Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 115 (Differential Geometry)

The tangent bundle is $TM={⨆}_{p\in M}{T}_{p}M$, with its smooth vector-bundle structure from coordinate charts. A vector field is a smooth section of this bundle. The Lie bracket of vector fields is their commutator as derivations:In coordinates, $[X,Y{]}^k=X^j{\partial }_j{Y}^k-Y^j{\partial }_j{X}^k$.

The four numbered conditions here are hypotheses of one argument, so the following sections assemble its proof. Let $j:\partial M\hookrightarrow M$. Because all $X_i$ are tangent to the boundary, evaluating $\eta ={\iota }_{X_1}\cdots {\iota }_{X_n}\omega$ on a tangent vector $Y$ there evaluates the volume form on $n+1$ vectors in the $n$-dimensional boundary tangent space. Alternation gives $\boxed{j^{\ast }\eta =0}$. This uses the interior product of a differential form and will make the boundary restriction of the de Rham cohomology class vanish.

At any point, the linearly independent vectors $X_1,\dots ,X_n$ can be completed by $Y$ to a basis of $T_{p}M$. A nowhere-zero volume form is nonzero on every basis, soThus $\boxed{\eta \text{ is nowhere zero}}$. In particular, if it is an exact differential form $df$, its potential has no critical points. This is the role of pointwise independence in the boundary obstruction for commuting volume-preserving vector fields.

Let ${\Phi }_t,{\Psi }_s$ be the local flows of $X,Y$. The derivative of the pullback of $Y$ by ${\Phi }_t$ is ${\Phi }_t^{\ast }[X,Y]$. Therefore $[X,Y]=0$ makes $Y$ invariant under the $X$ flow. For fixed $t$, the two curves ${\Phi }_t({\Psi }_s(p))$ and ${\Psi }_s({\Phi }_t(p))$ then solve the same initial-value problem for $Y$. Uniqueness gives commuting flows wherever both compositions are defined. Conversely, commutation implies invariance of $Y$ under ${\Phi }_t$, and differentiation at $t=0$ gives $\boxed{[X,Y]=0}$. This proves vanishing Lie bracket is equivalent to commuting local flows.

For the contraction $\eta$, commute each Lie derivative past the other interior products using $[{\mathcal{L}}_{X_i},{\iota }_{X_j}]={\iota }_{[X_i,X_j]}=0$. Each ${\mathcal{L}}_{X_i}\omega$ is zero by volume preservation. Cartan's magic formula $d{\iota }_X={\mathcal{L}}_X-{\iota }_{X}d$, applied successively, now givesThe last term vanishes since $\omega$ has top degree. Hence $\boxed{\eta \text{ is closed}}$.

By the preceding sections, $\eta$ is a nowhere-zero closed differential form and $j^{\ast }\eta =0$. Injectivity of $j^{\ast }:H_{\mathrm{dR}}^1(M)\to H_{\mathrm{dR}}^1(\partial M)$ gives $[\eta ]=0$. The definition of de Rham cohomology then givesfor a smooth $f$, with no critical points. If the boundary were connected, $j^{\ast }df=0$ would make $f$ constant there. A nonconstant function on a compact manifold attains both extrema; a value differing from the common boundary value gives an interior extremum, contradicting $df\ne 0$. If $f$ were constant everywhere there would be the same contradiction. Empty boundary is also impossible, because an extremum would necessarily be interior. Thus $\boxed{\partial M\text{ has at least two connected components}}$, the boundary obstruction for commuting volume-preserving vector fields.

The solid torus $S^1\times D^2$ retracts onto its first circle. Its degree-one de Rham cohomology is generated by that circle's angular one-form, which restricts to a nonzero class on $S^1\times S^1$; hence the restriction map is injective. Its boundary is connected. Extensions of the two specified coordinate vector fields would be tangent there and would satisfy all the forbidden conditions. Therefore no such pair of extensions exists, for any volume form.

A smooth rank-$r$ smooth distribution $D\subset TM$ is involutive distribution if the Lie bracket of vector fields of two local sections remains a section. It is an integrable distribution if every point lies on an immersed $r$-dimensional integral manifold with tangent spaces equal to $D$. The Frobenius theorem says these conditions are equivalent, and gives local coordinates with $D=\mathrm{span}({\partial }_1,\dots ,{\partial }_r)$.

For necessity, fields tangent to an integral manifold have brackets tangent to it: they annihilate functions vanishing on the manifold, and so does their commutator. For sufficiency, induct on $r$. The rank-zero case is immediate. Straighten a nonvanishing local section using the flow-box theorem to obtain ${\partial }_1\in D$. Choose a frame ${\partial }_1,Y_2,\dots ,Y_r$ with the $Y_i$ having no ${\partial }_1$ component. Involutivity gives ${\partial }_{1}Y=AY$ for the column of these fields and a smooth matrix $A$. Solve the matrix ordinary differential equation ${\partial }_{1}B=-BA$, with $B=I$ on $x_1=0$. It stays invertible, and $Z=BY$ satisfies ${\partial }_{1}Z=0$. On the transverse slice, the $Z_i$ span an involutive rank-$(r-1)$ distribution. The induction hypothesis supplies adapted slice coordinates; extend them independently of $x_1$. These give the required rank-$r$ coordinate distribution and its integral manifolds.

For the real Heisenberg group, multiplication isDifferentiating left translation at the identity gives its left-invariant frame of the real Heisenberg group:Then $[E_1,E_2]=E_3$ and the other basis brackets vanish. Since $E_3\notin \mathrm{span}(E_1,E_2)$, the Heisenberg horizontal distribution is not involutive, hence not integrable by the Frobenius theorem.

