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\title Operators with Singular Continuous Spectrum, VI. Graph Laplacians and
Laplace-Beltrami Operators
\endtitle
\rightheadtext{Graph Laplacians and Laplace-Beltrami Operators}
\author Barry Simon$^{*}$
\endauthor
\leftheadtext{Barry Simon}
\affil Division of Physics, Mathematics, and Astronomy \\
California Institute of Technology, 253-37 \\
Pasadena, CA 91125
\endaffil
\thanks $^{*}$ This material is based upon work supported by the
National Science Foundation under Grant No.~DMS-9101715. The
Government has certain rights in this material.
\endthanks
\thanks To appear in {\it{Proc.~Amer.~Math.~Soc.}}
\endthanks
\abstract Examples are constructed of Laplace-Beltrami operators and graph
Laplacians with singular continuous spectrum.
\endabstract
\endtopmatter
\document
\medpagebreak
\flushpar {\bf \S1.~Introduction}
\medpagebreak
In previous papers in this series [6,3,5,2,8], we have explored the
occurrence of singular continuous spectrum, particularly for
Schr\"odinger operators and Jacobi matrices. In this note, we'll
construct examples of graphs whose Laplacians and Riemannian manifolds
whose Laplace-Beltrami operators have singular continuous spectrum.
The idea will be to take models with considerable symmetry which
reduce to Jacobi matrices or Schr\"odinger operators, and so reduce
these two types of models to known cases.
The analysis is simple, almost trivial. However, since these are the
first examples I know of graph Laplacians or Laplace-Beltrami
operators with singular continuous spectrum, it seems worthwhile to
make it explicit.
\bigpagebreak
\flushpar {\bf \S2.~Graph Laplacians}
\medpagebreak
By a graph $G=(V,I)$, we will mean a countable set of vertices $V$,
and an incidence matrix $I_{ij}$ of zeros and ones obeying
\roster
\item"{(i)}" $I_{ij}=I_{ji}; \quad I_{ii}=0$.
\item"{(ii)}" $N_{i}=\{j\mid I_{ij}=1\}$ is finite for each $i$. Let
$r(i)=\#(N_{i})=\sum\limits_{j}I_{ij}$, the coordination number of
$r$.
\item"{(iii)}" $\sup\limits_{i}\,r(i)<\infty$.
\endroster
The graph is called regular if $r(i)$ is a constant.
Given a graph $G$, we can define two bounded operators on
$\ell_{2}(V)=\bigl\{\{u(n)\}_{n\in V}\mid\sum\limits_{n\in V}
|u(n)|^{2}<\infty\bigr\}$:
$$\align
(L_{1}u)(i) &=\sum_{j} I_{ij} u(j) =\sum_{j\in N_{i}} u(j) \\
(L_{2}u)(i) &=\sum_{j}I_{ij}(u(i)-u(j))=\sum_{j\in N_{i}} (u(i)-u(j)).
\endalign
$$
$L_1$ and $L_2$ define quadratic forms by
$$\align
(u, L_{1}u) &=\sum_{ij}I_{ij}\, \overline{u(i)}\,u(j) \\
(u, L_{2}u) &=\frac{1}{2}\sum_{ij} I_{ij}\, |u(i)-u(j)|^{2}.
\endalign
$$
They are related by
$$
L_{2}=R-L_{1}
$$
where $(Ru)(i)=r(i)u(i)$. In particular if $G$ is regular, $L_1$ and
$L_2$ differ by a constant and sign and their spectral types are the
same.
Our first example is a ladder with missing rungs. Let $V=\Bbb Z\times
\Bbb Z_{2}=\{(n,\alpha)\mid n\in\Bbb Z, \, \alpha=0\text{ or }1\}$.
\midinsert
\vspace{1.0in}
\botcaption{Figure 1} A ladder with missing rungs
\endcaption
\endinsert
\vskip 0.25in
\flushpar The incidence matrix will depend on the choice of a single
sequence $s\in\Bbb Z^{\Bbb Z}_{2}$; that is, $s=\{s_{n}\}_{n\in\Bbb
Z}$ with each $s_n$ 0 or 1. Given $s$, we define an incidence matrix
so $I_{ij}=1$ if and only if one of
\roster
\item"{(1)}" $i=(n,\alpha) \quad j=(n\pm 1,\alpha)$
\item"{(2)}" $i=(n,0) \quad j=(n,1)$ or vice-versa where $s_{n}=1$.
