Any oriented pseudo-Riemannian manifold $(X,g)$ has a natural volume form $\Omega_g$.
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 $e_1,\ldots,e_n$ be a positive orthonormal basis of $T_p X$. Then
$$ (\Omega_g)_p=e^1\wedge\cdots \wedge e^n $$where $\{e^i\}\subset T_p^*M$ is the dual basis, i.e., $e^i(e_j)=\delta^i_j$. It is easy to check that is well defined
In local coordinates $(x_1,\ldots,x_n)$ it can be expressed as
$$ \Omega_g=\sqrt{\left|g\right|} dx_1\wedge \cdots\wedge dx_n $$where $\left|g\right|$ is the absolute value of the determinant of the matrix $\{g_{ij}\}$ of the (pseudo)-Riemannian metric in these local coordinates: $g_{ij}=g(\partial_{x_i},\partial_{x_j})$. 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 $\{dx_i\}$ to $\{e^i\}$. Therefore
$$ (\Omega_g)_p=e^1\wedge\cdots \wedge e^n=det(A)dx_1\wedge \cdots\wedge dx_n $$But, finally, observe that $det(A)=\sqrt{det(A^tA)}=\sqrt{det(\{g_{ij}\})}$ .
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
$$ \Omega_{g_S}=N\,\lrcorner\,\Omega_g $$where $N$ is a unit normal vector field to $S$, and we are considering the orientation in $S$ induced by $N$.
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Author of the notes: Antonio J. Pan-Collantes
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