# Zhenghe's Blog

## February 21, 2011

### Notes 7: Examples (I)-Periodic Potentials

Filed under: Schrodinger Cocycles — Zhenghe @ 7:23 pm
Tags: , ,

From now on, I will mainly focus on the Schrodinger cocyles case.

Example 1: Constant Potentials

Let’s start with the simplest example: potential over fixed point. Equivalently, we consider constant $SL(2,\mathbb R)$– matrix $A$.

Since eigenvalues are invariant under conjugacy, we can up to a conjugacy assume $A$ is one of  the following

$\begin{pmatrix}\lambda&0\\ 0&\lambda^{-1}\end{pmatrix}, \begin{pmatrix}1&a\\ 0&1\end{pmatrix}, \begin{pmatrix}\cos2\pi\theta&-\sin2\pi\theta\\\sin2\pi\theta&\cos2\pi\theta\end{pmatrix}.$

Under mobius transformation, they have invariant directions in $\mathbb C\mathbb P^1$ as $\infty,\infty,i.$ Recall in Notes 3 we define functon

$\tau(A,m)$ such that $A\binom{m}{1}=\tau(A,m)\binom{A\cdot m}{1}.$

In particular if $u$ is the unstable invariant direction of $A,$ then $\ln\tau(A,u)=L(A)+i2\pi\rho(A).$ As the above cases, $u$ can be $\infty.$ But $\tau(A,u)$ (eigenvalue) is invariant under conjugacy. Thus we can always move the computation to unit disk to avoid $\infty.$ For simplicity let’s always denote it as $\tau(A,u).$ Then for above matrices, we get

$L(A)+i2\pi\rho(A)=\ln\tau(A,u)=\ln\lambda, 0, i2\pi\theta$ (note for each fixed $A, \rho(A)$ is well-defined but not uniquely determined).

Consider $A^{(E)}=\begin{pmatrix}E&-1\\1&0\end{pmatrix}$ and the function $\rho:\mathbb R\rightarrow\mathbb R, E\rightarrow\rho(E).$ Let’s use $Q=\frac{-1}{1+i}\begin{pmatrix}1&-i\\1&i\end{pmatrix}$ to move to unit disk $\mathbb D.$ Then $\mathbb R$ becomes the unit circle $\mathbb S.$ Let’s consider the function $N(E)$ (see notes 3) instead of $\rho(E).$ It’s easy to see that for nonelliptic matrices, $N$ is integer valued. By monotonicity of Notes 3, $N:\mathbb S\rightarrow\mathbb R$ is not well-defined. But it’s well-defined as $N:\mathbb S\rightarrow\mathbb R/\mathbb Z.$  By the proof of Lemma 2 of Notes 3, we know when $E$ changes from $-\infty$ to $\infty,$ so is some $\cot(\theta(E))=A^{(E)}\cdot y.$  Thus the change of correponding angle of the point in $\mathbb S$ is exactly $2\pi.$ This implies that $\deg N=1.$ WLOG, we can set $N(-\infty)=-1, N(\infty)=0.$ Thus $\rho(-\infty)=\frac{1}{2}, \rho(\infty)=0.$

The above argument can be applied to any $A^{(E-v)}=\begin{pmatrix}E-v&-1\\1&0\end{pmatrix},v\in R.$ Combining the equivalent description of spectrum in Notes 2 give the following description of the corresponding operator:

– the spectrum $\Sigma$ is $[v-2,v+2];$
$\rho|_{(-\infty,v-2]}=\frac{1}{2},$  $\rho|_{[v+2,\infty)}=0;$
$\rho$ is analytic and strictly decreasing in the interior of the spectrum $(v-2, v+2).$

Example 2: Periodic Potentials

Next let’s consider periodic potential case. Namely, $(X,f,\mu)=(\mathbb Z/(q\mathbb Z), f, \mu),$ where $f: \mathbb Z/(q\mathbb Z)\rightarrow \mathbb Z/(q\mathbb Z), x\mapsto x+1$ and $\mu$ is the averaged counting measure. Let $v:\mathbb Z/(q\mathbb Z)\rightarrow \mathbb R,$ hence $A^{(v)}:\mathbb Z/(q\mathbb Z)\rightarrow SL(2,\mathbb R).$

