$SO(1,3)$ has an infinite number of representations, classified by the Casimir invariant $p^2$.
$SO(3)$ also has an infinite number of representations, classified by the Casimir invariant $\mathbf J^2$.
Since representations are diffeomorphic if and only if their Casimir invariants are the same, we are justified in this method of classification.
In the case of $SO(3)$, the physical interpretation is:
$\mathbf J$ generates rotations of the particle’s rest frame.
$\mathbf J^2$, the total spin of a particle, is the dimension of the vector space in which we have chosen to embed the particle.
Now I am baffled by the fact that we use $\mathbf J^2$, i.e. total spin, to classify $SO(1,3)$. That's the wrong Lie group! How is this not nonsense?
$p^2$ is the correct Casimir invariant - what happened to that?
This post imported from StackExchange Physics at 2014-11-02 16:29 (UTC), posted by SE-user Dave
Why isn't $p^2$ sufficient? - it's a Casimir invariant, and so it should give us all the classification information (i.e. tell us if reps are diffeomorphic)!
Now, suppose that we do things correctly (i.e. discard $\mathbf J^2$) and use $p^2$ to classify representations.
Are there “fermions” or “bosons” corresponding to $m$ taking on half or whole integer values in this case?
Finally, the representation $m^2=3$ is not isomorphic to $m^2=\pi$ (because $p^2$ is a Casimir invariant). Same with $m^2=2$ and $m^2=2.00000001$. However, in most field theory textbooks, $m>0$ is treated as one case. It's all a blob to them. What?!!!