Liebe A.

This is only an attempt to answer how one might compute the Fibonacci *contents* whose existence you ask about; presently I don't know how to determine the $\tilde{J_k}$ operators. Let's begin by noticing that the quantity

\begin{equation}
\text{tr} \ \sigma_\lambda (t) \cdot \big| T \big|
\end{equation}

for $t \in T$ appears as the $\mu = 21^{n-2}$ contribution when expanding the Schur function $s_\lambda$ in terms of the power-symmetric functions $p_\mu$, specifically

\begin{equation}
\begin{array}{ll}
\displaystyle s_\lambda
&\displaystyle = \, {1 \over {n!}} \, \sum_{|\mu| = n} \,
\text{tr} \, \sigma_\lambda \big(c_\mu \big)
\cdot \big| C_\mu \big| \, p_\mu \\ \\
&\displaystyle = \, \sum_{|\mu| = n} \,
\text{tr} \, \sigma_\lambda \big(c_\mu \big)
\cdot {p_\mu \over {z(\mu)}} \end{array}
\end{equation}

where $c_\mu$ is any element in the conjugacy class $C_\mu$
indexed by the partition $\mu$ and

\begin{equation}
z(\mu) \, := \, {n! \over {\big| C_\mu \big|}}
\end{equation}

You're probably aware that
there are analogues of the Schur and power-symmetric functions
in Okada theory. Recall that for each word $w$ in the Young-Fibonacci
lattice a corresponding *Okada-Schur* function is constructed recursively using two infinite lists of variables $x_1, x_2, \dots$ and $y_1, y_2, \dots$
in the following way: For integers $n \geq 1$ define the following
$n \times n$ tridiagonal determinants

\begin{equation}
P_n \, := \, \det \begin{pmatrix} x_1 & y_1 & 0 & \cdots
\\ 1 & x_2 & y_2 & & \\
0 & 1 & x_3 & & \\
\vdots & & & \ddots \end{pmatrix}
\qquad \qquad
Q_{n-1} \, := \, \det \begin{pmatrix} y_1 & x_1y_2 & 0 & \cdots
\\ 1 & x_3 & y_3 & & \\
0 & 1 & x_4 & & \\
\vdots & & & \ddots \end{pmatrix}
\end{equation}

and by convention set $P_0 := 1$.
The *Schur-Okada* function $s_w$ corresponding to $w$ is
defined by the recursion

\begin{equation} s_w \, := \ \left\{ \begin{array}{ll} P_k &\text{if
$w= \, 1^k \, $ and $k \geq 0$} \\ \\
Q_k \, \big[+ |v| \big] \cdot s_v &\text{if $w= \, 1^k \, 2 \, v \, $
and $k \geq 0$}
\end{array} \right\} \end{equation}

The notation $Q_k \big[ + |v| \big]$ means perform the substitutions
$x_i \mapsto x_{i + |v|}$ and $y_i \mapsto y_{i + |v|}$ in the
expression for $Q_k$. Look in the paper by Goodman and Kerov (https://arxiv.org/pdf/math/9712266.pdf) for
a definition of the power-symmetric function analogues $p_w$. The crucial point is that

\begin{equation}
s_w \ = \ \sum_{|u| = n} \, X^w_u {p_u \over {z(u)}}
\end{equation}

where the $X^w_u$ are *Okada character values* for the Okada algebra $\mathcal{A}_n$ and

\begin{equation}
z(u) \, := \, k_0! \cdots k_t! \, \big(2 + k_{1} \big) \, \cdots \, \big(2 + k_t \big)
\end{equation}

where we have parsed $u$ as

\begin{equation}
u \, = \ 1^{k_0} \, 2 \, 1^{k_1} \, 2 \cdots \, 2 \, 1^{k_t}
\end{equation}

with $t$ equal to the number of occurrences of $2$ in $u$ and $k_j \geq 0$ for each $t \geq j \geq 0$. I would also point out that for $w$ with
$|w| = n$ the Okada character value $X^w_{1^n}$ equals the dimension of the irreducible $\mathcal{A}_n$-module $V_w$. So the correct analogue of

\begin{equation}
{ \text{tr} \ \sigma_\lambda (t) \over {\text{dim} V_\lambda}} \cdot \big| T \big|
\end{equation}

