---
abstract: |
    We consider a Riemannian cylinder $\Omega$ endowed with a closed potential $1$-form $A$ and study the magnetic Laplacian $\Delta_A$ with magnetic Neumann boundary conditions associated with those data. We establish a sharp lower bound for the first eigenvalue and show that the equality characterizes the situation where the metric is a product. We then look at the case of a planar domain bounded by two closed curves and obtain an explicit lower bound in terms of the geometry of the domain. We finally discuss sharpness of this last estimate.

    *2000 Mathematics Subject Classification. 58J50, 35P15.*

    *Key words and phrases. Magnetic Laplacian, Eigenvalues, Upper and lower bounds, Zero magnetic field*
author:
- Bruno Colbois and Alessandro Savo
bibliography:
- 'CS.bib'
title: 'Lower bounds for the first eigenvalue of the magnetic Laplacian\'
---

Introduction {#intro}
============

Let $(\Omega,g)$ be a compact Riemannian manifold with boundary. Consider the trivial complex line bundle $\Omega\times\bf C$ over $\Omega$; its space of sections can be identified with $C^{\infty}(\Omega,\bf C)$, the space of smooth complex valued functions on $\Omega$. Given a smooth real 1-form $A$ on $\Omega$ we define a connection $\nabla^A$ on $C^{\infty}(\Omega,\bf C)$ as follows: $$\label{connection}
\nabla^A_Xu=\nabla_Xu-iA(X)u$$ for all vector fields $X$ on $\Omega$ and for all $u\in C^{\infty}(\Omega,\bf C)$; here $\nabla$ is the Levi-Civita connection assocated to the metric $g$ of $\Omega$. The operator $$\label{magnetic laplacian}
\Delta_A=(\nabla^A)^{\star}\nabla ^A$$ is called the [*magnetic Laplacian*]{} associated to the magnetic potential $A$, and the smooth two form $$B=dA$$ is the associated [*magnetic field*]{}. We will consider Neumann magnetic conditions, that is: $$\label{mneumann}
\nabla^A_Nu=0\quad\text{on}\quad{\partial}\Omega,$$ where $N$ denotes the inner unit normal. Then, it is well-known that $\Delta_A$ is self-adjoint, and admits a discrete spectrum $$0\le {\lambda}_1(\Delta_A)\le {\lambda}_2(\Delta_A) \le ... \to \infty.$$

The above is a particular case of a more general situation, where $E\to M$ is a complex line bundle with a hermitian connection $\nabla^E$, and where the magnetic Laplacian is defined as $\Delta_E=(\nabla^E)^{\star}\nabla ^E$.

The spectrum of the magnetic Laplacian is very much studied in analysis (see for example [@BDP] and the references therein) and in relation with physics. For *Dirichlet boundary conditions*, lower estimates of its fundamental tone have been worked out, in particular, when $\Omega$ is a planar domain and $B$ is the constant magnetic field; that is, when the function $\star B$ is constant on $\Omega$ (see for example a Faber-Krahn type inequality in [@Er1] and the recent[@LS] and the references therein, also for Neumann boundary condition). The case when the potential $A$ is a closed $1$-form is particularly interesting from the physical point of view (Aharonov-Bohm effect), and also from the geometric point of view. For Dirichlet boundary conditions, there is a serie of papers for domains with a pole, when the pole approaches the boundary (see [@AFNN; @NNT] and the references therein). Last but not least, there is a Aharonov-Bohm approach to the question of nodal and minimal partitions, see chapter 8 of [@BH].

For *Neumann boundary conditions*, we refer in particular to the paper [@HHHO], where the authors study the multiplicity and the nodal sets corresponding to the ground state ${\lambda}_1$ for non-simply connected planar domains with harmonic potential (see the discussion below).

Let us also mention the recent article [@LLPP] (chapter 7) where the authors establish a *Cheeger type inequality* for ${\lambda}_1$; that is, they find a lower bound for ${\lambda}_1(\Delta_A)$ in terms of the geometry of $\Omega$ and the potential $A$. In the preprint [@ELMP], the authors approach the problem via the Bochner method.

Finally, in a more general context (see [@BBC]) the authors establish a lower bound for ${\lambda}_1(\Delta_A)$ in terms of the *holonomy* of the vector bundle on which $\Delta_A$ acts. In both cases, implicitly, the flux of the potential $A$ plays a crucial role.

[$\bullet\quad$]{}From now on we will denote by $\lambda_1(\Omega,A)$ the first eigenvalue of $\Delta_A$ on $(\Omega,g)$.

Main lower bound
----------------

Our lower bound is partly inspired by the results in [@HHHO] for plane domains. First, recall that if $c$ is a closed parametrized curve (a loop), the quantity: $$\Phi^A_c=\dfrac{1}{2\pi}\oint_{c}A$$ is called the [*flux*]{} of $A$ across $c$. (We assume that $c$ is travelled once, and we will not specify the orientation of the loop, so that the flux will only be defined up to sign: this will not affect any of the statements, definitions or results which we will prove in this paper). Let then $\Omega$ be a fixed plane domain with one hole, and let $\Phi^A$ be the flux of the harmonic potential $A$ across the inner boundary curve. In Theorem 1.1 of [@HHHO] it is first remarked that $\lambda_1(\Omega, A)$ is positive if and only if $\Phi^A$ is not an integer (but see the precise statement in Section 2.1 below). Then, it is shown that $\lambda_1(\Omega,A)$ is maximal precisely when $\Phi^A$ is congruent to $\frac 12$ modulo integers. The proof relies on a delicate argument involving the nodal line of a first eigenfunction; in particular, the conclusion does not follow from a specific comparison argument, or from an explicit lower bound.

In this paper we give a geometric lower bound of $\lambda_1(\Omega,A)$ when $\Omega$ is, more generally, a [*Riemannian cylinder*]{}, that is, a domain $(\Omega,g)$ diffeomorphic to $[0,1]\times{{\bf S}^{1}}$ endowed with a Riemannian metric $g$, and when $A$ is a closed potential $1$-form : hence, the magnetic field $B$ associated to $A$ is equal to $0$. The lower bound will depend on the geometry of $\Omega$ and, in an explicit way, on the flux of the potential $A$.

Let us write $\partial\Omega=\Sigma_1\cup\Sigma_2$ where $$\Sigma_1=\{0\}\times{{\bf S}^{1}}, \quad \Sigma_2=\{1\}\times{{\bf S}^{1}}.$$ We will need to foliate the cylinder by the (regular) level curves of a smooth function $\psi$ and then we introduce the following family of functions.

$$\begin{aligned}{\cal F}_{\Omega}=\{\psi:\Omega\to{{\bf R}}: \quad &\text{\it $\psi$ is constant on each boundary component}\\
&\text{\it and has no critical points inside $\Omega$}\}
\end{aligned}$$

As $\Omega$ is a cylinder, we see that ${\cal F}_{\Omega}$ is not empty. If $\psi\in{\cal F}_{\Omega}$, we set: $$K=K_{\Omega,\psi}=\dfrac{\sup_{\Omega}{\lvert{\nabla\psi}\rvert}}{\inf_{\Omega}{\lvert{\nabla\psi}\rvert}}.$$ It is clear that, in the definition of the constant $K$, we can assume that the range of $\psi$ is the interval $[0,1]$, and that $\psi=0$ on $\Sigma_1$ and $\psi=1$ on $\Sigma_2$. Note that the level curves of the function $\psi$ are all smooth, closed and connected; moreover they are all homotopic to each other so that the flux of a closed $1$-form $A$ across any of them is the same, and will be denoted by $\Phi^A$.

We say, briefly, that $\Omega$ is [*$K$-foliated by the level curves of $\psi$.*]{} We also denote by $d(\Phi^A,{\bf Z})$ the minimal distance between $\Phi^A$ and the set of integer $\bf Z$: $$d(\Phi^A,{\bf Z})^2=\min\Big\{(\Phi^A-k)^2: k\in\bf Z\Big\}.$$

Finally, we say that [*$\Omega$ is a Riemannian product*]{} if it is isometric to $[0,a]\times{{\bf S}^{1}}(R)$ for suitable positive constants $a,R$.

\[main3\]

a\) Let $(\Omega,g)$ be a Riemannian cylinder, and let $A$ be a closed $1$-form on $\Omega$. Assume that $\Omega$ is $K$-foliated by the level curves of the smooth function $\psi\in{\cal F}_{\Omega}$. Then: $$\label{cylinder}
\lambda_1(\Omega,A)\geq\dfrac{4\pi^2}{KL^2}\cdot d(\Phi^A,{\bf Z})^2,$$ where $L$ is the maximum length of a level curve of $\psi$ and $\Phi^A$ is the flux of $A$ across any of the boundary components of $\Omega$.

b\) Equality holds if and only if the cylinder $\Omega$ is a Riemannian product.