\endroster
Thus, as Figure 1 shows, the graph is a ladder with those rungs with
$s_{n}=0$ missing.
Let $X=\Bbb Z^{\Bbb Z}_{2}$, a compact metric space in the product
topology (and so complete). $s^{(m)}\to s^{(\infty)}$ if and only if
$s^{(m)}_{n}\to s^{\infty}_{n}$ for each $n$. For each $s\in X$, we
can define $L_{1}(s)$, $L_{2}(s)$, the two Laplacians on
$\ell_{2}(V)$. We say a set $S\subset X$ is generic if and only if $S$
is a dense $G_\delta$.
\proclaim{Theorem 2.1} For a generic $s\in X$, $L_{1}(s)$ has purely
singular continuous spectrum and $L_{2}(s)$ has a spectrum with a
non-zero purely singular component.
\endproclaim
As a preliminary, we'll need:
\proclaim{Lemma 2.2} Fix $c\neq 0$. Given $s\in X$, let $J(s)$ be
the Jacobi matrix on $\ell^{2}(\Bbb Z)$:
$$
(J(s)u)=u(n+1)+u(n-1)+cs_{n}u(n).
$$
For each fixed $c$, $\{s\in X\mid J(s)\text{\rom{ has spectrum }}
[-2,2]\cup [c-2, c+2]$ {\rom{and it is purely singular continuous}}$\}$
is a dense $G_\delta$.
\endproclaim
\demo{Proof} Because a countable intersection of dense $G_\delta$'s
is again a dense $G_\delta$ (by the Baire category theorem), we need
only show that each of
\roster
\item"{(i)}" $A_{1}=\{s\mid [-2,2]\subset\text{spec}(J(s))\text{ and
has no eigenvalues in }[-2,2]\}$
\item"{(ii)}" $A_{2}=\{s\mid [c-2, c+2]\subset\text{spec}(J(s))\text{
and has no eigenvalues in }[c-2, c+2]\}$
\item"{(iii)}" $A_{3}=\{s\mid J(s)\text{ has no a.c.~spectrum}\}$
\endroster
is a dense $G_\delta$. For writing $J=J_{0}+V$ where
$(Vu)(n)=cs_{n}(u(n))$, and noting $\text{spec}(V)\subset\{0, c\}$ and
$\|J_{0}\|\equiv 2$, we have $\text{spec}(J(s))\subset [-2,2]\cup [c-
2, c+2]$.
Each of the sets in (i)--(iii) is a $G_\delta$. That this is so for
point and a.c.~spectrum is a result of [6,7]. The argument that a
given interval is in $\text{spec}(A)$ on a $G_\delta$ is found in the
appendix.
To see (i) is dense, note that $\{s\mid s_{n}=0\text{ for $|n|$
large}\}\subset A_{1}$, and similarly, $\{s\mid s_{n}=1\text{ for $|n|$
large}\}\subset A_{2}$. But those sets are dense.
For $A_3$, put any non-trivial product measure on $X$. By the result
of Carmona, Klein, and Martinelli [1], for a.c.~$s\in X$ with respect
to this measure, $J(s)$ has pure point spectrum and, in particular,
has a.c.~spectrum. But the support of the product measure is all of
$X$, so $A_3$ is dense. \qed
\enddemo
\demo{Proof of Theorem \rom{2.1}} We'll use reflection symmetry about the
middle of the ladder. Let $\Cal H_{\pm}=\{u\in\ell^{2}(V)\mid u(n,1)=
\pm u(n,0)\}$ and let $W_{\pm}:\ell^{2}(\Bbb Z)\to\Cal H_{\pm}$ by
$$\align
(W_{\pm}f)(n,0) &=f(n)\big/\sqrt{2} \\
(W_{\pm}f)(n,1) &=\pm f(n)\big/\sqrt{2}.
\endalign
$$
Then $\Cal H_{\pm}$ is left invariant by both $L_{1}(s)$ and $L_{2}(s)$
and
$$\align
W^{-1}_{\pm} L_{1}(s)W_{\pm} &= J(\pm s) \\
W^{-1}_{+}L_{2}(s)W_{+} &= 2-J(0) \\
W^{-1}_{-} L_{2}(s)W_{-} &= 2-J(-2s_{n}).
