How Does a Diagonal Metric Affect the Symmetry and Tensor Equations in Space?

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Hey guys! I am considering a space with a diagonal metric, which is maximally symmetric.

It can be proven that in that case of a diagonal metric the following equations for the Christoffel symbols hold:
[tex]\Gamma^{\gamma}_{\alpha \beta} = 0[/tex]
[tex]\Gamma^{\beta}_{\alpha \alpha} = -(1/g_{\beta\beta})\partial_{\beta}g_{\alpha\alpha}[/tex]
[tex]\Gamma^{\beta}_{\alpha \beta} = \partial_{\alpha}\ln(\sqrt{|g_{\beta\beta}|})[/tex]
[tex]\Gamma^{\alpha}_{\alpha \alpha} = \partial_{\alpha}\ln(\sqrt{|g_{\alpha\alpha}|})[/tex]

Furthermore: for a maximally symmetric space we have for the Riemann tensor:
[tex]R_{\rho\sigma\mu\nu} = R/12(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu})[/tex], where R is the Ricci scalar.

Given these equations, I come across a contradiction. From the above equation for the Riemann tensor we easily see that if it has three different indices, it must be zero (IF the metric is diagonal). However, if I plug in the Christoffelsymbols into the definition of the Riemanntensor expressed in Christoffel symbols and their derivatives, I do not find that this is zero in general. Does anyone know what is going wrong here?

EDIT: For a maximally symmetric space, three indices cannot be unequal, but for a diagonal space this might not be the case. How is this all related?
 
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If I put the following diagonal line element into GrTensor, for coordinates (t,x,y,z),

-p(t,x,y,z)*d[t]^2+q(t,x,y,z)*d[x]^2+r(t,x,y,z)*d[y]^2+s(t,x,y,z)*d[z]^2;

I get the following (just a partial set).

[tex]\Gamma^t{}_{tt} = \frac{1}{2} \frac{\partial_t \, p}{p}[/tex]
[tex]\Gamma^x{}_{tt} = -\frac{1}{2} \frac{\partial_x \, p}{q}[/tex]
[tex]\Gamma^t{}_{xt} = \frac{1}{2} \frac{\partial_x \, p}{p}[/tex]

They look similar, except for a missing factor of 1/2. The ln's complicate things a bit - are they really that useful?

I was rather surprised to see an apparently non_zero term pop out for R_txty myself. I'm not sure if it's really nonzero, or just didn't simplify:

[tex]R_{txty} = 1/4\,{\frac {-2\, \left( {\frac {\partial ^{2}}{\partial x\partial y}}<br /> p \left( t,x,y,z \right) \right) p \left( t,x,y,z \right) q \left( t,<br /> x,y,z \right) r \left( t,x,y,z \right) + \left( {\frac {\partial }{<br /> \partial y}}p \left( t,x,y,z \right) \right) \left( {\frac {<br /> \partial }{\partial x}}p \left( t,x,y,z \right) \right) q \left( t,x,<br /> y,z \right) r \left( t,x,y,z \right) + \left( {\frac {\partial }{<br /> \partial x}}p \left( t,x,y,z \right) \right) \left( {\frac {<br /> \partial }{\partial y}}q \left( t,x,y,z \right) \right) p \left( t,x,<br /> y,z \right) r \left( t,x,y,z \right) + \left( {\frac {\partial }{<br /> \partial y}}p \left( t,x,y,z \right) \right) \left( {\frac {<br /> \partial }{\partial x}}r \left( t,x,y,z \right) \right) p \left( t,x,<br /> y,z \right) q \left( t,x,y,z \right) }{p \left( t,x,y,z \right) q<br /> \left( t,x,y,z \right) r \left( t,x,y,z \right) }}[/tex]
 
When you say "maximally symmetric space" and "diagonal space," you're talking about two different kinds of things. The symmetry is intrinsic, but the diagonal form of the metric is coordinate-dependent.

I think the only maximally symmetric spaces in 3+1 dimensions are Minkowski space, de Sitter space, and anti de Sitter space. Have you tried, for example, checking your calculations in the special case of de Sitter space?

Funzies said:
From the above equation for the Riemann tensor we easily see that if it has three different indices, it must be zero (IF the metric is diagonal).
What if the different components of the metric are unequal?

Funzies said:
EDIT: For a maximally symmetric space, three indices cannot be unequal, but for a diagonal space this might not be the case. How is this all related?
Are you thinking that the different components of the metric have to be equal because it's maximally symmetric? That's not true. The symmetry doesn't have to be manifest when you write the metric in certain coordinates.