characterizing semi-Riemannian spaces of constant curvature How does one characterize $n$-dimensional semi-Riemannian spaces of constant curvature? By "characterize," I mean giving both a definition and some insight into how the possibilities work out in low-dimensional spaces with signature $(1,n-1)$, which are the ones of interest in relativity. Googling is giving me lots of information on the Riemannian case, but not the semi-Riemannian one. I'm also having some trouble interpreting the info I find online for the Riemannian case because a lot of it is written in index-free notation, but I'm only really familiar with index-gymnastics notation. The sources that I'm finding give some criteria, but don't explain why they're valid or whether they're both necessary and sufficient.
Is the correct criterion the vanishing of the covariant derivative of the Riemann tensor, $\nabla_a R_{bcde}=0$? Is it sufficient for the covariant derivative of the Ricci tensor to vanish, $\nabla_a R_{bc}=0$? Why? Are these conditions equivalent to simply counting Killing vectors and getting $n(n+1)/2$ of them? (The WP article on the Riemannian case http://en.wikipedia.org/wiki/Constant_curvature seems to be saying this, but doesn't say why it's valid, or why $n(n+1)/2$ is the magic number.) What is the lowest $n$ for which there are constant-curvature spaces with signature $(1,n-1)$ that are not flat, and what do the possibilities look like for this $n$?
 A: I don't know anything about the semi-Riemannian case, but I can at least (hopefully) provide some insight in the Riemannian case.
First note that $\nabla_a R_{bcde} = 0$ does not guarantee that a space has constant curvature.  The simplest counterexample is probably $S^2\times S^2$ with the product of round metrics.  Since $S^2$ has positive curvature, some planes in $S^2\times S^2$ have positive curvature.  On the other hand, if $v$ is tangent to the first $S^2$ factor while $w$ is tangent to the second factor, then the sectional curvature of the plane spanned by $u$ and $v$ is $0$ (hence, not positive, so curvature isn't constant).
Spaces in which the covariant derivative of the Riemann tensor is $0$ are called Symmetric Spaces.  In the Riemannian case, all complete Riemannian manifolds of constant sectional curvature are symmetric spaces, but most symmetric spaces don't have constant curvature.
This also handles the case where the covariant derivative of the Ricci tensor is $0$.  (This is probably a much larger class of examples, but I'm not too familiar with them).
Constant curvature is equivalent to having $n(n+1)/2$ killing vector fields.  The number $n(n+1)/2$ is probably better expressed as $n + n(n-1)/2$.  The first "n" comes from the symmetry which allows you to move any point to any other point in an $n$-manifold of constant curvature ($n$ degrees of freedom there) and then $n(n-1)/2$ comes from the rotational symmetry.  If an isometry fixes a point $p$, it can change an orthonormal basis at $p$ to another orthonormal basis.  The collection of such changes is $O(n)$, the orthogonal group, which has dimension $n(n-1)/2$.
The intuition is that constant curvature is as symmetric as possible - you should be able to move any point to any other point and any orthonormal basis to any other.  Conversely, if you can move any point to any other point, then the space must be homogeneous, and if you can move every pair of orthonormal vectors to any other pair, it must then be constant curvature (and thus, you can move any $n$-tuple of orthonormal vectors to any other).
