1$ we have a partial result, namely we show that all self-adjoint extensions of $\Delta_{k,p}$ have pure point spectrum but only for the metric \eqref{mc} (see Proposition \ref{p:essDelta}). In a recent paper Antoci \cite{francesca1} studies the essential spectrum of a manifold $X$ with ends as above, with metric on the end \begin{equation}\label{anto} e^{-2(a+1)t}dt^2+e^{-2bt}h. \end{equation} Antoci assumes that $X$ is complete so $a\leq -1$. But with the change of variables $y=e^{(b-a-1)t}$ the metric \eqref{anto} is a constant multiple of \eqref{mc} under the assumption that $b-a-1> 0$ and $b <0$. Thus our results have some non-empty intersection with \cite{francesca1}; we improve one of Antoci's results by showing that $0$ cannot be isolated in the essential spectrum for a general $M$. (for $M=S^{n-1}$ this is proved in \cite[Theorem 6.1]{francesca1}). This issue is actually important, since the absence of $0$ from the essential spectrum is equivalent to the Hodge decomposition \[L^2(X,\Lambda^k X,g_p)=\ker(\Delta_{k,p})\oplus\im(\Delta_{k,p}).\] We also get somewhat more refined information on the nature of the essential spectrum. Namely for the metric \eqref{mc} we show the absence of singular continuous spectrum (a classical fact for $k=0$). The second topic of this article is the so-called magnetic Laplacian on a manifold $X$ as above. We develop this in Section \ref{sectionmagn}. A magnetic field $B$ is a smooth exact $2$-form on $X$. There exists a real $1$-form $A$, called magnetic potential, satisfying $dA=B$. Set $\dA:=d+iA\wedge: \cun(X)\to\cun(X,T^*X)$. The magnetic Laplacian on $\cun(X)$ is given by \begin{equation}\label{eq:intro1} \Delta_A:=\dA^*\dA \end{equation} When the manifold is complete, this operator is essentially self-adjoint \cite{shubin}. Given two magnetic potentials $A$ and $A'$ such that $A-A'$ is exact, the two magnetic Laplacians $\Delta_A$ and $\Delta_{A'}$ are unitarily equivalent (this property is called gauge invariance). Hence when $H^1(X,\rz)=0$, the spectral properties of the magnetic Laplacian do not depend on the choice of the magnetic potential $A$. We have found a strong relationship between the emptiness of the essential spectrum of $\Delta_A$ and the cohomology class $[B]$ inside the relative cohomology group $H^2(X,M)$. We show first that if the magnetic field does not vanish on the boundary, then the operator $\Delta_A$ has compact resolvent. In comparison, in $\rz^n$ with the Euclidean metric this happens when the magnetic field blows up at infinity with a certain control on its derivatives (see \cite{hm} and references therein). Furthermore, if $B$ does vanish on $M$ but it defines a non-integral cohomology class $[B]$ inside the relative cohomology group $H^2(X,M)$ (up to a factor of $2\pi$), then the same conclusion holds. In $\rz^n$, a compactly supported magnetic field does not affect the essential spectrum \cite{CFKS, MP} and furthermore the wave operators are asymptotically complete for a certain choice of gauge \cite{LT}. In contrast, on a manifold $X$ as above, in dimensions other than $3$ we find in particular that a generic magnetic field $B$ with compact support will turn off the essential spectrum of $\Delta$. It seems quite surprising on one hand that we get the (non)integrality condition, and on the other hand that there are no orientable examples in dimension $3$. We stress that it is not the size of the magnetic field which kills the essential spectrum, but a more subtle %arithmetic phenomenon which has yet to be interpreted physically. In the case of $H^1(X,\rz)\neq 0$, in quantum mechanics, it is known that the choice of a potential vector has a physical meaning. This is known as the Aharonov-Bohm effect \cite{AB}. In $\rz^n$ with a bounded obstacle, this phenomenon can be seen through a