eta,a)=\big{(}\begin{smallmatrix}h_{\gamma
a}&v^{\sharp}\alpha\\\ v^{\sharp}\bar{\alpha}&-\overline{h_{\gamma
a}}\end{smallmatrix}\big{)},\ \quad h_{\gamma
a}=-\Delta_{a}+v^{*}\gamma-v^{\sharp}\,\gamma\,,$ (1.9)

where, for the pair potential $v(x,y)$, the operator $v^{\sharp}$ (acting on
operators on $\mathfrak{h}$) is defined through the integral kernels as
$(v^{\sharp}\,\alpha)\,(x;y):=v(x,y)\alpha(x;y)$ and $(v^{*}\gamma)(x):=\int
v(x,y)\rho_{\gamma}(y)dy,$ with $\rho_{\gamma}(x):=\gamma(x;x)$. $v^{*}\gamma$
and $v^{\sharp}\,\gamma$ are the direct and exchange self-interaction
potentials, and $\Delta_{a}:=|\nabla_{a}|^{2},\nabla_{a}:=\nabla-ia$. (1.8) is
coupled to the Maxwell equation (Ampère’s law)

$\displaystyle\partial_{t}^{2}a$
$\displaystyle=-\operatorname{curl}^{*}\operatorname{curl}a+j(\eta,a),$ (1.10)

where $j(\eta,a)(x)\equiv j(\gamma,a)(x):=[-i\nabla_{a},\gamma]_{+}(x,x)$ is
the superconducting current. Here $[A,B]_{+}$ is the anti-commutator of $A$
and $B$, i.e. $[A,B]_{+}:=AB+BA$. (To see that (1.10) is Ampère’s law, one
recalls that in our gauge the electric field is $E=-\partial_{t}a$.) In what
follows, we assume that

$\displaystyle v(x;y)=v(x-y)\ \text{ and }\ v(-x)=v(x),$ (1.11)

so that $v^{*}\gamma=v*\rho_{\gamma}$. We specify below additional assumptions
on $v(x,y)$ and on the operators $\gamma$ and $\alpha$ so that all the terms
in (1.8)-(1.10) are well defined (at least weakly).

Connection with the BCS theory.  (1.8) can be reformulated as an equation on
the Fock space involving an effective quadratic Hamiltonian (see [11, 12, 20]
and [3], for the bosonic version). These are the effective BCS equations and
the effective BCS Hamiltonian.

Stationary equations. As was mentioned above, we are interested in stationary
solutions to (1.8)-(1.10). These solutions satisfy the stationary BdG
equations which, as we show below (see Proposition 2.1 and after), are of the
form

$\displaystyle\Lambda_{\eta a}-Tg^{\prime}(\eta)=0,$ (1.12)
$\displaystyle\operatorname{curl}^{*}\operatorname{curl}a-j(\eta,a)=0,$ (1.13)
$\displaystyle\Lambda_{\eta a}:=\left(\begin{array}[]{cc}h_{\gamma
a\mu}&v^{\sharp}\alpha\\\ v^{\sharp}\alpha^{*}&-\bar{h}_{\gamma
a\mu}\end{array}\right),$ (1.16)

with $h_{\gamma a\mu}:=h_{\gamma a}-\mu$, where $h_{\gamma a}$ is given in
(1.9). Here $\mu$ and $T\geq 0$ are the chemical potential and temperature
parameters. The physical function $g^{\prime}$ is selected by either a
thermodynamic limit (Gibbs states) or by a contact with a reservoir (or
imposing the maximum entropy principle) and is given by

$g^{\prime}(\lambda)=-\ln\frac{\lambda}{1-\lambda}.$ (1.17)

Its inverse, ${g^{\prime}}^{-1}$, is the Fermi-Dirac distribution

$\displaystyle f_{\rm FD}(h)=(1+e^{h})^{-1}.$ (1.18)

By inverting $g^{\prime}$ in (1.12), we can rewrite this equation as the
nonlinear Gibbs equation:

$\displaystyle\eta=f_{T}(\Lambda_{\eta a}),\ \text{ where }\
f_{T}(\lambda):=f_{\rm FD}(\frac{1}{T}\lambda).$ (1.19)

In Remark 1.15 below, we rewrite (1.12) in terms of the eigenfunctions of
$\Lambda_{\eta a}$ or $\eta$, the form common in physics literature and closer
in spirit to the PDEs.

###  1.3. Symmetries and conservation laws

(1.8)-(1.10) are invariant under the time-independent gauge transformations,

$T^{\rm gauge}_{\chi}:(\gamma,\alpha,a)\mapsto(e^{i\chi}\gamma
e^{-i\chi},e^{i\chi}\alpha e^{i\chi},a+\nabla\chi),$ (1.20)

for any sufficiently regular function $\chi:\mathbb{R}^{d}\to\mathbb{R}$, and
the translation, rotation and reflection transformations,

$\displaystyle T^{\rm trans}_{h}:(\gamma,\alpha,a)\mapsto(u_{h}\gamma
u_{h}^{-1},u_{h}\alpha u_{h}^{-1},u_{h}a),$ (1.21) $\displaystyle T^{\rm
rot}_{\rho}:(\gamma,\alpha,a)\mapsto(u_{\rho}\gamma
u_{\rho}^{-1},u_{\rho}\alpha u_{\rho}^{-1},\rho u_{\rho}a),$ (1.22)
$\displaystyle T^{\rm refl}:(\gamma,\alpha,a)\mapsto(u\gamma u^{-1},u\alpha
u^{-1},-ua),$ (1.23)

for any $h\in\mathbb{R}^{d}$ and $\rho\in O(d)$. Here $u_{h}\equiv u_{h}^{\rm
trans}$, $u_{\rho}\equiv u_{\rho}^{\rm rot}$ and $u\equiv u^{\rm refl}$ are
the standard translation, rotation and reflection transforms $u^{\rm
trans}_{h}:f(x)\mapsto f(x+h)$, $u^{\rm rot}_{\rho}:f(x)\mapsto f(\rho^{-1}x)$
and $u^{\rm refl}:f(x)\mapsto f(-x)$.

We also keep the notation $T^{\rm trans}_{h}$ for these operators restricted
to $\eta$’s.

(1.8)-(1.10) conserve the energy

$\displaystyle\mathcal{E}(\eta,a):=$
$\displaystyle\mathrm{Tr}\big{(}(-\Delta_{a})\gamma\big{)}+\frac{1}{2}\mathrm{Tr}\big{(}(v*\rho_{\gamma})\gamma\big{)}-\frac{1}{2}\mathrm{Tr}\big{(}(v^{\sharp}\gamma)\gamma\big{)}$
$\displaystyle+\frac{1}{2}\mathrm{Tr}\big{(}\alpha^{*}(v^{\sharp}\alpha)\big{)}+\frac{1}{2}\int(|\operatorname{curl}a|^{2}+|\partial_{t}a|^{2}),$
(1.24)

where the trace is taken in the one-particle space $\mathfrak{h}$. (Recall
that in our gauge, $-\partial_{t}a$ is the electric field.) Indeed, assuming
$(\eta,a)$ solves (1.8)-(1.10), taking the time derivative of (1.24), using
the expression

$d_{a}\mathrm{Tr}\big{(}(-\Delta_{a})\gamma\big{)}a^{\prime}=-\mathrm{Tr}\big{(}j(\eta,a)a^{\prime}\big{)}$

for the Gâteaux derivative in $a$ and using the notation
$\dot{f}\equiv\partial_{t}f$ and $\ddot{f}\equiv\partial_{t}^{2}f$ for
$f=\gamma,\alpha,a$, we find

$\displaystyle\partial_{t}\mathcal{E}(\eta,a)=$
$\displaystyle\mathrm{Tr}(h_{a\gamma}\dot{\gamma})+\mathrm{Tr}\big{(}\dot{\alpha}^{*}(v^{\sharp}\alpha)+\alpha^{*}(v^{\sharp}\dot{\alpha})\big{)}$
(1.25)
$\displaystyle+\langle\operatorname{curl}^{*}\operatorname{curl}-j(\eta,a),\dot{a}\rangle+\langle\ddot{a},\dot{a}\rangle,$
(1.26)

where the inner product is understood to be in $L^{2}(\Omega)$. A simple
computation shows that line (1.25)
$=\frac{1}{2}\mathrm{Tr}(\dot{\eta}\Lambda(\eta,a))$ and therefore by (1.8) is
$0$. Line (1.26) vanishes by (1.10). Hence $\partial_{t}\mathcal{E}=0$. $\Box$

Furthermore, the global gauge invariance implies the evolution conserves the
number of particles, $N:=\mathrm{Tr}\gamma$.

###  1.4. Free energy

The stationary BdG equations arise as the Euler-Lagrange equations for the
free energy (BCS) functional

$\displaystyle F_{T}(\eta,a):=E(\eta,a)-TS(\eta)-\mu N(\eta),$ (1.27)

where $S(\eta)=\mathrm{Tr}g(\eta)$, with
$g(\lambda):=-\lambda\ln\lambda-(1-\lambda)\ln(1-\lambda)$, an anti-derivative
of (1.17), is the entropy, $N(\eta):=\mathrm{Tr}\gamma$ is the number of
particles, and $E(\eta,a)$ is the energy functional (1.24) for $\eta$ and $a$
time-independent and is given by

$\displaystyle E(\eta,a)$
$\displaystyle=\mathrm{Tr}\big{(}(-\Delta_{a})\gamma\big{)}+\frac{1}{2}\mathrm{Tr}\big{(}(v*\rho_{\gamma})\gamma\big{)}-\frac{1}{2}\mathrm{Tr}\big{(}(v^{\sharp}\gamma)\gamma\big{)}$
$\displaystyle+\frac{1}{2}\mathrm{Tr}\big{(}\alpha^{*}(v^{\sharp}\alpha)\big{)}+\int|\operatorname{curl}a|^{2}.$
(1.28)

Note surprisingly, it turns out that $E(\eta,a):=\varphi(H_{a})$, where
$\varphi$ is a quasi-free state in question (see Appendix E) and $H_{a}$ is
the standard many-body given in (E.5), coupled to the vector potential $a$.

From now on, we assume that $d=2$. This is a typical case considered in
physics applications, where it is assumed that the solutions are independent
of the third coordinate (the cylindrical geometry in the physics terminology).

Let $a_{b}(x)$ be the vector potential with the constant magnetic field,
$\operatorname{curl}a_{b}=b$. Below, we work in the symmetric gauge:
$a_{b}(x)=\frac{b}{2}(-x_{2},x_{1})$.

###  1.5. Magnetic translations

Let $\mathcal{L}\subset\mathbb{R}^{2}$ be a Bravais lattice, fixed throughout
the paper. We define the magnetic translation operator

$\displaystyle u_{s}^{\mathcal{L}}:=u^{\rm
gauge}_{-\chi_{s}^{\mathcal{L}}}u^{\rm trans}_{s},$ (1.29)

where $u^{\rm gauge}_{\chi}:\phi(x)\mapsto e^{i\chi(x)}\phi(x)$ and, recall,
$u^{\rm trans}_{h}:\phi(x)\mapsto\phi(x+h)$ (cf. the definitions after (1.23))
and
$\chi_{\cdot}^{\mathcal{L}}:\mathcal{L}\times\mathbb{R}^{2}\rightarrow\mathbb{R}$
is given by

$\displaystyle\chi_{s}^{\mathcal{L}}(x):=\frac{b}{2}(s\wedge x)+c_{s},\ \quad
c_{s}:=\frac{b}{2}(s^{\prime\prime}\wedge s^{\prime}),$ (1.30)

where $s=k\omega_{1}+\ell\omega_{2}=:s^{\prime}+s^{\prime\prime},k,\ell\in Z$,
for any $s\in\mathcal{L}$, with $\\{\omega_{1},\omega_{2}\\}$ a basis in
$\mathcal{L}$, and $b$ is given by

$\displaystyle b=\frac{2\pi n}{|\mathcal{L}|},\ n\in\mathbb{Z}.$ (1.31)

Here $|\mathcal{L}|$ is the area of a fundamental cell of $\mathcal{L}$ (which
is independent of the choice of the cell). One can easily check that the
operator-family $u_{s}^{\mathcal{L}},s\in\mathcal{L},$ is a group
representation:

$\displaystyle u_{s}^{\mathcal{L}}u_{t}^{\mathcal{L}}=u_{s+t}^{\mathcal{L}}.$
(1.32)

###  1.6. One-particle spaces

With translations $u_{s}^{\mathcal{L}}$ given in (1.29), we define the
periodic one-particle state space

$\displaystyle\mathfrak{h}:=$ $\displaystyle\\{f\in
L^{2}_{loc}(\mathbb{R}^{2}):u_{s}^{\mathcal{L}}f=f\text{ for all
}s\in\mathcal{L}\\},$ (1.33)

which is a Hilbert space with the scalar product defined, for an arbitrary
fundamental cell $\Omega$ of $\mathcal{L}$, as

$\displaystyle\langle f,g\rangle_{\mathfrak{h}}:=\langle
f,g\rangle_{L^{2}(\Omega)}.$ (1.34)

Similarly, we consider the Sobolev space of vector potentials: let
$\vec{H}^{r}$ be given by

$\displaystyle\vec{H}^{r}:=$ $\displaystyle\\{a\in H^{r}_{\rm
loc}(\mathbb{R}^{2};\mathbb{R}^{2}):T_{s}^{\rm trans}a=a\ \forall
s\in\mathcal{L},\ \operatorname{div}a=0,\int a=0\\},$ (1.35)

with the Sobolev norm $\|a\|_{H^{r}}\equiv\|a\|_{H^{r}(\Omega)}$, for some
(and therefore for every) fundamental cell $\Omega$ of $\mathcal{L}$. (The
conditions $\operatorname{div}a=0$ and $\int a\equiv\int_{\Omega}a=0$ make the
operator $\operatorname{curl}^{*}\operatorname{curl}$ strictly positive.)
Finally, we define the affine space

$\displaystyle\vec{\mathfrak{h}}^{r}:=a_{b}+\vec{H}^{r}.$ (1.36)

(Recall that $a_{b}(x)$ is the vector potential with the constant magnetic
field.)

Now, we define spaces of $\gamma$’s and $\alpha$’s used below. Let $I^{r}$
denote the space of bounded operators satisfying
$\|A\|_{I^{r}}:=(\mathrm{Tr}(A^{*}A)^{r/2})^{1/r}<\infty$ (Schatten, or a
trace ideal, or non-commutative $L^{r}-$space) and let
$M_{b}:=\sqrt{-\Delta_{a_{b}}}$. We define Sobolev-type spaces for trace class
operators by

$\displaystyle
I^{s,1}:=\\{A:\mathfrak{h}\rightarrow\mathfrak{h}:\|A\|_{I^{s,1}}:=\|M_{b}^{s}AM_{b}^{s}\|_{I^{1}}<\infty\\},$
(1.37) $\displaystyle
I^{s,2}:=\\{A:\mathfrak{h}\rightarrow\mathfrak{h}:\,\|A\|_{I^{s,2}}:=\|AM_{b}^{s}\|_{I^{2}}<\infty\\}.$
(1.38)

Note that $I^{0,p}=I^{p}$. We will usually assume $\gamma\in I^{1,1}$ and
$\alpha\in I^{1,2}$. (Strictly speaking $\alpha$ acts from the dual space
$\mathfrak{h}^{*}$ to $\mathfrak{h}$ (see [4] for details), but for the sake
of notational simplicity we will ignore this subtlety, by identifying
$\mathfrak{h}^{*}$ with $\mathfrak{h}$.) We will use the notation
$\hat{I}^{s}$ for the space of $\eta$’s on $\mathfrak{h}\oplus\mathfrak{h}$
with $\gamma\in I^{s,1}$ and $\alpha\in I^{s,2}$ and the norm

$\displaystyle\|\eta\|_{(s)}:=\|\gamma\|_{I^{s,1}}+\|\alpha\|_{I^{s,2}}.$
(1.39)

Due to Lemma 5.4(2) below, $\gamma\in I^{1,1}\Rightarrow\alpha\in I^{1,2}$ and
$\|\eta\|_{(s)}\simeq\|\gamma\|_{I^{s,1}}$. Furthermore, we define

$\displaystyle\mathcal{D}^{s}_{\nu}=\big{\\{}\eta\in\hat{I}^{s},\ \eta\text{
satisfies \eqref{Gam-prop}},\ $ $\displaystyle\mathrm{Tr}\gamma=\nu\big{\\}}.$
(1.40)

###  1.7. Ground states of BdG equations

The static BdG equations (1.12)-(1.13) have the following key classes of
solutions which are candidates for the ground states:

  1. (1)

Normal state: $(\eta,a)$, with $\alpha=0$ (i.e. $\eta$ is diagonal).

  2. (2)

Superconducting state: $(\eta,a)$, with $\alpha\neq 0$ and $a=0$.

  3. (3)

Mixed state: $(\eta,a)$, with $\alpha\neq 0$ and $a\neq 0$.

We discuss the above states in more detail.

#### Superconducting states.

The existence of superconducting, translationally invariant solutions is
proven in [19]. (See the second part of Subsection 1.1 and [20] for the
references to earlier results.)

#### Normal states.

Since we assumed that the external fields are zero, the equations (considered
in $\mathbb{R}^{2}$) are translationally invariant. Because of the gauge
invariance, it is natural to consider the simplest, gauge (magnetically)
translationally invariant solutions, i.e. solutions invariant under the
transformations

$\displaystyle T_{bs}:=(T^{\rm gauge}_{\chi_{s}^{b}})^{-1}T^{\rm trans}_{s},$
(1.41)

for any $s\in\mathbb{R}^{2}$ and the function
$\chi_{s}^{b}(x),s,x\in\mathbb{R}^{2}$, given by

$\displaystyle\chi_{s}^{b}(x):=\frac{b}{2}(s\wedge x).$ (1.42)

For $b=0$, we can choose $a=0$. In this case, the existence of normal
translationally invariant solutions was proven in [10].

For $b\neq 0$, the simplest normal states are the magnetically translation
(mt-) invariant ones, i.e. ones satisfying

$\displaystyle T_{bs}(\eta,a)=(\eta,a),$ (1.43)

for any $s\in\mathbb{R}^{2}$, where $T_{bs}$ is defined in (1.41). Here the
operator $\eta$ acts on $L^{2}_{\rm loc}(\mathbb{R}^{2})\times L^{2}_{\rm
loc}(\mathbb{R}^{2})$.

#### Mixed states.

The main candidate for a mixed state is a vortex lattice, i.e. a state,
$(\eta,a)$, satisfying $\alpha\neq 0$ and the equivariance condition

$\displaystyle T^{\rm trans}_{s}(\eta,a)=T^{\rm
gauge}_{\chi_{s}^{\mathcal{L}}}(\eta,a)\ \text{ for every }\ s\in\mathcal{L},$
(1.44)

where $\chi_{s}^{\mathcal{L}}:\mathcal{L}\times\mathbb{R}^{2}\to\mathbb{R}$
are defined in (1.30), with $b$ satisfying the quantization condition (1.31).

######  Proposition 1.1 (Magnetic flux quantization).

If a vector potential $a$ satisfies (1.44), then for any fundamental cell
$\Omega$ of $\mathcal{L}$, we have

$\displaystyle\frac{1}{2\pi}\int_{\Omega}\operatorname{curl}a=c_{1}(\chi^{\mathcal{L}})\in\mathbb{Z},$
(1.45)

where $c_{1}(\chi^{\mathcal{L}})$, called the first Chern number of
$\chi_{s}^{\mathcal{L}}(x)$, is the integer $n$ entering (1.30)-(1.31).

One can show that $\chi^{\mathcal{L}}=\chi_{s}^{\mathcal{L}}(x)$ is a co-cycle
and $c_{1}(\chi^{\mathcal{L}})$ can be defined in terms of the function
$\chi^{\mathcal{L}}$ using its co-cycle property and without reference to the
explicit form (1.30) of $\chi^{\mathcal{L}}$ used, see Remark 1.18 for the
definitions and a discussion.

Recall that $a_{b}(x)$ is the magnetic potential with the constant magnetic
field $b$ ($\operatorname{curl}a_{b}=b$). In what follows we fix the gauge in
which $a_{b}(x)=\frac{b}{2}*x,*x:=(-x_{2},x_{1})$. Then $a_{b}$ satisfies
(1.44), provided $b$ satisfies the quantization condition (1.31).

###  1.8. Results

Let the reflection operator $\tau^{\rm refl}$ be given by conjugation by the
reflections, $u^{\rm refl}:f(x)\rightarrow f(-x)$. We say that a state
$(\eta,a)$ is even (or reflection symmetric) if and only if

$\displaystyle\tau^{\rm refl}\gamma=\gamma,\ \tau^{\rm refl}\alpha=\alpha\
\text{ and }\ u^{\rm refl}a=-a.$ (1.46)

The reflection symmetry of the BdG equations implies that if an initial
condition is even then so is the solution at every moment of time.

In what follows, we use the notation $A\lesssim B$ and $B\gtrsim A$ to signify
the inequalities $A\leq CB$ and $B\geq cA$, where $C$ and $c$ are positive
constants independent of the parameters involved.

