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The riemannian volume density is coordinate independent

Statement

The local Riemannian volume densities glue to a positive smooth density independent of coordinates.

Facts & Assumptions

Given: Overlapping charts with $J = \partial x / \partial y$ .

[F1]

Riemannian volume density: The Riemannian volume density is $μ_{g} = \sqrt{\det G_{x}} ∣ d x^{1} \dots d x^{n} ∣$ in coordinates. The matrix is that of prop-coordinate-criterion-for-a-riemannian-metric, so its determinant is positive and smooth. The density frames and their absolute-Jacobian law are def-density-bundle-and-smooth-density. In dimension zero take the empty determinant to be one, giving weight one at every point, independently of orientation. The compatibility of these local formulas is proved in lem-the-riemannian-volume-density-is-coordinate-independent.

[F2]

Coordinate criterion for a riemannian metric: A tensor $g = \sum_{i, j} g_{i j} d x^{i} \otimes d x^{j}$ is Riemannian exactly when its coordinate matrix $G = (g_{i j})$ has smooth entries and is symmetric positive definite. Under $J = \partial x / \partial y$ it transforms by $G_{y} = J^{T} G_{x} J$ .

Proof

technique · direct

1.1F2given

The metric matrix law gives $G_{y} = J^{T} G_{x} J$ , hence $\det G_{y} = (\det J)^{2} \det G_{x}$ . Taking positive square roots yields $\sqrt{\det G_{y}} = ∣ \det J ∣ \sqrt{\det G_{x}}$ .

2.1F1step 1.1∎

This is exactly the density coefficient change, since $∣ d x ∣ = ∣ \det J ∣ ∣ d y ∣$ . Thus the two local sections agree. Their positive smooth coefficients give a global positive smooth density. In dimension zero both determinants are one, and on empty $M$ gluing gives the unique section.

Source locator

Lee, Propositions 15.29–15.33 and Corollary 15.34, pp.389–391; density construction and integration pp.428–433.

Depends on

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, second edition