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A $d$ dimensional maximally symmetric spacetime is a spacetime with the maximum allowed number of Killing vectors. This number is $\frac{d(d+1)}{2}$.

Constant curvature spacetimes are spacetimes whose Weyl tensor is zero. Thus, the Riemann tensor $R_{abcd}$ is written as $$\frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc}).$$

If the metric is Euclidean, constant curvature spacetimes are either spherical, hyperbolic or flat and one can check explicitly for each of these examples that indeed spacetime is maximally symmetric. Similarly, if the metric is Lorentzian, constant curvature spacetimes are either deSitter, antideSitter or flat, and one can check that they are maximally symmetric. Thus, constant curvature spacetimes are maximally symmetric. Is the reverse statement also true? Also, is it possible to prove that constant curvature spacetimes are maximally symmetric without resorting to examples?

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  • $\begingroup$ Just as a comment: your title question and your question question are different; the title is "does maximal symmetry imply constant curvature?" while the question is "is there a way to prove that constant curvature implies maximal symmetry in general?"... Which question did you mean to ask? $\endgroup$
    – CR Drost
    Nov 28, 2015 at 19:40
  • $\begingroup$ Thanks. I edited the question to be more clear. I meant both questions, but I consider the one in the title to be more important ;) $\endgroup$ Nov 28, 2015 at 20:14
  • $\begingroup$ Wouldn't a constant curvature spacetime with a point removed still be of constant curvature but not maximally symmetric? $\endgroup$
    – Slereah
    May 31, 2016 at 12:46

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If the metric is Riemannian (positive) your conjecture (a maximally symmetric spacetime is a constant curvature spacetimes) is a known theorem: Theorem 3.1 in Transformation Groups in Differential Geometry by S. Kobayashi. From the proof, it seems to me that the result should hold in the Lorenzian case too, but without a closer scrutiny I am not completely sure.

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  • $\begingroup$ Thanks a lot! I took a look at the reference you provided, but, without further study, I couldn't really understand it. I might ask this question in the mathematics forum, just because I am curious if this is true (even though I might not understand the proofs). $\endgroup$ Nov 30, 2015 at 14:26
  • $\begingroup$ Just to clarify, there are multiple theorems 3.1 in the book, the referenced theorem is the one in chapter II. $\endgroup$
    – Inzinity
    Sep 29, 2019 at 13:41
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This is an old question but I'll expand Valter Moretti's answer. Yes, maximally symmetric pseudo-Riemannian spaces (in the sense that they have the maximum number of Killing vector fields) are constant curvature spaces.

The proof is in Weinberg: Gravitation and Cosmology at around page 376. Weinberg's arguments are not completely rigorous, but can be made so using the Frobenius integrability theorem.

A Killing vector field $\xi^\mu$ satisfies $$ \nabla_\mu\xi_\nu+\nabla_\nu\xi_\mu =0$$ and $$ \nabla_\mu\nabla_\nu\xi_\rho=-R^\sigma_{\ \mu\nu\rho}\xi_\sigma $$(copied this formula from Weinberg so depending on one's conventions it might be off by a sign).

Defining $\chi_{\mu\nu}=\nabla_\mu\xi_\nu$, this reduces to the overdetermined PDE system $$ \nabla_\mu\chi_{\nu\rho}=-R^\sigma_{\mu\nu\rho}\xi_\sigma \\ \nabla_\mu\xi_\nu=\chi_{\mu\nu} $$with the algebraic constraint $\chi_{\mu\nu}=-\chi_{\nu\mu}$.

If the manifold admits the maximal number of Killing vector fields, this means that the above system of equations can be solved for any initial value $\xi_\sigma(x_0)=a_\sigma,\quad\chi_{\mu\nu}(x_0)=b_{\mu\nu}$. Thus the equation must be completely integrable.

