Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 17 (Differential Geometry)

An integral curve of a vector field $X$ is a smooth curve $\gamma :I\to M$ satisfying $\dot{\gamma }(t)=X_{\gamma (t)}$ for every $t$ in its interval of definition. Passing through $p$ at time zero means $\gamma (0)=p$. The tangent-vector equality is intrinsic; in a coordinate chart $x:U\to {\mathbb{R}}^n$ it becomesSmoothness makes the coordinate right side locally Lipschitz continuous. The Picard-Lindelöf theorem supplies a locally unique solution through each coordinate initial value.

In fact, two integral curves of $X$ with the same initial point agree throughout their common time interval containing zero. To pass from local to global uniqueness, the set of coincidence times is nonempty and closed by continuity and the Hausdorff property of the smooth manifold. At a coincidence time, put both curves in one coordinate chart; local ordinary differential equation uniqueness makes them agree on a neighborhood of that time. Thus the coincidence set is open as well as closed in the connected common interval, so it is the whole interval. This argument applies around any prescribed time, not only zero.

By joining all extensions that agree on overlaps, one obtains the unique maximal integral curve of a vector field through $p$. Its maximal interval need not be all of $\mathbb{R}$: completeness is an additional property of the vector field.

The local flow of $X$ is the smooth map $\Phi :D\to M$, where $D\subset \mathbb{R}\times M$ is the maximal open domain of initial-point/time pairs, whose curve $t\mapsto \Phi (t,q)={\varphi }_t(q)$ is the maximal integral curve of a vector field through $q$, with ${\varphi }_0(q)=q$. Smooth dependence in the ordinary differential equation theorem gives joint smoothness and openness of $D$. Each time slice is a diffeomorphism between its domain and image, with inverse ${\varphi }_{-t}$.

Here is the group law, including its domain qualification. Fix $s,q$ where the relevant trajectories exist. Both curvessolve the same autonomous ordinary differential equation and have value ${\varphi }_s(q)$ at $t=0$. The uniqueness from the first part therefore giveson their common interval. These are local identities; neither the vector field nor its local flow has been assumed complete.

For the commuting local flows ${\varphi }_t$ and ${\psi }_s$, differentiate ${\varphi }_t({\psi }_s(q))={\psi }_s({\varphi }_t(q))$ with respect to $s$ at zero. The chain rule givesThus the pushforward of a vector field $Y$ by the $X$ flow equals $Y$. Interchanging the two fields gives the analogous identity for $X$ under ${\psi }_s$.

Now let ${\mathcal{D}}_q=\mathrm{span}(X_q,Y_q)$. Pointwise linear independence makes this a rank-two smooth distribution. The identity just proved implies $[X,Y]=0$: in coordinates, differentiating $D_q{\varphi }_t{Y}_q=Y_{{\varphi }_t(q)}$ at $t=0$ gives $DX(q)Y_q=DY(q)X_q$, precisely the vanishing Lie bracket of vector fields. The bracket of two local sections is also in $\mathcal{D}$, sinceand the analogous terms for $[fX,gX]$ and $[fY,gY]$ remain in the same span. Therefore $\mathcal{D}$ is an involutive distribution.

The Frobenius theorem says that a smooth constant-rank involutive distribution has local coordinates $(u^1,\dots ,u^n)$ in which it is spanned by ${\partial }_{u^1},{\partial }_{u^2}$, and has unique maximal connected integral manifolds through its points. These are the leaves of a regular foliation. Let $S$ be the maximal leaf through $p$. The global confinement to an immersed leaf gives the answer: $S$ is a two-dimensional immersed integral manifold and every smooth $\mathcal{D}$-tangent curve starting at $p$ stays in $S$.

For the curve assertion, in each Frobenius chart the transverse coordinate functions $u^3,\dots ,u^n$ have zero derivative along such a curve. Hence its segment in that chart stays in one plaque. On any compact subinterval of the curve's parameter interval, a finite chain of overlapping such segments connects the initial plaque to the last; all belong to the same maximal leaf. Exhausting the parameter interval proves the assertion for every time.

