# Must the isometry group of a spacetime be a subgroup of the Minkowski, dS or AdS isometry group?

There are three spacetimes that are, globally, maximally symmetric: Minkowski, de Sitter and anti-de Sitter.

(The global condition is important. For example, the flat torus given by $\mathbb M^N/\mathbb Z^N$ is maximally symmetric locally, but fails to be so globally because the identification $x^\mu \sim x^\mu + \ell^\mu$ isn't Lorentz-invariant.)

It is easy to construct new spacetimes that have fewer symmetries by starting with one of the maximal three and removing isometries while respecting group closure. Thus, you will be getting subgroups of Minkowski, de Sitter or anti-de Sitter.

For example, if you start with Minkowski, you can get the FLRW spacetime by getting rid of boosts and the time translation. On the other hand, throw away space translations and boosts and you get the Schwarzschild black hole. Etc.

The question then is, does a spacetime exist whose global isometry group is NOT a subgroup of Minkowski, de Sitter, or anti-de Sitter?

• I'd say probably not since the global isometries have to correspond locally to some of the local spacetime isometries? – Slereah Jul 25 '18 at 13:43
• although of course locally it's just Minkowski isometries, so the local isometries don't have to be exactly similar to the global ones, ie a spacetime with the $SO(4,2)$ isometry group of dS will still have the Poincaré group locally – Slereah Jul 25 '18 at 13:44
• Your own example, torus $\mathbb M^N/\mathbb Z^N$ has isometry group $U(1)^N$ and this group is not a subgroup of Poincaré group. – A.V.S. Jul 26 '18 at 16:33