Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 15 (Differential Geometry)

An integral curve of a vector field $X$ through $p$ is a smooth curve $c:I\to M$, on an interval containing zero, satisfying $c(0)=p$ and $\dot{c}(t)=X_{c(t)}$. The local flow is the smoothly varying family of these integral curves of a vector field:Here $D$ is an open neighborhood of $\{0\}\times M$; for the maximal local flow, its fibre over $p$ is the maximal existence interval of $c_p$. Local existence, uniqueness and smooth dependence for a smooth ordinary differential equation give this family. Uniqueness also gives ${\varphi }_{t+s}(p)={\varphi }_t({\varphi }_s(p))$ whenever the relevant curves and compositions are defined, and ${\varphi }_{-t}={\varphi }_t^{-1}$ on their common domains.

Reparametrizing the integral curves of a vector field givesThus the flow of $2X$ is $\boxed{{\tilde{\varphi }}_t={\varphi }_{2t}}$, with domain $\{(t,p):(2t,p)\in D\}$.

Commutation of the two local flows means ${\varphi }_t{\psi }_s={\psi }_s{\varphi }_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The chain rule givesIn a coordinate chart, differentiating again in $t$ at zero gives $D{X}_p{Y}_p=D{Y}_p{X}_p$. But the Lie bracket of vector fields has coordinate expressionso commuting local flows imply $[X,Y]=0$.

For the converse, suppose the Lie bracket of vector fields vanishes. Write $u(t)={\varphi }_t(p)$ in a coordinate chart, and put $J(t)=d{\varphi }_t{|}_p$. Differentiating the local flow equation with respect to the initial point gives the variational ordinary differential equationConsequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfieswhere the zero Lie bracket of vector fields was used in the second equality. Uniqueness for this linear ordinary differential equation gives $W=0$. This is a local argument and can be continued through successive coordinate charts, proving $d{\varphi }_t(Y)=Y\circ {\varphi }_t$ wherever the local flow is defined.

For fixed $t$, the curve $s\mapsto {\varphi }_t({\psi }_s(p))$ therefore has derivative $Y$ at its current point and starts at ${\varphi }_t(p)$. Uniqueness of the integral curve of a vector field $Y$ identifies it with $s\mapsto {\psi }_s({\varphi }_t(p))$. Hence $\boxed{{\varphi }_t{\psi }_s={\psi }_s{\varphi }_t⟺[X,Y]=0}$ locally. No complete vector field hypothesis is needed.

Using the coordinate formula for the Lie bracket of vector fields, the two components areThus $\boxed{[X,Y]=-(x^2/2){\partial }_x+xy{\partial }_y}$, which is not the zero vector field.

The integral curves of a vector field $X$ solve $\dot{x}=y$, $\dot{y}=0$, whereas those of $Y$ solve $\dot{x}=0$, $\dot{y}=x^2/2$. Their local flows areBoth formulas exist for all real parameters, so both are complete vector fields. Direct composition givesAt $(1,0)$ their first coordinates differ by $ts/2$. Hence these local flows do not commute when $s,t$ are nonzero.

For the Lie bracket of vector fields, start at $(2,0)$. An integral curve of a vector field $[X,Y]$ isIt solves $\dot{x}=-x^2/2$, $\dot{y}=xy$, but $x(t)$ diverges as $t\downarrow -1$. It cannot extend through that finite time as a curve in ${\mathbb{R}}^2$. Complete vector fields are therefore not closed under the Lie bracket.

For the sum $X+Y$, its integral curves of a vector field obey $\dot{x}=y$, $\dot{y}=x^2/2$. Substitution of $x=(1+\alpha t{)}^{-2}$ and $y=\beta (1+\alpha t{)}^{-3}$ requiresChoosing $\alpha =-1/(2\sqrt{3})$ gives the integral curve of a vector fieldIt starts at $(1,1/\sqrt{3})$ and escapes to infinity at the finite time $2\sqrt{3}$. Complete vector fields are not closed under addition either. Both failures use the same pair of complete vector fields.

