Recently, A. Gebarowski [1] has determined decomposable conformally recurrent Riemannian manifolds.Then, it is natural to ask how irreducible conformally recurrent Riemannian manifolds are determined.In fact, taking account of the de Rham decomposion of Riemannian manifolds, a complete simply connected Riemannian manifold is either irreducible or decomposable.
Jerzy PlebańskiMaciej Przanowski