# Symmetry v.s. isometry of Minkowski and AdS or dS spacetime

We know some nice spacetime have a lot of symmetries. It is said that

• Minkowski spacetime has $$ISO(d-1,1)/SO(d-1,1),$$

• de Sitter spacetime has $$SO(d,1)/SO(d-1,1)$$ and

• anti-de Sitter spacetime has $$SO(d-1,2)/SO(d-1,1).$$

Question: Is this correct that the above is the precise full symmetry of Minkowski, de Sitter spacetime, and anti-de Sitter spacetime? It this the same as the isometry of these spacetimes? How to show this is the complete symmetry?

• The proof of this can be found in Wolf's "Spaces of Constant Curvature", for instance. Sep 24, 2019 at 6:40

The examples listed in your question are spaces with a maximal amount of symmetries, i.e. they are homogeneous and isotropic (see this SE post). However, the space $$SO(d,1)/SO(d-1,1)$$ for example is not the symmetry group of the de Sitter space but the de Sitter space itself. It is not even a group because $$SO(d-1,1)$$ is not a normal subgroup of $$SO(d,1)$$. Thus, the quotient space is just a differential manifold with a transitive action of $$SO(d-1,1)$$. A similar case that is easier to imagine is the 2-sphere $$S_2$$. Similar to the above examples, it is a maximally symmetric space that can be defined as the quotient space $$SO(3)/SO(2).$$ Again, $$SO(2)$$ is not a normal subgroup and, thus, $$S_2$$ does not inherit the group structure of $$SO(3)$$. The same is true for the spaces you listed in your question. The quotient spaces are not the symmetry group but the respective space itself. However, the total group space, i.e. the "numerator" in the quotient, represents the isometry group whereas the "denominator" represents the isotropy group of the quotient space. The isotropy group is the stabilizer group of a point, for example the Lorentz group in a relativistic spacetime.