Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 115 (Differential Geometry)

A submersion is a smooth map between manifolds $F:X^n\to Y^m$ for whichis surjective at every $p\in X$.

The local submersion theorem says that around every $p\in X$ there are coordinates $(x^1,\dots ,x^n)$ centered at $p$ and $(y^1,\dots ,y^m)$ centered at $F(p)$ in whichTo prove it, surjectivity lets us choose $m$ source coordinates such that $d{F}^1,\dots ,d{F}^m$ are independent. Complete them by source coordinates $x^{m+1},\dots ,x^n$ and defineThe derivative of $G$ is invertible at $p$, so the inverse function theorem makes $G$ a local coordinate system; in these coordinates $F$ is the displayed projection.

For $q\in Y$, the fiber is locally given byThese are slice coordinates, so $F^{-1}(q)$ is an embedded submanifold of dimension $n-m$.

Choose a Riemannian metric on $X$, using a partition of unity if necessary. For each $p$, letSince $F$ is a submersion, $D_{p}F{|}_{H_p}:H_p\to T_{F(p)}Y$ is an isomorphism. Its inverse depends smoothly on $p$, sodefines a smooth vector field on $X$ satisfying $D_{p}F(v(p))=w(F(p))$. The formula also gives $v(p)=0$ whenever $w(F(p))=0$.

Choose a coordinate ball $W$ around $q_1$ and a smaller convex coordinate ball $U$ whose closure lies in $W$. For $q_2\in U$, let $a$ be the constant coordinate vector from $q_1$ to $q_2$. Choose a smooth bump function $\rho$ supported in $W$ and equal to one on $U$, and define in the chartextending it by zero outside $W$. This is a compactly supported smooth vector field. The segment $q_1+ta$ stays in $U$, where $w=a$, so uniqueness for ordinary differential equations givesIn particular, ${\Psi }^1(q_1)=q_2$.

Fix $q_1$ and choose $U$ as in part c. For $q_2\in U$, choose the compactly supported $w$ moving $q_1$ to $q_2$, and lift it by part b to $v$. The support of $v$ is contained inwhich is compact because $F$ is a proper map. Hence $v$ is compactly supported and complete. If ${\Phi }^t$ and ${\Psi }^t$ are the flows of $v$ and $w$, thenUniqueness of integral curves givesTherefore the diffeomorphism ${\Phi }^1$ maps $F^{-1}(q_1)$ onto $F^{-1}(q_2)$. Every equivalence class is open. Its complement, being a union of the other open classes, is also open; thus each class is clopen. If $Y$ is connected, there is only one class. This proves the fiber-diffeomorphism conclusion of the Ehresmann fibration theorem.

Properness is essential. The projectionis a submersion but is not proper. Its fiber over $x\ne 0$ is diffeomorphic to $\mathbb{R}$, whereas its fiber over zero is $\mathbb{R}\setminus \{0\}$ and has two connected components.

Merely requiring every fiber to be a submanifold is also insufficient. The surjective maphas every fiber finite and therefore a zero-dimensional embedded submanifold. Some regular values have three preimages and others have one, so the fibers are not all diffeomorphic. The map fails to be a submersion at $x=\pm 1/\sqrt{3}$.

In local coordinates, forthe exterior derivative isApplying $d$ again gives symmetric second partial derivatives contracted with the antisymmetric wedge $d{x}^k\wedge d{x}^j$, hence $d^2=0$. Expanding coefficients and moving $d\alpha$ past a degree-$r$ form gives the graded Leibniz rule

For a function $f$ and smooth $F:X\to Y$, the chain rule givesEvery differential form is locally a sum of products $f_{0}d{f}_1\wedge \cdots \wedge d{f}_r$. Since pullback preserves products and wedge products, the function case and the graded Leibniz rule implyfor every form.

The de Rham cohomology of $X$ isBecause pullback commutes with $d$, it sends closed forms to closed forms and exact forms to exact forms. Thus a smooth map induces

Let $H:X\times I\to Y$ be a smooth homotopy and write $H_t(x)=H(x,t)$. If $V={\partial }_t$, Cartan's magic formula givesIntegrating defines a degree-minus-one operator $K$ satisfyingFor closed $\omega$, the difference is exact, so smoothly homotopic maps induce the same map on de Rham cohomology.

If $F:X\to Y$ is a homotopy equivalence with inverse up to homotopy $G$, functoriality and homotopy invariance giveHence $F^{\ast }$ is an isomorphism.

