I'm following Carroll's book on General Relativity and in chapter 2, section 2.10, he claims that the volume element can be identified with
$$d^nx=dx^0\wedge\ldots \wedge dx^{n-1}.$$
I understand why this is a tensor density but I have troubles when I try to analyze the validity of this expression. Previously in section 2.9 he defined some differential forms, particularly, the exterior derivative $d$ and the wedge product $\wedge$. The wedge product is defined for forms, so I interpret that each $dx^0$, $dx^1$, $\ldots$, $dx^{n-1}$ is a form. My problem is that, by following the book, they should be exterior derivatives of $x^0, x^1, \ldots, x^{n-1}$, but how that would be possible if he defined the exterior derivative as an operator on forms? How $x^0, x^1, \ldots, x^{n-1}$ are forms if they are coordinates and $x^\mu$ transforms as a vector? Maybe the answer is that they are zero-forms but it is not even clear for me that they are scalars.
I don't know if my question is trivial in a general context, I'm only reading Carroll's book and I find it relevant according to what he had presented so far in the book.
Best Answer
For any given coordinate map $x^\mu$, you get a basis of differential 1-forms $\mathrm{d}x^\mu$ defined as the dual of the corresponding vector basis $\partial_\mu$, i.e. by $\mathrm{d}x^\mu(\partial_\nu) = \delta_\nu^\mu$. Note that this definition does not treat the coordinates as functions on the manifold or would require the coordinate map to have an exterior derivative - the notation $\mathrm{d}x^\mu$ is meant to imply that this is something similar to that, but it is not actually formalized as that. You can view a coordinate $x^\mu$ (or its inverse) just as a scalar function $M\supset U\to\mathbb{R}$ for some open $U\subset M$ on the manifold, but since coordinates do not extend to the full manifold in general, this doesn't give you a 0-form on the whole manifold (but "locally" the exterior derivative of this function would indeed be just $\mathrm{d}x^\mu$).
Therefore, for a fixed coordinate system $x^\mu$, $\omega[x] = \mathrm{d}x^0\wedge\dots\wedge \mathrm{d}x^{n-1} $ is a perfectly fine definition of a differential $n$-form. Note that, in general, this definition is dependent on the choice of $x$. The general definition of the volume element has to include the determinant of the metric as $$ \omega = \sqrt{\mathrm{det}(g)}\mathrm{d}x^0\wedge\dots\wedge\mathrm{d}x^{n-1},$$ since the way differential forms transform then means this definition looks the same in every coordinate system and this is therefore a coordinate-independent definition.