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The Okada algebra $\mathfrak{O}_n$ over a field $K$ has generators $E_1,\dots,E_{n-1}$ and relations $E_i^2=x_iE_i$, $E_{i+1}E_iE_{i+1}=y_i E_{i+1}$, and $E_iE_j=E_jE_i$ for $|i-j|\geq 2$, where $x_i,y_i\in K$. Okada (Algebras associated to the Young–Fibonacci lattice, Trans. Amer. Math. Soc. 346 (1994), 549–568) shows that $\mathfrak{O}_n$ is semisimple of dimension $n!$ when the $x_i$'s and $y_i$'s are generic. I computed (if I didn't make an error) that when $n=3$, the discriminant of $\mathfrak{O}_3$ is a constant times $x_1^2 y_1^4(x_1x_2-y_1)^4$. Can this result be extended to larger $n$?

Addendum. I managed to compute the discriminant of $\mathfrak{O}_4$. It is a constant times $$ x_1^{20} y_1^{10} y_2^{24} (x_1 y_2 - x_3 y_1)^6 (x_1 x_2 - y_1)^{10} (x_1 x_2 x_3 - x_1 y_2 - x_3 y_1)^6. $$

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    $\begingroup$ What do you mean by disriminant here? Is this related to the Gram determinant of the Markov trace on the Okada algebra? If so, it should factor into a product of (specialized) clone Schur functions. The examples that you give bear this out. I have code that calculates this. $\endgroup$
    – Jeanne Scott
    Commented Sep 22, 2023 at 19:49
  • $\begingroup$ For example: If we set all the $x$-parameters to $1$ (which still keeps the algebra semi-simple) the Gram determiants of $\frak{O}_3$ and $\frak{O}_4$ are given by $-y_1^4 y_2^4 (1-y_1)^4(y_1 - y_2)(1-y_1 -y_2)$ and $-y_1^{19} y_2^{24}y_3^{18} (1-y_1)^{19}(y_1-y_2)^6(1-y_1-y_2)^6(y_2-y_3)^4(y_1-y_2-y_1y_3)(1-y_1-y_2-y_3+y_1y_3)$. In addition to (powers of) clone Schur functions there could also be shifts of clone Schur functions (i.e. where the parameter indices of a clone Schur function are shifted by a common integer). $\endgroup$
    – Jeanne Scott
    Commented Sep 22, 2023 at 20:09
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    $\begingroup$ @JeanneScott: I don't know what is the Gram determinant of the Markov trace. I mean the determinant of the matrix $[\mathrm{tr}(B_i B_j)]$, where the $B_i$'s are a basis for $\mathfrak{O}_n$. The most natural basis is Okada's basis $\mathcal{B}_n$. Where does $y_2$ come from for $\mathfrak{O}_3$ and $y_3$ for $\mathfrak{O}_4$? They are not part of the defining relations. $\endgroup$
    – Richard Stanley
    Commented Sep 23, 2023 at 1:28
  • $\begingroup$ Instead of using the pairing $\mathrm{tr}(xy)$ for elements $x,y \in \frak{O}_n$ Okada defined a bilinear form $\mathrm{Tr}(xy) := \sum_{|w|=n} \mathrm{tr} \big[ \rho_w(xy) \big] s_w$ where the sum is taken over all fibonacci words $w$ of rank $n$, where $\rho_w$ is the corresponding irreducible representation of $\frak{O}_n$, and where $s_w$ is the associated specialized clone Schur function (a polynomial expressed in the parameters $x_1, \dots, x_n$ and $y_1, \dots, y_{n-1}$ determined by $w$). Okada's pairing has the property that it is a Markov trace. $\endgroup$
    – Jeanne Scott
    Commented Sep 23, 2023 at 1:44
  • $\begingroup$ Of course you can consider the Gram matrix $\big[ \varphi(B_iB_j) \big]$ for any trace functional $\varphi: \frak{O}_n \rightarrow \Bbb{C}$ together with any basis of $\frak{O}_n$. In general the determinant of this Gram matrix ought to be some product of specialized clone Schur functions because the simultaneous non-vanishing of these polynomials $s_w$ for all $|w| < n$ is equivalent to the semi-simplicity of $\frak{O}_n$. Florent Hivert and I are trying to find a closed formula for this Gram determinant in the case of Okada's Markov trace. $\endgroup$
    – Jeanne Scott
    Commented Sep 23, 2023 at 2:31

2 Answers 2

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Using some results in progress with Jeanne Scott, I can compute the few next one. Note that these computation rely on thing that are not yet fully proved. So it might differ from what you are actually computing. Here are the next two. I'm not sure what you want to do with them

$$\mathfrak{O}_5 = x_{1}^{52} \cdot y_{2}^{88} \cdot y_{1}^{98} \cdot y_{3}^{144} \cdot (-x_{3} y_{1} + x_{1} y_{2})^{22} \cdot (-x_{4} y_{2} + x_{2} y_{3})^{32} \cdot (-x_{1} x_{2} + y_{1})^{98} \cdot (-x_{3} x_{4} y_{1} + x_{1} x_{4} y_{2} + y_{1} y_{3})^{8} \cdot (-x_{1} x_{2} x_{3} + x_{3} y_{1} + x_{1} y_{2})^{22} \cdot (x_{1} x_{2} x_{3} x_{4} - x_{3} x_{4} y_{1} - x_{1} x_{4} y_{2} - x_{1} x_{2} y_{3} + y_{1} y_{3})^{8}$$

$$\mathfrak{O}_6 = y_{1}^{468} \cdot x_{1}^{552} \cdot y_{3}^{684} \cdot y_{4}^{960} \cdot y_{2}^{1008} \cdot (-x_{4} y_{2} + x_{2} y_{3})^{152} \cdot (-x_{5} y_{3} + x_{3} y_{4})^{180} \cdot (-x_{3} y_{1} + x_{1} y_{2})^{252} \cdot (-x_{1} x_{2} + y_{1})^{468} \cdot (-x_{3} x_{4} y_{1} + x_{1} x_{4} y_{2} + y_{1} y_{3})^{38} \cdot (-x_{4} x_{5} y_{2} + x_{2} x_{5} y_{3} + y_{2} y_{4})^{40} \cdot (-x_{1} x_{2} x_{3} + x_{3} y_{1} + x_{1} y_{2})^{252} \cdot (x_{3} x_{4} x_{5} y_{1} - x_{1} x_{4} x_{5} y_{2} - x_{5} y_{1} y_{3} - x_{3} y_{1} y_{4} + x_{1} y_{2} y_{4})^{10} \cdot (x_{1} x_{2} x_{3} x_{4} - x_{3} x_{4} y_{1} - x_{1} x_{4} y_{2} - x_{1} x_{2} y_{3} + y_{1} y_{3})^{38} \cdot (x_{1} x_{2} x_{3} x_{4} x_{5} - x_{3} x_{4} x_{5} y_{1} - x_{1} x_{4} x_{5} y_{2} - x_{1} x_{2} x_{5} y_{3} - x_{1} x_{2} x_{3} y_{4} + x_{5} y_{1} y_{3} + x_{3} y_{1} y_{4} + x_{1} y_{2} y_{4})^{10}$$

As Jeanne says, we are currently working on those, so I can't be more explicit right now. But We will definitely post more results (or conjectures) as soon as we have something to show.

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  • $\begingroup$ The recursive structure of the factors seems quite clear. $$\mathfrak{O}_6 = x_{1}^{552} \cdot y_{1}^{468} \cdot y_{2}^{1008} \cdot y_{3}^{684} \cdot y_{4}^{960} \cdot (-x_{1} x_{2} + y_{1})^{468} \cdot (-x_{3} y_{1} + x_{1} y_{2})^{252} \cdot (-x_{4} y_{2} + x_{2} y_{3})^{152} \cdot (-x_{5} y_{3} + x_{3} y_{4})^{180} \cdot (-x_{3} x_{4} y_{1} + x_{1} x_{4} y_{2} + y_{1} y_{3})^{38} \cdot (-x_{4} x_{5} y_{2} + x_{2} x_{5} y_{3} + y_{2} y_{4})^{40} \cdot (-x_{1} x_{2} x_{3} + x_{3} y_{1} + x_{1} y_{2})^{252} $$ (cont'd) $\endgroup$
    – Wolfgang
    Commented Sep 30, 2023 at 20:32
  • $\begingroup$ $$\cdot (\color{red}{- x_{5}}[ -x_{3} x_{4} y_{1} + x_{1} x_{4} y_{2} + y_{1} y_{3}] + \color{red}{y_{4}}[- x_{3} y_{1} + x_{1} y_{2} ])^{10} \cdot (\color{red}{-x_{4}}[-x_{1} x_{2} x_{3} + x_{3} y_{1} + x_{1} y_{2}] + \color{red}{y_{3}} [-x_{1} x_{2} + y_{1}] )^{38} \cdot (\color{red}{x_{5}}[x_{1} x_{2} x_{3} x_{4} - x_{3} x_{4} y_{1} - x_{1} x_{4} y_{2} - x_{1} x_{2} y_{3} + y_{1} y_{3}]+ \color{red}{y_{4}} [-x_{1} x_{2} x_{3} + x_{3} y_{1} + x_{1} y_{2}])^{10}$$ with all the terms in brackets occurring earlier. (and signs may be swapped to make it more consistent) $\endgroup$
    – Wolfgang
    Commented Sep 30, 2023 at 20:32
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Define $a_0:=1,a_1:=x_1$ and recursively for $k\ge2$ $$a_{k}:=x_{k}a_{k-1}-y_{k-1}a_{k-2}.$$ So $a_k$ is the factor starting with $x_1\cdots x_k$. (Note that most of them have their signs flipped compared with the above, but the outcome is the same as all exponents are even).

Further define $b_{1,\ell}:=y_\ell$ and recursively for $k\ge2$ $$b_{k,\ell}:=x_{k+\ell}b_{k-1,\ell}-y_{k+\ell-1}{b_{k-2,\ell}}.$$ If we put $b_{k,0}:=a_k$, the recursions are consistent, and for the given data, $\mathfrak{O}_n$ consists of exactly the factors $b_{k,\ell}$ where $$k=1,\dots,n-1, \ell=0,\dots,n-2\ \text{ and } \ k+\ell\le n-1$$ with exponents of the $b_{k,\ell}$ for $n=3,4,5,6$ $$ \begin{pmatrix} 2 & 4 \\ 4 \end{pmatrix},\begin{pmatrix} 20 &10 & 24 \\ 10&6 \\ 6 \end{pmatrix},\begin{pmatrix} 52&98&88&144 \\ 98&22&32 \\ 22&8\\8 \end{pmatrix}, \begin{pmatrix} 552&468&1008&684&960 \\ 468&252&152&180 \\ 252&38&40\\38&10\\10 \end{pmatrix}.$$ You may notice that the ratios for neighboring columns are $1, 4/1, 9/2, 16/3$. Now, supposing this pattern $k^2/(k-1)$ carries on, the first column (i.e. the exponents of the $a_k$'s) defines the whole matrix. But in order to come up with a conjecture what they are, more data would be needed...

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