I'm trying to understand how some characterizations of the Riemannian volume form $dV$ are equivalent on an oriented Riemannian manifold of dimension $n$. I'm a bit new to Riemannian geometry (and self-studying it), and I can tell that it should just come down to a straightforward calculation, but I'm stuck for some reason.

Characterization 1: If $(\omega^1,...,\omega^n)$ is a local oriented orthonormal coframe for the cotangent bundle, then $dV=\omega^1\wedge...\wedge\omega^n$.

Characterization 2: If $(y^1,...,y^n)$ are oriented local coordinates, then $dV=\sqrt{\det(g_{ij})}dy^1\wedge...\wedge dy^n$ where $g_{ij}$ is the representation of the Riemannian metric in local coordinates.

I'm guessing that the square root of the determinant factor shows up from something involving $\det(A)=\sqrt{\det(A A^t)}$ for a suitable matrix $A$. I have a hunch that using uniqueness of the form (which follows for e.g. characterization 1 from the fact that you can cover the manifold with charts that have local oriented orthonormal coframes) might be important? Any help is very much appreciated!

  • $\begingroup$ Hint: If your $\partial_{y_i}$ are orthogonal then the metric is represented by a diagonal matrix with the lengths of the $\partial_{y_i}$ on the diagonal. What is the determinant, then? $\endgroup$ Feb 2, 2019 at 16:18
  • $\begingroup$ You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right? $\endgroup$
    – Ben
    Feb 2, 2019 at 16:22
  • $\begingroup$ Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame! $\endgroup$ Feb 2, 2019 at 16:38
  • 2
    $\begingroup$ @AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame. $\endgroup$ Feb 2, 2019 at 16:43
  • $\begingroup$ @AmitaiYuval Oh, sorry! $\endgroup$ Feb 2, 2019 at 16:56

1 Answer 1


Let $dV$ denote the Riemannian volume form given by your first characterization, and let $y^1,\ldots,y^n$ be oriented local coordinates. Let $\frac{\partial}{\partial y^1},\ldots,\frac{\partial}{\partial y^n}$ denote the induced local frame of the tangent bundle, and let $e_1,\ldots,e_n$ be a local orthonormal frame of the tangent bundle. By definition, the matrix representation of the metric $g$ with respect to $e_1,\ldots,e_n$ at any point is the identity matrix. The matrix representation of $g$ with respect to $\frac{\partial}{\partial y^1},\ldots,\frac{\partial}{\partial y^n}$ is $(g_{ij})$.

Now, for a point $p$, let $A_p$ denote the transition matrix of the two above mentioned frames of the tangent space at $p$. Then it follows from characterization $1$ that $$dV\left(\frac{\partial}{\partial y^1},\ldots,\frac{\partial}{\partial y^n}\right)_p=\det A_p\cdot dV(e_1,\ldots,e_n)=\det A_p.$$On the other hand, by construction we have $$(g_{ij})_p=A_p^T\cdot I\cdot A_p=A_p^TA_p,$$and hence, $$\det A_p=\sqrt{\det(g_{ij})},$$ and it follows that $dV$ coincides with the volume form of your second characterization.

  • $\begingroup$ Thanks! This was exactly what I was looking for. $\endgroup$
    – Ben
    Feb 2, 2019 at 21:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .