# Expressing Volume Form in local coordinates on Riemannian Manifold

I'm just beginning to study Riemannian Geometry, and in particular the volume form on a Riemannian Manifold $$(M, g)$$. It was first introduced to me as a differential $$n$$-form $$dV$$ for which $$dV(e_1, \cdots, e_n) = 1$$ for any (positive) choice of orthonormal basis $$e_1, \cdots, e_n$$ on some tangent space $$T_p M$$. It's easy to see that in this local frame, with a dual basis $$\omega^1, \cdots, \omega^n$$, we have:

$$dV = \omega^1 \wedge \cdots \wedge \omega^n$$

Which is well-defined, as a change of basis to another positive orthonormal frame, say, $$\tilde{e}_1, \cdots, \tilde{e}_n$$ with dual basis $$\tilde{\omega}^1, \cdots, \tilde{\omega}^n$$ yields:

$$dV = \det(A) \tilde{\omega}^1 \wedge \cdots \wedge \tilde{\omega}^n$$

Where $$A$$ is the change of basis matrix, which must satisfy $$\det(A) = 1$$ because both bases are orthonormal and positive. However, I've seen now that in local coordinates, the volume form looks like:

$$dV = \sqrt{g_{ij}} dx^1 \wedge \cdots \wedge dx^n$$

And I'm not sure how to go from the previous representation to this one. I can see that we will have:

$$dV = \det(B) dx^1 \wedge \cdots \wedge dx^n$$

Where $$B$$ is the change of basis matrix with components $$\omega^i (\frac{\partial}{\partial x^j})$$, but I'm not sure how to relate $$B$$ in a sensible way to the metric. I guess we can also put $$g$$ into local coordinates as:

$$g = \omega^i \otimes \omega^i$$

But this doesn't seem tractable. I'm sure this will all boil down to rudimentary linear algebra in the end, but I cannot for the life of me seem to get it.

• Be careful. This should be $\sqrt g$, not $\sqrt{g_{ij}}$. This is the Gram determinant for the volume of a parallelepiped spanned by $\partial/\partial x^i$. $g$ here stands for the determinant of the matrix $[g_{ij}]$. Commented Feb 15, 2020 at 3:18
• researchgate.net/publication/… Commented Feb 15, 2020 at 3:44

We have $$dV = \sqrt{\det g}dx^1\wedge\dots\wedge dx^n$$ and for any matrix $$M$$ $$\det M = \sum_{\sigma\in S_n}\epsilon(\sigma)\prod_{i=1}^nM_{i\sigma(i)}.$$ So, for this to match your definition, we need to find $$f$$ such that $$\omega^a=f^a_{\mu}dx^\mu$$ is an orthonormal basis for $$T_p^*(M)$$, with \begin{aligned} \omega^1\wedge\dots\wedge\omega^n&=f^1_{\mu_1}\dots f^n_{\mu_n}dx^{\mu_1}\wedge\dots\wedge dx^{\mu_n}\\ &=\sum_{\sigma\in S_n}\epsilon(\sigma) \prod_{\nu=1}^n f^\nu_{\sigma(\nu)} dx^1\wedge\dots\wedge dx^n\\ &= \sqrt{\det g}dx^1\wedge\dots\wedge dx^n. \end{aligned}
Orthonormality would mean \begin{aligned} g^{-1}(\omega^{a_1}, \omega^{a_2}) &= g^{-1\mu_1\mu_2}f^{a_1}_{\mu_1} f^{a_2}_{\mu_2}=f^{a_1}_\mu f^{a_2\mu}=\delta^{a_1a_2}. \end{aligned}
At a particular point we can write orthonormality as the matrix multiplication $$fg^{-1}f=1_n,$$ which implies $$f^2=g$$.
At all points, $$g$$ is positive definite, so there exists a matrix $$h$$ such that $$h^2=g$$. We take $$f=h$$. This means that $$\sum_{\sigma\in S_n}\epsilon(\sigma)\prod_{a=1}^nf^a_{\sigma(a)}=\sqrt{\det g},$$ and that $$fg^{-1}f = 1_n.$$ So we see that we obey both conditions.
A change of basis such that $$\det A=1$$ clearly does not change the determinant equation and for orthonormality we would have $$fg^{-1}f\mapsto fA(A^{-1}g^{-1}A^{-1})Af=fg^{-1}f.$$