In GR, what is the most precise definition of a maximally symmetric spacetime?

Also, we study about the temporal boundary of dS space, and a spatial boundary of AdS space, but aren't these spaces maximally symmetric?

My point is does the condition of a spacetime being maximally symmetric imply it has to be homogeneous and isotropic?

  • $\begingroup$ Others could probably give a more definitive answer, but off the top of my head, I think we mean a space that has the greatest possible number of linearly independent killing vectors. $\endgroup$
    – user4552
    May 12, 2018 at 18:55
  • $\begingroup$ Related: physics.stackexchange.com/q/99882/2451 $\endgroup$
    – Qmechanic
    May 12, 2018 at 18:56
  • $\begingroup$ Okay...and by Killings equation which is symmetric...that implies that any space which has d(d+1)/2 Killing vectors is maximally symmetric. Is it? $\endgroup$ May 12, 2018 at 18:56

2 Answers 2


While the definition with Killing vectors works fine, it also suffers from some problems. If you take any maximally symmetric spacetime and remove a closed subset of it, it still remains a spacetime with the same Killing vectors, since those only encode local properties.

The proper definition of a maximally symmetric space is that it is homogeneous and isotropic.

Homogeneity says that for any two points $p$ and $q$ in $M$, there exists an isometry $\phi$ belonging to the group of all isometries $I(M)$ such that $\phi(p) = \phi(q)$.

Isotropy at a point $p$ implies that for any two tangent vectors $v, w \in T_pM$, such that $|v| = |w|$, then there is an isometry $\phi \in I(M)$ such that $\phi(p) = p$ (the point $p$ is the center of the isometry) and $\phi_*(v) = w$ (the isometry transports the vector $v$ to $w$).

Given this, it is possible to show that for a spacetime, this corresponds to the existence of $n(n+1) / 2$ Killing vectors, including $n$ translations and $n(n-1)/2$ rotations.

  • $\begingroup$ I think there is a typo. $\phi(p) = q$. $\endgroup$
    – James C
    May 13, 2022 at 9:13

Maximally symmetric space is a space that is both homogeneous and isotropic. Such a space possesses the largest possible number of Killing vectors which in an n-dimensional manifold equals $n(n+1)/2$. The following holds for a maximally symmetric space:

  1. The scalar curvature $R$ is a constant.
  2. The Ricci tensor is proportional to the metric tensor, i.e., $R_{\mu\nu}=\dfrac{1}{n}Rg_{\mu\nu}$.
  3. The Riemann curvature tensor is given by $R_{\mu\nu\lambda\rho}=\dfrac{R}{n(n-1)}(g_{\mu\lambda}g_{\nu\rho}-g_{\nu\lambda}g_{\mu\rho})$.


Distant galaxies are receding from us. So how do we account for the concepts of homogeneity and isotropy of space? The answer lies in the fact that the universe is apparently not static at all, but changing with time. So the idea that the universe is homogeneous and isotropic apply only to the spatial part of the space-time.


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