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Currently I'm reading through Sean Carroll's Spacetime and Geometry: an Introduction to General Relativity. According to Carroll, the symbol

$$dx^0 \wedge dx^1 \wedge \cdots \wedge dx^{n-1},$$

though it looks like an $n$-form, is not a tensor but rather a tensor density. However, I am confused as to how this can be the case, since as I understand it, this symbol is by definition an anti-symmetric tensor. In an effort to understand this, I worked out the case of two dimensions:

$$dx^1 \wedge dx^2 = dx^1 \otimes dx^2 - dx^2 \otimes dx^1. $$

If I take this and transform it to coordinates $y^1$ and $y^2$, I find that

$$dx^1 \otimes dx^2 - dx^2 \otimes dx^1 $$ $$\rightarrow (\frac{\partial x^1}{\partial y^1} dy^1 + \frac{\partial x^1}{\partial y^2} dy^2) \otimes (\frac{\partial x^2}{\partial y^1} dy^1 + \frac{\partial x^2}{\partial y^2} dy^2) - (\frac{\partial x^2}{\partial y^1} dy^1 + \frac{\partial x^2}{\partial y^2} dy^2) \otimes (\frac{\partial x^1}{\partial y^1} dy^1 + \frac{\partial x^1}{\partial y^2} dy^2) $$ $$= (\frac{\partial x^1}{\partial y^1}\frac{\partial x^2}{\partial y^2} - \frac{\partial x^1}{\partial y^2}\frac{\partial x^2}{\partial y^1}) dy^1 \otimes dy^2 - (\frac{\partial x^1}{\partial y^1}\frac{\partial x^2}{\partial y^2} - \frac{\partial x^1}{\partial y^2}\frac{\partial x^2}{\partial y^1}) dy^2 \otimes dy^1.$$

I see that there is a determinant of the jacobian of the coordinate change present, which is characteristic of a tensor density, but I do not see how this implies that the 2-form is not actually a tensor, but rather a tensor density. Can anyone help me reconcile this?

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First off: let me direct you to this answer. It is not an answer to your question, but just to make sure that we are on the same page. Then note that $$ dx^0 \wedge dx^1 \wedge \cdots \wedge dx^{n-1} = \epsilon_{i_1i_2\ldots i_n}dx^{i_1}dx^{i_2}\cdots dx^{i_n}, $$ where $\epsilon$ denotes the Levi-Civita symbol. Knowing the transformation of the Levi-Civita symbol, this is essentially what you showed in two dimensions extended to the $n$-dimensional case (alternatively, take a look at @mas' answer for an explicit proof of this fact). Tensors can be defined as to obey e.g. $$ T_{i_1i_2\ldots i_n} = \Lambda^{j_1}_{i_1}\Lambda^{j_2}_{i_2}\cdots\Lambda^{j_n}_{i_n} T_{j_1j_2\ldots j_n}, $$ for some frame transformation $\Lambda$, which should make it clear that Carroll's statement is correct, if you consider the symbol to be equivalent to the Levi-Civita symbol. However, if you consider $dx^0\wedge dx^1\wedge\cdots\wedge dx^{n-1}$ to be an $n$-form there is no reason to expect that the transformed quantity should be $dy^0\wedge dy^1\wedge\cdots\wedge dy^{n-1}$. I think the easiest way to explain this is to note that we can rewrite the Levi-Civita symbol as $$ \epsilon_{i_1i_2\ldots i_n} = n!\delta^0_{[i_1}\delta^1_{i_2}\cdots\delta^{n-1}_{i_n]}, $$ where $[i_1i_2\ldots i_n]$ denotes anti-symmetrization over the indices. Thus we can see that the Levi-Civita symbol components hide an antisymmetrization of Kroenecker deltas, and for illustrative purposes we can instead consider the 1-form locally defined by the components $\delta^0_i$, i.e. $dx^0$.

The point here is that $\Lambda^i_j\delta^0_i = \Lambda^0_j \neq \delta_j^0$, which is really saying that we cannot consider the components to be fixed to $\delta^0_i$ in any frame: the 1-form $dx^0$ is not equivalent to the symbol $dx^0$; it is only for a very restricted set of coordinates that $dx^0 = dy^0$, namely those where $x^0 = y^0 + C$, for some constant $C$; i.e. $x^0$ is just $y^0$ under some translation. Correspondingly $$ dx^0 \wedge dx^1 \wedge \cdots \wedge dx^{n-1} = dy^0 \wedge dy^1 \wedge \cdots \wedge dy^{n-1}, $$ if and only if the coordinate transformation has a Jacobian determinant of unity.

