# Characterizations of Riemannian Volume Form

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!

• 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? Feb 2, 2019 at 16:18
• You can only guarantee the existence of an orthonormal coordinate frame when the metric is flat though, right?
– Ben
Feb 2, 2019 at 16:22
• Locally (i.e. in a small enough open neighborhood of a point) you can always find such frame! Feb 2, 2019 at 16:38
• @AngeloBrillanteRomeo OP is referring to an orthonormal coordinate frame. Feb 2, 2019 at 16:43
• @AmitaiYuval Oh, sorry! Feb 2, 2019 at 16:56

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.