\endalign
$$
The result now follows by the lemma. \qed
\enddemo
Note that $L_{2}(s)$ always has an a.c.~component of the spectrum
since $J(0)$, the free Schr\"odinger Laplacian, has a.c.~spectrum.
This argument also shows that generically $\text{spec}(L_{1}(s))=
[-3,3]$ and $\text{spec}(L_{2}(s))\equiv [0,6]$ with a.c.~spectrum on
$[0,4]$ and singular spectrum on $[2,6]$.
We were able to find a cheap trick to reduce the first example to the
well-studied Jacobi case. That will not be true in the second
example, and so we'll settle for a partial result. A complete analysis
should be possible (I expect purely singular continuous spectrum for a
generic $s$) but will require some effort to extend Carmona et al.~[1] to
this example.
Our purpose in this second example is to find a regular lattice whose
Laplacian has singular spectrum; the coordination number in Example 1
is sometimes 2 and sometimes 3. Our lattice will again depend on
$s\in\Bbb Z^{\Bbb Z}_{2}$ and again be a basic ladder whose rungs are
of the two types shown in Figure 2 if $s_{n}=0$ or $s_{n}=1$.
\midinsert
\vspace{1.0 in.}
\botcaption{Figure 2} Rungs in Example 2
\endcaption
\endinsert
\vskip 0.25in
The corresponding lattices are regular with coordination number 3, so
we'll only look at $L_1$, which we'll call $L(s)$. As before, the
spaces $\Cal H_{\pm}$ are invariant and we'll look only at the
operator on $\Cal H_{-}$ (call the restriction of $L(s)$ to $\Cal H_{-}$,
$B(s)$). The space on which $B(s)$ acts is pairs of function
$u:\Bbb Z\to\Bbb C$ and $\varphi:\{n\mid s_{n}\equiv 1\}\equiv
Y(s)\to\Bbb C$ with $\sum\limits_{n\in\Bbb Z} |u(n)|^{2}+\sum\limits_
{n\in Y(s)} |\varphi(n)|^{2}<\infty$, and with
$$\alignat2
(B(s)u)(n) &= u(n-1)+u(n+1)-u(n) \qquad &&\text{if} \quad s_{n}=0 \\
&=u(n+1)+u(n-1)+\varphi(n) \qquad &&\text{if} \quad s_{n}=1 \\
(B(s)\varphi)(n) &=u(n).
\endalignat
$$
We want to compute $\text{spec}(B(s))$ for the two degenerate cases
$s_{n}\equiv 0$ and $s_{n}\equiv 1$. If $s_{n}\equiv 0$, we get a
Jacobi matrix whose spectrum is well known.
$B(s_{n}\equiv 0)$ has spectrum $[-3,1]$ with purely a.c.~spectrum.
For $s_{n}\equiv 1$, one looks at plane waves $u(n)=ce^{ikn}$, in which
case $\varphi(n)=cE^{-1}e^{ikn}$ and $E$ obeys $E=2\cos\, k+E^{-1}$,
which leads to
$$
E=\cos\,k\pm\sqrt{1+\cos^{2}k}\,.
$$
It is not hard to see that these continuous eigenfunctions are
complete and so
$$
\text{spec}(B(s_{n}\equiv 1))=[-1-\sqrt{2}, 1+\sqrt{2}\,] \text{ and is
purely absolutely continuous}.
$$
\proclaim{Theorem 2.3} For a generic $s\in X$, $B(s)$ contains $[-3,
-1-\sqrt{2}\,)$ in its spectrum and the spectrum is purely singular
continuous there.
\endproclaim
\demo{Proof} The spaces on which $B(s)$ act are $s$-dependent, but
there is still a natural sense in which $s^{(n)}\to s^{(\infty)}$ means
$(B(s^{(n)})-z)^{-1}\to (B(s^{(\infty)})-z)^{-1}$ strongly (apply to vectors
of compact support) and this notion is such that the results of [6,7]
still apply. Let
$$\align
A_{1}&=\{s\mid [-3, -1-\sqrt{2}\,)\subset\text{spec}(B(s))\text{ and has
no point eigenvalues there}\} \\
A_{2}&=\{s\mid B(s)\text{ has no a.c.~spectrum on }[-3, -1-\sqrt{2}\,)\}.
