# Tensor density from wedge product

1. Apr 4, 2015

### Ravi Mohan

Hi,
I am studying Sean Carroll's "Lecture notes on General Relativity". In the second chapter he identifies the volume element $d^nx$ on an n-dimensional manifold with
$$dx^0\wedge\ldots\wedge dx^{n-1}.$$

He then claims that this wedge product should be interpreted as a coordinate dependent object (he also proves that this object is in fact a tensor density). But, looking from an other point of view, if $dx^\mu$ is a well defined one-form, then how can the wedge product of these one-forms be a coordinate dependent object instead of being a well defined n-form?

2. Apr 4, 2015

### Staff: Mentor

Read the text starting at the bottom of p. 52, right before equation 2.44, where he explains how tensor densities apply to volume elements on manifolds. The wedge product of the $dx^\mu$ is actually a tensor density, not a tensor (i.e., not an n-form). To make it a tensor (i.e., an n-form), you have to multiply by $\sqrt{|g|}$. The text explains how this works.

3. Apr 4, 2015

### Ravi Mohan

I am sorry if I am not clear. I understand how the volume element on manifolds transform as density and also the proof that the wedge product of $dx^\mu$ also transforms as density. I am wondering why the wedge product itself not an n-form.

If we consider two functions $g$ and $f$, then the gradients of these functions are one-forms. And, furthermore, their wedge product is a two-form. But, according to the notes, for the chart $x^\mu$, the gradient $dx^\mu$ is not a tensor, but a density.

I am trying to understand the second paragraph on page 53. More specifically the line "But we would like to interpret the right hand side of (2.45) as a coordinate-dependent object which, in the $x^\mu$ coordinate system, acts like $dx^0\wedge\ldots\wedge dx^{n-1}$"

4. Apr 4, 2015

### Staff: Mentor

Because it doesn't transform like one. As you note, it transforms like a density. You might want to try comparing this transformation formula with the transformation formula for the gradients of ordinary functions.

What he means is that the object $\text{d}^nx$ is not actually an n-form (it's a tensor density, as above), but writing it as a wedge product makes it look (unfortunately) like an n-form. We know it is not actually an n-form because it doesn't transform like one, as above.

Last edited: Apr 4, 2015
5. Apr 4, 2015

### Ravi Mohan

Thanks. I really need to work on my communication skills :)

So, if I get it right, $dx^\mu$ is a well defined one-form and the wedge product of them is a tensor density. Therefore it is not always true that the wedge product of two tensors will be a tensor.

6. Apr 5, 2015

### Ben Niehoff

I think this is truly bizarre reasoning, and I'm afraid I have to disagree. The wedge product of 1-forms $\alpha, \beta$ is defined as

$$\alpha \wedge \beta = \alpha \otimes \beta - \beta \otimes \alpha$$
And in general, it's just the antisymmetrized tensor product. So of course the wedge product of the $n$ gradients $d x^\mu$ is a tensor!

I think Carroll has actually managed to confuse himself. I don't have a copy of his book, but I would assume he fixed this issue in his notes during the process of editing them for the book.

If you look at eq. (2.40), Carroll correctly points out that the Levi-Civita symbol (i.e. the object $\tilde \epsilon_{\mu_1 \ldots \mu_n}$ whose entries are always the numbers $0, \pm 1$) is a tensor density. In other words, $\tilde \epsilon_{\mu_1 \ldots \mu_n}$ is not a covariant object; it's just a bookkeeping device that is useful for writing down certain formulas.

Carroll then considers the object

$$d x^1 \wedge \ldots \wedge dx^n = \frac{1}{n!} \, \tilde \epsilon_{\mu_1 \ldots \mu_n} \, d x^{\mu_1} \wedge \ldots \wedge d x^{\mu_n}$$
and tries to argue what sort of object it is based on how it transforms. In the first line of eq. (2.47), Carroll has it completely right. If we call the components of this object $A_{\mu_1 \ldots \mu_n}$, then the first line of (2.47) shows

$$A_{\mu_1' \ldots \mu_n'} = A_{\mu_1 \ldots \mu_n} \frac{\partial x^{\mu_1}}{\partial x^{\mu_1'}} \cdots \frac{\partial x^{\mu_n}}{\partial x^{\mu_n'}}$$
which is the transformation law for a tensor! It just so happens that in one coordinate system (the unprimed $x^\mu$), the components $A_{\mu_1 \ldots \mu_n}$ happen to have values given by

$$A_{\mu_1 \ldots \mu_n} = \tilde \epsilon_{\mu_1 \ldots \mu_n}$$

The second line of eq. (2.47), however, is very misleading, because Carroll has plugged back in $\tilde \epsilon_{\mu_1' \ldots \mu_n'}$, which we have already agreed is not a covariant object. So the second line of (2.47) is really just reiterating the fact that the Levi-Civita symbol (whose entries are always the numbers $0, \pm 1$) fails to be a tensor.

