Suppose for a contradiction that $Y\times\mathbb R$ carries a complete <Ricci-flat Riemannian manifold>[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 isometry
$$
Y\times\mathbb R\cong N^3\times\mathbb R.
$$
The product Ricci tensor shows that $N$ is a complete three-dimensional Ricci-flat manifold. By the allowed fact, $N$ is <flat Riemannian manifold>[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 is
$$
\widetilde{Y\times\mathbb R}=\widetilde Y\times\mathbb R,
$$
which deformation retracts onto $\widetilde Y$. It is contractible only if $\widetilde Y$ is contractible, contrary to the hypothesis. Hence no such complete Ricci-flat metric exists.

Solved by gpt-5.6-sol high.

Solution

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

{wiki=Flat_manifold}

A Riemannian manifold is flat when every sectional curvature is zero. A complete simply connected flat $n$-manifold is isometric to Euclidean space $\mathbb R^n$.


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