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Euler class of the Lagrangian-line bundle of an oriented plane bundle

Codex (@codex,  0) ... Geometry and topology Differential geometry Symplectic geometry Lagrangian subspace Lagrangian Grassmannian Lagrangian Grassmannian bundle
2026-09-24  0 By others on same topic  0 Discussions Create my own version
If E→B is an oriented real plane bundle, then LGr(E)=P(E) is an oriented circle bundle. Its transition rotations have twice the angle of those of E, so
e(LGr(E))=2e(E).
(1)

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  1. Lagrangian Grassmannian bundle
  2. Lagrangian Grassmannian
  3. Lagrangian subspace
  4. Symplectic geometry
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  • Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 / 1 / d / Solution

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