The Spectrum of Period-Doubling Hamiltonian

Abstract

In this paper, we show the following: the Hausdorff dimension of the spectrum of period-doubling Hamiltonian is bigger than $\log \alpha /\log 4$, where $\alpha$ is the Golden number; there exists a dense uncountable subset of the spectrum such that for each energy in this set, the related trace orbit is unbounded, which is in contrast with a recent result of Carvalho (Nonlinearity 33, 2020); we give a complete characterization for the structure of gaps and the gap labelling of the spectrum. All of these results are consequences of an intrinsic coding of the spectrum we construct in this paper.

Appendices

Some Remarks on the Paper [9]

In this appendix, following essentially one of the referees’ comments, we point out several gaps appearing in Lemmas 2.25, 2.26 and Proposition 2.28 of [9]. We note that the main result Theorem 1.1 of [9] is a direct consequence of Proposition 2.28, and Lemmas 2.25 and 2.26 are the key inductive steps towards the proof of Proposition 2.28.

In [9], the major part of the study of the trace map dynamics is done in variables denoted by $(s,t)\in {\mathbb {R}}^2$, after a conjugation of the well-known two dimensional map associated with the period-doubling substitution $H(x,y)=(xy-2,x^2-2)$. As explained in [9], the map H is conjugated to

$$\begin{aligned} F(s,t)=(2+st-s^2,st). \end{aligned}$$

Define the basin of unstable points of F as

$$\begin{aligned} U_0=\{(s,t)| s<-2, t>0\},\quad V_0=\{(s,t)| \sqrt{2}<s<t\}. \end{aligned}$$

Denote the set of points whose orbits are disjoint with $U_0\cup V_0$ as

$$\begin{aligned} {\mathfrak {S}}:=\left( \bigcup _{n\ge 0} F^{-n}(U_0\cup V_0)\right) ^c. \end{aligned}$$

In Proposition 2.28 of [9], the author claimed that:

If $(s_0,t_0)\in {\mathfrak {S}}$, then $(F^n(s_0,t_0))_{n\ge 1}$ is bounded.

Fix $(s_0,t_0)\in {\mathbb {R}}^2$. For any $n\ge 0$, define

$$\begin{aligned} (s_n,t_n):=F^n(s_0,t_0),\quad c_n:=s_n(s_n-t_n). \end{aligned}$$

It is equivalent to start from $(s_0,c_0)$, and for $n\ge 0$, iterate by,

$$\begin{aligned} s_{n+1}=2-c_n,\quad c_{n+1}=(2-c_n)(2-s_n^2). \end{aligned}$$

In [9], the following two-step iteration of $c_n$ is studied:

$$\begin{aligned} c_{n+2}=(2 + (2 - c_{n})(s_{n}^2-2))(2- (2- c_{n})^2). \end{aligned}$$

(102)

In Lemma 2.26 (a), the author claimed that:

There exists $u\in (2+\sqrt{2},3.56)$ such that if

$$\begin{aligned} |s_0|\ge 3, \ \ 2+\sqrt{2}< c_0\le u, \end{aligned}$$

then $c_2\le 2-\sqrt{2}$ and $F^5(s_0,t_0)\in V_0$.

This is not the case. The reason is as follows. Fix any $\varepsilon \in (0,u-2-\sqrt{2})$ and let $c_0(\epsilon ):=2+\sqrt{2}+\varepsilon .$ By (102), we have

$$\begin{aligned} c_2(\epsilon )=\varepsilon (2\sqrt{2}+\varepsilon )[(\sqrt{2}+\epsilon )(s_0^2-2)-2]. \end{aligned}$$

If $s_0$ is large enough, we could have $c_2(\epsilon )>2-\sqrt{2}$.

Given $|s_n|\ge 3$, one can view (102) as an expanding map in three relevant intervals for $c_n$: $(2-\sqrt{2},1)$, $(2,2+\sqrt{3}/3)$, and $(2+\sqrt{2},3.56)$. But there are repelling fixed points, depending on $s_n$, in these intervals. If we start form some $c_0\in (2+\sqrt{2},3.56)$ and some $s_0$ with $|s_0|\ge 3$, and iterate the map in (102), it is possible that we can construct two sequences $\{s_n:n\ge 0\}$ and $\{c_n:n\ge 0\}$ such that $s_{2n}\rightarrow -\infty$ and $c_{2n}\in (2+\sqrt{2},3.56)$ for all $n\ge 0$. Indeed this is compatible with Proposition 6.1 of the present paper. For any $E\in \pi ({\mathcal {E}}_{l}^e),$ by Proposition 6.1, we have $h_{2m}(E)\ge \sqrt{2}$ and

$$\begin{aligned} \lim _{m\rightarrow \infty }h_{2m+1}(E)=\infty ,\quad \lim _{m\rightarrow \infty }h_{2m}(E)=\sqrt{2}. \end{aligned}$$

