= Solution
A <Levi-Civita connection> is a connection on the <tangent bundle> which is <torsion-free> and compatible with the <Riemannian metric>. Its defining identities are
$$
\nabla_XY-\nabla_YX=[X,Y],\qquad
Xg(Y,Z)=g(\nabla_XY,Z)+g(Y,\nabla_XZ).
$$
To obtain uniqueness, write the metric identity for $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and use the torsion identity to interchange the derivatives. This gives the <Koszul formula>
$$
\begin{aligned}
2g(\nabla_XY,Z)
={}&Xg(Y,Z)+Yg(Z,X)-Zg(X,Y)\\
&-g(X,[Y,Z])+g(Y,[Z,X])+g(Z,[X,Y]).
\end{aligned}
$$
The metric is nondegenerate, so these <inner products> determine $\nabla_XY$ uniquely.
For existence, let $K(X,Y,Z)$ denote the right-hand side. The bracket rules $[X,fY]=f[X,Y]+X(f)Y$ and $[fX,Y]=f[X,Y]-Y(f)X$ give, by cancellation of all extra derivatives,
$$
\begin{aligned}
K(fX,Y,Z)&=fK(X,Y,Z),\\
K(X,fY,Z)&=fK(X,Y,Z)+2X(f)g(Y,Z),\\
K(X,Y,fZ)&=fK(X,Y,Z).
\end{aligned}
$$
Thus the nondegenerate metric defines a unique smooth <vector field> $\nabla_XY$ from $2g(\nabla_XY,Z)=K(X,Y,Z)$, and these identities give the connection rules. In coordinates the construction reads
$$
\boxed{\Gamma^k_{ij}=\frac12g^{k\ell}
\big(\partial_i g_{j\ell}+\partial_jg_{i\ell}-\partial_\ell g_{ij}\big).}
$$
The coefficients are smooth and symmetric in $i,j$, proving zero torsion. Substituting them gives
$$
g_{\ell k}\Gamma^\ell_{ij}+g_{j\ell}\Gamma^\ell_{ik}=\partial_i g_{jk},
$$
which is exactly <metric compatibility>. Since the construction was given intrinsically by $K$, the local formulas agree on overlaps. This proves <existence and uniqueness of the Levi-Civita connection> on every <Riemannian manifold>.
Define the <Riemann curvature tensor> by
$$
R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.
$$
The connection rules make this expression tensorial in all three arguments. To fix the component convention, throughout this solution use
$$
R_{ijkl}=g\big(R(e_i,e_j)e_l,e_k\big).
$$
With this convention, the unit <sphere> has $R_{ijkl}=g_{ik}g_{jl}-g_{il}g_{jk}$ and positive <Ricci curvature>. In a coordinate frame, the components are obtained from
$$
\big(R(\partial_i,\partial_j)\partial_l\big)^a
=\partial_i\Gamma^a_{jl}-\partial_j\Gamma^a_{il}
+\Gamma^a_{ib}\Gamma^b_{jl}-\Gamma^a_{jb}\Gamma^b_{il}.
$$
The algebraic curvature symmetries are
$$
\boxed{R_{ijkl}=-R_{jikl}=-R_{ijlk},\qquad R_{ijkl}=R_{klij},}
$$
together with the <first Bianchi identity>
$$
R(X,Y)Z+R(Y,Z)X+R(Z,X)Y=0,
\qquad R_{ijkl}+R_{jlki}+R_{likj}=0.
$$
The first antisymmetry follows from the definition. <Metric compatibility> gives $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$, giving the second. Torsion-freeness reduces the cyclic sum to the <Jacobi identity> for <Lie brackets of vector fields>; pair interchange follows algebraically from these antisymmetries and the cyclic identity. These explain the identities rather than choosing unrelated sign conventions.
Define the <Ricci curvature> by the <tensor contraction>
$$
\operatorname{Ric}(Y,Z)=\operatorname{tr}\{X\mapsto R(X,Y)Z\},
\qquad \operatorname{Ric}_{jl}=g^{ik}R_{ijkl}.
$$
This is a <bilinear form>, since $R$ is tensorial and <trace> is linear. At a point choose an orthonormal <basis>. Pair interchange gives
$$
\operatorname{Ric}_{lj}=\sum_iR_{ilij}
=\sum_iR_{ijil}=\operatorname{Ric}_{jl},
$$
so \b[<Ricci curvature> is symmetric].
For the final determination in dimension three, an algebraic curvature tensor is determined by its entries in the three unordered index pairs $12,13,23$: antisymmetry removes repeated indices, and pair interchange makes the resulting $3\times3$ array symmetric. It therefore has at most six independent entries. Write
$$
K_{12}=R_{1212},\quad K_{13}=R_{1313},\quad K_{23}=R_{2323}.
$$
The diagonal Ricci components give
$$
\operatorname{Ric}_{11}=K_{12}+K_{13},\quad
\operatorname{Ric}_{22}=K_{12}+K_{23},\quad
\operatorname{Ric}_{33}=K_{13}+K_{23},
$$
so, for example, $K_{12}=(\operatorname{Ric}_{11}+\operatorname{Ric}_{22}-\operatorname{Ric}_{33})/2$, and the other two follow cyclically. The three off-diagonal entries are also recovered explicitly:
$$
R_{1323}=\operatorname{Ric}_{12},\qquad
R_{1223}=-\operatorname{Ric}_{13},\qquad
R_{1213}=\operatorname{Ric}_{23}.
$$
Thus every independent curvature entry is determined by Ricci; this proves injectivity, not merely a dimension count.
An invariant formula making the reconstruction explicit is, with $S=\operatorname{tr}_g\operatorname{Ric}$,
$$
\boxed{\begin{aligned}
R_{ijkl}={}&g_{ik}\operatorname{Ric}_{jl}+g_{jl}\operatorname{Ric}_{ik}
-g_{il}\operatorname{Ric}_{jk}-g_{jk}\operatorname{Ric}_{il}\\
&-\frac S2(g_{ik}g_{jl}-g_{il}g_{jk}).
\end{aligned}}
$$
To verify it, the right-hand side has the two antisymmetries, pair interchange and the cyclic identity by symmetry of $g$ and Ricci. Contracting its first and third indices in dimension three gives
$$
3\operatorname{Ric}_{jl}+Sg_{jl}-2\operatorname{Ric}_{jl}-Sg_{jl}
=\operatorname{Ric}_{jl}.
$$
Subtract it from the original tensor. The difference has zero Ricci contraction, and the six recovered-entry formulas above force it to vanish. This completes the proof of <three-dimensional curvature from the Ricci tensor> in arbitrary coordinates.
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