This is a doubt from Jelle Hartong's masters thesis on the geometry of dS spacetime. So basically I know that $$R= \frac{2d}{d-2} \Lambda$$ and $$R_{abcd}= \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc}).$$ Hartong argues that these equations and the fact that the Weyl tensor is zero implies that the metric is maximally symmetric. I know of examples worked out that verify this, for instance in Peter Szekeres, $\textit{A Course in Modern Mathematical Physics}$, Chapter 19 but that is using the Killing Equations and the metric components. I don't see an explicit link between the space being a constant curvature one. Can anyone shed some light on this?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Mar 24, 2019 at 19:15
  • $\begingroup$ This should be rewritten to make the question self contained, without reference to someone’s MSc thesis. $\endgroup$ Mar 24, 2019 at 20:42
  • $\begingroup$ @ZeroTheHero you are not familiar with this seminal work? $\endgroup$ Feb 15, 2021 at 20:39
  • $\begingroup$ @Oбжорoв the issue is more: if the general reader familiar with this work? $\endgroup$ Feb 15, 2021 at 21:19

3 Answers 3


You need to have the Riemann tensor of the form you wrote: $R_{abcd}= \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc})$, in order for a manifold to be maximally symmetric. But just having a constant Ricci scalar $R$ does not guarantee that the spacetime will be maximally symmetric. The easiest counterexample is Schwarzschild spacetime where $R$ is constant and zero but the Riemann tensor does not take the above form. In fact, it has a non-zero Weyl tensor as the only nonzero curvature (after Riemann tensor is decomposed into Weyl tensor, Ricci scalar and Ricci tensor).


Heuristically, the curvature being constant implies that all points are equivalent in terms of metric relations, and all directions are equivalent.

More precisely, I am somewhat confused about the exact terminology used in the mathematical literature, but generally speaking a (pseudo)-Riemannian space is maximally symmetric, if it is homogenous, i.e. for any two points $x,y\in M$ there is an isometry taking $x$ to $y$ and isotropic (about every points), i.e. for any point $x\in M$ and any pair of $1$-directions $u,v$ at $x$ (i.e. tangent vectors of some fixed length) there is an isometry that leaves $x$ fixed and maps $u$ to $v$.

This is a global notion, and iirc for each dimension and metric signature there are three canonical examples of such spaces (flat, spherical, hyperbolic), and the other maximally symmetric spaces arises as quotients of these spaces by discrete group actions (see Helgason: Symmetric Spaces).

In physics one often cares more about local aspects than global, so it is useful to introduce locally maximally symmetric (LMS) spaces. This might not be a standard terminology but whatever, I'm gonna use it.

An LMS space is a pseudo-Riemannian manifold $(M,g)$ which admits the maximal number of independent Killing vector fields (i.e. $(n^2+n)/2$ where $n=\dim M$). These can be considered to be locally homogenous (for any infinitesimally separated pair of points there is an infinitesimal isometry taking one into the other) and locally isotropic (for any infinitesimally close pair of $1$-directions at a point there is an infinitesimal isometry leaving the point fixed, mapping the $1$-directions into each other).

One can derive (see Weinberg - Gravitation and Cosmology for derivations, though personally I think his power series arguments are better replaced by a systematic use of the Frobenius integrability theorem) that $$ R_{\mu\nu\kappa\lambda}=K(g_{\mu\kappa}g_{\nu\lambda}-g_{\mu\lambda}g_{\nu\kappa}), $$ for $K$ a constant is a necessary and sufficient condition for this.

In addition, the following conditions are equivalent:

  • $ R_{\mu\nu\kappa\lambda}=K(g_{\mu\kappa}g_{\nu\lambda}-g_{\mu\lambda}g_{\nu\kappa})$ for constant $K$,
  • $\nabla_\rho R_{\mu\nu\kappa\lambda}=0$,
  • At a fixed point $x\in M$, the sectional curvature is constant for all $2$-directions at that point and the section curvature is further constant for all choices of points $x$.

Any of the above conditions can be called a space of constant curvature and thus spaces of constant curvature are equivalent to LMS spaces.

I am also noting that for each dimension and metric signature there are only three LMS line elements (flat, spherical, hyperbolic), however this only characterizes local geometry. As stated before if we look at global geometry/topology, then there are more than three maximally symmetric spaces (for each dimension/metric signature).

Proof of LMS space $\Leftrightarrow$ Constant curvature space is given in Weinberg using local integrability conditions. This does not prove that $\nabla R$ and constant sectional curvature are equivalent to that. Those are however proven iirc in Helgason and Lee - Introduction to Riemannian Manifolds. Maybe also in Kobayashi/Nomizu.


I guess a counterexample would be illuminating: BTZ black hole is of constant negative curvature but it is not maximally symmetric. (No curvature singularity at the origin.)


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