Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 14 (Differential Geometry)

A smooth vector field on a smooth manifold $M$ is a smooth section of a vector bundle $X:M\to TM$ of the tangent bundle projection $\pi$, so $\pi \circ X={\mathrm{id}}_M$. In a coordinate chart it has the form $X={\sum }_i{X}^i{\partial }_i$ with smooth component functions $X^i$.

Suppose a smooth vector bundle isomorphism $\Phi :M\times {\mathbb{R}}^n\to TM$ covering the identity is given. For the standard basis $e_1,\dots ,e_n$ of ${\mathbb{R}}^n$, put $X_i(p)=\Phi (p,e_i)$. These are smooth vector fields, and the linear isomorphism on each fibre makes $X_1(p),\dots ,X_n(p)$ a basis of $T_{p}M$.

Conversely, given such a global frame of a vector bundle, defineIt is smooth, covers the identity and is a linear isomorphism on each fibre. Its inverse is smooth as well: in any coordinate chart, the component columns of the $X_i$ form an invertible smooth matrix $C(p)$, and the fibre coordinates of the inverse are $C(p{)}^{-1}v$. matrix inversion is smooth on the invertible matrices. Thus the tangent bundle is trivial exactly when a global frame exists, the defining property of a parallelizable manifold.

For a Lie group $G$, write $L_h(g)=hg$. A left-invariant vector field is a field satisfyingSuch a field is determined by $v=X(e)\in T_{e}G$, since $X(g)=(d{L}_g{)}_{e}v$. Conversely, this formula defines a left-invariant field, by the chain rule and $L_h{L}_g=L_{hg}$.

It also proves smoothness even if smoothness was not initially assumed. Choose a smooth curve $c(t)$ with $c(0)=e$, $c^{\prime }(0)=v$. ThenMultiplication is a smooth map on a Lie group, so differentiating its coordinate functions in the second variable yields a smooth function of $g$. This gives smooth components for $X$ in every local chart.

Choose a basis $v_1,\dots ,v_m$ of $T_{e}G$, where $m=\dim G$, and left translate it. Each $L_g$ is a diffeomorphism, with inverse $L_{g^{-1}}$, so $(d{L}_g{)}_e$ is a linear isomorphism and the translated fields are a basis at every $g$. The explicit parallelization of a Lie group by left translations isIts inverse sends $w\in T_{g}G$ to $(g,(d{L}_{g^{-1}}{)}_{g}w)$. Identifying $T_{e}G$ with ${\mathbb{R}}^m$ gives the required product bundle. The differential facts used are linearity, the chain rule, $d(\mathrm{id})=\mathrm{id}$ and smooth dependence of the differential of a smooth map on its base point.

A smooth right group action is a smooth map $(p,g)\mapsto p\cdot g$ from $P\times G$ to $P$, with $p\cdot e=p$ and $(p\cdot g)\cdot h=p\cdot (gh)$. It is a free action of a group if $p\cdot g=p$ implies $g=e$.

A smooth principal bundle consists of such a free action and a surjective map $\pi :P\to B$ whose fibres are the orbits, together with equivariant local trivializationswhere $(x,g)\cdot h=(x,gh)$. The trivializations are diffeomorphisms; in particular the bundle projection is a submersion.

For two such trivializations, $H={\Phi }_2{\Phi }_1^{-1}$ fixes the base coordinate and is equivariant. Write $H(x,e)=(x,\psi (x))$. Equivariance immediately givesThe function $\psi$ is smooth because $x\mapsto (x,e)$ and $H$ are smooth. This proves the general transition function of a principal bundle formula; the multiplier is on the left, even though the original action is on the right.

For the explicit bundle, represent a point of ${\mathbb{RP}}^3$ by a nonzero vector $w=(w_1,w_2)\in {\mathbb{C}}^2$ modulo nonzero real scaling. The map sends $[w]$ to the complex line $[w_1:w_2]$, so it is well-defined and surjective. Its smoothness follows from the local ratio charts below. Ordinary scalar multiplication by $u\in U(1)$ would not be free, since $u=-1$ acts trivially on real projective classes. Instead defineThe two possible roots differ by a real sign, so give the same class. This is independent of the real representative of $[w]$. Products of chosen roots show the right action law, and the identity acts trivially. Locally on $U(1)$ a smooth square-root branch exists; these local definitions agree projectively and prove global smoothness of the action.

If $[\lambda w]=[w]$, then $\lambda w=rw$ for some nonzero real $r$. Since $w\ne 0$ and $|\lambda |=1$, this forces $\lambda =\pm 1$ and hence $u=1$. The action is free. Two points in the same fibre have representatives $w^{\prime }=cw$ for some $c\in {\mathbb{C}}^{\ast }$; absorb $|c|$ into the real scaling and use $u=(c/|c|{)}^2$. Thus the orbits are exactly the fibres.

