1. Nov 17, 2007

### Jim Kata

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 $$\mathbb{Z}_2$$
so picking your standard path a certain way you can get

$$U(\bar \Lambda )U(\Lambda ) = \pm U(\bar \Lambda \Lambda )$$

Now if you use the universal cover of SO(3,1), $$SL(2,\mathbb{C})$$ you can get

$$U(\bar \Lambda )U(\Lambda ) = U(\bar \Lambda \Lambda )$$

Now my question is by introducing the gradation are you basically just redefining your unitary transformations so that $$U(\bar \Lambda )U(\Lambda ) = U(\bar \Lambda \Lambda )$$ for SO(3,1) too?