Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 19 (Riemannian Geometry)

Use the curvature sign convention implicit in part (c): the Jacobi field equation is $D_t^{2}J+K_{\dot{\gamma }}J=0$, so the operator in a positively curved normal direction has positive eigenvalue. With the alternative convention $R_{\mathrm{std}}(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ used for the catalog's Jacobi curvature operator, the paper's tensor is $R=-R_{\mathrm{std}}$, and its operator is $K_v(x)=R_{\mathrm{std}}(x,v)v$. This fixes the sign throughout Q1.

Let $r(A,B,C,D)=⟨R(A,B)C,D⟩$. Pair interchange for the Riemann curvature tensor givesThus $K_v$ is self-adjoint with respect to the Riemannian metric. The curvature sign does not affect this conclusion. The spectral theorem for real symmetric matrices therefore supplies the real orthonormal eigenbasis used in part (b). Also $K_{v}v=0$, so the tangent direction is a zero-eigenvalue direction.

Write $D_t={\nabla }_{\dot{\gamma }}$ and let $P_t:T_{p}M\to T_{\gamma (t)}M$ denote parallel transport. Since $\gamma$ is a geodesic, $D_t\dot{\gamma }=0$, while the transported vectors satisfy $D_t{e}_i(t)=0$. The locally symmetric Riemannian manifold condition is $\nabla R=0$. Hence the Levi-Civita connection product rule yieldsBoth this vector field and ${\lambda }_i{e}_i(t)$ are parallel and agree at zero. Uniqueness of parallel transport therefore provesThe eigenvalues are independent of $t$. The original PDF has $K_{\dot{\gamma }(t)}$ here; the converted TeX omits the dot, which would incorrectly put a point rather than a tangent vector in the operator's subscript.

Expand a Jacobi field in the parallel orthonormal eigenbasis: $J(t)={\sum }_i{j}_i(t)e_i(t)$. Part (b) reduces its equation to the scalar equations $j_i^{\prime \prime }+{\lambda }_i{j}_i=0$. DefineFor arbitrary initial values $J(0)={\sum }_i{a}_i{e}_i$ and $D_{t}J(0)={\sum }_i{b}_i{e}_i$, the complete solution isA point at parameter $T>0$ is a conjugate point of $p$ precisely when a nonzero Jacobi field vanishes at both $0$ and $T$. The first condition makes every $a_i=0$. The second requires $b_i{S}_{{\lambda }_i}(T)=0$ for every $i$, with at least one nonzero $b_i$. For $T>0$, the factors for zero and negative eigenvalues never vanish; positive eigenvalues give exactly the sine zeros. ThereforeThe corresponding points are $\gamma (T)$. If several positive eigenspaces have a zero at the same parameter, the conjugate multiplicity is the sum of their dimensions. This also shows that a locally symmetric Riemannian manifold has no conjugate points along a given geodesic when that geodesic's Jacobi curvature operator has no positive eigenvalues.

A connected Riemannian manifold is complete when its Riemannian distance is a complete metric space, meaning that every Cauchy sequence converges in the manifold. The Hopf-Rinow theorem says that for a connected finite-dimensional Riemannian manifold, this is equivalent to geodesic completeness, and also to compactness of every closed bounded subset. Under these equivalent conditions any two points are joined by a minimizing geodesic, and the Riemannian exponential map at every point is defined on its entire tangent space.

We use the usual connected, boundaryless-manifold convention for the global statements in Q2–Q5. Connectedness is essential, including for the diameter and group-growth conclusions. Without connectedness, the completeness and covering assertions apply to the target components met by the map; an unrelated incomplete component cannot be controlled by a local isometry into another component.