#### Existence of normal and vortex lattice solutions.

We say an operator $A$ is magnetically translation invariant (mt-invariant,
for short), if it satisfies $\tau_{bh}A=A,\ \forall h\in\mathbb{R}^{2}$.

The existence of the mt-invariant normal states for $b\neq 0$ is stated in the
following theorem proven in Section 3:

######  Theorem 1.2.

Drop the exchange term $v^{\sharp}\gamma$ and let $v$ be
$\mathcal{L}$-periodic and either $\int v>0$, or $\|v\|_{\infty}<\infty$. Then
the BdG equations (1.12)-(1.13) on the space $I^{2,1}\times
I^{2,2}\times\vec{\mathfrak{h}}_{b}^{2}$ have a mt-invariant solution, unique
on the set of even (in the sense of the definition (1.46)) pairs $(\eta,a)$,
with $\eta$ mt-invariant (i.e. satisfying (1.43)).

Moreover, this solution is normal (i.e. $\alpha=0$) and is of the form
$(\eta_{T,b},\ a_{b})$ and

$\displaystyle\eta_{Tb}:=\left(\begin{array}[]{cc}\gamma_{Tb}&0\\\
0&\mathbf{1}-\bar{\gamma}_{Tb}\end{array}\right),$ (1.49)

with $\gamma_{Tb}$ solving the equation

$\displaystyle\gamma=f_{T}(h_{\gamma a_{b}}),$ (1.50)

where $f_{T}(\lambda)$ is given in (1.18).

The search for mt-invariant solutions is simplified by the following statement
proven in Section 3:

######  Proposition 1.3.

If $\eta$ is mt-invariant, then $\eta$ is normal, i.e. $\alpha=0$.

Recall that $b=\frac{2\pi n}{|\mathcal{L}|}$, see (1.31). For vortex lattices,
we have the following result

######  Theorem 1.4.

Drop the the exchange term $v^{\star}\gamma=v*\rho_{\gamma}$ in the definition
of $h_{\gamma a}$ in (1.9). Fix a lattice $\mathcal{L}$ and a value of the
Chern number $c_{1}(\chi^{\mathcal{L}})=n\in\mathbb{Z}$ (see (1.45)) and
assume that $v$ is $\mathcal{L}$-periodic and obeys $\|v\|_{\infty}<\infty$.
Then

(i) for any $T\geq 0$, there exists a (weak) solution
$(\eta,a)\in\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$ of the BdG
equations (1.12)-(1.13) (in particular, it satisfies $T^{\rm
trans}_{s}(\eta,a)=\hat{T}^{\rm gauge}_{\chi_{s}^{\mathcal{L}}}(\eta,a)$),
which minimizes the free energy $F_{T}$ (for the given
$c_{1}(\chi^{\mathcal{L}})=n\in\mathbb{Z}$);

(ii) for $T$ and $b$ sufficiently small, $(\eta,a)$ has $\alpha\not=0$, i.e.
this solution is a vortex lattice, provided
$v(x)\lesssim-(1+|x|)^{-\kappa},\kappa<2$, for $x$ in a fundamental cell
$\Omega$ centred at the origin. More generally, the latter holds if the
operator $L_{Tb}$, given in (1.56) and defined on $I^{2,2}$, has a negative
eigenvalue.

Statement (i) is proven in Section 5 and (ii) follows from Proposition 1.8.

#### Stability/instability of normal solutions.

We address the question of the energetic stability of the mt-invariant states.
To this end we define, in the standard way, the Hessian of the free energy in
$\eta$ as

$\displaystyle
F_{T}^{\prime\prime}(\eta_{*},a_{*}):=d_{\eta}\operatorname{grad}_{\eta}F_{T}(\eta_{*},a_{*}),$
(1.51)

where $d_{\eta}$ is the Gâteaux derivative w.r.to $\eta$ and
$\operatorname{grad}_{\eta}$ is the gradient w.r.to $\eta$, defined by the
equation

$\mathrm{Tr}(\operatorname{grad}_{\eta}F(\eta_{*},a_{*})\eta^{\prime})=d_{\eta}F(\eta_{*},a_{*})\eta^{\prime}.$

We consider $F_{T}^{\prime\prime}(\eta_{Tb},a_{b})$ along physically relevant
perturbations of the form $\eta^{\prime}=\phi(\alpha)$, where $\phi(\alpha)$
is the off-diagonal operator-matrix, defined by

$\displaystyle\phi(\alpha):=\left(\begin{array}[]{cc}0&\alpha\\\
\alpha^{*}&0\end{array}\right),$ (1.54)

with $\alpha$ a Hilbert-Schmidt operator on $\mathfrak{h}$. Moreover, we
require that perturbations $\eta^{\prime}=\phi(\alpha)$ satisfy the condition

$\displaystyle\alpha\alpha^{*}\lesssim[\gamma_{Tb}(1-\gamma_{Tb})]^{2}$ (1.55)

(which is equivalent to (2.11) at $\eta=\eta_{Tb}$, c.f. also Lemma 5.4(1)).

For an operator $h$, let $h^{L}$ and $h^{R}$ stand for the operators acting on
other operators by multiplication from the left by $h$ and from right by
$\bar{h}$, respectively, and recall $v^{\sharp}$ is defined after (1.8). We
have the following

######  Proposition 1.5.

For off-diagonal perturbations $\eta^{\prime}=\phi(\alpha)$,
$F_{T}^{\prime\prime}(\eta_{Tb},a_{b})\phi(\alpha)=\phi(L_{Tb}\alpha)$, where
the operator $L_{Tb}$ is given by

$\displaystyle L_{Tb}:=K_{Tb}+v^{\sharp},$ (1.56) $\displaystyle
K_{Tb}:=\frac{h_{Tb}^{L}+h_{Tb}^{R}}{\tanh(h_{Tb}^{L}/T)+\tanh(h_{Tb}^{R}/T)},$
(1.57)

on the space of Hilbert-Schmidt operators on $\mathfrak{h}$. Here
$h_{Tb}:=h_{\gamma_{Tb}a_{b}\mu}$ (with $h_{\gamma a\mu}:=h_{\gamma a}-\mu$,
where $h_{\gamma a}$ is given in (1.9)).

Let
$\langle\alpha,\alpha^{\prime}\rangle:=\mathrm{Tr}(\alpha^{*}\alpha^{\prime})$.
We say that $\eta^{\prime}$ is an off-diagonal perturbation if and only if
$\eta^{\prime}=\phi(\alpha)$ with $\alpha$ a Hilbert-Schmidt operator
satisfying (1.55). The next result generalizes that of [19] for $a=0$:

######  Proposition 1.6.

For off-diagonal perturbations $\eta^{\prime}=\phi(\alpha)$, we have

$\displaystyle F_{T}(\eta_{Tb}+\epsilon\eta^{\prime},a_{b})=$ $\displaystyle
F_{T}(\eta_{Tb},a_{b})+\epsilon^{2}\langle\alpha,L_{Tb}\alpha\rangle+O(\epsilon^{3}).$
(1.58)

Definition (1.28) of $E(\eta,a)$ implies that
$E(\eta_{Tb}+\epsilon\eta^{\prime},a_{b})=E(\eta_{Tb},a_{b})+\epsilon^{2}\mathrm{Tr}(\bar{\alpha}v^{\sharp}\alpha)$
for $\eta^{\prime}=\phi(\alpha)$. This together with Corollary A.3 and (A.17)
of Appendix A dealing with the entropy yields Propositions 1.5 and 1.6,
respectively. (The absence of the linear term is due to the fact that
$(\eta_{Tb},0)$ is a critical point of the energy function.)

The next two propositions are proven in Section 4.

######  Proposition 1.7.

$L_{Tb}\geq\frac{1}{2}T-\|v\|_{\infty}$ and consequently, for $T$ sufficiently
large, $L_{Tb}>0$ and the mt-invariant state $(\eta_{T,b},a_{b})$ is
energetically stable under off-diagonal perturbations.

On the other hand, for $T$ and $b$ sufficiently small, we have

######  Proposition 1.8.

Drop the exchange term $-v^{\sharp}\gamma$. Suppose that
$v(x)\lesssim-(1+|x|)^{-\kappa},\kappa<2$, for $x$ in a fundamental cell
$\Omega$ centred at the origin. Then, for $T$ and $b$ sufficiently small and
$T\lesssim b^{\sigma},\sigma>\kappa/2$, the operator $L_{Tb}$ acting on the
set of Hilbert-Schmidt operators on $\mathfrak{h}$ has a negative eigenvalue
and consequently the mt-invariant state $(\eta_{T,b},a_{b})$ is energetically
unstable under perturbations $\eta^{\prime}=\phi(\alpha)$.

Note that $L_{Tb}$ is discontinuous at $T=0$.

Let $T_{c}(b)$ (resp. $T_{c}^{\prime}(b)$) be the largest (resp. smallest)
temperature s.t. the normal solution is energetically unstable (resp. stable)
under off-diagonal perturbations $\eta^{\prime}=\phi(\alpha)$, for
$T<T_{c}(b)$ (resp. $T>T_{c}(b)^{\prime})$. Clearly, $\infty\geq
T_{c}^{\prime}(b)\geq T_{c}(b)\geq 0$. Propositions 1.7 and 1.8 imply

######  Corollary 1.9.

Under the conditions of Proposition 1.8, $T_{c}(b)>0$ for $b$ sufficiently
small and $T_{c}^{\prime}(b)=0$ for $b$ sufficiently large.

We conjecture that (a) mt-invariant states coincide with normal states, (b)
For $T\gg 1$, the free energy $F_{T}$ is minimized by normal states; (c)
normal states are stable if either $T\gg 1$ or $b\gg 1$ and unstable if $T\ll
1$ and $b\ll 1$; (d) $T_{c}(b)=T_{c}^{\prime}(b).$

The next corollary provides a convenient criterion for the determination of
$T_{c}(b)$ and $T_{c}(b)^{\prime}$.

######  Corollary 1.10.

At $T=T_{c}(b)$ and $T=T_{c}^{\prime}(b)$, zero is the lowest eigenvalue of
the operator $L_{Tb}$.

A proof of energetic stability under general perturbations for either $T$ or
$b$ sufficiently large is more subtle. For it, one has to use the full
linearized operator, $\operatorname{Hess}F_{T}(\eta_{Tb},a_{b})$. Our
computations suggest that $0$ is the lowest eigenvalue of
$\operatorname{Hess}F_{T}(\eta_{Tb},a_{b})$ if and only if $0$ is the lowest
eigenvalue of $L_{Tb}$ and, consequently, $T_{c}(b)$ and $T_{c}^{\prime}(b)$
apply also to the general perturbations.

The statement that $T_{c}=T_{c}(0)=T_{c}^{\prime}(0)>0$ for $a=0$ (and
therefore $b=0$) and for a large class of potentials is proven, by the
variational techniques, in [19].

In conclusion of this paragraph, we mention the proposition which follows from
a simple computation:

######  Proposition 1.11.

The operator $L_{Tb}$ commutes with the magnetic translations. The same is
true for the $\eta-$Hessian $F_{T}^{\prime\prime}(\eta_{Tb},a_{b})$ (see
(1.51)).

#### Remarks.

Here we collect the comments on various aspects covered in this introduction.

######  Remark 1.12 (BdG equations).

1) For $\alpha=0$, (1.8) becomes the time-dependent von Neumann-Hartree-Fock
equation for $\gamma$ (and $a$).

2) A formal derivation of (1.8) given in Appendix E can be extended to the
coupled system (1.8)-(1.10) by starting with the many-body quantum Hamiltonian
(E.5) coupled to the quantized electro-magnetic field.

######  Remark 1.13.

It is common in physics literature to drop the direct and exchange self-
interaction terms from $h_{\gamma a\mu}$ in (1.16). In this case,
$\Lambda_{\eta a}$ becomes independent of the diagonal part, $\gamma$, of
$\eta$ and equation (1.12) has always the solution

$\eta_{Ta}\ =f_{T}(\Lambda_{a}),\ \text{ where }\ \Lambda_{a}:=\Lambda_{\eta
a}\big{|}_{\eta=0}.$ (1.59)

Similarly, for equation (1.50) for normal states. In this case, it always has
the solution

$\displaystyle\gamma_{Tb}=f_{T}(h_{a_{b}}),\ \text{ where }\
h_{a}=-\Delta_{a}-\mu.$ (1.60)

######  Remark 1.14 (Particle density).

For a definition of $\rho_{\gamma}$ not relying on the integral kernels, see
(3.11).

######  Remark 1.15 (Particle-hole symmetry).

The evolution under the BdG equations (1.8)-(1.10) preserves the relations in
(1.7), i.e. if an initial condition has one of these properties, then so does
the solution. This follows from the relation

$\displaystyle J^{*}\Lambda J=-\overline{\Lambda},$ (1.61)

where $J$ is defined in (1.7). The second relation in (1.7) is called the
particle-hole symmetry.

######  Remark 1.16 (BdG equations in ‘coordinate’ form).

In physics literature, the BdG equations are written in terms of
eigenfunctions of the operator $\eta$, or $\Lambda_{\eta a}$ (assuming the
operator $-\Delta_{a}$ has purely discrete spectrum, see [4, 11, 12, 22, 33,
34], cf. [3]). In the case of stationary equations (1.12)-(1.13), we have

$\displaystyle\Lambda_{\eta a}\psi_{n}=\varepsilon_{n}\psi_{n},\
\quad\Lambda_{\eta a}J\bar{\psi}_{n}=-\varepsilon_{n}J\bar{\psi}_{n},$ (1.62)

with $\varepsilon_{n}>0$ and $J$ given in (1.7). The second equation above
follows from the first and relation (1.61). By (1.19), $\psi_{n}$ and
$J\bar{\psi}_{n}$ are also eigenfunctions of $\eta$ with the eigenvalues
$f_{n}:=f_{T}(\varepsilon_{n})$ and $1-f_{n}=f_{T}(-\varepsilon_{n})$,
respectively. As $\\{\psi_{n},J\bar{\psi}_{n}\\}$ form a complete orthonormal
set, we have

$\displaystyle\eta=\sum_{n}\big{(}f_{n}P_{\psi_{n}}+(1-f_{n})P_{J\bar{\psi}_{n}}\big{)},$
(1.63)

where $P_{\psi}$ stands for the rank-one orthogonal projection onto the
subspace spanned by $\psi$. (1.62)-(1.63) yield a nonlinear eigenvalue problem
replacing (1.12). Finally, one can express $j(\eta,a)$ in (1.13) in terms of
$\\{\psi_{n},J\bar{\psi}_{n}\\}$.

######  Remark 1.17 (Entropy).

Due to the symmetry (1.7) of $\eta$, we see that

$\displaystyle\mathrm{Tr}(\eta\ln\eta)=\mathrm{Tr}((1-\eta)\ln(1-\eta))$
(1.64)

which, recalling (2.9), implies that

$\displaystyle S(\eta):=\mathrm{Tr}(s(\eta))=\mathrm{Tr}(g(\eta)),$ (1.65)
$\displaystyle g(\eta):=-\eta\ln\eta-(1-\eta)\ln(1-\eta),\
s(\eta):=-2\eta\ln\eta.$ (1.66)

######  Remark 1.18 (Co-cycle equation).

One can readily show that the operator-family $u_{bs},s\in\mathcal{L},$
defined in (1.29) is a group representation (with the group law given in
(1.32)) if and only if the functions
$\chi_{s}^{\mathcal{L}}(x),s\in\mathcal{L},x\in\mathbb{R}^{2},$ satisfy the
co-cycle condition

$\displaystyle\chi_{s+t}^{\mathcal{L}}(x)-\chi_{s}^{\mathcal{L}}(x+t)-\chi_{t}^{\mathcal{L}}(x)\in
2\pi\mathbb{Z},\ \forall s,t\in\mathcal{L}.$ (1.67)

The function $\chi_{s}^{\mathcal{L}}$ defined in (1.30) satisfies this
relation, while the one in (1.42) does not.

Functions $\chi_{\cdot}:\mathcal{L}\times\mathbb{R}^{d}\rightarrow\mathbb{R}$
satisfying (1.67) are called the summands of automorphy, or co-cycles (see
[31] for a relevant discussion). (The map
$e^{i\chi}:\mathcal{L}\times\mathbb{R}^{2}\to U(1)$, where
$\chi(x,s)\equiv\chi_{s}(x)$ is called the factor of automorphy.) Every map
$\chi_{s}:\mathcal{L}\times\mathbb{R}^{2}\to\mathbb{R}$ satisfying (1.67) is
(gauge) equivalent to the one in (1.30), with $b$ satisfying (1.31).

With a summand of automorphy
$\chi_{s}^{\mathcal{L}}(x),s\in\mathcal{L},x\in\mathbb{R}^{2},$ one associates
the function

$c_{1}(\chi)=\frac{1}{2\pi}(\chi_{\nu_{2}}(x+\nu_{1})-\chi_{\nu_{2}}(x)-\chi_{\nu_{1}}(x+\nu_{2})+\chi_{\nu_{1}}(x)),$
(1.68)

for some basis $\\{\nu_{1},\nu_{2}\\}$ in $\mathcal{L}$. As it turns out, this
function is independent of $x$ and the choice of the basis and is an integer.
In physics literature, $c_{1}(\chi)$ is called the Chern number.

######  Remark 1.19 (Operator $L_{Tb}$).

1) The question of when $L_{Tb}$ has negative spectrum for a larger range of
$T$’s is a delicate one. For $T$ close to $T_{c}$, this depends, besides of
the parameters $T$ and $b$, also on $v$ and $\mu$ determining whether the
superconductor is of Type I or II. (As was discovered theoretically by A.A.
Abrikosov in his study of the Ginzburg-Landau equations ([1]) and confirmed
later experimentally, superconductors are divided, according to their basic
properties, into two groups, Type I or II superconductors. So far, there are
no results (rigorous, or not) establishing within the BdG theory these
properties for certain classes of potentials $v$ and $\mu$.)

2) Since the components of magnetic translations (1.41) do not commute, the
fiber decomposition of $L_{Tb}$ is somewhat subtle (see [2]).

#### Nomenclature.

In our most important results, we drop the exchange term $v^{\sharp}\gamma$.
By analogy with the Hartree-Fock equations, the resulting equations could be
called the reduced BdG equations.

An important and natural modification of the BdG equations would be replacing
the exchange term $v^{\sharp}\gamma$ by a density dependent exchange-
correlation term $xc(\rho_{\gamma})$ from the density functional theory. The
resulting equations could be called the density BdG equations.

Addressing the density or original BdG equations would be an important next
step.

The paper is organized as follows. In Sections 3 and 5, we prove Theorems 1.2
and 1.4, on existence of the normal and vortex lattice solutions,
respectively. These are our principal results. In Section 4, we prove
Propositions 1.7 and 1.8 on the stability/instability of the normal solutions.
In Appendix A we study the entropy functional. Results of this appendix imply
the proofs of Propositions 1.5 and 1.6 and are used in Appendix B in the proof
of technical Theorem 2.2. In Appendix C, we prove an elementary technical
result from Section 3. Finally, in Appendix D we prove some bounds on
functions relative to magnetic Laplacian and bounds on densities (both
elementary and probably well-known) and in Appendix E, we discuss the
derivation of the BdG equations from the quantum many-body problem.

##  2\. Stationary Bogoliubov-de Gennes equations

In this section we establish the connection between the time-dependent and
stationary BdG equations and show that the latter are the Euler-Lagrange
equations for the free energy, (1.27).

In terms of $\eta$, the gauge transformation, $T^{\rm gauge}_{\chi}$, could be
written as

$T^{\rm gauge}_{\chi}:\eta\rightarrow U^{\rm gauge}_{\chi}\eta(U^{\rm
gauge}_{\chi})^{-1},\ \text{ where }\ U^{\rm
gauge}_{\chi}=\left(\begin{array}[]{cc}e^{i\chi}&0\\\
0&e^{-i\chi}\end{array}\right).$ (2.1)

It is extended correspondingly to $(\eta,a)$ by $T_{\chi}^{\rm
gauge}(\eta,a)=(T_{\chi}^{\rm gauge}\eta,a+\nabla\chi)$. Notice the difference
in the action on the diagonal and off-diagonal elements of $\eta$.

The invariance under the gauge transformations follows from the relation

$\Lambda(T_{\chi}^{\rm gauge}(\eta,a))=T_{\chi}^{\rm gauge}(\Lambda(\eta,a)),$
(2.2)

shown by using the operator calculus.

We consider stationary solutions to (1.8), with $a$ independent of $t$, of the
form

$\displaystyle\eta_{t}:=T^{\rm gauge}_{\chi}\eta=U_{\chi}^{\rm
gauge}\eta(U_{\chi}^{\rm gauge})^{-1},$ (2.3)

where $\eta$ is independent of $t$. We have

######  Proposition 2.1.

The operator-family (2.3), with $\eta$ independent of $t$ and $\dot{\chi}$,
constant, say $\dot{\chi}\equiv-\mu$, is a solution to (1.8), if and only if
$\eta$ solves the equation

$[\Lambda_{\eta a},\eta]=0,$ (2.4)

where $\Lambda_{\eta a}\equiv\Lambda_{\eta a\mu}$ is given explicitly in
(1.16).

