I decided to keep my original question. However, I'm having trouble only in a part of it (check NOTE)
Let's consider a Riemannian manifold $(M,g)$, with the Levi-Civita connection $\nabla$. I would like to know why is it that if $M$ has constant sectional curvature, then $\nabla R=0$, where $R$ is the curvature tensor.
Moreover, I read that $\nabla R=0$ does not imply that necessarily that $M$ has constant curvature. However, if $\dim(M)=2$, that holds. (I read in an online set of exercises)
Can anyone give solution/hints/references for this questions ? Thanks in advance...
NOTE : Meanwhile, I figured it out how to prove that if $M$ has constant sectional curvature, then$\nabla R=0$. However, it is not clear to me the second question : If $\dim(M)=2$ and $M$ is such that $\nabla R=0$, then $M$ has constant curvature. How ?
If $(M,g)$ has constant sectional curvature, its curvature tensor $R$ can be built up from a constant function, the metric $g$ and a Kronecker-$\delta$ by tensorial operations, see e.g. http://en.wikipedia.org/wiki/Ricci_decomposition . Since all these ingredients are parallel, so is $R$. In general $\nabla R=0$ is equivalent to $(M,g)$ being a locally symmetric space, and there are many examples of such spaces which do not have constant curvature, for example Grassmannians. In dimension $2$, the Riemann curvature can be built up from the Gauss curvature, the metric and a Kronecker-$\delta$ by tensorial operations. If you look at the explicit form for this, you see that $\nabla R=0$ if and only if the Gauss curvature is constant.