Absolute value in pseudo Riemannian volume form

Question

This may once again be a naive question, but I am confused as to where the absolute value sign in the volume form for an oriented pseudo Riemannian manifold $(M,g)$. For one, in any coordinate chart with coordinates $(x^i)$ on an oriented Riemannian manifold, we have that the unique Riemannian volume form is given by:

$$\omega_g=\sqrt{\det g_{ij}}dx^1\wedge \cdots\wedge dx^n$$

where $g_{ij}$ are the components of the metric in these coordinates. I understand where this comes from, but I do not understand where the absolute value comes from in the pseudo Riemannian case, i.e in pseudo Riemannian geometry the above form is replaced with:

$$\omega_g=\sqrt{|\det g_{ij}|}dx^1\wedge \cdots\wedge dx^n$$

Is there away for one to arrive at this formula by appealing to the signature of the metric $(s,t)$? Or is it really just convention?

Suppose the signature of the metric leaves the $g$ with an odd number of negative values in the diagonal, then since the signature of the metric is invariant at each point, we can deduce the determinant of $g$ is negative everywhere. Then if $A$ is the matrix of smooth functions which takes an oriented coordinate frame to an orthonormal frame we can pretty quickly come to the realization that: $$\det(A)^2=\det(g)$$ while the volume form in these coordinates is given by: $$\omega_g=\det(A)dx^1\wedge \cdots \wedge dx^n$$ To we just take the absolute value of this to make it not complex or?

Answer 1

It's just not true that $\det(A)^2=\det(g)$ in general. Think of flat space $\mathbb{R}^{s,t}$ with signature $(s,t)$, with Cartesian coordinates and $A$ being the identity matrix.

I'm not sure how you came to the conclusion $\det(A)^2=\det(g)$; I suppose some positive-definiteness assumption is being smuggled in there. Perhaps you are arriving at it from something like $A^T A = g$. The problem is, an equation like that is supposed to have an $\eta$ in it, like $A^T \eta A = g$, where $\eta$ is a diagonal matrix with $s$ copies of +1 and $t$ copies of -1 on the diagonal. (Or, depending your sign convention, the opposite signs.) Then $\det(\eta)$ together with the $\det(g)$ results in $|\det(g)|$.