###### Proof.

We write $U_{\chi}^{\rm gauge}\equiv U_{\chi}$ and use that
$\partial_{t}U_{\chi}=i\dot{\chi}SU_{\chi}$, where

$S:=\left(\begin{array}[]{cc}1&0\\\ 0&-1\end{array}\right),$

and therefore $\partial_{t}(U_{\chi}\eta
U_{\chi}^{-1})=i\dot{\chi}[S,U_{\chi}\eta
U_{\chi}^{-1}]=i\dot{\chi}U_{\chi}[S,\eta]U_{\chi}^{-1}$. Plugging (2.3) into
(1.8), and using (2.2), the previous relation and that $\chi$ is independent
of $x$, we see that

$\displaystyle-\dot{\chi}U_{\chi}[S,\eta]U_{\chi}^{-1}=$
$\displaystyle[\Lambda(U_{\chi}\eta U_{\chi}^{-1},a),U_{\chi}\eta
U_{\chi}^{-1}]$ (2.5) $\displaystyle=$ $\displaystyle[\Lambda(U_{\chi}\eta
U_{\chi}^{-1},a+\nabla\chi),U_{\chi}\eta U_{\chi}^{-1}]$ (2.6)
$\displaystyle=$ $\displaystyle U_{\chi}[\Lambda(\eta,a),\eta]U_{\chi}^{-1}.$
(2.7)

Since $\dot{\chi}\equiv-\mu$, it follows then that $[\Lambda(\eta,a)-\mu
S,\eta]=0.$ Since $\Lambda(\eta,a)-\mu S$ is equal to (1.16), the last
equation is exactly (2.4) which gives the statement of the proposition. ∎

For any reasonable function $f$ and time-independent $a$, solutions of the
equation

$\displaystyle\eta=f(\frac{1}{T}\Lambda_{\eta a}),$ (2.8)

solve (2.4) and therefore give stationary solutions of (1.8). Conversely,
solutions to (2.4), s.t. the spectrum of $\Lambda_{\eta a}$ is simple, solve
(2.8). (The parameter $T>0$, the temperature, is introduced here for the
future reference.)

Inverting the function $f$, one can rewrite (2.8) as $\Lambda_{\eta
a}=Tf^{-1}(\eta)$. Let $f^{-1}=:g^{\prime}$. Then the stationary BdG equations
can be written (in the Coulomb gauge $\operatorname{div}a=0$) as
(1.12)-(1.13).

The physical function $f$ is selected by either a thermodynamic limit (Gibbs
states) or by coupling the system in question to a thermal reservoir (or
imposing the maximum entropy principle). It is given by the Fermi-Dirac
distribution (1.18) It follows from the equations $g^{\prime}=f^{-1}$ and
(1.18) that the function $g^{\prime}$ is given by (1.17) and the function $g$
is equal to

$g(\lambda)=-\lambda\ln\lambda-(1-\lambda)\ln(1-\lambda).$ (2.9)

From now on, we assume that $f(\lambda)$ and $g(\lambda)$ are given in (1.18)
and (2.9), respectively.

Our next result shows that the BdG equations arise as the Euler-Lagrange
equations for the free energy functional $F_{T}$ differentiated along the
perturbations (‘tangent vectors’) at $(\eta,a)$ from the following class (c.f.
also Lemma 5.4(1))

$\displaystyle\mathcal{P}_{\eta}:=\\{(\eta^{\prime},a^{\prime})\in\hat{I}^{1}\times\vec{H}^{1}:\text{
\eqref{Gam'-cond} holds }\\},$ (2.10) $\displaystyle
J^{*}\eta^{\prime}J=-\bar{\eta}^{\prime},\
(\eta^{\prime})^{2}\lesssim[\eta(1-\eta)]^{2},\text{ and
}\mathrm{Tr}(S_{1}\eta^{\prime})=0,$ (2.11)

where $S_{1}=\text{diag}(1,0)$ and $J$ is defined in (1.7). Conditions (2.11)
are designed to handle a delicate problem of non-differentiability of
$s(\lambda):=2\lambda\ln\lambda$ at $\lambda=0$, while providing a
sufficiently rich set, $\mathcal{P}_{\eta}$, to derive the BdG equations.

######  Theorem 2.2.

(a) The free energy functional $F_{T}$ is well defined on the space
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$.

(b) $F_{T}$ is continuously (Gâteaux or Fréchet) differentiable at
$(\eta,a)\in\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$, with respect
of perturbations $(\eta^{\prime},a^{\prime})\in\mathcal{P}_{\eta}$.

(c) Critical points of $F_{T}$, which are even (in the sense of the definition
(1.46)) and obey $0<\eta<1$, satisfy the BdG equations for some $\mu$
(determined by the constraint $\mathrm{Tr}\gamma=\nu$).

(d) Minimizers of $F_{T}$ over
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}_{b}^{1}$ satisfy the conditions
of statement (c) and therefore satisfy the BdG equations for some $\mu$.

This theorem is proven in Appendix B. For the translation invariant case, it
is proven in [19]. For $a=0$, but $\eta$ not necessarily translation
invariant, the fact that the BdG equations are the Euler-Lagrange equations of
the free energy functional is used in [13], but, it seems, with no proof
provided.

As a result of Theorem 2.2, we can write the BdG equations as

$\displaystyle F^{\prime}_{T}(\eta,a)=0,$ (2.12)

where the map $F^{\prime}_{T}(\eta,a)$ is defined by the r.h.s. of the static
BdG equations (1.12)-(1.13) and can be thought of as a gradient map of
$F_{T}$.

##  3\. The normal states: Proof of Theorem 1.2

We begin with some results about the magnetic translations. Recall the
operators $T_{bs}$ and $U^{\rm gauge}_{\chi}$, defined in (1.41) and (2.1).
Group properties of $T_{bs}$ are established in following

######  Lemma 3.1.

The operators $T_{bs}$, defined in (1.41) and restricted to $\eta$’s, satisfy

$\displaystyle T_{bs}T_{bt}=\hat{I}_{bst}T_{bs+t},$ (3.1)

where $\hat{I}_{bst}$ acts on bounded operators on $\mathfrak{h}$ as

$\displaystyle\hat{I}_{bst}A:=I_{bst}AI_{bst}^{-1},\ I_{bst}:=U^{\rm
gauge}_{\frac{b}{2}(s\wedge t)}.$ (3.2)

###### Proof of Lemma 3.1.

Let $U^{\rm trans}_{s}:=\mathrm{diag}(u_{s}^{\rm trans},u_{s}^{\rm trans})$ on
$L^{2}_{\rm loc}(\mathbb{R}^{2})\times L^{2}_{\rm loc}(\mathbb{R}^{2})$ and
define the transformations

$\displaystyle U_{bs}:=(U^{\rm gauge}_{\chi_{s}^{b}})^{-1}U^{\rm trans}_{s}.$
(3.3)

Then $T_{bs}\eta=U_{bs}\eta U_{bs}^{-1}$. Using the definition above and the
relation $U^{\rm trans}_{s}U^{\rm gauge}_{\chi}=U^{\rm gauge}_{u_{s}^{\rm
trans}\chi}U^{\rm trans}_{s}$, we compute

$U_{bs}U_{bt}=(U^{\rm gauge}_{\chi_{s}^{b}+u_{s}^{\rm
trans}\chi_{t}^{b}})^{-1}U^{\rm trans}_{s+t}=(U^{\rm
gauge}_{\chi_{s}^{b}+u_{s}^{\rm
trans}\chi_{t}^{b}-\chi^{b}_{s+t}})^{-1}U_{bs+t}.$

By definition (1.42), we have $\chi_{s}^{b}+u_{s}^{\rm
trans}\chi_{t}^{b}-\chi^{b}_{s+t}=-\frac{b}{2}(s\wedge t)$, which implies
$U_{bs}U_{bt}=I_{bst}U_{bs+t}=U_{bs+t}I_{bst}$. Hence, the result follows. ∎

Now, Lemma 3.1 implies Proposition 1.3:

###### Proof of Proposition 1.3.

(3.1)-(3.2) and the mt-invariance, (1.43), yield
$\eta=T_{bs}T_{bt}\eta=I_{bst}\eta I_{bst}^{-1}$, which implies that
$\alpha=e^{i\frac{b}{2}(s\wedge t)}\alpha$ for all $s,t\in\mathbb{R}^{2}$.
This yields that $\alpha=0$. ∎

Note that, ignoring $a$, the fact that $T_{bs}$ maps the set of bounded
diagonal operator-matrices on $\mathfrak{h}\times\mathfrak{h}$ into itself
follows from (3.1)-(3.2) and the fact that $\hat{I}_{bst}$ is an identity on
diagonal operator-matrices.

Proposition 1.3 shows that we can restrict our search of mt-invariant
solutions to normal states. For normal states, i.e. for $\alpha=0$, the BdG
equations (1.12)-(1.13) (with (1.12) replaced by (1.19)) reduce to the
following equations for $\gamma$ and $a$

$\displaystyle\gamma=f_{T}(h_{\gamma a\mu}),$ (3.4)
$\displaystyle\operatorname{curl}^{*}\operatorname{curl}a=j(\gamma,a),$ (3.5)

where, recall, $j(\gamma,a)(x):=([-i\nabla_{a},\gamma]_{+})(x,x)$. These are
the coupled Hartree-Fock and Ampére equations.

Recall that $a_{b}(x)$ is the magnetic potential with the constant magnetic
field $b$ ($\operatorname{curl}a_{b}=b$) in the gauge s.t.
$a_{b}(x)=\frac{b}{2}*x,*x:=(-x_{2},x_{1})$. First, we show that the second
equation is automatically satisfied for $a=a_{b}$ and $\gamma$ a magnetically
translation invariant operator, which is even in the sense of (1.46).

Define the magnetic translations $u_{bs}$ as (cf. (1.29) and (3.3))

$\displaystyle u_{bs}:=u^{\rm gauge}_{-\chi_{s}^{b}}u^{\rm trans}_{s},$ (3.6)

where, recall the operators $u^{\rm gauge}_{\chi}$ and $u^{\rm trans}_{s}$ are
defined are defined after (1.29) and
$\chi_{\cdot}^{b}:\mathcal{L}\times\mathbb{R}^{d}\rightarrow\mathbb{R}$ is
given in (1.42). Recall the definition of the space $\mathfrak{h}$ in (1.33).

######  Lemma 3.2.

With the definition $\tau_{bs}(\gamma)=u_{bs}\gamma{u_{bs}}^{-1}$, we have
(cf. (3.1))

$\displaystyle u_{bs}u_{bh}=e^{-ibh\wedge s}u_{bh}u_{bs},$ (3.7)
$\displaystyle\tau_{bs}\tau_{bh}=\tau_{bh}\tau_{bs},$ (3.8)
$\displaystyle\text{If $\gamma$ maps $\mathfrak{h}$ into itself, then so does
}\tau_{bh}\gamma.$ (3.9)

###### Proof.

(3.7) follows from $u_{bs}u_{bh}=e^{-i\frac{b}{2}h\wedge
s}e^{-i\chi_{bs}}u_{bh}\tau_{s}^{\rm trans}=e^{-i\frac{b}{2}h\wedge
s}e^{i\frac{b}{2}s\wedge h}u_{bh}u_{bs}$. To prove (3.8), we use that
$\tau_{s}^{\rm trans}e^{-i\chi_{bh}}=e^{-i\frac{b}{2}h\wedge
s}e^{-i\chi_{bs}}\tau_{h}^{\rm trans}$, where, recall, $\chi_{bs}$ is defined
in (1.30), and (3.7) which yield $(u_{bs}u_{bh})^{-1}=e^{ibh\wedge
s}(u_{bh}u_{bs})^{-1}$ and therefore, due to $\tau_{bh}\gamma:=u_{bh}\gamma
u_{bh}^{-1}$, (3.8) follows.

Finally, $\gamma$ maps the space $\mathfrak{h}$ into itself if and only if
$\tau_{bs}(\gamma)=\gamma,\ \forall s\in\mathcal{L}$. On the other hand, (3.8)
implies
$\tau_{bs}\tau_{bh}(\gamma)=\tau_{bh}\tau_{bs}(\gamma)=\tau_{bh}(\gamma),\
\forall s\in\mathcal{L},h\in\mathbb{R}^{2}$, so (3.9) follows. ∎

Note that (3.7) shows that $u_{bh}$ does not leave $\mathfrak{h}$ invariant.

The generators of the magnetic translations, $u_{bs}$, defined in (3.6), and
their properties are described in the following

######  Lemma 3.3 ([2]).

Let $p_{b}=-i\nabla_{a_{b}}$ and
$\pi_{b}=-\bar{p}_{b}:=-\mathcal{C}p_{b}\mathcal{C}$, with the components
$p_{bi}$ and $\pi_{bi}$. Then

  1. (1)

$-i\partial_{s_{j}}\mid_{s=0}u_{bs}=\pi_{bj}$;

  2. (2)

$[\pi_{bi},p_{bj}]=0$ and therefore $[u_{bs},p_{bj}]=0$.

To begin with, we define the operator $\gamma$ on $L^{2}_{\rm
loc}(\mathbb{R}^{2})$ which could be further specified as $\mathfrak{h}$.

For an operator $A$ on $\mathfrak{h}$, we can define the integral kernel
$A^{\prime}(x,y)$ by the relation

$\displaystyle\langle
g\otimes\bar{f},A^{\prime}\rangle_{\mathfrak{h}\otimes\mathfrak{h}}=\langle
g,Af\rangle_{\mathfrak{h}},\ \quad\forall f,g\in\mathfrak{h}.$ (3.10)

In this section, it is convenient to use the notation den$(A)$ for the one-
particle density $\rho_{A}$. For an operator $A$ on $\mathfrak{h}$, s.t. $fA$
and $Af$ are trace-class, the density den$(A)\equiv\rho_{A}$ obeys the
relation

$\displaystyle\int f\text{den}(A)=\mathrm{Tr}(fA),\ \quad\forall f\in
L^{\infty}.$ (3.11)

which can be also used as a definition of den$(A)$. (Recall our convention
$\int\equiv\int_{\Omega}$ for an arbitrary and fixed lattice cell $\Omega$.)
With this notation, we have e.g.
$j(\gamma,a):=\text{den}([-i\nabla_{a},\gamma]_{+})$.

We assume that all operators below are originally defined on $L^{2}_{\rm
loc}(\mathbb{R}^{2})$ (or on a local Sobolev space). This allows us to define
compositions and commutators of operators some of which do not leave
$\mathfrak{h}$ invariant.

Recall that we say an operator $A$ to be magnetically translation invariant
(mt-invariant, for short), if and only if it satisfies $\tau_{bh}A=A,\ \forall
h\in\mathbb{R}^{2}$. Our key result here is the following

######  Proposition 3.4.

(i) For a trace-class, mt-invariant operator $A$ on $\mathfrak{h}$, den$(A)$
is constant. (ii) If, in addition, $\tau^{\rm refl}A=-A$ (with the reflection
operator $\tau^{\rm refl}$ defined before (1.46)), then den$(A)=0$.

Recall that $\nabla_{a}:=\nabla-ia$. We derive Proposition 3.4 from the
following

######  Lemma 3.5.

For any linear vector field $a$ and any integral operator $A$ on
$\mathfrak{h}$, $[\nabla_{a},A]$ leaves $\mathfrak{h}$ invariant (though
$\mathfrak{h}$ is not invariant under $\nabla_{a}$) and

$\displaystyle\nabla\text{den}(A)=\text{den}([\nabla_{a},A])$ (3.12)

###### Proof.

Since $a$ is linear, the invariance of $\mathfrak{h}$ under $[\nabla_{a},A]$
is straightforward. We prove (3.12). We have

$\displaystyle\text{den}([\nabla_{a},A])=\text{den}([\nabla_{a_{b}},A])+i\text{den}([a_{b}-a,A]).$
(3.13)

Since $\nabla_{a_{b}}$ leaves $\mathfrak{h}$ invariant, we can use the
cyclicity of the trace to compute

$\displaystyle\int f\text{den}([\nabla_{a_{b}},A])$
$\displaystyle=\mathrm{Tr}_{\mathfrak{h}}(f[\nabla_{a_{b}},A])$
$\displaystyle=-\mathrm{Tr}_{\mathfrak{h}}(\nabla fA)=-\int\nabla
f\text{den}(A).$ (3.14)

For any (linear) vector field $c$, the integral kernel of $[c,A]$ is
$(c(x)-c(y))A(x,y)$ (see (3.10)). Hence, we see that $\text{den}([c,A])=0$.
Combining this with (3.13) and (3.14) gives $\int
f\text{den}([\nabla_{a},A])=-\int\nabla f\text{den}(A)$ for any $f\in
L^{\infty}(\mathbb{R}^{2})$ and $\mathcal{L}$-periodic. Hence
$\text{den}([\nabla_{a},A])=\nabla\text{den}[A]$. ∎

###### Proof of Proposition 3.4.

Since by Lemma 3.3, $\tau_{bh}$ is generated by the map $A\rightarrow
i[\pi_{b},A]$, the mt-invariance of $A$ implies that $[\pi_{b},A]=0$. (Though
$\pi_{b}$ does not leave $A$ invariant, $[\pi_{b},A]$ does.) This and (3.12)
yield that $\nabla\text{den}(A)=\text{den}([\pi_{b},A])=0$ and therefore den
$A$ is constant. ∎

######  Remark 3.6.

The argument above proving (3.12) establishes the intuitive fact that the
integral kernel of the operator $[\nabla_{a},A]$ acting on $\mathfrak{h}$ is
same as the integral kernel of this operator acting on $L^{2}(\Omega)$, which
is
$(\nabla_{ax}+\overline{\nabla_{ay}})A^{\prime}(x,y)=(\nabla_{x}+\nabla_{y})A^{\prime}(x,y)$
and consequently
$\text{den}([\nabla_{a},A])=(\nabla_{x}+\nabla_{y})A^{\prime}(x,y)|_{x=y}=\nabla\text{den}(A)$.

######  Lemma 3.7.

(3.4)-(3.5) have a solution $(\gamma_{b},a_{b})$, where $\gamma_{b}\geq 0$ and
$\gamma_{b}$ is an mt-invariant operator on the space $\mathfrak{h}$, if and
only if the fixed point problem

$\displaystyle\xi=(\int v)\text{den}(f_{T}(h_{a_{b}\mu}+\xi))(0),$ (3.15)

where $h_{a_{b}\mu}:=-\Delta_{a_{b}}-\mu$ and $\xi$ is a real number, has a
solution. (The operator $f_{T}(h_{a_{b}}+\xi)$ is well-defined and real since
$h_{a_{b}}$ is self-adjoint on the space $\mathfrak{h}$.)

We show in Appendix C that the fixed point problem (3.15) has a unique
solution, provided the first condition of Theorem 1.2 holds.

###### Proof.

Let $\gamma$ be an even, mt-invariant operator. Then
$\tilde{\gamma}=-i\nabla_{\\!\\!a_{b}}\gamma$ is an mt-invariant operator and
odd. Applying Proposition 3.4 to $\tilde{\gamma}=-i\nabla_{\\!\\!a_{b}}\gamma$
gives that $j(\gamma,a_{b})=0$ and therefore, since
$\operatorname{curl}^{*}\operatorname{curl}a_{b}=0$, the pair $(\gamma,a_{b})$
satisfies (3.5). Hence $(\gamma,a_{b})$ solves (3.4)-(3.5) if and only if
$\gamma$ solves

$\displaystyle\gamma=f_{T}(h_{\gamma a_{b}\mu}).$ (3.16)

Now, we solve (3.16) for magnetically translation invariant $\gamma$’s. We
treat this equation as a fixed point problem444If both the direct and exchange
self-interactions are dropped from $h_{\gamma a\mu}$, then the latter equation
gives $\gamma_{Tb}$ directly: $\gamma_{Tb}=f_{T}(h_{a_{b}\mu})$, where,
recall, $h_{a_{b}\mu}:=-\Delta_{a_{b}}-\mu$. . This problem simplifies
considerably since we dropped the exchange term, as in this case, $h_{\gamma
a_{b}\mu}$ becomes $h_{a_{b}\mu}+v*\rho$, where, recall,
$h_{a_{b}\mu}:=-\Delta_{a_{b}}-\mu$. Using this and applying den to (3.4)
gives the equation for $\rho=\text{den}\gamma$:

$\rho=\text{den}(f_{T}(h_{a_{b}\mu}+v*\rho)).$

Furthermore, by Proposition 3.4, $\rho=\text{den}\gamma$ and
$\text{den}(f_{T}(h_{a_{b}\mu}+v*\rho)$ are constant functions and therefore
$\xi=v*\rho=\int v\rho(0)$ is a real constant satisfying the fixed point
equation (3.15).

To summarize, we have shown that if an mt-invariant $\gamma$ solves (3.16)
(i.e. (3.4) with $a=a_{b}$), then $\xi=v*\text{den}\gamma=\int v\rho(0)$ is a
real constant solving (3.15).

