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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.

An end of a noncompact topological space describes a coherent way to escape every compact subset. For a cofinal compact exhaustion, when one exists, $K_1\subseteq K_2\subseteq \cdots$, it is represented by nested connected components of the complements. A connected noncompact manifold is disconnected at infinity when it has more than one end.

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.

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.

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.

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.