We induct on $n$. The claim is immediate for ${\mathbb{CP}}^0$. In the stated cover, $U=U_0\cong {\mathbb{C}}^n$ is contractible. The homotopydeformation retracts $V$ onto the hyperplane $z_0=0$, which is ${\mathbb{CP}}^{n-1}$. Moreover,deformation retracts onto $S^{2n-1}$.

The Mayer--Vietoris sequence for the de Rham complex containsFor odd $i>1$, the group on the right vanishes by contractibility and the induction hypothesis, while the group on the left vanishes because $i-1$ is positive and even and is neither $0$ nor $2n-1$. Thus the middle group vanishes. For $i=1$, the preceding mapis surjective because all three spaces are connected, so the connecting map into $H^1({\mathbb{CP}}^n)$ is zero. Therefore all odd de Rham cohomology groups of ${\mathbb{CP}}^n$ vanish.

In a local frame of $E$, a connection on a vector bundle has the formwhere $A$ is a matrix of one-forms. Under a frame change $g$, its matrix transforms asThe covariant exterior derivative on an $E$-valued $r$-form $\sigma$ isThe curvature form of a connection isUsing the graded Leibniz rule,

Regard an $E^{\vee }$-valued $r$-form $\lambda$ as a row vector and an $\mathrm{End}(E)$-valued $r$-form $\mu$ as a matrix. The dual connection and endomorphism connection areFor the curvature two-form,because substituting $F=dA+A\wedge A$ makes all terms cancel. This is the Bianchi identity.

For an endomorphism-valued $r$-form $\mu$ and an $E$-valued form $\sigma$, direct expansion gives the compatible Leibniz ruleThe cancellation of the two middle $A$-terms proves the identity.

For a path $\gamma :[0,1]\to B$, a section $s(t)$ of ${\gamma }^{\ast }E$ is parallel whenExistence and uniqueness for this linear ordinary differential equation define the parallel transportTransport along the reversed path solves the inverse initial-value problem, so

If $P(t)$ denotes transport from $0$ to $t$, thensatisfies the horizontal equation for the induced endomorphism connection. Uniqueness therefore gives

Let $b^{\prime }\in B$. Since $B$ is path-connected, choose a path $\gamma$ from $b$ to $b^{\prime }$. Horizontality of $\mu$ and part c implyThus $\mu (b^{\prime })$ is conjugate to the isomorphism $\mu (b)$ and is itself an isomorphism. Since $b^{\prime }$ was arbitrary, $\mu$ is fiberwise invertible everywhere.

The solder form on $X$ is the $TX$-valued one-formIn a coordinate frame it is $\theta =d{x}^i\otimes {\partial }_i$. The torsion form of $\nabla$ isEquivalently, for vector fields $U,V$,Writingand using $[{\partial }_i,{\partial }_j]=0$ givesHence $\nabla$ is torsion-free exactly whenfor every $i,j,k$.

The connection is orthogonal, or metric-compatible, when $\nabla g=0$; equivalently, its parallel transport preserves the Riemannian metric. Compatibility of the induced connections with tensor contraction givesfor all vector fields $U,V$. Taking $U={\partial }_j$ and $V={\partial }_k$ yieldsConversely, this coordinate identity makes every component of $\nabla g$ vanish, so it is equivalent to orthogonality.

For forms of the same degree, define the Hodge inner product byOn a $p$-form $\beta$, define the codifferentialequivalently $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ in dimension $n$. If $\alpha$ has degree $p-1$, Stokes theorem on the compact boundaryless manifold givesRearranging and using the definition of $\delta$ givesThus $\delta$ is the formal adjoint of $d$.

A form is harmonic whenAdjointness givesTherefore $\Delta \alpha =0$ implies $d\alpha =0$ and $\delta \alpha =0$; the converse follows immediately from the definition of $\Delta$.

The Hodge decomposition theorem states in particular that every de Rham cohomology class has a unique harmonic representative. Hence

For the stated flat metric and orientation,Because the metric coefficients and the coordinate one-forms are constant, the Hodge Laplacian acts coefficientwise:Thus a harmonic one-form $\alpha ={\sum }_i{\alpha }_{i}d{x}^i$ has harmonic coefficient functions. Every harmonic function on the compact connected torus is constant by the maximum principle for harmonic functions. HenceThe Hodge decomposition theorem now gives