Posted by: matheuscmss | August 22, 2017

## HD(M\L) < 0.986927

My friend Gugu and I have just uploaded to the arXiv our paper ${HD(M\setminus L) < 0.986927}$. This article continues our investigations of the Hausdorff dimension ${HD(M\setminus L)}$ of the complement of the Lagrange spectrum ${L}$ in the Markov spectrum ${M}$. More precisely, we showed in a previous paper (see also this blog post here) that ${HD(M\setminus L) > 0}$ and we prove now that ${HD(M\setminus L)<1}$.

The key dynamical idea to give upper bounds on ${HD(M\setminus L)}$ is to show that any sufficiently large element ${m\in M\setminus L}$ is realized by a sequence ${\underline{\theta}\in(\mathbb{N}^*)^{\mathbb{Z}}}$ whose past or future dynamics lies in the gaps of an appropriate horseshoe.

Qualitately speaking, this idea is explained by the following lemma.

Lemma 1 Fix ${\Lambda}$horseshoe of a surface diffeomorphism ${\varphi}$ and ${f}$ a height function. For simplicity, let us denote the orbits of ${\varphi}$ by ${x_n:=\varphi^n(x)}$. Denote by

$\displaystyle M=\{\sup\limits_{n\in\mathbb{Z}}f(x_n): x_0\in\Lambda\} \quad \textrm{ and } \quad L=\{\limsup\limits_{n\rightarrow\infty}f(x_n): x_0\in\Lambda\}$

the corresponding Markov and Lagrange spectra.Let ${\widetilde{\Lambda}}$ a subhorseshoe of ${\Lambda}$ and set

$\displaystyle m(\widetilde{\Lambda}) = \max\limits_{y\in\widetilde{\Lambda}} m(y) \quad (= \max\limits_{z\in\widetilde{\Lambda}} f(z) )$

where ${m(a):=\sup\limits_{n\in\mathbb{Z}}f(a_n)}$ is the Markov value of ${a}$. Consider ${m\in M\setminus L}$ such that ${m>m(\widetilde{\Lambda})}$, and denote by ${x_0\in\Lambda}$ a point with

$\displaystyle m=\sup\limits_{n\in\mathbb{Z}} f(x_n) = f(x_0)$

Then, either ${\alpha(x_0)\cap \overline{\Lambda}=\emptyset}$ or ${\omega(x_0)\cap \overline{\Lambda}=\emptyset}$ (where ${\alpha(x)}$ and ${\omega(x)}$ denote the ${\alpha}$ and ${\omega}$ limit sets of the orbit of ${x}$).

Proof: By contradiction, suppose that ${z\in\alpha(x)\cap\widetilde{\Lambda}}$ and ${w\in\omega(x)\cap\widetilde{\Lambda}}$.

Since ${m\in M\setminus L}$ and ${m>m(\widetilde{\Lambda})}$, we can select ${\varepsilon>0}$ and ${N\in\mathbb{N}}$ such that ${f(x_n) for all ${|n|\geq N}$, and ${m(\widetilde{\Lambda}). Also, the definitions allow us to take ${m_k\rightarrow-\infty}$ and ${n_k\rightarrow\infty}$ such that ${x_{m_k}\rightarrow z}$ and ${x_{n_k}\rightarrow w}$.

Fix ${y\in\widetilde{\Lambda}}$ with dense orbit and consider pieces ${y_{r_k}\dots y_{s_k}}$ of the orbit of ${y}$ with ${y_{r_k}\rightarrow w}$ and ${y_{s_k}\rightarrow z}$.

Consider the pseudo-orbits ${x_0\dots x_{n_k} y_{r_k}\dots y_{s_k}x_{m_k}\dots x_0}$. By the shadowing lemma, we obtain a sequence of periodic orbits accumulating ${x_0}$ whose Markov values converge to ${m}$. In particular, ${m\in L}$, a contradiction. $\Box$

In simple terms, this lemma says that an element ${m\in M\setminus L}$ with ${m>m(\widetilde{\Lambda})}$ is associated to an orbit ${(x_n)_{n\in\mathbb{Z}}}$ whose past dynamics (described by ${\alpha(x_0)}$) or future dynamics (described by ${\omega(x_0)}$) avoids ${\widetilde{\Lambda}}$. Thus, there exists ${k\in\mathbb{N}}$ such that either the piece ${(x_n)_{n\leq -k}}$ of past orbit or the piece ${(x_n)_{n\geq k}}$ of future orbit avoids a neighborhood of ${\widetilde{\Lambda}}$ in ${\Lambda}$ (i.e., one of these pieces of orbit lives in the gaps of ${\Lambda\setminus\widetilde{\Lambda}}$).

