Why must metric tensor be invertible? The metric can be written as a matrix, but why must this matrix be invertible? At the points where the matrix is singular, why is the metric not defined?
 A: This worried me at one time as well. The way I thought about it was by working at a fixed point and using the Gram-Schmidt process for inner products on the coordinate basis $\partial_1,...,\partial_n$ to produce an orthonormal basis $e_1,...,e_n$. It's a standard and easy fact that the matrices that represent these bilinear forms are related by $I=A^tgA$, where $I$ is the identity matrix (since $\{e_i\}$ is orthonormal), $g$ is the metric $g_{ij}$ with respect to the $\{\partial_i\}$ basis, and $A$ is the change of basis matrix taking the basis of partials to the orthonormal basis. Taking the determinant of both sides of this equation, and using the fact that $\det(A)=\det(A^t)\neq 0$, we see that $\det(g)\neq 0$ so $g$ is invertible.
A: Given a pseudo-Riemannian manifold $(\mathcal M,g)$ and some local tangent space $\mathrm T_p\mathcal M$ with coordinates $x^1,\dots, x^n$ , the volume form is given by
$$\mathrm dV=\sqrt{|\det g|}~\bigotimes_{i=1}^n \mathrm dx^i$$
So
$$g~\text{not invertible}\implies \det g=0\implies \mathrm dV=0$$
Which would imply our manifold is trivial, i.e, a single point.
