Tensor vs. Tensor Densities

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

What you have written is basically that the: 2 form in the old coordinates = det(J) *(the 2 form in the new coordinates)

Which is why it's a tensor density, it transformed by det of the jacobian. Had it been a tensor, the basis part would have transformed the opposite way to the component part, keeping it invariant.

In the case of the 2 form, the anti symmetrization introduces the levi civita symbol as the components to the tensor basis(when you expand the wedge product as outer product) . This stuff however, isn't a coordinate invariant, and changes as we saw before. Hope this explains it.