The (m, n) 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 u : G → GL(V), where V is finite-dimensional, be a continuous unitary representation of the non-compact connected simple Lie group G. Then u (G) ⊂ U(V) ⊂ GL(V) where U(V) is the compact subgroup of GL(V) consisting of unitary transformations of V. The kernel of u is a normal subgroup of G. Since G is simple, ker u is either all of G, in which case u is trivial, or ker u is trivial, in which case u is faithful. In the latter case u is a diffeomorphism onto its image, u (G) ≅ G and u (G) is a Lie group. This would mean that u (G) is an embedded non-compact Lie subgroup of the compact group U(V). This is impossible with the subspace topology on u (G) ⊂ U(V) since all embedded Lie subgroups of a Lie group are closed If u (G) were closed, it would be compact, and then G would be compact, contrary to assumption.