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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 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 $n$?

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Nice question. One easy class of examples would be $\mathbb{R} \times M$ where the spacetime is a cylinder with a time axis. The geometry of a constant time slice could be Riemannian. – James S. Cook Nov 5 '12 at 2:58

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).

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What are you taking to be the definition of constant curvature in a Riemannian space? Minkowski space in $n=4$ dimensions does have $n(n+1)/2=10$ Killing vectors, but it seems unlikely to me that any non-flat space with signature $(1,3)$ has that many. So maybe the notion of constant curvature just can't be generalized to semi-Riemannian spaces in any interesting way. – Ben Crowell Nov 8 '12 at 2:30
In the Riemannian case, constant curvature means that for all orthonormal vectors $u,v$ at any point we have $\langle R(u,v)u,v\rangle$ is the same constant number. (Independent of $u$ and $v$ and also of $p$). – Jason DeVito Nov 8 '12 at 2:58
I made a small error - the condition $\nabla_a R_{bcde} = 0$ is equivalent to being a locally symmetric space. In the simply connected case, symmetric and locally symmetric are the same, but in the nonsimply connected case, locally symmetric is slightly weaker than symmetric. – Jason DeVito Nov 11 '12 at 14:04
Hi Jason. Your example of $S^2\times S^2$ is not with the metric I normally consider (I normally take the flat metric on $S^2\times S^2$). With the round metric, is this still a smooth manifold? Does the condition $\nabla R = 0$ make sense at the transition lines from positive to zero curvature? It seems to me that there are regions with constant positive curvature, and constant zero curvature, and nothing in-between.... I'm probably being dense, but could you shed some light on this? – Glen Wheeler Nov 15 '12 at 4:06
@Glen: By "round" I mean the "usual" metric on $S^2$ - the one it gets from it's usual embedding into $\mathbb{R}^3$ as the unit sphere. It is conformally flat, but not flat. At least in the Riemannian case, $S^2\times S^2$ doesn't have a flat (meaning curvature $0$) metric. Flat manifolds (in fact, nonpositively curved) manifolds are covered by $\mathbb{R}^n$ (the Cartan-Hadamard Theorem). Also, the choice of metric doesn't affect the smoothness of $S^2\times S^2$, but yes, the metric is smooth. (continued) – Jason DeVito Nov 15 '12 at 13:24

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