Young-Fibonacci tableaux, content, and the Okada algebra

Using the French convention, the content of the $i \times j$ box in the Young diagram of a partition $\lambda$ is $i-j$. Now if $\lambda$ is partition of $n$ and $\sigma_\lambda: S_n \longrightarrow V_\lambda$ is the corresponding irreducible representation of the symmetric group $S_n$ then the sum of contents of all boxes in the young diagram of $\lambda$ equals

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

where $T$ is the conjugacy class consisting of all transpositions and $t$ is any choice of transposition. Moreover each Young tableau $Y_\lambda$ encodes an eigenvector in $V_\lambda$ for the operator $\sigma_\lambda \big( J_k \big)$ with eigenvalue $c_k$ where

\begin{equation} J_k \, := \, \sum_{i=1}^{k-1} \, \big(i,k \big) \, \in \Bbb{C} \big[ S_n \big] \end{equation}

and $c_k$ is the content of the box in $Y_\lambda$ labeled by $k$.

Consider now the Young-Fibonacci lattice whose elements consist of words $w = a_1 \cdots a_d$ taken from the alphabet $\{1,2 \}$ which can be visualised by stacking boxes into adjacient vertical columns going from left to right such that the number of boxes in the $i$-th column is $a_i$. The rank $|w|$ of a word is simply the number of boxes in such a picture; equivalently $|w|$ equals $a_1 \, + \, \cdots \, + \, a_d$. I won't describe the covering relations that give the lattice structure --- but suffice it to say each word $w$ or rank $n$ encodes an irreducible representation $V_w$ of the Okada algebra $\mathcal{A}_n$ and each complete chain $Y_w$ ending at $w$ indexes a basis vector in $V_w$.

Question: (1) Are there pairwise commuting operators $\tilde{J_1}, \dots, \tilde{J_n}$ within the Okada algebra $\mathcal{A}_n$ for which each complete chain $Y_w$ (viewed as a basis vector in $V_w$) is a simultaneous eigenvector and (2) is there a notion of content (a value for each covering relation in the Young-Fibonacci lattice) so that the $k$-th content $c_k$ of $Y_w$ (viewed as a complete chain) is the eigenvalue of $\tilde{J_k}$ corresponding to $Y_w$ and (3) will the sum of such contents along any complete chain $Y_w$ be constant ?

regards, A. Leverkühn

• So you are looking for analogues of the Young-Jucys-Murphy elements in the Okada algebra. Quite an interesting question. For starters, is there an analogue of the sum of all transpositions? – darij grinberg Aug 9 '17 at 1:10

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$} \end{array} \end{equation} • Sorry for the multiple corrections: I was having some difficulty with Okada's definition of$z(u)$--- I think it's settled now. best, Ines – Ines Institoris Aug 9 '17 at 1:22 Errors have been fixed In this second response let me try to address what might play the role of the operator sum$\tilde{J_1} + \cdots + \tilde{J_n}$which is the analogue of$\sum_{t \in T} \, t \ \in \Bbb{C}[S_n]$where$T$is the conjugacy class consisting of all transpositions. Given two sets of generic parameters$x_1, \dots, x_{n-1}$and and$y_1, \dots, y_{n-2}$the associated Okada algebra$\mathcal{F}_n$has a presentation given by generators$E_1, \dots, E_{n-1}$subject to the defining relations \begin{equation} \begin{array}{rll} E_i^2 &= \ x_i \, E_i & \\ E_i \, E_j &= \ E_j \, E_i &\text{whenever$|i-j| >1$} \\ E_j \, E_i \, E_j &= \ y_i \, E_j &\text{whenever$j = i+1$} \end{array} \end{equation} Given a word$w$the Okada power-symmetric functions$p_w$is defined by the recursion \begin{equation} p_w \ = \ \left\{ \begin{array}{ll} \displaystyle x_1 \cdots \, x_k &\text{if$\, w = 1^k \, $with$\, k \geq 0$} \\ \displaystyle q_k \cdot p_u \big[ + (k + 2) \big] &\text{if$\, w= u 2 1^k \,$with$\, k \geq 0$} \end{array} \right\} \end{equation} where \begin{equation} q_k \ := \ (x_1 \cdots \, x_k) \, \Big( x_{k+1} x_{k+2} \, - \, (k+2) \,y_{k+1} \Big) \end{equation} and$p_u \big[ +l \big]$means perform the substitutions$x_i \mapsto x_{i+l}$and$y_i \mapsto y_{i+l}$for all$i \geq 1$. For the purposes of this discussion, let's provisionally define the Okada power-symmetric function in terms of the Okada-Schur functions and the Okada characters values: \begin{equation} p_u \, = \ \sum_{|v| = |u|} \, X^v_u \, s_v \end{equation} Now let's define$\mathcal{F_n}$-valued versions$\Bbb{s}_w$and$\Bbb{p}_w$of the Okada-Schur and power-symmetric functions. For$n \geq k \geq 1$define the following$k \times k$tri-diagonal determinants whose values are in the Okada algebra$\mathcal{F_n}$\begin{equation} \begin{array}{ll} \mathcal{P}_k \, := &\det \, \begin{pmatrix} x_1 & x_2 E_1 & 0 & \cdots \\ 1 & x_2 & x_3 E_2 & & \\ 0 & 1 & x_3 & & \\ \vdots & & & \ddots \end{pmatrix} \\ \\ \\ \mathcal{Q}_{k-1} \, := &\det \, \begin{pmatrix} x_2 E_1 & x_1 x_3 E_2 & 0 & \cdots \\ 1 & x_3 & x_4 E_3 & & \\ 0 & 1 & x_4 & & \\ \vdots & & & \ddots \end{pmatrix} \end{array} \end{equation} where$\mathcal{P}_0 := 1$. Clearly an order must be observed when tabulating the determinant --- following the conventions of Kerov and Goodman, the$l$-th factor in the expansion will always be selected from the$l$-th column. The values of the determinants$\mathcal{P}_k$and$\mathcal{Q}_{k-1}$are in fact independent of the order in which products are taken in the Laplace expansion: This is because indices of the generators$E_1, \dots, E_{n-1}$which participate in any given monomial in the expansion of such a tri-diagonal determinant will always differ by at least two. Employ the same recursion above being mindful to place the accumulating$\mathcal{Q}$-factors to the left and in order: \begin{equation} \Bbb{s}_w \, := \ \left\{ \begin{array}{ll} \mathcal{P}_k &\text{if$w= \, 1^k \, $and$k \geq 0$} \\ \\ \mathcal{Q}_k \, \big[+ |v| \big] \cdot \Bbb{s}_v &\text{if$w= \, 1^k \, 2 \, v \, $and$k \geq 0$} \end{array} \right\} \end{equation} Once again$\mathcal{Q}_k \big[ + |v| \big]$means shift all indices by$|v|$. Play the same game and define the$\mathcal{F}_n$-valued power-symmetric functions by the expansion: \begin{equation} \Bbb{p}_u \, := \ \sum_{|v| = |u|} \, X^v_u \, \Bbb{s}_v \end{equation} I want to make use of Okada's trace functional$\text{Tr}: \mathcal{F}_n \longrightarrow \Bbb{C}$which is defined for an element$a \in \mathcal{F}_n$using the Okada-Schur values by \begin{equation} \text{Tr}(a) \ = \ {1 \over {(x_1 \cdots \, x_n)}} \, \sum_{|v|=n} \, s_v \cdot \text{tr} \Big[ \sigma_v(a) \Big] \end{equation} where$\sigma_v : \mathcal{F}_n \longrightarrow \text{End}(V_v)$is the irreducible representation of$\mathcal{F}_n$associated to a word$v$with$|v|=n$and where$\text{tr} \big[ \cdot \big]$denotes the usual trace. Okada proves in his paper that \begin{equation} \begin{array}{ll} \displaystyle \text{Tr} \, (1) &\displaystyle = \, 1 \\ \displaystyle \text{Tr} \, \big( ab \big) &\displaystyle = \, \text{Tr} \, \big(ba \big) \\ \displaystyle \text{Tr} \, \big( aE_i \big) &\displaystyle = \, {y_i \over {x_{i+1}}} \, \text{Tr} \, (a) \quad \text{when$a \in \mathcal{F}_i$} \quad \left( { \begin{array}{l} \text{As far as I can tell there seems to be} \\ \text{a missing$x_{i+1}$in the denominator of} \\ \text{part (4) of Proposition 2.7 in Okada's} \\ \text{paper which I have tried to correct here.} \end{array} }\right) \end{array} \end{equation} Using these multiplicative properties, a simple induction on the length$|v|$reveals that$\text{Tr} \, \big( \Bbb{s}_v \big) \, = \, s_v$and consequently$\text{Tr} \, \big( \Bbb{p}_u \big) \, = \, p_u$. Moreover \begin{equation} \begin{array}{c} \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{p}_u \big) \Big] \, = \, \big( x_1 \cdots x_n \big) \, X^v_u \\ \displaystyle \text{--- and ---} \\ \displaystyle \text{tr} \, \Big[ \pi_v \big( \Bbb{s}_u \big) \Big] \, = \, \big(x_1 \cdots x_n \big) \, \delta_{u,v} \end{array} \end{equation} Isn't this observation an indication that$\Bbb{p}_u$is an analogue of a characteristic function of a conjugacy class in the group setting ? However, pursuing this analogy, it's not immediately clear whether the elements$\mathcal{p}_u$are central in$\mathcal{F}_n$. yours, Ines. p.s. In fact the order of products taken in the expansion of the$\mathcal{F}_n$-valued determinants$\mathcal{P}_k$and$\mathcal{Q}_{k-1}$is irrelevant --- this is because all monomials which occur involve$E\$-generators whose subscripts differ by at least two.