Dual vector bundle - Knowlpedia

Let $\pi:E\to M$ be a smooth vector bundle (real or complex) over a smooth manifold. The dual vector bundle of $E$ is the vector bundle

\pi^*:E^*\to M

defined fiberwise by

E_x^* := \mathrm{Hom}_{\mathbb F}(E_x,\mathbb F),

with smooth structure characterized by the property that any fiberwise-linear local trivialization $E|_U\cong U\times \mathbb F^r$ induces a local trivialization

E^*|_U \cong U\times (\mathbb F^r)^*

by sending a linear functional on a fiber to its expression in those linear coordinates.

Functoriality

A vector bundle morphism $\Phi:E\to F$ over $\mathrm{id}_M$ induces a dual morphism $\Phi^*:F^*\to E^*$ over $\mathrm{id}_M$ by precomposition on each fiber.

Equivalent characterizations

Equivalently, if $(e_1,\dots,e_r)$ is a local frame on $U$ , then there is a uniquely determined dual local frame $(e^1,\dots,e^r)$ of $E^*|_U$ such that $e^i(e_j)=\delta^i_j$ pointwise. If the original frame changes by a matrix $A$ , then the dual frame changes by $A^{-T}$ over both $\mathbb R$ and $\mathbb C$ . An inverse conjugate transpose belongs instead to Hermitian duality, which uses conjugate-linear functionals.

Examples

Cotangent bundle. The cotangent bundle $T^*M$ is the dual vector bundle of the tangent bundle $TM$ .

Dual of a trivial bundle. $(M\times \mathbb F^r)^*\cong M\times (\mathbb F^r)^*$ canonically.

Dual line bundle. If $L\to M$ is a real or complex line bundle, then $L^*$ is again a line bundle; fiberwise, it consists of linear functionals on $L_x$ . The tensor product $L\otimes L^*$ has a canonical nowhere-zero section given by evaluation, so it is canonically trivial.