The volume form $\epsilon$ on a pseudo-Riemannian manifold is a top-degree form given in coordinates by $\epsilon = \sqrt{|g|}dx^1 \wedge \cdots \wedge dx^n$, where $|g|$ is the absolute value of the determinant of the matrix representation of the metric $g$ in coordinates $(x^1, \ldots, x^n)$.

Is there a concise form for the $n$-contravariant tensor $\epsilon^{a_1\cdots a_n}$ obtained by raising all the indices on $\epsilon$? I have seen that it should be related to $1/\sqrt{|g|}$ but cannot find a proof.


1 Answer 1


With upper indices, the expression in coordinates is $$ \epsilon^{a_1\cdots a_n} = \frac{(-1)^s}{\sqrt{\lvert g \rvert}} \mathrm{sgn}(k \mapsto a_k) $$ where $(-1)^s$ is $g/\lvert g \rvert$ (e.g. $1$ if Riemannian, $-1$ if Lorentzian) and $\mathrm{sgn}(k \mapsto a_k)$ is the sign of the permutation.

One way to think of things is that the volume form $\epsilon$ (a.k.a. Levi-Civita tensor) is normalized so that $$ \frac{1}{n!} g^{a_1 b_1} \cdots g^{a_n b_n} \epsilon_{a_1 \cdots a_n} \epsilon_{b_1 \cdots b_n} = (-1)^s. $$ We can solve this equation to find that $\epsilon_{1\cdots n} = \sqrt{\lvert g \rvert}$ and $\epsilon_{a_1 \cdots a_n} = \sqrt{\lvert g \rvert} \mathrm{sgn}(k \mapsto a_k)$; the determinant $g$ makes its appearance because we're summing over permutations. And since the left-hand side of the equation can also be though of as $$ \frac{1}{n!} \epsilon^{a_1 \cdots a_n} \epsilon_{a_1 \cdots a_n}, $$ we immediately also find $\epsilon^{a_1 \cdots a_n}$.

(Wald, General Relativity, Appendix B essentially takes the view just outlined, but stops short of explicitly stating the formula $\epsilon^{a_1 \cdots a_n}$. Carroll, Spacetime and Geometry, Section 2.8 states the formula, although with not as much proof. Note that the lecture notes which formed the basis of Carroll's book are openly available on his website.)


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