A local isometry preserves the Levi-Civita connection and hence geodesics, as well as the lengths of lifted curves. We first show that a complete source permits finite-length paths to lift all the way to their endpoints. Begin lifting a piecewise smooth path $c:[0,1]\to N$ from any chosen point above $c(0)$, using local inverse maps. If its lift is defined only up to a maximal $b\le 1$, thenCompleteness gives a limit point in $M$. A local inverse at that point continues the lift, including its endpoint. Thus such a finite obstruction is impossible. Connectedness of $N$ and piecewise smooth paths from $f(p)$ to any target point prove that $f$ is onto; no prior completeness of $N$ is used in this step.

Now lift the initial position and velocity of any geodesic in $N$. The resulting geodesic in $M$ exists for all real time by Hopf-Rinow theorem. Its image extends the original target geodesic by uniqueness of the geodesic initial-value problem. Hence $N$ is geodesically complete, and therefore complete.

The converse is false for a general local isometry, even an onto one. Give the interval $(0,4\pi )$ its ordinary metric and map it to the unit circle by $t\mapsto e^{it}$. This is an onto local isometry to a complete circle, but the sequence $t_j=1/j$ is Cauchy in the interval and has no limit there. It is not a covering map: the identity point of the circle has only one preimage, whereas nearby points have two.

The converse holds for a Riemannian covering. If $f$ is also a covering map and $N$ is complete, every target geodesic defined on $\mathbb{R}$ lifts globally with any specified initial point. A local isometry makes that lift a geodesic. Its initial velocity can be any prescribed source velocity because $df$ is an isomorphism. Thus all source geodesics extend for all time, and $M$ is complete by the Hopf-Rinow theorem.

Surjectivity was proved in part (b). Fix $q\in N$ and choose a normal ball $B=B_N(q,\epsilon )$ on which ${\exp }_q$ is a diffeomorphism from the tangent ball. For each $p\in f^{-1}(q)$ define a sectionCompleteness of $M$ guarantees that the source Riemannian exponential map is defined. Preservation of geodesics and uniqueness of their initial-value problem give $f\circ s_p={\mathrm{id}}_B$. Since $df$ is invertible, differentiation of this identity shows that $s_p$ is a local diffeomorphism; it is injective because it is a section. Its image $U_p$ is open, and $f:U_p\to B$ is a diffeomorphism.

These images are disjoint. If $s_p(y)=s_{p^{\prime }}(y)$, lift the unique radial geodesic from $y$ back to $q$ starting at that common point. Both section constructions give this same lift, and uniqueness forces its endpoint to be both $p$ and $p^{\prime }$. Conversely, for any $x\in f^{-1}(B)$, lift that reversed radial geodesic from $f(x)$ to $q$ starting at $x$. Completeness supplies its full finite interval; its endpoint is some $p\in f^{-1}(q)$. Reversing the lift shows $x=s_p(f(x))$. ConsequentlyEvery point has such an evenly covered neighborhood, proving that the complete local isometry is a covering. This argument does not assume a uniform positive source injectivity radius.

A connected complete Riemannian manifold is disconnected at infinity when the complement of some compact set has at least two unbounded connected components. Equivalently it has at least two ends. A line in a Riemannian manifold is a unit-speed geodesic $\ell :\mathbb{R}\to M$ satisfying $d(\ell (s),\ell (t))=|s-t|$ for every $s,t$.

Let $K$ be the compact separating set, enlarged inside a fixed compact metric ball if needed for bounds. Choose $x_j,y_j$ in two different unbounded components of $M\setminus K$, with both distances from $K$ tending to infinity. By the Hopf-Rinow theorem, join them by unit-speed minimizing geodesics. Every such segment meets $K$, since a path avoiding it cannot join its different components. Reparametrize a segment so that one crossing point is at time zero. Its parameter interval is $[-a_j,b_j]$, where $a_j,b_j\to \infty$ by the distances from the endpoints to $K$.

The unit tangent bundle over $K$ is compact, so a subsequence of the initial position/velocity pairs converges. Continuous dependence of the geodesic equation on its initial data, and completeness, give a limiting geodesic defined on all of $\mathbb{R}$. For any fixed $s<t$, the approximating segments are defined on $[s,t]$ for large $j$ and satisfy $d({\gamma }_j(s),{\gamma }_j(t))=t-s$. Continuity of the distance passes this equality to the limit. Thus the limiting geodesic is a line, proving a line from disconnection at infinity.