Let $N_1=\{[z:1]\}$ and $N_2=\{[1:\zeta ]\}$. Using these base coordinates, defineBoth formulas are invariant under real nonzero scaling, including negative scaling. Their inverses areAgain the sign of the root does not matter. Direct substitution shows they are two-sided inverses. The forward maps are smooth, and local square-root branches prove that the inverses are smooth. Under the right action, each squared phase is multiplied by $u$, so the maps are equivariant. They are therefore genuine principal trivializations of the principal circle bundle on real projective three-space.

On the overlap $z\ne 0$, we have $\zeta =1/z$ andConsequentlyWhen the base point itself, rather than its chart coordinate, is written as $x$, its coordinate remains $x$ in the general transition formula. The opposite transition has reciprocal multiplier. The doubled phase is essential to freeness and is not the ordinary Hopf-bundle phase.

A connection on a vector bundle $E\to M$ is a linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ with $D(fs)=df\otimes s+fDs$. Write a local frame as a row $e=(e_1,\dots ,e_r)$, and a section as $s=eu$ with a coefficient column $u$. If the connection one-form is the matrix $A={\sum }_i{A}_{i}d{x}^i$, thenThis fixes the matrix convention for all subsequent signs.

For an $E$-valued differential form of degree $k$, represented locally by a column $\alpha$ of ordinary forms, its covariant exterior derivative isThe defining graded Leibniz rule with a scalar form $\omega$ of degree $j$ isIt follows by moving the one-form $A$ past $\omega$, which introduces $(-1{)}^j$. It uniquely extends the connection on sections.

The induced endomorphism bundle connection is characterized on an endomorphism section $T$ by $(D^{\mathrm{End}}T)s=D(Ts)-T(Ds)$. For a matrix-valued form $\beta$ of degree $k$, the extension isIndeed it is exactly the formula that makesThe endomorphism-valued exterior product uses matrix composition as well as the ordinary exterior product, so factors cannot in general be interchanged.

Define the curvature form of a connection by $D^2$ acting on sections, or equivalently byIt is tensorial in both vector arguments and in $s$: the derivatives of multiplying functions cancel by the connection rule. For example $D^2(fs)=D(dfs+fDs)=f{D}^{2}s$, since the two $df\wedge Ds$ terms cancel and $d^{2}f=0$. Thus curvature is a two-form with values in the endomorphism bundle. Locally, applying the graded rule givessoFor a change of frame of a vector-bundle connection $e^{\prime }=eg$, the coefficients satisfy $u^{\prime }=g^{-1}u$, and differentiation givesSquaring yields $F^{\prime }=g^{-1}Fg$, precisely the transformation rule of an endomorphism-valued two-form. This independently proves that the local expressions define one global curvature form of a connection.

For the Bianchi identity, use the endomorphism formula with degree two:Every term cancels with its indicated sign. Thereforewhere $d_A$ here means the induced derivative on the endomorphism bundle.

For a line bundle, the endomorphism bundle is canonically the trivial scalar bundle: every fibre endomorphism is a scalar times the identity. Scalar multiplication commutes, so $A\wedge A=0$ and the endomorphism covariant derivative on scalar-valued forms reduces to $d$. The Bianchi identity consequently says $dF=0$.

If $D^{\prime }$ is another connection, their difference is $C^{\infty }$-linear in sections because the Leibniz terms cancel. Hence $D^{\prime }-D=\eta$ is a global endomorphism-valued one-form; for a line bundle it is an ordinary global scalar one-form. Locally $A^{\prime }=A+\eta$, soThus the two closed curvatures differ by an exact form, proving the connection-independent curvature class of a line bundle:No global trivialization of the line bundle itself was assumed; it is its endomorphism bundle that has the canonical scalar identification.

A geodesic is an affinely parametrized curve satisfying ${\nabla }_{\dot{\gamma }}\dot{\gamma }=0$ for the Levi-Civita connection of the Riemannian metric. In coordinates, write ${\nabla }_{{\partial }_i}{\partial }_j={\Gamma }_{ij}^k{\partial }_k$. For a vector field $V(t)=V^k(t){\partial }_k$ along the curve,Putting $V=\dot{\gamma }$ gives the geodesic equationThis is a smooth second-order ordinary differential system, so the Picard-Lindelöf theorem gives local existence and uniqueness for prescribed position and velocity.

To prove local extension of geodesic velocity, first suppose $\dot{\gamma }(0)=0$. The constant curve through $\gamma (0)$ solves this initial value problem, so uniqueness makes $\gamma$ constant locally. The zero vector field is the desired extension.

Otherwise choose a chart about $\gamma (0)$ in which the first component of the coordinate velocity is nonzero. Write the coordinate curve as $c(t)\in {\mathbb{R}}^n$ and defineAt $(0,0)$ its derivative has columns $c^{\prime }(0),e_2,\dots ,e_n$ and is invertible, because the first component of $c^{\prime }(0)$ is nonzero. The inverse function theorem therefore supplies coordinates $(t,z_2,\dots ,z_n)$ on an open neighbourhood after shrinking the parameter interval to $|t|<\delta$. Along the curve these coordinates are $(t,0,\dots ,0)$. The smooth coordinate field $\partial /\partial t$, pushed back through $\Psi$ and the original chart, agrees with $\dot{\gamma }(t)$. This proves the extension without assuming that a long geodesic has no self-intersections.

