ProofThatTransitionFunctionsOfCotangentBundleAreValid 2004-12-10 03:41:37 2004-12-20 03:12:25 Proof cotangent bundle 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here In this entry, we shall verify that the transition functions proposed for the cotangent bundle \PMlinkescapeword{satisfy} 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 $$\bigg({\sigma'}_{\alpha \alpha} (x_1, \ldots, x_{2n}) \bigg)^i = \bigg(\sigma_{\alpha \alpha} (x_1, \ldots, x_n) \bigg)^i = x^i \qquad 1 \le i \le n $$ $$\bigg({\sigma'}_{\alpha \alpha} (x_1, \ldots, x_{2n}) \bigg)^{i+n} = \sum_{j = 1}^n {\partial \bigg(\sigma_{\alpha \alpha} (x_1, \ldots, x_n) \bigg)^i \over \partial x_j} x^{j+n} = \sum_{j = 1}^n {\partial x^i \over \partial x_j} x^{j+n} = x^{i+n} \qquad 1 \le i \le n$$ Thus we see that ${\sigma'}_{\alpha \alpha}$ is the identity map, as required. Next, we turn our attention to the third criterion --- showing that ${\sigma'}_{\beta \gamma} \circ {\sigma'}_{\alpha \beta} = {\sigma'}_{\alpha \gamma}$ . For clarity of notation let us define $y^i = ({\sigma'}_{\alpha \beta})^i (x^1, \ldots x^{2n})$. Then we have \begin{eqnarray*} ({\sigma'}_{\beta \gamma} \circ {\sigma'}_{\alpha \beta})^i (x^1, \dots, x^{2n}) &=& ({\sigma'}_{\beta \gamma})^i (y^1, \dots, y^{2n}) \\ &=& (\sigma_{\beta \gamma})^i (y^1, \dots, y^n) \\ &=& (\sigma_{\beta \gamma} \circ \sigma_{\alpha \beta})^i (x^1, \dots, x^n) \\ &=& (\sigma_{\alpha \gamma})^i (x^1, \dots, x^n) \\ &=& ({\sigma'}_{\alpha \gamma})^i (x^1, \dots, x^{2n}) \\ \end{eqnarray*} when $1 \le i \le n $. \begin{eqnarray*} ({\sigma'}_{\beta \gamma} \circ {\sigma'}_{\alpha \beta})^{i+n} (x^1, \dots, x^{2n}) &=& ({\sigma'}_{\beta \gamma})^{i+n} (y^1, \dots, y^{2n}) \\ &=& \sum_{j = 1}^n {\partial \bigg(\sigma_{\beta \gamma} (y_1, \ldots, y_n) \bigg)^i \over \partial y_j} y^{j+n} \\ &=& \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} \\ &=& \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} \\ &=& \sum_{k = 1}^n {\partial \bigg(\sigma_{\alpha \gamma} (x_1, \ldots, x_n) \bigg)^i \over \partial x_k} x^{n+k} \\ &=& {\sigma'}_{\alpha \gamma} (x^1, \dots, x^{2n}) \\ \end{eqnarray*} when $1 \le i \le n $. Finally, the second criterion does not need to be checked because it is a consequence of the first and third criteria.