Symplectic form

See Hamiltonian mechanics#Symplectic form.

It is a closed non-degenerate differential 2-form. It defines a symplectic manifold.

In finite dimensional manifolds we can use, I think, the Pfaff-Darboux theorem to obtain coordinates which provide a particularly simple expression for the symplectic form. According to @olver86 page 390 it is not true in the infinite dimensional case.

Discussion

In any cotangent bundle $T^*Q$ we have a canonical 1-form known as tautological 1-form, Poincaré 1-form, Liouville 1-form, symplectic potential or canonical 1-form. Let's see how is defined. We are looking for an element $\lambda \in \Omega^1(T^*Q)$, so consider an element (in any local chart, it doesn't have to be the special one provide by the lagrangian) $(q,p)\in T^*Q.$ We want a definition for a map

$$\lambda_{(q,p)}: T_{(q,p)}(T^*Q)\longmapsto \mathbb{R} $$

Since in local chart we see elements in the form

$$ (q,p,v,\eta)\in T_{(q,p)}(T^*Q) $$

it seems natural to forget $\eta$ and take

$$ \lambda_{(q,p)}(q,p,v,\eta)=p(v) $$

But we can define it without coordinates.

Consider

$$ \pi: T^*Q\longmapsto Q $$

and

$$ d\pi: T(T^*Q)\longmapsto TQ $$

So we can define

$$ \lambda: T^*Q \longmapsto T^*(T^*Q) $$

by

$$ \lambda_{\alpha}=\pi^*(\alpha) $$

Now, we can take the exterior derivative, $\omega:=d\lambda$. This is called the Poincaré 2-form, or symplectic 2-form. It is canonical for every cotangent bundle and verifies:

Any manifold equipped with a 2-form satisfying both conditions is called a symplectic manifold.

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es


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