The split decomposition of a tridiagonal pair

Details The split decomposition of a tridiagonal pair The split decomposition of a tridiagonal pair

2006-12-16

arxiv.org/abs/math/0612460v1

Mathematics

Rings and Algebras

7 pages

Scientific paper

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 $A^*:V \to V$ that satisfy (i)--(iv) below: (i) Each of $A$, $A^*$ is diagonalizable. (ii) There exists an ordering $V_{0},V_{1},...,V_{d}$ of the eigenspaces of $A$ such that $A^* V_i \subseteq V_{i-1} + V_{i} + V_{i+1}$ for $0 \leq i \leq d$, where $V_{-1}=0$, $V_{d+1}=0$. (iii) There exists an ordering $V^*_{0},V^*_{1},...,V^*_{\delta}$ of the eigenspaces of $A^*$ such that $A V^*_i \subseteq V^*_{i-1} + V^*_{i} + V^*_{i+1}$ for $0 \leq i \leq \delta$, where $V^*_{-1}=0$, $V^*_{\delta+1}=0$. (iv) There is no subspace $W$ of $V$ such that both $AW \subseteq W$, $A^* W \subseteq W$, other than W=0 and $W=V$. We call such a pair a tridiagonal pair on $V$. In this note we obtain two results. First, we show that each of $A,A^*$ is determined up to affine transformation by the $V_i$ and $V^*_i$. Secondly, we characterize the case in which the $V_i$ and $V^*_i$ all have dimension one. We prove both results using a certain decomposition of $V$ called the split decomposition.

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