Since Minkowski and anti de Sitter are homogeneous Lorentzian manifolds, it is natural to ask if de Sitter is too, but nobody ever discusses this. In Riemannian geometry a globally symmetric complete Riemannian manifold is homogeneous but I have no idea if this extends to semi Riemannian geometry and hence to Lorentzian manifolds.

My guess is that it is not homogeneous since de Sitter can be embedded in 5-dimensional Minkowski as a hyperboloid and I would guess the neck of the hyperboloid is cannot be moved by isometries (meaning no point on the neck can be sent on another point on the manifold). Physically I would describe this as there is a "canonical" (coordinate independent) time zero where the expansion of the universe is inverted. However, I realize this intuition is quite Riemannian and I am not sure whether weird Lorentz-boost-like symmetries could actually behave differently.

  • $\begingroup$ These might help answer your question: physics.stackexchange.com/q/353148 and physics.stackexchange.com/q/328422 $\endgroup$ Commented May 8, 2020 at 22:35
  • $\begingroup$ The latter question you linked might help even though it looks to be concerning de Sitter in the category of smooth manifolds, although when I say homogeneous I mean it in the category of Lorentzian manifolds. Is there a standard metric on $SL(2, \mathbb{C})$? I don't really know how to deal with metrics on non-compact Lie groups! $\endgroup$ Commented May 8, 2020 at 23:04
  • $\begingroup$ Maybe this is better: "Les Houches Lectures on De Sitter Space" (arxiv.org/abs/hep-th/0110007), section 2, which describes $d$-dimensional de Sitter spacetime as an explicitly-homogeneous submanifold of a $d+1$-dimensional Minkowski spacetime. I think this is what you referenced in the question, and it is explicitly homogeneous because the submanifold is invariant under arbitrary Lorentz transformations of the ambient Minkowski spacetime, which can move any point to any other point in the submanifold. $\endgroup$ Commented May 8, 2020 at 23:12
  • $\begingroup$ How do you prove the Lorentz group is transitive on the submanifold? I agree the Riemannian equivalent is obvious but I do not yet have any intuition for the Lorentzian case! $\endgroup$ Commented May 8, 2020 at 23:59
  • $\begingroup$ A $n$ dimensional DeSitter space is the Lorentzian manifold analog of an $n$-sphere. See "en.wikipedia.org/wiki/De_Sitter_space" $\endgroup$ Commented May 9, 2020 at 1:43

1 Answer 1


Thanks to the comments above I realized it is just a linear algebra problem. The issue is that in linear algebra courses people often go through the inner product spaces theory only, while, before doing some mathematical relativity, one should get familiar with what results transfer to the indefinite scalar product theory, which I did not do up to now. The answer to my question was given in the comments above, but here I want to provide a more abstract linear algebra proof that works for more general cases.

Let $V$ be an $\mathbb{R}$ vector space with a scalar product $g$ with signature $(p, q)$. Then the wanted result is a simple corollary of the following analogous result to the inner product spaces theory:

Proposition: Let $T$ be a linear operator from $(V, g)$ to $(V', g')$. Then the followings are equivalent:

  1. $T$ preserves the scalar product, that is to say, is orthogonal
  2. $T$ sends any orthonormal basis to an orthonormal basis, preserving time and space components
  3. $T$ sends one orthonormal basis to an orthonormal basis, preserving time and space components

where a basis $\{e_i\}$ is orthonormal if $|g(e_i, e_j)| = \delta_{ij}$.

Proof: $1 \Rightarrow 2$ and $2 \Rightarrow 3$ are obvious so let us prove $3 \Rightarrow 1$. Let $\{e_i\}$ be the orthonormal basis in question, which is sent to $\{e'_j\}$ and write $v = v^i e_i, w = w^i e_i$ for $v, w \in V$. Then $$ g(Tv, Tw) = g\left(T(v^i e_i), T(w^j e_j)\right) = v^i w^j g(T e_i, T e_j) = v^i w^j g(e'_i, e'_j) = v^i w^j g(e_i, e_j) = g(v, w) . $$ Now, on a space $V$ with a non-definite scalar product, there are two standard spheres: the set of vectors of squared norm $1$ and the set of vectors of squared norm $-1$.

Corollary: The orthogonal group of a scalar product vector space preserves both spheres and is transitive on both of them.

Proof: It is immediate that an orthogonal operator preserves light, null, and space vectors and hence it preserves each of the two spheres. Let us see it is orthogonal on the time sphere: for the space sphere it is analogous. Take $e$ and $e'$ to be two vectors in the time sphere. Complete each of the two vectors to an orthonormal basis. The two bases clearly have the same number of time and space vectors by Sylvester's theorem and hence there is a linear operator sending one basis to the other, which respects time and space components of the basis. This operator is orthogonal by the above proposition.


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