1
$\begingroup$

Let $\mathcal{Y}$ be a vector $\mathbb{Q}$-space of all Young diagrams. Denote by $\delta_\lambda$ the Young diagram of the partition $\lambda$ and $c(\square)$ be the content of the square $\square$: $$ c(\square) = {\rm column}(\square) - {\rm row}(\square)\, $$ where ${\rm column}(\square)$ denotes the column number of the square $\square$.

The paper

A. Okounkov, SL(2) and z-measures, Random matrix models and their applications (PM Bleher and AR Its, eds). Mathematical Sciences Research Institute Publications 40 (2001), pp.407–420

says (without proof) that the following Kerov's operators ( $z, z'$ are parametrs)

\begin{alignat*}{2} U\, \delta_\lambda &= \sum_{\mu=\lambda+\square} &(z+c(\square)) \, &\delta_\mu \,\\ L\, \delta_\lambda &= &(zz'+2|\lambda|)\, &\delta_\lambda\,\\ D\, \delta_\lambda &= \sum_{\mu=\lambda-\square} &(z'+c(\square)) \, &\delta_\mu \,, \end{alignat*}

define a $\mathfrak{sl}_2$ action on $\mathcal{Y}$, i.e.: $$ \begin{equation}\label{comm} [D,U]=L\,, \quad [L,U]=2U\,, \quad [L,D]=-2D\,. \end{equation} $$

Is there any proof of the statement or it is a trivial fact?

$\endgroup$
1
  • 2
    $\begingroup$ I think it shouldn't be too hard to prove directly. It is just a slight extension of Stanley's differential poset identity. $\endgroup$ May 29, 2022 at 12:15

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.