6
$\begingroup$

On the one hand the determinant (technically the absolute value of the determinant) represents the "volume distortion" experienced by a region after being transformed.

On the other hand the scalar curvature represents the amount by which the volume of a small geodesic ball in a Riemannian manifold deviates from that of the standard ball in Euclidean space.

Is there a link between the two? Can the determinant of a matrix be interpreted as a scalar curvature? If I makes a difference, I am mostly interested in the context of general relativity.

https://math.stackexchange.com/q/4481076/

$\endgroup$

1 Answer 1

8
$\begingroup$

The determinant of the metric tensor $\det g $ is a scalar density of weight $+2$, and thus, when expressed in terms of local coordinates, transforms with the square of the Jacobian of the mapping.

It follows then that its value at a point provides no invariant measure of curvature content, since you can always e.g. rescale your coordinates and obtain larger and larger values of the determinant.

Nevertheless, it's extremely important - when its square root $\sqrt{-\det g} $ (density weight +1) is multiplied with a tensor density, a Levi-Civita symbol $\epsilon_{\alpha \beta \gamma \delta} $(symbol!), of opposite density weight, the result is an antisymmetric tensor field - a volume 4-form, also called a pseudo-Riemannian volume form.

$$ \eta = \sqrt{-\det g} dx^{0} \wedge dx^{1} \wedge dx^{2} \wedge dx^{3} $$

$$ \eta = \frac{1}{4!} \sqrt{-\det g} \;\epsilon_{\alpha\beta\gamma\delta} dx^{\alpha} \wedge dx^{\beta} \wedge dx^{\gamma} \wedge dx^{\delta} $$

If you use a definition of the exterior product with a suitable factor in front, the above is equal to:

$$ \eta = \sqrt{-\det g} \;\epsilon_{\alpha\beta\gamma\delta} dx^{\alpha} \otimes dx^{\beta} \otimes dx^{\gamma} \otimes dx^{\delta} $$

Now, such an entity can be integrated in a coordinate invariant manner to provide a measure of volume.

$$ V = \int \eta$$

EDIT:

It has been recognised in the comment that it's not really a good answer, and I agree.

I will therefore derive the only correspondence that I can think of now. The other one would relate the second Lie derivative of the volume form $\mathcal{L}_{X}\mathcal{L}_{X}\eta$, which could be massaged into a form that involves a Ricci tensor, but this is pretty arbitrary, as anything that involves second covariant derivatives could have that.

The expansion of the metric determinant in normal coordinates around a point $p$ involves the Ricci tensor. This is essentially the claim present in your question.

To derive this, we shall be using the known result of the expansion of the metric tensor in normal (Riemann) coordinates. The result can be found stated here, here, or here.

The statement is that in normal coordinates around a point p, the following expansion holds:

$$ g_{\mu\nu}(\gamma(t)) = \delta_{\mu\nu} - \frac{1}{3}t^{2} R_{\mu\alpha\nu\beta}\dot{\gamma}^{\alpha}\dot{\gamma}^{\beta} - \frac{1}{6}t^{4}R_{\mu\alpha\nu\beta;\sigma}\dot{\gamma}^{\alpha}\dot{\gamma}^{\beta}\dot{\gamma}^{\sigma} + \mathcal{O}(t^{6})$$

If we expand the metric determinant along a curve $x=\gamma(t)$, we will get:

$$ \det g_{\mu\nu} (\gamma(t)) = \det g_{\mu\nu} (\gamma(0)) + t (\partial_{t}\det g_{\mu\nu})(\gamma(0)) + \frac{1}{2}t^{2}(\partial^{2}_{t}\det g_{\mu\nu})(\gamma(0)) + \mathcal{O}(t^{3})$$

The rule for the derivative of the determinant is:

$$ \partial_{t}\det g_{\mu\nu}(\gamma(t)) =(\det g_{\mu\nu})(\gamma(t))\; g^{\mu\nu}((\gamma(t))\;(\partial_{t}g_{\mu\nu})(\gamma(t)) $$

and thus, evaluating at $t=0$:

$$ (\partial_{t}\det g_{\mu\nu})(\gamma(0)) = (\det g_{\mu\nu})(\gamma(0))\; g^{\mu\nu}((\gamma(0))\;(\partial_{t}g_{\mu\nu})(\gamma(0))=0$$

because $\partial_{t}g_{\mu\nu}(\gamma(0))=0$.

EDIT2: One needs to take the second derivative wrt. to the parameter $t$. See the powers of $t$ provided in the blog post and copied by me above.

Second derivative reads:

$$ \partial^{2}_{t}\det g_{\mu\nu}(\gamma(t)) =\partial_{t}(\det g_{\mu\nu})(\gamma(t))\; g^{\mu\nu}((\gamma(t))\;(\partial_{t}g_{\mu\nu})(\gamma(t)) + \\ (\det g_{\mu\nu})(\gamma(t))\; (\partial_{t}g^{\mu\nu})((\gamma(t))\;(\partial_{t}g_{\mu\nu})(\gamma(t)) \\ + (\det g_{\mu\nu})(\gamma(t))\; g^{\mu\nu}((\gamma(t))\;(\partial^{2}_{t}g_{\mu\nu})(\gamma(t))$$

If we evaluate second derivative at $t=0$, we see that the $\partial_{t}\det g_{\mu\nu}$ contribution vanishes by virtue of the previous calculation. Same holds for $\partial_{t}g^{\mu\nu}$, which is (using $g_{\mu\nu}g^{\nu\sigma} = \delta^{\sigma}_{\mu}$ and differentiating both sides) equal to:

$$\partial_{t}g^{\mu\nu} = -g^{\mu\alpha}g^{\nu\beta}\partial_{t}g_{\alpha\beta}$$

and, involving $\partial_{t}g_{\alpha\beta}$ as well, vanishes at $t=0$.

Therefore the only contribution to the $\partial^{2}_{t}\det g_{\mu\nu}$ that can be non-zero comes from the part with $\partial^{2}_{t}g_{\mu\nu}$. From the metric expansion, differentating twice and evaluating at $t=0$ leaves only the term with the Riemann tensor contracted with the curve velocity vectors, and the factor of $2$ cancels the $\frac{1}{2}$ in the $\det g_{\mu\nu}$ Taylor expansion.

At long last it holds that up to quartic terms:

$$ \det g_{\mu\nu} (\gamma(t)) = 1 - \frac{1}{3}\,t^{2}\,R_{\alpha\beta}\dot{\gamma}^{\alpha}\dot{\gamma}^{\beta} + \mathcal{O}(t^{4})$$

$\endgroup$
1
  • 3
    $\begingroup$ This is a very nice answer but it doesn't mention the Ricci scalar at all $\endgroup$ Jun 27 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.