Let $(M,g)$ be a pseudo-Riemannian manifold. We say that this is a space of constant curvature if the sectional curvature $K(p,S)=\frac{\langle R(X,Y)X,Y\rangle}{X^2Y^2-\langle X,Y\rangle^2}$ is constant for all nondegenerate $2$-directions $S\le T_pM$ for all $p\in M$ ($X,Y\in S$ is any basis for the $2$-direction).
It is well-known and is quite easy to show that an equivalent condition to this is that $$ R_{ijkl}=K\left(g_{ik}g_{jl}-g_{il}g_{jk}\right), $$where $K$ is a constant.
It follows from this that $\nabla R=0$ or in coordinates $R_{ijkl;m}=0$, i.e. the curvature tensor is parallel.
I always had the impression that this is also an equivalent condition, i.e $$ R_{ijkl;m}=0\Longleftrightarrow (M,g)\text{ is of constant curvature}, $$ however I do not know any proof and I skimmed through quite a few books on Riemannian geometry both modern and old-school and I did not see this statement or a proof of it.
So my question is, if one is given a pseudo-Riemannian space with parallel curvature tensor ($R_{ijkl;m}=0$) does it follow that the space is also a space of constant sectional curvature?
If so I'd like to know a reference where this is proven in this direction.
Best Answer
The implication does not go the other way. All locally symmetric spaces have parallel curvature and there are locally symmetric spaces of nonconstant sectional curvature, such as the Fubini-Study metric on $\mathbb{CP}^n$ for $n\ge 2$.