Is Minkowski space the only Poincare invariant space?

[Rearden]

Hi everyone,

I was wondering: if a space is invariant under Poincare transformations, does that mean it has to be Minkowski space? Or could it have some further isometries?
By the same token, if a space is invariant under the orthogonal transformations, does it have to be Euclidean?

I hope I've been clear; I don't really know how to start tackling this problem, so any help will be much appreciated.

Thanks!

 

[Fredrik]

You're asking if a smooth manifold with a metric is determined up to diffeomorphism by its isometry group? Hm, I don't know, but I would guess "yes". (Edit: Temporary insanity. See #7).

 

[Rearden]

Neat way of phrasing it! I'm also fairly sure it's a "yes" given the original context of the problem:

Poisson's equation (i.e. Newtonian gravity) is invariant under the Euclidean transformations. Because the space we see around us is manifestly Euclidean, I don't think the equation has any other isometries. But I can't prove it...

(Note: I'm trying my hardest to avoid deriving the symmetry group using brute-force Lie Theory)

 

[bcrowell]

You're asking if a smooth manifold with a metric is determined up to diffeomorphism by its isometry group? Hm, I don't know, but I would guess "yes".

Huh? I don't see how this can be right, unless I'm missing something. Let a manifold M have a totally empty isometry group, no symmetries at all. Then clearly it's not identical up to diffeomorphism to every other manifold with the same complete lack of symmetry.

if a space is invariant under Poincare transformations, does that mean it has to be Minkowski space? By the same token, if a space is invariant under the orthogonal transformations, does it have to be Euclidean?

I think so. The kind of possible exception that I could imagine would be the Euclidean plane versus a cylinder. Locally, they have the same group of isometries, but globally they don't.

The Poincare group contains transformations like reflections that can't be generated by series of infinitesimal actions, so they take distant points to distant points; because of this, when someone talks about the Poincare group, I would generally take that as being explicitly global. But experimentalists certainly do test Poincare invariance in the lab, which is local, and then, e.g., if they want to test full parity inversion, they have to test inversion in a point that lies inside the lab.

Or could it have some further isometries?

I doubt it. The plane/cylinder type of example involves *reducing* the global symmetry, not increasing it. Isn't 10 simply the maximum number of Killing vectors a spacetime can have? That doesn't prove that there can't be additional discrete symmetries, but I'm having a hard time imagining what they could be.

 

[bcrowell]

This http://en.wikipedia.org/wiki/Doubly_special_relativity could be an example of adding further symmetries to the symmetries of Minkowski space, but I guess it requires generalizing from regular geometry to noncommutative geometry, which I know nothing about.

 

[bcrowell]

"In a general N dimensional manifold with metric gab, the maximum num-
ber of independent Killing vectors is N(N + 1)/2. It can be shown that
maximally symmetric spaces (spaces that allow the maximum number of in-
dependent Killing vectors) are unique, which means they only differ by some coordinate transformation." -- http://www.math.toronto.edu/~colliand/426_03/Papers03/J_Bauer.pdf

 

[Fredrik]

Huh? I don't see how this can be right, unless I'm missing something. Let a manifold M have a totally empty isometry group, no symmetries at all. Then clearly it's not identical up to diffeomorphism to every other manifold with the same complete lack of symmetry.

Good point.

 

[atyy]

Huh? I don't see how this can be right, unless I'm missing something. Let a manifold M have a totally empty isometry group, no symmetries at all. Then clearly it's not identical up to diffeomorphism to every other manifold with the same complete lack of symmetry.

Can't you use the diffeomorphism to generate isometric manifolds (pullback)?

 

[bcrowell]

Hmm...I think Bauer's statement in #6 is false without adding another condition. Manifolds don't have to be connected. So for example, I can make a manifold M out of two nonintersecting lines. It has two independent Killing vectors, not N(N+1)/2=1 as he claims. So I guess this is an unexpected answer to the OP's question: "if a space is invariant under Poincare transformations, does that mean it has to be Minkowski space? Or could it have some further isometries?" If I make a 4-dimensional manifold out of two disconnected copies of Minkowski space, then it is invariant under Poincare transformations (in fact, under two independent copies of the Poincare transformations), but it's not Minkowski space, and it does have further isometries, such as an exchange of identities of the two disconnected spaces. I would imagine that Bauer's statement is true if you restrict to connected spaces.

 

[atyy]

Not sure this is relevant, but just in case bcrowell and Fredrik would like to comment
http://arxiv.org/abs/hep-th/9803165
http://arxiv.org/abs/math-ph/9904022

 

[arkajad]

Hi everyone,

I was wondering: if a space is invariant under Poincare transformations, does that mean it has to be Minkowski space?

That depends on what you mean by "space". If you mean just "a manifold", then you can take any homogeneous space P/H, where P is the Poincare group and H its closed subgroup.