Now, in the opposite direction, suppose that a real $\xi$ solves (3.15) and
define

$\displaystyle\gamma:=f_{T}(h_{a_{b}\mu}+\xi).$ (3.17)

Since $f_{T}>0$, $h_{a_{b}}$ is self-adjoint and $\xi$ is real, we have that
$\gamma\geq 0$. Since
$\xi=v*\text{den}(f_{T}(h_{a_{b}\mu}+\xi))=v*\rho_{\gamma}$, where, recall,
$\rho_{\gamma}\equiv\text{den}(\gamma)$, (3.17) becomes

$\displaystyle\gamma$ $\displaystyle=f_{T}(h_{a_{b}\mu}+v*\rho_{\gamma}).$
(3.18)

Hence $\gamma$ satisfies (3.16), i.e. (3.4) with $a=a_{b}$. ∎

Since, as we show in Appendix C, the fixed point problem (3.15) has a unique
solution, provided the potential $v$ satisfies the first condition of Theorem
1.2, we obtain an unique magnetic translation invariant solution of
(3.4)-(3.5), under the same condition.

The uniqueness part of Lemma 3.7 is strengthened and extended to the second
condition in the following

######  Lemma 3.8.

A solution of (3.4)-(3.5) is unique among pairs $(\gamma,a)$ with $\gamma$ mt-
invariant.

###### Proof.

First observe that, by Proposition 3.4, $\gamma(x,x)$ is constant and the term
$\mathrm{Re}(-i\nabla_{a_{b}}\gamma)(x,x)$ vanishes. We decompose
$a=a_{b}+a^{\prime}$, where $a^{\prime}$ is defined by this expression. Using
(3.5) and $\operatorname{curl}^{*}\operatorname{curl}a_{b}=0$, we see that

$\displaystyle\operatorname{curl}^{*}\operatorname{curl}a^{\prime}=-\gamma(0,0)a^{\prime}.$
(3.19)

Since $h_{a_{b}}+\xi$ is bounded below and $f_{T}$ is strictly positive and
monotonically increasing, we see that $\gamma=f_{T}(h_{a_{b}}+\xi)\geq c>0$.
Thus $\gamma(0,0)>0$. Multiplying both sides of (3.19) by $a^{\prime}$ and
integrating, we find

$\displaystyle\int|\operatorname{curl}a^{\prime}|^{2}+\gamma(0,0)\int|a^{\prime}|^{2}=0.$
(3.20)

Since $\gamma(0,0)>0$, this implies that $a^{\prime}=0$. Hence $a=a_{b}$ and
therefore (3.4) shows that $\gamma$ is a function of $-\Delta_{a_{b}}$ and
therefore mt-invariant. Hence, we can conclude uniqueness by Lemma 3.7 and the
proof is complete. ∎

Lemmas 3.7, 3.8 and C.1 of Appendix C imply Theorem 1.2, under the first
condition.

To prove Theorem 1.2 under the second condition, i.e. for
$\|v\|_{\infty}<\infty$, we use the variational approach. Let
$D^{s}_{\nu}=\big{\\{}\gamma\in I^{s,1},\ 0\leq\gamma=\gamma^{*}\leq 1,\
\gamma\text{ satisfies \eqref{even}},\\\ \mathrm{Tr}\gamma=\nu\big{\\}}$ (see
(1.1) and (1.37)). Recall, that $S(\gamma)$ is the entropy defined in
(1.65)-(1.66) and define the free energy functional

$\displaystyle
F_{1}(\gamma)=\mathrm{Tr}\big{(}h_{a_{b}}\gamma\big{)}+\frac{1}{2}\int\rho_{\gamma}(v*\rho_{\gamma})-TS(\gamma),$
(3.21)

on $D^{1}_{\nu}$, setting $F(\gamma)=-\infty$, if $S(\eta)=\infty$. (To
compare with (1.27)-(1.28),
$\int\rho_{\gamma}(v*\rho_{\gamma})=\mathrm{Tr}\big{(}(v*\rho_{\gamma})\gamma\big{)}$.)
We show below the following

######  Proposition 3.9.

Assume $v$ is $\mathcal{L}$-periodic and satisfies $\|v\|_{\infty}<\infty$.
Then the functional $F_{1}(\gamma)$ on $D^{1}_{\nu}$ has a minimizer;
minimizers of $F_{1}(\gamma)$ satisfy $0<\gamma<1$.

It follows then, by a special case of Theorem 2.2, that minimizers of
$F_{1}(\gamma)$ satisfy the Euler-Lagrange equation, which is equivalent to
(3.16). Using the latter equation, the solution could be bootstrapped from
$I^{1,1}$ to $I^{2,1}$, giving Theorem 1.2. $\Box$

###### Proof of Proposition 3.9.

First, we show that $F_{1}(\gamma)$ is coercive, namely, that

$\displaystyle F_{1}(\gamma)\geq\frac{1}{2}\|\gamma\|_{I^{1,1}}-C_{\nu,T}.$
(3.22)

Indeed, by the definition, $\|\gamma\|_{I^{1,1}}=\mathrm{Tr}(h_{a_{b}}\gamma)$
and the elementary estimate

$\displaystyle|\int\rho(v*\rho)|\leq\|v\|_{\infty}\|\rho\|_{1}^{2},$ (3.23)

we have

$\displaystyle
F_{1}(\gamma)\geq\frac{1}{2}\|\gamma\|_{I^{1,1}}+f(\gamma)-\|v\|_{\infty}\|\rho_{\gamma}\|_{1}^{2},$
(3.24)

where $f(\gamma):=\frac{1}{2}\|\gamma\|_{I^{1,1}}-TS(\gamma)$. We minimize the
functional $f(\gamma)$ on the r.h.s. on the set $I^{1,1}_{\nu}:=\\{\gamma\in
I^{1,1}:\mathrm{Tr}\gamma=\nu\\}$. Since $f(\gamma)$ is convex and the
constraint $\mathrm{Tr}\gamma=\nu$ is linear, each solution to the standard
Euler-Lagrange equation $df(\gamma)-\mu d\mathrm{Tr}\gamma=0$ (written in
terms of the Gâteaux derivatives), where $\mu$ is the Lagrange multiplier, is
a global minimizer. The latter equation is computed to be
$\frac{1}{2}h_{a_{b}}-T\ln\big{(}\frac{\gamma}{\mathbf{1}-\gamma}\big{)}-\mu\mathbf{1}=0$.
Solving this equation gives the minimizer

$\displaystyle\gamma_{\mu,T}=f_{T}(\frac{1}{2}h_{a_{b}}-\mu),$ (3.25)

for $\mu$ such that $\mathrm{Tr}\gamma_{\mu,T}=\nu$. By the inverse function
theorem, the latter equation has a solution, $\mu=\mu(T,\nu)$, for $\mu$. This
shows $-C_{\nu,T}:=\inf\\{f(\gamma):\ \gamma\in
I^{1,1}_{\nu}\\}=f(\gamma_{\mu,T})>-\infty$, with $\mu=\mu(T,\nu)$, which,
together with (3.24), implies (3.22).

Next, we show that $F_{1}(\gamma)$ is weak lower semi-continuous. To this end,
we pass from the positive, trace class operators $\gamma$ to the Hilbert-
Schmidt ones, $\kappa:=\sqrt{\gamma}$. Note that $\gamma\in I^{1,1},\gamma\geq
0\Longleftrightarrow\kappa:=\sqrt{\gamma}\in I^{1,2}$. Thus we consider
$F_{1}(\gamma)=:F^{\prime}(\kappa)$ on the space
$\tilde{D}^{1}_{\nu}:=\big{\\{}\kappa\in I^{1,2},\ 0\leq\kappa=\kappa^{*}\leq
1,\ \kappa\text{ satisfies \eqref{even}},\
\mathrm{Tr}\kappa^{2}=\nu\big{\\}}$. The first term on the r.h.s. of (3.21)
satisfies $\mathrm{Tr}(h_{a_{b}}\gamma)=\|\kappa\|_{I^{1,2}}^{2}$ and is
quadratic in $\kappa$. Hence, it is $\|\cdot\|_{I^{1,2}}$-weakly lower semi-
continuous. For the second term, we use the inequalities (3.23) and

$\displaystyle\|\rho_{\gamma^{\prime}}-\rho_{\gamma}\|_{1}\leq\|\kappa^{\prime}-\kappa\|_{I^{2}}(\|\kappa^{\prime}\|_{I^{2}}+\|\kappa\|_{I^{2}})$
(3.26)

to show that it is also lower semi-continuous. The third term on the r.h.s. of
(3.21), $-TS(\gamma)$, is lower semi-continuous, by Lemma A.4 of Appendix A.
Hence $F_{1}(\gamma)=:F^{\prime}(\kappa)$ is lower semi-continuous.

Finally, we observe that the set $\tilde{D}^{1}_{\nu}$ is closed in $I^{1,2}$
under the weak convergence.

With the results above, the proof of existence of a minimizer is standard. To
avoid repetitions, we refer to the second paragraph after Lemma 5.5, where
this is done in somewhat more complicated notation.

Now, we establish properties of minimizers $\gamma_{*}=\kappa_{*}^{2}$. By
(3.22), we have $S(\gamma_{*})=\mathrm{Tr}g(\gamma_{*})<\infty$. This and the
fact that $g(\gamma)\geq 0$ imply that $g(\gamma_{*})$ is trace class.

Finally, a simplified version of the proof of Lemma 5.6 shows that $0$ and $1$
are not eigenvalues of $\gamma_{*}$ and therefore, $0<\gamma_{*}<1$.∎

##  4\. Stability/instability of the normal states for small $T$ and $b$:
Proof of Propositions 1.7 and 1.8

###### Proof of Proposition 1.7.

Recall that $K_{Tb}=Tf(h_{x}/T,h_{y}/T)$, where
$f(u,v):=\frac{u+v}{\tanh(u)+\tanh(v)}$ and $h_{z}$ is the operator $h_{Tb}$,
defined in Proposition 1.5, acting on the variable $z$. By Lemma 4.1 below
$f(u,v)\geq 1$. (A weaker bound $f(u,v)\geq\frac{1}{4}$ which suffices for us
could be easily proved directly.) Hence

$\displaystyle K_{Tb}\geq T,$ (4.1)

which implies that $L_{Tb}\geq T-\|v\|_{\infty}$ and consequently, Proposition
1.7 follows. ∎

###### Proof of Proposition 1.8.

We use the Birman-Schwinger principle to show that $L_{Tb}$ has a negative
eigenvalue. Set $w^{2}=-v\geq 0$ so that $L_{Tb}=K_{Tb}-(w^{\sharp})^{2}$,
where, recall, $(w^{\sharp}\,\alpha)\,(x;y):=w(x-y)\alpha(x;y)$.

By the Birman-Schwinger principle, $L_{Tb}$ has a negative eigenvalue $-E$ if
and only if $G_{Tb}(E):=w^{\sharp}(K_{Tb}+E)^{-1}w^{\sharp}$ has the
eigenvalue $1$ for some $E>0$ (see e.g. [18]). By (4.1), we have
$G_{Tb}(E)\geq 0$ for all $E\geq 0$. Moreover, since $(K_{Tb}+E)^{-1}$ is
continuous and monotonically decreasing in $E\geq 0$ and vanishing as
$E\rightarrow\infty$, so is $G_{Tb}(E)$. Hence, it suffices to show that
$G_{Tb}:=G_{Tb}(0)$ satisfies the estimate $\|G_{Tb}\|>1$, which we now prove.

Recall that $K_{Tb}=Tf(h_{x}/T,h_{y}/T)$, where
$f(s,t):=\frac{s+t}{\tanh(s)+\tanh(t)}$ and $h_{z}$ is the operator $h_{Tb}$,
defined in Proposition 1.5, acting on the variable $z$. Since $h_{Tb}$
satisfies $h_{Tb}\geq-\mu^{\prime}$, for some $\mu^{\prime}>\mu$, it suffices
to consider $f(s,t)$ for $s,t\geq-\mu^{\prime}$. A simple estimate

$\displaystyle f(s,t)\lesssim 1+|s+t|,$ (4.2)

for $s,t\geq-\mu^{\prime}$, which follows from Lemma 4.1 below, yields
$K_{Tb}\lesssim T+|h_{x}+h_{y}|$. This implies the inequality

$G_{Tb}\geq w^{\sharp}(T+|h_{x}+h_{y}|)^{-1}w^{\sharp}\geq 0.$

Since we omit the exchange term $-v^{\sharp}\gamma$, the operator $h_{Tb}$ is
of the form $h_{a_{b}\mu}:=-\Delta_{a_{b}}+v*\rho_{\gamma_{Tb}}-\mu$. By
Proposition 3.4, $\rho_{\gamma_{Tb}}$ is a constant function and therefore
$\xi:=v*\rho=\int v\rho(0)$ is a real constant. Hence
$h_{a_{b}\mu}:=-\Delta_{a_{b}}+\xi-\mu$.

Since the gaps between the eigenvalues $\lambda_{n}=b(2n+1)$ of
$-\Delta_{a_{b}}$ on $\mathfrak{h}$ are equal to $b$, we can choose $m$ s.t.
$|\lambda_{m}+\xi-\mu|\lesssim b$.

Recall that $L_{Tb}$ acts on the space of the Hilbert-Schmidt operators which
can be identified through their integral kernels with
$\mathfrak{h}\otimes\mathfrak{h}$. Let $\phi_{m}$ be the normalized
eigenfunctions of $-\Delta_{a_{b}}$ corresponding to the eigenvalues
$\lambda_{m}=b(2m+1)$. We take
$u:=c(w^{\sharp})^{-1}(\phi_{m}\otimes\phi_{m})$, where
$c=\|(w^{\sharp})^{-1}(\phi_{m}\otimes\phi_{m})\|^{-1}$, so that $\|u\|=1$. By
analyzing $f(s,t)$, $s,t\geq-\mu^{\prime}$, separately in several domains, we
obtain

$\displaystyle\langle u,G_{Tb}u\rangle$
$\displaystyle\gtrsim(T+|\lambda_{m}+\xi-\mu|)^{-1}\|(w^{\sharp})^{-1}(\phi_{m}\otimes\phi_{m})\|^{-2}$
$\displaystyle\gtrsim(T+b)^{-1}\|(w^{\sharp})^{-1}(\phi_{m}\otimes\phi_{m})\|^{-2}.$
(4.3)

((4) also follows from the stronger, but more involved, Lemma 4.1 below.) Now,
write $\phi_{m}(x)=\sqrt{b}\phi_{m}^{0}(\sqrt{b}x)$, where $\phi_{m}^{0}(x)$
is independent of $b$. Furthermore, by the assumption on $v=-w^{2}$, we have
$w(x-y)\gtrsim(1+|x-y|)^{-\kappa/2},\kappa<2$. The last two relations,
together with the inequality $(a+b)^{\kappa}\leq
2^{\kappa}(a^{\kappa}+b^{\kappa})$, imply

$\|(w^{\sharp})^{-1}(\phi_{m}\otimes\phi_{m})\|^{2}\lesssim\int(1+|x-y|^{\kappa})|\sqrt{b}\phi_{m}^{0}(\sqrt{b}x)\sqrt{b}\phi_{m}^{0}(\sqrt{b}y))|^{2}dxdy.$

Changing the variables of integration as
$x^{\prime}=\sqrt{b}x,y^{\prime}=\sqrt{b}y$, we find
$\|(w^{\sharp})^{-1}(\phi_{m}\otimes\phi_{m})\|\lesssim b^{-\kappa/4}$, which
in turn gives $\langle
u,G_{Tb}u\rangle\gtrsim(T+b)^{-1}b^{\kappa/2}\rightarrow\infty$ as
$b\rightarrow 0,$ provided $T\lesssim b^{\sigma},\sigma>\kappa/2$.

Thus we have shown that $\|G_{Tb}\|$, or the largest eigenvalue of $G_{Tb}$,
can be made arbitrarily large if $T$ and $b$ are sufficiently small, which, by
the Birman-Schwinger principle, proves Proposition 1.8. ∎

Bounds on the function $f(s,t):=\frac{s+t}{\tanh(s)+\tanh(t)}$ used in the
proof above could be proved directly; they also follow from the following

######  Lemma 4.1.

The function $f(s,t):=\frac{s+t}{\tanh(s)+\tanh(t)}$ has the minimum $1$
achieved at $s=t=0$.

###### Proof.

To find minimum of $f$, we look for its critical points. We let
$g(s,t)=\tanh(s)+\tanh(t)$ and compute

$\displaystyle\nabla
f=\frac{1}{g}(1-f(s,t)\text{sech}^{2}(s),1-f(s,t)\text{sech}^{2}(t)).$ (4.4)

Setting $\nabla f=0$ yields that

$\displaystyle f(s,t)^{-1}=\text{sech}^{2}(s)\ \text{ and }\
f(s,t)^{-1}=\text{sech}^{2}(t).$ (4.5)

It follows that $\text{sech}^{2}(s)=\text{sech}^{2}(t)$ and therefore either
$s=t$ or $s=-t$. If $s=-t$, then $f(s,t)=\text{sech}^{-2}(s)$, which has a
single critical point - minimum - $s=0$. Hence in this case we have a single
critical point - minimum - at $s=-t=0$ and $f(0,0)=1$. If $s=t$, then (4.5)
becomes

$\displaystyle\tanh(s)=s\,\text{sech}^{2}(s),\text{ or equivalently, }\
\sinh(u)\cosh(s)=s.$ (4.6)

This is equivalent to $\sin(2s)=2s$ which implies that $s=0$. Hence the
minimum is reached at $s=0$ and $f(0,0)=1$. ∎

##  5\. The existence of the vortex lattices

In this section, we prove Theorem 1.4 on existence of the vortex lattice
solutions to the BdG equations with arbitrary but fixed lattice $\mathcal{L}$
and first Chern (vortex) number $c_{1}(\chi^{\mathcal{L}})=n\in\mathbb{Z}$,
i.e. the integer $n$ entering (1.31) (see (1.45) and (1.68)). Recall that
$b=\frac{2\pi n}{|\mathcal{L}|},\ n\in\mathbb{Z}$, see (1.31).

Recall that we drop the exchange self-interaction term $v^{\sharp}\gamma$. We
minimize the resulting energy for $\mathrm{Tr}(\gamma)$ fixed. Hence we omit
the term $-\mu\mathrm{Tr}(\gamma)$ in (1.27). Also, in this section, we
display the domain of integration $\Omega$ which is an arbitrary but fixed
fundamental cell of the lattice $\mathcal{L}$. Thus, with the notation
$h_{a}:=-\Delta_{a}$, the free energy functional $F_{T}(\eta,a)$ in (1.27)
becomes

$\displaystyle\mathcal{F}(\eta,a)=\mathrm{Tr}\big{(}h_{a}\gamma\big{)}$
$\displaystyle+\frac{1}{2}\int\rho_{\gamma}(v*\rho_{\gamma})$
$\displaystyle+\frac{1}{2}\mathrm{Tr}\big{(}\alpha^{*}v^{\sharp}\alpha\big{)}+\int_{\Omega}dx|\operatorname{curl}a|^{2}-TS(\eta),$
(5.1)

where, recall, the entropy $S(\eta)$ is defined in (1.65)-(1.66). The
functional $\mathcal{F}$ is defined on
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$, if $S(\eta)<\infty$.
Otherwise, we set $\mathcal{F}(\eta,a)=\infty$.

Theorem 1.4(i) follows from Theorem 2.2 and the following

######  Theorem 5.1.

Fix a lattice $\mathcal{L}$ and a Chern number $c_{1}(\chi^{\mathcal{L}})=n$.
Assume that $T>0$ and $\|v\|_{\infty}<\infty$. There exists a finite energy
minimizer
$(\eta_{*},a_{*})\in\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$ of the
functional $\mathcal{F}(\eta,a)$ on the set
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$. This minimizer has the
equivariance and the flux quantization properties, (1.44) and (1.45),
satisfies $0<\eta_{*}<1$ and is s.t. $g(\eta_{*})$ (see (1.65)) is trace
class. Furthermore, the minimizer $(\eta_{*},a_{*})$ can be chosen to be even,
i.e. satisfying (1.46).

###### Proof of Theorem 5.1.

We will use standard minimization techniques proving that
$\mathcal{F}(\eta,a)$ is coercive and weakly lower semi-continuous, and
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$ weakly closed.

Part 1: coercivity. The main result of this step is the following proposition:

######  Proposition 5.2.

Let $T>0$, $e:=a-a_{b}$, $\gamma\in I^{1,1}$ and $\alpha\in I^{1,2}$, with
$\mathrm{Tr}\gamma=\nu$ and $0\leq\eta\leq\mathbf{1}$. Then

$\displaystyle\mathcal{F}(\eta,a_{b}+e)$ $\displaystyle\geq
c^{\prime}[\|\gamma\|_{I^{1,1}}/\nu]^{r}+c\|e\|_{H^{1}}^{2}-C$ (5.2)
$\displaystyle\geq\frac{1}{4}c^{\prime}[(\|\gamma\|_{I^{1,1}}+\|\alpha\|_{I^{1,2}}^{2})/\nu]^{r}+c\|e\|_{H^{1}}^{2}-C,$
(5.3)

for any $0<r<1$ and for suitable constants $c^{\prime},c,C>0$, with
$c^{\prime},c>0$ independent of $\nu$, $T,v$ and $\mathcal{L}$ and $C$
depending on $\nu$, $T,\|v\|_{\infty}$ and $|\mathcal{L}|$.

