A Ricci-semi-symmetric hypersurface of Euclidean space which is not semi-symmetric (Q2781352)

Let \((M,g)\) be a Riemannian manifold. Let \(R\) be the Riemann-Christoffel curvature tensor and \(S\) the Ricci curvature tensor of \(g\). Then \(R\) assigns any \((0,k)\) tensor \(T\) on \(M\) to a \((0,k+2)\) tensor \(R\cdot T\) defined by NEWLINE\[NEWLINE\begin{multlined} -(R\cdot T)(X_1,X_2, \dots, X_k,X,Y)=\\ =T\bigl(R(X,Y) X_1,X_2, \dots,X_k\bigr) +\cdots+ T\bigl(X_1,X_2, \dots, R(X,Y)X_k\bigr). \end{multlined}NEWLINE\]NEWLINE A Riemannian manifold \((M,g)\) is called semi-symmetric, if \(R\cdot R=0\). \((M,g)\) is called Ricci-semi-symmetric, if \(R\cdot S=0\). It is known that any locally symmetric manifold is semi-symmetric and that any semi-symmetric Riemannian manifold must be Ricci-semi-symmetric. A long-standing open problem is that whether for hypersurfaces in Euclidean space the two conditions \(R\cdot R=0\) and \(R\cdot S=0\) are equivalent. F. Defever shows that there exists a 5-dimensional Ricci-semi-symmetric hypersurface in Euclidean space which is not semi-symmetric. In this paper the authors give a simple and explicit example of such a hypersurface in Euclidean space \(E^{n+1}\) for every \(n>4\).