The two commuting local flows also give the leaf parametrization directly near $p$:Its partial derivatives are ${\partial }_{s}F=X_F$ and ${\partial }_{t}F=Y_F$, by the pushforward identity. Its differential has rank two. The constant rank theorem makes a sufficiently small image an embedded local plaque with tangent plane $\mathcal{D}$. This is a concrete local construction of the integral manifold.

There is a global distinction in the word “submanifold”. If it is required to mean an embedded submanifold, the assertion for all times is not valid in general. On the three-dimensional torus ${\mathbb{R}}^3/{\mathbb{Z}}^3$, takeThe fields are independent and commute, and their dense immersed cylinder in a three-dimensional torus through zero isIt is an injectively immersed cylinder $\mathbb{R}\times (\mathbb{R}/\mathbb{Z})$ and is dense in the torus. Indeed, fixing $x\mathrm{mod}1$ and varying $s$ by integers makes $\alpha s\mathrm{mod}1$ dense by an irrational rotation of the circle; $t$ is unrestricted. Every point of this leaf is reached by a tangent curve $r\mapsto (rs,rt,\alpha rs)\mathrm{mod}{\mathbb{Z}}^3$. An embedded surface containing it would, in a submanifold chart near one of its points, be a closed coordinate plane containing a dense subset of the ambient open chart, an impossibility. Thus the global conclusion uses an immersed leaf; the embedded conclusion is local, for curve segments remaining in a suitable chart.

The space of smooth differential forms of degree $p$ isThus a form assigns smoothly to each point an alternating $p$-linear function on its tangent space. Here ${\Omega }^0(M)=C^{\infty }(M)$, and forms of degree greater than $\dim M$ vanish. The wedge product of differential forms gives the graded exterior algebra of forms.

The exterior derivative is an $\mathbb{R}$-linear map of degree one satisfying the following axioms: $df(X)=X(f)$ for functions; for a $p$-form $\alpha$,and $d^2=0$. The sign is the graded Leibniz rule.

To construct it, write in a coordinate chartThis formula is linear, agrees with differentiation on functions and obeys the graded Leibniz rule. Its square vanishes because symmetric second partial derivatives are contracted against antisymmetric wedge products.

The construction is independent of the chart. For coordinates $y^j=y^j(x)$, the chain rule gives $d{y}^j={\sum }_i({\partial }_i{y}^j)d{x}^i$, and the coordinate formula gives $d(d{y}^j)=0$ by equality of mixed partial derivatives. Consequently, expressing a form as ${\sum }_I{b}_{I}d{y}^I$ and applying the $x$-coordinate formula gives exactly ${\sum }_{I}d{b}_I\wedge d{y}^I$, the $y$-coordinate formula. The local constructions agree on overlaps and therefore define a global exterior derivative.

For uniqueness, first derive locality from the axioms. If $\alpha$ vanishes near $p$, choose a smooth bump function $\chi$ supported in that neighborhood and equal to one near $p$. Then $\chi \alpha =0$, andimplies $(d\alpha {)}_p=0$. Thus the value depends only on the local germ of $\alpha$. Local coordinate functions can be extended using bump functions, so the same axioms apply to their germs. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forces precisely the coordinate formula above. The exterior derivative exists and is unique, as expressed by uniqueness of the exterior derivative from its axioms.

Write the differential one-form as $\omega ={\sum }_i{\omega }_{i}d{x}^i$, and the vector fields as $X=X^i{\partial }_i$, $Y=Y^i{\partial }_i$. The coordinate definition of the exterior derivative givesExpand the right side of the invariant formula:The last line is $\omega ([X,Y])$, by the coordinate formula for the Lie bracket of vector fields. Subtract it and interchange the two indices in the second sum. The remaining terms are exactly $d\omega (X,Y)$. ThereforeThis is the exterior derivative of a one-form evaluated on vector fields; it is valid in any frame, not just a commuting coordinate frame.

Define the pointwise representation of one-forms by vector-field functionals byIt is $\mathbb{R}$-linear, its value is a smooth function, and $X_p=0$ implies $\theta (\omega )(X)(p)=0$. Moreoverso it is a map of $C^{\infty }(M)$-modules. This is the properly typed scalar-multiplication property: the same symbol for a form and its associated functional is understood through $\theta$.

To construct its inverse, let $\alpha \in W$ and $v\in T_{p}M$. Choose a global smooth vector field $X$ with $X_p=v$; a local coordinate field multiplied by a smooth bump function supplies such an extension. SetThis is well-defined: if $X_p=Y_p$, then $X-Y$ vanishes at $p$, so $\alpha (X-Y)(p)=0$. Real linearity of $\alpha$ makes ${\omega }_p$ a covector. In fact the pointwise vanishing assumption also forcesbecause $(f-f(p))X$ vanishes at $p$. Thus $\alpha (fX)=f\alpha (X)$ for every smooth $f$, even though only real linearity was initially imposed.

It remains to prove smoothness, rather than assume continuity of an abstract linear map. Near a fixed point, choose global fields $E_1,\dots ,E_n$ equal to the coordinate basis on a smaller neighborhood, using a bump function equal to one there. On that neighborhood,All coefficients are smooth by the definition of $W$, so $\omega$ is a smooth differential one-form. The construction holds near every point, hence gives $\omega \in {\Omega }^1(M)$ with $\alpha (X)=\omega (X)$ globally.

Every tangent vector can be realized by a global field, so $\theta (\omega )=0$ forces $\omega =0$. The inverse just constructed gives surjectivity. The evaluation map is a natural $C^{\infty }(M)$-module isomorphism ${\Omega }^1(M)\cong W$. No auxiliary metric or connection was chosen.

Let $\omega ={\theta }^{-1}(\alpha )$. The preceding invariant formula for the exterior derivative identifies the expression in the question withIts vanishing for every pair of global vector fields is equivalent to $d\omega =0$: global fields with arbitrary prescribed tangent values are available by bump-function extension. Thus it is exactly the condition that $\omega$ be a closed differential one-form.

By definition, the first de Rham cohomology is closed one-forms modulo exact differential forms. The hypothesis $H_{\mathrm{dR}}^1(M)=0$ makes every closed one-form exact. Hence $d\omega =0$ gives a globally smooth function $g$ with $\omega =dg$, andConversely, if $\alpha (X)=X(g)$, thenby the definition of the Lie bracket of vector fields. Equivalently, $d(dg)=0$. The bracket compatibility condition is necessary and, when $H_{\mathrm{dR}}^1(M)=0$, sufficient for a global potential. The potential is unique up to a constant on each connected component, since the difference of two potentials has zero differential.

The Lie algebra of a Lie group $G$ is the vector space $\mathfrak{g}=T_{e}G$ equipped with the bracket transferred from left-invariant vector fields. For $a\in G$, write $L_a(g)=ag$. A field is left-invariant when $(D_g{L}_a)X_g=X_{ag}$ for all $a,g$. Given $v\in T_{e}G$, defineThis is smooth and left-invariant, because $L_a{L}_g=L_{ag}$ and the chain rule applies. Conversely a left-invariant field is determined by its value at $e$, through this formula. Thusis a vector-space isomorphism, whose inverse is evaluation at the identity.

The Lie bracket of vector fields is preserved by diffeomorphisms: on functions this follows by transporting the commutator of derivations. Therefore the bracket of two left-invariant fields is again left-invariant. Define $[v,w]=[X_v,X_w{]}_e$. Bilinearity, antisymmetry and the Jacobi identity follow from the same properties of commutators of derivations. This supplies the asserted Lie algebra structure.

For the general linear group $G{L}_n(\mathbb{R})$, invertible matrices form an open subset of $M_n(\mathbb{R})$, so $T_{I}G{L}_n(\mathbb{R})=M_n(\mathbb{R})$. The left-invariant field determined by $A$ is $X_A(g)=gA$. Its ambient derivative is $D_g{X}_A(H)=HA$. Therefore differentiating left-invariant matrix fields givesHenceThis is the general linear Lie algebra, with the ordinary matrix commutator; right invariance with the same identification would give the opposite sign.