###### Proof of Proposition 5.2.

We begin with estimating the entropy term $-TS(\eta)$ (cf. [20]). Recall
definition (1.65)-(1.66) of $S(\eta)$, which we reproduce here

$\displaystyle S(\eta):=\mathrm{Tr}(s(\eta))=\mathrm{Tr}(g(\eta)),$ (5.4)
$\displaystyle g(\eta):=-\eta\ln\eta-(1-\eta)\ln(1-\eta),\
s(\eta):=-2\eta\ln\eta.$ (5.5)

Note that since $s(\eta)\geq 0$ for $0\leq\eta\leq\mathbf{1}$, we have that
$S(\eta)\geq 0$. We define the relative entropy

$\displaystyle S(A|B)=\mathrm{Tr}(s(A|B)),\quad$ $\displaystyle s(A|B):=A(\ln
A-\ln B).$ (5.6)

We define the diagonal operator-matrix $\eta_{0}$ and recall the off-diagonal
one $\phi$:

$\displaystyle\eta_{0}:=\left(\begin{array}[]{cc}\gamma&0\\\
0&\mathbf{1}-\bar{\gamma}\end{array}\right),\
\phi(\beta):=\left(\begin{array}[]{cc}0&\beta\\\
\beta^{*}&0\end{array}\right).$ (5.11)

######  Lemma 5.3 (cf. [13]).

We have for $\eta=\eta_{0}+\phi(\alpha)$,

$\displaystyle S(\eta)=S(\eta_{0})-S(\eta|\eta_{0})\leq S(\eta_{0}).$ (5.12)

###### Proof.

We note that for $\eta:=\eta_{0}+\phi(\alpha)$,

$\displaystyle\eta\ln\eta$
$\displaystyle-\eta_{0}\ln\eta_{0}=\eta\ln\eta-\eta\ln\eta_{0}+\eta\ln\eta_{0}-\eta_{0}\ln\eta_{0}$
(5.13) $\displaystyle=s(\eta,\eta_{0})+(\eta-\eta_{0})\ln\eta_{0}$ (5.14)
$\displaystyle=s(\eta,\eta_{0})+\phi(\alpha)\ln\eta_{0}.$ (5.15)

Since the last term, $\phi(\alpha)\ln\eta_{0}$, in (5.15) is off-diagonal and
therefore has zero trace, the first equation in (5.12) follows.

The inequality in (5.12) follows from Klein’s inequality and the fact that
$\mathrm{Tr}\eta=\mathrm{Tr}\eta_{0}$.∎

With the definitions (5.4)-(5.5) and (5.11), we have that (5.12) implies

$\displaystyle S(\eta)\leq S(\eta_{0})=\mathrm{Tr}(g(\gamma)),\ g(x)=-[x\ln
x+(1-x)\ln(1-x)].$ (5.16)

Next, we estimate the second ($\alpha$-) term on the r.h.s. of (5). To this
end, we bound $\alpha$ by $\gamma$ via the constraint $0\leq\eta\leq 1$ using
the following result (see [3] and references therein):

######  Lemma 5.4.

The constraint $0\leq\eta\leq 1$ implies that

  1. (1)

$\alpha^{*}\alpha\leq\bar{\gamma}(1-\bar{\gamma})$ and
$\alpha\alpha^{*}\leq\gamma(1-\gamma)$.

  2. (2)

$\mathrm{Tr}(M\alpha\alpha^{*}M^{*})\leq\|M\gamma M\|_{1}$ for any operator
$M$.

###### Proof.

Since $0\leq\eta\leq 1$, then $0\leq 1-\eta\leq 1$ as well and therefore

$0\leq\eta(1-\eta)\leq 1.$

From (1.4), we see that the 1,1-entry of $\eta(1-\eta)$ is
$0\leq\gamma(1-\gamma)-\alpha\alpha^{*}$. By considering the 2,2-entry, we
have that $\alpha^{*}\alpha\leq\bar{\gamma}(1-\bar{\gamma})$. Finally, since
$1-\gamma\leq 1$, we see that $M\alpha\alpha^{*}M^{*}\leq M\gamma M^{*}$,
which completes the proof. ∎

Since $v$ is bounded, Lemma 5.4(1) gives

$\displaystyle|\mathrm{Tr}(\alpha^{*}v^{\sharp}\alpha)|\leq\|v\|_{\infty}\mathrm{Tr}(\alpha^{*}\alpha)\leq\|v\|_{\infty}\mathrm{Tr}(\gamma).$
(5.17)

Now, using estimates (5.16) and (5.17) in expression (5) and introducing the
notation $S(\gamma):=\mathrm{Tr}(g(\gamma))$, we find

$\displaystyle\mathcal{F}(\eta,a)\geq\mathrm{Tr}\big{(}h_{a}\gamma\big{)}$
$\displaystyle+\frac{1}{2}\int\rho_{\gamma}(v*\rho_{\gamma})$
$\displaystyle+\int_{\Omega}|\operatorname{curl}a|^{2}-TS(\gamma)-\|v\|_{\infty}\mathrm{Tr}(\gamma).$
(5.18)

The r.h.s. depends only on $\gamma$ and $a=:a_{b}+e$ with the only constraint
$\mathrm{Tr}\gamma=\nu$ and we estimate it next.

Next, we pass from the vector potential $a$ to $e:=a-a_{b}$. Using
$a=a_{b}+e\in\vec{\mathfrak{h}}^{1}$, $\operatorname{curl}a_{b}=b$ and
$\int_{\Omega}\operatorname{curl}e=0$ (see (1.35)-(1.36)), we compute

$\displaystyle\int_{\Omega}|\operatorname{curl}a|^{2}=$
$\displaystyle\int_{\Omega}|\operatorname{curl}e|^{2}+b^{2}|\Omega|.$ (5.19)

Since $\int_{\Omega}e=0$ and $\operatorname{div}e=0$, the Poincaré’s
inequality shows that, for some $c>0$,

$\displaystyle
2c\|e\|_{H^{1}}^{2}\leq\int_{\Omega}|\operatorname{curl}e|^{2}.$ (5.20)

Now, we estimate $\mathrm{Tr}(h_{a}\gamma)$, where, recall, $a=a_{b}+e$. Using
$\operatorname{div}e=0$, we write

$\displaystyle\mathrm{Tr}(h_{a}\gamma)$
$\displaystyle=\mathrm{Tr}(h_{a_{b}}\gamma)+2i\mathrm{Tr}(e\cdot\nabla_{a_{b}}\gamma)+\mathrm{Tr}(|e|^{2}\gamma)$
$\displaystyle\geq$
$\displaystyle(1-\epsilon)\mathrm{Tr}(h_{a_{b}}\gamma)-(\epsilon^{-1}-1)\mathrm{Tr}(|e|^{2}\gamma),$
(5.21)

for any $\epsilon>0$. For the last term, we claim, for any $r\in(0,1)$ and
some constant $C$, the estimate

$\displaystyle 0\leq\mathrm{Tr}(|e|^{2}\gamma)\leq
C\|\gamma\|_{I^{1,1}}^{1-r}(\mathrm{Tr}\gamma)^{r}\|e\|_{H^{1}}^{2},$ (5.22)

where we used that $\mathrm{Tr}(h_{a_{b}}\gamma)=\|\gamma\|_{I^{1,1}}$. We
prove this estimate for $r=1/2$, which suffices for us. For general
$r\in(0,1)$, see Lemma D.3 of Appendix D. Recall the definition
$M_{b}:=\sqrt{h_{a_{b}}}$. We use relative bound (D.3) of Appendix D to find

$\displaystyle
0\leq\mathrm{Tr}(|e|^{2}\gamma)\lesssim\||e|^{2}M_{b}^{-s}\|\|M_{b}^{s}\kappa\|_{I^{2}}\|\kappa\|_{I^{2}}\lesssim\|e\|_{H^{v}}^{2}\|\kappa\|_{I^{s,2}}\|\kappa\|_{I^{0,2}}$
(5.23)

with $s>2(1-v)$. The last estimate gives (5.22) with $r=1/2$.

Let $\xi(\gamma):=c/(C(\epsilon^{-1}-1)\|\gamma\|_{I^{1,1}}^{1-r}\nu^{r})$,
where, recall, $\nu=\mathrm{Tr}\gamma$ and $r\in(0,1)$, for the constant $C$
coming from (5.22), set $\epsilon=\frac{1}{4}$. Then, (5.21) and (5.22), the
inequality $\mathrm{Tr}(h_{a}\gamma)\geq\delta\mathrm{Tr}(h_{a}\gamma),$ for
$\delta\leq 1$, and the relation
$\|\gamma\|_{I^{1,1}}=\mathrm{Tr}(h_{a_{b}}\gamma)$ imply

$\displaystyle\mathrm{Tr}(h_{a}\gamma)\geq$
$\displaystyle\delta[\frac{3}{4}\|\gamma\|_{I^{1,1}}-c\xi(\gamma)^{-1}\|e\|_{H^{1}}^{2}].$
(5.24)

Now, taking $\delta=\xi(\gamma)$ and using that
$\xi(\gamma)\mathrm{Tr}(h_{a_{b}}\gamma)=c/(3C)(\|\gamma\|_{I^{1,1}}/\nu)^{r}$
gives

$\mathrm{Tr}(h_{a}\gamma)\geq
2c^{\prime}[\|\gamma\|_{I^{1,1}}/\nu]^{r}-c\|e\|_{H^{1}}^{2},$

where $c^{\prime}:=c/(8C)$. The latter equation implies, in turn,

$\displaystyle\mathrm{Tr}(h_{a}\gamma)-TS(\gamma)\geq
c^{\prime}E^{\prime}(\gamma)+c^{\prime}[\|\gamma\|_{I^{1,1}}/\nu]^{r}-c\|e\|_{H^{1}}^{2},$
(5.25) $\displaystyle
E^{\prime}(\gamma):=[\|\gamma\|_{I^{1,1}}/\nu]^{r}-TS(\gamma).$ (5.26)

To estimate $E^{\prime}(\gamma)$ from below, let $e_{k}$ be an orthonormal
eigenbasis of $h_{a_{b}}$ and $\lambda_{k}$ be the corresponding eigenvalues.
Using that
$\|\gamma\|_{I^{1,1}}=\mathrm{Tr}(h_{a_{b}}\gamma)=\sum_{k}\lambda_{k}\langle
e_{k},\gamma e_{k}\rangle$, regarding $\langle e_{k},\gamma
e_{k}\rangle/\mathrm{Tr}\gamma$ as a probability measure and applying Jensen’s
inequality (for $0<r<1$ so that $x^{r}$ is concave), we find

$\displaystyle(\mathrm{Tr}(h_{a_{b}}\gamma)/\mathrm{Tr}\gamma)^{r}=$
$\displaystyle\big{(}\sum_{k}\lambda_{k}\langle e_{k},\gamma
e_{k}\rangle/\mathrm{Tr}\gamma\big{)}^{r}$ (5.27)
$\displaystyle\geq\sum_{k}\lambda_{k}^{r}\langle e_{k},\gamma
e_{k}\rangle/\mathrm{Tr}\gamma.$ (5.28)

Since $\sum_{k}\lambda_{k}^{r}\langle e_{k},\gamma
e_{k}\rangle=\mathrm{Tr}(h_{a_{b}}^{r}\gamma)$, we have

$\displaystyle(\mathrm{Tr}(h_{a_{b}}\gamma)/\mathrm{Tr}\gamma)^{r}$
$\displaystyle\geq\mathrm{Tr}(h_{a_{b}}^{r}\gamma)/\mathrm{Tr}\gamma.$ (5.29)

Let $I^{1,1}_{\nu}:=\\{\gamma\in I^{1,1}:\mathrm{Tr}\gamma=\nu\\}$. The above
estimate, together with (5.26), restricted to the se $I^{1,1}_{\nu}$, gives

$\displaystyle E^{\prime}(\gamma)\geq$
$\displaystyle\mathrm{Tr}(h_{a_{b}}^{r}\gamma)/\nu-
TS(\gamma)=:E_{\nu}(\gamma).$ (5.30)

We minimize the functional $E_{\nu}(\gamma)$ on the r.h.s. on $I^{1,1}_{\nu}$.
Since $E_{\nu}(\gamma)$ is convex and the constraint $\mathrm{Tr}\gamma=\nu$
is linear, each solution to the standard Euler-Lagrange equation
$dE_{\nu}(\gamma)-\mu^{\prime}d\mathrm{Tr}\gamma=0$ (written in terms of the
Gâteaux derivatives), where $\mu^{\prime}$ is the Lagrange multiplier, is a
global minimizer. The latter equation is computed to be
$h_{a_{b}}^{r}/\nu-T\ln\big{(}\frac{\gamma}{\mathbf{1}-\gamma}\big{)}-\mu^{\prime}\mathbf{1}=0$.
Solving this equation and setting $\mu^{\prime}=\mu/\nu$ gives the minimizer

$\displaystyle\gamma_{\mu,T\nu}=f_{T}((h_{a_{b}}^{r}-\mu)/\nu)$ (5.31)

for $\mu$ such that $\mathrm{Tr}\gamma_{\mu,T\nu}=\nu$. By the implicit
function theorem, the latter equation has a solution, $\mu=\mu(T,\nu)$, for
$\mu$. This shows $-C_{\nu,T}/c^{\prime}:=\inf\\{E_{\nu}(\gamma):\gamma\in
I^{1,1}_{\nu}\\}=E_{\nu}(\gamma_{\mu,T\nu})>-\infty$, with $\mu=\mu(T,\nu)$,
which, together with (5.30), implies

$\displaystyle\inf_{\gamma\in
I^{1,1}_{\nu}}E^{\prime}(\gamma)\geq\inf_{\gamma\in
I^{1,1}_{\nu}}E_{\nu}(\gamma)=-C_{\nu,T}/c^{\prime}>-\infty,$ (5.32)

which, together with (5.25), implies

$\displaystyle\mathrm{Tr}(h_{a}\gamma)-TS(\gamma)\geq$ $\displaystyle
c^{\prime}\ [\|\gamma\|_{I^{1,1}}/\nu]^{r}-C_{\nu,T}-c\|e\|_{H^{1}}^{2},$
(5.33)

with, recall, $c^{\prime}:=c/(8C)$. Estimates (5), (5.19), (5.20), (3.23) and
(5.33), restricted to $I^{1,1}_{\nu},\nu\geq 1,$ give bound (5.2). This bound
and Lemma 5.4(2) imply bound (5.3). This completes the proof of Proposition
5.2. ∎

We continue with the proof of Theorem 5.1. Eq. (5.33) shows that it suffices
to consider $\gamma$ with $S(\gamma)<\infty$ and therefore, by (5.16), $\eta$
with

$\displaystyle S(\eta)<\infty.$ (5.34)

Part 2: weak lower semi-continuity. We pass from the positive trace class
operator $\gamma$ to the Hilbert-Schmidt one, $\kappa:=\sqrt{\gamma}$. Note
that $\gamma\in I^{1,1},\gamma\geq
0\Longleftrightarrow\kappa:=\sqrt{\gamma}\in I^{1,2}$. Now, instead of the
free energy functional (5), we consider the equivalent functional

$\displaystyle F(\kappa,\alpha,e)$
$\displaystyle:=\mathcal{F}(\eta,a_{b}+e)\big{|}_{\gamma=\kappa^{2}}-b^{2}|\Omega|$
(5.35) $\displaystyle=\mathrm{Tr}\big{(}\kappa
h_{a_{b}+e}\kappa\big{)}+\frac{1}{2}\int\rho_{\gamma}(v*\rho_{\gamma})$
$\displaystyle\qquad+\frac{1}{2}\mathrm{Tr}\big{(}\alpha^{*}v^{\sharp}\alpha\big{)}+\int_{\Omega}|\operatorname{curl}e|^{2}-TS(\eta)\big{|}_{\gamma=\kappa^{2}}$
(5.36)

on the space $I^{1,2}\times I^{1,2}\times\vec{H}^{1}$ with the norm
$\|(\kappa,\alpha,e)\|_{(1)}:=\|\kappa\|_{I^{1,2}}+\|\alpha\|_{I^{1,2}}+\|e\|_{H^{1}}$
and with the side conditions $0\leq\eta\big{|}_{\gamma=\kappa^{2}}\leq 1$ and
$\mathrm{Tr}\kappa^{2}=\nu$. We will keep the notation $\mathcal{D}^{1}_{\nu}$
for $I^{1,2}\times I^{1,2}$ with these side conditions.

By Proposition 5.2, we find that, for $\mathrm{Tr}\gamma=\nu$ and
$0\leq\eta\leq\mathbf{1}$,

$\displaystyle F(\kappa,\alpha,e)$
$\displaystyle\geq\frac{1}{4}c^{\prime}\min([\|(\kappa,\alpha)\|_{1}^{2}/\nu]^{r},\|(\kappa,\alpha)\|_{1}^{2})+c\|e\|_{H^{1}}^{2}-C,$
(5.37)

where
$\|(\kappa,\alpha)\|_{1}^{2}:=\|\kappa\|_{I^{1,2}}^{2}+\|\alpha\|_{I^{1,2}}^{2}$,
and for suitable constants $c,C>0$, with $C$ depending on $\nu$,
$T,\|v\|_{\infty}$ and $|\mathcal{L}|$.

######  Lemma 5.5.

The functional $F(\kappa,\alpha,e)$ is weakly lower semi-continuous in
$I^{1,2}\times I^{1,2}\times\vec{\mathfrak{h}}^{1}$.

###### Proof.

We study the functional $F(\kappa,\alpha,e)$ term by term. For the first term
on the r.h.s. of (5), with $a=a_{b}+e$ and $h_{a}:=-\Delta_{a}$, we write

$\displaystyle\mathrm{Tr}(h_{a}\gamma)=\mathrm{Tr}((-\Delta_{a_{b}})\gamma)+2i\mathrm{Tr}(e\cdot\nabla_{a_{b}}\gamma)+\mathrm{Tr}(|e|^{2}\gamma).$
(5.38)

Since the first term on the r.h.s. of (5.38) satisfies
$\mathrm{Tr}((-\Delta_{a_{b}})\gamma)=\|\kappa\|_{I^{1,2}}^{2}$ and is
quadratic in $\kappa$, it is $\|\cdot\|_{I^{1,2}}$-weakly lower semi-
continuous.

For the second term on the r.h.s. of (5.38), we let
$e,e^{\prime}\in\vec{H}^{1}$ and estimate the difference
$\mathrm{Tr}(e\cdot\nabla_{a_{b}}\gamma)-\mathrm{Tr}(e^{\prime}\cdot\nabla_{a_{b}}\gamma^{\prime})$.
We write

$\displaystyle\mathrm{Tr}(e\cdot\nabla_{a_{b}}\gamma)-\mathrm{Tr}(e^{\prime}\cdot\nabla_{a_{b}}\gamma^{\prime})$
$\displaystyle=\mathrm{Tr}((e-e^{\prime})\cdot\nabla_{a_{b}}\gamma)-\mathrm{Tr}(e^{\prime}\cdot\nabla_{a_{b}}(\gamma-\gamma^{\prime})).$
(5.39)

For the first term on the r.h.s., letting $c:=e-e^{\prime}$, we claim that

$\displaystyle|\mathrm{Tr}(c\cdot\nabla_{a_{b}}\gamma)|\lesssim\|c\|_{H^{s}}\|\gamma\|_{I^{1,1}},\
s<1.$ (5.40)

To prove this inequality, we recall that $M_{b}:=\sqrt{-\Delta_{a_{b}}}$ and
write
$\mathrm{Tr}(c\cdot\nabla_{a_{b}}\gamma)=\mathrm{Tr}(M_{b}^{-1}c\cdot\nabla_{a_{b}}M_{b}^{-1}M_{b}\gamma
M_{b})$ and use a standard trace class estimate to obtain
$|\mathrm{Tr}(c\cdot\nabla_{a_{b}}\gamma)|\lesssim\|M_{b}^{-1}c\cdot\nabla_{a_{b}}M_{b}^{-1}\|\|M_{b}\gamma
M_{b}\|_{I^{0,1}}$. Next, we use the boundedness of
$\nabla_{a_{b}}M_{b}^{-1}$, the relative bound
$\|M_{b}^{-1}c\|\lesssim\|c\|_{H^{s}},s<1$ (see (D.1) of Appendix D), and the
relation $\|M_{b}\gamma M_{b}\|_{I^{0,1}}=\|\gamma\|_{I^{1,1}}$ to find
(5.40). (Recall that $\|\kappa\|_{I^{s,2}}=\|\gamma\|_{I^{s,1}}^{1/2}$.)