Remark 1 As it turns out, this qualitative lemma is not sufficient for our purposes and, for this reason, Gugu and I end up using a quantitative version of this lemma (called Lemma 3.1) in our paper.

Once we got some constraints on the dynamics of orbits generating elements of ${M\setminus L}$, our strategy to estimate ${HD(M\setminus L)}$ consists in careful choices of ${\widetilde{\Lambda}}$ and ${\Lambda}$.

For the sake of exposition, let us explain how our strategy yields some bounds for ${HD((M\setminus L)\cap [3.06, \sqrt{12}])}$.

Perron proved that any ${m\in M\cap(-\infty, \sqrt{12}]}$ has the form ${m=m(\underline{\theta})}$ where ${\underline{\theta}\in\{1,2\}^{\mathbb{Z}}=\Lambda}$.

Consider the subhorseshoe ${\widetilde{\Lambda} = \{11,22\}^{\mathbb{Z}}}$ (of sequences formed by concatenations of two consecutive 1’s and two consecutive 2’s).

By applying (a quantitative version of) Lemma 1 (cf. Remark 1), one concludes that if ${m=m(\underline{\theta})\in (M\setminus L)\cap [3.06, \sqrt{12}]}$, then the past or future dynamics of ${\underline{\theta}}$ lives in the gaps of ${\widetilde{\Lambda}}$.

This means that, up to replacing ${\underline{\theta}=(\theta_n)_{n\in\mathbb{Z}}}$ by ${(\theta_{-n})_{n\in\mathbb{Z}}}$, for all ${n\in\mathbb{N}}$ sufficiently large:

• either there is an unique extension of ${\dots\theta_0\dots\theta_n}$ giving a sequence whose Markov value in ${(M\setminus L)\cap [3.06,\sqrt{12}]}$;
• or there are two continuations ${\dots\theta_0\dots\theta_n1\alpha_{n+2}}$ and ${\dots\theta_0\dots\theta_n2\beta_{n+2}}$ of ${\dots\theta_0\dots\theta_n}$ so that the interval ${[[0;2\beta_{n+1}], [0;1\alpha_{n+1}]]}$ is disjoint from the Cantor set

$\displaystyle K(\{11,22\}):=\{[0;\gamma]:\gamma\in\{11, 22\}^{\mathbb{N}}\}$

associated to ${\widetilde{\Lambda} = \{11,22\}^{\mathbb{Z}}}$.

(Here, ${[a_0;a_1,\dots] = a_0+\frac{1}{a_1+\frac{1}{\ddots}}}$ denotes continued fraction expansions.)

Note that this dichotomy imposes severe restrictions on the future ${(\theta_n)_{n\geq 0}}$ of ${\underline{\theta}}$ because there are not many ways to build sequences associated to ${(M\setminus L)\cap[3.06,\sqrt{12}]}$. More precisely, we claim that at each sufficiently large step ${n}$,

• either we get a forced continuation ${\dots\theta_0\dots\theta_n\rightarrow \dots\theta_0\dots\theta_n\theta_{n+1}}$;
• or our possible continuations are ${\dots\theta_0\dots\theta_n112\alpha_{n+4}}$ and ${\dots\theta_0\dots\theta_n221\beta_{n+4}}$.

Indeed, suppose that we have two possible continuations ${\dots\theta_0\dots\theta_n1\alpha_{n+2}}$ and ${\dots\theta_0\dots\theta_n2\beta_{n+2}}$. If ${I(a_1,\dots,a_n)}$ denotes the interval of numbers in ${[0,1]}$ whose continued fraction expansion starts by ${[0;a_1,\dots, a_n,\dots]}$, then the intervals ${I(a_1a_2)}$, ${(a_1,a_2)\in\{1,2\}^2}$ appear in the following order on the real line:

$\displaystyle I(21), I(22), I(11), I(12)$

Thus:

• the continuation ${\dots\theta_0\dots\theta_n12\alpha_{n+3}}$ is not possible (otherwise ${[[0;2\beta_{n+2}]], [0;12\alpha_{n+3}]]}$ would contain ${I(11)}$ and, a fortiori, intersect ${K(\{11,22\})}$;
• the continuation ${\dots\theta_0\dots\theta_n21\beta_{n+3}}$ is not possible (otherwise ${[[0;21\beta_{n+3}]], [0;1\alpha_{n+2}]]}$ would contain ${I(22)}$ and, a fortiori, intersect ${K(\{11,22\})}$

so that our continuations are ${\dots\theta_0\dots\theta_n11\alpha_{n+3}}$ and ${\dots\theta_0\dots\theta_n22\beta_{n+3}}$. Now, we observe that the intervals ${I(11a_3)}$ and ${I(22a_3)}$, ${a_3\in\{1,2\}}$, appear in the following order on the real line:

$\displaystyle I(222), I(221), I(112), I(111)$

Hence:

• the continuation ${\dots\theta_0\dots\theta_n111\alpha_{n+4}}$ is not possible (otherwise ${[[0;22\beta_{n+3}]], [0;111\alpha_{n+4}]]}$ would contain ${I(112)}$ and, a fortiori, intersect ${K(\{11,22\})}$;
• the continuation ${\dots\theta_0\dots\theta_n222\beta_{n+3}}$ is not possible (otherwise ${[[0;222\beta_{n+4}]], [0;11\alpha_{n+3}]]}$ would contain ${I(221)}$ and, a fortiori, intersect ${K(\{11,22\})}$

so that our claim is proved.

