Symmetry of the Riemannian curvature tensor

Question

The Riemannian curvature tensor, in local coordinates, $R_{ijkl}$, has the following symmetries: $$R_{ijkl}+R_{jikl}=0;$$ $$R_{ijkl}+R_{ijlk}=0;$$ $$R_{ijkl}=R_{klij};$$ $$R_{ijkl}+R_{jkil}+R_{kijl}=0.$$ These algebraic identities give the degree of the freedom of curvature to be $\frac{1}{12}n^2(n^2-1)$. where $n$ is the dimension of manifold. I believe, but have trouble to show, that these identities completely describe the pointwise symmetries of the curvature tensor, i.e. given any $\frac{1}{12}n^2(n^2-1)$ numbers, we can find a Riemannian manifold such that the curvature tensor $R_{ijkl}$ evaluated at certain point, in some local coordinate, is of these particular numbers.

Answer 1

A positive answer (and much more!) can be found in

"Riemannian Metrics with the Prescribed Curvature Tensor and all Its Covariant Derivatives at One Point" by M. Berger and O. Kowalski, Mathematische Nachrichten, Volume 168 (1994) Issue 1, p. 209-225.

If your library does not have the journal, you can probably get it through some form of an interlibrary loan.