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A connected complete Riemannian manifold that is disconnected at infinity contains a line in a Riemannian manifold. Choose points escaping in two components outside a fixed compact set and join them by minimizing geodesics. Each segment crosses the compact set. Parametrize from a crossing point; both parameter endpoints tend to infinity in opposite directions. Compactness of the unit tangent vectors over the crossing set gives a subsequence converging to an entire geodesic. Every finite subsegment is minimizing by continuity of distance, so the limit is a line.

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.

A line in a Riemannian manifold is a unit-speed geodesic $\gamma :\mathbb{R}\to M$ that minimizes globally:for all $s,t\in \mathbb{R}$. A connected noncompact manifold is disconnected at infinity if some compact set $K$ has a complement with at least two unbounded connected components.

Choose points $p_j$ and $q_j$ in two such components withThe Hopf-Rinow theorem supplies a length-minimizing geodesic ${\gamma }_j$ from $p_j$ to $q_j$. Its image must meet $K$, since otherwise it would connect the two different components of $M\setminus K$. Reparametrize so that ${\gamma }_j(0)=x_j\in K$. After taking a subsequence, compactness gives $x_j\to x\in K$ and the unit tangent vectors ${\dot{\gamma }}_j(0)$ converge to some unit $v\in T_{x}M$.

Both endpoint parameters tend to infinity because their distances from $K$ do. Smooth dependence of geodesics on initial data therefore makes ${\gamma }_j$ converge on every compact parameter interval to the complete geodesicEvery finite segment of every ${\gamma }_j$ minimizes length. Passing to the limit gives $d_g(\gamma (s),\gamma (t))=|s-t|$, so $\gamma$ is a line.

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

The Hadamard-Cartan theorem states that if a complete simply connected Riemannian manifold has nonpositive sectional curvature, then for every point $p$ its exponential mapis a diffeomorphism. In particular, the manifold is diffeomorphic to Euclidean space and is contractible.

Suppose for a contradiction that $Y\times \mathbb{R}$ carries a complete Ricci-flat metric. Since $Y$ is closed, the two subsets $Y\times (A,\infty )$ and $Y\times (-\infty ,-A)$ are different unbounded components outside the compact set $Y\times [-A,A]$. Thus $Y\times \mathbb{R}$ is disconnected at infinity and, by part (a), contains a line in a Riemannian manifold.

Its Ricci curvature is zero, so the Cheeger-Gromoll splitting theorem gives an isometryThe product Ricci tensor shows that $N$ is a complete three-dimensional Ricci-flat manifold. By the allowed fact, $N$ is flat, and hence so is $Y\times \mathbb{R}$.

The universal cover of a complete flat manifold is complete, simply connected, and has zero sectional curvature. The Hadamard-Cartan theorem therefore identifies it diffeomorphically with ${\mathbb{R}}^4$, so it is contractible. On the other hand, the universal cover of the product iswhich deformation retracts onto $\tilde{Y}$. It is contractible only if $\tilde{Y}$ is contractible, contrary to the hypothesis. Hence no such complete Ricci-flat metric exists.

If a connected closed manifold $M^3$ has a noncontractible universal cover, then $M\times \mathbb{R}$ has no complete metric making it a Ricci-flat Riemannian manifold. The product has two ends and hence contains a line in a Riemannian manifold for any complete metric. The Cheeger-Gromoll splitting theorem would split it isometrically as $N^3\times \mathbb{R}$. Its three-dimensional factor is Ricci-flat and therefore flat by three-dimensional curvature from the Ricci tensor. The complete simply connected cover of the product would be Euclidean four-space. But that cover is $\tilde{M}\times \mathbb{R}$, which retracts onto $\tilde{M}$, contradicting noncontractibility.