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.

  • 1
    $\begingroup$ 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 $[g_{ij}]$. $\endgroup$ Feb 15, 2020 at 3:18
  • $\begingroup$ researchgate.net/publication/… $\endgroup$ Feb 15, 2020 at 3:44

1 Answer 1


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{equation}\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}\end{equation}

Orthonormality would mean \begin{equation}\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}\end{equation}

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 $$ fg^{-1}f\mapsto fA(A^{-1}g^{-1}A^{-1})Af=fg^{-1}f. $$


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