Riemann volume forms: Absolute value of the determinant of the metric, or not

differential-geometryriemannian-geometry

When using differential geometry in the context of General Relativity, you are usually taught that the general invariant scalar volume element reads

$$ \mathrm{d}V=\sqrt{|g|}\mathrm{d}x^n $$

for an n-dimensional Riemannian manifold. With $g$ being the determinant of the metric tensor. Cf. for example [1,2] and also the Wikipedia article on volume forms [3].

Now, my question is: Is it just a convention that we use the $\textit{absolute value}$ of the determinant when defining the volume form?

In the Wikipedia article [3] it is mentioned that volume forms are in general not unique. Does this imply that we could equally say that $\mathrm{d}V=\sqrt{g}\mathrm{d}x^n$ is the volume form of choice?

PS: Note that this would imply for the Minkowski metric ($g=\mathrm{diag}(-1,1,1,1)$), that in one incident that volume form would be real and in the other imaginary. See for example this article [4] and the consequences for the complex structure of Quantum Mechanics.

[1] http://www.blau.itp.unibe.ch/newlecturesGR.pdf Eq. 4.47 and 4.51

[2] https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll2.html Eq. 2.48

[3] https://en.wikipedia.org/wiki/Volume_form#Riemannian_volume_form

[4] Lindgren, Liukkon 2019: https://www.nature.com/articles/s41598-019-56357-3

Best Answer

The absolute value is "just convention" in some sense, but it is better motivated than your choice of definition might suggest.

In $\mathbb{R}^n$, one typically defines a notion of "volume" such that the volume of the box is given by the product of its side lengths. $$ \text{volume}\left([a_1,b_1]\times\cdots\times[a_n,b_n]\right)=\prod_{i=1}^n|b_i-a_i| $$ This (eventually) leads to the Lebesgue measure on $\mathbb{R}^n$. One could just as easily define the volume of such a box to be some other (negative, or even complex) number, but this would be a rather unusual choice, generally speaking. If one were to make such a choice, it would probably be best to not call the result "volume" to avoid confusion.

On an pseudo-Riemannian $n$-manifold $(M,g)$, there is an analogue of the Labesgue measure which has essentially all the same properties. In the oriented case*, it is most easily described by a differential $n$-form $\omega_g$ known as the Riemann volume form. The analogue of the "volume of a box" condition is an "infinitesimal cube" condition $$\tag{1} \omega_g(e_1,\cdots,e_n)=1 \\ e_1,\cdots,e_n\in T_pM\ \ \ \text{any oriented orthonormal basis} $$ This condition uniquely determines $\omega_g$, and implies the coordinate definition: in oriented coordinates $x^1,\cdots,x^n$, we have $\omega_g=\sqrt{|g|}dx^1\wedge\cdots\wedge dx^n$.

One could just as easily use some other (complex) number in place of $1$ in $(1)$, but the result wouldn't be the Riemann volume form (as it is commonly understood), but some scalar multiple of it.

*There is actually no need for an orientation, provided one uses an $n$-density in place of an $n$-form.

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