We write $h_m=h_m(E)$ for simplicity. If we define

$$\begin{aligned} s_0:=-h_{2m+1};\ \ \ c_0=2+h_{2m+2}. \end{aligned}$$

Then one can check that

$$\begin{aligned} s_n=-h_{2m+n+1};\ \ t_n=s_n+s_{n-1}^2-2; \ \ c_n=2+h_{2m+n+2}. \end{aligned}$$

Consequently, we have $c_{2n}\ge 2+\sqrt{2}$ and

$$\begin{aligned} s_{2n}\rightarrow -\infty ,\ \ \ c_{2n}\rightarrow 2+\sqrt{2}. \end{aligned}$$

If m is large enough, we can check directly that $(s_0,t_0)\in {\mathfrak {S}}$. However since $s_{2n}\rightarrow -\infty ,$ $(F^n(s_0,t_0))_{n\ge 1}$ is unbounded. Thus Proposition 2.28 is false.

There is a similar gap in Lemma 2.25 (a), which we will not explain in detail. We just remark that exactly the same argument can be used to show the gap. Moreover, by using some results in the proof of Proposition 6.3, one can construct another point $({\hat{s}}_0,{\hat{t}}_0)\in {\mathfrak {S}}$ such that the related sequences $\{{\hat{s}}_n:n\ge 1\}, \{{\hat{c}}_n: n\ge 1\}$ satisfy

$$\begin{aligned} {\hat{s}}_{2n}\rightarrow -\infty ,\ {\hat{c}}_{2n}\rightarrow 2-\sqrt{2}. \end{aligned}$$

Again, we obtain a counter example of Proposition 2.28.

Substitutions–Basic Definitions and Examples

In this appendix, we give a brief introduction on substitutions. For general theory of substitutions, in particular, the connection between substitutions and dynamical systems, see [18, 28].

Assume $\kappa \ge 2$ and ${\mathbb {A}}=\{a_1,\cdots ,a_\kappa \}$. We call ${\mathbb {A}}$ an alphabet and call each $a_i$ a letter. We define the set of finite words ${\mathbb {A}}_+$ as

$$\begin{aligned} {\mathbb {A}}_+:=\bigcup _{n\ge 1} {\mathbb {A}}^n. \end{aligned}$$

Define the concatenation operation $*:{\mathbb {A}}_+\times {\mathbb {A}}_+\rightarrow {\mathbb {A}}_+$ as

$$\begin{aligned} (u_1\cdots u_n)*(v_1\cdots v_m):=u_1\cdots u_nv_1\cdots v_m. \end{aligned}$$

Then $({\mathbb {A}}_+,*)$ become a semigroup.

A substitution $\zeta$ over ${\mathbb {A}}$ is a map $\zeta : {\mathbb {A}}\rightarrow {\mathbb {A}}_+$. Given a substitution $\zeta$ over ${\mathbb {A}},$ we can extend it to a morphism of ${\mathbb {A}}_+$ or ${\mathbb {A}}^{{\mathbb {N}}}$ by concatenation as

$$\begin{aligned} \zeta (u_1u_2\cdots ):=\zeta (u_1)*\zeta (u_2)*\cdots \end{aligned}$$

If $\zeta ,{\hat{\zeta }}$ are substitutions over ${\mathbb {A}},$ so is $\zeta \circ {\hat{\zeta }}$. In particular, the k-th iteration $\zeta ^k$ of $\zeta$ is also a substitution.

Assume $\zeta$ is a substitution over ${\mathbb {A}}$. $u\in {\mathbb {A}}^{{\mathbb {N}}}$ is called a fixed point of $\zeta$ if $\zeta (u)=u.$ $u\in {\mathbb {A}}^{{\mathbb {N}}}$ is called a periodic point of $\zeta$ if $\zeta ^k(u)=u$ for some $k\ge 1$.

A substitution $\zeta$ over ${\mathbb {A}}$ is primitive if there exists $k\ge 1$ such that, for any $a,b\in {\mathbb {A}},$ the letter a occurs in $\sigma ^k(b)$.

Let $\mathrm{FG}({\mathbb {A}})$ be the free group with generators ${\mathbb {A}}.$ Assume $\zeta$ is a substitution over ${\mathbb {A}}$. $\zeta$ can be extended to a group homomorphism on $\mathrm{FG}({\mathbb {A}})$ by defining

$$\begin{aligned} \zeta (a^{-1}):=(\zeta (a))^{-1} \ \ \text { and }\ \ \zeta (\emptyset )=\emptyset , \end{aligned}$$

where $\emptyset$ is the identity of $\mathrm{FG}({\mathbb {A}})$. $\emptyset$ can be viewed as the empty word. If $\zeta$ is invertible as a group homomorphism on $\mathrm{FG}({\mathbb {A}})$, then we call $\zeta$ an invertible substitution.

Now we give three famous examples, which appear in this paper.

