# $\mathfrak{sl}_2$-action on Young diagrams

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.: $$$$\label{comm} [D,U]=L\,, \quad [L,U]=2U\,, \quad [L,D]=-2D\,.$$$$

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

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