Totally antisymmetric Riemann tensor

Question

Is it possible for a Riemannian manifold to have a non-vanishing Riemann tensor that is totally antisymmetric in the four indices? Of course, the antisymmetry would imply that the Ricci tensor vanishes, i.e. the manifold is Ricci flat.

Answer 1

The Riemannian curvature tensor $R$ is defined via $R(X,Y,Z,W) = g(R(X,Y)Z,W)$, where $g$ is a Riemannian metric. If $R$ was totally skew-symmetric, the first Bianchi identity would imply \begin{align*} 0 & = R(X,Y,Z,{}\cdot{})+R(Y,Z,X,{}\cdot{})+R(Z,X,Y,{}\cdot{}) \\ & = 3R(X,Y,Z,{}\cdot{}) \\ & = 3g(R(X,Y)Z,{}\cdot{}). \end{align*} Non-degeneracy of the metric $g$ would then give $R(X,Y)Z = 0$ for all $X,Y,Z$, i.e. $R=0$.