Proving $T^*M$ is a symplectic manifold

330 Views Asked by At

I kind of jump right into the middle of some differential geometry, and don't fully understand the following. Given $\mathbb{R^{2n}}$ with standard coordinates $(x^1,\dots,x^n,y^1,\dots,y^n)$ (i.e., I suppose, if $(e_i)$ is the standard basis then $x^i(e_j) = \delta_{ij}$ and $y^i(e_j) = \delta_{(n+i)j}$) then the following is a standard example:

$(\mathbb{R^{2n}},\omega^2)$ is a symplectic manifold, with structure $\omega^2 = \sum_{i=1}^n dx^i \wedge dy^i$.

Fine, it's easy to show that the above form is indeed a symplectic form.

Then comes the cotangent bundle. We choose ''natural coordinates'' $(x^1,\dots x^n,\xi_1,\dots,\xi_n)$ with $(x^1)$ some local coordinates on our base manifold $M$ and $(\xi_i)$ component functions of the $1$-forms, corresponding to $(x^i)$ (i.e. every $1$-form $\omega^1$ can locally be written $\xi_idx^i$.) The quest is now to prove that $T^*M$ is a symplectic manifold. This is done by constructing the tautological $1$-form $\omega^1 = \xi dx^i$ on $T^*M$ and then forming the two form $\omega^2 = -d\omega^1 = \sum_{i=1}^n dx^i \wedge d\xi^i$. Then the argument is that the form we got is just as the one for $(\mathbb{R}^{2n},\omega^2)$ and hence symplectic. But surely something must be hidden here? Because given a basis $(x^i,\xi_i)$ for $T^*M$ one could immediately declare $\omega^2 = \sum_{i=1}^n dx^i \wedge d\xi^i$ and be done with it, just as in the case $\mathbb{R^{2n}}$? Why do we need to go through the construction of the tautological $1$-form?

EDIT: I guess the question boils down to the following: given local coordinates $(x^1,\dots,x^n,y^1,\dots,y^n)$ on some manifold $M$, why isn't $\sum_{i=1}^n dx^i \wedge dy^i$ always a symplectic form?

1

There are 1 best solutions below

1
On BEST ANSWER

Given local coordinates $(x^{1}, \dots, x^{n}, y^{1}, \dots, y^{n})$ in some open set $U$ on a manifold $M$, the form $$ \omega = \sum_{i=1}^{n} dx^{i} \wedge dy^{i} $$ is a symplectic form, but only in $U$, not on all of $M$.

The tautological $1$-form in $T^{*}M$, despite being defined in local coordinates, is globally-defined, so its exterior derivative is a global symplectic structure.