On the unit sphere $S^n\subseteq\mathbb R^{n+1}$ choose the outward unit normal $N(x)=x$. For tangent vector fields $X,Y$,
$$
0=X\langle Y,N\rangle
=\langle\overline\nabla_XY,N\rangle+\langle Y,X\rangle,
$$
so
$$
A(X,Y)=-\langle X,Y\rangle N.
$$
If $X,Y$ are orthonormal, the <Gauss equation> gives
$$
\langle R(X,Y)Y,X\rangle
=\langle A(X,X),A(Y,Y)\rangle-|A(X,Y)|^2=1.
$$
Thus the round unit sphere has <sectional curvature> one. Tracing over an orthonormal basis gives its <scalar curvature>
$$
\operatorname{Scal}_{S^n}=n(n-1).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 115 / 1 / c / Solution

Give $S^2\times\mathbb R$ the <Riemannian product> of the unit round metric and the Euclidean metric. It is complete and has infinite <diameter of a metric space>[diameter]. The round sphere has <scalar curvature> $2$, while the line has scalar curvature $0$; scalar curvature is additive under Riemannian products, so
$$
\operatorname{Scal}_{S^2\times\mathbb R}=2.
$$
Thus this manifold has a strictly positive uniform lower bound on scalar curvature but violates the conclusion of the <Bonnet-Myers theorem>. Its <Ricci curvature> vanishes in the $\mathbb R$ direction, showing precisely why a scalar-curvature bound is insufficient.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 131 / 2 / c / Solution

{title2=$\operatorname{Scal}$}
{wiki}

Scalar curvature is the complete trace of the Riemann curvature tensor:
$$
\operatorname{Scal}=\sum_{i,j}\langle R(e_i,e_j)e_j,e_i\rangle.
$$
An $n$-manifold of constant sectional curvature $K$ has $\operatorname{Scal}=n(n-1)K$.


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