Now let $F:G\to H$ be a smooth Lie group homomorphism. It sends the identity to the identity and induces the linear mapSince $F\circ L_g=L_{F(g)}\circ F$, differentiation givesThese fields are $F$-related; no injectivity or surjectivity of $F$ is needed. For every smooth function $f$ on $H$,Apply this first with $w$ and then with $v$, and subtract the reversed order. The result isAt $e$, smooth functions detect tangent vectors, soThis proves that the differential of a Lie group homomorphism preserves Lie brackets, and hence that $F_{\ast }$ is a Lie algebra homomorphism.

For the exponential identity, let ${\eta }_v$ be the integral curve of $X_v$ through $e$. Left invariance and uniqueness show ${\eta }_v(t+s)={\eta }_v(t){\eta }_v(s)$ for small times: translating the curve through $e$ gives the curve through ${\eta }_v(s)$. Repeating this identity extends the curve to all real times. This proves completeness of left-invariant vector fields without assuming that an arbitrary smooth field is complete. Rescaling the parameter also gives ${\eta }_v(t)={\eta }_{tv}(1)$. By the definition of the Exponential map of a Lie group, ${\exp }_G(v)={\eta }_v(1)$.

The field-related identity shows that $F({\eta }_v(t))$ is an integral curve of $X_{F_{\ast }v}$ starting at $e_H$. By uniqueness it equals ${\eta }_{F_{\ast }v}(t)$ for all times. At time one,This is the naturality of the Lie group exponential.

Apply this to the determinant homomorphism $G{L}_n(\mathbb{R})\to {\mathbb{R}}^{\times }$. The derivative of the determinant at $I$ issince the permutation formula gives $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$. The left-invariant field on ${\mathbb{R}}^{\times }$ with initial tangent $c$ is $y\mapsto cy$; its curve through one solves $y^{\prime }=cy$, hence is $e^{ct}$. The matrix exponential determinant identity therefore follows from naturality:This argument uses only the flow definition of the exponential and the scalar exponential, not an unproved matrix formula.

To identify it explicitly with the usual matrix exponential, the series $E(t)={\sum }_{k\ge 0}{t}^k{A}^k/k!$ converges in operator norm, uniformly with its differentiated series on bounded intervals. Termwise differentiation gives $E^{\prime }=EA$ and $E(0)=I$. The series commutes with $A$, and differentiating $E(t)E(-t)$ gives zero, so $E(t)E(-t)=I$. Thus $E(t)$ stays in $G{L}_n(\mathbb{R})$ and, by uniqueness, is the integral curve of $X_A$ through $I$. The matrix series is exactly the flow-defined exponential, justifying either notation in the determinant identity.

A connection on a vector bundle $E\to M$ is an $\mathbb{R}$-linear mapsatisfying $\nabla (fs)=df\otimes s+f\nabla s$. Equivalently its directional version is real-bilinear and satisfiesFor a complex bundle it is complex-linear in the section variable and has the same Leibniz property.

Every smooth vector bundle over an ordinary Hausdorff, second-countable smooth manifold has such a connection. Choose trivializing open sets $U_a$, local frames $e_{a,i}$ and a subordinate locally finite partition of unity ${\rho }_a$. In each frame define the flat local connectionThen set $\nabla s={\sum }_a{\rho }_a{\nabla }^{(a)}(s{|}_{U_a})$, extending each weighted term by zero outside $U_a$. The supports lie inside the corresponding trivializing sets, making the extensions smooth, and local finiteness makes the sum smooth. Since ${\sum }_a{\rho }_a=1$, its Leibniz rule isThis is the construction of a vector bundle connection by a partition of unity.

The difference $A={\nabla }^{\prime }-\nabla$ of two connections is $C^{\infty }(M)$-linear in both its vector-field argument and its section argument. The derivative terms cancel, so $A_{X}s$ depends only on $X_p,s_p$ at each point. Smooth local coefficients therefore identify it withConversely, any such $A$ added to a connection preserves its axioms. Thus connections form an affine space modeled on ${\Omega }^1(M;\mathrm{End}E)$. Choosing a base connection ${\nabla }^0$ gives the noncanonical bijection $A\mapsto {\nabla }^0+A$; there is no distinguished zero connection and hence no canonical vector-space identification. This is the affine space of vector-bundle connections.