A smooth rank-$r$ vector bundle over a smooth manifold $M$ consists of a smooth total space $E$, a smooth projection $\pi :E\to M$, and an $r$-dimensional real vector space structure on each fibre $E_p={\pi }^{-1}(p)$. Each point has a neighborhood $U$ with a vector bundle trivializationthat is a diffeomorphism, commutes with projection to $U$, and is a linear isomorphism on every fibre. The overlap maps have the form $(p,v)\mapsto (p,h(p)v)$ with $h:U\cap V\to G{L}_r(\mathbb{R})$ smooth. For a complex vector bundle, replace $\mathbb{R}$ by $\mathbb{C}$ and require complex-linear fibre maps.

A connection on a vector bundle is an $\mathbb{R}$-linear mapwhere $\Gamma (E)$ is the module of smooth sections, ${\Omega }^1(M;E)$ consists of vector-bundle-valued differential forms of degree one, and $f$ is a smooth function. Equivalently, evaluating on a vector field $X$ gives operators ${\nabla }_X$ with ${\nabla }_{fX}s=f{\nabla }_{X}s$ and ${\nabla }_X(fs)=X(f)s+f{\nabla }_{X}s$. A connection on a vector bundle is local: its value at a point depends on the germ of the section of a vector bundle, so the same definition applies to local sections.

Realize the pullback vector bundle asIts fibre over $p$ is canonically $E_{\varphi (p)}$, with the same vector space operations. If a vector bundle trivialization over $U\subset M$ is ${\tau }_U(v)=(\pi (v),\hat{v})$, give the pullback vector bundle the local trivializationDeclare these maps to be local homeomorphisms and use coordinate charts on $M^{\prime }$ to obtain a smooth atlas on the total space. On overlaps the smooth transition maps are $(p,w)\mapsto (p,h(\varphi (p))w)$, hence smooth and fibrewise linear. They satisfy the same cocycle identities as the original transition maps. This gives a well-defined smooth vector bundle, with smooth projection and rank $r$. It also gives the subspace topology inherited from $M^{\prime }\times E$: inside each product ${\varphi }^{-1}(U)\times {\pi }^{-1}(U)$ the defining condition is the graph of $\varphi$ in the base coordinates. Thus the total space is Hausdorff and second countable, as required for a smooth manifold.

For a smooth section of a vector bundle $s$ over $U$, its pulled-back section is more precisely $p\mapsto (p,s(\varphi (p)))$. If $s$ has component column $u:U\to {\mathbb{R}}^r$ in the original vector bundle trivialization, its component column in $T_U$ is $u\circ \varphi$, which is smooth. The pullback is a smooth rank-$r$ vector bundle, and every local section pulls back smoothly.

The expression $d\varphi (X)$ is in general a vector field along a map, rather than a vector field on $M$. Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form $\nabla s$:using the canonical identification $({\varphi }^{\ast }E{)}_p=E_{\varphi (p)}$. This interpretation works for every smooth map, including a constant map or a map with self-intersections.

Choose a frame of a vector bundle over $U$, written as a row $e=(e_1,\dots ,e_r)$. Define its connection one-form $\Omega$ byfor component columns $u$. The pulled-back frame $e^{\varphi }=({\varphi }^{\ast }{e}_1,\dots ,{\varphi }^{\ast }{e}_r)$ is a frame of a vector bundle over ${\varphi }^{-1}U$. For any component column $v$ of smooth functions on that open set, defineHere the matrix entries of ${\varphi }^{\ast }\Omega$ are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.

To prove these local operators glue, change the original frame of a vector bundle to $\tilde{e}=eh$. Expanding $\nabla (ehu)$ gives the change of frame of a vector-bundle connectionThe pullback of a differential form and the chain rule giveThis is precisely the required transformation law for the pulled-back frame ${\tilde{e}}^{\varphi }=e^{\varphi }(h\circ \varphi )$. Thus the definitions agree and give a global pullback connection.

