In my Quantum Field Theory class we were discussing the finite dimensional representations of the Lorentz group. We discussed the vector representation which acts on 4 vectors, and then also the spinor representation that acts on spinors. My issue is that both of these representations are 4 dimensional.
I'm used to dealing with SO(3) where the representations are labeled by half integers and there is a unique mapping from each label to the dimension of the representation. Spin (1/2) gives 2 dimensional matrices, Spin (1) gives 3 dimensional matrices, and so on. But it seems we have lost that uniqueness in the Lorentz group. The vector representation and the spinor representation have the same dimension. Does this have to do with the non compactness of the lorentz group?
Thanks.
Best Answer
Indeed, you have lost some sort of uniqueness of dimension, but not between the vector and the spinor representation: The vector representation of $\mathrm{SO}(1,3)$ is irreducible, while the four-dimensional Dirac-spinor representation is not - it is the sum of a left-chiral and a right-chiral Weyl representation.
In general, the finite-dimensional representations of the (connected component of the) Lorentz group are in bijection to the finite-dimensional representations of $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$. For the precise relation between $\mathrm{SO}(1,3)$ and $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$, see this answer by Qmechanic. The representation theory of $\mathfrak{su}(2)$ is precisely that of spin as we know it, and therefore a finite-dimensional representation of the Lorentz group is labeled by two half-integers $(s_1,s_2)$. If one examines the way the $\mathfrak{su}(2)$ algebras actually related to the Lorentz algebra, one finds that the total spin of such a representation should be $s_1+s_2$.
The representation space associated to $(s_1,s_2)$ is just $\mathbb{C}^{2s_1 +1}\otimes\mathbb{C}^{2s_2+1}$, i.e. we tensor the spin-$s_i$ representations with each other. Of course, this shows you that the dimension of the space is no longer unique for a given representation, even if its irreducible.
However, this has nothing to do with the non-compactness of the Lorentz group, it's simply because it's a little more complicated than the "easy" $\mathrm{SO}(3)$. For instance, the compact $\mathrm{SU}(2)\times\mathrm{SU}(2)$ shares the same finite-dimensional representation theory.