Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 131 / 4

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.

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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.

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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.

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