By metric compatibility, differentiation along the curve givesThe local extension above permits the ordinary vector-field version of the metric identity to be used along the arc, or the same identity follows directly from the coordinate derivative. Hence an affinely parametrized geodesic has constant speed, including the zero-speed case.

Now let $S^n$ be the unit sphere in ${\mathbb{R}}^{n+1}$. Its induced Levi-Civita derivative is the tangential projection of the ordinary ambient derivative: projection preserves the metric identity, and the commuting ambient coordinate derivatives give zero torsion. By uniqueness of the Levi-Civita connection, proved in Question 5, this is the induced connection.

Thus a curve on the sphere is a geodesic exactly when its ambient acceleration has no tangential component. The normal space is spanned by $\gamma$, so ${\gamma }^{\prime \prime }=a(t)\gamma$. Differentiating $\gamma \cdot \gamma =1$ twice gives $\gamma \cdot {\gamma }^{\prime }=0$ andwhere $c\ge 0$ is the constant speed. Therefore ${\gamma }^{\prime \prime }+c^2\gamma =0$. With initial data $\gamma (0)=p$ and ${\gamma }^{\prime }(0)=v$, all solutions areand, when $v=0$, $\boxed{\gamma (t)=p}$. The orthogonality and length conditions show directly that the displayed curve remains on the sphere, and its acceleration is $-c^2\gamma$, so it is indeed a geodesic. The nonconstant geodesics are the great circles traversed at any constant speed; arbitrary parameter intervals give their restrictions, and shifting the initial parameter simply shifts the formula. There are no other geodesics.

A Levi-Civita connection is a connection on the tangent bundle which is torsion-free and compatible with the Riemannian metric. Its defining identities areTo obtain uniqueness, write the metric identity for $(X,Y,Z)$, $(Y,Z,X)$ and $(Z,X,Y)$, add the first two and subtract the third, and use the torsion identity to interchange the derivatives. This gives the Koszul formulaThe metric is nondegenerate, so these inner products determine ${\nabla }_{X}Y$ uniquely.

For existence, let $K(X,Y,Z)$ denote the right-hand side. The bracket rules $[X,fY]=f[X,Y]+X(f)Y$ and $[fX,Y]=f[X,Y]-Y(f)X$ give, by cancellation of all extra derivatives,Thus the nondegenerate metric defines a unique smooth vector field ${\nabla }_{X}Y$ from $2g({\nabla }_{X}Y,Z)=K(X,Y,Z)$, and these identities give the connection rules. In coordinates the construction readsThe coefficients are smooth and symmetric in $i,j$, proving zero torsion. Substituting them giveswhich is exactly metric compatibility. Since the construction was given intrinsically by $K$, the local formulas agree on overlaps. This proves existence and uniqueness of the Levi-Civita connection on every Riemannian manifold.

Define the Riemann curvature tensor byThe connection rules make this expression tensorial in all three arguments. To fix the component convention, throughout this solution useWith this convention, the unit sphere has $R_{ijkl}=g_{ik}{g}_{jl}-g_{il}{g}_{jk}$ and positive Ricci curvature. In a coordinate frame, the components are obtained fromThe algebraic curvature symmetries aretogether with the first Bianchi identityThe first antisymmetry follows from the definition. Metric compatibility gives $g(R(X,Y)Z,W)+g(Z,R(X,Y)W)=0$, giving the second. Torsion-freeness reduces the cyclic sum to the Jacobi identity for Lie brackets of vector fields; pair interchange follows algebraically from these antisymmetries and the cyclic identity. These explain the identities rather than choosing unrelated sign conventions.

Define the Ricci curvature by the tensor contractionThis is a bilinear form, since $R$ is tensorial and trace is linear. At a point choose an orthonormal basis. Pair interchange givesso Ricci curvature is symmetric.

For the final determination in dimension three, an algebraic curvature tensor is determined by its entries in the three unordered index pairs $12,13,23$: antisymmetry removes repeated indices, and pair interchange makes the resulting $3\times 3$ array symmetric. It therefore has at most six independent entries. WriteThe diagonal Ricci components giveso, for example, $K_{12}=({\mathrm{Ric}}_{11}+{\mathrm{Ric}}_{22}-{\mathrm{Ric}}_{33})/2$, and the other two follow cyclically. The three off-diagonal entries are also recovered explicitly:Thus every independent curvature entry is determined by Ricci; this proves injectivity, not merely a dimension count.

An invariant formula making the reconstruction explicit is, with $S={\mathrm{tr}}_g\mathrm{Ric}$,To verify it, the right-hand side has the two antisymmetries, pair interchange and the cyclic identity by symmetry of $g$ and Ricci. Contracting its first and third indices in dimension three givesSubtract it from the original tensor. The difference has zero Ricci contraction, and the six recovered-entry formulas above force it to vanish. This completes the proof of three-dimensional curvature from the Ricci tensor in arbitrary coordinates.