• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,348 answers , 22,725 comments
1,470 users with positive rep
818 active unimported users
More ...

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

+ 3 like - 0 dislike

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 $$SO(d-1,2)/SO(d-1,1).$$

e.g. see https://physics.stackexchange.com/a/75604/42982

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?

This post imported from StackExchange Physics at 2020-11-26 15:41 (UTC), posted by SE-user annie marie heart
asked Sep 24, 2019 in Theoretical Physics by annie marie heart (1,205 points) [ no revision ]
The proof of this can be found in Wolf's "Spaces of Constant Curvature", for instance.

This post imported from StackExchange Physics at 2020-11-26 15:41 (UTC), posted by SE-user Slereah

1 Answer

+ 2 like - 0 dislike

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.

However, these are in the first instance topological properties, but we are interested in the isometries and thus the (pseudo-)Riemannian structure. The connection to this is given by the natural metric provided by the Killing form if the total group (numerator group) is a semi-simple Lie group. This induces a metric on the quotient space invariant under the total group, raising the total group to the isometry group. Thus, this quotient construction naturally provides a (pseudo-)Riemannian space symmetric under the total group.

I hope this can help you with this issue. For a more detailed treatment of homogeneous spaces see for example this. Cheers!

This post imported from StackExchange Physics at 2020-11-26 15:41 (UTC), posted by SE-user Johnny Longsom
answered Oct 5, 2020 by Johnny Longsom (20 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights