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Two commuting operators associated with a tridiagonal pair (1110.3434v1)

Published 15 Oct 2011 in math.RA

Abstract: Let \K denote a field and let V denote a vector space over \K with finite positive dimension. We consider an ordered pair of linear transformations A:V\to V and A*:V \to V that satisfy the following four conditions: (i) Each of A,A* is diagonalizable; (ii) there exists an ordering {V_i}{i=0}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 and V_{d+1}=0; (iii) there exists an ordering {V*i}{i=0}{\delta} of the eigenspaces of A* such that AV*i\subseteq V*{i-1}+V*i+V*{i+1} for 0\leq i\leq\delta, where V*{-1}=0 and V*{\delta+1}=0; (iv) there does not exist a subspace W of V such that AW\subseteq W, A*W\subseteq W, W\neq0, W\neq V. We call such a pair a TD pair on V. It is known that d=\delta; to avoid trivialities assume d\geq 1. We show that there exists a unique linear transformation \Delta:V\to V such that (\Delta -I)V*i\subseteq V*_0+V*_1+...+V*{i-1} and \Delta(V_i+V_{i+1}+...+V_d)=V_0 +V_{1}+...+V_{d-i} for 0\leq i \leq d. We show that there exists a unique linear transformation \Psi:V\to V such that \Psi V_i\subseteq V_{i-1}+V_i+V_{i+1} and (\Psi-\Lambda)V*i\subseteq V*_0+V*_1+...+V*{i-2} for 0\leq i\leq d, where \Lambda=(\Delta-I)(\theta_0-\theta_d){-1} and \theta_0 (resp \theta_d) denotes the eigenvalue of A associated with V_0 (resp V_d). We characterize \Delta,\Psi in several ways. There are two well-known decompositions of V called the first and second split decomposition. We discuss how \Delta,\Psi act on these decompositions. We also show how \Delta,\Psi relate to each other. Along this line we have two main results. Our first main result is that \Delta,\Psi commute. In the literature on TD pairs there is a scalar \beta used to describe the eigenvalues. Our second main result is that each of \Delta{\pm 1} is a polynomial of degree d in \Psi, under a minor assumption on \beta.

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