# To Reconcile Two Different Descriptions of the Dual Bundle

$\newcommand{\mc}{\mathcal}$ Let $\pi:E\to M$ be a smooth vector bundle with typical fibre a $k$-dimensional vector space $\mc V$. There are (at least) two ways to construct the dual bundle of $E$.

Direct Approach: Define $E^*$ as the disjoint uinon $\bigcup_{p\in M}E^*_p$. Define $\pi^*:E^*\to M$ as $\pi(E^*_p)=\{p\}$. For each smooth local trivialization $\Phi:\pi^{-1}(U)\to U\times \mc V$ of $E$ over $U\subseteq M$, define $\Phi^*:{\pi^*}^{-1}(U)\to U\times \mc V^*$ as $\Phi^*(v)=(p, \Phi_p^{-t}v)$ for all $v\in E^*_p\cap {\pi^*}^{-1}(U)$. Then using the vector bundle construction theorem, one can establish that there is a unique topology and smooth structure such that $\pi^*:E^*\to M$ is a smooth vector bundle over $M$ with typical fibre $\mc V^*$.

Associated Bundle Approach: Consider the frame bundle $F(E)$ which is known to be a principal $GL(\mc V)$-bundle. Define a representation of $GL(\mc V)$ on $\mc V^*$ by defining a map $\rho:GL(\mc V)\to GL(\mc V^*)$ as $\rho(T)=T^{-t}$ (Here $T^{-t}$ means the inverse of the transpose of $T$). Then we can define the dual bundle of $\pi:E\to M$ as the associated bundle of $F(E)$ with respect to the representation $\rho$.

These two constructions look quite different. One way to show that they are same is by coming up with a bundle isomorphism between them.

Can somebody please try to give some intuition as to why these two constructions are same?

Thank you.

Let's first describe the associated bundle $F(E) \times_{\rho} \mathcal{V}^*$ to the frame bundle $\pi_{F} \colon F(E) \to M$ with respect to the representation $\rho$. It is constructed by taking the product $F(E) \times \mathcal{V}^*$ and modding out by the diagonal action of $GL(\mathcal{V})$, so the fibre above the point $p \in M$ consists of equivalence classes of pairs $(B, \nu) \in F(E)_p \times \mathcal{V}^*$, where $(B_1,\nu_1) \sim (B_2,\nu_2)$ if there is $T \in GL(V)$ such that $T \cdot B_1 = B_2$ and $T^{-t} \nu_1 = \nu_2$. (Here, $B_1$ and $B_2$ are ordered bases of $E_p$.)
How does this relate to the fibre $E_p^*$? An element of $\left( F(E) \times_{\rho} \mathcal{V}^* \right)_p$ is precisely the data one needs to define a linear functional $E_p \to \mathbb{R}$! Indeed, given a pair $(B,\nu) \in F(E)_p \times \mathcal{V}^*$, identify $E_p$ with $\mathcal{V}$ with the ordered basis $B$, and then compose with $\nu$ to get a linear map $E_p \simeq \mathcal{V} \stackrel{\nu}{\to} \mathbb{R}$. Moreover, if $(B_1,\nu_1)$ and $(B_2,\nu_2)$ are equivalent in the fibre $\left( F(E) \times_{\rho} \mathcal{V}^* \right)_p$, then they determine the same linear functional $E_p \to \mathbb{R}$. Therefore, the two characterizations of the dual bundle coincide on fibres.
Furthemore, if $(U,\Phi)$ is a local trivialization of $\pi \colon E \to M$, then it also trivializes both the dual bundle $\pi^* \colon E^* \to M$ (in the sense of the direct approach in the question) and the frame bundle $\pi_F \colon F(E) \to M$. It follows that the transition functions of $F(E) \times_{\rho} \mathcal{V}^*$ coincide with those of $E^*$, since the transition functions of the associated bundle to $\pi_F \colon F(E) \to M$ are the transition functions of $\pi_F$ post-composed with the representation $\rho$ (i.e. we are taking the `inverse-transpose' of the map $\Phi$).