- #1
Jim Kata
- 204
- 10
Sorry, I really hate reading, and I do better by just asking stupid questions.
The lorentz group SO(3,1) is not simply connected so its unitary representation is in a projective space. It's fundamental group is [tex]\mathbb{Z}_2[/tex]
so picking your standard path a certain way you can get
[tex]U(\bar \Lambda )U(\Lambda ) = \pm U(\bar \Lambda \Lambda )[/tex]
Now if you use the universal cover of SO(3,1), [tex]SL(2,\mathbb{C})[/tex] you can get
[tex]U(\bar \Lambda )U(\Lambda ) = U(\bar \Lambda \Lambda )[/tex]
Now my question is by introducing the gradation are you basically just redefining your unitary transformations so that [tex]U(\bar \Lambda )U(\Lambda ) = U(\bar \Lambda \Lambda )[/tex] for SO(3,1) too?
The lorentz group SO(3,1) is not simply connected so its unitary representation is in a projective space. It's fundamental group is [tex]\mathbb{Z}_2[/tex]
so picking your standard path a certain way you can get
[tex]U(\bar \Lambda )U(\Lambda ) = \pm U(\bar \Lambda \Lambda )[/tex]
Now if you use the universal cover of SO(3,1), [tex]SL(2,\mathbb{C})[/tex] you can get
[tex]U(\bar \Lambda )U(\Lambda ) = U(\bar \Lambda \Lambda )[/tex]
Now my question is by introducing the gradation are you basically just redefining your unitary transformations so that [tex]U(\bar \Lambda )U(\Lambda ) = U(\bar \Lambda \Lambda )[/tex] for SO(3,1) too?