This claim allows us to bound the Hausdorff dimension of

$\displaystyle K:=\{[\theta_0;\theta_1,\dots]: 3.06

In fact, the claim says that we refine the natural cover of ${K}$ by the intervals ${I(\theta_1,\dots, \theta_n)}$ by replacing it by a “forced” ${I(\theta_1,\dots, \theta_n,\theta_{n+1})}$ or by the couple of intervals

$\displaystyle I(\theta_1,\dots, \theta_n, 1, 1, 2) \quad \textrm{ and } \quad I(\theta_1,\dots, \theta_n, 2, 2, 1)$

Therefore, it follows from the definition of Hausdorff dimension that

$\displaystyle HD(K)\leq s_0$

for any parameter ${0\leq s_0\leq 1}$ such that

$\displaystyle |I(\theta_1,\dots,\theta_n,1,1,2)|^{s_0} + |I(\theta_1,\dots,\theta_n,2,2,1)|^{s_0} \leq |I(\theta_1,\dots,\theta_n)|^{s_0} \ \ \ \ \ (1)$

Because we can assume that ${m=m(\underline{\theta}) = [\theta_0;\theta_1,\dots]+[0;\theta_{-1},\dots]}$ with ${[\theta_0;\theta_1,\dots]\in K}$ and ${[0;\theta_{-1},\dots]\in C(2) := \{[0;\gamma]: \gamma\in\{1,2\}^{\mathbb{N}}\}}$, our discussion so far can be summarized by following proposition:

Proposition 2 ${(M\setminus L)\cap[3.06,\sqrt{12}]}$ is contained in the arithmetic sum

$\displaystyle C(2)+K$

where ${HD(K)\leq s_0}$ for any parameter ${0\leq s_0\leq 1}$ satisfying (1).

Since the arithmetic sum ${C(2)+K}$ is the projection ${\pi(C(2)\times K)}$, ${\pi(x,y)=x+y}$, of the product set ${C(2)\times K}$, this proposition implies the following result:

Corollary 3 ${HD((M\setminus L)\cap[3.06,\sqrt{12}])\leq HD(C(2))+s_0}$ where ${0\leq s_0\leq 1}$ satisfies (1).

The Hausdorff dimension of ${C(2)}$ was computed with high accuracy by Hensley among other authors: one has ${HD(C(2))<0.531291}$. In particular,

$\displaystyle HD((M\setminus L)\cap[3.06,\sqrt{12}])<0.531291+s_0$

where ${s_0}$ verifies (1).

Closing this post, let us show that (1) holds for ${s_0=0.174813}$ and, consequently,

$\displaystyle HD((M\setminus L)\cap[3.06,\sqrt{12}])<0.706104$

For this sake, recall that

$\displaystyle |I(b_1,\dots,b_n)|=\frac{1}{q_n(q_n+q_{n-1})},$

where ${q_j}$ is the denominator of ${[0;b_1,\dots,b_j]}$.

Hence, if we set

$\displaystyle g(s) := \frac{|I(a_1,\dots,a_n,1,1,2)|^s+|I(a_1,\dots,a_n,2,2,1)|^s}{|I(a_1,\dots,a_n)|^s}$

then the recurrence formula ${q_{j+2}=a_{j+2}q_{j+1}+q_j}$ implies that

$\displaystyle g(s) = \left(\frac{r+1}{(3r+5)(4r+7)}\right)^s + \left(\frac{r+1}{(3r+7)(5r+12)}\right)^s$

where ${r=q_{n-1}/q_n\in (0,1)}$.

Because ${\frac{r+1}{(3r+5)(4r+7)}\leq \frac{1}{35}}$ and ${\frac{r+1}{(3r+7)(5r+12)}<\frac{1}{81.98}}$ for all ${0\leq r\leq 1}$, we have

$\displaystyle g(s)<\left(\frac{1}{35}\right)^s + \left(\frac{1}{81.98}\right)^s$

This completes the argument because ${\left(\frac{1}{35}\right)^{0.174813} + \left(\frac{1}{81.98}\right)^{0.174813} < 1}$.