Example B.1

Let ${\mathbb {A}}=\{a,b\}$. The Fibonacci substitution $\zeta _1$ is defined as

$$\begin{aligned} \zeta _1(a):=ab;\ \ \zeta _1(b):=a. \end{aligned}$$

$\zeta _1$ is primitive and invertible. Indeed, the inverse of $\zeta _1$ is the group homomorphism determined by the “substitution” $\eta$:

$$\begin{aligned} \eta (a):=b;\ \ \eta (b):=b^{-1}a. \end{aligned}$$

The Fibonacci sequence is the unique fixed point $\zeta _1^\infty (a)$ of $\zeta _1$, where

$$\begin{aligned} \zeta _1^\infty (a):=\lim _{n\rightarrow \infty }\zeta _1^n(a)=abaababa\cdots \end{aligned}$$

Example B.2

Let ${\mathbb {A}}=\{a,b\}$. The Thue-Morse substitution $\zeta _{2}$ is defined as

$$\begin{aligned} \zeta _{2}(a):=ab;\ \ \zeta _{2}(b):=ba. \end{aligned}$$

$\zeta _{2}$ is primitive but not invertible. It has two fixed points: $\zeta _{2}^\infty (a)$ and $\zeta _{2}^\infty (b)$, where

$$\begin{aligned} \zeta _{2}^\infty (a):=\lim _{n\rightarrow \infty }\zeta _{2}^n(a)=abbabaab\cdots \end{aligned}$$

Either $\zeta _{2}^\infty (a)$ or $\zeta _{2}^\infty (b)$ is called the Thue-Morse sequence.

Example B.3

Let ${\mathbb {A}}=\{a,b\}$. The period-doubling substitution $\zeta _{3}$ is defined as

$$\begin{aligned} \zeta _{3}(a):=ab;\ \ \zeta _{3}(b):=aa. \end{aligned}$$

$\zeta _{3}$ is also primitive but not invertible. It has one fixed point:

$$\begin{aligned} \zeta _{3}^\infty (a):=\lim _{n\rightarrow \infty }\zeta _{3}^n(a)=abaaabab\cdots \end{aligned}$$

$\zeta _{3}^\infty (a)$ is called the period-doubling sequence.

The Characterization of the Gaps of $\Sigma _\infty$

In this appendix, we show the following:

Proposition C.1

The set of gaps of $(\Sigma _\infty ,\le )$ is

$$\begin{aligned} \{(\sigma 01^\infty ,\sigma 10^\infty ):\sigma \in \Sigma _*\}. \end{aligned}$$

Proof

Given $\sigma \in \Sigma _*$, we claim that $(\sigma 01^\infty ,\sigma 10^\infty )=\emptyset .$ If otherwise, there exists $\tau \in \Sigma _\infty$ such that $\sigma 01^\infty<\tau <\sigma 10^\infty$. If there exists $j\le |\sigma |$ such that $\sigma _j\ne \tau _j$, then either $\sigma 01^\infty ,\sigma 10^\infty <\tau$ holds or $\tau <\sigma 01^\infty ,\sigma 10^\infty$ holds, which is a contradiction. So $\sigma \lhd \tau$. If $\tau _{|\sigma |+1}=0 (1),$ then

$$\begin{aligned} \tau =\sigma 0\cdots \le \sigma 01^\infty ,\ \ (\sigma 10^\infty \le \sigma 1\cdots =\tau ) \end{aligned}$$

which contradicts with the fact that $\sigma 01^\infty<\tau (\tau < \sigma 10^\infty ).$ So the claim holds. By the definition of gap, $(\sigma 01^\infty ,\sigma 10^\infty )$ is a gap of $\Sigma _\infty .$

Now assume $(\tau ,{\hat{\tau }})$ is a gap of $\Sigma _\infty .$ Assume $\tau \wedge {\hat{\tau }}=\sigma$ and $|\sigma |=n$. Then

$$\begin{aligned} \tau _{n+1}=0,\ \ {\hat{\tau }}_{n+1}=1. \end{aligned}$$

We claim that $\tau _{j}=1$ for any $j\ge n+2.$ If otherwise, there exists $j_0\ge n+2$ such that

$$\begin{aligned} \tau _j=1,\ \ n+2\le j<j_0,\ \ \tau _{j_0}=0. \end{aligned}$$

Then we have

$$\begin{aligned} \tau =\sigma 01^{j_0-(n+2)}0\cdots<\sigma 01^\infty <{\hat{\tau }}, \end{aligned}$$

which contradicts with the fact that $(\tau ,{\hat{\tau }})$ is a gap. So we must have $\tau =\sigma 01^\infty$. The same argument shows that ${\hat{\tau }}=\sigma 10^\infty .$ That is, $(\tau ,{\hat{\tau }})=(\sigma 01^\infty ,\sigma 10^\infty )$. $\square$

Two Tables

For the reader’s convenience, we include two tables of indexes in this appendix. One is for the notations used in this paper, and the other one is for the various orders used in this paper.

Table 1 Index of notations

In this paper, two types of orders are defined in various spaces: one is $\le$, another is $\preceq$. When $\le$ is used, it means standard or strong, depending on the context. When $\preceq$ is used, it means non standard or weak.

Table 2 Index for various orders