cotangent bundle is a bundle

Verifying the first criterion is simply a matter of writing it out:

 $(x^{1},\ldots,x^{2n})\mapsto(\pi(x^{1},\ldots,x^{2n}),\phi_{\alpha}(x^{1},% \ldots,x^{n}))=((x^{1},\ldots,x^{n}),(x^{n+1},\ldots,x^{2n}))$

This is obviously a homeomorphism.

As for the second criterion,

 $\displaystyle\sum_{j=1}^{n}{g_{\alpha\beta}}^{i}_{j}(x^{1},\ldots,x^{2n}){\phi% _{\beta}}^{j}(x^{1},\ldots,x^{2n})$ $\displaystyle=$ $\displaystyle\sum_{j=n}^{2n}{\partial\big{(}\sigma_{\alpha\beta}(x^{1},\ldots x% ^{n})\big{)}^{i}\over\partial x^{j}}x^{j+n}$ $\displaystyle=$ $\displaystyle{{\sigma^{\prime}}_{\alpha\beta}}^{i+n}(x^{1},\ldots,x^{2n})$ $\displaystyle=$ $\displaystyle{\phi_{\alpha}}^{i}(x^{1},\ldots,x^{2n})$

The third criterion follows from the chain rule:

 $\displaystyle{g_{\alpha\beta}}^{i}_{j}{g_{\beta\gamma}}^{j}_{k}={\partial\big{% (}\sigma_{\alpha\beta}(x^{1},\ldots x^{n})\big{)}^{i}\over\partial x^{j}}{% \partial\big{(}\sigma_{\beta\gamma}(x^{1},\ldots x^{n})\big{)}^{j}\over% \partial x^{k}}$
Title cotangent bundle is a bundle CotangentBundleIsABundle 2013-03-22 14:54:31 2013-03-22 14:54:31 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Proof msc 58A32