For any $A:\mathbb Z/(q\mathbb Z)\rightarrow SL(2,\mathbb R),$ we have for any $j, A_q(j+1)A(j)=A(j)A_q(j).$ This implies that

-the eigenvalue $\tau(A_q(j), u_q(j))$ of $A_q(j)$ is independent of $j.$ Denote it by $\tau(A_q, u_q).$
$L(A)=\frac{1}{q}\ln\delta(A_q).$

Here for any bounded linear operator $A$ on any Banach space, $\delta(A)=\lim\limits_{n\rightarrow\infty}\|A^n\|^{\frac{1}{n}}=\inf_{n\leq1}\|A^n\|^{\frac{1}{n}}$ is the spectral radius of $A.$  From the above facts we get

$L(A)+i2\pi\rho(A)=\frac{1}{q}\ln\tau(A_q,u_q).$

We also have the following obvious equivalent relations

$L(A)>0\Leftrightarrow |trA_q|>2\Leftrightarrow (f,A)\in\mathcal U\mathcal H$ and  $|trA_q|<2\Leftrightarrow$ Elliptic,

where $tr$ stands for trace. For the second case we also have $tr(A_q)=2\cos(2\pi \rho(A)q).$

Back to Schrodinger case, there is a nice graph of the function $\psi(E)=trA^{(E-v)}_q.$ (which is also $2\cos(2\pi\rho(E)q)$ for elliptic case). It’s obviously a polynomial in $E$ of degree $q.$ It has exactly $q-1$ critical points with critical values $y$ satisfying $|y|\geq2.$ We have in fact the following description of the spectrum $\Sigma$ of the correponding operator $H_v$

Theorem: $\Sigma$ consists of $q$ bands and there are $q-1$ spectral gaps ( bands may touch at the boundary points, or equivalently, gap may collapse).

Proof: By the same argument for Schrodinger cocycle over fixed point above and the fact $\rho(AB)=\rho(A)+\rho(B)$, we have the following easy facts for $\rho(E)$ as a function on real line:

-it’s continuous on $\mathbb R$ and is analytic in the interior of spectrum;
-it’s $\frac{k}{2q}$-valued outside the interior of spectrum thus constant in each connected components of the resolvent set;
$\rho(-\infty)=\frac{1}{2}$ and $\rho(\infty)=0$ and nonincreasing.

These together implies that:

$\psi^{-1}[-2,2]$ consists of $q$ compact intervals. Some of them may touch at the boundary points. These are spectrum bands. $\rho'(E)<0$ on $\psi^{-1}(-2,2);$
-between each two bands is the so-called spectrum gap and there are $q-1$ of them (some of them may collasped). On each of them $\rho(E)=\frac{k}{2q},k=q-1,\cdots,1.$ These are labelings of spectral gaps. $\square$

The proof obviously implies the properties of the polynomial $\psi(E).$

A natural question is that when are all gaps open (not collapsed). Obviously by the proof of Theorem, all gaps are open if and only if all the roots of polynomials $\psi(E)-2, \psi(E)+2$ are simple. Let’s give another description of these polynomials. Let’s restrict the operator $H_v$ to $l^2(\mathbb Z/(q\mathbb Z))$ with three different type of boundary conditions. Let $u\in l^2(\mathbb Z/(q\mathbb Z))$

1. First let’s restrict to any subinterval $[j,j+n-1]\subset[0,q-1]$ and consider Dirichlet boundary condition, i.e. $u(j-1)=u(j+n)=0.$ Denote it by $H^{n}_{v,j}$. In this case it can be represented as a $n\times n$ symmetric matrix

$H^{n}_{v,j}=\begin{pmatrix}v(j)&1&0&0\\1&v(j+1)&1&\vdots\\ 0&1&v(j+2)&\vdots\\ 0&0&1&\vdots\\\vdots&\vdots&0&1\\\vdots&\vdots&\vdots&v(j+n-1)\end{pmatrix}$

