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!
Best Answer
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 $\nabla_b \epsilon_{1\cdots n} = 0.$