For my topology class I have to show the following:

Let $g = g_{ij} \; dx^i \otimes dx^j$ be a metric on a differentiable manifold. Show that the volume form $dV = \sqrt{\det g} \; dx^1 \land ... \land dx^n$ is invariant under coordinate transformations.

Now, I am not quite sure what I have to do here. If $x = (x^1, ..., x^n) $ is a set of coordinates and $\tilde{x} = (\tilde{x}^1, ..., \tilde{x}^n)$ is another set of coordinates, then do I have to show that

$$d\tilde{V}( \tilde{x} ) \enspace = \enspace dV(x) \quad ?$$

But how do I know how $d\tilde{V}(\tilde{x})$ looks like? Does anyone have a hint for me?

  • 1
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Dec 12 '20 at 18:42
  • $\begingroup$ Well, I guess since differential geometry is essential for e.g. general relativity I thought it would be fruitful to post this question here. Frankly though, I posted it in mathematics too. $\endgroup$
    – Octavius
    Dec 12 '20 at 18:48
  • $\begingroup$ Possible duplicate of Proof of volume density transformation under infinitesimal diffeomorphisms using Levi-civita/ determinant $\endgroup$
    – pglpm
    Dec 12 '20 at 18:56
  • $\begingroup$ You have to show that the general formula for the component of the volume element, that is, the prescription "take the square root of the determinant of the metric's components" is a prescription valid in all coordinates. $\endgroup$
    – pglpm
    Dec 12 '20 at 19:01
  • $\begingroup$ To be honest I don't like how they worded the problem in your class, though. Any tensor or form, per se, is coordinate invariant, since it's the same object no matter which coordinates we choose – if we choose any. What may be invariant are particular expressions for the tensor components. $\endgroup$
    – pglpm
    Dec 12 '20 at 19:06

Using basic coordinate transformation rules for tensors ($g_{ij}$ and $dx^i$), as well as antisymmetric properties of the wedge product, show that the wedge product
$dx^1\wedge\dots dx^n$ transforms as follows under coordinate transformations:

$$ dx^1\wedge\dots dx^n=\frac{\partial(x)}{\partial(\tilde{x})}\,d\tilde{x}^1\wedge\dots d\tilde{x}^n $$

Where $\frac{\partial(x)}{\partial(\tilde{x})}$ is the determinant of the Jacobian matrix. Then, again using coordinate transformations, show that

$$ \mbox{det}\left[\tilde{g}\right]=\left(\frac{\partial(x)}{\partial(\tilde{x})}\right)^2\,\mbox{det}\left[g\right] $$

Conclude that, under coordinate transformations:

$$ dx^1\wedge\dots dx^n=\sqrt{\frac{\mbox{det}\left[\tilde{g}\right]}{\mbox{det}\left[g\right]}}\frac{\partial(x)}{\partial(\tilde{x})}\,d\tilde{x}^1\wedge\dots d\tilde{x}^n $$

Finally conclude that:

$$ \sqrt{\mbox{det}\left[g\right]}\cdot dx^1\wedge\dots dx^n = \sqrt{\mbox{det}\left[\tilde{g}\right]}\cdot d\tilde{x}^1\wedge\dots d\tilde{x}^n $$

i.e. the form of this quantity does not change as you change coordinates.

Finally, finally, note that any manifold is locally a Cartesian space, and there, in Cartesian coordinates $\mbox{det}\left[g\right]=1$ and $dx^1\wedge \dots dx^n=d^nV$, the standard volume form. From this conclude that $d^n V=\sqrt{\mbox{det}\left[g\right]}dx^1\wedge \dots dx^n$ is the correct volume form in any coordinate system and at all points on the manifold

Adding more specifics. I would suggest https://www.amazon.com/Tensors-Differential-Variational-Principles-Mathematics-ebook/dp/B00A735HK8#reader_B00A735HK8. It deals with all that you want in a rigorous and hands-on-way.

To compute the determinant of the metric tensor you need a way of expressing this determinant using tensors. At this point on usually reaches for Levi-Civita, but the problem with that is that Levi-Civita is not a tensor (it is a relative tensor, in the language of Lovelock and Rund). Fortunately, there is something called generalized Kroenecker delta [https://en.wikipedia.org/wiki/Kronecker_delta#Definitions_of_the_generalized_Kronecker_delta], which is a tensor (to prove this note that generalized Kroenecker delta can be written as products of usual Kronecker delta's $\delta^{\mu}_{\nu}$, so it is sufficient to prove that latter is a tensor).

