Does determinant of a metric depend only on coordinate system used?

Question

Determinant of Schwarzschild metric is $-r^4\sin^2\theta$ which is also the determinant of flat spacetime represented in the same coordinates. Is this just coincidence or is it always so that the determinant of a metric depends only on the coordinate system used and not the manifold itself?

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EDIT: Determinant of interior Schwarzschild metric is not $-r^4\sin^2\theta$ even though that metric is represented in same coordinates as exterior Schwarzschild metric. So is it only in vacuum when determinant does not depend on geometry?

Answer 1

is it always so that the determinant of a metric depends only on the coordinate system used and not the manifold itself?

Yes, this is always true. The equivalence principle says that locally, spacetime is always equivalent to flat spacetime. That means that locally, we can always choose Minkowski coordinates, and the metric will have the Minkowski form. Therefore we can always make the determinant of the metric be $-1$. By rescaling the coordinates, we can also make the determinant of the metric have any other desired (negative) value.

Answer 2

A somewhat higher-brow answer is that volume forms on a manifold have no local invariants.

Let the "model space" for a "volumed" geometry be $(\mathbb R^n,\mu)$, where $\mu=\mathrm dx^1\wedge...\wedge\mathrm dx^n$.

Let $M$ be an $n$ dimensional, smooth, orientable manifold with positively oriented volume form $\omega$. Then any $x\in M$ has some open neighborhood $U\subseteq M$ such that there is a smooth local diffeomorphism $\psi:U\rightarrow\mathbb R^n$ for which $\omega=\psi^\ast\mu$.

This can be restated in "lowbrow" coordinate language that one may always find a coordinate system where $\omega=\mathrm dx^1\wedge...\wedge\mathrm dx^n$.

So essentially all volume forms "look like one another" locally, and there is no analogue of the Riemann curvature tensor for volume forms.