Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 115 (Differential Geometry)

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).

The integral of an exact differential form over the closed curve $S^1$ is zero by Stokes theorem, so the map is well defined on de Rham cohomology. It is surjective because the angular form $d\theta$ has integral $2\pi$. If a closed one-form $\alpha$ has zero integral, defineThe zero period makes this definition $2\pi$-periodic, and $dF=\alpha$. Thus the kernel is zero andis an isomorphism.

The overlap $U_0\cap U_1$ is ${\mathbb{C}}^{\ast }$ in the coordinate $z$, and radial projection $z\mapsto z/|z|$ is a deformation retraction onto $S^1$. The homotopy invariance of de Rham cohomology identifies its first de Rham cohomology with that of $S^1$, and the identification is compatible with integration around the unit circle. Hence the same integral map is an isomorphism.

Each $U_j$ is isomorphic to $\mathbb{C}$ and is therefore contractible. Every complex line bundle over a contractible paracompact space is trivial, so $L{|}_{U_0}$ and $L{|}_{U_1}$ admit the required trivializations.

Since $\beta =f^{-1}df$,The fundamental theorem of calculus and the chain rule then giveThus $F$ is constant.

Parametrize the unit circle by a loop $\gamma :[0,2\pi ]\to {\mathbb{C}}^{\ast }$. Part (d) and $f(\gamma (0))=f(\gamma (2\pi ))$ implyThe kernel of the complex exponential function is $2\pi i\mathbb{Z}$, sofor some $k\in \mathbb{Z}$. This integer is the winding number of $f{|}_{S^1}$.

For $h=z^{-k}f$,The scalar-valued form $h^{-1}dh$ is closed, andBy the period isomorphism from part (b), $[\alpha ]=0$ in complexified de Rham cohomology. Hence there is a smooth complex-valued function $\phi$ on $U_0\cap U_1$ with $\alpha =d\phi$.

Using $d\phi =h^{-1}dh$,The overlap ${\mathbb{C}}^{\ast }$ is connected, so $h{e}^{-\phi }=c$ for some $c\in {\mathbb{C}}^{\ast }$. Choose $C\in \mathbb{C}$ with $e^C=c$ and replace $\phi$ by $\phi +C$. Then $h=e^{\phi }$.

Let $({\rho }_0,{\rho }_1)$ be a smooth partition of unity subordinate to $(U_0,U_1)$. On $U_0$, extend ${\rho }_1\phi$ by zero away from the overlap, and on $U_1$ extend ${\rho }_0\phi$ similarly. DefineOn the overlap,and therefore

Changing the two local frames by the nowhere-zero functions ${\psi }_j$ replaces the transition function $f$ byWith the standard convention that the tautological bundle ${\mathcal{O}}_{{\mathbb{CP}}^1}(-1)$ has transition function $z$, its $k$th tensor power ${\mathcal{O}}_{{\mathbb{CP}}^1}(-k)$ has transition function $z^k$. ThusThis is the smooth classification of complex line bundles on the complex projective line by their winding number.

A principal connection on $P$ is a $G$-equivariant smooth splittingwhere $V_p$ is tangent to the $G$-orbit. Equivalently, it is a $\mathfrak{g}$-valued one-form $\mathcal{A}$ satisfying $\mathcal{A}({\xi }_P)=\xi$ and $R_g^{\ast }\mathcal{A}={\mathrm{Ad}}_{g^{-1}}\mathcal{A}$. Its curvature of a principal connection is

On $S^n\times S^n$, consider $q(x,y)=⟨x,y⟩$. At a point of $q^{-1}(0)$, its differential isThe tangent vector $(y,x)$ lies in $T_x{S}^n\oplus T_y{S}^n$ and has $dq(y,x)=2$, so zero is a regular value. The regular level set theorem makes $P=q^{-1}(0)$ a smooth submanifold. Its tangent space is

The points $(x,y)$ are ordered orthonormal two-frames, so $P$ is the Stiefel manifold $V_2({\mathbb{R}}^{n+1})$. Changing an orthonormal basis of the same plane gives the displayed free right $O(2)$-action. Since $O(2)$ is compact, the action is proper, and the quotient is the Grassmannian of unoriented two-planes. The Free proper Lie-group action theorem makesa principal $O(2)$-bundle.

LetThe vertical space is spanned by the fundamental vector $(y,-x)$ corresponding to $J\in \mathfrak{o}(2)$. DefineIt sends $(y,-x)$ to $J$, is $O(2)$-equivariant, and therefore is a principal connection. Its kernel consists exactly of those $(u,v)$ for whichThese are precisely the velocities satisfying the stated horizontality condition. Every tangent vector has the unique decompositioninto vertical and horizontal parts, proving uniqueness. This is the Canonical principal connection on the Stiefel bundle over a Grassmannian.

Write $\mathcal{A}=aJ$ with the scalar one-formThe Lie algebra $\mathfrak{o}(2)$ is abelian, so $[\mathcal{A}\wedge \mathcal{A}]=0$. MoreoverConsequentlyas required.

As a vector space, the Lie algebra of $G$ is $\mathfrak{g}=T_{e}G$. For $\xi \in \mathfrak{g}$, definewhere $L_g$ is left translation on a Lie group. This vector field is smooth and left-invariant because $(d{L}_h{)}_g{l}_{\xi }(g)=l_{\xi }(hg)$. The assignment $\xi \mapsto l_{\xi }$ is linear and injective by evaluation at $e$. Conversely, every left-invariant vector field $V$ satisfies $V(g)=(d{L}_g{)}_{e}V(e)$, so it equals $l_{V(e)}$. Hence the map is an isomorphism.

The Lie bracket of vector fields is bilinear, alternating, satisfies the Jacobi identity, and is natural under diffeomorphisms. Therefore the bracket of two left-invariant vector fields is left-invariant. DefineThenand the inherited bilinearity, alternation, and Jacobi identity make $T_{e}G$ a Lie algebra.

The one-parameter subgroup ${\gamma }_{\xi }(t)=\exp (t\xi )$ is the integral curve through the identity of $l_{\xi }$. Its defining initial-value problem is

The Levi-Civita connection is a torsion-free connection. Its torsion form isSince $T=0$,In coordinates this is the symmetry ${\Gamma }_{jk}^i={\Gamma }_{kj}^i$ of the Christoffel symbols.

The Levi-Civita connection is also a metric connection, soand similarly for the two cyclic permutations. Add the identities with leading derivatives $u$ and $v$, subtract the one with leading derivative $w$, and useCancellation and symmetry of $g$ giveThis is the Koszul formula written with the bracket terms on the right.

Take $u=v=l_{\xi }$ and $w=l_{\eta }$ in the identity from part (e). For a left-invariant metric, all three scalar products are constant and $[l_{\xi },l_{\xi }]=0$, soThe curve ${\gamma }_{\xi }$ has velocity $l_{\xi }$, hence is a geodesic exactly when ${\nabla }_{l_{\xi }}{l}_{\xi }=0$. Nondegeneracy of the inner product now givesThis is the geodesic-vector criterion for a left-invariant metric.