But the object $dx^1 \wedge \ldots \wedge dx^n$ is absolutely a tensor. When you change it to another coordinate system $x^{\mu'}$, it will become

$$dx^1 \wedge \ldots \wedge dx^n = \frac{1}{n!} \, \frac{\partial x^{1}}{\partial x^{\mu_1'}} \cdots \frac{\partial x^{n}}{\partial x^{\mu_n'}} \, dx^{\mu_1'} \wedge \ldots \wedge dx^{\mu_n'}$$
as expected. This notion that it should somehow be a tensor density comes from the false expectation that you should be able to simply replace $x^\mu \to x^{\mu'}$. Clearly you cannot do that, because the $x^\mu$ are sitting behind partial derivative operators (that is what $d$ is, after all).

If you disagree, let me ask you this: If $dx^1 \wedge \ldots \wedge dx^n$ is a tensor density rather than a proper $n$-form, then what is

$$dx^1 \wedge \ldots \wedge dx^{n-1} + dx^2 \wedge \ldots \wedge dx^n$$
? (Each term has $n-1$ factors).

By the way, the mathematical definition of a tensor is a map from $T : V^m \times (V^*)^n \to \mathbb{R}$ which is function-linear; i.e. for some vectors $X, Y \in V^m \times (V^*)^n$, we have

$$T(f X + g Y) = f \, T(X) + g \, T(Y)$$
where $f, g$ are functions, not just numbers. You should be able to show that this definition is equivalent to the "transforms as such-and-such" definition. Moreover, it should be obvious that the wedge product of a bunch of one-forms is a tensor.

7. Apr 5, 2015

### Ravi Mohan

I agree that the conclusion in post #5 is indeed bizarre which resulted due to little ambiguity in the notes. I do agree that the wedge product of one-forms better be a tensor itself (hence the point of creating this thread :) ).

I am actually comfortable with this step which follows from the equation 2.40.

Now I have a question. Would it be alright if we identify the volume element $d^nx$ with the object $\frac{1}{!n}\tilde{\epsilon}_{\mu_1\ldots\mu_n}dx^{\mu_1}\wedge\ldots dx^{\mu_n}$? This way, the volume element will be a coordinate dependent object which transforms as a density as it should and rest of the arguments will follow.

Thanks.

8. Apr 5, 2015

### Ben Niehoff

Yes, that will work. I think the issue here is the notation $d^n x$, which always means $dx^1 \wedge \ldots \wedge dx^n$, and thus it is $d^nx$ which fails to be a tensor. The point being that $dx^1 \wedge \ldots \wedge dx^n$ is a tensor, but its coordinate expression in another coordinate system $y^\mu$ is

$$\frac{\partial x^1}{\partial y^1} \cdots \frac{\partial x^n}{\partial y^n} \, dy^1 \wedge \ldots \wedge dy^n$$
rather than simply

$$d^n y \equiv dy^1 \wedge \ldots \wedge dy^n$$
Frankly, I find it less confusing to integrate with respect to the invariant volume form $\omega \equiv \sqrt{|g|} \, d^n x$, which does have the same expression in every coordinate system. Then an integral can be expressed as

$$\int_M f \, \omega$$
where $f$ is a proper scalar function, not a scalar density.

9. Apr 5, 2015

### Staff: Mentor

But that's exactly the point of Carroll's discussion: he wants to find an expression for an invariant volume element for use in integration. So maybe a better way of phrasing the question would be: the wedge product $\text{d} x^1 \wedge \ldots \wedge \text{d} x^n$ is a tensor, and physically it represents an "egg-crate" type structure formed by the intersection of the $n$ basis 1-forms. Those 1-forms are perfectly well-defined geometric objects, even if we switch charts; and so their wedge product is also a perfectly well-defined geometric object, even if we switch charts. So why isn't the volume of a cell of this geometric object invariant? Why do we have to stick a factor $\sqrt{|g|}$ on it to make it invariant?

10. Apr 5, 2015

### Ben Niehoff

Any non-vanishing top form is a volume form, so in that sense, $dx^1 \wedge \ldots \wedge dx^n$ is a perfectly legitimate choice.

When you have a metric $g$, the particular top form $\omega = \sqrt{|g|} \, dx^1 \wedge \ldots \wedge dx^n$ is singled out by the added constraint that its action on any $n$-vector (at a point) gives a volume that agrees with $g$ (at that point, for all points).

Stated another way, if you write $g$ in orthonormal frames,

$$ds^2 = \delta_{ab} \, e^a \otimes e^b,$$
then $\omega$ is the unique volume form that assigns a volume of 1 to the $n$-cube spanned by the orthonormal basis, at every point of the manifold.

11. Apr 5, 2015

### Staff: Mentor

Ah, I see: the wedge product by itself gives an invariant at a given point of the manifold, but the volume represented by this invariant can be different at different points of the manifold. Adding the factor $\sqrt{|g|}$ compensates for this so that the volume represented by the form is the same at all points, which is what we want to make integration simple.