PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Coordinate criterion for a riemannian metric

Statement

A covariant two-tensor $g$ on $M$ is Riemannian exactly when, in every smooth coordinate chart $x$ , its coordinate matrix $G_{x} = (g_{i j})$ has smooth entries and is symmetric positive definite. On overlapping charts, under $J = \partial x / \partial y$ , the matrices transform by $G_{y} = J^{T} G_{x} J$ .

Facts & Assumptions

Given: A covariant two-tensor and overlapping smooth coordinate systems.

[F1]

Riemannian metric and riemannian manifold: A Riemannian metric on a Hausdorff second-countable smooth manifold $M$ is a smooth symmetric covariant two-tensor $g$ such that $g_{p} (v, v) > 0$ for every point $p$ and every nonzero $v \in T_{p} M$ . A Riemannian manifold is the pair $(M, g)$ . This is a def-smooth-tensor-field giving a def-smooth-bundle-metric on $T M$ . Dimension zero is allowed: its zero bilinear form is positive definite because there are no nonzero vectors. The empty manifold has its unique empty metric. Boundaries are allowed where stated, with smoothness understood up to the boundary.

[F2]

Smoothness of a tensor field is equivalent to smooth coordinate components: A type $(r, s)$ tensor field is smooth if and only if, in every smooth chart, its coordinate component functions are smooth.

Proof

technique · direct

1.1

If $g$ is Riemannian, in every smooth chart $g_{i j} = g (\partial_{i}, \partial_{j})$ is smooth and symmetric. For a nonzero coordinate vector $v$ , $v^{T} G_{x} v = g (v, v) > 0$ . Conversely, smooth entries in every smooth chart make the tensor smooth, and the displayed quadratic equality makes symmetry and positive definiteness of every $G_{x}$ precisely those of $g$ .

F1F2given

2.1

Since $\partial / \partial y^{a} = \sum_{i} J_{i a} \partial / \partial x^{i}$ , bilinearity gives $(G_{y})_{a b} = \sum_{i, j} J_{i a} (G_{x})_{i j} J_{j b}$ . The Jacobian is invertible; hence $J v \neq 0$ for $v \neq 0$ , and $v^{T} G_{y} v = (J v)^{T} G_{x} (J v) > 0$ . The condition is coordinate independent. Empty charts and zero-dimensional matrices give vacuous positivity.

F1step 1.1∎

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Chapter 13, pp.328–332 and 341–342.

Depends on

Used by

Riemannian volume density Definition
The euclidean metric and its musical maps Example
The flat torus metric from periodic euclidean coordinates Example
The hyperbolic upper half space metric Example
The product riemannian metric Example
The round metric on the sphere as an induced metric Example
Local comparison of a riemannian metric with the euclidean metric Lemma
The riemannian volume density is coordinate independent Lemma
The musical maps are smooth inverse bundle isomorphisms Theorem

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, second edition (standard reference, not scraped)
Ved Datar, Lectures on Riemannian Geometry, September 2025 (standard reference, not scraped)