For a connection on $TM$, the torsion tensor is the alternating operation $\tau (X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. The connection and bracket Leibniz rules giveAlternation then gives $\tau (X,fY)=f\tau (X,Y)$ as well. Hence it is tensorial, of type one contravariant and two covariant indices:It is equivalently a $TM$-valued two-form, a torsion form.

For the local frame $e_i$ with dual coframe ${\varphi }_j$, defineThe connection is $C^{\infty }$-linear in $X$, so these are smooth differential one-forms. Expansion in the frame gives ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$, and uniqueness follows by applying each ${\varphi }_j$. Here the first index is the input frame vector and the second the output coefficient; this fixes the connection one-form convention.

Expand $Y={\sum }_i{\varphi }_i(Y)e_i$ and apply the connection Leibniz rule:Subtract the corresponding expression with $X,Y$ exchanged, and then subtract ${\varphi }_j([X,Y])$. By the exterior derivative of a one-form evaluated on vector fields,Since all pairs of tangent vectors can be tested, equality of the two-forms follows:This is the Cartan first structure equation with input-first indices. Its plus sign is correct with the stated index order and wedge order; the common output-first convention writes the same identity with a minus sign and reversed wedge factors.

A Riemannian metric gives a smooth fibrewise isomorphism $g^{♭}:TM\to T^{\ast }M$, defined by $g^{♭}(v)=g(v,\cdot )$. Its inverse is smooth because the inverse of a smoothly varying invertible matrix is smooth. Define the Riemannian gradient byIt is a smooth vector field satisfying $g(V_{\sigma },Y)=d\sigma (Y)=Y(\sigma )$. Nondegeneracy also gives uniqueness. In coordinates its components are $V_{\sigma }^i={\sum }_j{g}^{ij}{\partial }_j\sigma$.

Multiplication of $g$ by the globally defined smooth positive function $e^{2\sigma }$ gives a smooth symmetric two-tensor $\tilde{g}$. For any nonzero tangent vector $v$,Thus $\tilde{g}=e^{2\sigma }g$ is a Riemannian metric. This conformal rescaling of a Riemannian metric changes lengths but preserves angles.

The Levi-Civita connection is the unique connection on a vector bundle $TM$ that is torsion-free and metric-compatible. These conditions areExistence and uniqueness follow from the Koszul formula; equivalently, these are the defining properties that characterize the connection.

Define the correction tensorand set $D_{X}Y={\nabla }_{X}Y+A(X,Y)$. All terms of $A$ are $C^{\infty }(M)$-linear in both $X$ and $Y$, so $D$ is a connection. The symmetry $A(X,Y)=A(Y,X)$ preserves vanishing of the torsion tensor.

To check compatibility with $\tilde{g}$, computeOn the other hand,The cross terms cancel, so their sum is $2X(\sigma )g(Y,Z)$. Multiplying by $e^{2\sigma }$ proves $X(\tilde{g}(Y,Z))=\tilde{g}(D_{X}Y,Z)+\tilde{g}(Y,D_{X}Z)$. Hence $D$ is torsion-free and compatible with $\tilde{g}$, so uniqueness of the Levi-Civita connection givesThis is the Levi-Civita connection under conformal rescaling.

On a nonempty compact manifold without boundary, $\sigma$ attains a maximum at some $p$. The derivative of a smooth function vanishes at an interior extremum, so $(d\sigma {)}_p=0$ and $(V_{\sigma }{)}_p=0$. The correction tensor therefore vanishes at $p$, givingThis is equality for all vector fields at the same point, not an assertion that either connection vanishes there. More generally, the two connections agree at any critical point of $\sigma$. Conversely, if their correction vanishes for all $X,Y$ at a point of positive dimension, evaluating $g(A(v,v),v)=g(v,v)d\sigma (v)$ for every $v$ shows that the point is critical. This is the critical-point criterion for equal conformal connections.

The boundaryless hypothesis is the usual convention for “manifold” here. If manifolds with boundary were included, compactness alone would not suffice: on $[0,1]$ with $g=d{x}^2$ and $\sigma =x$, one has ${\tilde{\nabla }}_{{\partial }_x}{\partial }_x={\partial }_x$ everywhere while ${\nabla }_{{\partial }_x}{\partial }_x=0$. An extremum at the boundary need not be a critical point.