Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 (Differential Geometry)

An $m$ by $n$ matrix has matrix rank $m$ exactly when at least one of its $m$ by $m$ minors has nonzero determinant. For each multi-index $I$, the setis open in the vector space of all real $m$ by $n$ matrices, because the determinant is continuous. Their union is $X_{m,n}$, so $X_{m,n}$ is itself open in ${\mathbb{R}}^{mn}$. It therefore inherits the standard smooth manifold structure and has dimension $mn$. This is the full-row-rank matrix manifold.

Define the smooth mapAt $(M,v)$ its derivative in the direction $(A,w)$ isThe restriction to variations $(0,w)$ is the surjective linear map $M:{\mathbb{R}}^n\to {\mathbb{R}}^m$, since $M$ has full row rank. Thus $0$ is a regular value of $\Phi$. The regular level set theorem now shows thatis an embedded submanifold of $X_{m,n}\times {\mathbb{R}}^n$, of dimension $mn+n-m$.

Fix $I$ and write $I^{\prime }$ for the complementary column indices. Over $U_I$, decompose $M=(M_I,M_{I^{\prime }})$ and $v=(v_I,v_{I^{\prime }})$. The equation $Mv=0$ is equivalent toMatrix inversion is smooth on the invertible matrices, sois a smooth, fiberwise-linear trivialization $U_I\times {\mathbb{R}}^{n-m}\to {\pi }^{-1}(U_I)$. The sets $U_I$ cover the base, proving that $\pi :E_{m,n}\to X_{m,n}$ is a vector bundle of rank $n-m$. It is the kernel bundle of a constant-rank family of linear maps, and its rank also follows from the rank-nullity theorem.

For $x\in S^2\subset {\mathbb{R}}^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a smooth map $F:S^2\to X_{1,3}$. Its pulled-back fiber isso the evident fiberwise identity gives an isomorphism of vector bundles $F^{\ast }{E}_{1,3}\cong T{S}^2$.

If $E_{1,3}$ were a trivial vector bundle, its pullback would be trivial. A trivial rank-two bundle has a nowhere-zero section, whereas the assumed form of the Hairy ball theorem says that the tangent bundle $T{S}^2$ does not. Hence $E_{1,3}$ is nontrivial.

Give $\partial X$ the outward-normal-first boundary orientation. The Generalized Stokes theorem states that, for every compactly supported $(n-1)$-form $\omega$,

Choose an oriented coordinate cover by charts into ${\mathbb{R}}^n$ or the half-space ${\mathbb{H}}^n=\{x^1\ge 0\}$, and choose a partition of unity $({\rho }_i)$ subordinate to it. Since the family is locally finite and $\omega$ has compact support, only finitely many ${\rho }_i\omega$ are nonzero. It is therefore legitimate to write both integrals as finite sums and prove the identity for a form supported in one chart.

In an interior chart the integral of an exact compactly supported top form is zero by the fundamental theorem of calculus. In a boundary chart writeIntegrating $d\omega =({\sum }_j{\partial }_j{a}_j)d{x}^1\wedge \cdots \wedge d{x}^n$ coordinate by coordinate kills every tangential derivative. The normal derivative leaves precisely the restriction to $x^1=0$, with the sign selected by the outward-normal-first convention. This is ${\int }_{\partial X}{F}^{\ast }\omega$, proving the theorem.

Let $({\theta }^1,\dots ,{\theta }^n)$ be the dual basis of the positively oriented orthonormal basis $(v_1,\dots ,v_n)$. By the definition of the Riemannian volume form,The interior product of a differential form with the outward unit normal isThe vectors $(v_2,\dots ,v_n)$ form a positive orthonormal frame of $T\partial X$ by the outward-normal-first boundary orientation. Consequently the pullback of the last display is the positive unit boundary volume form:

On the Euclidean unit ball $D^n$,The outward unit normal along $S^{n-1}$ is the radial vector field $N={\sum }_i{x}^i{\partial }_i$, so part b givesLet $R={\sum }_i{x}^i{\partial }_i$ on the ball and put $\beta ={\iota }_R{\omega }_X$. Direct use of the exterior derivative gives $d\beta =n{\omega }_X$. Therefore the Generalized Stokes theorem yieldsThis proves the volume of a Euclidean unit sphere formula.

