# proof that transition functions of cotangent bundle are valid

In this entry, we shall verify that the transition functions proposed for the cotangent bundle the three criteria required by the classical definition of a manifold.

The first criterion is the easiest to verify. If $\alpha=\beta$, then $\sigma_{\alpha\alpha}$ reduces to the identity and we have

 $\bigg{(}{\sigma^{\prime}}_{\alpha\alpha}(x_{1},\ldots,x_{2n})\bigg{)}^{i}=% \bigg{(}\sigma_{\alpha\alpha}(x_{1},\ldots,x_{n})\bigg{)}^{i}=x^{i}\qquad 1% \leq i\leq n$
 $\bigg{(}{\sigma^{\prime}}_{\alpha\alpha}(x_{1},\ldots,x_{2n})\bigg{)}^{i+n}=% \sum_{j=1}^{n}{\partial\bigg{(}\sigma_{\alpha\alpha}(x_{1},\ldots,x_{n})\bigg{% )}^{i}\over\partial x_{j}}x^{j+n}=\sum_{j=1}^{n}{\partial x^{i}\over\partial x% _{j}}x^{j+n}=x^{i+n}\qquad 1\leq i\leq n$

Thus we see that ${\sigma^{\prime}}_{\alpha\alpha}$ is the identity map, as required.

Next, we turn our attention to the third criterion — showing that ${\sigma^{\prime}}_{\beta\gamma}\circ{\sigma^{\prime}}_{\alpha\beta}={\sigma^{% \prime}}_{\alpha\gamma}$ . For clarity of notation let us define $y^{i}=({\sigma^{\prime}}_{\alpha\beta})^{i}(x^{1},\ldots x^{2n})$. Then we have

 $\displaystyle({\sigma^{\prime}}_{\beta\gamma}\circ{\sigma^{\prime}}_{\alpha% \beta})^{i}(x^{1},\dots,x^{2n})$ $\displaystyle=$ $\displaystyle({\sigma^{\prime}}_{\beta\gamma})^{i}(y^{1},\dots,y^{2n})$ $\displaystyle=$ $\displaystyle(\sigma_{\beta\gamma})^{i}(y^{1},\dots,y^{n})$ $\displaystyle=$ $\displaystyle(\sigma_{\beta\gamma}\circ\sigma_{\alpha\beta})^{i}(x^{1},\dots,x% ^{n})$ $\displaystyle=$ $\displaystyle(\sigma_{\alpha\gamma})^{i}(x^{1},\dots,x^{n})$ $\displaystyle=$ $\displaystyle({\sigma^{\prime}}_{\alpha\gamma})^{i}(x^{1},\dots,x^{2n})$

when $1\leq i\leq n$.

 $\displaystyle({\sigma^{\prime}}_{\beta\gamma}\circ{\sigma^{\prime}}_{\alpha% \beta})^{i+n}(x^{1},\dots,x^{2n})$ $\displaystyle=$ $\displaystyle({\sigma^{\prime}}_{\beta\gamma})^{i+n}(y^{1},\dots,y^{2n})$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{n}{\partial\bigg{(}\sigma_{\beta\gamma}(y_{1},\ldots,% y_{n})\bigg{)}^{i}\over\partial y_{j}}y^{j+n}$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}{\partial\bigg{(}\sigma_{\beta\gamma}% (y_{1},\ldots,y_{n})\bigg{)}^{i}\over\partial y_{j}}{\partial\bigg{(}\sigma_{% \alpha\beta}(x_{1},\ldots,x_{n})\bigg{)}^{j}\over\partial x_{k}}x^{n+k}$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{n}{\partial\bigg{(}\sigma_{\beta\gamma}\circ\sigma_{% \alpha\beta}(x_{1},\ldots,x_{n})\bigg{)}^{i}\over\partial x_{k}}x^{n+k}$ $\displaystyle=$ $\displaystyle\sum_{k=1}^{n}{\partial\bigg{(}\sigma_{\alpha\gamma}(x_{1},\ldots% ,x_{n})\bigg{)}^{i}\over\partial x_{k}}x^{n+k}$ $\displaystyle=$ $\displaystyle{\sigma^{\prime}}_{\alpha\gamma}(x^{1},\dots,x^{2n})$

when $1\leq i\leq n$.

Finally, the second criterion does not need to be checked because it is a consequence of the first and third criteria.

Title proof that transition functions of cotangent bundle are valid ProofThatTransitionFunctionsOfCotangentBundleAreValid 2013-03-22 14:52:25 2013-03-22 14:52:25 rspuzio (6075) rspuzio (6075) 13 rspuzio (6075) Proof msc 58A32