difference of phase of the waves arising from two non-homotopic paths that circumvent the obstacle. Note that the essential spectrum is independent of this choice. In contrast, in our setting we show that the essential spectrum itself can be turned off by a suitable choice of magnetic potential. In all these instances when the operators have empty essential spectra, the eigenvalues turn out to satisfy an asymptotic law that we prove using results from \cite{wlom}. We state here only one such result and we refer to the body of the paper for the others. \begin{te}\label{th3} In the setting of Theorem \ref{th:0}, assume that $h^k(M)=h^{k-1}(M)=0$. Let $N_{k,p}$ be the counting function of the eigenvalues of $\Delta_{k,p}$. Then \begin{equation}\label{rg} N_{k,p}(\lambda) \approx \begin{cases} C_1\lambda^{n/2}& \text{for $1/n< p<\infty$,}\\ C_2\lambda^{n/2}\log \lambda &\text{for $p=1/n$,}\\ C_3\lambda^{1/2p}&\text{for $0

0$, $g_p$ a metric on $X$ given by \eqref{pme} near $\partial X$ and $A\in\cun(X,T^*X)$ a complex-valued $1$-form satisfying near $\partial X$ \begin{equation}\label{aft} A=\varphi(x)\frac{dx}{x^2}+\theta(x)\end{equation} where $\varphi\in\cun(\oX)$ and $\theta\in\cun([0,\varepsilon)\times M, \Lambda^1(M))$. Assume that, on each connected component $M_0$ of $M$, either $\varphi_0:=\varphi(0)$ is not constant, or that $\theta_0:=\theta(0)$ is not closed, or the cohomology class $[\theta_0]\in H^1_{\mathrm{dR}}(M_0)$ does not belong to the image of \[2\pi H^1(M_0,\zz)\to H^1(M_0,\cz)\simeq H^1_{\mathrm{dR}}(M_0)\otimes\cz.\] Then $\Delta_A$ has pure-point spectrum. Its counting function satisfies \begin{equation*} N_{A,p}(\lambda) \approx \begin{cases} C_1\lambda^{n/2}& \text{for $1/n< p<\infty$,}\\ C_2\lambda^{n/2}\log \lambda &\text{for $p=1/n$,}\\ C_3\lambda^{1/2p}&\text{for $0

0$ the form $\theta'_x$ is exact. By the Hodge decomposition theorem, the space of exact forms on $M$ is closed, so the limit $\theta'_0$ is also exact. Now $dw$ is an exact cusp form, in particular it is closed. This implies that $x^2\px\theta'=d^M \varphi'$. Setting $x=0$ we deduce $d^M(\varphi')_{x=0}=0$, or equivalently $\varphi'_{|x=0}$ is constant. Hence the conditions from Theorem \ref{thmag} on the vector potential are satisfied simultaneously by $A$ and $A'$. \subsection{On the non-stability of the essential spectrum} In $\rz^n$, with the flat metric, it is shown in \cite{MP} that the essential spectrum of a magnetic Laplacian (for a bounded magnetic field) is determined by the restriction of the magnetic field to the boundary of a suitable compactification of $\rz^n$. In other words, only the behavior of the magnetic field at infinity plays a r\^ole in the computation of the essential spectrum. Moreover, \cite[Theorem 4.1]{KS} states that the non-emptiness of the essential spectrum is preserved by the addition of a bounded magnetic field, again for the Euclidean metric on $\rz^n$. In contrast with these examples, Theorem \ref{thmag} indicates that in general the essential spectrum is not stable (and may even vanish) under compact perturbation of the magnetic field. This phenomenon is quite remarkable. \subsection{The case $H^1(X)\neq 0$} The gauge invariance property does not hold in this case. Then the operators $\Delta_A$ and $\Delta_{A'}$ may have different essential spectra even when $d(A-A')=0$. We give below an example where the essential spectrum of $\Delta_{A'}$ is empty, while that of $\Delta_{A}$ is not. \begin{example} Consider the manifold $\oX=(S^1)^{n-1}\times [0,1]$ with a metric $g_p$ as in \eqref{pme} near the two boundary components. Let $\theta_j$ be variables on the torus $(S^1)^{n-1}$, so $e^{i\theta_j}\in S^1$. Choose the vector potentials $A$ to be $0$, and $A'$ to be the closed form $\mu d\theta_1$ for some $\mu\in\rz$. In this example the