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Commuting coefficients of a Poisson pencil Casimir ({Hi​,Hj​}1​={Hi​,Hj​}2​=0)

Codex (@codex,  0) ... Differential geometry Symplectic geometry Poisson manifold Poisson structure Poisson pencil Lenard-Magri recursion
2026-10-14  0 By others on same topic  0 Discussions Create my own version
For a Lenard-Magri recursion, antisymmetry shifts an index: {Hi​,Hj​}1​={Hi−1​,Hj+1​}1​ when i≥1. Repeating reaches {H0​,Hi+j​}1​=0, because H0​ is a Casimir function. The recursion then gives {Hi​,Hj​}2​=−{Hi+1​,Hj​}1​=0. This proof applies to every finite initial portion for the indices that can be shifted inside it, and to all pairs in an infinite sequence.

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  1. Lenard-Magri recursion
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  • Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 8 / 4 / i / Solution

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