For $\gamma$ and $\gamma^{\prime}$ non-negative, we claim the following
estimate for the second term on the r.h.s. of (5) with $c=e^{\prime}$:

$\displaystyle|\mathrm{Tr}(c\cdot\nabla_{a_{b}}$
$\displaystyle(\gamma-\gamma^{\prime}))|$
$\displaystyle\lesssim\|c\|_{H^{t}}(\|\kappa\|_{I^{1,2}}+\|\kappa^{\prime}\|_{I^{1,2}})\|\kappa-\kappa^{\prime}\|_{I^{s,2}},\
s,t<1,$ (5.41)

where $\kappa:=\gamma^{1/2}$ and $\kappa^{\prime}:=(\gamma^{\prime})^{1/2}$.
To prove this estimate, we write
$\gamma=\kappa^{2},\gamma^{\prime}={\kappa^{\prime}}^{2}$ to expand

$\displaystyle(\gamma-\gamma^{\prime})$
$\displaystyle=\kappa(\kappa-\kappa^{\prime})+(\kappa-\kappa^{\prime})\kappa^{\prime}.$
(5.42)

Now, we use the boundedness of $\nabla_{a_{b}}M_{b}^{-1}$ and the relative
bound $\|M_{b}^{-s}c\|\lesssim\|c\|_{H^{t}},s,t<1$ (see (D.1) of Appendix D),
and the relations $\|M_{b}^{s}\kappa\|_{I^{0,2}}=\|\kappa
M_{b}^{s}\|_{I^{0,2}}=\|\kappa\|_{I^{s,2}}$, to find

$\displaystyle|\mathrm{Tr}(c\cdot\nabla_{a_{b}}\kappa(\kappa-\kappa^{\prime}))|$
$\displaystyle=|\mathrm{Tr}(M_{b}^{-s}c\cdot\nabla_{a_{b}}M_{b}^{-1}M_{b}\kappa(\kappa-\kappa^{\prime})M_{b}^{s})|$
$\displaystyle\lesssim\|c\|_{H^{t}}\|\kappa\|_{I^{1,2}}\|\kappa-\kappa^{\prime}\|_{I^{s,2}},\
s,t<1.$ (5.43)

For the second term on the r.h.s. of (5.42), we use the relative bound
$\|M_{b}^{-1}c\cdot\nabla_{a_{b}}M_{b}^{-s}\|\lesssim\|c\|_{H^{1}},$ for any
$s<1$ (see (D.2) of Appendix D) to find

$\displaystyle|\mathrm{Tr}(c\cdot\nabla_{a_{b}}(\kappa-\kappa^{\prime})\kappa^{\prime})|$
$\displaystyle=|\mathrm{Tr}(M_{b}^{-1}c\cdot\nabla_{a_{b}}M_{b}^{-s}M_{b}^{s}(\kappa-\kappa^{\prime})\kappa^{\prime})M_{b})|$
$\displaystyle\lesssim\|c\|_{H^{1}}\|\kappa^{\prime}\|_{I^{1,2}}\|\kappa-\kappa^{\prime}\|_{I^{s,2}}.$
(5.44)

The last two estimates yield (5).

Applying (5.40) and (5) to the terms on the r.h.s. of (5), we find, for
$3/4<s<1$,

$\displaystyle|\mathrm{Tr}(c\cdot\nabla_{a_{b}}\gamma)$
$\displaystyle-\mathrm{Tr}(c^{\prime}\cdot\nabla_{a_{b}}\gamma^{\prime})|\lesssim\|c-c^{\prime}\|_{\vec{\mathfrak{h}}^{s}}\|\gamma\|_{I^{1,1}}$
$\displaystyle\quad\quad+\|c^{\prime}\|_{\vec{H}^{s}}(\|\kappa\|_{I^{1,2}}+\|\kappa^{\prime}\|_{I^{1,2}})\|\kappa-\kappa^{\prime}\|_{I^{s,2}},\
s,t<1,$ (5.45)

where $\kappa:=\gamma^{1/2}$ and $\kappa:=(\gamma^{\prime})^{1/2}$.

Now, we use a standard result for Sobolev spaces, $\vec{H}^{s^{\prime}}$ is
compactly embedded in $\vec{H}^{s}$, for any $s^{\prime}>s$, and, perhaps, a
less standard one, that $I^{s^{\prime},2}$ is compactly embedded in $I^{s,2}$,
for any $s^{\prime}>s$. (One shows the latter fact by passing to the integral
kernels and using a standard Sobolev embedding result.)

Now, let $\\{(\kappa_{n},\alpha_{n},e_{n})\\}$ be a weakly convergent sequence
in $I^{1,2}\times I^{1,2}\times\vec{\mathfrak{h}}^{1}$ and denote its limit by
$(\kappa_{*},\alpha_{*},a_{*})$. Then, by above, it converges strongly in
$\mathcal{D}^{s}_{\nu}\times\vec{\mathfrak{h}}^{s},\ s<1$. Hence, we have by
(5),

$\displaystyle|\mathrm{Tr}(e_{n}\cdot\nabla_{a_{b}}\gamma_{n})$
$\displaystyle-\mathrm{Tr}(e_{*}\cdot\nabla_{a_{b}}\gamma_{*})|\rightarrow 0,\
n\rightarrow\infty,$ (5.46)

where, as usual, $\gamma_{n}=\kappa_{n}^{2}$ and $\gamma_{*}=\kappa_{*}^{2}$.

Finally, we consider the difference
$\mathrm{Tr}(|e|^{2}\gamma)-\mathrm{Tr}(|e^{\prime}|^{2}\gamma^{\prime})$ due
to the last term in (5.38). We decompose

$\displaystyle\mathrm{Tr}(|e|^{2}\gamma)-$
$\displaystyle\mathrm{Tr}(|e^{\prime}|^{2}\gamma^{\prime})=\mathrm{Tr}(|e|^{2}(\gamma-\gamma^{\prime}))+\mathrm{Tr}((|e|^{2}-|e^{\prime}|^{2})\gamma^{\prime}).$
(5.47)

For the first term on the r.h.s. we claim the following estimate

$\displaystyle|\mathrm{Tr}(|e|^{2}(\gamma-\gamma^{\prime}))|$
$\displaystyle\lesssim\|e\|_{H^{t}}^{2}(\|\kappa\|_{I^{1,2}}+\|\kappa^{\prime}\|_{I^{1,2}})\|\kappa-\kappa^{\prime}\|_{I^{s,2}},\
s<1,$ (5.48)

where $\kappa:=\gamma^{1/2}$ and $\kappa^{\prime}:=(\gamma^{\prime})^{1/2}$.
We use again (5.42) and the relative bound
$\|M_{b}^{-s}|e|^{2}M_{b}^{-t}\|\lesssim\|e\|_{H^{r}},s+t>2(1-r),$ (see (D.3)
of Appendix D) to find, similarly to (5) and (5),

$\displaystyle|\mathrm{Tr}(|e|^{2}(\gamma-\gamma^{\prime}))|$
$\displaystyle=|\mathrm{Tr}(|e|^{2}(\kappa(\kappa-\kappa^{\prime})+(\kappa-\kappa^{\prime})\kappa^{\prime})|$
(5.49)
$\displaystyle\leq|\mathrm{Tr}(M_{b}^{-s}|e|^{2}M_{b}^{-1}M_{b}\kappa(\kappa-\kappa^{\prime})M_{b}^{s})|$
$\displaystyle\qquad+|\mathrm{Tr}(M_{b}^{-1}|e|^{2}M_{b}^{-s}M_{b}^{s}(\kappa-\kappa^{\prime})\kappa^{\prime}M_{b})|$
$\displaystyle\lesssim\|e\|_{H^{t}}(\|\kappa\|_{I^{1,2}}+\|\kappa^{\prime}\|_{I^{1,2}})\|\kappa-\kappa^{\prime}\|_{I^{s,2}},\
s<1,$ (5.50)

which gives (5.48).

Finally, similarly to (5.40), we find for the second term on the r.h.s. of
(5.47),

$\displaystyle|\mathrm{Tr}((|e|^{2}-|e^{\prime}|^{2})\gamma^{\prime})|\lesssim(\|e\|_{H^{s}}+\|e^{\prime}\|_{H^{s}})\|e-e^{\prime}\|_{H^{s}}\|\gamma\|_{I^{1,1}},\
s<1.$ (5.51)

Now, Eqs (5.47), (5.48) and (5.51) imply

$\displaystyle|\mathrm{Tr}(|e_{n}|^{2}\gamma_{n})-\mathrm{Tr}(|e_{*}|^{2}\gamma_{*})|$
$\displaystyle\lesssim\|e_{n}\|_{H^{1}}^{2}(\|\kappa_{n}\|_{I^{1,2}}+\|\kappa_{*}\|_{I^{1,2}})\|\kappa_{n}-\kappa_{*}\|_{I^{s,2}}$
$\displaystyle\quad+\|\gamma_{*}\|_{I^{1,1}}(\|e_{n}\|_{H^{s}}+\|e_{*}\|_{H^{s}})\|e_{n}-e_{*}\|_{H^{s}},$
(5.52)

for $s<1$, and therefore, as above the r.h.s. converges to $0$. Hence, the
first term on the r.h.s. of (5) is is weakly lower semi-continuous.

For the second term on the r.h.s. of (5), as in the proof of Proposition 3.9,
we use the inequalities (3.23) and (3.26) to show that it is weakly lower
semi-continuous in $I^{1,2}$. The third term on the r.h.s. of (5) is quadratic
in $\alpha$ and therefore it is continuous in $I^{1,2}\times
I^{1,2}\times\vec{\mathfrak{h}}^{1}$ since $v\in L^{\infty}$. It follows that
it is weakly lower semi-continuous in $I^{1,2}\times
I^{1,2}\times\vec{\mathfrak{h}}^{1}$.

The third term on the r.h.s. of (5),
$\int_{\Omega}|\operatorname{curl}e|^{2}$, is clearly convex. So its norm
lower semi-continuity is equivalent to weak semi-continuity. Since it is
clearly $\vec{\mathfrak{h}}^{1}-$norm continuous, it is
$\vec{\mathfrak{h}}^{1}-$weakly lower semi-continuous.

Hence all the terms on the r.h.s. of the expression (5) for
$F(\kappa,\alpha,e)$, save $-TS(\eta)$, are lower semi-continuous under the
convergence indicated. The lower semi-continuity of the latter term is proven
in Lemma A.4 of Appendix A. Hence $F(\kappa,\alpha,e)$ is lower semi-
continuous, which completes the proof of Lemma 5.5.∎

Finally, we observe that the set
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$ is closed in
$I^{1,2}\times I^{1,2}\times\vec{\mathfrak{h}}^{1}$ under the weak
convergence.

We continue with the proof of Theorem 5.1. With the results above, the proof
of existence of a minimizer is standard. Let
$\\{(\kappa_{n},\alpha_{n},e_{n})\\}$ be a weakly convergent sequence in
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$ which is a minimizing
sequence for $F(\kappa,\alpha,e)$. By (5.37), the norm
$\|(\kappa_{n},\alpha_{n},e_{n})\|_{(1)}(:=\|\kappa_{n}\|_{I^{1,2}}+\|\alpha_{n}\|_{I^{1,2}}+\|e_{n}\|_{\vec{\mathfrak{h}}^{1}})$
(see the line after (5)) is bounded uniformly in $n$. By Sobolev-type
embedding theorems, $(\kappa_{n},\alpha_{n},e_{n})$ converges strongly in
$\mathcal{D}^{s}_{\nu}\times\vec{\mathfrak{h}}^{s}$ for any $s<1$ and by the
Banach-Alaoglu theorem, $(\kappa_{n},\alpha_{n},e_{n})$ converges weakly in
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}^{1}$. Denote the limit by
$(\kappa_{*},\alpha_{*},e_{*})$. Since, by Lemma 5.5, $F$ is lower semi-
continuous, we have

$\displaystyle\liminf_{n\rightarrow\infty}F(\kappa_{n},\alpha_{n},e_{n})\geq
F(\kappa_{*},\alpha_{*},e_{*}).$ (5.53)

Hence, $(\kappa_{*},\alpha_{*},e_{*})$ is indeed a minimizer. This proves the
existence of a minimizer of the functional (5.35) and therefore of
$\mathcal{F}(\eta,a)$.

Now, we establish properties of the minimizer $(\kappa_{*},\alpha_{*},e_{*})$.
By (5.34), we have $\mathrm{Tr}g(\eta_{*})<\infty$, where $\eta_{*}$
corresponds to $(\kappa_{*},\alpha_{*})$. The statement, that $g(\eta_{*})$ is
trace class, follows from (5.34) and the fact that $g(\eta)\geq 0$.

Furthermore, if we restrict ourselves to even $(\kappa,\alpha,e)$, i.e. to
$(\kappa,\alpha,e)$ satisfying (1.46), then the minimizer
$(\kappa_{*},\alpha_{*},e_{*})$ is also even.

Now, since a minimizing sequence $e_{n}$ converges to $e_{*}$ strongly in
$\vec{\mathfrak{h}}^{s}$ for any $s<1$, we have, by the magnetic flux
quantization (1.45) for $e_{n}$, the convergence of $e_{n}$ to $e_{*}$ and by
the Stokes theorem, that
$\frac{1}{2\pi}\int_{\Omega}\operatorname{curl}a_{*}=c_{1}(\rho)\in\mathbb{Z}$,
where, recall, $a_{*}=a_{b}+e_{*}$.

Finally, the last property of the minimizer $(\kappa_{*},\alpha_{*},e_{*})$ is
shown in the following

######  Lemma 5.6.

$0$ and $1$ are not eigenvalues of $\eta_{*}$. Consequently, $0<\eta_{*}<1$.

###### Proof.

We assume for the sake of contradiction that $\eta_{*}$ has the eigenvalue
$0$. Hence $1$ is also an eigenvalue, since, if $\eta_{*}x=0$, then (1.7)
implies $\eta_{*}J\bar{x}=J\bar{x}$. Let $P_{x}$ and $P_{J\bar{x}}$ be the
orthogonal projections onto the subspaces spanned by $x$ and $J\bar{x}$.
Define

$\displaystyle\eta^{\prime}:=P_{x}-J\bar{P}_{x}J^{*}=P_{x}-P_{J\bar{x}}.$
(5.54)

Let $\eta_{\epsilon}:=(1+2\epsilon/\nu)^{-1}(\eta_{*}+\epsilon\eta^{\prime})$.
Since $x$ and $J\bar{x}$ are orthogonal, it is straightforward to see that
$\mathrm{Tr}\eta_{\epsilon}=\nu$, $\eta_{\epsilon}\in\mathcal{D}^{1}_{\nu}$
and $0\leq\eta_{\epsilon}\leq 1$.

Below, we write $F(\eta)$ for $F(\kappa,\alpha,e)$ and compute
$F(\eta_{\epsilon})$. By (5), we have

$\displaystyle
F(\eta_{\epsilon})-F(\eta_{*})=-T(S(\eta_{\epsilon})-S(\eta_{*}))+O(\epsilon).$
(5.55)

Using that $S(\eta)=\mathrm{Tr}(s(\eta))$, with $s(\eta)=-2\eta\ln\eta$, and
writing $\mathrm{Tr}(s(\eta))$ in an orthonormal eigen-basis which includes
the eigenvectors $x$ and $J\bar{x}$, we find

$S(\eta_{\epsilon})-S(\eta_{*})=\langle x,s(\eta_{\epsilon})x\rangle+\langle
J\bar{x},s(\eta_{\epsilon})J\bar{x}\rangle.$

Then we use Jensen’s inequality to find

$\displaystyle S(\eta_{\epsilon})-S(\eta_{*})\geq s(\langle
x,\eta_{\epsilon}x\rangle)+s(\langle
J\bar{x},\eta_{\epsilon}J\bar{x}\rangle).$ (5.56)

Now, using the definition
$\eta_{\epsilon}:=(1+2\epsilon/\nu)^{-1}(\eta_{*}+\epsilon\eta^{\prime})$ and
the relations $\eta_{*}x=0$ and $\eta_{*}J\bar{x}=J\bar{x}$, we compute
$\langle x,\eta_{\epsilon}x\rangle=(1+2\epsilon/\nu)^{-1}\epsilon$ and
$\langle
J\bar{x},\eta_{\epsilon}J\bar{x}\rangle=(1+2\epsilon/\nu)^{-1}(1+\epsilon)$.
This, together with (5.56) and the definition
$s(\lambda):=-2\lambda\ln\lambda$, yields

$\displaystyle S(\eta_{\epsilon})-S(\eta_{*})$ $\displaystyle\geq
s((1+2\epsilon/\nu)^{-1}\epsilon)+s((1+2\epsilon/\nu)^{-1}(1+\epsilon))$
$\displaystyle=-2\epsilon\ln\epsilon+O(\epsilon).$ (5.57)

(The nonlinear $\epsilon\ln\epsilon$ term in (5) is due to the fact that
$x\ln(x)$ is not differentiable at $0$.) This, together with (5.55), implies

$\displaystyle F(\eta_{\epsilon})$ $\displaystyle\leq
F(\eta_{*})+2T\epsilon\ln\epsilon+O(\epsilon).$ (5.58)

Since $\ln\epsilon<0$ and $|\epsilon\ln\epsilon|\gg\epsilon$, we conclude that
$F(\eta_{\epsilon})<F(\eta_{*})$ which contradicts minimality of $\eta_{*}$.
We conclude that $\eta_{*}$ has a trivial kernel; hence, it has a trivial
$1$-eigenspace. Consequently, $0<\eta_{*}<1$. ∎

This completes the proof of Theorem 5.1.∎

### Acknowledgments

The first author is very grateful for Almut Burchard’s kind help and
suggestions. The second author is grateful to Volker Bach, Sébastien Breteaux,
Thomas Chen and Jürg Fröhlich for enjoyable collaboration and both authors
thank Rupert Frank and Christian Hainzl, for stimulating discussions.

The authors are indebted to the anonymous referees for many useful suggestions
and remarks.

The authors’ research is supported in part by NSERC Grant No. NA7901. During
the work on the paper, the authors enjoyed the support of the NCCR SwissMAP.

##  Appendix A Entropy

In this appendix we prove the differentiability and expansion of the entropy
functional, which we recall here

$\displaystyle S(\eta):=\mathrm{Tr}(s(\eta))=\mathrm{Tr}(g(\eta)),$ (A.1)
$\displaystyle g(\eta):=-\eta\ln\eta-(1-\eta)\ln(1-\eta),\
s(\eta):=-2\eta\ln\eta.$ (A.2)

This is used in the next two appendices in order to prove Theorem 2.2 and
Propositions 1.5 and 1.6. Let
$dS(\eta)\eta^{\prime}:=\partial_{\epsilon}S(\eta+\epsilon\eta^{\prime})\mid_{\epsilon=0}$.
We have

######  Proposition A.1.

Let $\eta\in\mathcal{D}^{1}_{\nu}$ be such that
$g(\eta):=-\eta\ln\eta-(1-\eta)\ln(1-\eta)$ is trace class and $\eta^{\prime}$
satisfy (2.11). Then $S$ is $C^{1}$ and its derivative is given by

$\displaystyle
dS(\eta)\eta^{\prime}=\mathrm{Tr}(g^{\prime}(\eta)\eta^{\prime})=\mathrm{Tr}(s^{\prime}(\eta)\eta^{\prime}).$
(A.3)

###### Proof.

By (A.1), it suffices for us to prove the proposition for
$s(\eta)=-\eta\ln(\eta)$. Denote
$\eta^{\prime\prime}:=\eta+\epsilon\eta^{\prime}$. We write

$\displaystyle S(\eta^{\prime\prime})-S(\eta)$
$\displaystyle=-\mathrm{Tr}(\eta(\ln\eta^{\prime\prime}-\ln\eta)-\epsilon\eta^{\prime}(\ln\eta^{\prime\prime}-\ln\eta)-\epsilon\eta^{\prime}\ln\eta)$
(A.4) $\displaystyle=:A+B-\epsilon\mathrm{Tr}(\eta^{\prime}\ln\eta).$ (A.5)

Using the formula $\ln a-\ln b=\int_{0}^{\infty}[(b+t)^{-1}-(a+t)^{-1}]dt$ and
the second resolvent equation, we compute

$\displaystyle A$
$\displaystyle:=-\mathrm{Tr}(\eta(\eta^{\prime\prime}-\ln\eta))$
$\displaystyle=\int_{0}^{\infty}\mathrm{Tr}\\{\eta[(\eta^{\prime\prime}+t)^{-1}-(\eta+t)^{-1}]\\}dt$
$\displaystyle=-\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt$
$\displaystyle=-\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\\}dt$
$\displaystyle\quad-\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.6)

Similarly, we have

$\displaystyle B$
$\displaystyle:=-\mathrm{Tr}(\epsilon\eta^{\prime}(\eta^{\prime\prime}-\ln\eta))$
$\displaystyle=\int_{0}^{\infty}\mathrm{Tr}\\{\epsilon\eta^{\prime}[(\eta^{\prime\prime}+t)^{-1}-(\eta+t)^{-1}]\\}dt$
$\displaystyle=-\int_{0}^{\infty}\mathrm{Tr}\\{\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.7)

Combining the last two relations with (A.5), we find

$\displaystyle S(\eta+\epsilon\eta^{\prime})-S(\eta)=\epsilon
S_{1}+\epsilon^{2}R_{2},$ (A.8) $\displaystyle
S_{1}:=-\mathrm{Tr}\eta^{\prime}\ln\eta-\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\\}dt,$
(A.9) $\displaystyle
R_{2}:=\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}$
$\displaystyle\qquad-\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.10)

The estimates below show that the integrals on the r.h.s. converge. Computing
the integral
$\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\\}dt=\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-2}\eta^{\prime}\\}dt=\mathrm{Tr}\eta^{\prime}$
in the expression for $S_{1}$ and transforming the expression for $R_{2}$, we
obtain

$\displaystyle S_{1}:=$
$\displaystyle-\mathrm{Tr}\\{\eta^{\prime}\ln\eta+\eta^{\prime}\\},$ (A.11)
$\displaystyle R_{2}:=$
$\displaystyle-\int_{0}^{\infty}\mathrm{Tr}\\{t(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.12)

The proofs of convergence of (A.9) and (A.12) are similar. We consider the
case of (A.12). We estimate the integrand on the r.h.s. of (A.12). We have

$\displaystyle|\mathrm{Tr}\\{\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}|$
$\displaystyle\qquad\qquad\qquad\leq\|\eta^{\prime}(\eta+t)^{-1}\|_{I^{2}}\|\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\|_{I^{2}}$
(A.13)

Now, we show that the factors on the r.h.s. are $L^{2}(dt)$. By the second
condition in (2.11) on $\eta^{\prime}$, we have

$\displaystyle\|\eta^{\prime}(\eta^{\\#}+t)^{-1}\|_{I^{2}}$
$\displaystyle\leq\|\eta(\mathbf{1}-\eta^{\\#})(\eta^{\\#}+t)^{-1}\|_{I^{2}}$
$\displaystyle\leq\|\xi^{\\#}(\xi^{\\#}+t)^{-1}\|_{I^{2}},$ (A.14)

where $\eta^{\\#}$ is either $\eta$ or $\eta^{\prime\prime}$ and
$\xi^{\\#}:=\eta^{\\#}(\mathbf{1}-\eta^{\\#})$. Let $\mu_{n}$ be the
eigenvalues of the operator $\xi^{\\#}:=\eta^{\\#}(\mathbf{1}-\eta^{\\#})$.
Then we have

$\displaystyle\|\xi^{\\#}(\xi^{\\#}+t)^{-1}\|_{I^{2}}^{2}=\sum_{n}\mu_{n}^{2}(\mu_{n}+t)^{-2},$
(A.15)

and therefore

$\displaystyle\int_{0}^{\infty}\|\xi^{\\#}(\xi^{\\#}+t)^{-1}\|_{I^{2}}^{2}dt$
$\displaystyle=\int_{0}^{\infty}\sum_{n}\mu_{n}^{2}(\mu_{n}+t)^{-2}dt$
$\displaystyle=\sum_{n}\mu_{n}=\mathrm{Tr}\xi^{\\#}.$ (A.16)

Since $\eta(\mathbf{1}-\eta)$ and
$\eta^{\prime\prime}(\mathbf{1}-\eta^{\prime\prime})$ are trace class
operators, this proves the claim and, with it, the convergence of the integral
on the l.h.s.. Similarly, one shows the convergence of the other integrals.

