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[parent] cotangent bundle is a bundle (Proof)

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, \begin{eqnarray*} \sum_{j = 1}^n {g_{\alpha \beta}}^i_j (x^1, \ldots, x^{2n}) {\phi_\beta}^j (x^1, \ldots, x^{2n}) &=& \sum_{j = n}^{2n} {\partial \big( \sigma_{\alpha \beta} (x^1, \ldots x^n) \big)^i \over \partial x^j} x^{j+n} \\ &=& {{\sigma'}_{\alpha \beta}}^{i+n} (x^1, \ldots, x^{2n}) \\ &=& {\phi_\alpha}^i (x^1, \ldots, x^{2n}) \end{eqnarray*} The third criterion follows from the chain rule: \begin{eqnarray*} {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} \end{eqnarray*}



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Cross-references: chain rule, homeomorphism

This is version 3 of cotangent bundle is a bundle, born on 2004-12-21, modified 2004-12-21.
Object id is 6593, canonical name is CotangentBundleIsABundle.
Accessed 1168 times total.

Classification:
AMS MSC58A32 (Global analysis, analysis on manifolds :: General theory of differentiable manifolds :: Natural bundles)

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