Calculating Ricci tensor in AdS space

[HamOnRye]

Consider the AdS metric in D+1 dimensions
ds^{2}=\frac{L^{2}}{z^{2}}\left(dz^{2}+\eta_{\mu\nu}dx^{\mu}dx^{\nu}\right)
I wanted to calculate the Ricci tensor for this metric for D=3. (\eta_{\mu\nu} is the Minkowski metric in D dimensions)
I have found the following Christoffel symbols
\Gamma^{t}_{tz}=\frac{L^{2}}{z^{3}}, \Gamma^{x}_{xz}=\Gamma^{y}_{yz}=\Gamma^{z}_{zz}=-\frac{L^{2}}{z^{3}}
From this point I wanted to determine the Riemann tensor in order to finally determine the Ricci tensor.
What I've got the following contributing Riemann tensors
R^{x}_{zxz}, R^{y}_{zyz}, R^{t}_{ztz}
I also noticed that if I have a z-coordinate in the upper index for the Riemann tensor it will be zero no matter what I choose for the lower indices.
My problem is as follows, based on symmetry, the above Riemann tensors should also be zero but I can't see how. Did I make a mistake with my Christoffel symbols or anywhere else?
Any help is appreciated!

Tim

 

[Paul Colby]

Is $\eta_{\mu\mu} = (1,-1,-1)$ or $\eta_{\mu\mu} = (-1,1,1)$? I would think this would matter.

 

[HamOnRye]

The signature is given by $\eta_{\mu\mu} = (-1,1,1)$
Thanks for responding!

 

[Paul Colby]

Okay, that makes sense. I'm just going to plug and grind with maxima. So if I write things out the metric is,

$\left(\begin{array}[cccc] --f(z) & 0 & 0 & 0 \cr 0 & f(z) & 0 & 0 \cr 0 & 0 & f(z) & 0 \cr 0 & 0 & 0 & f(z) \end{array}\right)$

where $f(z) = L^2/z^2$. Using the ctensor package in maxima I get 12 non-zero Riemann tensor component all $\pm 1/z^2$. The constant $L$ seems to drop out.

 

[HamOnRye]

Okay, that makes sense. I'm just going to plug and grind with maxima. So if I write things out the metric is,

$\left(\begin{array}[cccc] --f(z) & 0 & 0 & 0 \cr 0 & f(z) & 0 & 0 \cr 0 & 0 & f(z) & 0 \cr 0 & 0 & 0 & f(z) \end{array}\right)$

where $f(z) = L^2/z^2$. Using the ctensor package in maxima I get 12 non-zero Riemann tensor component all $\pm 1/z^2$. The constant $L$ seems to drop out.

Thanks! I made some mistakes with the $L^{2}$ terms.
But I think my main problem, or misunderstanding, still remains.
For the Riemann tensor, if we have a $z$ in the upper index there will be no contribution. But the Riemann tensor I found have a $z$ in the lower indices. But via symmetry we can place this lower index in the upper index, so any contribution with a $z$ in the lower index should also be zero.
I would like to thank you for the help so far and I hope you, or some one else, can point out the flaw in my reasoning/

Tim

 

[Paul Colby]

The symmetries I'm aware of are

$R_{\alpha\beta\mu\nu} = -R_{\beta\alpha\mu\nu} = R_{\mu\nu\alpha\beta}$​

plus whatever you can get by applying these permutations. So, $R_{\alpha\alpha\mu\nu} = 0$ and so on.

 

[HamOnRye]

The symmetries I'm aware of are

$R_{\alpha\beta\mu\nu} = -R_{\beta\alpha\mu\nu} = R_{\mu\nu\alpha\beta}$​

plus whatever you can get by applying these permutations. So, $R_{\alpha\alpha\mu\nu} = 0$ and so on.

I'll check the permutations. Thank you for the quick response and the patientce.

Tim

 

[robphy]

Possibly useful:
http://pages.pomona.edu/~tmoore/grw/Resources/DiagMetricb.pdf

 

[HamOnRye]

Possibly useful:
http://pages.pomona.edu/~tmoore/grw/Resources/DiagMetricb.pdf

Thanks!
I'm sure that will be usefull to me.

Tim