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.


1 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.

  • $\begingroup$ "This definition does not treat the coordinates as functions on the manifold". I don't think this is quite correct. A chart $x: U \rightarrow \mathbb{R}^n$ where $U$ is an open in a manifold $M$. Now $Tx: TU \rightarrow T\mathbb{R}^n$. Then applying the natural projection $\pi$ of the codomain to its tangent space, we get $dx:= \pi.Tx: TU \rightarrow \mathbb{R}^n$. This is a chart on $TM$ and in fact, a bundle chart. Moreover, we get $dx^i: TM \rightarrow \mathbb{R}^n$. And it can be shown that $dx^i.\partial x_j = \delta^i_j$. $\endgroup$ Commented Dec 20, 2021 at 1:53
  • $\begingroup$ Moreover, "... or would require the coordinate map to have an exterior derivative." I don't think that this is quite right either. $dx^i: TM \rightarrow \mathbb{R}$ is a differential form as you have already observed. But $dx: TM \rightarrow \mathbb{R}^n$ can be equivalently be written as $dx: \mathbb{R} \rightarrow T^*M \otimes \mathbb{R}^n$ and this shows that dx is a $\mathbb{R}^n$-valued differential 1-form. Here, $d$ is the exterior covariant differential which, as we're valuing in a flat bundle, will satisfy $d^2=0$. $\endgroup$ Commented Dec 20, 2021 at 2:07
  • $\begingroup$ @MoziburUllah I don't think we're really disagreeing - that's more or less what I had in mind with "locally the exterior derivative of this function would indeed just be $\mathrm{d}x^\mu$". All I'm saying is that the definition of $\mathrm{d}x^\mu$ as the dual of $\partial_\mu$ suffices and does not require us to do any of that. $\endgroup$
    – ACuriousMind
    Commented Dec 20, 2021 at 12:31

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