[$\bullet\quad$]{}It is clear that we can also state the lower bound as follows: $$\lambda_1(\Omega,A)\geq\dfrac{4\pi^2}{\tilde K_{\Omega}}\cdot d(\Phi^A,{\bf Z})^2,$$ where $\tilde K_{\Omega}$ is an invariant depending only $\Omega$: $$\tilde K_{\Omega}=\inf_{\psi\in{\cal F}_{\Omega}}K_{\Omega,\psi}L_{\psi}^2\quad\text{and}\quad L_{\psi}=\sup_{r\in {\rm range}(\psi)}{\lvert{\psi^{-1}(r)}\rvert}.$$ It is is not always easy to estimate $K$. In Section \[estimate K\] we will show how to estimate $K$ in terms of the metric tensor. Note that $K\geq 1$; we will see that in many interesting situations (for example, for revolution cylinders, or for smooth embedded tubes around a closed curve) one has in fact $K=1$.

Doubly connected planar domains
-------------------------------

We now estimate the constant $K$ above when $\Omega$ is an annular region in the plane, bounded by the inner curve $\Sigma_1$ and the outer curve $\Sigma_2$.

[$\bullet\quad$]{}We assume that the inner curve $\Sigma_1$ is convex.

From each point $x\in\Sigma_1$, consider the ray $\gamma_x(t)=x+tN_x$, where $N_x$ is the exterior normal to $\Sigma_1$ at $x$ and $t\geq 0$. Let $Q(x)$ be the first intersection of $\gamma_x(t)$ with $\Sigma_2$, and let $$r(x)=d(x,Q(x)).$$ We say that $\Omega$ is [*starlike with respect to $\Sigma_1$*]{} if the map $x\to Q(x)$ is a bijection between $\Sigma_1$ and $\Sigma_2$; equivalently, if given any point $y\in\Sigma_2$, the geodesic segment which minimizes distance from $y$ to $\Sigma_1$ is entirely contained in $\Omega$.

For $x\in\Sigma_1$, we denote by $\theta_x$ the angle between $\gamma'_x$ and the outer normal to $\Sigma_2$ at the point $Q(x)$, and we let $$m\doteq\min_{x\in\Sigma_1}{\cos\theta_x}.$$ Note that as $\Omega$ is starlike w.r.t. $\Sigma_1$, one has $\theta_x\in [0,\frac{\pi}2]$ and then $m\geq 0$.

[$\bullet\quad$]{}To have a positive lower bound, we will assume that $m>0$ (that is, $\Omega$ is [*strictly*]{} starlike w.r.t. $\Sigma_1$).

We also define $$\label{annulus}
\twosystem
{\beta=\min\{r(x): x\in\Sigma_1\}}
{B=\max\{r(x): x\in\Sigma_1\}}$$

We then have the following result.

\[main2\] Let $\Omega$ be an annulus in ${{\bf R}^{2}}$, which is strictly-starlike with respect to its inner (convex) boundary component $\Sigma_1$. Assume that $A$ is a closed potential having flux $\Phi^A$ around $\Sigma_1$. Then: $$\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta m}{B} d(\Phi^A,{\bf Z})^2$$ where $\beta$ and $B$ are as in , and $L$ is the length of the outer boundary component. If $\Sigma_2$ is also convex, then $m\geq \beta/B$ and the lower bound takes the form: $$\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta^2}{B^2} d(\Phi^A,{\bf Z})^2.$$

In section \[sharpness\], we will explain why we need to control $\dfrac{\beta}{B}$, $L$, and why we need to impose the starlike condition. If $\beta=B$ and $\Sigma_2$ is the circle of length $L$ we get the estimate $$\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} d(\Phi^A,{\bf Z})^2$$ which is the first eigenvalue of the magnetic Laplacian on the circle with potential $A$ (see section \[riemannian circle\]). If $\Sigma_2$ and $\Sigma_1$ are two concentric circles of respective lengths $L$ and $L_{\epsilon} \to L$, the domain is a thin annulus with $\lambda_1 \to \dfrac{4\pi^2}{L^2} d(\Phi^A,{\bf Z})^2$ which shows that our estimate is sharp.

Our aim is to use these estimates on cylinders as a basis stone in order to study the same type of questions on compact surfaces of higher genus.

Proof of the main theorem
=========================

Preliminary facts and notation {#preliminary}
------------------------------

First, we recall the variational definition of the spectrum. Let $\Omega$ be a compact manifold with boundary and $\Delta_A$ the magnetic Laplacian with Neumann boundary conditions. One verifies that $$\int_{\Omega}(\Delta_Au)\bar u=\int_{\Omega}{\lvert{\nabla^Au}\rvert}^2,$$ and the associated quadratic form is then $$Q_A(u)=\int_{\Omega}{\lvert{\nabla^Au}\rvert}^2.$$ The usual variational characterization gives:

$$\lambda_1(\Omega,A)= \min\Big\{ \frac{Q_A(u)}{\Vert u\Vert^2}:\ u\in C^{1}(\Omega,\mathbb C) / \{0\}\Big\}$$

The following proposition (which is well-known) expresses the [*gauge invariance*]{} of the spectrum of the magnetic Laplacian.

\[basic facts\] a The spectrum of $\Delta_A$ is equal to the spectrum of $\Delta_{A+d\phi}$ for all smooth real valued functions $\phi$; in particular, when $A$ is exact, the spectrum of $\Delta_A$ reduces to that of the classical Laplace-Beltrami operator acting on functions (with Neumann boundary conditions if ${\partial}\Omega$ is not empty).

b If $A$ is a closed $1$-form, then $A$ is gauge equivalent to a unique (harmonic) $1$-form $\tilde A$ satisfying $$\twosystem
{d\tilde A=\delta\tilde A=0\quad\text{on}\quad \Omega}
{\tilde A(N)=0\quad\text{on}\quad {\partial}\Omega}$$ The form $\tilde A$ is often called the [Coulomb gauge]{} of $A$. Note that $\tilde A$ is the harmonic representative of $A$ for the absolute boundary conditions.

a This comes from the fact that $
\Delta_A e^{-i\phi}=e^{-i\phi} \Delta_{A+d\phi}
$ hence $\Delta_A$ and $\Delta_{A+d\phi}$ are unitarily equivalent.

b Consider a solution $\phi$ of the problem: $$\twosystem{\Delta\phi=\delta A \quad\text{on}\quad \Omega,}
{{\dfrac{{\partial}\phi}{\bdN}}=A(N) \quad\text{on}\quad {\partial}\Omega.}$$ Then one checks that $\tilde A=A-d\phi$ is a Coulomb gauge of $A$. As $\phi$ is unique up to an additive constant, $d\phi$, hence $\tilde A$, is unique.

We now focus on the first eigenvalue. Clearly, if $A=0$, then $\lambda_1(\Omega,A)=0$ simply because $\Delta_A$ reduces to the usual Laplacian, which has first eigenvalue equal to zero and first eigenspace spanned by the constant functions. If $A$ is exact, then $\Delta_{A}$ is unitarily equivalent to $\Delta$, hence, again, $\lambda_1(\Omega,A)=0$. In fact one checks easily from the definition of the connection that, if $A=d\phi$ for some real-valued function $\phi$ then $
\nabla^{A}e^{i\phi}=0,
$ which means that $u=e^{i\phi}$ is $\nabla^A$-parallel hence $\Delta_A$-harmonic. On the other hand, if the magnetic field $B=dA$ is non-zero then $\lambda_1(\Omega,A)>0$.

It then remains to examine the case when $A$ is closed but not exact. The situation was clarified in [@Sh] for closed manifolds and in [@HHHO] for Neumann boundary conditions.

\[shikegawa\]The following statements are equivalent:

a\) $\lambda_1(\Omega,A)=0$;

b\) $dA=0$ and $\Phi^A_c\in\bf Z$ for any closed curve $c$ in $\Omega$.

Thus, the first eigenvalue vanishes if and only if $A$ is a closed form whose flux around every closed curve is an integer; equivalently, if $A$ has non-integral flux around at least one closed loop, then $\lambda_1(\Omega,A)>0$.