Using the left-invariant frame of the real Heisenberg group, horizontality givesFor an arbitrary starting point $(x_0,y_0,z_0)$,This is the control equation for the Heisenberg horizontal distribution. The final coordinate records a signed area-like integral, so changing the path in the $(x,y)$ plane can change the endpoint in the central direction.

The printed hint is impossible: if ${\int }_0^{1}f=0$, thenChanging the order of integration gives the same zero, so the additional condition involving $\int tf$ cannot repair it. Using the same function in both proposed controls cannot generate an arbitrary third coordinate.

Instead, put $\Delta =P_z-P_x{P}_y/2$, $a=2(\Delta +P_x)$, and choose two different oscillatory controls:Their integrals give $x(t)=P_{x}t+a\sin (2\pi t)/(2\pi )$ and $y(t)=P_{y}t+1-\cos (2\pi t)$, hence $x(1)=P_x$ and $y(1)=P_y$. The third endpoint isThe resulting curve is smooth and horizontal at every time, proving smooth horizontal reachability in the real Heisenberg group.

If the Heisenberg horizontal distribution were an integrable distribution, every horizontal curve from the identity would remain in its maximal connected integral leaf. Reachability would force that two-dimensional leaf to be the whole three-dimensional group, contradicting the local leaf coordinates of the Frobenius theorem. The bracket computation in part (i) gives the same obstruction directly.

In coordinates, expanding $d\alpha$ and the Lie bracket of vector fields showsThe terms differentiating components of $X,Y$ cancel against the bracket, leaving $({\partial }_i{\alpha }_j-{\partial }_j{\alpha }_i)X^i{Y}^j$. This is the exterior derivative of a one-form evaluated on vector fields.

A connection on a vector bundle is a linear map $\nabla :\Gamma (E)\to {\Omega }^1(E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$; evaluation on $X$ defines ${\nabla }_{X}s$. Its curvature form of a connection is $d_A^2$, locally $F_A=dA+A\wedge A$. Direct expansion, using the displayed exterior-derivative identity, givesThe derivative terms on a scalar multiplying $s$ cancel, and the expression is also linear over smooth functions in $X,Y$. Thus it is an alternating tensor with values in $\mathrm{End}(E)$, namely an element of ${\Omega }^2(\mathrm{End}(E))$.

The dual connection and tensor product connection induce the endomorphism bundle connection on $E^{\ast }\otimes E$:Expanding twice cancels the cross terms and givesHere $\varphi ,s$ are local smooth sections; derivatives are not defined for isolated fiber elements without extensions. Subtract the corresponding ${\tilde{\nabla }}_{[X,Y]}$ identity to get the curvature of an endomorphism bundle connection:This vanishes for all $\varphi$ exactly when each $F_A(X,Y)$ is central in the full matrix algebra, hence scalar. For rank $r>0$, $\omega =r^{-1}\mathrm{tr}{F}_A$ is a smooth two-form andConversely scalar curvature commutes with every endomorphism. This is the scalar-curvature criterion for a flat endomorphism connection; it permits a nonflat connection on $E$ itself.

Use the energy of a curve normalizationFor a variation $F(s,t)$ put $T={\partial }_{t}F$ and $V={\partial }_{s}F$. The Levi-Civita connection is torsion free, so ${\nabla }_{s}T={\nabla }_{t}V$. Differentiating energy once gives $E^{\prime }=[⟨V,T⟩{]}_a^b-\int ⟨V,{\nabla }_{t}T⟩$. Along a geodesic, ${\nabla }_{t}T=0$. Differentiating again, commuting covariant derivatives, and integrating by parts gives the second variation of geodesic energy:The curvature convention is $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, consistent with the specified positive sectional curvature. Indeed ${\nabla }_s{\nabla }_{t}V={\nabla }_t{\nabla }_{s}V+R(V,T)V$, and $⟨R(V,T)V,T⟩=-⟨R(V,T)T,V⟩$. The endpoint term vanishes for fixed endpoints or for periodic variations of a closed geodesic. The integral is the Riemannian index form $I(V,V)$.

Let $\dim M=2m$. Parallel transport around the closed geodesic preserves the metric and orientation, so lies in the special orthogonal group. It fixes the nonzero tangent $T$, and its restriction to $T^{\perp }$ is an orientation-preserving orthogonal map of odd dimension $2m-1$. An odd-dimensional special orthogonal transformation has a fixed vector: nonreal eigenvalues occur in conjugate pairs, while an odd-dimensional real orthogonal map with determinant one must have an eigenvalue $+1$. Choose a nonzero fixed vector $v$ normal to $T$ and parallel-transport it along the curve. It gives a nonzero smooth periodic normal field $V$ with ${\nabla }_{t}V=0$.

For the exponential variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$, the endpoints match periodically. Strictly positive sectional curvature givesThus $E({\gamma }_s)<E(\gamma )$ for small nonzero $s$. By the Cauchy-Schwarz inequality,where the final equality uses the geodesic's constant speed. This proves the instability of a closed geodesic in positive even-dimensional curvature.

For an embedded closed geodesic, sufficiently small variations remain embeddings, hence give a smooth isotopy with strictly shorter curves. For a nonembedded closed geodesic the construction gives a smooth deformation through immersions; an isotopy class of embeddings is not literally defined for such a curve. The stated isotopy conclusion therefore uses the usual embedded-curve interpretation, while the shorter-loop variation holds without it.