On the shape of a tridiagonal pair

Abstract

Let $K$ denote a field and let $V$ denote a vector space over $K$ with finite positive dimension. We consider a pair of linear transformations $A:V\to V$ and $A^{\ast }:V\to V$ that satisfy the following conditions: (i) each of $A,A^{\ast }$ is diagonalizable; (ii) there exists an ordering $\{V_i{\}}_{i=0}^d$ of the eigenspaces of $A$ such that $A^{\ast }{V}_i\subseteq V_{i-1}+V_i+V_{i+1}$ for $0\le i\le d$, where $V_{-1}=0$ and $V_{d+1}=0$; (iii) there exists an ordering $\{V_i^{\ast }{\}}_{i=0}^{\delta }$ of the eigenspaces of $A^{\ast }$ such that $A{V}_i^{\ast }\subseteq V_{i-1}^{\ast }+V_i^{\ast }+V_{i+1}^{\ast }$ for $0\le i\le \delta$, where $V_{-1}^{\ast }=0$ and $V_{\delta +1}^{\ast }=0$; (iv) there is no subspace $W$ of $V$ such that $AW\subseteq W$, $A^{\ast }W\subseteq W$, $W\ne 0$, $W\ne V$. We call such a pair a {\it tridiagonal pair} on $V$. It is known that $d=\delta$ and for $0\le i\le d$ the dimensions of $V_i$, $V_i^{\ast }$, $V_{d-i}$, $V_{d-i}^{\ast }$ coincide; we denote this common dimension by ${\rho }_i$. In this paper we prove that ${\rho }_i\le {\rho }_0\left(\genfrac{}{}{0ex}{}{d}{i}\right)$ for $0\le i\le d$. It is already known that ${\rho }_0=1$ if $\K$ is algebraically closed.

30 pages

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