Proof of the lower bound
------------------------

From now on we assume that $\Omega$ is a Riemannian cylinder. Fix a first eigenfunction $u$ associated to $\lambda_1(\Omega, A)$ and fix a level curve $$\Sigma_r=\{\psi=r\}, \quad\text{where $r\in [0,1]$.}$$ As $\psi$ has no critical points, $\Sigma_r$ is isometric to ${{\bf S}^{1}}(\frac{L_r}{2\pi})$, where $L_r$ is the length of $\Sigma_r$. The restriction of $A$ to $\Sigma_r$ is a closed $1$-form denoted by $\tilde A$; we use the restriction of $u$ to $\Sigma_r$ as a test-function for the first eigenvalue $\lambda_1(\Sigma_r,\tilde A)$ and obtain: $$\label{level}
\lambda_1(\Sigma_r,\tilde A)\int_{\Sigma_r}{\lvert{u}\rvert}^2\leq\int_{\Sigma_r}{\lvert{\nabla^{\tilde A}u}\rvert}^2.$$ By the estimate on the eigenvalues of a circle done in Section \[sectioncircle\] below we see : $$\lambda_1(\Sigma_r,\tilde A)=\dfrac{4\pi^2}{L_r^2}d(\Phi^{\tilde A},{\bf Z})^2,$$ where $\Phi^{\tilde A}$ is the flux of $\tilde A$ across $\Sigma_r$. Now note that $\Phi^{\tilde A}=\Phi^{A}$, because $\tilde A$ is the restriction of $A$ to $\Sigma_r$; moreover $L_r\leq L$ by the definition of $L$. Therefore: $$\label{llower}
\lambda_1(\Sigma_r,\tilde A)\geq \dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2$$ for all $r$. Let $X$ be a unit vector tangent to $\Sigma_r$. Then: $$\begin{aligned}
\nabla^{\tilde A}_{X}u&=\nabla_{X}u-i\tilde A(X)u\\
&=\nabla_{X}u-iA(X)u\\
&=\nabla^A_{X}u.
\end{aligned}$$ The consequence is that: $$\label{energy}
{\lvert{\nabla^{\tilde A}u}\rvert}^2={\lvert{\nabla^{\tilde A}_{X}u}\rvert}^2={\lvert{\nabla^{A}_{X}u}\rvert}^2\leq {\lvert{\nabla^{A}u}\rvert}^2.$$ [$\bullet\quad$]{}[*Note that equality holds in iff $\nabla^A_{N}u=0$ where $N$ is a unit vector normal to the level curve $\Sigma_r$ (we could take $N=\nabla\psi/{\lvert{\nabla\psi}\rvert}$).*]{}

For any fixed level curve $\Sigma_r=\{\psi=r\}$ we then have, taking into account , and : $$\dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2\int_{\psi=r}{\lvert{u}\rvert}^2\leq \int_{\psi=r}{\lvert{\nabla^Au}\rvert}^2.$$ Assume that $B_1\leq{\lvert{\nabla\psi}\rvert}\leq B_2$ for positive constants $B_1,B_2$. Then the above inequality implies: $$\dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2\cdot B_1\int_{\psi=r}\dfrac{{\lvert{u}\rvert}^2}{{\lvert{\nabla\psi}\rvert}}\leq B_2\int_{\psi=r}\dfrac{{\lvert{\nabla^Au}\rvert}^2}{{\lvert{\nabla\psi}\rvert}}.$$ We now integrate both sides from $r=0$ to $r=1$ and use the coarea formula. Conclude that $$\dfrac{4\pi^2}{L^2}d(\Phi^{ A},{\bf Z})^2\cdot B_1\int_{\Omega}{{\lvert{u}\rvert}^2}\leq B_2\int_{\Omega}{\lvert{\nabla^Au}\rvert}^2.$$ As $u$ is a first eigenfunction, one has: $$\int_{\Omega}{\lvert{\nabla^Au}\rvert}^2=\lambda_1(\Omega,A)\int_{\Omega}{\lvert{u}\rvert}^2.$$ Recalling that $K=\frac{B_2}{B_1}$ we finally obtain the estimate .

Proof of the equality case {#equalitycase}
--------------------------

If the cylinder $\Omega$ is a Riemannian product then it is obvious that we can take $K=1$ and then we have equality by Proposition \[cyl\] below. Now assume that we do have equality: we have to show that $\Omega$ is a Riemannian product. Going back to the proof, we must have the following facts.

[**F1.**]{} [*All level curves of $\psi$ have the same length $L$*]{}.

[**F2.**]{} [*${\lvert{\nabla\psi}\rvert}$ must be constant and, by renormalization, we can assume that it is everywhere equal to $1$.* ]{}Then, $\psi:\Omega\to [0,a]$ for some $a>0$ and we set $$N\doteq\nabla\psi.$$

[**F3.**]{} [*The eigenfunction $u$ on $\Omega$ restricts to an eigenfunction of the magnetic Laplacian of each level set $\Sigma_r=\{\psi=r\}$, with potential given by the restriction of $A$ to $\Sigma_r$.*]{}

[**F4.**]{} [*One has $\nabla^A_Nu=0$ identically on $\Omega$.* ]{}

### First step: description of the metric

$\Omega$ is isometric to the product $[0,a]\times {{\bf S}^{1}}(\frac{L}{2\pi})$ with metric $$\label{metric}
g={{\begin{pmatrix}}{1&0\\}{0&\theta^2(r,t)\\}{\end{pmatrix}}}, \quad (r,t)\in [0,a]\times [0,L]$$ where $\theta(r,t)$ is positive and periodic of period $L$ in the variable $t$. Moreover $\theta(0,t)=1$ for all $t$.

We first show that the integral curves of $N$ are geodesics; for this it is enough to show that $
\nabla_NN=0
$ on $\Omega$. Let $e_1(x)$ be a vector tangent to the level curve of $\psi$ passing through $x$. Then, we obtain a smooth vector field $e_1$ which, together with $N$, forms a global orthonormal frame. Now $${\langle{\nabla_NN},{N}\rangle}=\dfrac 12 N\cdot{\langle{N},{N}\rangle}=0.$$ On the other hand, as the Hessian is a symmetric tensor: $${\langle{\nabla_NN},{e_1}\rangle}=\nabla^2\psi(N,e_1)=\nabla^2\psi(e_1,N)={\langle{\nabla_{e_1}N},{N}\rangle}=\dfrac 12e_1\cdot{\langle{N},{N}\rangle}=0.$$ Hence $\nabla_NN=0$ as asserted. As each integral curve of $N=\nabla\psi$ is a geodesic meeting $\Sigma_1$ orthogonally, we see that $\psi$ is actually the distance function to $\Sigma_1$. We introduce coordinates on $\Omega$ as follows. For a fixed point $p\in\Omega$ consider the unique integral curve $\gamma$ of $N$ passing through $p$ and let $x\in\Sigma_1$ be the intersection of $\gamma$ with $\Sigma_1$ (note that $x$ is the foot of the unique geodesic which minimizes the distance from $p$ to $\Sigma_1$). Let $r$ be the distance of $p$ to $\Sigma_1$. We then have a map $
\Omega\to [0,a]\times\Sigma_1
$ which sends $p$ to $(r,x)$. Its inverse is the map $F: [0,a]\times\Sigma_1\to\Omega$ defined by $$F(r,x)=\exp_x(rN).$$ Note that $F$ is a diffeomeorphism; we call the pair $(r,x)$ the [*normal coordinates*]{} based on $\Sigma_1$. We introduce the arc-length $t$ on $\Sigma_1$ (with origin in any assigned point of $\Sigma_1$) and recall that $L$ is length of $\Sigma_1$ (which is also the length of $\Sigma_2)$). Let us compute the metric $g$ in normal coordinates. Since $N={\dfrac{{\partial}}{{\partial}r}}$ one sees that $g_{11}=1$ everywhere; for any fixed $r=r_0$ we have that $F(r_0,\cdot)$ maps $\Sigma_1$ diffeomorphically onto the level set $\{\psi=r_0\}$ so that ${\dfrac{{\partial}}{{\partial}r}}$ and ${\dfrac{{\partial}}{{\partial}t}}$ will be mapped onto orthogonal vectors, and indeed $g_{12}=0$. Setting $\theta(r,t)^2={\langle{{\dfrac{{\partial}}{{\partial}t}}},{{\dfrac{{\partial}}{{\partial}t}}}\rangle}$ one sees that the metric takes the form . Finally note that $
\theta(0,t)=1
$ for all $t$, because $F(0,\cdot)$ is the identity.

### Second step : Gauge invariance

Let $\Omega$ be any Riemannian cylinder and $A=f(r,t)\,dr+h(r,t)\,dt$ a closed $1$-form on $\Omega$. Then, there exists a smooth function $\phi$ on $\Omega$ such that $$A+d\phi=H(t)\,dt$$ for a smooth function $H(t)$ depending only on t. Hence, by gauge invariance, we can assume from the start that $A=H(t)\,dt$.

Consider the function $
\phi(r,t)=-\int_0^rf(x,t)\,dx.
$ Then: $$A+d\phi=\tilde h(r,t)\,dt$$ for some smooth function $\tilde h(r,t)$. As $A$ is closed, also $A+d\phi$ is closed, which implies that ${\dfrac{{\partial}\tilde h}{\bdr}}=0$, that is, $
\tilde h(t,r)
$ does not depend on $r$; if we set $H(t)\doteq\tilde h(t,0)$ we get the assertion.

