In the context of (semi-)Riemannian geometry, the following fact is well-known: if a (semi-)Riemannian manifold $(M,g)$ is oriented, then the unique volume form $\epsilon = \mathrm{vol}_g$, induced by the metric together with the orientation, is parallel with respect to the Levi-Civita connection $\nabla$. That is, $$ \nabla \epsilon = 0$$ or, using indices, $\nabla_b \epsilon _{a_1 \cdots a_n}=0$ where $n = \mathrm{dim}(M)$.

I am aware of a few different ways of proving this result and most make good sense to me. My problem is that I can't seem to completely follow the logic in one particular proof which I found in Robert M. Wald's "General Relativity" (Appendix B, page 432).

The argument there goes as follows: $\epsilon$ is uniquely specified by the choice of orientation together with the condition $$ \epsilon^{a_1 \cdots a_n} \epsilon_{a_1 \cdots a_n} = (-1)^s n! $$ where $s$ is the number of negative eigenvalues of $g$ (so $s=0$ for a Riemannian metric) and indices are raised and lowered using $g$. Taking covariant derivatives, since the RHS is constant, one has $$ 0 = \nabla_b (\epsilon^{a_1 \cdots a_n} \epsilon_{a_1 \cdots a_n}) = (\nabla_b \epsilon^{a_1 \cdots a_n}) \epsilon_{a_1 \cdots a_n} + \epsilon^{a_1 \cdots a_n} \nabla_b \epsilon_{a_1 \cdots a_n} = 2\epsilon^{a_1 \cdots a_n} \nabla_b \epsilon_{a_1 \cdots a_n}$$ using the fact that the metric is parallel with respect to $g$ in the last step. So far so good, we have obtained that $\epsilon^{a_1 \cdots a_n} \nabla_b \epsilon_{a_1 \cdots a_n} = 0$.

Question. But then it is argued that this, "in turn, implies that $\nabla_b \epsilon_{a_1 \cdots a_n}=0$ since $\epsilon_{a_1 \cdots a_n}$ is totally antisymmetric in its last $n$ indices and $\epsilon^{a_1 \cdots a_n}$ is non-vanishing". Can anyone expand on the logic of how this implication works?

I have tried to interpret this by thinking of $\epsilon^{a_1 \cdots a_n}$ and of $\nabla_j \epsilon_{a_1 \cdots a_n}$, for each fixed $j$, as analogous to two antisymmetric matrices $A$ and $B$ respectively, and the desired statement is then something like $$ \mathrm{Tr}(A^TB) = 0 \ \Longrightarrow \ B = 0,$$ but I can't seem to get very far with this reasoning. Thanks in advance for your help!

  • $\begingroup$ You said you were aware of a few different proofs for this fact. Would you be able to tell me where to find any of them? I've been looking around the internet and can't find any. $\endgroup$ – eigenchris Oct 17 '19 at 4:28

Since $\epsilon$ and $\nabla_b \epsilon$ (for fixed $b$) are both antisymmetric, the only non-zero terms in the sum $$\epsilon^{a_1 \cdots a_n} \nabla_b \epsilon_{a_1 \cdots a_n}$$ will be those where $a_1\cdots a_n$ is a permutation of $1 \cdots n$. Moreover, if we permute the indices back to this order in each term, the signs that each $\epsilon$ pick up will be the same, so they will always cancel. Thus all the terms are in fact the same; i.e.

$$\epsilon^{a_1 \cdots a_n} \nabla_b \epsilon_{a_1 \cdots a_n} = n!\ \epsilon^{1\cdots n} \nabla_b \epsilon_{1 \cdots n}.$$

The RHS is now a genuine product, and we know $\epsilon^{1 \cdots n}$ is non-zero (since it is a non-degenerate volume form); so for the product to be zero we must have $\nabla_b \epsilon_{1\cdots n} = 0.$

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    $\begingroup$ Great, this answers my question in full and removes any doubt. Many thanks! $\endgroup$ – JahvedM Apr 18 '16 at 12:06

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