# 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?

## 1 Answer

Yes. That unique differentiable structure is given by Lang in Proposition 2, p. 43 which applies by Theorem 1, p. 54. (A later edition has these and the cotangent bundle example on pp. 45, 59 and 61.)