Is the volume form a tensor?

Question

The definition of the volume form confused me whenever I read about it through a general relativity textbook. Perhaps it is better explained in Mathematics text, but I haven't taken a look at them. The volume form in general relativity is introduced through something called a tensor density. As far as I understand, tensor densities aren't really much different from tensors - they just come from looking at tensors differently. When physicists mean the form $dx^1 \wedge dx^2 \wedge ...\wedge dx^n$ is a tensor density what they mean is that in a new coordinate system $y^\alpha$, the form becomes $J dy^1 \wedge dy^2 \wedge ... \wedge dy^n$ where $J$ is the Jacobian determinant. However, the n-form is actually a tensor; the only reason it picks up the factor of the Jacobian is because we represent the tensor in terms of wedge products and not tensor products. We multiply the the square root of the metric to compensate for this factor. I understand this fairly well.

However, is the final volume form given by $\sqrt{-g} \text{ } dx^1 \wedge dx^2 \wedge ...\wedge dx^n$a tensor? It seems that the $\sqrt{-g}$ messes with the transformation law because it cancels the partial derivatives that come from transforming the n-form.

Answer 1

However, the n-form is actually a tensor; the only reason it picks up the factor of the Jacobian is because we represent the tensor in terms of wedge products and not tensor products.

Yes, every $n$-form is a tensor, but what we actually need in this case is a tensor field, and note that the $n-$form you mentioned is defined for each point $p$. We need to consistently assign a tensor to each point of our manifold, and that is not an automatic feature of the wedge product of two functions.I think that is the source of the problem. $\sqrt{|g|}dx^{1}\wedge...\wedge dx^n$ (I'm using the absolute value because $g$ might not be negative for an arbitrary $n$) is also an $n$-form, so it will give you a tensor as well. But since, as you said, the $\sqrt{|g|}$ factor compensates the change induced by the jacobian, this may actually be used over a finite volume for integration (you need a measure for integration, and the integral of this measure can't depend on your coordinate choice). As I understand it, the formal statement is that the differential $n-$form you wrote is a density, and as such they may be used to define a measure on an orientable manifold, which is what is being done here.

Answer 2

Yes and no.

If the volume form $\omega$ were well defined, it would be independent of the choice of coordinates used to define it, or equivalently changing coordinates would give an identical expression. Considering two coordinates $x^a$ and $y^\alpha$ with transformation Jacobian $\Phi^\alpha_a=\frac{\partial y^\alpha}{\partial x^a}$, we have the transformation rules $dy^1\wedge\cdots\wedge dy^n=\det(\Phi^\alpha_a)dx^1\wedge\cdots\wedge dx^n$ and $g_{ab}=\Phi^\alpha_a\Phi^\beta_bg_{\alpha\beta}$. The factors involving $\Phi$ almost cancel properly, but there will be a minus sign if the coordinte change is orientation-reversing. This tells us that simply defining $\omega:=\sqrt{|\det(g_{ab})|}dx^1\wedge\cdots\wedge dx^n$ in any coordintes is not adequate. There are two ways of alleviating this issue.

The first is to restrict attention to oriented manifolds, and define $\omega:=\sqrt{|\det(g_{ab})|}dx^1\wedge\cdots\wedge dx^n$ only in oriented local coordinates $x^i$. This means that the sign issue won't arise, and $\omega$ is a globally well defined $n$-form, and one can do integration, Stokes' theorem, etc using the language of differential forms. However, on nonorientable manifolds, this approach is not possible.

Instead, one can use densities, objects which behave like the absolute value of a $n$-forms. These are not $n$-forms, nor are they tensor fields of any kind, since they are not linear. Still, one can define integration, divergence theorem, etc. on manifolds without the need for an orientation using densities, and things end up looking very similar (with some sign differences here and there). Given an $n$-form $\lambda$, one can define the corresponding density $|\lambda|$ by $|\lambda|(X_1,\cdots,X_n)=|\lambda(X_1,\cdots,X_n)|$. If we then replace the form with the volume density $\omega:=\sqrt{|\det(g_{ab})|}|dx^1\wedge\cdots\wedge dx^n|$, we have a well-defined density which does not depend on the coordinates used to define it. This density exists on all pseudo-Reiamnnian manifolds, no orientation needed.

Which of these two conventions is in effect is often difficult to determine at a glance, and the notation is far from set in stone. Regardless, when dealing with integration on manifolds it's important to be mindful of the signs and orientations.

---

Edit: some clarification on definition of a tensor

Some of the confusion seems to come from a divide in how tensors (tensor fields, more precisely) are defined. Mathematicians will generally define tensors as sections of a tensor power of the (co)tangent bundle, or as multilinear maps on tangent spaces. These definitions won't make direct reference to coordintes, and tensors defined this way do not change when changing coordinates. More precisely, given coordintes $x^a$, we have a canonical basis for the space of $(k,l)$ tensors, and so every $(k,l)$ tensor $T$ can be written as: $$ T=T^{a_1\cdots a_k}_{b_1\cdots b_l}\partial_{a_1}\otimes\cdots\otimes\partial_{a_k}\otimes dx^{b_1}\otimes\cdots\otimes dx^{b_l} $$ The tenor $T$ does not depend on coordinates and does not change at all when changing coordinates, but the basis elements $\partial_a$ and $dx^a$ will change, and so the component functions $T^{a_1\cdots a_k}_{b_1\cdots b_l}$ will also change to compensate.

Some (mostly phisicists) will instead define a tensor as a coordinte-dependent collection of numbers with a particular transformation rule. In terms of the previous definition, this such an object is not tensor, but rather the components of a tensor; I'll refer to them this way.

The story is the same with $n$-forms which are just antisymmetrized $(0,n)$ tensors. The space of $n$-forms has rank one, so for coordintes $x^a$ there is a single basis element $dx^1\wedge\cdots\wedge dx^n$, and every $n$-form can be written as $\omega=\omega_{1\cdots n}^{(x)}dx^1\wedge\cdots\wedge dx^n$ with a single component function $\omega_{1\cdots n}^{(x)}$. The components should transform as $\omega^{(y)}_{1\cdots n}=\det(\Phi^\alpha_a)^{-1}\omega^{(x)}_{1\cdots n}$ to compensate for the change in this basis element. In the case of the volume form (restricting to oriented coordinates), the tensor itself is $\omega:=\sqrt{|\det(g_{ab})|}dx^1\wedge\cdots\wedge dx^n$, which should not change when changing coordinates, while the component function is $\omega_{1\cdots n}=\sqrt{|\det(g_{ab})|}$ which should obey the appropriate transformation rule (and both do, up to the aforementioned sign issues).