cotangent bundle is a bundle

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

$$(x^1,\mathrm{\dots },x^{2n})\mapsto (\pi (x^1,\mathrm{\dots },x^{2n}),{\varphi }_{\alpha }(x^1,\mathrm{\dots },x^n))=((x^1,\mathrm{\dots },x^n),(x^{n+1},\mathrm{\dots },x^{2n}))$$

This is obviously a homeomorphism.

As for the second criterion,

${\displaystyle \sum _{j=1}^n}g_{\alpha \beta }{}_j{}^i(x^1,\mathrm{\dots },x^{2n})\varphi _{\beta }{}^j(x^1,\mathrm{\dots },x^{2n})$ $=$ ${\displaystyle \sum _{j=n}^{2n}}{\displaystyle \frac{\partial {\left({\sigma }_{\alpha \beta }(x^1,\mathrm{\dots }{x}^n)\right)}^i}{\partial x^j}}{x}^{j+n}$

$=$ $\sigma ^{\prime }{}_{\alpha \beta }{}^{i+n}(x^1,\mathrm{\dots },x^{2n})$

$=$ $\varphi _{\alpha }{}^i(x^1,\mathrm{\dots },x^{2n})$

The third criterion follows from the chain rule:

$g_{\alpha \beta }{}_j{}^{i}g_{\beta \gamma }{}_k{}^j={\displaystyle \frac{\partial {\left({\sigma }_{\alpha \beta }(x^1,\mathrm{\dots }{x}^n)\right)}^i}{\partial x^j}}{\displaystyle \frac{\partial {\left({\sigma }_{\beta \gamma }(x^1,\mathrm{\dots }{x}^n)\right)}^j}{\partial x^k}}$

Title	cotangent bundle is a bundle
Canonical name	CotangentBundleIsABundle
Date of creation	2013-03-22 14:54:31
Last modified on	2013-03-22 14:54:31
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	6
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 58A32