# Symmetry of Ricci Tensor

1. Dec 28, 2013

### ProfDawgstein

Hey,

I have been doing a few proofs and stumbled across this little problem.

Trying to show the symmetry of the Ricci tensor by using the Riemann tensor definition

$R^m_{\ ikp} = \partial_k \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ ki} + \Gamma^a_{\ ip} \Gamma^m_{\ ak} - \Gamma^a_{\ ik} \Gamma^m_{\ ap}$

Now set m = k

$R^m_{\ \,imp} = R_{ip} = \partial_m \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ mi} + \Gamma^a_{\ ip} \Gamma^m_{\ am} - \Gamma^a_{\ im} \Gamma^m_{\ ap}$

Checking every term for symmetry (i <-> p)

1. symmetric because $\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}$

2. see below

3. symmetric because $\Gamma^{m}_{\ ip} = \Gamma^{m}_{\ pi}$

4. symmetric because $\Gamma^a_{\ im} \Gamma^m_{\ ap} = \Gamma^m_{\ pa} \Gamma^a_{\ mi}$ (interchanging a & m, and just rotate) and $\Gamma^{a}_{\ im} = \Gamma^{a}_{\ mi}$

Now for the 2nd term

$\partial_p \Gamma^m_{\ mi} == \partial_i \Gamma^m_{\ mp}$

This is a contracted Christoffel symbol...

using $\Gamma^k_{ij} = \frac{1}{2} g^{kl} \left(\partial_i g_{jl} + \partial_j g_{il} - \partial_l g_{ij}\right)$

$\Gamma^{m}_{\ mi} = \frac{1}{2} g^{ml} \frac{\partial g_{ml}}{\partial {x^i}}$

Now the derivative

$\partial_p \Gamma^m_{\ mi} = \frac{1}{2} \left( \frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} + g^{ml} \frac{\partial g_{ml}}{\partial {x^i} \partial {x^p}}\right)$

The 2nd term in this expression is symmetric (i <-> p), because order of partial differentiation doesn't matter.

For the first term I am not sure though.

Is $\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}$?

The inverse is something totally different, but you also contract it, this is the point where I am slightly confused.

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I know there are other ways to "derive" the symmetry of the Ricci tensor, but I wanted to try this one :)

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Thank you in advance!

It's probably so obvious that I don't see it

Hope you guys had a nice Christmas ;)

2. Dec 28, 2013

### WannabeNewton

Well you're definitely torturing yourself needlessly by doing it this way :tongue2:

If you just use the real definition of the Riemann tensor, $2\nabla_{[a}\nabla_{b]}\xi^{c} = -R_{abd}{}{}^{c}\xi^{d}$, then it immediately follows that $R_{ab} = g^{cd}R_{dabc} = g^{cd}R_{cbad} = R_{ba}$.

But if you really want to do it using the coordinate-dependent definition (ick!)* of the Riemann tensor then make it easier for yourself and go to a locally inertial coordinate system at an arbitrary event $p$ in space-time so that $\Gamma^{\mu}_{\nu\gamma}(p) = 0$ hence $R_{\mu\nu}(p) = 2\Gamma^{\gamma}_{\mu[\nu,\gamma]}(p)$. Now all you have to do is plug in for $\Gamma^{\mu}_{\nu\gamma}(p)$.

*Coordinate-independent ("abstract index") calculations are infinitely more elegant than coordinate-dependent calculations

3. Dec 28, 2013

### Bill_K

Here's the fact you need: δgμν = - gμα gνβ δgαβ. Thus

∂gmℓ/∂xp = - gma gℓb ∂gab/∂xp

and therefore

∂gmℓ/∂xp ∂gmℓ/∂xi = - gma gℓb ∂gab/∂xp ∂gmℓ/∂xi

which is symmetric.

4. Dec 28, 2013

### ProfDawgstein

It wasn't much work, actually.

But that still boils down to $\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}$, doesn't it?

Using your definition, I get the first term to be symmetric and the 2nd one is

$\Gamma^{\gamma}_{\ \mu \gamma , \nu}$

which should be equal to

$\Gamma^{\gamma}_{\ \nu \gamma , \mu}$.

And this still leads to

$\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}$.

This is the only remaining thing I need to show and I don't know how :/

Oh yes, I totally forgot this thing. Thank you!

5. Dec 28, 2013

### WannabeNewton

But keep in mind that if you do this in a locally inertial coordinate system like I told you to, $\partial_{\gamma}g_{\mu\nu}(p) = 0$ so you would just have the term with the second derivative of the metric and that's trivially symmetric.

The point I'm trying to convey is: if you're going to do a coordinate-dependent calculation then you may as well pick the coordinate system that makes the calculation as simple as possible. However I personally like to do coordinate-free calculations whenever possible.

Last edited: Dec 28, 2013
6. Dec 28, 2013

### ProfDawgstein

Ok, done :)

Just plug in, replace ab by ml, use symmetry of g_ml.

Thank you both.

Now I know what to prove next - or maybe I should finally start a serious book about this

I could do that for sure, but I haven't been that much into this, that I think about things like
-lets use this to simplify this
-use trick 123
-...

At the moment I am still doing this for fun, or when I am bored.
I am planning to do "rigorous" tensor/GR from the beginning very soon, so that will work out I hope :)

7. Dec 28, 2013

### WannabeNewton

It's a ton of fun so you'll enjoy it without a doubt, especially if you like index gymnastics. Good luck!

8. Dec 28, 2013

### ProfDawgstein

Thanks!

Schutz GR, Carrol GR, Schutz MM, Cahill, Lovelock, Ohanian, ... should do the trick :)