Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 25H (Differential Geometry)

An embedded smooth $n$-dimensional manifold $X\subseteq {\mathbb{R}}^N$ has, around every point, a parametrization by an open subset of ${\mathbb{R}}^n$ which is a homeomorphism onto a relative neighbourhood, has rank $n$, and has smooth coordinate changes. Equivalently, after smooth ambient coordinates it is locally a coordinate $n$-plane. For a manifold with boundary use relatively open subsets of ${\mathbb{H}}^n=\{u_n\ge 0\}$ instead, with parametrizations and transitions smoothly extendible across the boundary. Points corresponding to $u_n=0$ form the boundary. Interior charts are ordinary manifold charts; a boundary point is not a corner.

A regular value $y$ of a smooth map $f:X\to Y$ is one for which the differential $d{f}_x:T_{x}X\to T_{y}Y$ is surjective at every $x\in f^{-1}(y)$. A value outside the image is regular vacuously. The condition concerns all preimages, not just one.

The regular level set theorem says that a regular level of a smooth map from an $m$-manifold to an $r$-manifold is an $(m-r)$-manifold. The corresponding sublevel-set theorem says that $\{F\le c\}$ is a manifold with boundary $F^{-1}(c)$ when $c$ is a regular value of a smooth real-valued function, with the interior $F<c$.

Apply this to $F(x)=|x{|}^2$ on ${\mathbb{R}}^{n+1}$. On $F^{-1}(1)$ its gradient $2x$ is nonzero. Thus $S^n$ is an $n$-manifold and $B^{n+1}$ is an $(n+1)$-manifold with boundary $\boxed{\partial B^{n+1}=S^n}$. Locally the implicit function theorem supplies $F-1$ as one coordinate; the ball occupies its nonpositive side, which is a half-space after reversing that coordinate. This also covers $n=0$: the sphere consists of two points and the ball is the interval with those endpoints.

The antipodal map is the restriction of the smooth linear map $-I$, so is smooth. For $n\ge 1$ orient the sphere by the outward-normal convention. Its degree is $(-1{)}^{n+1}$: the ambient determinant is $(-1{)}^{n+1}$, and the outward normal is carried to the outward normal at the antipodal point.

Choose a regular value of $f$, available by Sard theorem; if it is outside the image, the degree is zero already. Otherwise its finite preimage is partitioned into distinct antipodal pairs. Differentiating $f\circ h=f$ shows that their local degree signs obeyFor even $n$, the signs in each pair cancel, so $\mathrm{deg}f=0$. For odd $n$, each pair contributes $2$ or $-2$, so $\mathrm{deg}f$ is even. In neither case can its degree equal the identity's degree one. Since the degree of a map between oriented manifolds is invariant under smooth homotopy, $f$ is not smoothly homotopic to the identity. For $n=0$, the symmetry makes $f$ constant on the two-point discrete sphere; a continuous homotopy into this discrete space cannot change either point's image, so the conclusion still holds.