# Dual Bundles' Local Trivializations Confusion

$$\newcommand{\R}{\mathbb R}$$ Let $\pi:E\to M$ be a rank $k$ smooth vector bundle over a smooth manifold $M$. I will below describe how to form the dual bundle, wherein lies my question.

Let $E_p=\pi^{-1}(p)$ for each $p\in M$, and define $E^*=\bigsqcup_{p\in M}E_p^*$. Further define $\pi^*:E^*\to M$ as $\pi(E^*_p)=\{p\}$ for all $p\in M$.

The main part is to define the local trivializations. For each local trivialization $\Phi:\pi^{-1}(U)\to U\times \R^k$ of $E$ over $U$, define $\Phi^*:\pi^{*-1}(U)\to U\times \R^{k*}$ as $$\Phi^*(\omega)=(p, (\Phi|_{E_p})^{-t}\omega)$$ for all $\omega\in E_p^*$, and all $p\in U$. (The '$t$' denotes the transpose).

The problem with the above definition is that a local trivialization is supposed to map $\pi^{*-1}(U)$ to $U\times \R^k$ and not $U\times \R^{k*}$. Now I know that $\R^k$ can be canonical identified with $\R^{k*}$ provided we have an inner product on $\R^k$.

But then different inner products will lead to different $\Phi^*$ and consequently different transition functions.

Is there a preferred inner product with respect to which the identification of $\R^k$ with $\R^{k*}$ is made?

The identification of $$V$$ with $$V^*$$ via a scalar product is done for vector spaces in which there is no preferred basis. This is not the case of $$\mathbb R ^n$$ which has the usual basis (in terms of which it is constructed, after all) that allows you to define the identification $$e_i \mapsto e^i$$ where $$e^i (v) = e^i (\sum v^j e_j) =v^i$$. So, the idea is that $$\mathbb R^n$$ identifies with $$(\mathbb R^n)^*$$ purely algebraically, with no need for a scalar product.

The fibers $p^{-1}(\{x\})$ of a vector bundle $p : E \to X$ are just finite-dimensional vector spaces (say, over $\mathbb{R}$ here). No bases are chosen. So it is "wrong" to write $\mathbb{R}^n$ here. $\mathbb{R}^n$ is not "the" $n$-dimensional vector space, but it is the "the" $n$-dimensional vector space with a chosen basis.

The dual of the trivial vector bundle $X \times V \to X$ (where $V$ is a finite-dimensional vector space) is defined by the trivial vector bundle $X \times V^* \to X$. Any isomorphism $X \times V \to X \times W$ of vector bundles over $X$ is induced by a unique isomorphism $V \to W$ of vector spaces. This may be dualized to an isomorphism $W^* \to V^*$ of vector spaces. The inverse of this is an isomorphism of vector spaces $V^* \to W^*$, which thus corresponds to an isomorphism $X \times V^* \to X \times W^*$ of vector bundles over $X$.

Thus, when $E \to X$ is any vector bundle, glued from trivial vector bundles, along isomorphisms on the intersections, we may also glue the corresponding dual trivial vector bundles together and obtain the dual vector bundle $E^* \to X$.

Notice that $E^* = \coprod_{x \in X} (E_x)^*$ is wrong (or at least, utterly confusing); we only have $|E^*| = \coprod_{x \in X} |(E_x)^*|$, where $|-|$ means the underlying set here.

• I am not sure if you approve of the dual bundle construction I have posted or not. When I wrote $E^*=\bigsqcup_{p\in M} E^*_p$, I just meant a set theoretic union. May 17, 2015 at 15:14
• I don't completely approve it. a) I would not write $\mathbb{R}^n$. b) I would not confuse a space with its underlying set. c) In general, I don't think that fibers are the best way of thinking about vector bundles. May 18, 2015 at 7:45
• Thank you for your answer. I am reading Lee's Introduction to Smooth Manifolds. And the way I have done is pretty much how lee asks to do it in one of the exercises. I don't know how vector bundles are treated in other texts. May 18, 2015 at 8:12