\endalign
$$
Each is a $G_\delta$ by [6,7].
We claim that if $s_{n}\equiv 0$ near infinity, $s\in A_{1}$; but if
$s_{n}\equiv 1$ near infinity, then $s\in A_2$. This implies $A_1$
and $A_2$ are dense so their intersection is a dense $G_\delta$ and
Theorem 2.3 is proven.
The claim for $s_{n}\equiv 0$ near infinity follows by using plane
waves near infinity as a Weyl sequence (to see that $[-3,1]\subset
\text{spec}(B(s))$) and then noting that eigenvalues must have compact
support and then be zero by a simple argument. The claim for
$s_{n}\equiv 1$ near infinity follows because $B(s)$ differs from
$B(s_{n}\equiv 1\text{ all }n)$ by a finite rank operator, so $\text
{spec}(B(s))$ is thus discrete outside $[-1-\sqrt{2}, 1+\sqrt{2}\,]$.
\qed
\enddemo
\bigpagebreak
\flushpar {\bf \S3.~Some Laplace-Beltrami Operators}
\medpagebreak
In this section, we'll find metrics on $\Bbb R\times S^{1}=M$ for
which the Laplace-Beltrami operator has purely singular spectrum. Use
coordinates $(x,\theta)$ on $M$ and consider metrics of the form
$dx^{2}+f(x)^{2}\,d\theta$. We'll consider $1\leq f(x)\leq 2$ and is
$C^{\infty}$ with the metric topology associated to $\bigl\{\sup\limits_
{|x|\leq N}\,\|\frac{d^{m}f}{dx^{m}}\|_{\infty}\bigr\}\Sb m=0,1,\dots \\
N=1,2,\dots \endSb$. Denote by $X$ the set of such $f$'s. Then:
\proclaim{Theorem 3.1} For a generic $f$ in $X$, the Laplacian
operator $H_f$ has spectrum $[0,\infty)$ and is purely singular
continuous.
\endproclaim
\demo{Proof} By a simple calculation (see, e.g., [4]), under the
decomposition $L^{2}(\Bbb R\times S^{1})\cong\operatornamewithlimits
{\oplus}\limits^{\infty}_{n=-\infty}\,L^{2}(\Bbb R)$ (with $g(z)=
\sum g_{n}(x)e^{in\theta}$) $H_{f}\cong\oplus H_{f;n}$
and $H_{f;n}$ is unitarily equivalent to the operator $L_{f;n}$ on
$L^{2}(\Bbb R, dx)$ with
$$
L_{f;n}=-\frac{d^2}{dx^2}+V(x)+\frac{n^2}{f^{2}(x)}\equiv -\frac{d^2}
{dx^2}+V_{n}(x)
$$
where
$$
V(x)=\frac{1}{4} \biggl(\frac{f'}{f}\biggr)^{2}+\frac{1}{2}\biggl(
\frac{f'}{f}\biggr)'.
$$
We'll show for each $n$ that $L_{f;n}$ generically has purely singular
spectrum on $[\frac{n^2}{4},\infty)$ and that implies the full result.
If $f\equiv 2$ near infinity, $L_{f;n}$ is $\frac{n^2}{4}$ plus a
Schr\"odinger operator with potential of compact support. This
operator has no point spectrum in $[\frac{n^2}{4}, \infty)$. Since
$L_{f;n=0}\geq 0$ and $f\leq 2$, in general $L_{f;n}\geq\frac{n^2}
{4}$. Moreover, the functions $f\equiv 2$ near $\infty$ are dense in
the topology on $X$. Thus, generically $L_{f;n}$ has spectrum $[\frac
{n^2}{4},\infty)$ with no point spectrum. We'll need to show that
generically there is no a.c.~spectrum either.