Ok, with that:

$$ \det\left[\tilde{g}\right]=\tilde{\delta}^{\alpha_1\dots\alpha_n}_{1\dots n}\tilde{g}_{1\alpha_1}\dots\tilde{g}_{n\alpha_n} $$

To check that, note that $\tilde{\delta}^{1\dots n}_{1\dots n}=1$ and then picks up a factor of -1 on every exchange of $\alpha$-s. Now you can apply change of coordinates to the RHS, since there you only have tensors:

$$ \begin{align} \det\left[\tilde{g}\right]=\frac{\partial x^{\mu_1}}{\partial \tilde{x}^1}\dots\frac{\partial x^{\mu_n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1\dots\alpha_n}_{\mu_1 \dots \mu_n} \cdot \frac{\partial x^{\sigma_1}}{\partial \tilde{x}^1}\dots \frac{\partial x^{\sigma_n}}{\partial \tilde{x}^n} \cdot g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n} \end{align} $$

Firstly deal with sum over $\mu$-s. Let's look at just two terms, for example:

$$ \begin{align} \frac{\partial x^{\mu_1}}{\partial \tilde{x}^1}\dots\frac{\partial x^{\mu_n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1\dots\alpha_n}_{\mu_1 \dots \mu_n} =& \frac{\partial x^{1}}{\partial \tilde{x}^1}\frac{\partial x^{2}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n} + \frac{\partial x^{2}}{\partial \tilde{x}^1}\frac{\partial x^{1}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{2, 1\dots n}+\dots\\ =& \frac{\partial x^{1}}{\partial \tilde{x}^1}\frac{\partial x^{2}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n} - \frac{\partial x^{2}}{\partial \tilde{x}^1}\frac{\partial x^{1}}{\partial \tilde{x}^2}\dots\frac{\partial x^{n}}{\partial \tilde{x}^n}\cdot \delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n}+\dots \end{align} $$

Note the swap of indices on the least delta. In this way we can show that the sum over $\mu$-s consist of a product $\delta^{\alpha_1 \dots\alpha_n}_{1, 2\dots n}$ and a completely anti-symmetrized product of partial derivatives - a Jacobian determinant:

$$ \begin{align} \det\left[\tilde{g}\right]=\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\cdot \delta^{\alpha_1\dots\alpha_n}_{1 \dots n} \cdot \frac{\partial x^{\sigma_1}}{\partial \tilde{x}^1}\dots \frac{\partial x^{\sigma_n}}{\partial \tilde{x}^n} \cdot g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n} \end{align} $$

Next consider:

$$ w_{\sigma_1\dots\sigma_n}=\delta^{\alpha_1\dots\alpha_n}_{1 \dots n} g_{\sigma_1\alpha_1}\dots g_{\sigma_n\alpha_n} $$

And note that:

$$ w_{1\dots n}=det\left[g\right] $$

and any exchange of $\sigma$ indices simply adds a factor of -1 to $w_{\sigma_1\dots\sigma_n}$. We can therefore repeat the trick above to get

$$ \begin{align} \det\left[\tilde{g}\right]=&\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\cdot w_{1\dots n} \cdot \frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\\ =&\left(\frac{\partial\left(x\right)}{\partial\left(\tilde{x}\right)}\right)^2\cdot\det\left[g\right] \end{align} $$

  • 1
    $\begingroup$ For the det.: $$g = g_{ij} d\tilde{x}^i \otimes d\tilde{x}^j = g_{ij} \frac{\partial d\tilde{x}^i}{\partial dx^k}\frac{\partial d\tilde{x}^j}{\partial dx^{\ell}} dx^k \otimes dx^{\ell}$$ The Jacobian Matrix is $J_{ij} = \frac{\partial d\tilde{x}^i}{\partial dx^j}$ and thus $$ g = g_{ij} J_{ik} J_{j\ell} \; dx^k \otimes dx^{\ell} = (J^T g J)_{k\ell} \; dx^k \otimes dx^{\ell} $$ But how do I calculate the det.? Is is just $\det g(\tilde{x}) = (\det J)^2 \det g (x)$? How do I handle the tensor product mathematically rigorous when calculating the det.? Is my way correct so far? $\endgroup$
    – Octavius
    Dec 13 '20 at 17:28
  • $\begingroup$ @Octavius. Firstly $\partial d x^k$ makes no sense. Either $\partial$ or $d$, not both. Secondly don't denote Jacobian matrix with downstairs indices, this will mess up your equations, should be $J^i_j$. Thirdly, use $\mathbf{g}$to denote a non-scalar value, in most texts I have seen, $g$ stands for the determinant of the metric tensor. I will try to deal with the bigger question in the main post $\endgroup$
    – Cryo
    Dec 13 '20 at 17:53
  • $\begingroup$ Certainly, you are perfectly right to object $\partial dx^k$. This was just some kind of typo that I carelessly committed. However, thank you for your effort so far. $\endgroup$
    – Octavius
    Dec 13 '20 at 18:33

You do not know how $dV$ looks like for a specific case. So, you have to use the general form of $dV$ which is what you have written.

You need to show that if you perform a coordinate transformation on the wedge products and on the determinant of the metric at the same time, everything will be cancelled and you will recover $dV$ again.


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