Since the Lorentz group $SO(3,1)$ is non-compact, it doesn't have any finite dimensional unitary irreducible representation. Is this theorem really valid?
One can take complex linear combinations of hermitian angular momentum generator $J_i^\dagger=J_i$ boost generator $K_i^\dagger=-K_i$ to construct two hermitian generators $N_i^{\pm}=J_i\pm iK_i$. Then, it can be easily shown that the complexified Lie algebra of $SO(3,1)$ is isomorphic to that of $SU(2)\times SU(2)$. Since, the generators are now hermitian, the exponentiation of $\{iN_i^+,iN_i^-\}$ with real coefficients should produce finite dimensional unitary irreducible representations. The finite dimensional representations labeled by $(j_+,j_-)$ are therefore unitary.
$\bullet$ Does it mean we have achieved finite dimensional unitary representations of $SO(3,1)$?
$\bullet$ If the $(j_+,j_-)$ representations, are for some reason are non-unitary (why I do not understand), what is the need for considering such representations?
$\bullet$ Even if they are not unitary (for a reason I don't understand yet), they tell how classical fields transform such as Weyl fields, Dirac fields etc. So what is the problem even if they are non-unitary?