A local flow of $v$ is a smooth family ${\Phi }^t$ defined near $\{0\}\times X$ such that ${\Phi }^0=\mathrm{id}$, ${\partial }_t{\Phi }^t(x)=v({\Phi }^t(x))$, and ${\Phi }^{s+t}={\Phi }^s\circ {\Phi }^t$ whenever defined. The Lie derivative of a differential form isCartan's magic formula is

For $\xi$ in the Lie algebra $\mathfrak{g}$, set ${\Phi }^t=R_{\exp (t\xi )}$. The one-parameter subgroup law gives the flow law, andso this is the global flow of the left-invariant vector field $l_{\xi }$.

For $f(g)=R_g^{\ast }\alpha$, every tangent vector at $g$ is the initial velocity of $g\exp (t\xi )$ for some $\xi$. Since $R_{g\exp (t\xi )}=R_{\exp (t\xi )}\circ R_g$,Thus all ${\mathcal{L}}_{l_{\xi }}\alpha$ vanish exactly when every derivative of $f$ vanishes. This is equivalent to $f$ being a locally constant function.

Suppose first that the left-invariant 1-form $\alpha$ is closed. The scalar function $\alpha (l_{\xi })$ is constant for every $\xi$, because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that $g\mapsto R_g^{\ast }\alpha$ is locally constant. Since $G$ is connected, it is constant, and its value at the identity is $\alpha$. Hence $R_g^{\ast }\alpha =\alpha$ for every $g$, so $\alpha$ is bi-invariant.

Conversely, if $\alpha$ is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of $\alpha (l_{\xi })$ give ${\iota }_{l_{\xi }}d\alpha =0$. The left-invariant vector fields span every tangent space, so $d\alpha =0$.

Connectedness cannot be omitted. The orthogonal group $O(2)$ has an Abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.

The differential-forms version of the Frobenius theorem says that a constant-rank distribution $D={\bigcap }_i\ker {\alpha }^i$ is integrable exactly whenfor suitable 1-forms ${\beta }^i{}_j$. For a plane distribution $D=\ker \alpha$ on ${\mathbb{R}}^3$, this reduces to the integrability criterion for a plane distribution $\alpha \wedge d\alpha =0$.

For example, $\ker dz$ is integrable: its integral surfaces are the horizontal planes $z=\text{constant}$. In contrast, for $\alpha =dz-xdy$,Thus $\ker (dz-xdy)$ is not integrable; it is the standard contact structure on ${\mathbb{R}}^3$.

For a principal connection with connection form $\mathcal{A}$, the horizontal distribution of a principal connection is $H=\ker \mathcal{A}$, the complement of the tangent spaces to the $G$-orbits. Its curvature of a principal connection isIf $X,Y$ are horizontal vector fields, then $\mathcal{A}(X)=\mathcal{A}(Y)=0$, and henceThe Frobenius theorem says that $H$ is integrable exactly when $[X,Y]$ is horizontal for all horizontal $X,Y$. The displayed identity makes this equivalent to the vanishing of the horizontal two-form $\mathcal{F}$, hence to $\mathcal{F}=0$. Thus the horizontal distribution is integrable exactly for a flat principal connection.

The vertical tangent space of the trivial principal $\mathbb{R}$-bundle is spanned by ${\partial }_z$. Sinceprojects isomorphically onto the tangent space of $S^1\times \mathbb{R}$, it is always complementary to the vertical direction. It is the horizontal distribution of a principal connection precisely when it is invariant under the principal translations $z\mapsto z+a$. The horizontal lifts of ${\partial }_{\theta }$ and ${\partial }_t$ are unique, so this invariance is equivalent toThe functions must also be smooth and periodic in $\theta$, as is already required for them to be functions on the cylinder.

Under these conditions the connection form isIt sends ${\partial }_z$ to $1$, is translation-invariant, and has kernel $D$, proving sufficiency as well. Since the structure group $\mathbb{R}$ is abelian, the bracket term vanishes and

For $f=t\cos \theta +1$ and $g=\sin \theta$,The connection is therefore flat, so the Frobenius theorem gives horizontal sections locally.

A global section has the form $z=h(\theta ,t)$ and is horizontal exactly whenThe second equation gives $h=t\sin \theta +k(\theta )$, and the first then forces $k^{\prime }(\theta )=1$. No such $k$ is periodic on $S^1$, so no global horizontal section exists. Equivalently, the horizontal lift of one positive circuit in the $\theta$ direction changes $z$ bywhich is nontrivial holonomy.