Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 / 1

An orientable smooth manifold is a smooth $n$-manifold that admits a smoothly varying orientation of its tangent spaces. Equivalently, it has an atlas whose coordinate-transition Jacobian determinants are positive, or a nowhere-vanishing smooth top-degree differential form.

The zero section $C=\{t=0\}\subset M$ is a circle. Its normal bundle is the real line bundle obtained fromso its clutching map reverses sign once around $C$. Its mod-two Euler class of a real line bundle, equivalently its first Stiefel–Whitney class, therefore satisfiesIf $M$ were orientable, the splittingand the orientation of the circle would orient ${\nu }_C$, forcing this mod-two Euler number to vanish. This contradiction proves that the Möbius band is nonorientable.

Choose a locally finite cover $(U_{\alpha })$ such that on every chart meeting $Y$ there is a smooth defining function $f_{\alpha }$ withand take $f_{\alpha }=1$ on charts disjoint from $Y$. On an overlap, the supplied division lemma extendssmoothly across $Y$. After shrinking the charts, this extension is nowhere zero. The identities $g_{\alpha \beta }{g}_{\beta \gamma }=g_{\alpha \gamma }$ make these functions transition functions for a real line bundle $L\to X$.

Choose local frames $e_{\alpha }$ with $e_{\beta }=g_{\alpha \beta }{e}_{\alpha }$. Then the local sectionsagree on overlaps and define a global section. Its zero set is exactly $Y$. Along $Y$, its vertical derivative is represented by the nonzero covector $d{f}_{\alpha }$, so $s$ is transverse to the zero section, as in the transverse intersection theorem. This is the defining line bundle of a properly embedded hypersurface.

Real line bundles over a paracompact space $X$ are classified byEuclidean space ${\mathbb{R}}^n$ is a contractible space, so its first cohomology vanishes and every real line bundle on it is trivial. Apply this to the defining bundle from part (c). In a global trivialization, $s$ is a smooth real function with $Y=s^{-1}(0)$ and $ds{|}_Y\ne 0$. Thus $ds$, or a metric-dual normal vector field, gives a global orientation of the normal line. Combining this with the standard orientation of ${\mathbb{R}}^n$ gives an orientation of $Y$ using the same normal-first convention as the outward-normal-first boundary orientation. Hence every properly embedded hypersurface in ${\mathbb{R}}^n$ is orientable.

Yes. For any $R>1$, the formuladefines the standard smooth embedding of the open Möbius band into ${\mathbb{R}}^3$. Replacing $(x,t)$ by $(x+2\pi ,-t)$ leaves the displayed point unchanged. The image is the interior of a compact Möbius strip, so the embedding fails to be a proper map; this is why it does not contradict part (d).