Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 26G (Differential Geometry)

A point $y_0\in Y$ is a regular value of the smooth map $f:X\to Y$ if, for every $x\in f^{-1}(y_0)$, the derivativeis surjective. A value outside the image is regular vacuously.

For a nonvacuous example, takeThe value $1$ is regular: its fibre is the unit circle, and at every $(x,y)$ on that circle,is a nonzero linear functional and hence surjective.

Sard theorem states that the set of critical values of a smooth map between smooth manifolds has measure zero in the target. Equivalently, almost every target point is a regular value.

Assume first that $Y$ is connected; otherwise the definition is made separately on each connected component. Choose a regular value $y\in Y$ and define the degree modulo two by

The following course results make this well-defined.

The same transverse-preimage argument applied to a smooth homotopy proves homotopy invariance. Consequently the definition depends only on $f$, rather than on the chosen regular value.

For $v\in T_{p}S$, let ${\gamma }_v$ be the maximal geodesic satisfyingThe exponential map ison the domain of the exponential mapThis is an open, star-shaped neighbourhood of zero. In local coordinates the geodesic equation is a smooth ordinary differential equation, so smooth dependence on its initial position and velocity proves that $(p,v)\mapsto {\exp }_p(v)$ is smooth on its domain.

For both requested phenomena consider the embedded surfacewith $p=(1,0,0)$. Its geodesics are Euclidean straight lines for as long as they remain in $S$. The initial vector $v=(-2,0,0)$ would givewhich reaches the missing origin at $t=1/2$. Thus $v\notin {\mathcal{D}}_p$, and the domain is not all of $T_{p}S$.

The point $q=(-1,0,0)$ is not in the image of ${\exp }_p$. Any geodesic from $p$ to $q$ would have to be their unique Euclidean straight line, which passes through the omitted origin. Hence this exponential map is also not surjective, as summarized by exponential map of the punctured plane.

The induced metric makes the compact surface $S$ a compact Riemannian manifold. By the Hopf-Rinow theorem it is geodesically complete, so every ${\exp }_p$ is defined on all of $T_{p}S$. In particular, ${\exp }_p(V(p))$ is defined for every $p$.

The global exponential mapis smooth by smooth dependence of geodesics on initial data. Since $p\mapsto (p,V(p))$ is a smooth section of $TS$, their compositionis smooth.

For $0\le t\le 1$, defineCompleteness makes this a well-defined smooth homotopy, with ${\varphi }_0={\mathrm{id}}_S$ and ${\varphi }_1=\varphi$. By homotopy invariance of degree modulo two,This is the exponential displacement map on a compact surface.