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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.

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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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