TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01

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Tensor transition laws define a smooth vector bundle

Statement

For every smooth manifold $M$ and integers $r, s \geq 0$ , the tensor-coordinate change rules define a smooth vector bundle whose fibre over $p$ is the space of type $(r, s)$ tensors on $T_{p} M$ .

Facts & Assumptions

Given: A smooth manifold $M$ with overlapping charts $(U, x)$ and $(V, y)$ .

[F1]

The fibre of $T_{s}^{r} M$ at $p$ is the space of type $(r, s)$ tensors on $T_{p} M$ (The type $(r, s)$ tensor bundle).

[L1]

Tangent bases transform by the Jacobian, and cotangent bases transform by the inverse transpose Jacobian (Change-of-coordinate formula for tangent bases, Cotangent coordinate changes use the inverse transpose Jacobian).

[L2]

A smooth cocycle of fibrewise linear transition maps defines a smooth vector bundle (Construction of a vector bundle from a smooth cocycle).

Proof

technique · direct

1.1

On a chart domain $U$ , the coordinate bases $\partial / \partial x^{i}$ and [F1, given, construct] $d x^{i}$ identify each fibre in [F1] with the fixed finite-dimensional vector space $Mult (((R^{m})^{*})^{r} \times (R^{m})^{s}, R)$ of type $(r, s)$ tensors on $R^{m}$ . This gives local trivializations $T_{s}^{r} M ∣_{U} ≅ U \times Mult (((R^{m})^{*})^{r} \times (R^{m})^{s}, R) .$

F1givenconstruct

2.1

On an overlap, [L1] shows that each contravariant slot picks up one inverse [L1, step 1.1, algebra] Jacobian factor and each covariant slot picks up one Jacobian factor. Hence the tensor-coordinate change map is fibrewise linear, smooth in the base point, and satisfies the cocycle law because Jacobians and inverse Jacobians do.

L1step 1.1algebra

3.1

Therefore [L2] applies to these local transition maps and produces a smooth [F1, L2, step 2.1] vector bundle. By construction its fibre over $p$ is the tensor space from [F1].

F1L2step 2.1

4.1

Thus the tensor transition laws define the smooth tensor bundle $T_{s}^{r} M$ . [step 3.1]

step 3.1∎

Depends on

Used by

Tensor components do not transform as independent scalar functions False statement
Smoothness of a tensor field is equivalent to smooth coordinate components Proposition
Symmetric and alternating images are smooth subbundles Theorem

Cited to discharge well-definedness by The type (r,s) tensor bundle.

Dependency tree · two levels

16 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)
Will J. Merry, Differential Geometry (standard reference, not scraped)