Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 4 / 24H (Differential Geometry)

The global Gauss-Bonnet theorem for the induced Riemannian metric isHere $K$ is the Gaussian curvature, $dA$ the induced surface-area element, and $\chi (X)$ the Euler characteristic. For a connected orientable compact surface of genus $g$, $\chi (X)=2-2g$. For several components, sum their contributions. There is no boundary-curvature term because $X$ has no boundary.

A parametrized curve is a geodesic when its covariant acceleration vanishes, ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$, for the induced Levi-Civita connection. In coordinates this isEquivalently, for an embedded surface its ordinary acceleration is normal to the tangent plane. The parameter is affine; arbitrary reparametrizations need not remain geodesics.

The local existence and uniqueness theorem says that for every $p\in X$ and $v\in T_{p}X$ there is a unique geodesic with $\gamma (0)=p$, $\dot{\gamma }(0)=v$, on a maximal open interval containing zero, depending smoothly on $(p,v)$. This follows from the local existence and uniqueness theorem for the smooth first-order system in $(\gamma ,\dot{\gamma })$. The exponential map is ${\exp }_p(v)={\gamma }_v(1)$ wherever the solution exists through time one. Its domain contains a neighborhood of zero, and $d({\exp }_p{)}_0=I$, since ${\exp }_p(tv)={\gamma }_v(t)$. Thus it is a local diffeomorphism near zero. Completeness is not assumed for a general surface; on a complete surface it is defined on all of $T_{p}X$.

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.