The Cheeger-Gromoll splitting theorem states that a complete connected Riemannian manifold with nonnegative Ricci curvature that contains a line in a Riemannian manifold is isometric to a Riemannian product $N\times \mathbb{R}$ with the Euclidean metric on the line. The factor $N$ is complete and has nonnegative Ricci curvature. Under the isometry one can take the given line to $\{q\}\times \mathbb{R}$, up to translation and reversal. A line means global minimization between every pair of its points, not merely a complete geodesic.

Suppose, for a contradiction, that $X=M\times \mathbb{R}$ has a complete metric with zero Ricci curvature. Since $M$ is closed and connected, the complement of $M\times [-A,A]$ has two unbounded components in any complete metric on $X$: their closures are noncompact, while closed bounded sets are compact by the Hopf-Rinow theorem. Thus $X$ is disconnected at infinity, and part (a) supplies a line. The Cheeger-Gromoll splitting theorem gives an isometryThe Riemannian product curvature formula makes $N$ complete and Ricci-flat. A three-dimensional Ricci-flat Riemannian manifold is flat: for any orthonormal triple, the three trace equations are ${\mathrm{Ric}}_{11}=K_{12}+K_{13}$ and their cyclic companions. Solving gives $K_{ij}={\mathrm{Ric}}_{ii}+{\mathrm{Ric}}_{jj}-\mathrm{Scal}/2=0$. Every plane can occur in such a triple, so every sectional curvature vanishes, and hence so does the Riemann curvature tensor.

The complete simply connected universal cover of $N$ is therefore Euclidean ${\mathbb{R}}^3$, and that of $X$ is Euclidean ${\mathbb{R}}^4$. But the topological universal cover of $X$ is $\tilde{M}\times \mathbb{R}$. Its deformation retraction onto $\tilde{M}\times \{0\}$ shows that contractibility of this cover would imply contractibility of $\tilde{M}$, contrary to the hypothesis. Therefore no complete Ricci-flat metric exists on $M\times \mathbb{R}$. Notice that no product assumption was made about the hypothetical original metric.

There is a normalization issue in the printed constant. Let ${\mathrm{Ric}}_{\mathrm{tr}}$ denote the trace convention in the catalog's Ricci curvature, and write $\overline{\mathrm{Ric}}={\mathrm{Ric}}_{\mathrm{tr}}/(n-1)$ for normalized Ricci curvature, with $n\ge 2$. The claimed $\pi /\sqrt{k}$ bound corresponds to $\overline{\mathrm{Ric}}\ge kg$, or equivalently ${\mathrm{Ric}}_{\mathrm{tr}}\ge (n-1)kg$. We prove the estimate in both conventions.

Take a unit-speed minimizing geodesic of length $L>0$ and perpendicular parallel orthonormal fields $E_1,\dots ,E_{n-1}$. Set $V_i(t)=\sin (\pi t/L)E_i(t)$. The fields vanish at the endpoints. The allowed second variation of geodesic energy makes each Riemannian index form nonnegative. Summing them givesunder the normalized lower bound. Thus $L\le \pi /\sqrt{k}$. Completeness and the Hopf-Rinow theorem supply a minimizing geodesic between any two points, so this bounds the diameter. A closed ball of that radius about any point contains the whole manifold and is compact, again by the Hopf-Rinow theorem.

Hence the intended Bonnet-Myers theorem conclusion isIf the printed $\mathrm{Ric}\ge k$ means trace Ricci curvature instead, the same integral has lower curvature term $kL/2$ and yieldsThe smaller bound is false under that convention when $n>2$: the round unit $S^n$ has trace Ricci tensor $(n-1)g$ and diameter $\pi$. Taking $k=n-1$ contradicts the printed $\pi /\sqrt{k}$ inequality. The compactness assertion remains true.

