A hypersurface normal is locally proportional to the gradient of a defining function. The <Frobenius theorem> therefore implies
$$
h_a{}^ch_b{}^d\nabla_{[c}n_{d]}=0.
$$
Taking the antisymmetric part of the formula in part i gives $K_{[ab]}=0$, hence
$$
\boxed{K_{ab}=K_{ba}}.
$$
Equivalently, torsion freedom gives $K(X,Y)-K(Y,X)=-n\mathbin\cdot[X,Y]=0$ because the bracket of tangent vector fields is tangent.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-309/2/b/ii/solution