Formulating the Poincare group and its double cover

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  • #1
redtree
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As direct/semi-direct products
Given the full Lorentz group is and the restricted Lorentz group is , the full Poincare group is and the restricted Poincare group is .

Given that , how might one formulate the full Poincare group as the direct/semi-direct product of ?

Would it be any of the following?:




For the full Lorentz group the double cover is , and for the restricted Lorentz group is , the double cover is .

And a similar question for the double cover of the full Poincare group?:

 
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  • #2
I don't think your characterization of as isn't quite correct. There is no direct action of on .

Rather, I believe your semi-direct product should be reversed as it is that is the normal subgroup. Recall that two inversions will result in a (regular or pseudo)rotation. Hence, the adjoint action of an inversion on a rotation will yield a rotation.
Rather: .
 
  • #3
My understanding is that is abelian and therefore a normal subgroup.
 
  • #4
There are several 's in you must be careful which one you are considering. Let me think out loud and see where this goes:

Simple example: . The action of the rotation group(s) on the translation subgroup (normal subgroup) is to rotate it as a vector. The quotient structure is defined by the short exact sequence: hence . The normality of is why the quotient is a group. All standard group theory.

Now when you reverse the quotient structure you get a manifold:
however this is not a group (the cosets fail to close) but rather the manifold of subgroups of identifiable with each fixed center point for a given rotation in space. This reversed quotient is in a sense dual to the translation subgroup as it is what gets translated by the adjoint action of the translators on rotators.

If we quotent out the central subgroup of we get the projective orthogonal group .

This is not how the special orthogonal subgroup is defined. Rather the special orthogonal group is obtained by removing all inversions (reflections in one direction in the vector irrep). This is a whole manifold of subgroups none of which are central or normal but are rotated by the subgroup.

But note that inversions acting adjointly on rotations are again rotations. Hence is a normal subgroup of .

Yep, is necessarily the image of the inverting and non-inverting orthogonal transformations with the non-inverting mapping to the identity (exactness of the sequence).

While we're here, we can consider the reverse sequence:

Here the quotient is not a group since the first mapping is necessarily (as a reversial of the above) not a whole-space inversion but it maps to one of the vector inversion subgroups in O(n) it is not a normal subgroup. I'm not clear on what is; it must incorporate the manifold of inversion subgroups in O(n) but also, I think, the rotations leaving it invariant. I think that will end up being homeomorphic to SO(n) but I'm guessing at this point as I'm now reaching a bit beyond my understanding of the topic.
 
  • #5
redtree said:
My understanding is that is abelian and therefore a normal subgroup.

Abelian subgroups do not need to be normal. For example, if is a transposition in with then is abelian but not normal in

On the other hand, index 2 subgroups, like are always normal.
 

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