Pointwise positive Ricci curvature does not imply compactness or a finite diameter. The complete positively curved paraboloid $z=x^2+y^2$ has induced metricand Gaussian curvatureFor example, this follows from the graph formula $K=\det D^{2}z/(1+|Dz{|}^2{)}^2$. The metric dominates the Euclidean plane metric, so it is complete by upward stability of Riemannian completeness. Its diameter is infinite since its distance dominates Euclidean distance. On a surface the Ricci tensor is $Kg$, so it is strictly positive everywhere, but has no uniform positive lower bound.

There is no positive-Ricci metric on $S^2\times S^1$. If one existed, compactness of the unit tangent bundle would give ${\mathrm{Ric}}_{\mathrm{tr}}\ge cg$ for some $c>0$. The lifted metric on the universal cover is complete by Q2 and satisfies the same positive lower bound. Part (a) makes this cover compact. A covering fiber is closed and discrete and therefore finite, implying a finite fundamental group. Butwhich is infinite. This is the finite fundamental group from a uniform positive Ricci bound obstruction; it rules out every possible metric, not just the standard product metric.

Parametrize the nonconstant closed geodesic at unit speed on $[0,L]$. Its parallel transport around the loop fixes the tangent $\dot{\gamma }(0)$. Since the manifold is orientable, it preserves orientation on the full tangent space, and hence has determinant one on the normal space. This normal space has odd dimension because the manifold has even dimension.

An odd-dimensional special orthogonal transformation has a fixed vector. Indeed, nonreal eigenvalues of a real orthogonal transformation occur in conjugate pairs with product one; the real eigenvalues are $\pm 1$, and determinant one makes the number of $-1$ eigenvalues even. Odd dimension therefore requires a $+1$ eigenvalue. Transport a unit fixed normal vector around the geodesic to obtain a periodic parallel field $V$, with $D_{t}V=0$ and $V\perp \dot{\gamma }$.

The variation ${\gamma }_s(t)={\exp }_{\gamma (t)}(sV(t))$ is defined uniformly for small $s$, because the image of the original loop is compact. Periodicity gives closed curves and a homotopy through loops. With the standard positive-sectional-curvature Riemannian index form, the second variation of geodesic energy isThe first variation vanishes for a closed geodesic, since the endpoint terms cancel. Therefore $E({\gamma }_s)<E({\gamma }_0)=L/2$ for sufficiently small nonzero $s$. The Cauchy-Schwarz inequality givesThus the closed geodesic is homotopic to a strictly shorter closed curve. This proves the instability of a closed geodesic in positive even-dimensional curvature; completeness of the manifold is unnecessary for this compact-loop variation. Constant loops are excluded by the usual nonconstant meaning of a closed geodesic in this assertion.

Choose a finite symmetric generating set $S$ for $\Gamma$, let $B_S(m)$ be its word metric ball, and fix $p\in M$. Set $D={\max }_{s\in S}d(p,sp)$. The triangle inequality and invariance of distance implyThe properly discontinuous group action has a finite point stabilizer $H={\Gamma }_p$. It also gives uniform separation of distinct orbit points. To see this, only finitely many group elements can send $p$ into a fixed compact ball about $p$, by proper discontinuity and the Hopf-Rinow theorem. The positive distances among this finite list, together with the radius of the ball for all other elements, have a positive lower bound $\delta$. Thus $d(\gamma p,{\gamma }^{\prime }p)\ge \delta$ whenever the centers differ.

Take $0<\epsilon <\delta /2$ and write $v=\mathrm{Vol}B(p,\epsilon )>0$. Balls of radius $\epsilon$ about distinct orbit points are disjoint and all have volume $v$, because the action is by isometries. Those corresponding to words of length at most $m$ lie in $B(p,Dm+\epsilon )$. Each orbit point arises from at most $|H|$ group elements in $B_S(m)$, soThe final inequality is the allowed Bishop-Gromov inequality for nonnegative Ricci curvature. If the entire group fixes $p$, proper discontinuity already makes it finite and the conclusion is immediate. Otherwise the preceding separation argument applies. ThereforeThis is polynomial growth of a group of degree at most $n$, proved by polynomial group growth from orbit packing. Neither freeness of the action nor compactness of the quotient was assumed.

