9
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – darij grinberg Aug 9 '17 at 1:10
2
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Ines Institoris Aug 9 '17 at 1:22
2
$\begingroup$

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( { \scriptstyle \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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.