The Lie algebra representation is not Hermitian. Accordingly, the corresponding (projective) representation of the group is never unitary. This is due to the non-compactness of the Lorentz group. In fact, a connected simple non-compact Lie group cannot have "any" nontrivial unitary finite-dimensional representations. There is a topological proof of this. Let , where is finite-dimensional, be a continuous unitary representation of the non-compact connected simple Lie group . Then where is the compact subgroup of consisting of unitary transformations of . The kernel of is a normal subgroup of . Since is simple, is either all of , in which case is trivial, or is trivial, in which case is faithful. In the latter case is a diffeomorphism onto its image, and is a Lie group. This would mean that is an embedded non-compact Lie subgroup of the compact group . This is impossible with the subspace topology on since all "embedded" Lie subgroups of a Lie group are closed If were closed, it would be compact, and then would be compact, contrary to assumption.