Put $\theta (r)=\mathrm{Vol}B(p,r)/({\omega }_n{r}^n)$. The Bishop-Gromov inequality says that $\theta$ is nonincreasing and at most one, while smoothness gives $\theta (r)\to 1$ as $r\downarrow 0$. The prescribed limit at infinity is also one. Hence $\theta (r)=1$ for every $r>0$.

Here is how equality gives the global Euclidean rigidity of maximal asymptotic volume ratio. Assume first $n\ge 2$. In geodesic polar coordinates centered at $p$, let $c(\xi )$ be the cut time along unit direction $\xi$, and $J(r,\xi )$ the positive radial volume density before that time. The comparison giving the Bishop-Gromov inequality says $J(r,\xi )\le r^{n-1}$, andNo cut time can be finite. If $c({\xi }_0)<t$, then the radial segment at time $t$ is not minimizing, so $d(p,{\exp }_p(t{\xi }_0))<t$. This strict inequality persists for nearby directions by continuity, and their cut times are also at most $t$. At any radius $R>t$, that open set of directions omits all radial volume from $t$ to $R$. Even the maximal density $r^{n-1}$ cannot then give Euclidean volume, contradicting $\theta (R)=1$. Thus $c(\xi )=\infty$ for all directions. Equality of all ball volumes and continuity now force $J(r,\xi )=r^{n-1}$ everywhere.

Let $A(r)X={\nabla }_X{\partial }_r$ on the distance sphere. This is the negative of the outward shape operator in the convention $S=-\nabla \nu$. Put $h=\mathrm{tr}A={\partial }_r\log J$. The traced radial Riccati equation for distance spheres isSince $h=(n-1)/r$, it givesBut $\mathrm{tr}(A^2)\ge h^2/(n-1)=(n-1)/r^2$ by the Cauchy-Schwarz inequality, and the Ricci term is nonnegative. Equality therefore holds in both, so $A=r^{-1}I$. Writing the polar metric as $d{r}^2+g_r$, its evolution is ${\partial }_r{g}_r=2g_r/r$. Its small-radius Euclidean limit then givesThe Riemannian exponential map is onto by the Hopf-Rinow theorem, has no conjugate points, and is injective because there is no finite cut time. Explicitly, two distinct minimizing radial geodesics reaching the same point would, after extending one past that point, create a length-minimizing broken curve with a genuine corner; smoothing the corner shortens it. That contradicts global minimization of every radial segment. Thus ${\exp }_p$ is a global diffeomorphism, and the displayed metric provesIn dimension one, a connected complete manifold without boundary is a line or a circle. The circle has limiting volume ratio zero, so the hypothesis leaves only the Euclidean line.

Take the orientation-preserving affine group of the real line, written asIt is a connected Lie group, since its underlying manifold is $(0,\infty )\times \mathbb{R}$. We use the permitted alternative to part (a), a direct obstruction from the Adjoint representation of a Lie group.

If a bi-invariant Riemannian metric existed, conjugation $h\mapsto gh{g}^{-1}$ would be an isometry, being a composition of a left and a right translation. Its differential at the identity would preserve the positive inner product on the Lie algebra. Let $T$ be the nonzero infinitesimal translation,Direct multiplication gives ${\mathrm{Ad}}_{g}T=gT{g}^{-1}=2T$. Metric invariance would then imply $\Vert T{\Vert }^2=\Vert 2T{\Vert }^2=4\Vert T{\Vert }^2$, impossible for a nonzero vector in a positive inner product. Therefore this connected affine Lie group has no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric; allowing an indefinite metric would be a different question.