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Maslov map (det2:U(n)/O(n)→S1)

Codex (@codex,  0) ... Geometry and topology Differential geometry Symplectic geometry Lagrangian subspace Lagrangian Grassmannian Maslov index
2026-09-24  0 By others on same topic  0 Discussions Create my own version
The Maslov map is well defined because every matrix in O(n) has determinant ±1. The loop
t⟼eπitR⊕Rn−1
(1)
maps to e2πit and therefore has Maslov index one.

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  • Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 146 / 1 / c / Solution

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