magnetic field $B$ vanishes. We have mentioned in the introduction that the Laplacian on functions has non-empty essential spectrum. However, the class $[i_M^*(A')]$ is an integer multiple of $2\pi$ if and only if $\mu\in\zz$. Therefore by Theorem \ref{thmag}, for $\mu\in\rz\setminus\zz$ the essential spectrum of $\Delta_{A'}$ is empty. \end{example} This effect is known as the Aharonov-Bohm effect \cite{AB}. The phenomenon we point out is surprising in light of what is known from the case of $\rz^n$ with an bounded obstacle. Indeed, the most you can expect in this case is a phase difference of the waves due to the choice of two non-homotopic paths that circumvent the obstacle. Note that the essential spectrum will not depend on the choice of $A$ in the case. \subsection{The case $H^1(X)= 0$} Assume that $H^1(X,\rz)=0$ so $\Delta_A$ is determined by $B$ up to unitary equivalence (by gauge invariance). Assume that $B$ is a smooth $2$-form on $\oX$. which is closed on $X$ and such that its restriction to $X$ is exact. Then there exists $A\in \cun(\oX,\Lambda^1(\oX))$ such that $B=dA$ (since the cohomology of the de Rham complex on $\oX$ equals the singular cohomology of $\oX$, hence that of $X$). Assume moreover that the pull-back of $B$ to $M$ vanishes. Then $B$ defines a relative cohomology class in $H^2(\oX,M,\rz)$. Besides, the vanishing of $i_M^*(B)$ means that $i^*_M(A)$ is closed. It is easy to see that $B$ determines a $1$-cohomology class $[i_M^*(A)]$ on $M$. Notice that $\partial[i^*_M(A)]=[dA]=[B]$, where $\partial:H^1(M,\rz)\to H^2(\oX,M,\rz)$ is the connecting homomorphism in the cohomology long exact sequence of the pair $(\oX,M)$ with real coefficients. Consider the same map but for integer coefficients, and the natural arrows from $\zz$- to $\rz$-cohomology, which are just tensoring with $\rz$. \begin{equation}\label{cohoz} \begin{CD} H^{1}(M,\rz)@>{\partial}>>H^2(X,M,\rz)\\ @AA{\otimes\rz}A@AA{\otimes\rz}A\\ H^{1}(M,\zz)@>{\partial}>> H^2(X,M,\zz) \end{CD} \end{equation} We see that $[B]$ lives in the image of $\otimes\rz$ if $[A]$ does (conversely, if we assume additionally that $H_1(X,\zz)=0$, which is the case for instance if $\pi_1(X)=0$, then $[B]$ is integer if and only if $[A]$ is; this holds because $H^2(X,\zz)$ is torsion-free). Thus from Theorem \ref{thmag} we get the following \begin{cor} Assume that the first Betti number of $X$ vanishes. Let $B$ be a smooth exact $2$-form on $\oX$. Assume that either $i_M^* B$ does not vanish identically, or that $i_M^* B=0$ and the cohomology class $[B]\in H^2(\oX,M,\rz)$ is not an integer multiple of $2\pi$ (in the sense described above). Then the essential spectrum of the associated magnetic Laplacian with respect to the metric \eqref{pme} is empty. \end{cor} \begin{example} Suppose that $B=df\wedge dx/x^2$ where $f$ is a function on $X$ smooth down to the boundary $M$ of $X$. Assume that $f$ is not constant on $M$. Then the essential spectrum of the magnetic operator (which is well-defined by $B$ if $H^1(X)=0$) is empty. \end{example} Note that in this example the pull-back to the border of the magnetic field is null. For such magnetic fields in flat $\rz^n$, the essential spectrum would be $[0,\infty)$. \begin{example}\label{contreex} There exist Riemannian manifolds $X$ with gauge invariance (i.e., $H^1(X,\rz)=0$) and a smooth magnetic field $B$ with compact support in $X$ such that the essential spectrum of the magnetic Laplacian is empty. We choose $X$ to be the interior of a smooth manifold with boundary. Assume that $H^1(X)=0$, for instance $X$ is simply connected. Assume moreover that $H^1(M,\rz)\neq 0$ where $M$ is the boundary of $X$. Note that these two assumptions are impossible to fulfill simultaneously for orientable manifolds in dimension $3$ (see Remark \ref{rem3}). We construct $A$ like in (\ref{aft}) satisfying the hypotheses of Theorem \ref{thmag}. We take $\varphi_0$ to be constant. Let $\psi\in\cun([0, \varepsilon))$ be a cut-off function such that $\psi(x)=0$ for $x\in[3\varepsilon/4,\varepsilon)$ and $\psi(x)=1$ for $x\in[0, \varepsilon/2)$. Since $H^1(M)\neq 0$, there exists a closed $1$-form $\beta$ on $M$ which is not exact. Up to multiplying $\beta$ by a real constant, we can assume that the cohomology class $[\beta]\in H^1_{\mathrm{dR}}(M_0)$ does not belong to the image of $2\pi H^1(M_0,\zz)\to H^1(M_0,\rz)\simeq H^1_{\mathrm{dR}}(M_0)$. Choose $\theta_x$ to be $\psi(x)\beta$ for $\varepsilon>x>0$ and extend it by $0$ to $X$. By Theorem \ref{thmag}, $\Delta_A$ has pure point spectrum. On the other hand, the magnetic field $B=dA=(d\psi/dx)\wedge\beta$ has compact support in $X$. \end{example} \begin{example}\label{ex12} The simplest particular case of Example \ref{contreex} is $X=\rz^2$, but of course not with its Euclidean metric. The metric \eqref{pme} could be for instance (in polar coordinates) \[r^{-2p}(dr^2+d\sigma^2).\] Here $M$ is the circle at infinity and $x=1/r$ for large $r$. The product of this manifold with a closed, connected, simply connected manifold $Y$ of dimension $k$ yields an example of dimension $2+k$. Indeed, by the K\"unneth formula, the first cohomology group of $\rz^2\times Y$ vanishes, while $H^1(S^1\times Y)\simeq H^1(S^1)=\zz$. \end{example} Clearly $k$ cannot be $1$ since the only closed manifold in dimension $1$ is the circle. Thus dimension $3$ is actually exceptional. \begin{ob}\label{rem3} For orientable $X$ of dimension $3$ (the most interesting for Physical purposes) the assumptions $H^1(X)=0$ and $H^1(M)\neq 0$ cannot be simultaneously fulfilled. Indeed, we have the following long exact sequence (valid actually regardless of the dimension of $X$) \[H^1(X)\to H^1(M)\to H^2(X,M).\] If $\dim(X)=3$, the spaces $H^1(X)$ and $H^2(X,M)$ are dual by Poincar\'e duality, hence $H^1(X)=0$ implies $H^2(X,M)=0$ and so (by exactness) $H^1(M)=0$. \end{ob} \begin{example} In the setting of Example \ref{ex12}, the relative cohomology group $H^2(X,M)$ is one-dimensional. The integrality condition becomes particularly nice in this case. Namely, let $B$ be a compactly supported magnetic field (or more generally, a smooth exact $2$-form on $\oX$ vanishing at $M$) which is not zero in $H^2(X,M)$. Then $\lambda B$ is integral precisely for $\lambda$ in an infinite discrete subgroup of $\rz$. Thus the multiples of $B$ for which the magnetic Laplacian has non-empty essential spectrum are discrete and periodic. The fact that there is essential spectrum for integral $[B]$ follows from Proposition \ref{thema}. \end{example} \section{Schr\"odinger operators}\label{schro} We now consider Schr\"odinger operators with magnetic potentials on $k$-forms. An easy and well-known fact, which follows from Proposition \ref{p:cut}, is the compactness of the resolvent of \eqref{eq:op} for any self-adjoint potential $V$ such that $V\rightarrow+\infty$ at infinity. Using the cusp algebra we get much stronger results. We will apply our methods for $V$ belonging to $x^{-2p}\cun(\oX)$ but not necessarily positive. Unlike in the Euclidean case, for $p\geq 1/n$ the eigenvalue asymptotics do not depend on $V$. The result holds moreover for endomorphism-valued potentials $V$. \begin{p}\label{thschr} Let $p>0$, $g_p$ the metric on $X$ given by \eqref{pme} near $\partial X$ and $V\in x^{-2p}\cun(\oX, \End(\Lambda^k \oX))$ a self-adjoint potential. Set $V_0:=(x^{2p}V)_{|M}\in\cun(M, \End(\Lambda^k \oX)_{|M})$. Assume that $V_0\geq 0$ (i.e., $V_0$ is semi-positive definite), and in each connected component of $M$ there exists $z$ with $V_0(z)>0$. Let $A$ be any vector potential of the form \eqref{aft} near $M=\partial X$. Then the magnetic