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The canonical one-form on a cotangent bundle as a covariant tensor

Example

On $T^{*} R^{2}$ with coordinates $(x, y, ξ_{1}, ξ_{2})$ , the canonical $1$ -form is

$λ = ξ_{1} d x + ξ_{2} d y .$

At a point $(p, ξ)$ and a tangent vector $v$ to $T^{*} R^{2}$ , it satisfies

$λ_{(p, ξ)} (v) = ξ (d π (v)),$

where $π : T^{*} R^{2} \to R^{2}$ is the bundle projection.

Facts & Assumptions

Given: The cotangent bundle projection $π : T^{*} R^{2} \to R^{2}$ and a point $(x, y, ξ_{1}, ξ_{2})$ .

[F1]

A differential $1$ -form is a smooth section of the cotangent bundle (A smooth differential $k$ -form).

[F2]

The cotangent bundle fibre at $(x, y)$ consists of covectors on $T_{(x, y)} R^{2}$ (Cotangent space and cotangent bundle as a disjoint union).

Verification

technique · direct

1.1F2givenalgebra

Write $v = a \partial_{x} + b \partial_{y} + c \partial_{ξ_{1}} + d \partial_{ξ_{2}}$ at $(x, y, ξ_{1}, ξ_{2})$ . Then $d π (v) = a \partial_{x} + b \partial_{y}$ , so the covector $ξ = ξ_{1} d x + ξ_{2} d y$ gives $ξ (d π (v)) = ξ_{1} a + ξ_{2} b$ .

2.1F1step 1.1algebra

The form $λ = ξ_{1} d x + ξ_{2} d y$ takes the same value on $v$ , namely $λ (v) = ξ_{1} a + ξ_{2} b$ . Thus $λ_{(p, ξ)} (v) = ξ (d π (v))$ . Its coefficients are smooth coordinate functions, so [F1] makes it a smooth $1$ -form.

3.1step 2.1∎

Therefore the canonical one-form is a concrete covariant tensor on the cotangent bundle.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)