The Fibonacci lattice $\mathcal{F}$ is the poset of all finite words consisting of 1's and 2's where a word $v$ covers a word $u$ if $v$ is obtained from $v$ by either (a) inserting a 1 in $u$ prior to its leftmost 1 or (b) changing the leftmost 1 occurring in $u$ into a 2. Let $\mathcal{L}$ be the partially ordered set obtained by "vetting" the Young-Fibonacci lattice of all "constant" words --- i.e. all words of length at least two of the form $11 \dots 1$ and $22 \dots 2$. Interpret the remaining words as purely periodic continued fractions; e.g. the word $1121$ would correspond to the quadratic surd $\tau$ defined by

\begin{equation} \tau \ = \ {1 \over {1 + {\displaystyle 1 \over {\displaystyle 1 + {\displaystyle 1 \over {\displaystyle 2 + {\displaystyle 1 \over {\displaystyle 1 + \tau} }}}}} }} \end{equation}

Recall that the Lagrange number $L(\tau)$ of a positive real number $\tau$ is defined as

\begin{equation} \text{sup} \, \Bigg\{ L \ : \ \Big| \tau - {p \over q} \Big| < {1 \over {Lq^2}} \ \, \begin{array}{l} \text{for infinitely may reduced} \\ \text{postive rational numbers ${p \over q}$} \end{array} \Bigg\} \end{equation}

Finite words consisting of 1's and 2's correspond, under the recipe described above, to congruence classes of quadratic surds $\tau$ whose Lagrange numbers $L(\tau)$ are strictly less than $3$, where two real numbers $\alpha$ and $\beta$ are understood to be congruent if there exists $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \text{GL}_2 \big( \Bbb{Z} \big)$ with $\det(g) = \pm 1$ such that

\begin{equation} \alpha \ = \ {a \beta + b \over {c \beta +d}} \end{equation}

Question: Can the order relation on $\mathcal{L}$ be meanfully interpreted as a kind of closure relation on the congruence classes of the Lagrange spectrum beneath 3?

regards, Ines Institoris.


It seems there is some confusion in what you wrote. There are purely periodic continued fractions consisting only of 1's and 2's whose Lagrange numbers are greater than 3 --- for example the Lagrange number of $\Big[ \overline{121} \Big]$ is $\sqrt{10}$. See Martin Aigner's book "Markov's Theorem and 100 Years of the Uniqueness Conjecture" page 27 for this computation. However the converse is true, namely: If $\tau$ is a real number whose Lagrange number is strictly less than 3 then $\tau$ is congruent (in the sense you explain) to a continued fraction consisting only of 1's and 2's. So it seems to me that the Young-Fibonacci lattice ought to be further "vetted" in order to exclude surds like Aigner's example.

The question about $G$-orbit "closures" and covering relations (presumably in the spirit of how one defines the Bruhat order ?) where $G$ is the group of all $2 \times 2$ integer matrices with determinant $\pm 1$ is compelling.

My apologies for not adding this to the comment section; I don't have enough "points" to exercise this privilege yet :P

best, A. Leverkühn


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