To sum up, we proved the expansion (A.8) with $S_{1}$ given by (A.11), which
is the same as (A.3), and $R_{2}$ bounded as $|R_{2}|\lesssim 1$. In
particular, this implies that $S$ is $C^{1}$ and its derivative is given by
(A.3). ∎

######  Proposition A.2.

$S(\eta):=\mathrm{Tr}(g(\eta))$ is $C^{3}$ at $\eta_{Tb}$ w.r.t. perturbations
$\eta^{\prime}$ satisfying (2.11). Moreover, we have

$\displaystyle S(\eta_{Tb}+\epsilon\eta^{\prime})=$ $\displaystyle
S(\eta_{Tb})+\epsilon
S^{\prime}(\eta_{Tb})\eta^{\prime}+\frac{1}{2}\epsilon^{2}S^{\prime\prime}(\eta^{\prime},\eta^{\prime})+O(\epsilon^{3}),$
(A.17)

where
$S^{\prime}(\eta_{Tb})\eta^{\prime}:=\mathrm{Tr}(g^{\prime}(\eta_{Tb})\eta^{\prime})$,
$S^{\prime\prime}(\eta^{\prime},\eta^{\prime})$ is a quadratic form given by

$\displaystyle S^{\prime\prime}(\eta^{\prime},\eta^{\prime})=$
$\displaystyle\frac{1}{2}\int_{0}^{\infty}\mathrm{Tr}\\{(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}\\}dt$
(A.18)

and the error term is uniform in $\eta^{\prime}$ and is bounded by
$\epsilon^{3}\mathrm{Tr}(\eta_{Tb}(1-\eta_{Tb}))$. For
$\eta^{\prime}=\phi(\alpha)$, the quadratic term becomes

$\displaystyle
S^{\prime\prime}(\eta^{\prime},\eta^{\prime})=-\mathrm{Tr}\left(\bar{\alpha}K_{Tb}\alpha\right),$
(A.19) $\displaystyle
K_{Tb}:=\frac{1}{T}\frac{h_{Tb}^{L}+h_{Tb}^{R}}{\tanh(h^{L}_{Tb}/T)+\tanh(h^{R}_{Tb}/T)},$
(A.20)

where $h_{Tb}:=h_{\gamma_{Tb}a_{b}\mu}\equiv h_{\gamma_{Tb}a_{b}}-\mu$, with
$h_{\gamma a}$ defined in (1.9).

###### Proof.

For the duration of the proof we omit the subindex $Tb$ in $\eta_{Tb}$. Recall
(A.5)-(A.7) and continuing computing $A$ and $B$ in (A.6)-(A.7) in the same
fashion as in the derivation of these equations, we find

$\displaystyle A=$
$\displaystyle\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\\}dt$
$\displaystyle-\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\\}dt$
$\displaystyle+\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt,$
(A.21)

and

$\displaystyle B=$
$\displaystyle\int_{0}^{\infty}\mathrm{Tr}\\{\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\\}dt$
$\displaystyle-\int_{0}^{\infty}\mathrm{Tr}\\{\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta+t)^{-1}\epsilon\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.22)

Combining the last two relations with (A.5) and recalling the computation of
$S_{1}$, we find

$\displaystyle S(\eta+\epsilon\eta^{\prime})$ $\displaystyle-S(\eta)=\epsilon
S_{1}+\epsilon^{2}S_{2}+\epsilon^{3}R_{3}$ (A.23) $\displaystyle S_{2}:=$
$\displaystyle-\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}$
$\displaystyle-\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\\}dt,$
$\displaystyle R_{3}:=$
$\displaystyle\int_{0}^{\infty}\mathrm{Tr}\\{\eta(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}$
$\displaystyle-\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.24)

Transforming the expressions for $S_{2}$ and $R_{3}$, we obtain

$\displaystyle S_{2}=$
$\displaystyle\int_{0}^{\infty}\mathrm{Tr}\\{t(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\\}dt,$
(A.25) $\displaystyle R_{3}=$
$\displaystyle-\int_{0}^{\infty}\mathrm{Tr}\\{t(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}(\eta^{\prime\prime}+t)^{-1}\\}dt.$
(A.26)

Estimates similar to those done after (A.12) show that the integrals on the
r.h.s. converge. This proves the expansion (A.23) with $S_{1}$ and $S_{2}$
given by (A.11), which is the same as (A.3), and (A.25) and $R_{3}$ bounded as
$|R_{3}|\lesssim 1$. Identifying the quadratic form $S_{2}$ with
$S^{\prime\prime}(\eta^{\prime},\eta^{\prime})$, we arrive at the expansion
(A.17).

Before computing $S_{2}\equiv S^{\prime\prime}(\eta^{\prime},\eta^{\prime})$,
we find a simpler representation for it. Integrating the r.h.s. of (A.25) by
parts, we find

$\displaystyle S^{\prime\prime}=$
$\displaystyle\int_{0}^{\infty}\mathrm{Tr}\\{t(\eta+t)^{-2}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}\\}dt$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathrm{Tr}\\{(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}\\}dt$
$\displaystyle-\int_{0}^{\infty}\mathrm{Tr}\\{t(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-2}\eta^{\prime}\\}dt.$
(A.27)

But by the cyclicity of the trace the last integral is equal to the first one
and therefore we have (A.18). Eq (1.64) gives

$\displaystyle S^{\prime\prime}(\eta^{\prime},\eta^{\prime})=$
$\displaystyle\frac{1}{4}\int_{0}^{\infty}\mathrm{Tr}\\{[(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}$
$\displaystyle+(\mathbf{1}-\eta+t)^{-1}\eta^{\prime}(\mathbf{1}-\eta+t)^{-1}]\eta^{\prime}\\}dt.$
(A.28)

Now, we use (A) to compute to $S^{\prime\prime}$ for
$\eta^{\prime}=\phi(\alpha)$. First, we recall that $\eta=\eta_{Tb}$ and
observe that for $\eta^{\prime}=\phi(\alpha)$,

$\displaystyle\mathrm{Tr}((\eta_{Tb}+t)^{-1}\eta^{\prime}(\eta_{Tb}+t)^{-1}\eta^{\prime})$
$\displaystyle=2\mathrm{Tr}((\gamma_{Tb}+t)^{-1}\alpha(\mathbf{1}-\bar{\gamma}_{Tb}+t)^{-1}\alpha)$
$\displaystyle=2\mathrm{Tr}(((x+t)^{-1}(y+t)^{-1}\alpha)\bar{\alpha})$ (A.29)

where $x$ and $y$ are regarded as operators acting on $\alpha$ from the left
by multiplying by $\gamma_{Tb}$ and from the right, by $1-\bar{\gamma}_{Tb}$.
Putting this together with a similar expression for the second term on the
r.h.s. of (A) and performing the integral in $t$, we obtain

$\displaystyle
S^{\prime\prime}(\eta^{\prime},\eta^{\prime})=-\mathrm{Tr}\left[\bar{\alpha}K(\alpha)\right],$
(A.30) $\displaystyle
K:=\frac{\log(x)-\log(y)}{x-y}+\frac{\log(1-x)-\log(1-y)}{(1-x)-(1-y)},$
(A.31)

with $x$ acting on the left and $y$ acting on the right. (A.31) can be written
as

$\displaystyle K$ $\displaystyle=-\frac{\log(x^{-1}-1)-\log(y^{-1}-1)}{x-y}.$
(A.32)

Recalling that $\gamma_{Tb}=f_{T}(h_{Tb})=(1+e^{2h_{Tb}/T})^{-1}$ and
therefore $x^{-1}-1=e^{2h^{L}_{Tb}/T}$ and
$y^{-1}-1=e^{-2\bar{h}^{R}_{Tb}/T}$, we see that

$\displaystyle K=$
$\displaystyle\frac{1}{T}\frac{h_{Tb}^{L}+\bar{h}_{Tb}^{R}}{(1+e^{h^{L}_{Tb}/T})^{-1}+(1+e^{\bar{h}^{R}_{Tb}/T})^{-1}},$
(A.33)

which, together the hyperbolic functions identities,
$(1+e^{h})^{-1}=\frac{1}{2}(1-\tanh h)$ and
$(1+e^{-h})^{-1}=\frac{1}{2}(1+\tanh h)$, gives (A.19). ∎

By the definition of the Gâteaux derive and the Hessian and Proposition A.2,
we have

######  Corollary A.3.

We have

$\displaystyle
dS(\eta_{Tb})\eta^{\prime}:=\mathrm{Tr}(g^{\prime}(\eta_{Tb})\eta^{\prime}),$
(A.34)

and, for $\eta^{\prime}=\phi(\alpha)$ and with
$h_{Tb}:=h_{\gamma_{Tb},a_{b}}$,

$\displaystyle S^{\prime\prime}(\eta_{Tb})\phi(\alpha)=-\phi(K_{Tb}\alpha),\
K_{Tb}:=\frac{1}{T}\frac{h_{Tb}^{L}+h_{Tb}^{R}}{\tanh(h^{L}_{Tb}/T)+\tanh(h^{R}_{Tb}/T)}.$
(A.35)

Our next result on the entropy is the following

######  Lemma A.4.

The functional $-S(\eta)$ is weakly lower semi-continuous in
$\mathcal{D}^{1}_{\nu}$.

###### Proof.

We use an idea from [25] which allows to reduce the problem to a finite-
dimensional one. We use (5.12), to pass from $-S(\eta)$ to the relative
entropy, $S(\eta|\eta_{0})$, defined in (5.6), with $\eta_{0}$ of the form
(5.11), with $\mathrm{Tr}\gamma_{0}<\infty$ and s.t. $S(\eta_{0})<\infty$. By
(5.12), $S(\eta|\eta_{0})\geq 0$. Moreover,

$\displaystyle
S(\eta)=S(\eta_{0})-S(\eta|\eta_{0})-\mathrm{Tr}[(\eta-\eta_{0})\ln\eta_{0}].$
(A.36)

We choose $\eta_{0}$ so that $(\eta-\eta_{0})\ln\eta_{0}$ is trace class and
the term $\mathrm{Tr}[(\eta-\eta_{0})\ln\eta_{0}]$ is weakly lower semi-
continuous. We take

$\displaystyle\eta_{0}=f_{T}(M),\,M:=\text{diag}(\sqrt{-\Delta_{a_{b}}},-\overline{\sqrt{-\Delta_{a_{b}}}}).$
(A.37)

Since $f_{T}(h)=\left(e^{h/T}+1\right)^{-1}$, we see that that

$\displaystyle 0\leq-\ln\eta_{0}=\ln(1+e^{M/T})\lesssim$ $\displaystyle
1+M/T.$ (A.38)

This estimate and (A.37) show that $\mathrm{Tr}[(\eta-\eta_{0})\ln\eta_{0}]$
is $\hat{I}^{1}$-norm continuous ($\hat{I}^{1}-$norm is defined in (1.39)).
Indeed, writing $\ln\eta_{0}=(1+M/T)(1+M/T)^{-1}\ln\eta_{0}$ and using (A.38),
we find

$\displaystyle\|(\eta-\eta_{0})\ln\eta_{0}\|_{I^{1}}\leq\|(\eta-\eta_{0})(1+M/T)\|_{I^{1}}$
$\displaystyle\|(1+M/T)^{-1}\ln\eta_{0}\|$
$\displaystyle\lesssim\|\eta-\eta_{0}\|_{I^{1,1}}.$ (A.39)

This completes the proof of the claim that
$\mathrm{Tr}[(\eta-\eta_{0})\ln\eta_{0}]$ is $\hat{I}^{1}$-norm continuous.

Furthermore, since this term is affine in $\eta$, it is convex. Thus it is
weakly lower semicontinuous.

Now, following [25], let $s_{\lambda}(A|B)=\lambda^{-1}(s(\lambda
A+(1-\lambda)B)-\lambda s(A)-(1-\lambda)s(B))$ and write

$\displaystyle S(\eta_{n}|\eta_{0})+$
$\displaystyle\mathrm{Tr}(\eta_{0}-\eta_{n})=\sup_{\lambda\in(0,1)}\mathrm{Tr}(s_{\lambda}(\eta_{n}|\eta_{*})).$
(A.40)

Since the entropy function $s$ is concave, $s_{\lambda}(A|B)\geq 0$ for any
$A,B$. For any non-negative operator $T$ on $L^{2}(\Omega)$,
$\mathrm{Tr}_{L^{2}(\Omega)}T=\sup_{P}\mathrm{Tr}_{L^{2}(\Omega)}PT$ where the
sup is taken over all finite rank projections. Hence, we may write

$\displaystyle S(\eta_{n}|\eta_{0})+$
$\displaystyle\mathrm{Tr}(\eta_{0}-\eta_{n})=\sup_{\lambda\in(0,1)}\sup_{P}\mathrm{Tr}(Ps_{\lambda}(\eta_{n}|\eta_{0})$
(A.41)

where the $\sup_{P}$ is taken over all finite rank projections $P$. It follows
that for any $\lambda\in(0,1)$ and any finite rank projection $P$,

$\displaystyle
S(\eta_{n}|\eta_{0})+\mathrm{Tr}(\eta_{0}-\eta_{n})\geq\mathrm{Tr}(Ps_{\lambda}(\eta_{n}|\eta_{0})).$
(A.42)

Since $\eta_{n}\rightarrow\eta_{*}$ in $\|\cdot\|_{(0)}$ (hence in operator
norm) and $-x\ln x$ is continuous on $[0,1]$, we see that

$\displaystyle s_{\lambda}(\eta_{n}|\eta_{0})\rightarrow
s_{\lambda}(\eta_{*}|\eta_{0})$ (A.43)

in the operator norm. In particular, for any finite dimensional projection
$P$,

$\displaystyle\mathrm{Tr}(Ps_{\lambda}(\eta_{n}|\eta_{0}))\rightarrow\mathrm{Tr}(Ps_{\lambda}(\eta_{*}|\eta_{0})).$
(A.44)

Consequently,

$\displaystyle\liminf_{n\rightarrow\infty}S(\eta_{n}|\eta_{0})+$
$\displaystyle\mathrm{Tr}(\eta_{0}-\eta_{n})\geq\mathrm{Tr}(Ps_{\lambda}(\eta_{*}|\eta_{0})).$
(A.45)

Now taking $\sup_{\lambda\in(0,1)}$ and $\sup_{P}$ and using that
$\mathrm{Tr}(\eta_{0}-\eta_{n})=0$, by condition (2.11), we see that

$\displaystyle\liminf_{n\rightarrow\infty}S(\eta_{n}|\eta_{0})\geq\mathrm{Tr}(s(\eta_{*}|\eta_{0})),$
(A.46)

which implies the desired statement. ∎

As an aside not used in this paper, we compute the Hessian,
$\partial_{\gamma\gamma}S(\gamma_{Tb})$, of $S$ w.r.t. diagonal perturbations,

$\displaystyle
d(\gamma^{\prime}):=\left(\begin{array}[]{cc}\gamma^{\prime}&0\\\
0&-\bar{\gamma}^{\prime}\end{array}\right).$ (A.49)

$\partial_{\gamma\gamma}S(\gamma_{Tb})$ is defined by

$\displaystyle[\langle\gamma^{\prime},\partial_{\gamma\gamma}S(\gamma_{Tb})\gamma^{\prime}\rangle:=\partial^{2}_{\epsilon}S(\eta_{Tb}+\epsilon
d(\gamma^{\prime}))\big{|}_{\epsilon=0}.$ (A.50)

We have

######  Proposition A.5.

The hessian operator $\partial_{\gamma\gamma}S(\gamma_{Tb})$ is given by

$\displaystyle\partial_{\gamma\gamma}S(\gamma_{Tb})=\frac{1}{T}\frac{h_{Tb}^{L}-h_{Tb}^{R}}{\tanh(h^{L}_{Tb}/T)-\tanh(h^{R}_{Tb}/T)}.$
(A.51)

###### Proof.

Our starting point in the formula (A.18) and hence we begin with the
computation of the term
$\int_{0}^{\infty}\mathrm{Tr}[(\eta+t)^{-1}\eta^{\prime}(\eta+t)^{-1}\eta^{\prime}]dt$,
with $\eta^{\prime}=d(\gamma^{\prime})$, where $d(\gamma^{\prime})$ denotes
the perturbation in $\gamma$ given by

$\displaystyle
d(\gamma^{\prime}):=\left(\begin{array}[]{cc}\gamma^{\prime}&0\\\
0&-\bar{\gamma}^{\prime}\end{array}\right).$ (A.54)

First, we recall that $\eta=\eta_{Tb}$ and observe that for
$\eta^{\prime}=d(\gamma^{\prime})$,

$\displaystyle\mathrm{Tr}($
$\displaystyle(\eta_{Tb}+t)^{-1}\eta^{\prime}(\eta_{Tb}+t)^{-1}\eta^{\prime})=\mathrm{Tr}((\gamma_{Tb}+t)^{-1}\gamma^{\prime}(\gamma_{Tb}+t)^{-1}\gamma^{\prime}$
(A.55)
$\displaystyle+(\mathbf{1}-\bar{\gamma}_{Tb}+t)^{-1}\bar{\gamma}^{\prime}(\mathbf{1}-\bar{\gamma}_{Tb}+t)^{-1}\bar{\gamma}^{\prime})$
(A.56)
$\displaystyle=\mathrm{Tr}(([(x+t)^{-1}(x^{\prime}+t)^{-1}+(\mathbf{1}-x+t)^{-1}(\mathbf{1}-x^{\prime}+t)^{-1}]\gamma^{\prime})\gamma^{\prime}),$
(A.57)

where the last follows from $\mathrm{Tr}(A)=\mathrm{Tr}(\bar{A})$ for self-
adjoint operators, and $x$ and $x^{\prime}$ are regarded as operators acting
on $\gamma^{\prime}$ from the left by multiplying by $\gamma_{Tb}$ and from
the right, by $\gamma_{Tb}$. Performing the integral in $t$, we obtain
$S^{\prime\prime}(\eta^{\prime},\eta^{\prime})=-\mathrm{Tr}\left[\bar{\gamma^{\prime}}K^{\prime}(\gamma^{\prime})\right],$
where the operator $K^{\prime}$ is given by

$\displaystyle
K^{\prime}:=\frac{\log(x)-\log(x^{\prime})}{x-x^{\prime}}+\frac{\log(\mathbf{1}-x)-\log(\mathbf{1}-x^{\prime})}{(\mathbf{1}-x)-(\mathbf{1}-x^{\prime})},$
(A.58)

with $x$ acting on the left and $x^{\prime}$ acting on the right. Clearly,
$K^{\prime}$ is identified with $\partial_{\gamma\gamma}S(\gamma_{Tb})$.
Rewrite the operator $K^{\prime}$ as

$\displaystyle K^{\prime}$
$\displaystyle=-\frac{\log(x^{-1}-1)-\log({x^{\prime}}^{-1}-1)}{x-x^{\prime}}.$
(A.59)

Recalling that $\gamma_{Tb}=g^{\sharp}(h_{Tb}/T)=(1+e^{2h_{Tb}/T})^{-1}$,
where $h_{Tb}:=h_{\gamma_{Tb},a_{b}}$, we see that

$\displaystyle K=$
$\displaystyle\frac{1}{T}\frac{h_{Tb}^{L}-h_{Tb}^{R}}{(1+e^{h^{L}_{Tb}/T})^{-1}-(1+e^{h^{R}_{Tb}/T})^{-1}},$
(A.60)

which, together with (A.59) and the hyperbolic functions identities,
$(1+e^{h})^{-1}=\frac{1}{2}(1-\tanh h)$ and
$(1+e^{-h})^{-1}=\frac{1}{2}(1+\tanh h)$, gives (A.51). ∎

##  Appendix B Energy functional: Proof of Theorem 2.2

The proof of Theorem 2.2 consists of three parts: 1) differentiability of
$F_{T}$, 2) identification of the BdG equations with the Euler-Lagrange
equation of $F_{T}$, and 3) showing minimizers of $F_{T}$ among the set
$\mathcal{D}^{1}_{\nu}\times\vec{\mathfrak{h}}_{b}^{1}$ are critical points.

