# What would be the cotangent bundle of a Banach manifold?

Reference: Lang - Differential manifolds p.123

Quick question: Lang defines the cotangent bundle as the dual vector bundle of the tangent bundle, but shouldn't there be additionally a somewhat canonical differentiable structure on the cotangent bundle? (Or does the dual vector bundle naturally determine the differentiable structure?) How do I define such canonical differentiable structure?

Let me illustrate my approach in detail below:

Let $X$ be a Banach $C^k$-manifold.

Define $C_p(X)$ as the class of $(c,v)$ where $c=(U,\phi,E)$ is a chart at $p$ and $v\in E$. (Note that $U$ denotes a coordinate domain and $\phi$ denotes a coordinate map and $E$ denotes a Banach space where $\phi$ is mapping to)

Now, define an equivalence relation on $C_p(X)$ as follows:

$((U,\phi,E),v)\sim((V,\psi,F),w)$ iff $D(\psi\circ \phi^{-1})(\phi(p))(v)=(w)$

Define a map $\mathscr{O}_c^p:E \rightarrow C_p(X)/\sim: v\mapsto [(c,v)]$.

Then, it can be shown that $\mathscr{O}_c^p$ is a bijection. Using this map, we can isomorphically give a vector space structure on $C_p(X)/\sim$, and this structure can be shown independent of choice of charts $c$. Let $T_p(x)$ denote this set equipped with the natural vector space structure.

Moreover, we can homeomorphically give a Banachable topology on $T_p(X)$ using those maps $\mathscr{O}_c^p$.

Now, define $T_p^*(X)$ as the collection of continuous linear functionals on $T_p(X)$. So that $T_p^*(X)$ is a Banachable space too.

With these terminologies (which are standard as far as I know), let's define the cotangent bundle.

Define $T^*X:=\bigcup_{p\in X} (\{p\}\times T_p^*(X))$, and let $\pi^*:T^*X\rightarrow X$ be the map such that $\pi^*(p,y)=p$.

Let $c=(U,\phi,E)$ be a chart of $X$.

Define $\mathscr{D}_c ^p:T_p^*(X)\rightarrow E^*: A\mapsto A\circ \mathscr{O}_c^p$. Note that this map is linear homeomorphism.

Define $U^*:=(\pi^*)^{-1}(U)$, and $\phi^*(p,A)=(\phi(p),\mathscr{D}_c^p(A))$ for each $(p,A)\in U^*$.

Now define $c^*:=(U^*,\phi^*,E\times E^*)$. Then, $\phi^*:U^*\rightarrow E\times E^*$ is an injection and $\phi^*(U^*)=\phi(U)\times E^*$ is open and $E\times E^*$ is Banachable. Hence, $c^*$ is a chart on the set $T^*X$.

Take the unique differentiable structure generated by those $c^*$'s, and call the set $T^*X$ together with this differentiable structure the cotangent bundle of $X$. Would it be the correct one?