Curvature tensor and symetries

Question

Let $M$ be a Riemannian manifold and $X,Y,Z,T \in \mathfrak{X}(M)$. Is true the next expression? $$ (X,Y,Z,T):=<R(X,Y)Z,T> = <R(X,Z)Y,T> =: (X,Z,Y,T). $$ I think that is not true in general, because, by Bianchi's Identity $$ (X,Y,Z,T) + (Y,Z,X,T) + (Z,X,Y,T) = 0. $$ We know that $(Z,X,Y,T) = -(X,Z,Y,T)$, so, $$ (X,Y,Z,T) + (Y,Z,X,T) = (X,Z,Y,T). $$

Answer 1

Try $M$ a surface with $K\ne 0$, and take $X=e_1$, $Y=e_2$, $Z=e_1$, $T=e_2$.