Volume form in terms of the Levi-Civita Symbol I'm currently reading Sean Carroll's book on General Relativity, and at some point he writes:

First notice that the definition of the wedge product allows us to write
  \begin{equation}
\mathrm{d}x^0\wedge\cdots\wedge\mathrm{d}x^{n-1} = \frac{1}{n!}\tilde{\epsilon}_{\mu_1\dots\mu_n}\mathrm{d}x^{\mu_1}\wedge\cdots\wedge\mathrm{d}x^{\mu_n},
\end{equation}
  since both the wedge product and the Levi-Civita symbol are completely antisymmetric.

How do I notice that? I am struggling to see how the fact that both objects are completely antisymmetric implies the above equation. Does it mean that any completely antsymmetric tensor can be written as a contraction over the Levi-Civita symbol (I hope this is the right terminology; correct me if I'm wrong)? 
 A: A totally antisymmetric symbol in $n$ dimension with $n$ indices has only one independent component. Eg. if $\rho_{a_1...a_n}$ is a totally antisymmetric symbol, then $$\rho_{a_1...a_n}=\rho_{12...n}\tilde{\epsilon}_{a_1...a_n},$$ where $\tilde\epsilon$ is the Levi-Civita symbol. It is because in $n$ dimensions an antisymmetric symbol of $k$ indices has $\left(\begin{matrix}n \\ k\end{matrix}\right)$ components, and $\left(\begin{matrix}n \\ n\end{matrix}\right)=1$.
Which means that $$ \mathrm dx^{a_1}\wedge...\wedge\mathrm dx^{a_n}=\mathrm dx^{1}\wedge...\wedge\mathrm dx^{n}\tilde{\epsilon}^{a_1...a_n}. $$ Now, we know that $\tilde{\epsilon}_{a_1...a_n}\tilde{\epsilon}^{a_1...a_n}=n!$, thus $$ \mathrm dx^{a_1}\wedge...\wedge\mathrm dx^{a_n}\tilde{\epsilon}_{a_1...a_n}=\mathrm dx^{1}\wedge...\wedge\mathrm dx^{n}\tilde{\epsilon}^{a_1...a_n}\tilde{\epsilon}_{a_1...a_n}=n!\mathrm dx^{1}\wedge...\wedge\mathrm dx^{n}, $$ which, after division by $n!$ produces the required formula.
A: This question was already answered some time ago, however I am itching to clearify something since it sometimes leads to confusion. Essentially, the Levi-Civita tensor is the volume form. They are the very same object. This becomes clear as soon as you go to a coordinate free describtion. From this point, the equation $$\epsilon=\text{d}x^0\wedge…\text{d}x^{n-1} = \frac{1}{n!} \epsilon_{\mu_1...\mu_n} \text{d}x^{\mu_1}…\text{d}x^{\mu_n}$$ is nothing but the coordinate representation of $\epsilon$.
Cheers!
A: This is in fact true for any fully antisymmetric tensor. To see this Consider the tensor contraction $T^{ijk}\tilde\epsilon_{ijk}$ with $T$ a fully antisymmetric tensor. For 3 dimensions we can fully expand the sum:
\begin{align}\tilde\epsilon_{ijk}T^{ijk}=&\tilde\epsilon_{123}T^{123}+\tilde\epsilon_{231}T^{231}+\tilde\epsilon_{312}T^{312}+\\
&\tilde\epsilon_{132}T^{132}+\tilde\epsilon_{321}T^{321}+\tilde\epsilon_{213}T^{213}\end{align}
Notice that the top row consists of even permutations and the bottom row of odd permutations. Since $T$ is fully antisymmetric you can equate each term to $T_{123}$ by permuting the indices at the possible cost of a minus sign. The top row doesn't get a minus sign since its permutations are even. The bottom row does get minus signs, but you can permute the Levi-Civita the same way to introduce another minus sign. Since $\tilde\epsilon_{123}=1$ you get 6 times the same term:
$$\tilde\epsilon_{ijk}T^{ijk}=6\cdot T^{123}$$
Where 6 is ofcourse the number of permutations for $n=3$. This only works if the number of indices is equal to the dimension of the indices.
A: Multiplying an antisymmetric object by another one results in a symmetric object.
(This is true for functions---the product of an odd function like sin with another odd function, like sin, is a symmetric function, like $\sin  ^ 2$.  Same for tensors.)
Since the right hand side is completely symmetric, it is unchanged by each of the
permutations you are applying to it.  So you are summing the same thing n! times,
and then dividing by n!.
So, yes, it's obvious.  It has nothing to do with the volume form or anything.  It would be true for sin(x) or any other odd functions as well.
