# Absolute value in pseudo Riemannian volume form

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?

• You need the absolute value so that the square root even makes sense…afterall we’re talking about real analysis, not complex analysis. Take a look at the following question and answer of mine: Existence and Uniqueness of a Volume Element on an Oriented, Pseudo Inner Product Space?. There I talk about a single vector space, but of course on an oriented pseudo-Riemannian manifold, you just do things at each tangent space. Sep 13, 2022 at 8:41

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 $$|\det(g)|$$.