In Carroll's book on general relativity ("Spacetime and Geometry"), I found the following remark:

In two dimensions, finding that $R$ is a constant suffices to prove that the space is maximally symmetric [...] In higher dimensions you have to work harder

Here, $R$ is the Ricci scalar. This begs the following questions:

  • How does ones prove that a $d$-dimensional spacetime is maximally symmetric?
  • If the general case is very complicated, then is there a simpler way to obtain the result in four dimensions?
  • 1
    $\begingroup$ To disprove the assertion that a spacetime is maximally symmetric, one technique would be to find a curvature invariant that isn't constant. For example, the Schwarzschild spacetime has $R=0$ everywhere, because it's a vacuum solution, but there are other curvature invariants such as the Kretschmann invariant that are varying. $\endgroup$
    – user4552
    May 12, 2018 at 19:11

1 Answer 1


Two general methods come to mind:

  1. Prove that the Riemann tensor takes the form of equation 3.191, i.e. $$R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc}) $$ If you are handed a metric, this should in principle be a straightforward calculation. If the metric is actually maximally symmetric, the calculation of the Riemann tensor usually turns out to be easier than usual, especially if you use a high-tech method like the Cartan formalism with vielbeins and spin connections (see Appendix J of Carroll).
  2. Find the maximal number of Killing vectors. For a manifold with dimension $d$, it admits a maximum of $\frac12 d(d+1)$ Killing vectors (these are explained in section 3.8 of Carroll). This technique is usually easier if you have a good sense of what the isometries of the metric are and can basically guess all the Killing vectors. For example, in flat Minkowski space it is fairly obvious that boosts, rotations and translations are all symmetries of the metric. So you write down the vectors corresponding to flowing in the direction of these transformations, and can easily check that they satisfy Killing's equation, $\nabla_{(a}\xi_{b)}=0$, or $£_\xi g_{ab}=0$, where $\xi^a$ is the Killing vector.

In practice, constructing maximally symmetric spaces is easier than it sounds. Generally, you start with the manifold $\mathbb{R}^{n,m}$ with a flat metric of signature $(n,m)$ (i.e. there are n spacelike coordinates with a $+dx_i^2$ contribution to the line element and $m$ coordinates with a $-dy_j^2$ contribution). It is fairly straightforward to prove that this is maximally symmetric. Then, you define a submanifold $\mathcal{S}$ as the locus of points that are some fixed distance from the origin. For example, if we started with $\mathbb{R}^{3,0}$, which is just Euclidean $3$-space, we say $\mathcal{S}$ is the set of all points such that $$x^2+y^2+z^2=r_0^2$$ for some fixed radius $r_0$. This of course defines a 2-sphere, which is a maximally symmetric space of one less dimension from $\mathbb{R}^{3,0}$. You can infer from this construction that the submanifold $\mathcal{S}$ is going to be maximally symmetric, because we broke the full group of isometries of $\mathbb{R}^{n,m}$ down into the ones that leave the origin fixed. So we started with $\frac12 d(d+1)$ isometries ($d=n+m$), and lost all the translations, of which there are $d$, so we are left with $\frac12 d(d+1)-d = \frac12(d-1)d$ isometries, which is the maximal number for $d-1$ dimensions.

Since you asked about 4D maximally symmetric spacetimes, there are basically three things you can do. The trivial one is just Minkowski space, $\mathbb{R}^{3,1}$. The next thing we can do is start with five dimensional Minkowski space, $\mathbb{R}^{4,1}$, and pick out all the points that are a fixed spacelike distance from the origin, $$x^2+y^2+z^2+w^2-t^2 = r_0^2 $$ (here $(x,y,z,w)$ are the spatial coordinates). The induced metric on the submanifold is de Sitter space, $dS_4$, the 4D spacetime with constant positive curvature.

Finally, you can begin with $\mathbb{R}^{3,2}$ the Euclidean space with $3$ spacelike directions $(x,y,z)$ and $2$ timelike directions $(t_1,t_2)$. This time we consider all points that are a fixed timelike distance from the origin, $$x^2+y^2+z^2-t_1^2-t_2^2=-r_0^2 $$ The induced metric for this submanifold is anti-de Sitter space, $AdS_4$, which is the 4D spacetime of negative constant curvature.

Locally I believe any maximally symmetric space will look like one of the spaces constructed using this embedding technique. In some cases however, there can be nontrivial topological features that cause the maximally symmetric space to differ from the embedded manifolds we just constructed. One example is for $AdS_4$: as it stands, the manifold we constructed has closed timelike curves (arising from moving around in the $t_1$-$t_2$ plane). These can be removed by "unwinding the time direction," which mathematically means we go to the simply connected universal cover, which is topologically different from the $AdS_4$ we just constructed, but locally looks exactly the same.

  • $\begingroup$ Sorry if its a trivial question but I don't understand why you didn't consider the possibilities $x^2+y^2+z^2+w^2-t^2=-r_0^2$ and $x^2+y^2+z^2-t_1^2-t_2^2=r_0^2$? $\endgroup$
    – user10001
    Jul 21, 2015 at 16:51
  • 2
    $\begingroup$ I was just listing the examples that give you the Lorentzian signature maximally symmetric spaces. The possibilities you listed correspond to the negatively curved Euclidean space, and I think a positively curved space with signature (2,2). $\endgroup$
    – asperanz
    Jul 21, 2015 at 17:02

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