Frobenius' theorem gives the necessary and sufficient conditions for complete integrability. We can actually use a fully covariant version of the theorem as follows:

  1. We first form the second covariant derivatives of the equation: $$ \nabla_\mu\nabla_\nu\chi_{\rho\sigma}=-\nabla_\mu(R^\tau_{\ \nu\rho\sigma}\xi_\tau)=-\nabla_\mu R^{\tau}_{\ \nu\rho\sigma}\xi_\sigma-R^\tau_{\ \nu\rho\sigma}\nabla_\mu\xi_\sigma \\ \nabla_{\mu}\nabla_\nu\xi_\rho=\nabla_\mu\chi_{\nu\rho}. $$
  2. We substitute from the equation here: $$ \nabla_\mu\nabla_\nu\chi_{\rho\sigma}=-\nabla_\mu R^{\tau}_{\ \nu\rho\sigma}\xi_\sigma-R^\tau_{\ \nu\rho\sigma}\chi_{\mu\sigma} \\ \nabla_\mu \nabla_\nu\xi_\rho=-R^\sigma_{\ \mu\nu\rho}\xi_\sigma. $$
  3. The functions $\xi_\sigma$ and $\chi_{\rho\sigma}$ must satisfy the Ricci identities, thus if we substract from the above equations themselves with $\mu,\nu$ interchanged, we get $$ -R^\tau_{\ \rho\mu\nu}\chi_{\tau\sigma}-R^\tau_{\ \sigma\mu\nu}\chi_{\rho\tau}=-\nabla_\mu R^{\tau}_{\ \nu\rho\sigma}\xi_\sigma+\nabla_\nu R^{\tau}_{\ \mu\rho\sigma}\xi_\sigma-R^\tau_{\ \nu\rho\sigma}\chi_{\mu\sigma}+R^\tau_{\ \mu\rho\sigma}\chi_{\nu\sigma} \\ -R^\tau_{\ \rho\mu\nu}\xi_\tau=-R^\tau_{\ \mu\nu\rho}\xi_\tau+R^\tau_{\ \nu\mu\rho}\xi_\tau. $$

After manipulation [there might be sign errors, because I haven't checked Weinberg's conventions but used my own with the Ricci identity], this reduces to the system of equations [I am copying this from Weinberg] $$ [-R^\lambda_{\ \rho\sigma\nu}\delta^\kappa_\mu+R^\lambda_{\ \mu\sigma\nu}\delta^\kappa_\rho-R^\lambda_{\ \sigma\rho\mu}\delta^\kappa_{\nu}+R^\lambda_{\ \nu\rho\mu}\delta^\kappa_\sigma]\chi_{\kappa\lambda}=[\nabla_\nu R^\lambda_{\ \sigma\rho\mu}-\nabla_\sigma R^\lambda_{\ \nu\rho\mu}]\xi_\lambda. $$

These equations must be valid for any values of $\xi_\lambda$ and any antisymmetric values of $\chi_{\kappa\lambda}$.

I am not going to write this out explicitly, but setting $\xi_\lambda=0$ and antisymmetrizing on $\kappa,\lambda$, then "cancelling" $\chi_{\kappa\lambda}$ gives a set of equations (can be found in Weinberg). If we take the trace on an appropriate pair of indices, we find $$ (m-1)R_{\lambda\rho\sigma\nu}=R_{\nu\rho}g_{\lambda\sigma}-R_{\sigma\rho}g_{\lambda\nu}. $$ Antisymmetrizing on $\lambda,\rho$ then taking a second trace gives $$ R_{\mu\nu}=\frac{1}{m}Rg_{\mu\nu}. $$ Inserting this back gives $$ R_{\mu\nu\rho\sigma}=\frac{R}{m(m-1)}(g_{\mu\rho}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\rho}), $$ so the manifold is isotropic.

When $m>2$ ($m$ is the dimension of the space) the usual Schur lemma gives that $R$ must be a constant. When $m=2$ we can use the second set of integrability conditions (setting $\chi_{\kappa\lambda}=0$ and $\xi_\lambda$ to arbitrary) together with the Bianchi identity to show that $R$ is constant for $m=2$ as well.

Thus maximally symmetric spaces are also constant curvature spaces. The metric signature plays no role here.

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