Fix the convention
$$
(\nabla_c\nabla_d-\nabla_d\nabla_c)X^a
=R^a{}_{bcd}X^b.
$$
Although a single covariant derivative of a vector is tensorial, the second derivative contains connection-dependent terms. Their antisymmetric difference cancels every second derivative of a coordinate change. More intrinsically, the map
$$
R(U,V)X=\nabla_U\nabla_VX-\nabla_V\nabla_UX-\nabla_{[U,V]}X
$$
is $C^\infty(M)$-linear in each of $U,V,X$, so it defines the <Riemann curvature tensor>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 309 / 2 / a / i / Solution

 Solution

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