[$\bullet\quad$]{}We point out the following consequence. If $u=u(r,t)$ is an eigenfunction, we know from [**F4**]{} above that $\nabla^A_Nu=0$, where $N={\dfrac{{\partial}}{\bdr}}$. As $
\nabla^A_Nu={\dfrac{{\partial}u}{\bdr}}-iA({\dfrac{{\partial}}{\bdr}})u
$ and $A=H(t)\,dt$ we obtain $A({\dfrac{{\partial}}{\bdr}})=0$ hence $
{\dfrac{{\partial}u}{\bdr}}=0
$ at all points of $\Omega$. This implies that $$\label{uoft}
u=u(t)$$ depends only on $t$.

### Third step : spectrum of circles and Riemannian products {#sectioncircle}

In this section, we give an expression for the eigenfunctions of the magnetic Laplacian on a circle with a Riemannian metric $g$ and a closed potential $A$. Of course, we know that any metric $g$ on a circle is always isometric to the canonical metric $g_{\rm can}=\,dt^2$, where $t$ is arc-length. But our problem in this proof is to reconstruct the global metric of the cylinder and to show that it is a product, and we cannot suppose a priori that the restricted metric of each level set of $\psi$ is the canonical metric. The same is true for the restricted potential: we know that it is Gauge equivalent to a potential of the type $a\,dt$ for a scalar $a$, but we cannot suppose a priori that it is of that form.

We refer to Appendix \[riemannian circle\] for the complete proof of the following fact.

\[circle\] Let $(M,g)$ be the circle of length $L$ endowed with the metric $
g=\theta(t)^2\,dt^2
$ where $t\in [0,L]$ and $\theta(t)$ is a positive function, periodic of period $L$. Let $A=H(t)\,dt$. Then, the eigenvalues of the magnetic Laplacian with potential $A$ are: $$\lambda_k(M,A)=\dfrac{4\pi^2}{L^2}(k-\Phi^A)^2, \quad k\in\bf Z$$ with associated eigenfunctions $$u_k(t)=e^{i\phi(t)}e^{\frac{2\pi i (k-\Phi^A)}{L}s(t)}, \quad k\in\bf Z.$$ where $\phi(t)=\int_0^tH(\tau)\,d\tau$ and $s(t)=\int_0^t\theta(\tau)\,d\tau$.

In particular, if the metric is the canonical one, that is, $g=dt^2$, and the potential $1$-form is harmonic, so that $A=\frac{2\pi \Phi^A}{L}dt$, then the eigenfunctions are simply : $$u_k(t)=e^{\frac{2\pi i k}{L}t}, \quad k\in\bf Z.$$

We remark that if the flux $\Phi^A$ is not congruent to $1/2$ modulo integers, then the eigenvalues are all simple. If the flux is congruent to $1/2$ modulo integers, then there are two consecutive integers $k,k+1$ such that $
\lambda_{k}=\lambda_{k+1}.
$ Consequently, the lowest eigenvalue has multiplicity two, and the first eigenspace is spanned by $$e^{i\phi(t)}e^{\frac{\pi i}{L}s(t)}, \, e^{i\phi(t)}e^{-\frac{\pi i}{L}s(t)}.$$ The following proposition is an easy consequence (for a proof, see also Appendix \[riemannian circle\]).

\[cyl\] Consider the Riemannian product $\Omega=[0,a]\times{{\bf S}^{1}}(\frac{L}{2\pi})$, and let $A$ be a closed $1-$form on $\Omega$. Then, the spectrum of $\Delta_A$ is given by $$\dfrac{\pi^2 h^2}{a^2}+\dfrac{4\pi^2}{L^2}(k-\Phi^A)^2, \quad h, k\in{\bf Z}, h\geq 0.$$ In particular, $$\lambda_1(\Omega,A)=\dfrac{4\pi^2}{L^2}d(\Phi^A,{\bf Z})^2.$$

### Fourth step : a calculus lemma

In this section, we state a technical lemma which will allow us to conclude. The proof is conceptually simple, but perhaps tricky at some points; then, we decided to put it in Appendix \[technical lemma\].

\[calculus\] Let $s:[0,a]\times [0,L]\to {{\bf R}}$ be a smooth, non-negative function such that $$s(0,t)=t,\quad s(r,0)=0, \quad s(r,L)=L \quad\text{and}\quad {\dfrac{{\partial}s}{\bdt}}(r,t)\doteq\theta(r,t)>0.$$ Assume that there exist smooth functions $p(r),q(r)$ with $p(r)^2+q(r)^2>0$ such that $$p(r)\cos(\frac{\pi}{L}s(r,t))+q(r)\sin(\frac{\pi}{L}s(r,t))=F(t)$$ where $F(t)$ depends only on $t$. Then $p$ and $q$ are constant and $
{\dfrac{{\partial}s}{\bdr}}=0
$ so that $$s(r,t)=t$$ for all $(r,t)$.

### End of proof of the equality case

Assume that equality holds. Then, if $u$ is an eigenfunction, we know that $u=u(t)$ by the discussion in and $u$ restricts to an eigenfunction on each level circle $\Sigma_r$ for the potential $A=H(t)\,dt$ above (see Fact 3 at the beginning of Section \[equalitycase\] and the second step above).

We assume that $\Phi^A$ is congruent to $\frac 12$ modulo integers. This is the most difficult case; in the other cases the proof is a particular case of this, it is simpler and we omit it.

Recall that each level set $\Sigma_r$ is a circle of length $L$ for all $r$, with metric $g=\theta(r,t)^2\,dt$. As the flux of $A$ is congruent to $\frac 12$ modulo integers, we see that there exist complex-valued functions $w_1(r),w_2(r)$ such that $$u(t)=e^{i\phi(t)}\Big(w_1(r)e^{\frac{\pi i}{L}  s(r,t)}+w_2(r)e^{-\frac{\pi i}{L}  s(r,t)}\Big),$$ which, setting $f(t)=e^{-i\phi(t)}u(t)$, we can re-write $$\label{rewrite}
f(t)=w_1(r)e^{\frac{\pi i}{L}  s(r,t)}+w_2(r)e^{-\frac{\pi i}{L} s(r,t)}.$$ Recall that here $\phi(t)=\int_0^tH(\tau)\,d\tau$ and $$s(r,t)=\int_0^t\theta(r,\tau)\,d\tau.$$ We take the real part on both sides of and obtain smooth real-valued functions $F(t), p(r),q(r)$ such that $$F(t)=p(r)\cos({\frac{\pi}{L}}s(r,t))+q(r)\sin(\frac{\pi}{L} s(r,t)).$$ Since $\theta(0,t)=1$ for all $t$, we see $$s(0,t)=t.$$ Clearly $s(r,0)=0$; finally, $s(r,L)=\int_0^L\theta(r,\tau)\,d\tau=L$, being the length of the level circle $\Sigma_r$. Thus, we can apply Lemma \[calculus\] and conclude that $s(r,t)=t$ for all $t$, that is, $$\theta(r,t)=1$$ for all $(r,t)$ and the metric is a Riemannian product.

It might happen that $p(r)=q(r)\equiv 0$. But then the real part of $f(t)$ is zero and we can work in an analogous way with the imaginary part of $f(t)$, which cannot vanish unless $u\equiv 0$.

General estimate of $K_{\Omega,\psi}$ {#estimate K}
-------------------------------------

We can estimate $K_{\Omega,\psi}$ for a Riemannian cylinder $\Omega=[0,a]\times{{\bf S}^{1}}$ if we know the explicit expression of the metric in the normal coordinates $(r,t)$, where $t\in [0,2\pi]$ is arc-length : $$g=
\left(
    \begin{array}{cc}
    g_{11} &  g_{12}  \\
     g_{21} & g_{22}
     \end{array}
         \right).$$ If $g^{ij}$ is the inverse matrix of $g_{ij}$, and if $\psi=\psi(r,t)$ one has: $${\lvert{\nabla\psi}\rvert}^2=g^{11}\Big({\dfrac{{\partial}\psi}{\bdr}}\Big)^2+2g^{12}{\dfrac{{\partial}\psi}{\bdr}}{\dfrac{{\partial}\psi}{\bdt}}+
g^{22}\Big({\dfrac{{\partial}\psi}{\bdt}}\Big)^2.$$ The function $\psi(r,t)=r$ belongs to ${\cal F}_{\Omega}$ and one has: $
{\lvert{\nabla\psi}\rvert}^2=g^{11},
$ which immediately implies that we can take $$K_{\Omega,\psi}\leq \dfrac{\sup_{\Omega}g^{11}}{\inf_{\Omega}g^{11}}.$$ Note in particular that if $\Omega$ is rotationally invariant, so that the metric can be put in the form: $$g=\left(
    \begin{array}{cc}
    1 &  0 \\
    0 & \alpha(r)^2
     \end{array}
         \right),$$ for some function $\alpha(r)$, then $K_{\Omega,\psi}=1$. The estimate becomes $$\label{simple}
\lambda_1(\Omega,A)\geq\dfrac{4\pi^2}{L^2}\cdot d(\Phi^A,{\bf Z})^2,$$ where $L$ is the maximum length of a level curve $r={\rm const}$.

