Published online by Cambridge University Press: 22 January 2016
Let (M,g) be a Riemannian manifold and let R be its Riemannian curvature tensor. If (M, g) is a locally symmetric space, we have
(*) R(X,Y)·R = 0 for all tangent vectors X,Y
where the endomorphism R(X,Y) (i.e., the curvature transformation) operates on R as a derivation of the tensor algebra at each point of M. There is a question: Under what additional condition does this algebraic condition (*) on R imply that (M,g) is locally symmetric (i.e., ∇R = 0)? A conjecture by K. Nomizu [5] is as follows : (*) implies ∇R = 0 in the case where (M, g) is complete and irreducible, and dim M ≥ 3. He gave an affirmative answer in the case where (M,g) is a certain complete hypersurface in a Euclidean space ([5]).