Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 309 / 2 / b

The projection $X\mapsto X_{\parallel }$ is tensorial. Although a covariant derivative is not $C^{\infty }$-linear in its second argument, the extra term in ${\nabla }_{X_{\parallel }}(f{Y}_{\parallel })$ is proportional to $Y_{\parallel }$, whose contraction with the normal vanishes. Thus $K(X,Y)$ is $C^{\infty }$-linear in both arguments and defines a $(0,2)$ tensor on the hypersurface.

Since $n_a{Y}_{\parallel }^a=0$,Therefore the extrinsic curvature is

A hypersurface normal is locally proportional to the gradient of a defining function. The Frobenius theorem therefore impliesTaking the antisymmetric part of the formula in part i gives $K_{[ab]}=0$, henceEquivalently, torsion freedom gives $K(X,Y)-K(Y,X)=-n\cdot [X,Y]=0$ because the bracket of tangent vector fields is tangent.

The affine geodesic equation givesUnit normalization implies $n^b{\nabla }_a{n}_b=0$, and decomposition of the first index givesConsequently, at the intersection point,

Choose any tangent vector $V^a\in T_p\Sigma$ and let the affinely parametrized geodesic with initial tangent $V^a$ start at $p$. If $\Sigma$ is totally geodesic, then $U^a{n}_a$ remains zero. Its initial derivative is therefore zero. Part iii, with $U^a{n}_a=0$ at $p$, givesThis holds for every tangent $V$. Since $K$ is symmetric, the polarization identity implies