Yet more generally, one can fix a smooth closed curve $\gamma$ on a Riemannian surface $M$ and consider the tube of radius $R$ around $\gamma$: $$\Omega=\{x\in M: d(x,\gamma)\leq R\}.$$ It is well-known that if $R$ is sufficiently small (less than the injectivity radius of the normal exponential map) then $\Omega$ is a cylinder with smooth boundary which can be foliated by the level sets of $\psi$, the distance function to $\gamma$. Clearly ${\lvert{\nabla\psi}\rvert}=1$ and holds as well.

A concrete example where we could estimate the width $R$ is the case of a compact surface $M$ of genus $\ge 2$ and curvature $-a^2\le K\le -b^2$, $a \ge b >0$. Let $\gamma$ be a simple closed geodesic. Then, using the Gauss-Bonnet theorem, one can show that $R$ is bounded below by an explicit positive constant $R=R(\gamma,a)$, hence the $R$-neighborhood of $\gamma$ is diffeomorphic to the product $S^1 \times (-1,1)$ (see for example [@CF]). If we take $\Omega$ as the Riemannian cylinder of width $R(\gamma,a)$ having one boundary component equal to $\gamma$ then we can foliate $\Omega$ with the level sets of the distance function to $\gamma$ and so $K=1$ and holds, with $L$ given by the length of the other boundary component.

Proof of Theorem \[main2\]: plane annuli {#convex}
========================================

Let $\Omega$ be an annulus in ${{\bf R}^{2}}$, which is starlike with respect to its inner convex boundary component $\Sigma_1$. Assume that $A$ is a closed potential having flux $\Phi^A$ around $\Sigma_1$. Recall that we have to show: $$\label{annuliestimate}
\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta m}{B} d(\Phi^A,{\bf Z})^2$$ where $\beta, B$ and $m$ will be recalled below and $L$ is the length of the outer boundary component. If we assume that $\Sigma_2$ is also convex, then we show that $m\geq \beta/B$ and the lower bound takes the form: $$\label{annuliestimatetwo}
\lambda_1(\Omega,A)\geq \dfrac{4\pi^2}{L^2} \dfrac{\beta^2}{B^2} d(\Phi^A,{\bf Z})^2.$$

Before giving the proof let us recall notation. For $x\in\Sigma_1$, the ray $\gamma_x$ is the geodesic segment $\gamma_x(t)=x+tN_x$, where $N_x$ is the exterior normal to $\Sigma_1$ at $x$ and $t\geq 0$. The ray $\gamma_x$ meets $\Sigma_2$ at a first point $Q(x)$, and we let $
r(x)=d(x,Q(x)).
$ For $x\in\Sigma_1$, we denote by $\theta_x$ the angle between the ray $\gamma'_x$ and the outer normal to $\Sigma_2$ at the point $Q(x)$, and we let $$m\doteq\min_{x\in\Sigma_1}{\cos\theta_x}.$$ We assume that $\Omega$ is strictly starlike, that is, $m>0$; in particular $Q(x)$ is unique. Recall also that: $$\label{annulus}
\beta=\min_{x\in\Sigma_1}r(x), \quad B=\max_{x\in\Sigma_1}r(x).$$ We construct a suitable smooth function $\psi$ and estimate the constant $K=K_{\Omega,\psi}$ with respect to the geometry of $\Omega$. The starlike assumption implies that each point in $\Omega$ belongs to a unique ray $\gamma_x$. Then we can define a function $\psi:\Omega\to [0,1]$ as follows: $$\psi=\threesystem
{0\quad\text{on}\quad\Sigma_1}
{1\quad\text{on}\quad\Sigma_2}
{\text{linear on each ray from $\Sigma_1$ to $\Sigma_2$}.}$$ Estimates and now follow from Theorem \[main3\] together with the following Proposition.

\[estimate doubly convex\] a At all points of $\Omega$ one has: $
\frac{1}{B}\leq{\lvert{\nabla\psi}\rvert}\leq\frac{1}{\beta m}.
$ Therefore: $$K_{\Omega,\psi}=\dfrac{\sup_{\Omega}{\lvert{\nabla\psi}\rvert}}{\inf_{\Omega}{\lvert{\nabla\psi}\rvert}}\leq\dfrac{B}{\beta m}.$$ b One has $$\sup_{r\in [0,1]}{\lvert{\psi^{-1}(r)}\rvert}=L={\lvert{\Sigma_2}\rvert}.$$ c If $\Sigma_2$ is also convex, then $m\geq \beta/B$ hence we can take $K=\beta^2/B^2$.

The proof of the Proposition \[estimate doubly convex\] depends on the following steps.

[**Step 1.**]{} [*On the ray $\gamma_x$ joining $x$ to $Q(x)$, consider the point $Q_t(x)$ at distance $t$ from $x$, and let $\theta_x(t)$ be the angle between $\gamma'_x$ and $\nabla\psi(Q_t(x))$. Then the function $$h(t)=\cos(\theta_x(t))$$ is non-increasing in $t$. As $\theta_x(r(x))=\theta_x$ we have in particular: $$\cos(\theta_x(t))\geq \cos(\theta_x)\geq m$$ for all $t\in [0,r(x)]$ and $x\in\Sigma_1$.*]{}

[**Step 2.**]{} [*The function $r\to{\lvert{\psi^{-1}(r)}\rvert}$ is non-decreasing in $r$.*]{}

[**Step 3.**]{} [*If $\Sigma_2$ is also convex we have $m\geq \beta/B$.*]{}

We will prove Steps 1-3 below.

[**Proof of Proposition \[estimate doubly convex\]**]{}. a) At any point of $\Omega$, let $\nabla^R\psi$ denote the radial part of $\nabla\psi$, which is the gradient of the restriction of $\psi$ to the ray passing through the given point. As such restriction is a linear function, one sees that $$\dfrac{1}{B}\leq{\lvert{\nabla^R\psi}\rvert}\leq \dfrac{1}{\beta}.$$ Since ${\lvert{\nabla\psi}\rvert}\geq{\lvert{\nabla^R\psi}\rvert}$ one gets immediately $${\lvert{\nabla\psi}\rvert}\geq\dfrac{1}{B}.$$ Note that $\theta_x(t)$, as defined above, is precisely the angle between $\nabla\psi$ and $\nabla^R\psi$, so that, using Step 1,

$${\lvert{\nabla^R\psi}\rvert}={\lvert{\nabla\psi}\rvert}\cos\theta_x(t)\geq m{\lvert{\nabla\psi}\rvert}$$ hence: $${\lvert{\nabla\psi}\rvert}\leq \dfrac{1}{m}{\lvert{\nabla^R\psi}\rvert}\leq\dfrac{1}{\beta m}.$$ as asserted. It is clear that b) and c) are immediate consequences of Steps 2-3.

