TITLE: Equations satisfied by the Riemann curvature tensor
QUESTION [17 upvotes]: It is well known that the Riemann curvature tensor of a metric satisfies
\begin{eqnarray}
R_{jikl}=-R_{ijkl}=R_{ijlk},(1)\\
R_{klij}=R_{ijkl},(2)\\
R_{i[jkl]}=0 \mbox{(1st Bianchi identity)}.(3)
\end{eqnarray}
The question is whether there are more (may be non-linear and/or differential) equations satisfied by $R_{ijkl}?$ 
(I do not mention here the 2nd Bianchi identity $R_{ij[kl;m]}=0$ since it involves covariant derivatives and hence is equation not only on the tensor $R$ but also on the original metric $g$).
One can ask this question in the following more precise form.
Question 1. Let $R_{ijkl}$ be a collection of numbers satisfying (1)-(3) with the range of indices from 1 to $n$. Does there exist a metric $g$ in a neighborhood of 0 in $\mathbb{R}^n$ such that its curvature tensor at 0 is equal to the given collection of numbers?
Question 2. Let $R_{ijkl}(x)$ be a collection of smooth (real analytic?) functions in a neighborgood of 0 in $\mathbb{R}^n$ satisfying (1)-(3). Does there exist a metric in a neighborhood of 0 such that its curvature tensor is equal to the given collection of functions $R_{ijkl}(x)$?
UPDATE: As mentioned by Peter Michor below, the answer to Q1 is positive. As mentioned by him and Vladimir Matveev, in dimensions $n>3$, $g\mapsto R(g)$ is an overdetermined system of PDE. Hence the answer to Q2 is 'No' for $n>3$, though it is not clear to me how to write explicitly any constrain on the image of the map. In $n=2$ the answer is 'Yes' according to Vladimir Matveev. 
ANOTHER UPDATE: According to Robert Bryant's answer below, in $n=3$ the answer to Q2 is 'No' in general. However under some non-degeneracy assumptions it is 'Yes'.

REPLY [8 votes]: EDIT: Attempted to insert correct constants and fix indices. 
Might as well write down the explicit answer to Question 1. Everything below is written with respect to local co-ordinates.
Suppose you are given a curvature tensor $R_{ijkl}$ at one point. This satisfies the following symmetries:
$$
R_{ijkl} = R_{klij} = -R_{jikl} = -R_{ijlk}
$$
and the first Bianchi identity
$$
R_{ijkl} + R_{iklj} + R_{iljk} = 0.
$$
It is straightforward using the definition of the curvature tensor to show that if the Christoffel symbols vanish at the point $p$, then the curvature tensor is given (up to a not necessarily positive constant factor) at $p$ by
$$
R_{ijkl} = \frac{1}{2}(\partial^2_{ik}g_{jl} + \partial^2_{jl}g_{ik} - \partial^2_{il}g_{jk} - \partial^2_{jk}g_{il}).
$$
It is also straightforward to check that this is satisfied (up to a constant factor), if you set
$$
\partial^2_{ik}g_{jl}(p) = \frac{1}{3}(R_{ijkl} + R_{ilkj}).
$$
Note that this is not the unique solution. I leave the determination of the correct constant factors to the reader. Note that you do not even need to assume that $g_{ij}(p) = \delta_{ij}$. Only that $\partial_kg_{ij}(0) = 0$, which is equivalent to having the Christoffel symbols vanish at $0$.
To summarize, given any tensor $R_{ijkl}$ satisfying the symmetries stated above, any smooth metric of the form
$$
g_{jl}(x) = g_{jl}(0) + \frac{1}{3}(R_{ijkl} + R_{ilkj})x^ix^k + O(|x|^3)
$$
has curvature tensor $R_{ijkl}$ at $x = 0$.