Two Dimensional Ricci curvature

[craigthone]

I want to know if there is some simple metric form for Ricci curvature in dimensions generally.
In this paper https://arxiv.org/abs/1402.6334 ,
formula (5.21), the authers seem had a simple formula for Ricci curvature like this
$R= -\frac{1}{\sqrt{-g}} \partial^\mu \big[\sqrt{-g}(g_{\mu\rho} \partial_\sigma-g_{\rho\sigma}\partial_\mu )g^{\sigma\rho} \big]$
I konw the formula from
https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Riemann_curvature_tensor
I wonder how to derive this and if it is a general formula in any dimensions.

 

[JMz]

The same WP article you cite gives you the answer, a few paragraphs later: Ricci is defined as a contraction of Riemann, so the derivation is a one-liner. And Ricci is essentially unique, up to a minus sign: There's just one way to do it.

So if you can define Riemann in any space (which you can, in any space with an affine connection, including any differentiable manifold of any dimensionality), then Ricci is defined in that space: See https://en.wikipedia.org/wiki/Riemann_curvature_tensor

 

[romsofia]

2D curvature?

In another way, you can define the Ricci Curvature Tensor with differential forms as $\Omega^k_i(\hat{e}_k)$ where $\Omega$ is the curvature two form. But that's the thing... you won't get any components in 2D. I'll give an example on how to use this in 2D with the polar line element $ds^2=dr^2+r^2d\theta^2$

I wrote these notes for myself a few months back, so maybe they might help you.Let's recall our curvature equation, and the Levi-Cievta connection.

$\Omega^i_j = d\omega_{ij} + \omega^i_k \wedge \omega^k_j$

$\omega_{ij} = -\omega_{ji}$

$d\sigma^i+\omega^i_j \wedge \sigma^j = 0$So before we can calculate the curvature, we need to know what our connections are! So, let's get going. We will use the 3rd equation from above (recall it's called the Torsion free condition). Using our line element, we know that our basis is $\omega = \left\{dr, rd\theta \right\}$If i = 1, we see that it become $0 = d\sigma^1+\omega^1_1 \wedge \sigma^1 + \omega^1_2 \wedge \sigma^2 = d(dr)++\omega^1_1 \wedge dr + \omega^1_2 \wedge rd\theta = 0 + 0 + \omega^1_2 \wedge rd\theta = 0$

Recall that $d^2ANYTHING = 0$ that's why $d(dr) = 0$ and that $\omega^1_1 = 0$ due to metric compatibility. So, we're left with $\omega^1_2 \wedge rd\theta = 0$ and this implies that $\omega^1_2 = A d\theta$ where A is a constant. Why? Remember that $d\sigma^i \wedge d\sigma^j = 0$ if $i = j$Now let i = 2, the torsion free condition becomes $d\sigma^2+\omega^2_1 \wedge \sigma^1 + \omega^2_2 \wedge \sigma^2 = d(rd\theta)+\omega^2_1 \wedge dr + \omega^2_2 \wedge rd\theta = dr \wedge d\theta + rd^2\theta + \omega^2_1 \wedge dr + 0 =$ $dr \wedge d\theta + \omega^2_1 \wedge dr = 0$Now, we use what we learned from the first basis being used, that is $\omega^1_2 = A d\theta$ and use metric compatibility to say that $\omega^1_2 = -\omega^2_1 = -Ad\theta$ so let's substitute this in, to see that

$dr \wedge d\theta + \omega^2_1 \wedge dr = dr \wedge d\theta -Ad\theta \wedge dr = 0$We will have to flip the order of our wedge product, and recall in order to do this, we have to multiply by -1. Thus, our equation becomes $dr \wedge d\theta +Adr \wedge d\theta = 0$ and the only way this equation becomes 0 is if $A = -1$. Thus, $\omega^1_2 =-d\theta$ and $\omega^2_1 = d\theta$

So, now we can find curvature! Using $\Omega^i_j = d\omega_{ij} + \omega^i_k \wedge \omega^k_j$Now, this is where it differs from the Euclidean Flat space, because now we need to pick i AND j, because our connections are not zero. But, as we predicted, the curvature should just be zero in any direction.

Let i = j = 1, we will see the curvature equation look like $\Omega^1_1 = d\omega_{11} + \omega^1_1 \wedge \omega^1_1 + \omega^1_2 \wedge \omega^2_1 = d(0)+0 \wedge 0 + -d\theta \wedge d\theta = 0$ The same thing happens with i = j = 2.

Let i = 1, and j = 2. We will see the curvature equation look like $\Omega^1_2 = d\omega_{12} + \omega^1_1 \wedge \omega^1_2 + \omega^1_2 \wedge \omega^2_2 = d(d\theta) + 0 + 0 = 0 + 0 + 0$ Why is $\omega^1_1 \wedge \omega^1_2 = 0$? Because the wedge product can be thought of as multiplication, so anything times 0 is 0. So, the curvature in this case is ALSO zero! The same thing will occur with i = 2, and j = 1. Which you should try on your own.Thus, the curvature is zero!

Hopefully this helps you compute the Ricci Curvature Tensor if given the line element. I'm not sure about 2D curvature, so test it out and post your results!

 

[craigthone]

The same WP article you cite gives you the answer, a few paragraphs later: Ricci is defined as a contraction of Riemann, so the derivation is a one-liner. And Ricci is essentially unique, up to a minus sign: There's just one way to do it.

So if you can define Riemann in any space (which you can, in any space with an affine connection, including any differentiable manifold of any dimensionality), then Ricci is defined in that space: See https://en.wikipedia.org/wiki/Riemann_curvature_tensor

You mean we do have such a simple formula? For the simple metric $-e^{2\omega} dx^+dx^-$, the formula seems not to work.

 

[JMz]

You mean we do have such a simple formula? For the simple metric $-e^{2\omega} dx^+dx^-$, the formula seems not to work.

Can you define the Riemann tensor? If so, then, yes, we have such a simple formula. Of course, for any particular space, it may happen that Ricci = 0, or even that the whole Riemann = 0.

 

[craigthone]

For the spacetime discussed in AP's paper:
$$ds^2=-\exp(2\omega)dx^+dx^-=\exp(2\omega)(-dt^2+dx^2)$$
Suppose $\omega=\omega(z)$, and Riemann tensor is defined as
https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry#Riemann_curvature_tensor
Then we can calculate Riemann tensor and Ricci scalar using the above defination
$$R_{+-+-}=-\exp(2\omega)\partial_+\partial_-\omega $$
$$R=8\exp(-2\omega)\partial_+\partial_-\omega=-2\exp(-2\omega)\partial^2_z \omega$$
If we use the above simple formula,
$$R=8\exp(-2\omega)\partial_+\partial_-\omega-16\exp(-2\omega)\partial_+\omega\partial_-\omega$$

 

[kent davidge]

@romsofia very good and clear explanation, thanks for sharing with us

I wrote these notes for myself

I have this habit, too