Riemannian volume form

Any oriented pseudo-Riemannian manifold $(X, g)$ has a natural volume form $Ω_{g}$ .
Is the differential $n$ -form whose integral over pieces of $(X, g)$ computes the volume as measured by the metric $g$ . Explicitly, is defined as follows. For $p \in X$ let $e_{1}, \dots, e_{n}$ be a positive orthonormal basis of $T_{p} X$ . Then

(Ω_{g})_{p} = e^{1} \land \dots \land e^{n}

where ${e^{i}} \subset T_{p}^{*} M$ is the dual basis, i.e., $e^{i} (e_{j}) = δ_{j}^{i}$ . It is easy to check that is well defined

In local coordinates $(x_{1}, \dots, x_{n})$ it can be expressed as

Ω_{g} = \sqrt{| g |} d x_{1} \land \dots \land d x_{n}

where $| g |$ is the absolute value of the determinant of the matrix ${g_{i j}}$ of the (pseudo)-Riemannian metric in these local coordinates: $g_{i j} = g (\partial_{x_{i}}, \partial_{x_{j}})$ . This follows from the following idea. Given an orthonormal basis ${e_{i}}$ consider the matrix $A$ whose columns are the coordinates of $\partial_{x_{i}}$ in this basis. Then it is the matrix to go from the dual basis ${d x_{i}}$ to ${e^{i}}$ . Therefore

(Ω_{g})_{p} = e^{1} \land \dots \land e^{n} = d e t (A) d x_{1} \land \dots \land d x_{n}

But, finally, observe that $d e t (A) = \sqrt{d e t (A^{t} A)} = \sqrt{d e t ({g_{i j}})}$ .

Induced volume in submanifolds

Suppose $S \subset X$ is a submanifold of codimension 1. We have an induced Riemannian metric $g_{S}$ on $S$ . Therefore we have a volume form on $S$ given by

Ω_{g_{S}} = N ⌟ Ω_{g}

where $N$ is a unit normal vector field to $S$ , and we are considering the orientation in $S$ induced by $N$ .