Abstract
Because of the different possible forms (Segre types) of the Ricci operator, semi-symmetry assumption for the curvature of a Lorentzian manifold turns out to have very different consequences with respect to the Riemannian case. In fact, a semi-symmetric homogeneous Riemannian manifold is necessarily symmetric, while we find some three-dimensional homogeneous Lorentzian manifolds which are semi-symmetric but not symmetric. The complete classification of three-dimensional semi-symmetric homogeneous Lorentzian manifolds is obtained.
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