Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 24H / c / Solution

An isometry preserves the Levi-Civita connection. To justify this, transport the connection on $X$ by $d\psi$. It is torsion-free and preserves the metric on $Y$, since $\psi$ preserves the metric and Lie brackets. These two properties characterize the Levi-Civita connection uniquely: if two such connections differ by $A$, the tensor $a(U,V,W)=⟨A(U,V),W⟩$ is symmetric in its first two arguments and antisymmetric in its last two. Cycling these symmetries yields $a=-a$, hence $A=0$. ThereforeApplying this along $\gamma$ shows that the covariant acceleration of $\psi \circ \gamma$ is zero, so an isometry carries affinely parametrized geodesics to geodesics.

The same identity and the definition $R(U,V)W={\nabla }_U{\nabla }_{V}W-{\nabla }_V{\nabla }_{U}W-{\nabla }_{[U,V]}W$ show that $d\psi$ intertwines the Riemann curvature tensor. If $e_1,e_2$ is an orthonormal tangent basis, its image is orthonormal and

Both proposed converses are false. Take $X=Y={\mathbb{R}}^2$ with the Euclidean metric and $\psi (x)=2x$. A geodesic $p+tv$ maps to the geodesic $2p+2tv$, and both Gaussian curvatures vanish identically. Nevertheless $\psi$ multiplies lengths by two, so is not an isometry. This one counterexample answers both unheaded continuation requests.