# Proof of volume density transformation under infinitesimal diffeomorphisms using Levi-civita/ determinant

Given I diffeomorphism $$x^\mu \rightarrow y^\mu = y^\mu(x)$$, I want to show that the volume density is invariant, i.e. $$\sqrt{-g(x)}\,\mathrm d^4x \rightarrow \sqrt{-g(y)}\,\mathrm d^4y$$. The first step I took was finding how $$\mathrm d^4x$$ transforms. We were given a hint to use the definition of the determinant, given as :

$$\mathrm d^4x = \frac{1}{4!} \epsilon_{\mu\nu\rho\sigma} \mathrm dx^\mu dx^\nu dx^\rho dx^\sigma \tag{1}$$

From here, I thought about plugging in the transformation $$\mathrm dx^\mu = \frac{\partial x^\mu}{\partial y^\nu}\mathrm dy^\nu$$ to give me

$$\frac{1}{4!} \epsilon_{\mu\nu\rho\sigma} \mathrm dx^\mu dx^\nu dx^\rho dx^\sigma = \frac{1}{4!} \epsilon_{ijkl} \frac{\partial x^\mu}{\partial y^i}\mathrm dy^i \frac{\partial x^\nu}{\partial y^j}\mathrm dy^j \frac{\partial x^\rho}{\partial y^k}\mathrm dy^k \frac{\partial x^\sigma}{\partial y^l}\mathrm dy^l = \mathrm{Det}\left( \frac{\partial x^\mu}{\partial y^\nu} \right) \mathrm d^4y \tag{2}$$

I am not sure if what I did here is mathematically sound, and I believe my problem is that I don't understand equation (1). In the past, I have used the levi-civita symbol/determinant to find the volume enclosed by a set of vectors. For example, given column vectors $$v$$, $$u$$, $$w$$ and a matrix $$M$$ = [$$v$$ $$u$$ $$w$$], then we had the Volume = $$\mathrm{Det}(M) = \epsilon_{ijk}v^i u^j w^k$$. Comparing this to equation (2), are we saying that $$\mathrm dx^\mu$$ and $$\mathrm dx^\nu$$ are distinct vectors with components $$\mu$$\ $$\nu = 0,1,2,3$$? If so, where is the $$4!$$ term coming from?

• Are you treating $dx$, etc. in (1) as differential forms with an implied wedge product between them? Without the wedge product you’re contracting an antisymmetric tensor with a symmetric tensor, which gives zero. Nov 2, 2020 at 1:39
• To be honest, I am not familiar with this equation or wedge products. If an implied wedge product is the only way this makes sense, then I think its safe to assume its there. Nov 2, 2020 at 1:49
• The way you are being taught this, is $dx\,dy=-dy\,dx$? Nov 2, 2020 at 4:30
• He's summing over permutations, with the sign of the permutation, etc., so obviously he is antisymmetrizing a tensor product. The result is the wedge product. Dec 7, 2020 at 9:47
• You're almost there. The metric components in the new coordinates are $g'_{ab} = g_{\mu\nu}\frac{\partial x^{\mu}}{\partial y^a}\frac{\partial x^{\nu}}{\partial y^b}$. You can interpret this formula as the product of three matrices: $(g_{\mu\nu})$, $(\frac{\partial x^{\mu}}{\partial y^a})$, $(\frac{\partial x^{\nu}}{\partial y^b})$, the latter two identical. Take the determinant of both sides: $\det(g'_{ab}) = \det(g_{\mu\nu})\ \det(\frac{\partial x^{\mu}}{\partial y^a})^2$. Take the square root, combine with the determinant you got already, noting that's the inverse of the squared one. Done. Dec 7, 2020 at 10:29

This will help you get started. For simplicity, and due to some of your questions about components, it is better to use x,y,z than the subscripts and superscripts. Since you do not know about wedge products, perhaps we'd better not go into that in depth, but just for the record, the two-dimensional analogue to your Eq. 1 is $$dx \wedge dy =\frac 1{2!} (dx\otimes dy - dy \otimes dx).$$

Now you have plugged in your transformation correctly, but you have to expand it out and collect like terms. $$\frac 1{2!} (dx\otimes dy - dy \otimes dx) =$$

$$\frac 1{2}\bigg\{\bigg(\frac {\partial x} {\partial x'} dx' + \frac {\partial x} {\partial y'} dy'\bigg) \otimes \bigg(\frac {\partial y} {\partial x'} dx' + \frac {\partial y} {\partial y'} dy'\bigg) -$$ $$\bigg(\frac {\partial y} {\partial x'} dx' + \frac {\partial y} {\partial y'} dy'\bigg) \otimes \bigg(\frac {\partial x} {\partial x'} dx' + \frac {\partial x} {\partial y'} dy' \bigg)\bigg\}.$$

Now expand, using the distributive law, but you must preserve the order since dx $$\otimes$$ dy is noncommutative multiplication, it is not equal to dy $$\otimes dx$$.
Then there is quite a lot of cancellation and some like terms to collect, and you will see that you do indeed get the determinant (in this case, the 2x2 determinant) of the transformation, i.e., the Jacobian.

Now you ask where is the 4! coming from?!? In my analogue, you will see that you get a 2 out in front, just like you got a 4. It comes because you put it in there in the first place. This volume form transforms by the Jacobian, but it isn't equal to the Jacobian. And, if you evaluate this volume form $$\frac 1{2!} (dx\otimes dy - dy \otimes dx)$$ on an orthonormal basis of the tangent space, you don't get the volume of the square, you get the volume of the triangle. So you are getting the volume of the tetrahedron, and there's nothing wrong with that. It transforms according to the determinant, but it does not give the value of the determinant when evaluated on vectors.

Now you ask about the components. dx and dy are indeed distinct vectors, and precisely for that reason they are NOT components of a big supervector. That is where the superscripts have misled you. I hope you see it more clearly when I write x and y for the coordinates. That is, x is a scalar function, and does not have any components. same for y. As a coordinate system, they determine dx, which is a covector, and has two components, and dy, which is a different covector, and also has two components. Also $$\frac {\partial } {\partial x}$$ and $$\frac {\partial } {\partial y}$$ are two contravariant tangent vectors, each of which has two components, too.

The volume form is only defined up to a scalar. You put a 1/24 in the definition, as do many people. But others want it to be just a 1, and that way the value of the volume form on the ordered pair of basis vectors $$(\frac {\partial } {\partial x} , \frac {\partial } {\partial y})$$ will be exactly the determinant, and the volume of the parallelipiped instead of the tetrahedron.

It is obvious that the choice of normalising factor you put in the volume form does not affect its transformation law. It is remarkable, and may seem paradoxical, that what value you put in the normalising factor does not affect the values you get by integrating the volume form over a region in space. Strange but true. See The value of the volume form on an orthonormal frame