I think that this is what Carroll wants to warn the reader about, but to me it seems a confusing way to do it. Though admittedly I have not read the book.

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  • $\begingroup$ Let me test my understanding with an example. Suppose I have an object $D$ which I define by saying that in any coordinates, it takes the form $dx^1 \wedge dx^2$. For example in $y$ coordinates it takes the form $dy^1 \wedge dy^2$, and in $z$ coordinates it takes the form $dz^1 \wedge dz^2$. $D$ is of course not an tensor, but it still "looks like" a 2-form. Is Carroll warning against thinking of this object as a tensor? $\endgroup$ – bittermania Oct 5 '16 at 15:18
  • $\begingroup$ @bittermania As far as I can tell, yes. Alternatively, he is letting you know how an object that looks like that in some coordinates transform under a change-of-coordinates. $\endgroup$ – Erik Jörgenfelt Oct 5 '16 at 15:21
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Since both the wedge product and the Levi-Civita symbol are completely antisymmetric, we can rewrite $dx^0 \wedge dx^1 \wedge ... \wedge dx^{n-1}$ as

$$dx^0 \wedge dx^1 \wedge ... \wedge dx^{n-1} =\frac{1}{n!}\tilde{\epsilon}_{\mu_{1}\mu_{2}\cdots\mu_n} dx^{\mu_{1}} \wedge dx^{\mu_{2}} \wedge ... \wedge dx^{\mu_n}\tag{1}$$ Where there is a summation over repeated indices following Einstein summation convention and $$\epsilon_{\mu_{1}\mu_{2}\cdots\mu_{n}}=\sqrt{{|g|}}\tilde{\epsilon}_{\mu_{1}\mu_{2}\cdots\mu_{n}}$$

Now under the coordinate transformations $x^{\mu}\rightarrow x'^{\mu}$, $\tilde{\epsilon}_{\mu_{1}\mu_{2}\cdots\mu_{n}}$ remain same, while the basis one form transforms as $$\mathrm{d}x^{\mu'}=\frac{\partial x^{\mu'}}{\partial x^{\mu}}\mathrm{d}x^{\mu}\tag{2}$$ Using (2) in (1) yields \begin{eqnarray} \epsilon_{\mu_{1}\mu_{2}\cdots\mu_n} dx^{\mu_{1}} \wedge dx^{\mu_{2}} \wedge ... \wedge dx^{\mu_n} & = & \left(\tilde{\epsilon}_{\mu_{1}\mu_{2}\cdots\mu_n}\frac{\partial x^{\mu_{1}}}{\partial x^{\mu'_{1}}}\cdots\frac{\partial x^{\mu_{n}}}{\partial x^{\mu'_{n}}}\right)dx^{\mu'_{1}}\wedge dx^{\mu'_{2}} \wedge ... \wedge dx^{\mu'_n}\notag\\ \end{eqnarray} Therefore \begin{eqnarray} \tilde{\epsilon}_{\mu_{1}\mu_{2}\cdots\mu_n} dx^{\mu_{1}} \wedge dx^{\mu_{2}} \wedge ... \wedge dx^{\mu_n} & = & \left|\frac{\partial x^{\mu}}{\partial x^{\mu'}}\right|\tilde{\epsilon}_{\mu'_{1}\mu'_{2}\cdots\mu'_n}dx^{\mu'_{1}}\wedge dx^{\mu'_{2}} \wedge ... \wedge dx^{\mu'_n}\tag{3} \end{eqnarray} Expression (3) proves the claim that $dx^0 \wedge dx^1 \wedge ... \wedge dx^{n-1}$ is tensor density.

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  • $\begingroup$ As evident by the two-dimensional case, the OP is using the convention where the wedge product is $k!$ times the convention you are using. Perhaps it would be more enlightening to follow the OP convention. Furthermore, the $\epsilon$ should be taken to mean the Levi-Civita symbol in your (1), which does not transform into the Levi-Civita symbol under a general coordinate transformation. This is no problem in the sense that you are actually transforming the wedge product, not the components, but it is misleading to make the claim that you do. $\endgroup$ – Erik Jörgenfelt Oct 5 '16 at 14:25
  • $\begingroup$ Thanks for pointing the mistake, You can check the answer now. $\endgroup$ – AMS Oct 5 '16 at 14:46
  • $\begingroup$ I think the best way to do it in your case would be to skip any talk about transformation of the Levi-Civita symbol (and tensor), since you are not using it later on. When working explicitly with the frame forms you don't apply any transformation to the component functions. Rather you transform the frame, and identify the transformed components by collecting the transformed expression into the standard form. This is precisely what you do later on. $\endgroup$ – Erik Jörgenfelt Oct 5 '16 at 14:50

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