Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 14 / 5 / Solution

A Riemannian metric is a smooth positive-definite symmetric inner product $g_x$ on every tangent space. A Levi-Civita connection is a connection on the tangent bundle which is torsion-free, ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$, and metric-compatible,Apply this identity with the cyclic triples $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and replace differences of covariant derivatives by Lie brackets. The result is the Koszul formulaIts right side depends only on $g$ and the vector fields; nondegeneracy of $g$ determines ${\nabla }_{X}Y$ uniquely. This proves uniqueness.

For existence, in a coordinate chart defineThis is linear over smooth functions in $X$ and obeys the Leibniz rule in $Y$. Symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ proves zero torsion, and direct substitution giveswhich proves metric compatibility. On chart overlaps the two resulting connections satisfy the same intrinsic Koszul identity and therefore agree by uniqueness. They glue globally. Thus the existence and uniqueness of the Levi-Civita connection holds on every Riemannian manifold.

Fix the curvature conventionFor coordinate vectors $e_i$, define the Riemann curvature tensor components by $R_{ij,kl}=g(R(e_i,e_j)e_l,e_k)$. In these indices the Ricci and scalar contractions areExplicitly, $R_{ij,kl}=g_{ka}{R}^a{}_{lij}$, whereThese expressions define Ricci curvature and scalar curvature without ambiguity about which indices are contracted. With this convention a round unit $m$-sphere has $\mathrm{Ric}=(m-1)g$ and $s=m(m-1)$.

If $\mathrm{Ric}=\lambda g$ with a constant real $\lambda$, tracing gives $\boxed{s=m\lambda }$, so every Einstein metric has constant scalar curvature. The converse is not valid in arbitrary dimension. On $S^2\times S^1$ with the product Riemannian metric of the unit round sphere and a circle, the connection splits between factors. The mixed curvatures and circle curvature vanish, while the sphere has curvature one. HenceA scalar multiple of the whole metric would require $\lambda =1$ on the sphere directions and $\lambda =0$ on the circle direction, which is impossible. This proves that constant scalar curvature does not imply an Einstein metric, and supplies a counterexample to the unrestricted requested equivalence.

In dimension two, the curvature tensor is determined by the Gaussian curvature $K$:There the valid equivalence isIn dimension one both curvature and scalar curvature vanish identically. Without a two-dimensional restriction or another added geometric hypothesis, the general converse cannot be proved.