Consider a metric of the form $f(x)=\alpha+\lambda\,\cos(kx)$ where
$1<\alpha <2$ and $\lambda$ is small. Then
$$
V_{n}(x)=\frac{n^2}{\alpha^2}+\lambda\biggl[\frac{k^2}{2\alpha} -
\frac{n^2}{\alpha^2}\biggr]\,\cos(kx)+0(\lambda^{2}).
$$
By the standard theory of periodic Schr\"odinger operators, for
$\lambda$ small, $-\frac{d^2}{dx^2}+V_{n}(x)$ has a gap of order
$\lambda$ about every $\frac{n^2}{\alpha^2}+\big(\frac{k}{2}\bigr)^{2}$.
If $f(x)\equiv\alpha +\lambda\,\cos(kx)$ near infinity, $-\frac{d^2}
{dx^2}+V_{n}(x)$ has only point spectrum in that interval. Thus, for
any $E_{0}\in[\frac{n^2}{4}, \infty)$, we can find $\delta$ and a
dense set of $f$'s with no a.c.~spectrum in $(E_{0}-\delta,
E_{0}+\delta)$. By a compactness argument, we conclude that
generically there is no a.c.~spectrum. \qed
\enddemo
Is it true that for any non-compact manifold, a generic metric leads
to purely singular continuous spectrum?
\bigpagebreak
\flushpar {\bf Appendix: $\boldkey G_{\boldsymbol\delta}$'s and the
Spectrum of Operators}
\medpagebreak
Recall that a regular space of operators is a family of self-adjoint
operators indexed by $x\in X$ a complete metric space so that
$x_{n}\to x$ in the metric topology implies that $A_{x_n}\to A_{x}$ in
strong resolvent sense.
Our purpose here is to prove that
\proclaim{Theorem A.1} Let $[a,b]$ be a fixed closed interval and $X$
a regular space of operators. Then $\{x\mid [a,b]\subset
\text{\rom{spec}}(A_{x})\}$ is a $G_\delta$.
\endproclaim
Let $\{\varphi_{n}\}^{\infty}_{n=1}$ be an orthonormal basis for the
underlying Hilbert space and let $d\mu_{n}(x)$ be the spectral measure
defined by
$$
(\varphi_{n}, e^{-itA}\varphi_{n})=\int e^{-ixE}\,d\mu_{n}(E)
$$
and $d\mu=\sum\limits^{\infty}_{n=1}2^{-n}\,d\mu_{n}$. Then
$\text{spec}(A)=\text{supp}(d\mu)$ so Theorem A.1 is a consequence of
\proclaim{Lemma A.2} Let $M^{R}_{+,1}$ denote the family of
probability measures on $[-R, R]\subset\Bbb R$. Fix $[a,b]\subset [-R,
R]$. Then $\{\mu\in M^{R}_{+,1}\mid [a,b]\subset\text{\rom{supp}}
(d\mu)\}$ is a $G_\delta$ in the topology of weak convergence.
\endproclaim
\remark{Remark} Thus, if $S\subset M^{R}_{+,1}$ is closed, $S\cap
\{\mu\mid [a,b]\subset\text{supp}(d\mu)\}$ is a $G_\delta$ in $S$.
\endremark
\demo{Proof of lemma} Let $I_n$ be a counting of open intervals $I_{n}
\subset [a,b]$ with rational endpoints. Let $f_{n}(x)=\text{dist}(x,
\Bbb R\backslash I_{n})$. Then $[a,b]\subset\text{supp}\,\mu$ if and only
if for all $n$, $\int f_{n}(x)\,d\mu>0$, that is,
$$
\{\mu\mid [a,b]\subset\text{supp}(d\mu)\}=\bigcap\limits^{\infty}_{n=1}
\biggl\{\mu\mid \int f_{n}(x)\,d\mu >0\biggr\}
$$
is clearly a $G_\delta$. \qed
\enddemo
\remark{Remark} Using $\lim\limits_{\epsilon\downarrow 0}\frac{1}{\pi}
\int\limits^{d}_{c} \text{Im}\,F_{\mu}(x+i\epsilon)\,dx =\frac{1}{2}\,
\mu((c,d))+\frac{1}{2}\mu([c,d])$ with $F_{\mu}(z)=\int d\mu(x)/(x-z)$,
it is easy to show that $\{\mu\mid[a,b]\cap\text{supp}\,d\mu=\emptyset\}$
is also a $G_\delta$.
\endremark
\vskip 0.3in
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\ref \key 4 \by V.~Jaksic, S.~Molchanov, and B.~Simon \paper
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\endRefs
\enddocument