Part 1: differentiability. We consider first the variation
$\eta+\epsilon\eta^{\prime}$ for $\epsilon>0$ small and perturbations
satisfying (2.11), Note that such $\eta^{\prime}$ satisfies, for $\epsilon$
small enough,

$\displaystyle 0\leq\eta+\epsilon\eta^{\prime}\leq 1.$ (B.1)

Let
$d_{\eta}F_{T}(\eta,a)\eta^{\prime}:=\partial_{\epsilon}F_{T}(\eta+\epsilon\eta^{\prime},a)\mid_{\epsilon=0}$,
if the r.h.s. exists. From (1.28), it is easy to see that $E(\eta,a)$ is
Fréchet differentiable and

$\displaystyle
d_{\eta}E(\eta,a)\eta^{\prime}=\mathrm{Tr}(\Lambda(\eta,a)\eta^{\prime}).$
(B.2)

Hence it suffices to prove the Fréchet differentiability of $S(\eta)$. This is
done in Appendix A above.

Differentiability of $F_{T}$ with respect to $a$ is standard and can be easily
done. The only two terms in $F_{T}$ that depend on $a$ are
$\mathrm{Tr}((-\Delta_{a})\gamma)$ and
$\frac{1}{2}\int|\operatorname{curl}a|^{2}$. The first term can be
differentiated by using
$-\Delta_{a_{0}+a^{\prime}}=(-\Delta_{a_{0}})^{2}-2a^{\prime}(-i\nabla-
a_{0})+|a^{\prime}|^{2}$ while the second term is differentiable by standard
variational calculus. Hence the differentiability of $F_{T}$ follows from
(1.27), (B.2) and Proposition A.1.

Part 2: Euler-Lagrange equation.

Now, we show that if $0<\eta<1$ and $d_{\eta}F_{T}(\eta,a)\eta^{\prime}=0$ and
$d_{a}F_{T}(\eta,a)a^{\prime}=0$ for all $\eta^{\prime}$ on
$\mathfrak{h}\times\mathfrak{h}$ satisfying (2.11) and
$a^{\prime}\in\vec{\mathfrak{h}}^{1}$, then $(\eta,a)$ satisfies the BdG
equations (1.8)-(1.10).

We start with $d_{\eta}F_{T}\eta^{\prime}=0$ for all $\eta^{\prime}$
satisfying (2.11). First, we construct explicitly a dense subset of
perturbations $\eta^{\prime}$ satisfying (2.11). For a critical point
$(\eta,a),0<\eta<1$, we define a reference unit vector
$v_{0}=(1,0)^{T}\in\mathfrak{h}\times\mathfrak{h}$. We note that the
difference in norm of $v_{0}$’s two components is simply

$\displaystyle
0\not=1=\|(v_{0})_{1}\|_{\mathfrak{h}}^{2}-\|(v_{0})_{2}\|_{\mathfrak{h}}^{2}=\langle
v_{0},Sv_{0}\rangle=\mathrm{Tr}(SP_{v_{0}}).$ (B.3)

For simplicity and without loss of generality, we assume that $v_{0}$ is in
the image of $\eta(1-\eta)$ since $0<\eta<1$ (i.e. its range is dense). We
define $V\subset\mathfrak{h}\times\mathfrak{h}$ as

$\displaystyle V=\\{v:$ $\displaystyle\,\|v\|_{2}=1,\,v=\eta(1-\eta)\xi,\
\xi\in\mathfrak{h}\times\mathfrak{h}\\}.$ (B.4)

For each $v\in V$, we define

$\displaystyle\eta^{\prime}_{v}=(P_{v}-P_{J\bar{v}})-\frac{\mathrm{Tr}SP_{v}}{\mathrm{Tr}SP_{v_{0}}}(P_{v_{0}}-P_{J\bar{v}_{0}})\,,$
(B.5)

where $P_{x}$ is the orthogonal projection onto $x$ and $J$ is the complex
structure in (1.7).

######  Lemma B.1.

$\eta^{\prime}_{v}$ satisfies (2.11).

###### Proof.

To prove the first condition of (2.11), we only prove it for
$P_{v}-P_{J\bar{v}}=P_{v}-JP_{\bar{v}}J^{*}$ since this condition is real
linear. (Note that $S$ is self-adjoint, so $\langle
v,Sv\rangle=\mathrm{Tr}SP_{v}$ is real for all $v$.) We note that
$J^{*}=J^{-1}=-J$ and has only real number components. Let $\mathcal{C}$
denote the complex conjugation. It follows then that

$\displaystyle J^{*}(P_{v}-JP_{\bar{v}}J^{*})J=$ $\displaystyle
J^{*}P_{v}J-P_{\bar{v}}$ (B.6) $\displaystyle=$
$\displaystyle-\mathcal{C}(P_{v}-J^{*}\bar{P}_{v}J)\mathcal{C}$ (B.7)
$\displaystyle=$
$\displaystyle-\mathcal{C}(P_{v}-JP_{\bar{v}}J^{*})\mathcal{C}\,.$ (B.8)

This proves the first condition in (2.11).

To prove the second condition in (2.11), it suffices to show that $P_{v}$
satisfies this condition for every $v\in\mathfrak{h}\times\mathfrak{h}$ since
$J$ is unitary. For any
$v=\eta(1-\eta)\xi,x\in\mathfrak{h}\times\mathfrak{h}$, we note that

$\displaystyle\|P_{v}x\|=|\langle
v,x\rangle|=|\langle\eta(1-\eta)\xi,x\rangle|\lesssim\|\xi\|_{2}\|\eta(1-\eta)x\|_{2}.$
(B.9)

This shows that $(\eta^{\prime})^{2}\leq C[\eta(1-\eta)]^{2}$. This proves the
second condition in (2.11).

Finally, we prove the last condition in (2.11). For any unit norm
$v\in\mathfrak{h}\times\mathfrak{h}$,

$\displaystyle\mathrm{Tr}S_{1}(P_{v}-JP_{\bar{v}}J^{*})=$
$\displaystyle\mathrm{Tr}S_{1}P_{v}-\mathrm{Tr}S_{1}JP_{\bar{v}}J^{*}$ (B.10)
$\displaystyle=$
$\displaystyle\mathrm{Tr}SP_{v}-\mathrm{Tr}J^{*}SJP_{\bar{v}}.$ (B.11)

We note that $J^{*}S_{1}J=S_{2}:=\text{diag}(0,1)$. Hence,

$\displaystyle\mathrm{Tr}S_{1}(P_{v}-JP_{\bar{v}}J^{*})=$
$\displaystyle\mathrm{Tr}SP_{v}.$ (B.12)

It follows that

$\displaystyle\mathrm{Tr}S_{1}\eta^{\prime}_{v}=\mathrm{Tr}SP_{v}-\frac{\mathrm{Tr}SP_{v}}{\mathrm{Tr}SP_{v_{0}}}\mathrm{Tr}SP_{v_{0}}=0.$
(B.13)

This proves that $\eta^{\prime}_{v}$ satisfies the last condition in (2.11). ∎

We show that $0<\eta<1$ and $d_{\eta}F_{T}(\eta,a)\eta^{\prime}=0$ for all
$\eta^{\prime}$ satisfying (2.11) imply $\Lambda(\eta,a)-\mu
S-Tg^{\prime}(\eta)=0$ for some $\mu$ and $S=\text{diag}(1,-1)$ (see
Proposition 2.1). First note that (1.27), (B.2), (A.3) and (1.64) yield that

$\displaystyle
d_{\eta}F_{T}(\eta,a)\eta^{\prime}=\mathrm{Tr}\left[A\eta^{\prime}\right],$
(B.14)

where $A:=\Lambda(\eta,a)-Tg^{\prime}(\eta)$. If $(\eta,a)$ is a critical
point, then, for all $v\in V$, it satisfies

$\displaystyle\mathrm{Tr}(A\eta^{\prime}_{v})=0\,.$ (B.15)

Since $A$ is in the tangent space of all the $\eta$ such that $J^{*}\eta
J=1-\bar{\eta}$. We note that $A$ also satisfies the first condition in
(2.11). It follows that

$\displaystyle 0=$
$\displaystyle\mathrm{Tr}(A\eta^{\prime}_{v})=\mathrm{Tr}(AP_{v})-\mathrm{Tr}(AJ\bar{P}_{v}J^{*})$
(B.16)
$\displaystyle-\frac{\mathrm{Tr}SP_{v}}{\mathrm{Tr}SP_{v_{0}}}(\mathrm{Tr}(AP_{v})-\mathrm{Tr}(AJ\bar{P}_{v}J^{*}))$
(B.17) $\displaystyle=$
$\displaystyle\mathrm{Tr}(AP_{v})-\mathrm{Tr}(J^{*}AJ\bar{P}_{v})$ (B.18)
$\displaystyle-\frac{\mathrm{Tr}SP_{v}}{\mathrm{Tr}SP_{v_{0}}}(\mathrm{Tr}(AP_{v_{0}})-\mathrm{Tr}(J^{*}AJ\bar{P}_{v_{0}}))$
(B.19) $\displaystyle=$
$\displaystyle\mathrm{Tr}(AP_{v})+\mathrm{Tr}(\bar{A}\bar{P}_{v})$ (B.20)
$\displaystyle-\frac{\mathrm{Tr}SP_{v}}{\mathrm{Tr}SP_{v_{0}}}(\mathrm{Tr}(AP_{v_{0}})+\mathrm{Tr}(\bar{A}\bar{P}_{v_{0}}))$
(B.21) $\displaystyle=$ $\displaystyle
2\mathrm{Tr}(AP_{v})-\frac{\mathrm{Tr}SP_{v}}{\mathrm{Tr}SP_{v_{0}}}(2\mathrm{Tr}AP_{v_{0}}).$
(B.22)

We conclude that

$\displaystyle\mathrm{Tr}AP_{v}=\frac{\mathrm{Tr}AP_{v_{0}}}{\mathrm{Tr}SP_{v_{0}}}\mathrm{Tr}SP_{v}=:\mu\mathrm{Tr}SP_{v}$
(B.23)

for all $v\in V$. We note that $\mu$ is real since $A$ and $S$ are self-
adjoint. Since $0<\eta<1$, the linear space spanned $V$ is dense. We conclude
that $A$ is a multiple of $S$, which we denote by $\mu$. This shows that

$\displaystyle 0=A-\mu S.$ (B.24)

We conclude that $(\eta,a)$ solves the first BdG equation, (1.12).

Now, we consider the equation $d_{a}F_{T}(\eta,a)a^{\prime}=0$. As was
mentioned above, one can easily show that

$\displaystyle d_{a}F_{T}(\eta,a)a^{\prime}=2\int dxv\cdot a^{\prime},$ (B.25)

where $v:=\operatorname{curl}^{*}\operatorname{curl}a-j(\gamma,a)$ and the
perturbation $a^{\prime}\in\vec{\mathfrak{h}}^{1}$ is divergence free and mean
zero. Hence, to conclude that $v=0$, we have to show that
$v:=\operatorname{curl}^{*}\operatorname{curl}a-j(\gamma,a)$ is divergence
free and mean zero. (Indeed, any vector field $e$ can be written as
$e=a^{\prime}+\nabla g+c$, where $a^{\prime}$ is divergence free and mean zero
and $c$ is a constant, and therefore

$\int_{\Omega}ve=\int_{\Omega}va^{\prime}-\int_{\Omega}\operatorname{div}vg+c\int_{\Omega}v.$

So, if $v$ is divergence free and mean zero and $\int_{\Omega}va^{\prime}=0$,
for every $a^{\prime}$ divergence free and mean zero, then $v=0$.)

Clearly, the term $\operatorname{curl}^{*}\operatorname{curl}a$ is divergence
free and mean zero. So we show that $j(\gamma,a)$ is divergence free and mean
zero. For the first property, we use the fact that our free energy functional
is invariant under gauge transformation. In fact, it suffices to use the gauge
invariance of the first line in (1.28), $E_{1}(T^{\rm
gauge}_{t\chi}(\eta,a))=E_{1}(\eta,a)$, where
$E_{1}(\eta,a):=\mathrm{Tr}\big{(}(-\Delta_{a})\gamma\big{)}+\mathrm{Tr}\big{(}(v*\rho_{\gamma})\gamma\big{)}-\frac{1}{2}\mathrm{Tr}\big{(}(v^{\sharp}\gamma)\gamma\big{)}$.
It gives

$\displaystyle 0=\partial_{t}\mid_{t=0}E_{1}(T^{\rm gauge}_{t\chi}(\eta,a)),$
(B.26)

for all $\chi\in H^{1}_{\rm loc}$ which are $\mathcal{L}$-periodic. Using the
cyclicity of trace, we compute this explicitly

$\displaystyle 0=$
$\displaystyle\mathrm{Tr}(\mathrm{Re}(2i\nabla_{a}\gamma)\cdot\nabla\chi)+\mathrm{Tr}([\gamma,h_{\gamma,a}]\chi).$
(B.27)

Since $(\eta,a)$ solves the BdG equation, we have that
$[\Lambda(\eta,a),\eta]=0$. Taking the upper left component of this operator-
valued matrix equation, we see that

$\displaystyle[h_{\gamma,a},\gamma]+(v^{\sharp}\alpha)\bar{\alpha}-\alpha(v^{\sharp}\bar{\alpha})=0.$
(B.28)

Since $v(x)=v(-x)$, we conclude that the integral kernel of
$(v^{\sharp}\alpha)\bar{\alpha}-\alpha(v^{\sharp}\bar{\alpha})$,

$\displaystyle\int(v(x-z)-v(z-y))\alpha(x,z)\bar{\alpha}(z,y)dz,$ (B.29)

is zero on the diagonal. Thus, the same conclusion holds for
$[\gamma,h_{\gamma,a}]$. Consequently,
$\mathrm{Tr}([\gamma,h_{\gamma,a}]\chi)=0$ and we conclude, by (B.27), that

$\displaystyle
0=-\mathrm{Tr}(\mathrm{Re}(2i\nabla_{a}\gamma)\cdot\nabla\chi)=\int_{\Omega}j(\gamma,a)\cdot\nabla\chi=0.$
(B.30)

Since this is true for every $\chi\in H^{1}_{\rm loc}$ which are
$\mathcal{L}$-periodic, it follows that $\operatorname{div}j(\gamma,a)=0$.

To show that $j(\gamma,a)$ is mean zero, we use, that by our assumptions,
$\gamma$ is even and $a$ is odd. Since for any operator $A$, $u^{\rm
refl}\text{den}[A]=\text{den}[u^{\rm refl}Au^{\rm refl}]$ where $(u^{\rm
refl}f)(x)=f(-x)$, this shows that $j(\gamma,a)$ is odd. Hence so is
$v:=\operatorname{curl}^{*}\operatorname{curl}a-j(\gamma,a)$ and therefore
$v:=\operatorname{curl}^{*}\operatorname{curl}a-j(\gamma,a)=0$.

Since $\operatorname{div}a=0$, we may replace
$\operatorname{curl}^{*}\operatorname{curl}$ by $-\Delta$. Hence, the elliptic
regularity theory shows that $a\in\vec{\mathfrak{h}}_{b}^{2}$. This completes
the proof. $\Box$

Part 3: minimizers are critical points. For a minimizer $(\eta,a)$, we have
that $d_{\eta}F_{T}(\eta,a)\eta^{\prime}$, $d_{a}F_{T}(\eta,a)a^{\prime}\geq
0$. Since $\vec{H}^{1}$ is linear, $a^{\prime}\in\vec{\mathfrak{h}}^{1}$ if
and only if $-a^{\prime}\in\vec{H}^{1}$. So $d_{a}F_{T}(\eta,a)a^{\prime}=0$
for all $a\in\vec{\mathfrak{h}}^{1}$. Similarly, we note that $\eta^{\prime}$
satisfies the assumption (2.11) if and only if $-\eta^{\prime}$ satisfies the
same requirement. Hence we conclude that $0=dF_{T}(\eta,a)\eta^{\prime},$
which completes the proof. $\Box$

##  Appendix C Proof of the existence of solution to (3.15)

######  Lemma C.1.

Assume $\int v\geq 0$. Then, for each $T>0$ and $b=\frac{2\pi n}{|\Omega|}$,
the fixed point problem (3.15) has a unique solution.

###### Proof.

Let $\lambda:=\int v$. We define the real function
$g_{T}:\mathbb{R}\rightarrow\mathbb{R}$ by

$\displaystyle
g_{T}(\xi):=\lambda\text{den}(f_{T}(-\Delta_{a_{b}}-\mu+\xi))(0),$ (C.1)

where, recall, $\Delta_{a_{b}}$ acts on the space $\mathfrak{h}$ and is self-
adjoint. Then (3.15) can be rewritten as

$\displaystyle\xi=g_{T}(\xi).$ (C.2)

First, we derive a more convenient formula for the function $g_{T}$. Note
that, since, recall $f_{T}(s)=(1+e^{s/T})^{-1}$ and, for $b=\frac{2\pi
n}{|\Omega|}$, the operator $-\Delta_{a_{b}}$ on $\mathfrak{h}$ has the
eigenvalues $b(2m+1),m=0,1,\dots$, each of the same multiplicity $n$, the
trace $\mathrm{Tr}f_{T}(-\Delta_{a_{b}}-\mu+\xi)$ is finite and is smooth in
$\xi$. Since, by Proposition 3.4, $\text{den}[f_{T}(-\Delta_{a_{b}}-\mu+\xi)]$
is constant and since $\int_{\Omega}\text{den}A=\mathrm{Tr}A$, we have that

$\displaystyle
g_{T}(\xi)=\lambda\frac{1}{|\Omega|}\mathrm{Tr}f_{T}(-\Delta_{a_{b}}-\mu+\xi)$
(C.3)

and that $g_{T}$ is a smooth function.

Since $f_{T}(s)=(1+e^{s/T})^{-1}$ is positive and
$f_{T}^{\prime}(s)=-T^{-1}e^{s/T}(1+e^{s/T})^{-2}$ is negative, (C.3) shows
that $\pm g_{T}(\xi)>0$ and $\pm g_{T}^{\prime}(\xi)<0$, if $\pm\lambda>0$.
Hence, for $\lambda>0$, (C.2) has a unique solution for every $T>0$ and
$b=\frac{2\pi n}{|\Omega|}$. ∎

##  Appendix D Relative bounds and estimates on density

In this appendix we prove bounds on functions relative to the operator $M_{b}$
and estimates on density $\rho_{\gamma}$. Our first result is the following

######  Lemma D.1.

We have the following Sobolev-type inequalities

$\displaystyle\|M_{b}^{-s}cM_{b}^{-t}\|\lesssim\|c\|_{H^{r}},s+t>1-r,$ (D.1)
$\displaystyle\|\|M_{b}^{-1}c\cdot\nabla_{a_{b}}M_{b}^{-t}\|\lesssim\|c\|_{H^{r}},t>1-r,$
(D.2)
$\displaystyle\|M_{b}^{-s}|c|^{2}M_{b}^{-t}\|\lesssim\|c\|_{H^{r}}^{2},s+t>2(1-r).$
(D.3)

where in the second estimate we assumed $\operatorname{div}c=0$.

###### Proof.

We use the diamagnetic inequality $|M_{b}^{-s}f|\leq M_{b=0}^{-s}|f|$ (see
[2]) to reduce the problem to the $b=0$ case. To estimate the r.h.s. we write
$M_{0}^{-s}$ as the convolution, $M_{0}^{-s}u=G_{s}*u$, where $G_{s}(x)$ is
the Fourier transform of $(1+|k|^{2})^{-s/2}$, and use that $G_{s}(x)$ decays