Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 115 (Differential Geometry)

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.

Let $\Omega$ orient $M\times \mathbb{R}$ and write $t$ for the coordinate on $\mathbb{R}$. Along $M\times \{0\}$, contraction with the transverse vector ${\partial }_t$ givesFor every basis of $T_{p}M$, adjoining ${\partial }_t$ gives a basis of $T_{(p,0)}(M\times \mathbb{R})$, so $\omega$ never vanishes. The top form $\omega$ orients $M$.

Assume $M\times N$ is oriented. Fix $q\in N$ and an ordered basis $(w_1,\dots ,w_n)$ of $T_{q}N$. Contracting the product orientation successively with the constant vectors $w_i$ along $M\times \{q\}$ produces a nowhere-zero top form on $M$. Hence $M$ is orientable. Fixing a point and a basis in $M$ gives an orientation of $N$ in the same way. This is the orientability of factors of a product manifold.

An immersed submanifold of $M$ is a manifold $N$ with an injective immersion $\iota :N\to M$. It is an embedded submanifold when $\iota$ is also a homeomorphism onto its image with the subspace topology, equivalently when it is a smooth embedding.

Because the inclusion $\iota :N\hookrightarrow M$ has constant rank $n$, the constant rank theorem gives coordinates $y^1,\dots ,y^n$ on $N$ and $x^1,\dots ,x^m$ on $M$ in whichSince $\iota$ is an embedding, the ambient chart can be shrunk so that it meets no other local sheet of $N$. It then satisfiesand $(x^1,\dots ,x^n)$ restricts to the required chart on $N$. This is a slice chart for an embedded submanifold.

Suppose the immersed subset is not embedded. Using the assumed embedded neighborhoods, there are $p\in N$, a relatively small coordinate neighborhood $U\ni p$, and points $p_j\in N\setminus U$ with $p_j\to p$ in $M$. Choose a bump function $g$ on $N$, supported in $U$, with $g(p)=1$. Then $g(p_j)=0$.

If $g=f{|}_N$ for some smooth $f$ on $M$, continuity gives both $f(p_j)\to f(p)=1$ and $f(p_j)=0$, a contradiction. Thus the extension hypothesis forces the subspace and manifold topologies to agree locally, and the immersion is an embedding. This proves the smooth extension criterion for an immersed submanifold.

If $X$ is tangent to $N$ and $f{|}_N=0$, then the restriction of $f$ to every curve in $N$ is zero, so $(Xf){|}_N=0$.

Conversely, use a slice chart for an embedded submanifold. The functions $x^{n+1},\dots ,x^m$ vanish on $N$. Writing $X=X^i{\partial }_i$, the hypothesis givesThus $X$ has no normal component and is tangent to $N$. Equivalently, $X$ preserves the vanishing ideal of an embedded submanifold.

The Lie bracket of vector fields is the commutator of the corresponding derivations:For $X=X^i{\partial }_i$ and $Y=Y^i{\partial }_i$,

Let $I(N)$ be the vanishing ideal of an embedded submanifold. Tangency says $X(I(N)),Y(I(N))\subseteq I(N)$. Hence, for $h\in I(N)$,so the criterion from part (d) makes $[X,Y]$ tangent to $N$.

In adapted coordinates, the tangential coefficients of the displayed bracket use only the restrictions of the tangential coefficients of $X,Y$ and their derivatives along $N$; all normal coefficients vanish there. Consequentlyso the restriction depends only on $X{|}_N$ and $Y{|}_N$. This is tangency under the Lie bracket.

For $\pi :E\to B$, the vertical space at $p\in E$ is $V_{p}E=\ker (d\pi {)}_p$. A horizontal subspace is a complementary subspaceA connection is a smooth choice of such complements, linear with respect to the vector-bundle structure. Equivalently, it is a smooth horizontal distribution as in horizontal subspace of a vector bundle connection.

Put $b=\pi (p)$ and $u=(d\pi {)}_{p}v$. Choose a vector field $X$ near $b$ with $X(b)=u$, multiply it by a bump function, and extend it by zero to $B$. At each $q\in E$, the restrictionis an isomorphism. Define $V(q)$ as the unique horizontal lift of $X(\pi (q))$. Smoothness of the horizontal distribution makes $V$ smooth, and uniqueness gives $V(p)=v$.

Take bundle charts ${\Phi }_i:E{|}_{U_i}\to U_i\times {\mathbb{R}}^r$ with transition functions $g_{ij}:U_i\cap U_j\to GL(r,\mathbb{R})$. Over $f^{-1}(U_i)$ defineTheir transition functions are $g_{ij}\circ f$, so they give the fibre producta smooth manifold and make its projection to $M$ a rank-$r$ pullback vector bundle.

A local section $s:U\to E$ determines the sectionof $f^{\ast }E$ over $f^{-1}(U)$. It is unique with second component $s\circ f$.

A covariant derivative is an $\mathbb{R}$-linear mapsatisfying $d_A(fs)=df\otimes s+f{d}_{A}s$. It is local: its value over an open set depends only on the restriction of the section there. A horizontal connection differentiates a section, projects its derivative vertically, and identifies the vertical tangent with the fibre.

In a local frame write $d_{A}s=ds+A_{i}sd{y}^i$. If $x^a$ are coordinates on $M$, the defining identity in the question forceswithThus $A^{\prime }=f^{\ast }A$, the pullback connection.

For local coordinates $x^1,\dots ,x^n$, a curve is an affinely parametrized geodesic exactly when it satisfies the geodesic equationwhere the Christoffel symbols of the Levi-Civita connection are

For $v$ near $0\in T_{p}M$, let ${\gamma }_v$ be the unique geodesic with ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The exponential map is ${\exp }_p(v)={\gamma }_v(1)$.

The constant initial velocity $0$ gives ${\exp }_p(0)=p$. Varying the initial velocity through $sv$ yields ${\exp }_p(sv)={\gamma }_v(s)$, soThus $(d{\exp }_p{)}_0$ is the identity. The inverse function theorem makes ${\exp }_p$ a diffeomorphism from a neighborhood of $0$ onto a neighborhood of $p$. Coordinates from an orthonormal basis of $T_{p}M$ are the geodesic normal coordinates.

In geodesic coordinates built from an orthonormal basis, $g_{ij}(0)={\delta }_{ij}$. Every radial curve has coordinates $x(t)=ta$, and substitution into the geodesic equation givesTaking $t=1$ proves $x^i{x}^j{\Gamma }_{ij}^k(x)=0$.

Conversely, assume the metric and Christoffel-symbol conditions. For every $a$ in the star domain, $x(t)=ta$ satisfies the geodesic equation and has initial velocity $a$ in an orthonormal coordinate frame. Uniqueness of solutions to ordinary differential equations gives ${\varphi }^{-1}(ta)={\exp }_p(ta)$ wherever defined. The star-domain assumption covers all of $U$, so $\varphi$ is precisely a geodesic coordinate chart. This is the Radial Christoffel-symbol criterion for geodesic coordinates.

The metric induces an inner product on decomposable $r$-forms byextended bilinearly. On an oriented Riemannian manifold, the Hodge star operator is characterized byOn $r$-forms in dimension $n$,

Put $\alpha =\delta \eta$. The hypothesis says that the exact two-form $\beta =d\alpha$ is anti-self-dual. Since $M$ is compact without boundary, Stokes theorem givesHence $d\alpha =0$ by the exact anti-self-dual form on a compact four-manifold argument. Since $\delta$ is the formal $L^2$ adjoint of $d$,Therefore $\boxed{\delta \eta =0}$.