**Proof of Step 1.** We use a suitable parametrization of $\Omega$. Let $l$ be the length of $\Sigma_1$ and consider a parametrization $\gamma:[0,l]\to \Sigma_1$ by arc-length $s$ with origin at a given point in $\Sigma_1$. Let $N(s)$ be the outer normal vector to $\Sigma_1$ at the point $\gamma(s)$. Consider the set: $$\tilde\Omega=\{(t,s)\in [0,\infty)\times [0,l): t\leq \rho(s)\}$$ where we have set $\rho(s)=r(\gamma(s))$. The starlike property implies that the map $
\Phi:\tilde\Omega\to \Omega
$ defined by $$\Phi(t,s)=\gamma(s)+tN(s)$$ is a diffeomorphism. Let us compute the Euclidean metric tensor in the coordinates $(t,s)$. Write $\gamma'(s)=T(s)$ for the unit tangent vector to $\gamma$ and observe that $N'(s)=k(s)T(s)$, where $k(s)$ is the curvature of $\Sigma_1$ which is everywhere non-negative because $\Sigma_1$ is convex. Then: $$\twosystem
{d\Phi(\dfrac{{\partial}}{{\partial}t})=N(s)}
{d\Phi(\dfrac{{\partial}}{{\partial}s})=(1+tk(s))T(s)}$$ If we set $\Theta(t,s)=1+t k(s)$ the metric tensor is: $$g={{\begin{pmatrix}}{1&0\\}{0&\Theta^2\\}{\end{pmatrix}}}$$ and an orthonormal basis is then $(e_1,e_2)$, where $$e_1=\dfrac{{\partial}}{{\partial}t}, \quad e_2=\dfrac{1}{\Theta}\dfrac{{\partial}}{{\partial}s}.$$ In these coordinates, our function $\psi$ is written: $$\psi(t,s)=\dfrac{t}{\rho(s)}.$$ Now $$\twosystem
{{\langle{\nabla\psi},{e_1}\rangle}={\dfrac{{\partial}\psi}{\bdt}}=\dfrac{1}{\rho(s)}}
{{\langle{\nabla\psi},{e_2}\rangle}=\dfrac{1}{\Theta}{\dfrac{{\partial}\psi}{\bds}}=-\dfrac{t\rho'(s)}{\Theta(t,s)\rho(s)^2}}.$$ It follows that $${\lvert{\nabla\psi}\rvert}^2=\dfrac{1}{\rho^2}+\dfrac{t^2\rho'^2}{\Theta^2\rho^4}=
\dfrac{\Theta^2\rho^2+t^2\rho'^2}{\Theta^2\rho^4}.$$ Recall the radial gradient, which is the orthogonal projection of $\nabla\psi$ on the ray, whose direction is given by $e_1$. If we fix $x\in\Sigma_1$, we have $$\theta_x(t)=\text{angle between $\nabla\psi$ and $e_1$}$$ and we have to study the function $$h(t)=\cos\theta_x(t)=\dfrac{{\langle{\nabla\psi},{e_1}\rangle}}{{\lvert{\nabla\psi}\rvert}}=\dfrac{1}{\rho(s){\lvert{\nabla\psi}\rvert}}$$ for a fixed $s$. From the above expression of ${\lvert{\nabla\psi}\rvert}$ and a suitable manipulation we see $$h(t)^2=\dfrac{\Theta^2}{\Theta^2+t^2g^2}$$ where $g=\rho'(s)/\rho(s)$. Now $$\begin{aligned}
\dfrac{d}{dt}\dfrac{\Theta^2}{\Theta^2+t^2g^2}&=\dfrac{2t\Theta g^2}{(\Theta^2+t^2g^2)^2}
(t{\dfrac{{\partial}\Theta}{\bdt}}-\Theta)\\
\end{aligned}$$ As $\Theta(t,s)=1+tk(s)$ one sees that $t{\dfrac{{\partial}\Theta}{\bdt}}-\Theta=-1$ hence $$\dfrac{d}{dt}h(t)^2=-\dfrac{2t\Theta g^2}{(\Theta^2+t^2g^2)^2}\leq 0$$ Hence $h(t)^2$ is non-increasing and, as $h(t)$ is positive, it is itself non-increasing.

[**Proof of Step 2.**]{} In the coordinates $(t,s)$ the curve $\psi^{-1}(r)$ is parametrized by $\alpha:[0,l]\to\tilde\Omega$ as follows: $$\alpha(u)=(r\rho(u),u)\quad u\in [0,l].$$ Then: $$\begin{aligned}
{\lvert{\psi^{-1}(r)}\rvert}&=\int_0^l\sqrt{g(\alpha'(u),\alpha'(u))}\,du\\
&=\int_0^l\sqrt{r^2\rho'(u)^2+(1+rk(u)\rho(u))^2}\,du
\end{aligned}$$ Convexity of $\Sigma_1$ implies that $k(u)\geq 0$ for all $u$; differentiating under the integral sign with respect to $r$ one sees that indeed $\frac{d}{dr}{\lvert{\psi^{-1}(r)}\rvert}\geq 0$ for all $r\in [0,1]$.

[**Proof of Step 3.**]{} Let $T_x$ be the tangent line to $\Sigma_2$ at $Q(x)$ and $H(x)$ the point of $T_x$ closest to $x$. As $\Sigma_2$ is convex, $H(x)$ is not an interior point of $\Omega$, hence $$d(x,H(x))\geq\beta.$$ The triangle formed by $x, Q(x)$ and $H(x)$ is rectangle in $H(x)$, then we have: $$r(x)\cos\theta_x=d(x,H(x)).$$ As $r(x)\leq B$ we conclude: $$B \cos\theta_x\geq \beta,$$ which gives the assertion.

Sharpness of the lower bound {#sharpness}
============================

An upper bound
--------------

In this short paragraph, we give a simple way to get an upper bound when the potential $A$ is *closed*. Then, we will use this in different kinds of examples, in order to show that the assumptions of Theorem \[main2\] are sharp. The geometric idea is the following: if we have a region $D \subset \Omega$ such that the first absolute cohomology group $H^1(D)$ is $0$, then we can estimate from above the spectrum of $\Delta_A$ in $\Omega$ in terms of the spectrum of the usual Laplacian on $D$. The reason is that the potential $A$ is $0$ on $D$ up to a gauge transformation; then, on $D$, $\Delta_A$ becomes the usual Laplacian and any eigenfunction of the Laplacian on $D$ may be extended by $0$ on $\Omega$ and thus used as a test function for the magnetic Laplacian on the whole of $\Omega$.

Let us give the details. Let $D$ be a closed subset of $\Omega$ such that, for some (small) $\delta>0$ one has $H^1(D^{\delta},{{\bf R}})=0$, where $D^{\delta}=\{p\in \Omega: {\rm dist}(p,D) < \delta\}$. This happens when $D^{\delta}$ has a retraction onto $D$. We write $$\partial D= (\partial D\cap \partial \Omega) \cup (\partial D \cap \Omega)=\partial^{\rm ext}D\cup\partial^{\rm int}D$$ and we denote by $(\nu_j(D))_{j=1}^{\infty}$ the spectrum of the Laplacian acting on functions, with the Neumann boundary condition on $\partial^{\rm ext}D$ (if non empty) and the Dirichlet boundary condition on $\partial^{\rm int}D$.

\[upperharmonic\] Let $\Omega$ be a compact manifold with smooth boundary and $A$ a closed potential on $\Omega$. Assume that $D\subset \Omega$ is a compact subdomain such that $H^1(D,\textbf R)=H^1(D^{\delta},\textbf R)=0$ for some $\delta>0$. Then we have $$\lambda_k(\Omega,A) \le \nu_k(D)$$ for each $k\geq 1$.

**Proof.** We recall that for any function $\phi$ on $\Omega$, the operator $\Delta_A$ and $\Delta_{A+d\phi}$ are unitarily equivalent and have the same spectrum. As $A$ is closed and, by assumption, $H^1(D^{\delta},\textbf R)=0$, $A$ is exact on $D^{\delta}$ and there exists a function $\tilde \phi$ on $D^{\delta}$ such that $A+d\tilde\phi=0$ on $D^{\delta}$.

We consider the restriction of $\tilde\phi$ to $D$ and extend it differentiably on $\Omega$ by using a partition of unity $(\chi_1,\chi_2)$ subordinated to $(D^{\delta},\Omega/D)$. Then, setting $$\phi\doteq\chi_1 \tilde\phi$$ we see that $\phi$ is a smooth function on $\Omega$ which is equal to $\tilde\phi$ on $D$ so that, on $D$, one has $A+d\phi=0$. We consider the new potential $\tilde A=A+d\phi$ and observe that $\tilde A=0$ on $D$.

Now consider an eigenfunction $f$ for the mixed problem on $D$ (Neumann boundary conditions on $\partial^{\rm ext}D$ and Dirichlet boundary conditions on $\partial^{\rm int}D$), and extend it by $0$ on $\Omega\setminus D$. As $\tilde A=0$ on $D$, we see that $${\lvert{\nabla^{\tilde A}f}\rvert}^2={\lvert{\nabla f}\rvert}^2,$$ and we get a test function having the same Rayleigh quotient as that of $f$. Thanks to the usual min-max characterization of the spectrum, we obtain, for all $k$: $$\lambda_k(\Omega,A) = \lambda_k(\Omega,\tilde A)\le \nu_k(D).$$

Sharpness
---------

We will use Proposition \[upperharmonic\] to show the sharpness of the hypothesis in Theorem \[main2\]. Let us first show that we need to control the ratio $\frac{BL}{\beta}$.

\[Blarge\] In the first situation, we give an example where the ratio $\frac{BL}{\beta} \to \infty$ and the distance $\beta$ between the two components of the boundary is uniformly bounded from below. We want to show that $\lambda_1\to 0$. We consider an annulus $\Omega$ composed of two concentric balls of radius $1$ and $R+1$ and same center, with $R\to \infty$. We have $B=\beta=R$ and $L \to \infty$.