For $s=eu$, substitute $v=u\circ \varphi$ in the defining formula and use $d(u\circ \varphi )(X)=du(d\varphi (X))$. This proves the required identity. Conversely that identity fixes ${\nabla }_X^{\prime }({\varphi }^{\ast }{e}_a)$; the Leibniz rule then fixes the derivative of every ${\sum }_a{v}^a{\varphi }^{\ast }{e}_a$, proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in $U$, and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

The curvature form of a connection is the endomorphism-valued differential two-form defined byIt is alternating in $X,Y$. To see its tensoriality, replacing $X$ by $fX$ produces an extra $-Y(f){\nabla }_{X}s$ from the second term and an extra $+Y(f){\nabla }_{X}s$ from $[fX,Y]=f[X,Y]-Y(f)X$, so these cancel. Alternation gives linearity over smooth functions in $Y$ too. Replacing $s$ by $fs$ produces the additional coefficient $(X(Yf)-Y(Xf)-[X,Y]f)s=0$; all terms involving one derivative of $s$ also cancel. Thus $R^{\nabla }$ belongs to ${\Omega }^2(M;\mathrm{End}E)$.

In the coefficient-column convention of part (b), expanding the definition givesThe first three terms are $d\Omega (X,Y)$, and the last two are $(\Omega \wedge \Omega )(X,Y)$. This proves the Cartan curvature matrix equation $F=d\Omega +\Omega \wedge \Omega$, with matrix multiplication combined with the wedge product of differential forms.

The pullback connection has matrix ${\Omega }^{\prime }={\varphi }^{\ast }\Omega$. Since the pullback of a differential form commutes with the exterior derivative and the wedge product of differential forms, its curvature form of a connection has matrixChanges of frame of a vector bundle conjugate both sides by the same pulled-back transition matrix, so this identity is intrinsic. In fibre notation the answer isBoth sides act on the same fibre $E_{\varphi (p)}$. No claim that $d\varphi$ preserves Lie brackets of vector fields for an arbitrary map is needed.

An oriented atlas is a smooth atlas whose smooth transition maps have positive Jacobian determinants at every point of their domains. Such an atlas supplies an orientation of a vector space on every tangent space, and these choices agree on overlaps.

Let $\omega$ be a nowhere-vanishing smooth top-degree differential form. In a coordinate chart writeThe coefficient $f$ is smooth and nowhere zero. Restrict the coordinate chart to connected neighborhoods; the intermediate value theorem makes the sign of $f$ constant on each. If $f$ is negative, replacing $x^1$ by $-x^1$ changes the sign of the coordinate volume form. Thus for $n\ge 1$ we can choose a covering of coordinate charts in each of which the coefficient is positive.

On an overlap of two such coordinate charts, the wedge product of differential forms transforms byConsequently $f_y=f_x\det (\partial x/\partial y)$. Both coefficients are positive, so the Jacobian determinant is positive, as is that of the inverse transition. A nowhere-vanishing top form therefore determines an oriented atlas. The zero-dimensional case has only zero-dimensional transitions, whose determinant is one, and is automatically orientable. Connectedness of $M$ is not needed for the positive-dimensional construction; only the local sign choice matters.

Use coordinate charts from the chosen oriented atlas. In each, the matrix $G_x=(g_{ij})$ of the Riemannian metric is smooth and positive-definite, so $\det G_x>0$ and its positive square root is smooth. Define the local Riemannian volume form byIf $y$ is another positive coordinate chart and $J=\partial x/\partial y$, the transformation rule for the metric tensor is $G_y=J^T{G}_{x}J$. Taking determinants givesThe oriented atlas has $\det J>0$, so the wedge product of differential forms transformation from part (a) givesThe local expressions therefore glue to a global smooth top-degree differential form. Its local coefficient never vanishes, so it is a volume form. The global answer is $\boxed{{\omega }_g=\sqrt{\det (g_{ij})}d{x}^1\wedge \cdots \wedge d{x}^n}$ in positive charts. In an arbitrary negatively oriented chart the corresponding expression needs a minus sign; the orientation is essential to the form, although a positive metric volume density exists without it.

The integral identity requires compactness of $M$, compact support of $X$, or suitable conditions eliminating flux at infinity. The printed question does not state such a hypothesis. We prove the Riemannian divergence theorem for a compact manifold, and more generally for a compactly supported smooth vector field on an oriented Riemannian manifold with boundary.

Write $j:\partial M\hookrightarrow M$ for the inclusion. Give $\partial M$ the outward-normal-first boundary orientation, and let $N$ be the outward unit normal vector. The Riemannian volume form of the induced Riemannian metric isIndeed, if $(e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis tangent to the boundary, $(N,e_1,\dots ,e_{n-1})$ is a positively oriented orthonormal basis in $M$. Both sides therefore evaluate to one on the boundary basis.

At a boundary point decompose $X=g(X,N)N+X^{\top }$, where $X^{\top }$ is tangent to $\partial M$. Evaluating on $n-1$ tangent vectors, the term with $X^{\top }$ vanishes: all $n$ arguments of the alternating volume form then lie in the $(n-1)$-dimensional boundary tangent space. Thus the interior product of a differential form satisfiesThe general Stokes theorem for an oriented manifold with boundary states ${\int }_{M}d\eta ={\int }_{\partial M}{j}^{\ast }\eta$ for any smooth differential form $\eta$ of degree $n-1$ with compact support; compactness of $M$ makes the support condition automatic. Applying it to $\eta ={\iota }_X{\omega }_g$ and using the defining equation for the divergence of a Riemannian vector field proves

For a concrete failure without the extra hypothesis, take $M=[0,\infty )$ with Riemannian metric $d{x}^2$ and $X=(\arctan x){\partial }_x$. The divergence of a Riemannian vector field is $\mathrm{div}X=(1+x^2{)}^{-1}$, sobecause $X$ vanishes at the only boundary point. Both integrals are finite. The missing $\pi /2$ is the limiting flux at infinity, so mere integrability of the divergence cannot repair the unrestricted assertion.

A subset $Z$ of an $m$-dimensional smooth manifold $M$ is an $r$-dimensional embedded submanifold if, for each $p\in Z$, some coordinate chart $x:U\to V\subset {\mathbb{R}}^m$ containing $p$ satisfiesThe induced topology on $Z$ is the subspace topology, and the first $r$ coordinates restricted to $Z$ give its smooth atlas. To see their smooth transition maps agree, restrict the ambient smooth transitions to the coordinate planes; their inverses restrict in the same way. The inclusion then has injective differential of a smooth map and is a homeomorphism onto its image. Conversely this is the local coordinate description of a smooth smooth embedding. An embedded submanifold need not be a closed subset of $M$.

Put $m=\dim M$ and $n=\dim N$. The hypothesis says that $q$ is a regular value of $F$. Since its preimage is nonempty and the differential of a smooth map is surjective there, $m\ge n$.

We use the following finite-dimensional inverse function theorem: if a smooth map $H:O\to {\mathbb{R}}^m$, with $O\subset {\mathbb{R}}^m$ open, has invertible derivative at $a$, there are open neighborhoods of $a$ and $H(a)$ on which $H$ is a diffeomorphism, with smooth inverse.