Let $\det(H^{n}_{v,j}-E)=P_n(j,E).$ Then by induction it’s easy to see that

$A^{(v-E)}_n(j)=\begin{pmatrix}P_n(j,E)&-P_{n-1}(j+1,E)\\P_{n-1}(j,E)&-P_{n-2}(j+1,E)\end{pmatrix}.$

Hence $\psi(E)=P_{q}(0,E)-P_{q-2}(1,E).$ In case 2 and 3 we will only restrict to $[0,q-1].$

2. Periodic boundary condition, i.e. $u(-1)=u(q-1), u(q)=u(0).$ Denote it by $H^{p}_v,$ then it is

$H^{p}_v=\begin{pmatrix}v(0)&1&0&1\\1&v(1)&1&0\\ 0&1&v(2)&\vdots\\ 0&0&1&0\\ 0&\vdots&0&1\\1&0&\vdots&v(q-1)\end{pmatrix}$

3. Antipeiodic boundary condition, i.e. $u(-1)=-u(q-1), u(q)=-u(0).$ Denote it by $H^{a}_v,$ then it is

$H^{a}_v=\begin{pmatrix}v(0)&1&0&-1\\1&v(1)&1&0\\ 0&1&v(2)&\vdots\\ 0&0&1&0\\ 0&\vdots&0&1\\-1&0&\vdots&v(q-1)\end{pmatrix}$

Then by case 1 it’s not difficult to see that for some integer $l,$

$\psi(E)-2=\begin{cases}\det(H^{p}_v-E)&\text{ if }q=2l\\\det(H^{a}_v-E)&\text{ if }q=2l-1\end{cases}$ and $\psi(E)+2=\begin{cases}\det(H^{a}_v-E)&\text{ if } q=2l\\\det(H^{p}_v-E)&\text{ if } q=2l-1\end{cases}.$

Thus all the spectral gaps are open if and only if all eigenvalues of the operators $H^{p}_v$ and $H^{a}_v$ are simple. An easy case is that assume $v(j), j=0,\cdots,q-1$ are sufficiently large and distinct. Then after scaling, all nondigonal coefficients of $H^{p}_v$ and $H^{a}_v$ are sufficiently small. Namely, $H^{p}_v$ and $H^{a}_v$ are small perturbation of digonal matrices with distinct eigenvalues. Then so are $H^{p}_v$ and $H^{a}_v$ themself. Which implies that all gaps are open.

For simplicity, consider $q$ even. Denote eigenvalues of $H^{p}_v$ by $\mu_1<\cdots<\mu_q$ and $H^{a}_v$ by $\lambda_1<\cdots<\lambda_q.$ Then spectrum bands are

$[\mu_j,\lambda_j], j$ odd; $[\lambda_j,\mu_j], j$ even.

And spectral gaps are

$(\mu_j,\mu_{j+1}), j$ even; $(\lambda_j,\lambda_{j+1}), j$ odd.

Finally let’s consider $L(E)+i2\pi\rho(E):\mathbb H\rightarrow\mathbb C.$ Then it’s easy to see that it can be analytically extends through interior of each bands and through gaps. But they cannot be globally defined, since there are nontrivial winding of the invariant direction $u_q(E)$ around each $\mu_j, \lambda_j.$ This winding comes from the parabolicity. Indeed, if we instead consider the eigenvalue $\lambda(E)$ of $A^{(E-v)}_q,$ then

$\frac{1}{q}\ln\lambda(E)=L(E)+i2\pi\rho(E)$ and $\lambda(E)=\frac{1}{2}(\psi(E)\pm\sqrt{\psi(E)^2-4}),$

of which the derivative has singularity at parabolicity. This also explain why $L(E)$ and $\rho(E)$ as functions on the whole real line can at most be $\frac{1}{2}-H\ddot older$ continous.

Note for periodic potential and in spectrum bands, it’s always $L(E)=0$ since they are just parabolic and elliptic matrices. Thus by Theorem 2 of Notes 4, all the spectrum are purely absolutely continous.

Assume potentials are uniformly bounded. Then it’s interesting to note that as $q\rightarrow\infty,$ there are more and more bands which are also thiner and thiner. Thus in quasiperiodic potential case, it’s natural to expect the spectrum is Cantor set under some assumption. We will give an example in next post.