(up to some uniform factor in $n$) ought to be

\begin{equation}
{X^w_{u(n)} \over {X^w_{1_n}}} \cdot {1 \over {z \big( u(n) \big)}}
\quad (\dagger) \end{equation}

for some family of words $u(n)$ with lengths $|u(n)| = n$. Having made a guess for this family, label the nodes of the Young-Fibonacci lattice by
these values ($\dagger$) and then try to recursively solve for the *contents* associated to each covering relation in the lattice by taking
differences.

your, Ines.

p.s. Here's a guess: $u(n) := \, 21^{n-2}$ for $n \geq 2$ with $u(1) := 1$.

p.p.s. According to Okada's paper $\displaystyle {|u|! \over {z(u)}} \in \Bbb{Z}$ and $\displaystyle \sum_{|u|=n} \, {n! \over {z(u)}} \, = \, n!$ so my guess is that the

\begin{equation} \text{$\dagger \dagger$-values} \quad
{X^w_{u(n)} \over {X^w_{1_n}}} \cdot {n! \over {z \big( u(n) \big)}}
\end{equation}

be used instead of the previously defined $\dagger$-values

p.p.p.s Just to be clear let me elaborate with an example. Suppose we select $u(n) = 21^{n-2}$ for all $n\geq 2$ (this will be our choice to play the role of the conjugacy class consisting of all reflections in $S_n$ for $n\geq 2$). Suppose we want the fibonacci *content* associated to the covering relation (i.e. edge in the Hasse diagram) between $121$ and $1121$. So we need to compute the respective $\dagger \dagger$-values,
namely:

\begin{equation}
{X^{121}_{211} \over {X^{121}_{111}}} \cdot {4! \over {z(211)}} \, = \, 3
\quad \quad {X^{1121}_{2111} \over {X^{1121}_{1111}}} \cdot {5! \over {z(2111)}} \, = \, 4
\end{equation}

So the *content* ought to be $4-3 = 1$.

Continuing with the same example but for higher rank, please note that
for elements $u$ and $v$ of respective ranks $|u|=n-1$ and $|v|=n-2 one can use Okada's recursive formula to obtain

\begin{equation}
{X^{1u}_{21^{n-2}} \over {X^{1u}_{1^n}}} \, = \, 1 \quad \quad
{X^{2v}_{21^{n-2}} \over {X^{2v}_{1^n}}} \, = \, {1 \over {1-n}}
\end{equation}

In addition

\begin{equation}
{n! \over {z\big(21^{n-2} \big)}} \, = \, {n! \over {{(n-2)}! \, n }} \, = \, n-1
\end{equation}

There are four types of covering relations $u \sqsubset v$ between the elements within levels of rank $n$ and rank $n+1$ depending on whether the prefixes ($\text{pf}$ for short) of $u$ and $v$ are $1$ or $2$. I list them together with the corresponding contents in following table:

\begin{equation}
\begin{array}{cl}
\text{contents:}
&\text{types of covering relations:} \\
1
&\text{if $\text{pf}(u) = 1 \, $ and $\, \text{pf}(v) = 1$} \\
-n
&\text{if $\text{pf}(u) = 1 \, $ and $\, \text{pf}(v) = 2$} \\
n+1
&\text{if $\text{pf}(u) = 2 \, $ and $\, \text{pf}(v) = 1$} \\
0
&\text{if $\text{pf}(u) = 2 \, $ and $\, \text{pf}(v) = 2$}
\end{array}
\end{equation}