From the assumptions we get the existence of a point $x\in \Omega$ such that the ball $B(x,\frac{R}{2})$ of center $x$ and radius $\frac{R}{2}$ is contained in $\Omega$. Proposition \[upperharmonic\] implies that $\lambda_1(\Omega,A)$ is bounded from above by the first eigenvalue of the Dirichlet problem for the Laplacian of the ball, which is proportional to $\frac{1}{R^2}$ and tends to zero because $R\to\infty$.

\[example1\] Next, we construct an example to show that if the distance $\beta$ tends to $0$ and $B$ and $L$ are uniformly bounded from below and from above, then again $\lambda_1 \to 0$. We again use Proposition \[upperharmonic\]. Fix the rectangles : $$R_2 = [-4,4]\times [0,4], \quad R_{1,\epsilon}=[-3,3]\times [\epsilon,2]$$ and consider the region $\Omega_{\epsilon}$ given by the closure of $R_2\setminus R_{1,\epsilon}$. Note that $\Omega_{\epsilon}$ is a planar annulus whose boundary components are convex and get closer and closer as $\epsilon\to 0$.

![$\lambda_1 \to 0$ as $\epsilon \to 0$](PVP.jpg){width="70mm"}

We show that, for any closed potential $A$ one has: $$\label{small}
\lim_{\epsilon\to 0}\lambda_1(\Omega_{\epsilon},A)=0.$$ Consider the simply connected region $D_{\epsilon}\subset\Omega_{\epsilon}$ given by the complement of the rectangle $[-1,1]\times [0,\epsilon]$. Now $D_{\epsilon}$ has trivial $1$-cohomology; by Proposition \[upperharmonic\], to show it is enough to show that $$\label{enough}
\lim_{\epsilon\to 0}\nu_1(D_{\epsilon})=0.$$ By the min-max principle : $$\nu_1(D_{\epsilon})=\inf\Big\{ \frac{\int_{D_{\epsilon}}\vert \nabla f\vert^2}{\int_{D_{\epsilon}}f^2} : f=0\,\,\text{on}\,\, {\partial}D_{\epsilon}^{\rm int} \Big\}$$ where $${\partial}D_{\epsilon}^{\rm int} =\{(x,y)\in\Omega_{\epsilon}:x=\pm 1, y\in [0,\epsilon]\}.$$ Define the test-function $f:D_{\epsilon}\to{{\bf R}}$ as follows. $$f=\threesystem{1\quad\text{on the complement of $[-2,2]\times[0,\epsilon]$}}
{x-1\quad\text{on $[1,2]\times [0,\epsilon]$}}
{-x-1\quad\text{on $[-2,-1]\times [0,\epsilon]$}}$$ One checks easily that, for all $\epsilon$: $$\int_{D_{\epsilon}}\vert \nabla f\vert^2=2\epsilon, \quad \int_{D_{\epsilon}}f^2\geq {\rm const}>0$$ Then follows immediately by observing that the Rayleigh quotient of $f$ tends to $0$ as $\epsilon\to 0$

\[example small\] In the example we constructed previously the two boundary components approach each other along a common set of positive measure (precisely, a segment of total length $6$). In the next example we sketch a construction showing that, in fact, this is not necessary.

So, let us fix the outside curve $\Sigma_2$ and choose a family of inner convex curves $\Sigma_1$ such that $B$ is bounded below (say, $B\ge 1$) and $\beta \to 0$ (no other assumption is made). Then, we want to show that $\lambda_1(\Omega,A)\to 0$.

Fix points $x\in \Sigma_2$, $y\in \Sigma_1$ such that $d(x,y)=\beta$. We take $b=2\beta$ and introduce the balls of center $x$ and radius $b$ and $\sqrt b$, denoted by $B(x,b)$ and $B(x,\sqrt b)$, respectively. Then the set $D=\Omega\setminus (B(x,b)\cap\Omega)$ is simply connected so that, by Proposition \[upperharmonic\]: $$\lambda_1(\Omega,A)\leq \nu_1(D)$$ and it remains to show that $\nu_1(D)\to 0$ as $b\to 0$.

Introduce the function $F(r)$ ( $r$ being the distance to $x$): $$F(r)=\threesystem{1\quad\text{on the complement of $B(x,\sqrt b)$}}
{0\quad\text{on $B(x,b)$}}
{\frac{-2}{\ln b} (\ln r -\ln b)\quad\text{on $B(x,\sqrt b)-B(x,b)$}}$$ and let $f$ be the restriction of $F$ to $D$. As $f=0$ on ${\partial}^{\rm int}D={\partial}B(x,b)\cap\Omega$, we see that $f$ is a test function for the eigenvalue $\nu_1(D)$. A straightforward calculation shows that, as $b\to 0$, we have $$\int_D \vert \nabla f\vert^2 \to 0;$$ on the other hand, as $B \ge 1$, the volume of $D$ is uniformly bounded from below, which implies that $$\int_D f^2 \ge C >0.$$ We conclude that the Rayleigh quotient of $f$ tends to $0$ as $b \to 0$, which shows the assertion.

\[example2\] The following example shows that we need to impose some condition on the outer curve in order to get a positive lower bound as in Theorem \[main2\].

It is an easy and classical fact that, in order to create a small eigenvalue for the Neumann problem, it is sufficient to deform a domain locally, near a boundary point, as indicated by the mushroom-shaped region shown in the figure below. Up to a gauge transformation, we can suppose that the potential $A$ is locally $0$ in a neigborhood of the mushroom, and we have to estimate the first eigenvalue of the Laplacian with Dirichlet boundary condition at the basis of the mushroom (which is a segment of length $\epsilon$) and Neumann boundary condition on the remaining part of its boundary, as required by Proposition \[upperharmonic\].

![A local deformation implying $\lambda_1 \to 0$](Deformation.jpg){width="70mm"}

The only point is to take the value of the parameter $\epsilon$ much smaller than $\delta$ as $\delta \to 0$. Take for example $\epsilon =\delta^4$ and consider a function $u$ taking value $1$ in the square of size $\delta$ and passing linearly from $1$ to $0$ outside the rectangle of sizes $\epsilon,\delta$. The norm of the gradient of $u$ is $0$ on the square of size $\delta$ and $\frac{1}{\delta}$ in the rectangle of size $\delta,\epsilon$.

Then the Rayleigh quotient is $$R(u) \le \frac{\frac{1}{\delta^2}\delta \epsilon}{\delta^2}=\frac{\epsilon}{\delta^3}$$ which tends to $0$ as $\delta \to 0$.

Moreover, we can make such local deformation keeping the curvature of the boundary uniformly bounded in absolute value (see Example 2 in [@CGI]).

Appendix
========

Spectrum of circles and Riemannian products {#riemannian circle}
-------------------------------------------

We first prove Proposition \[circle\].

Let then $(M,g)$ be the circle of length $L$ with metric $g=\theta(t)^2dt^2$, where $t\in [0,L]$ and $\theta(t)$ is periodic of period $L$. Given the $1$-form $A=H(t)dt$ we first want to find the harmonic $1$-form $\omega$ which is cohomologous to $A$; that is, we look for a smooth function $\phi$ so that $
\omega=A+d\phi
$ is harmonic. Now a unit tangent vector field to the circle is $$e_1=\dfrac{1}{\theta} \dfrac{d}{dt}.$$ Write $\omega=G(t)\,dt$. Then $$\delta\omega=-\dfrac 1{\theta}\Big(\dfrac{G}{\theta}\Big)'.$$ As any $1$-form on the circle is closed, we see that $\omega$ is harmonic iff $G(t)=c\theta(t)$ for a constant $c$. We look for $\phi$ and $c\in{{\bf R}}$ so that $$\phi'=-H+c\theta.$$ As $\phi$ must be periodic of period $L$, we must have $\int_0^L\phi'=0$. As the volume of $M$ is $L$, we also have $\int_0^L\theta=L$. This forces $$c=\dfrac{1}{L}\int_0^LH(t)\,dt.$$ On the other hand, as the curve $\gamma(t)=t$ parametrizes $M$ with velocity $\frac{d}{dt}$, one sees that the flux of $A$ across $M$ is given by $$\Phi^A=\dfrac{1}{2\pi}\int_0^LH(t)\,dt.$$ Therefore $
c=\frac{2\pi}{L} \Phi^A
$ and a primitive could be $$\phi(t)=-\int_0^tH+c\int_0^t\theta.$$ Conclusion:

[$\bullet\quad$]{}[*The form $A=H(t)dt$ is cohomologous to the harmonic form $\omega=c\theta\, dt$ with $c=\frac{2\pi}{L} \Phi^A$*]{}.

