Volume element of manifolds via wedge products

Question

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.

Answer 1

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.