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 Quote by Messenger Ahhhhh. Perhaps I would be less confused just reading the forum instead of asking questions. Thanks for the link Mentz.
Just to clear things up, this is the metric (b is a constant > 0)
$$ds^2={d\phi}^{2}\,{r}^{2}\,{sin\left( \theta\right) }^{2}+{dt}^{2}\,\left( b\,{r}^{2}-1\right) +\frac{{dr}^{2}}{1-b\,{r}^{2}}+{d\theta}^{2}\,{r}^{2}$$
and the Einstein tensor is $-3b\ diag(g_{00},g_{11},g_{22},g_{33})$. In any local frame the metric g can be replaced by η, the Minkowski metric.

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 Quote by Messenger So the metric for the Lambdavacuum solution is not $diag(-\Lambda,\Lambda,\Lambda,\Lambda)$?
No. The cosmological constant $\Lambda$ *multiplies* the metric in the Einstein Field Equation:
$$G_{\mu \nu} + \Lambda g_{\mu \nu} = 8 \pi T_{\mu \nu}$$
To find out what the metric $g_{\mu \nu}$ actually is, you have to *solve* the above equation, given some stress-energy tensor $T_{\mu \nu}$. The Lambdavacuum solution is the solution for which $T_{\mu \nu} = 0$; there is no "ordinary" stress-energy present. (Some people prefer to move the $\Lambda$ term to the other side of the equation and call it a form of "stress energy"--"dark energy" or something like that. That's a matter of terminology and doesn't change the physics.)