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I started studying Hamiltonian Fields and Symplectic Geometry, and I have a question about how we can construct a bundle isomorphism induced by the symplectic form.

Let $(M,\omega)$ be a symplectic manifold and consider the bundle homomorphism $\widehat{\omega}: TM \rightarrow T^*M$. How can I prove that $\widehat{\omega}$ is a bundle isomorphism?

I know that this is useful to define the Hamiltonian Vector Field of a function $f\in \mathcal C^{\infty}(M)$, but I cannot see why Symplectic Manifolds are the natural way to define Hamiltonian Systems rather than Riemannian Manifolds. What are the main differences between them?

ayphyros
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    For the isomorphism: use the nondegeneracy. This gives you pointwise an isomorphism, which is what you need (if you are picky about smoothness, you probably need to write $\omega$ in coordinates and check everything). As for your second question, the Hamiltonian vector field gives you a geometric way to write down Hamilton's eqs (the integral curves of the Hamiltonian vector field satisfy Hamilton's eqs). – WishYouTheBest Dec 10 '23 at 12:43
  • @WishYouTheBest thanks so much for your answer :) – ayphyros Dec 10 '23 at 18:25
  • In your case, because you’re dealing with linear things, you can get away with a variant of Cramer’s formula, but for the general case here is a good-to-know theorem about smoothness and inverses in the bundle setting. – peek-a-boo Dec 11 '23 at 20:37

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