Fix $p\in F^{-1}(q)$. Choose coordinate charts centered at $p$ and $q$, and shrink the source neighborhood so that its image under $F$ lies in the target chart. Its coordinate expression is $f=(f^1,\dots ,f^n)$ with $f(0)=0$ and derivative of rank $n$. Choose $n$ domain coordinates giving an invertible $n\times n$ minor, and relabel them as $x^1,\dots ,x^n$. DefineIts derivative at zero has block formso it is invertible. The inverse function theorem makes $u=H(x)$ a new coordinate chart near $p$. In this chart $F=q$ is exactly $u^1=\cdots =u^n=0$, because the target chart sends $q$ to zero. These are slice charts for an embedded submanifold; reordering coordinates if desired puts the free coordinates first. Giving the level set the subspace topology, these restricted charts are smoothly compatible, as in part (a). This proves the regular level set theorem rather than merely invoking it.

The level set is an embedded submanifold of dimension $\boxed{m-n}$. The same coordinates show $T_{p}Z=\ker d{F}_p$: tangent vectors to the slice have zero first $n$ components, which are exactly the components of $d{F}_p$. The empty-dimensional case $m=n$ gives a discrete zero-dimensional embedded submanifold.

Let $m=\dim M$, $n=\dim N$ and $k={\mathrm{codim}}_{N}S$. Fix $p\in f^{-1}(S)$ and choose the allowed slice chart for an embedded submanifold on a neighborhood $U$ of $f(p)$, so that $S\cap U$ is given by $x^1=\cdots =x^k=0$. On $f^{-1}(U)$ define the smooth mapThe derivative $L=d(x^1,\dots ,x^k)$ at a point of $S\cap U$ is surjective and has kernel $TS$, by the slice description. The printed sum-of-spaces condition givesIndeed any target tangent vector is a sum of an image vector and a tangent vector to $S$, and $L$ kills the latter. By the chain rule, $d{G}_r=L\circ d{f}_r$ is therefore surjective. This condition is precisely transversality of a map to a submanifold; the original sum need not be direct.

We have $G^{-1}(0)=f^{-1}(S)\cap f^{-1}(U)$. Part (b), applied to $G$, gives slice charts for an embedded submanifold of codimension $k$ around every point of this zero set. Since the construction is available around every $p\in f^{-1}(S)$, the charts give its global structure as an embedded submanifold with the subspace topology. Its dimension is $\boxed{m-k=m-n+\dim S}$. Moreover,This follows either from the slice coordinates or from $T_p{G}^{-1}(0)=\ker d{G}_p$ and $\ker L=T_{f(p)}S$. No injectivity of $df$, closedness of $S$, or direct-sum hypothesis is required.

An affine connection $\nabla$ is a metric connection for the Riemannian metric $g$ whenEquivalently the induced covariant derivative of the metric tensor satisfies $\nabla g=0$. This is metric compatibility; it imposes no condition on the torsion tensor. A Levi-Civita connection has the additional property of being a torsion-free connection.

A vector field along a map $\gamma$ assigns $V(t)\in T_{\gamma (t)}M$ smoothly to each parameter value. Equivalently, it is a section of a vector bundle ${\gamma }^{\ast }TM$. The pullback connection defines its covariant derivative along a curve, $D_{t}V$. To make the definition explicit, use connection coefficients with the conventionFor $V=V^k(t){\partial }_k{|}_{\gamma (t)}$ the derivative isThe change of frame of a vector-bundle connection ensures this expression is independent of the coordinate chart. It remains meaningful when $\dot{\gamma }=0$ and does not require extending $V$ to one ambient vector field.

The field is parallel exactly when $\boxed{D_{t}V=0}$. A geodesic for $\nabla$, with its specified affine parameter, is a smooth curve whose own tangent is parallel:This definition applies to any affine connection, whether or not it is a metric connection. A non-affine change of parameter generally gives a pregeodesic instead of this parametrized geodesic.

We use the following existence and uniqueness statement for parallel transport: for a smooth affine connection, a smooth curve $\gamma :[0,1]\to M$, an initial time $t_0$ and a vector $v\in T_{\gamma (t_0)}M$, there is a unique smooth vector field along a map $V$ on the whole curve satisfying $D_{t}V=0$ and $V(t_0)=v$. In a frame of a vector bundle this is the linear ordinary differential equation $\dot{V}=-\Gamma (\dot{\gamma })V$. Its smooth coefficient matrix is bounded on every compact time subinterval in a trivializing neighborhood, and the linear existence theorem gives a solution throughout that subinterval. A finite subdivision of $[0,1]$ into such neighborhoods and uniqueness patch the solutions. No completeness hypothesis on $M$ is needed.

