Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 19 / 3

A connected complete Riemannian manifold is disconnected at infinity when the complement of some compact set has at least two unbounded connected components. Equivalently it has at least two ends. A line in a Riemannian manifold is a unit-speed geodesic $\ell :\mathbb{R}\to M$ satisfying $d(\ell (s),\ell (t))=|s-t|$ for every $s,t$.

Let $K$ be the compact separating set, enlarged inside a fixed compact metric ball if needed for bounds. Choose $x_j,y_j$ in two different unbounded components of $M\setminus K$, with both distances from $K$ tending to infinity. By the Hopf-Rinow theorem, join them by unit-speed minimizing geodesics. Every such segment meets $K$, since a path avoiding it cannot join its different components. Reparametrize a segment so that one crossing point is at time zero. Its parameter interval is $[-a_j,b_j]$, where $a_j,b_j\to \infty$ by the distances from the endpoints to $K$.

The unit tangent bundle over $K$ is compact, so a subsequence of the initial position/velocity pairs converges. Continuous dependence of the geodesic equation on its initial data, and completeness, give a limiting geodesic defined on all of $\mathbb{R}$. For any fixed $s<t$, the approximating segments are defined on $[s,t]$ for large $j$ and satisfy $d({\gamma }_j(s),{\gamma }_j(t))=t-s$. Continuity of the distance passes this equality to the limit. Thus the limiting geodesic is a line, proving a line from disconnection at infinity.

The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product $N\times \mathbb{R}$ with the Euclidean metric on the line. The factor $N$ is complete and has nonnegative Ricci curvature. Under the isometry one can take the given line to $\{q\}\times \mathbb{R}$, up to translation and reversal. A line means global minimization between every pair of its points, not merely a complete geodesic.

Suppose, for a contradiction, that $X=M\times \mathbb{R}$ has a complete metric with zero Ricci curvature. Since $M$ is closed and connected, the complement of $M\times [-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are compact by the Hopf-Rinow theorem. Thus $X$ is disconnected at infinity, and part (a) supplies a line. The Cheeger-Gromoll splitting theorem gives an isometryThe Riemannian product curvature formula makes $N$ complete and Ricci-flat. A three-dimensional Ricci-flat Riemannian manifold is flat: for any orthonormal triple, the three trace equations are ${\mathrm{Ric}}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2=0$. Every plane can occur in such a triple, so every sectional curvature vanishes, and hence so does the Riemann curvature tensor.

The complete simply connected universal cover of $N$ is therefore Euclidean ${\mathbb{R}}^3$, and that of $X$ is Euclidean ${\mathbb{R}}^4$. But the topological universal cover of $X$ is $\tilde{M}\times \mathbb{R}$. Its deformation retraction onto $\tilde{M}\times \{0\}$ shows that contractibility of this cover would imply contractibility of $\tilde{M}$, contrary to the hypothesis. Therefore no complete Ricci-flat metric exists on $M\times \mathbb{R}$. Notice that no product assumption was made about the hypothetical original metric.