Schr\"odinger Laplacian \begin{equation}\label{eq:op} \Delta_{A,V}:=\dA^*\dA+\dA \dA^*+V \end{equation} is essentially self-adjoint and has pure-point spectrum. Its eigenvalue counting function satisfies \begin{equation*} N_{A,V}(\lambda) \approx \begin{cases} C_1\lambda^{n/2}& \text{for $1/n< p<\infty$,}\\ C_2\lambda^{n/2}\log \lambda &\text{for $p=1/n$} \\ C_3 \lambda^{\frac{1}{2p}} &\text{for $p<1/n$} \end{cases} \end{equation*} with $C_1,C_2$ given by \eqref{c12}. \end{p} \begin{ob} Note that for $p\geq 1/n$, the asymptotic behavior of the eigenvalues depends neither on $V$ nor on $A$, i.e., $C_1,C_2$ depend only on the metric! \end{ob} \begin{proof} We claim that the operator $\Delta_{A,V}$ is fully elliptic. Indeed, we write first $\cN(x^{2p}\Delta_{A,V})=\cN(x^{2p}\Delta_A)+V_0$. For all $\xi\in\rz$, the operators $\cN(x^{2p}\Delta_A)(\xi)$ and $V_0$ are non-negative, so a solution of $\cN(x^{2p}\Delta_{A,V})(\xi)$ must satisfy \begin{align*} \cN(x^{2p}\Delta_A)(\xi)\phi=0,&& V_0\phi=0. \end{align*} By unique continuation, solutions of the elliptic operator $\cN(x^{2p}\Delta_A)(\xi)$ which are not identically zero on a given connected component of $M$ do not vanish anywhere on that component. But then $V_0\phi=0$ and the assumption on $V_0$ show that there are no non-zero solutions, in other words $\Delta_{A,V}$ is fully elliptic. By \cite[Lemma 10 and Corollary 13]{wlom}, it follows that $\Delta_{A,V}$ is essentially self-adjoint with domain $x^{2p}H^2_c(M)$, and has pure-point spectrum. Note that $\Delta_{A,V}$ is not necessarily positive since $V$ may take negative values. Write $V=V^++V^-$, where $V^+_0=V_0$ while $V^-\in x^{1-2p}\cun(\oX)$. Then $V^-$ is compact from $H^{2,2p}=\Dc(\Delta_{A,V^+})$ to $L^2$, so $\Delta_{A,V}$ is a compact perturbation of a strictly positive operator and therefore it is bounded below. By \cite[Proposition 14]{wlom}, for $p\geq 1/n$, the asymptotic behaviour of the function $N(\Delta_{A,V})$ as $\lambda\to\infty$ is given by \eqref{c12}, in particular it only depends on the principal symbol. For $0

0$.
Let $V\in x^{-2p}\cun(\oX)$ be such that $V_0=(x^{2p}V)_{|M}$ is non-negative
and not identically zero on $M$. Then Proposition \ref{thschr} implies that
the Schr\"odinger operator $\Delta+V$ has pure-point spectrum and obeys
a generalized Weyl law. There exist such $V$ which tend to $-\infty$ towards
some part of $M$. For instance, assume $\supp(V_0)$ does not cover $M$; this
implies that near $M\setminus \supp(V_0)$, $V/x^{1-2p}$ has a limit, which
can be very well negative. Now if $p>1/2$, then $x^{1-2p}$ goes to $\infty$
so $V$ tends to $-\infty$ as we approach $M\setminus \supp(V_0)$.
\end{example}
\begin{ob}
This example is quite surprising in light of what happens in Euclidean
$\R^n$. Let $\overline{\R^n}$ be its spherical compactification, and
$V\in \cC({\R^n};{\R})$ a potential which extends continuously to
$\overline{\rz^n}$ with values in $\overline\rz$.
Then $\sigma_{\rm ess}(\Delta +V)=\emptyset$ if and only if
$V(x)=+\infty$ for $x\notin \R^n$. One can nicely see this in the setting of
$C^*$-algebras using the formalism from \cite{GI}.
The same remains true for the Laplacian on a tree with potentials that extend
to the hyperbolic compactification of the tree (see \cite{G}), and
for the natural generalization of these operators on a total Fock space
\cite{GG2}.
\end{ob}
\begin{example} Assume that the magnetic potential $A$
satisfies the hypotheses of Theorem \ref{thmag}. Then we can control
the essential spectrum of the magnetic Schr\"odinger operator
$\Delta_{A,V}$ even for potentials which tend to $-\infty$ at infinity
in \emph{all} directions. Namely, there exists a strictly positive constant $c$
depending on $g_p$ and $i^*_M A$, such that for any real potential
$V\in x^{-2p} \cun(\oX)$ satisfying
\[|V_0|