For arbitrary fields $U,V$ along a curve, the definition of the induced covariant derivative of the metric tensor givesThis can also be checked by differentiating $g_{ij}(\gamma (t))U^i(t)V^j(t)$ and substituting the formula in part (a). If $\nabla$ is a metric connection and $V$ is parallel, it follows that $d(g(V,V))/dt=0$. Positivity of the Riemannian metric makes $|V|=\sqrt{g(V,V)}$ constant. Thus parallel transport preserves lengths.

Conversely assume every parallel field along every smooth curve has constant length. Fix $p\in M$ and arbitrary $x,v\in T_{p}M$. Choose a curve through $p$ with velocity $x$ at an interior time $t_0$. Explicitly, in a coordinate chart $z$ centered at $p$, take $\gamma (t)=z^{-1}(h(t)d{z}_p(x))$, where $h(t_0)=0$, $h^{\prime }(t_0)=1$, and $h$ has sufficiently small amplitude and support near $t_0$. A smooth bump function produces such $h$, so the curve is defined on all of $[0,1]$ and remains inside the chart. By the stated parallel transport theorem there is a parallel field with $V(t_0)=v$. Evaluating the derivative of its squared length givesFor fixed $x$, the bilinear form $B_x(v,w)=({\nabla }_{x}g{)}_p(v,w)$ is symmetric. Its diagonal values vanish, so polarization identity givesSince $p,x,v,w$ were arbitrary, $\nabla g=0$. A connection is metric-compatible if and only if all parallel fields have constant length. Polarization also shows that its parallel transport preserves all inner products, not just lengths.

First verify the difference of affine connections is a tensor. Linearity over smooth functions in $X$ follows from the affine connection axioms. In $Y$, the two derivative terms cancel:Thus $A$ is a smooth $(1,2)$-tensor field, and $A_p(u,v)$ depends only on the tangent vectors $u,v$ at $p$. For any smooth curve, its two covariant accelerations satisfyIf $A(X,Y)=-A(Y,X)$, its diagonal values vanish. The two geodesic equations are therefore identical, and the connections have the same parametrized geodesics.

For necessity, the required local geodesic existence theorem is this: for every smooth affine connection, $p\in M$ and $v\in T_{p}M$, a unique geodesic exists on some interval about zero with $\gamma (0)=p$ and $\dot{\gamma }(0)=v$. Indeed the geodesic equation is the smooth first-order ordinary differential equation on position and velocityso local existence and uniqueness apply. If the two connections have the same parametrized geodesics, apply the acceleration identity to this geodesic at zero to obtain $A_p(v,v)=0$ for every $p,v$. Expanding $A_p(u+v,u+v)=0$ gives $A_p(u,v)+A_p(v,u)=0$. ThereforeThis is the criterion that parametrized geodesics determine the symmetric part of an affine connection. It concerns equality with the same affine parameters. Agreement merely of unparametrized images is weaker: adding $\alpha (X)Y+\alpha (Y)X$ to a connection, for a smooth one-form $\alpha$, changes acceleration by a multiple of the velocity, which can be absorbed by reparametrization. Such a difference need not be antisymmetric.

Finally, since ${\nabla }_X^{\prime }Y={\nabla }_{X}Y-A(X,Y)$, expand the derivative of the metric tensor:Under the assumed metric compatibility of $\nabla$, the first term is zero. ConsequentlyEquivalently, for each fixed $X$, the endomorphism $A(X,\cdot )$ is skew-adjoint for the Riemannian metric. This metric criterion does not require the two connections to have the same geodesics.