I Formulating the Poincare group and its double cover

[redtree]

TL;DR Summary

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?:
$$
$$

 

[jambaugh]

I don't think your characterization of O(3,1) as \mathbb{Z}_2\rtimes SO(3,1) isn't quite correct. There is no direct action of SO(3,1) on \mathbb{Z}_2.

Rather, I believe your semi-direct product should be reversed as it is SO(3,1) 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: O(3,1) = SO(3,1)\rtimes \mathbb{Z}_2.

 

[redtree]

My understanding is that $$ is abelian and therefore a normal subgroup.

 

[jambaugh]

There are several \mathbb{Z}_2's in O(p,n) you must be careful which one you are considering. Let me think out loud and see where this goes:

Simple example: ISO(3) = \mathbb{R}^3 \rtimes SO(3). 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: 1 \to \mathbb{R}^3 \to ISO(3) \to SO(3) \to 1 hence SO(3)=ISO(3)/\mathbb{R}^3. The normality of \mathbb{R}^3 is why the quotient is a group. All standard group theory.

Now when you reverse the quotient structure you get a manifold:
1 \to SO(3)\to ISO(3)\to \mathbb{R}^3 however this ISO(3)/SO(3) =\mathbb{R}^3 is not a group (the cosets fail to close) but rather the manifold of SO(3) subgroups of ISO(3) 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 \mathbb{Z}_2 subgroup of O(n) we get the projective orthogonal group PO(n).
1\to \mathbb{Z}_2 \to O(n)\to PO(n)\to 1
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 {Z}_2 subgroups none of which are central or normal but are rotated by the SO(n) subgroup.

But note that inversions acting adjointly on rotations are again rotations. Hence SO(n) is a normal subgroup of O(n).
1 \to SO(n)\to O(n)\to O(n)/SO(n) \to 1
Yep, O(n)/SO(n)= \mathbb{Z}_2 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:
1 \to \mathbb{Z}_2 \to O(n)\to Q \to 1
Here the quotient Q is not a group since the first mapping is necessarily (as a reversial of the above) not a whole-space inversion but it maps \mathbb{Z}_2 to one of the vector inversion subgroups in O(n) it is not a normal subgroup. I'm not clear on what Q 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.

 

[Infrared]

My understanding is that $\mathbb{Z}_2$ 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.