# Does de Sitter space have a preferred frame?

Consider the flat, expanding coordinates for de Sitter space: $$ds^2=-dt^2+e^{2Ht}d\vec{x}^2\quad .$$ This is clearly not invariant under the ordinary Lorentz transformations. Does this mean that if I move on a worldline with velocity $$\vec{v}=d\vec{x}/dt$$ (or $$\vec{v}=e^{Ht}d\vec{x}/dt$$, whichever is correct, see comment below by Yukterez), that the universe will look like it is "moving"?

• the universe will look like it is "moving"? What does that mean? Feb 23, 2020 at 18:00
• If $v$ is meant to be the local velocity (in your coordinates relative to a comoving buoy which is flowing with the Hubbe flow), i.e. the contracted distance by the dilated (proper) time, you have to use $\sqrt{1-v^2/c^2}=\sqrt{g^{tt}}/\dot{t}$, if you differentiate by the coordinate time you get the shapirodelayed velocity Feb 24, 2020 at 0:43
• @G.Smith That means, "Is there an experiment I can do in a vacuum that gives a different result in both frames?". Same as in special relativity. What else could it mean? Feb 24, 2020 at 11:30
• @Yukterez Thanks, for pointing that out. Feb 24, 2020 at 11:31
• You seem to have a different question in the title and in the body.
– user21299
Feb 27, 2020 at 15:13

I'll answer the question in the title.

No, ($$n$$-dimensional) de Sitter does not have a preferred frame. The reason is that it is a maximally symmetric spacetime (with isometry group $$O(n,1)$$*). Similarly to Minkowski space which is also maximally symmetric, there is no preferred frame because for every two points there exists an isometry transformation that maps one point to the other and also orthonormal frames over the points to each other. Therefore unless the symmetry is broken (by, for instance, matter) there is no experiment that can distinguish between different frames.

(Perhaps this is easier to swallow for Minkowski, where the claim is that there exist a combination of translations and Lorentz transformations that will map any event into any other event, which is intuitively obvious.)

*The isometry group can depend on what we precisely mean by "de Sitter" space, as in whether we are taking a universal cover or not. But the lie algebra is always $$so(n,1)$$.

• This answer is formally correct but it gives the impression that under the same logic even the galilean group would not have any preferred frame, which is not (historically) what one would say and, I guess, what the OP intended with the question. Feb 29, 2020 at 20:19
• I'd say Galilean spacetime also doesn't have a preferred frame. (It has a preferred class of frames, as specified by newton's first law I suppose.) The group theory I mentioned is fundamental to the answer. If you had the same metric ansatz except you replace $e^{2Ht}$ with some other function of time, you no longer have $O(n,1)$ invariance, and suddenly the universe looks different depending on the spacetime event you're observing from.
– user21299
Mar 1, 2020 at 2:49
• @TwoBs Galilean physics indeed does not have any preferred frame. And this is what was known historically from the times Galilei. But the Maxwell equations are not Galilean invariant and this is what spurred all this preferred frame discussion. I.e. the preferred frame appeared not in the Galilean-Newtonian physics but when people have encountered its violation
– OON
Mar 1, 2020 at 7:45
• @OON I agree, I have said already the answer is correct and that Galilean group does not give rise to any preferred frame, but I brought this up to highlight that this is probably not what the OP intended. Mar 1, 2020 at 12:47
• @alexarvanitakis Thank you I know dS has dS symmetry. My question is whether this symmetry corresponds (locally) to observers moving with different velocities. Hence the connection between the title and the body. Mar 2, 2020 at 12:26