We derive the most basic laws of physics from several fundamental symmetries (those from Noether's theorems, gauge symmetries, Lorentz symmetry...). But are there any types of spacetime where no symmetries, no matter how fundamental, would hold? Any special kind of metric, geometry or shape?


2 Answers 2


I assume you're actually asking about isometries of the metric $\phi^* g = g$ in the context of GR (physical spacetime symmetries). The 'gauge symmetry' of GR, diffeomorphism invariance, is an intrinsic part of the theory and always present.

But yes, the majority of spacetimes would have no such symmetries. (You could write down any metric with this property: e.g. think about systems of multiple matter sources not confined to a plane, binary systems of rotating BH's, chaotic systems, etc). The problem is that these metrics are incredibly hard to solve, and so we often want to make symmetry assumptions about the systems we're dealing with. If any physically interesting examples of these types of metrics with exact solutions come to mind, I'll edit it into this question.

  • 2
    $\begingroup$ Overall, great answer, +1. Just to clarify, in your last sentence, do you mean that you "any physically interesting examples of these types of metrics that can be solved analytically?" Because a binary system of two rotating black holes (or a rotating black hole and neutron star) is a perfect example of a physically relevant spacetime that does not have symmetries for which Einstein's equations can be solved numerically. $\endgroup$
    – Andrew
    May 21 at 16:57
  • $\begingroup$ @Andrew yes, thank you for the clarification! I will make this clearer. Actually, I don't actually know the specific metrics for these cases (I've seen examples where additional symmetries are assumed, but not without them), so this may be a useful seperate answer for the OP. $\endgroup$
    – Eletie
    May 21 at 17:09

Let me rephrase your question.

You are asking if there exists a Riemannian (or a Lorentzian) manifold $(M,g)$, such that any smooth vector field $X$ on $M$ satisfying the Lie derivative equation $$\mathcal{L}_{X}g=0$$

is trivial (i.e vanishes everywhere on $M$).

The answer to your question is yes.

  1. For a Riemannian manifold, there's a mathematical theorem proven by Salomon Bochner that no non-trivial Killing vector field exists on a compact Riemannian manifold of negative Ricci curvature.

  2. For a semi-Riemannian manifold (aka a Lorentzian manifold), one can proof that if there exists a point $p\in M$, such that $X_{p}=0$, and $(DX)_{p}=0$, then $X\equiv 0$ everywhere on $M$. The proof can be found in Semi-Riemannian Geometry With Applications to Relativity written by Barrett O'Neill. In other words, if for any killing vector field $X$ on $M$, $X$ and $DX$ vanish at the same point, then $X$ is trivial. Now the problem is reduced to finding such a manifold. I'll keep looking for such a manifold.

  • $\begingroup$ It appears though that this theorem only applies to riemannian manifolds (positive definite metric), not to pseudo-riemannian manifolds (spacetimes), as asked by the OP. Am I correct? $\endgroup$
    – magma
    May 22 at 0:45
  • $\begingroup$ @magma Yes you are right. I've been thinking about this problem. There must be a generalization in the Lorentzian case, but the paper from that link is quite difficult to read. $\endgroup$ May 22 at 1:25

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