2 votes

Darboux's theorem in other words

This answer expands on @J.V.Gaiter's comment. Let $\varphi\colon U \subset \Bbb R^{2n} \to \varphi(U) \subset M$ and $\varphi' \colon U' \subset \Bbb R^{2n} \to \varphi'(U')\subset M'$ be Darboux ...
Didier's user avatar
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1 vote

The Liouville form pulled back to the Lagrangian on the cotangent bundle vanishes

The claim simply isn't true. The defining property of the Liouville form is that $\alpha^*\lambda=\alpha$ for any one form $\alpha$ on $M$ (viewed as a map $M\to T^*M$). This means that the graph of ...
J.V.Gaiter's user avatar
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