OurBigBook About$ Donate
 Sign in Sign up

Obstruction to a Lenard-Magri recursion

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
The next step of a Lenard-Magri recursion requires −Π2​dHn−1​ to be a Hamiltonian vector field for Π1​. It must be tangent to the symplectic leaves, and its contraction with the leaf symplectic form must have a global potential, with a compatible smooth extension between leaves. Even the first step may fail: on R2, take Π1​=0, Π2​=∂x​∧∂y​ and H0​=x. Both Poisson structures are compatible, but Π1​dH1​=−Π2​dx cannot hold, because its left side is zero and its right side is nonzero.

 Ancestors (10)

  1. Lenard-Magri recursion
  2. Poisson pencil
  3. Poisson structure
  4. Poisson manifold
  5. Symplectic geometry
  6. Differential geometry
  7. Geometry and topology
  8. Area of mathematics
  9. Mathematics
  10.  Home

 Incoming links (1)

  • Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 8 / 4 / ii / Solution

 View article source

 Discussion (0)

New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.
  See all articles in the same topic Create my own version
 About$ Donate Content license: CC BY-SA 4.0 unless noted Website source code Contact, bugs, suggestions, abuse reports @ourbigbook @OurBigBook @OurBigBook