We first compute the eigenvalues. By gauge invariance, we can use the potential $\omega$. In that case $$\Delta_{\omega}=-\nabla^{\omega}_{e_1}\nabla^{\omega}_{e_1}.$$ Now $$\nabla^{\omega}_{e_1}u=\dfrac{u'}{\theta}-icu$$ hence $$\nabla^{\omega}_{e_1}\nabla^{\omega}_{e_1}u=\dfrac{1}{\theta}\Big(\dfrac{u'}{\theta}-icu\Big)'-ic\Big(\dfrac{u'}{\theta}-icu\Big).$$ After some calculation, the eigenfunction equation $\Delta_{\omega}u=\lambda u$ takes the form: $$-u''+\dfrac{\theta'}{\theta}u'+2ic \theta u'+c^2\theta^2 u=\lambda\theta^2 u.$$ Recall the arc-length function $s(t)=\int_0^t\theta(\tau)\,d\tau$. We make the change of variables: $$u(t)=v(s(t)), \quad\text{that is}\quad v=u\circ s^{-1}.$$ Then: $$\twosystem
{u'=v'(s)\theta}
{u''=v''(s)\theta^2+v'(s)\theta'}$$ and the equation becomes: $$-v''+2ic v'+c^2v=\lambda v$$ with solutions : $$v_k(s)=e^{\frac{2\pi i k}{L}s}, \quad \lambda=\dfrac{4\pi^2}{L^2}(k-\Phi^A)^2, \quad k\in\bf Z.$$ Now Gauge invariance says that $$\Delta_{A+d\phi}=e^{i\phi}\Delta_Ae^{-i\phi};$$ and $v_k$ is an eigenfunction of $\Delta_{A+d\phi}$ iff $e^{-i\phi}v_k$ is an eigenfunction of $\Delta_A$. Hence, the eigenfunctions of $\Delta_A$ (where $A=H(t)\,dt$) are $$u_k=e^{-i\phi}v_k,$$ where $\phi(t)=-\int_0^tH+c\, s(t)$ and $c=\frac{2\pi}{L}\Phi^A$. Explicitly: $$\label{eigenfunctions}
u_k(t)=e^{i\int_0^tH}e^{\frac{2\pi i(k-\Phi^A)s(t)}{L}}$$ as asserted in Proposition \[circle\].

Let us know verify the last statement. If the metric is $g=dt^2$ then $\theta(t)=1$ and $s(t)=t$. If $A$ is a harmonic $1$-form then it has the expression $A=\frac{2\pi \Phi^A}{L}dt$. Taking into account we indeed verify that $u_k(t)=e^{\frac{2\pi i k}{L}t}$.

[$\bullet\quad$]{}We now prove Proposition \[cyl\].

Here we assume that $\Omega$ is a Riemannian product $[0,a]\times{{\bf S}^{1}}(\frac{L}{2\pi})$ with coordinates $(r,t)$ and the canonical metric on the circle. We fix a closed potential $A$ on $\Omega$. By gauge invariance we can assume that $A$ is a Coulomb gauge, and by what we said above we have easily $$A=\dfrac{2\pi \Phi^A}{L}\,dt.$$ Then $A$ restrict to zero on $[0,a]$; as $A(N)=0$ on ${\partial}\Omega$ the magnetic Neumann conditions reduce simply to ${\dfrac{{\partial}u}{\bdN}}=0$. At this point we apply a standard argument of separation of variables; if $\phi(r)$ is an eigenfunction of the usual Neumann Laplacian on $[0,a]$, and $v(t)$ is an eigenfunction of $\Delta_A$ on ${{\bf S}^{1}}(\frac{L}{2\pi})$, we see that the product $u(r,t)=\phi(r)v(t)$ is indeed an eigenfunction of $\Delta_A$ on $\Omega$. As the set of eigenfunctions we obtain that way is a complete orthonormal system in $L^2(\Omega)$, we see that each eigenvalue of the product is the sum of an eigenvalue in the Neumann spectrum of $[0,a]$ and an eigenvalue in the magnetic spectrum of the circle, as computed before. We omit further details.

Proof of Lemma \[calculus\] {#technical lemma}
---------------------------

For simplicity of notation, we give the proof when $a=L=1$. This will not affect generality. Then, assume that $s : [0,1]\times [0,1]\to{{\bf R}}$ is smooth, non-negative and satisfies $$s(0,t)=t,\quad s(r,0)=0,\quad s(r,1)=1\quad\text{and}\quad {\dfrac{{\partial}s}{\bdt}}(r,t)\doteq \theta(r,t)>0.$$ Assume the identity $$\label{identity}
F(t)=p(r)\cos(\pi s(r,t))+q(r)\sin(\pi s(r,t))$$ for real-valued functions $F(t),p(r),q(r)$, such that $p(r)^2+q(r)^2>0$. Then we must show: $$\label{sr}
{\dfrac{{\partial}s}{\bdr}}=0$$ everywhere.

Differentiate with respect to $t$ and get: $$\label{fprime}
F'(t)=-\pi p(r)\theta(r,t)\sin(\pi s)+\pi q(r)\theta(r,t)\cos(\pi s)$$ and we have the following matrix identity $${{\begin{pmatrix}}{\cos(\pi s)&\sin(\pi s)\\}{-\pi\theta\sin(\pi s)&\pi\theta\cos(\pi s)\\}{\end{pmatrix}}}{\begin{pmatrix}p\\#2\end{pmatrix}}={\begin{pmatrix}F\\#2\end{pmatrix}}.$$ We then see: $$p(r)=F(t)\cos(\pi s)-\dfrac{F'(t)}{\pi\theta}\sin(\pi s).$$ Set $t=0$ so that $s=0$ and $p(r)=F(0)\doteq p$ is constant; the previous identity becomes $$\label{p}
p=F(t)\cos(\pi s)-\dfrac{F'(t)}{\pi\theta}\sin(\pi s).$$ Observe that: $$\label{changes}
\twosystem
{F'(0)=\pi q(r)\theta(r,0)}
{F'(1)=-\pi q(r)\theta(r,1)}$$ [$\bullet\quad$]{}Assume $F'(0)=0$. Then, as $\theta(t,r)$ is positive one must have $q(r)=0$ for all $r$, hence $p\ne 0$ and $
F(t)=p\cos(\pi s),
$ from which, differentiating with respect to $r$, one gets easily $
{\dfrac{{\partial}s}{\bdr}}=0
$ and we are finished.

[$\bullet\quad$]{}We now assume that $F'(0)\ne 0$: then we see from that $q$ is not identically zero and the smooth function $F':[0,1]\to{{\bf R}}$ changes sign. This implies that

[$\bullet\quad$]{}[*there exists $t_0\in (0,1)$ such that $F'(t_0)=0$.*]{}

Now evaluated at $t=t_0$ gives: $$p=F(t_0)\cos(\pi s(r,t_0))$$ for all $r$. Differentiate w.r.t. $r$ and get, for all $r\in [0,1]$: $$0=\sin(\pi s(r,t_0)){\dfrac{{\partial}s}{\bdr}}(r,t_0).$$ Since $s(r,t)$ is increasing in $t$, we have $$0<s(r,t_0)<s(r,1)=1.$$ Hence $\sin(\pi s(r,t_0))>0$ and we get $${\dfrac{{\partial}s}{\bdr}}(r,t_0)=0.$$ writes: $$F(t)=p\cos(\pi s)+q(r)\sin(\pi s),$$ and then, differentiating w.r.t. $r$: $$0=-p\pi \sin(\pi s){\dfrac{{\partial}s}{\bdr}}+q'(r)\sin(\pi s)+\pi q(r)\cos(\pi s){\dfrac{{\partial}s}{\bdr}}.$$ Evaluating at $t=t_0$ we obtain $
0=q'(r)\sin(\pi s(r,t_0))
$ which implies $$q'(r)=0$$ hence $q(r)=q$, a constant. We conclude that $$F(t)=p\cos(\pi s)+q\sin(\pi s)$$ for constants $p,q$. We differentiate the above w.r.to $r$ and get: $$0=\Big(-\pi p \sin(\pi s)+\pi q\cos(\pi s)\Big){\dfrac{{\partial}s}{\bdr}}$$ for all $(r,t)\in [0,1]\times [0,1]$. Now, the expression inside parenthesis is non-zero a.e. on the square. Then one must have ${\dfrac{{\partial}s}{\bdr}}=0$ everywhere and the final assertion follows.

Bruno Colbois

Université de Neuchâtel, Institut de Mathématiques\
Rue Emile Argand 11\
CH-2000, Neuchâtel, Suisse

bruno.colbois@unine.ch

Alessandro Savo

Dipartimento SBAI, Sezione di Matematica\
Sapienza Università di Roma, Via Antonio Scarpa 16\
00161 Roma, Italy

alessandro.savo@sbai.uniroma1.it
