False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-01

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Tensor components do not transform as independent scalar functions

Statement

False claim: tensor coordinate components transform as if each were an independent scalar function.

Facts & Assumptions

Given: The Euclidean metric $g = d x \otimes d x + d y \otimes d y$ on $R^{2}$ and the polar chart $(r, θ) \mapsto (r \cos θ, r \sin θ)$ on ${r > 0}$ .

[L1]

Tensor-coordinate changes involve Jacobian factors on each slot (Tensor transition laws define a smooth vector bundle).

Refutation

technique · direct

1.1given

In Cartesian coordinates, the coefficient matrix of $g$ is $(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ .

2.1L1step 1.1algebra

In polar coordinates, $d x = \cos θ d r - r \sin θ d θ$ and $d y = \sin θ d r + r \cos θ d θ$ , so $g = d r \otimes d r + r^{2} d θ \otimes d θ .$ Thus the coefficient matrix becomes $(\begin{array}{cc} 1 & 0 \\ 0 & r^{2} \end{array})$ , not the unchanged scalar pair from step 1.1. This is exactly the Jacobian action described in [L1].

3.1step 2.1∎

Therefore tensor components do not transform as independent scalar functions.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

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