Cotangent bundle of a $n$-dimensional differentiable manifold is a $2n$-dimensional manifold

Question

How to prove that cotangent bundle of a $n$-dimensional differentiable manifold is a $2n$-dimensional manifold? Detailed explanation is welcome. Thanks in advance.

Answer 1

I suppose you are given a definition of tangent space to every point (with curves or in an algebraic way). As set the cotangent bundle is defined as $\bigcup_{p \in M} T_{p}^{*}$, where the union is disjointed. Now consider the atlas you have, call it $\lbrace \left( U_{i} , \varphi_{i} \right) \rbrace_{i \in I}$. First, you define the topology on the cotangent bundle of each open set of the covering. For every $i$ you have a map $\left( \varphi_{i}, d\varphi_{i} \right): T^{*}U_{i} \rightarrow \varphi_{i}\left(U_{i}\right)\times \mathbb{R}^{n}$. Since you know it is a bijection, you simply impose it is an homeomorphism. Then you impose that the whole topology is the one generated by the sets which are "obliged" to be open by the maps $\left( \varphi_{i}, d\varphi_{i}\right)$.It is easy to show it is Hausdorff, since you separate covectors which "live" in different points with the topology of the manifold itself, while if two covectors "Live" on the same point you separate them in the $\mathbb{R}^{n}$ where they live, since you have endowed every cotangent space with the euclidean topology. Then the maps we have used to define the topology are themselves the charts, since $\varphi_{i}\left(U_{i}\right) \times \mathbb{R}^{n}$ is an open set in $\mathbb{R}^{2n}$. What is left is the change of charts. By construction the atlas is $\lbrace \left( V_{i} , \left(\varphi_{i}, d\varphi_{i}\right) \right) \rbrace_{i \in I}$, with the projection on the manifold $\pi : V_{i} \rightarrow U_{i}$. By construction $V_{i} \bigcap V_{j} \neq \emptyset$ iff $U_{i} \bigcap U_{j} \neq \emptyset$. Now the change on charts is $\left(\varphi_{i} \circ\varphi_{j}^{-1},d\varphi_{i} \circ d \varphi_{j}^{-1}=d\left(\varphi_{i} \circ\varphi_{j}^{-1} \right)\right):\varphi_{j}\left(U_{j}\right) \times \mathbb{R}^{n} \rightarrow \varphi_{i}\left(U_{i}\right) \times \mathbb{R}^{n}$. You already know $\varphi_{i} \circ\varphi_{j}^{-1}$ is smooth; the map on the second $n$ coordinates is linear, since it is the differential of a smooth function between open sets of $\mathbb{R}^{n}$. Hence the function is smooth itself and the structure of manifold is given.