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On the unit sphere $S^n\subseteq {\mathbb{R}}^{n+1}$ choose the outward unit normal $N(x)=x$. For tangent vector fields $X,Y$,soIf $X,Y$ are orthonormal, the Gauss equation givesThus the round unit sphere has sectional curvature one. Tracing over an orthonormal basis gives its scalar curvature

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.