2
votes
Accepted
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 ...
- 18k
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 ...
- 1,944
Only top scored, non community-wiki answers of a minimum length are eligible