The Lagrange and Markov Spectra from the Dynamical Point of View

Abstract

This text grew out of my lecture notes for a 4-h minicourse delivered on October 17 and 19, 2016 during the research school “Applications of Ergodic Theory in Number Theory”—an activity related to the Jean-Molet Chair project of Mariusz Lemańczyk and Sébastien Ferenczi—realized at CIRM, Marseille, France. The subject of this text is the same as my minicourse, namely, the structure of the so-called Lagrange and Markov spectra (with a special emphasis on a recent theorem of C.G. Moreira).

Appendices

Appendix 1: Proof of Hurwitz Theorem

Given $\alpha \notin \mathbb {Q}$, we want to show that the inequality

$$\displaystyle \begin{aligned}\left|\alpha-\frac{p}{q}\right|\leqslant\frac{1}{\sqrt{5}q^2}\end{aligned}$$

has infinitely many rational solutions.

In this direction, let α = [a ₀;a ₁, … ] be the continued fraction expansion of α and denote by [a ₀;a ₁, …, a _n ] = p _n ∕q _n . We affirm that, for every $\alpha \notin \mathbb {Q}$ and every $n\geqslant 1$, we have

$$\displaystyle \begin{aligned}\left|\alpha-\frac{p}{q}\right|<\frac{1}{\sqrt{5}q^2}\end{aligned}$$

for some $\frac {p}{q}\in \{\frac {p_{n-1}}{q_{n-1}}, \frac {p_n}{q_n}, \frac {p_{n+1}}{q_{n+1}}\}$.

Remark 14.60

Of course, this last statement provides infinitely many solutions to the inequality $\left |\alpha -\frac {p}{q}\right |\leqslant \frac {1}{\sqrt {5}q^2}$. So, our task is reduced to prove the affirmation above.

The proof of the claim starts by recalling Perron’s Proposition 14.20:

$$\displaystyle \begin{aligned}\alpha-\frac{p_n}{q_n} = \frac{(-1)^n}{(\alpha_{n+1}+\beta_{n+1})q_n^2}\end{aligned}$$

where α _n+1 := [a _n+1;a _n+2, … ] and $\beta _{n+1} = \frac {q_{n-1}}{q_n} = [0;a_n,\dots ,a_1]$.

For the sake of contradiction, suppose that the claim is false, i.e., there exists $k\geqslant 1$ such that

$$\displaystyle \begin{aligned} \max\{(\alpha_k+\beta_k), (\alpha_{k+1}+\beta_{k+1}), (\alpha_{k+2}+\beta_{k+2})\}\leqslant \sqrt{5} \end{aligned} $$

(14.7)

Since $\sqrt {5}<3$ and $a_m\leqslant \alpha _m+\beta _m$ for all $m\geqslant 1$, it follows from (14.7) that

$$\displaystyle \begin{aligned} \max\{a_k,a_{k+1},a_{k+2}\}\leqslant 2 \end{aligned} $$

(14.8)

If a _m  = 2 for some $k\leqslant m\leqslant k+2$, then (14.8) would imply that $\alpha _m+\beta _m\geqslant 2+[0;2,1] = 2+\frac {1}{3}>\sqrt {5}$, a contradiction with our assumption (14.7).

So, our hypothesis (14.7) forces

$$\displaystyle \begin{aligned} a_k=a_{k+1}=a_{k+2}=1 \end{aligned} $$

(14.9)

Denoting by $x=\frac {1}{\alpha _{k+2}}$ and $y=\beta _{k+1} = q_{k-1}/q_k\in \mathbb {Q}$, we have from (14.9) that

$$\displaystyle \begin{aligned}\alpha_{k+1}=1+x, \quad\alpha_k = 1+\frac{1}{1+x}, \quad\beta_{k} = \frac{1}{y}-1, \quad\beta_{k+2} = \frac{1}{1+y}\end{aligned}$$

By plugging this into (14.7), we obtain

$$\displaystyle \begin{aligned} \max\left\{\frac{1}{1+x}+\frac{1}{y}, 1+x+y, \frac{1}{x}+\frac{1}{1+y}\right\}\leqslant \sqrt{5} \end{aligned} $$

(14.10)

On one hand, (14.10) implies that

$$\displaystyle \begin{aligned}\frac{1}{1+x}+\frac{1}{y}\leqslant \sqrt{5} \quad\mathrm{and} \quad1+x\leqslant \sqrt{5}-y.\end{aligned}$$

Thus,

$$\displaystyle \begin{aligned}\frac{\sqrt{5}}{y(\sqrt{5}-y)} = \frac{1}{\sqrt{5}-y}+\frac{1}{y}\leqslant \frac{1}{1+x}+\frac{1}{y}\leqslant\sqrt{5},\end{aligned}$$

and, a fortiori, $y(\sqrt {5}-y)\geqslant 1$, i.e.,

$$\displaystyle \begin{aligned} \frac{\sqrt{5}-1}{2}\leqslant y\leqslant \frac{\sqrt{5}+1}{2} \end{aligned} $$

(14.11)

On the other hand, (14.10) implies that

$$\displaystyle \begin{aligned}x\leqslant \sqrt{5}-1-y \quad\mathrm{and} \quad\frac{1}{x}+\frac{1}{1+y}\leqslant \sqrt{5}.\end{aligned}$$

Hence,

$$\displaystyle \begin{aligned}\frac{\sqrt{5}}{(1+y)(\sqrt{5}-1-y)} = \frac{1}{\sqrt{5}-1-y}+\frac{1}{1+y}\leqslant \frac{1}{x}+\frac{1}{1+y}\leqslant\sqrt{5},\end{aligned}$$

and, a fortiori, $(1+y)(\sqrt {5}-1-y)\geqslant 1$, i.e.,

$$\displaystyle \begin{aligned} \frac{\sqrt{5}-1}{2}\leqslant y\leqslant \frac{\sqrt{5}+1}{2} \end{aligned} $$

(14.12)

It follows from (14.11) and (14.12) that $y=(\sqrt {5}-1)/2$, a contradiction because $y=\beta _{k+1}= q_{k-1}/q_k\in \mathbb {Q}$. This completes the argument.

Appendix 2: Proof of Euler’s Remark

Denote by $[0; a_1, a_2,\dots , a_n] = \frac {p(a_1,\dots ,a_n)}{q(a_1,\dots ,a_n)} = \frac {p_n}{q_n}$. It is not hard to see that

$$\displaystyle \begin{aligned} &q(a_1)=a_1, \quad q(a_1,a_2) = a_1a_2+1,\\ &q(a_1,\dots,a_n) = a_n q(a_1,\dots,a_{n-1}) + q(a_1,\dots,a_{n-2}) \,\,\,\,\forall\,\,n\geqslant 3. \end{aligned} $$

From this formula, we see that q(a ₁, …, a _n ) is a sum of the following products of elements of {a ₁, …, a _n }. First, we take the product a ₁…a _n of all a _i ’s. Secondly, we take all products obtained by removing any pair a _i a _i+1 of adjacent elements. Then, we iterate this procedure until no pairs can be omitted (with the convention that if n is even, then the empty product gives 1). This rule to describe q(a ₁, …, a _n ) was discovered by Euler.

It follows immediately from Euler’s rule that q(a ₁, …, a _n ) = q(a _n , …, a ₁). This proves Proposition 14.47.