Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 16 (Global Riemannian Geometry)

Write $T=\dot{\gamma }$ along an affinely parametrized geodesic and $D_t={\nabla }_T$ for the covariant derivative from the Levi-Civita connection. We use the Riemann curvature tensor convention $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$, for which positive sectional curvature means $⟨R(X,Y)Y,X⟩>0$ on an orthonormal pair. A Jacobi field is a smooth vector field along $\gamma$ satisfyingSuch a field describes the first-order separation of nearby geodesics, as the geodesic variation construction in part (c) shows.

For $t_0>0$, $\gamma (t_0)$ is a conjugate point of $\gamma (0)$ along this geodesic if a nonzero Jacobi field satisfies $J(0)=J(t_0)=0$. The multiplicity of a conjugate point isIt is equivalently the dimension of the kernel of the differential of the Riemannian exponential map at $t_0\dot{\gamma }(0)$, by Jacobi fields vanishing at their initial point. Conjugacy concerns the specified geodesic and its two parameter values, even when its endpoints represent the same point of the Riemannian manifold.

Suppose $J$ is a Jacobi field vanishing at both endpoints of a nonconstant geodesic segment $[0,t_0]$. Its tangential component $h(t)=⟨J(t),T(t)⟩$ satisfiesbecause $D_{t}T=0$ and the Riemann curvature tensor has the usual skew-adjoint symmetry. The two zero endpoint values imply $h\equiv 0$. Both $J$ and $D_{t}J$ are therefore perpendicular to $T$.

The linear map from these Jacobi fields to $T(0{)}^{\perp }$ given by $J\mapsto D_{t}J(0)$ is injective: $J(0)=0$ together with $D_{t}J(0)=0$ forces $J\equiv 0$ by uniqueness for the linear ordinary differential equation defining a Jacobi field. Its target has dimension $n-1$, soThis proves the endpoint-vanishing normal Jacobi fields have dimension at most n-1 bound. For a constant geodesic the equation is $J^{\prime \prime }=0$, and two zero values instead force $J=0$, so there is no conjugate point.

Put $p=\gamma (0)$, $v=\dot{\gamma }(0)$, $a=J(0)$ and $b=D_{t}J(0)$. Choose the short curve $c(s)={\exp }_p(sa)$ and let $P_s:T_{p}M\to T_{c(s)}M$ be parallel transport along $c$. The tangent vectorssatisfy $V(0)=v$ and ${\nabla }_{s}V(0)=b$. Define a geodesic variation byFor each fixed $s$ this is an affinely parametrized geodesic with initial point $c(s)$ and initial velocity $V(s)$. Although the Riemannian manifold need not be complete, smooth dependence for the geodesic equation on a neighborhood of the compact reference segment provides a single $\epsilon >0$ on which $F$ exists for all $|s|<\epsilon$ and $0\le t\le 1$. Thus completeness is unnecessary.

Set $X(t)={\partial }_{s}F(0,t)$. The torsion-free Levi-Civita connection gives ${\nabla }_s{\partial }_{t}F={\nabla }_t{\partial }_{s}F$. Differentiating ${\nabla }_t{\partial }_{t}F=0$ with respect to $s$, and commuting the two covariant derivatives, yieldsAlso $X(0)=c^{\prime }(0)=a$ and $D_{t}X(0)={\nabla }_{s}V(0)=b$. Uniqueness for the Jacobi field equation now gives $X(t)=J(t)$ on the whole segment. This is realization of Jacobi fields by geodesic variations.

The printed formula switches the order of its two variables. The consistent notation here is $F(0,t)=\gamma (t)$ and $J(t)={\partial }_{s}F(0,t)$. A parametrized surface here means a smooth variation map; it cannot be required to be an immersion, since the allowed Jacobi field $J=0$ would already violate that condition along the reference curve.

Consider the subsetIt is nonempty by the given initial data. A local isometry preserves the Levi-Civita connection and therefore affinely parametrized geodesics. On a sufficiently small normal neighborhood of $x\in A$, this givesThe right sides agree, so $f_1=f_2$ on that neighborhood and their differentials agree there as well. Consequently $A$ is open.

It is also closed. If $x_j\in A$ converges to $x$, continuity first gives $f_1(x)=f_2(x)$. Choose coordinate charts around $x$ and this common image. The coordinate matrices of the two differentials depend continuously on the base point; their equalities at $x_j$ therefore pass to $x$. Thus $x\in A$. Since $M$ is connected, the nonempty subset $A$ that is both open and closed equals $M$. Hence $f_1=f_2$ everywhere. This proves local isometries are determined by first-order data without requiring completeness.

Let $\pi :\tilde{M}\to M$ be the universal cover and equip it with the pullback Riemannian metric $\tilde{g}={\pi }^{\ast }g$. The covering map is a local isometry, so $\tilde{M}$ has constant sectional curvature one.

We must check completeness of the lifted Riemannian metric. A geodesic on $\tilde{M}$ projects to a geodesic on $M$. By geodesic completeness of $M$, the projected geodesic extends for all real time. Its unique lift with a specified initial point extends the original lifted geodesic, because $\pi$ is a local isometry. Hence $\tilde{M}$ is geodesically complete, and the Hopf-Rinow theorem also gives metric completeness.

The allowed constant-curvature form of Cartan-Ambrose-Hicks theorem says that a linear tangent-space isometry between complete simply connected Riemannian manifolds of the same constant sectional curvature extends to a global isometry. The unit round sphere $S^n$ is complete, simply connected for $n\ge 2$, and has sectional curvature one. Comparing it with $\tilde{M}$ therefore yieldsThis is the covering step in the classification of complete positive constant-curvature manifolds. As usual for this assertion, $n\ge 2$ is understood: in dimension one there are no sectional two-planes, and the universal cover of the circle is the line, not the circle.

By part (b), $M=S^n/\Gamma$, where the group of deck transformations acts freely by round isometries. Every round isometry is the restriction of an orthogonal transformation of ${\mathbb{R}}^{n+1}$: it preserves the inner product $⟨x,y⟩=\cos d(x,y)$, so its images of an orthonormal basis determine its linear extension.

Suppose $n$ is even. Every $A\in SO(n+1)$ has eigenvalue one. Indeed, the nonreal eigenvalues come in conjugate pairs with product one, while the remaining real eigenvalues are $\pm 1$. Their number is odd and their product is one, forcing at least one $+1$. Such a transformation fixes a unit vector. Freeness of the deck transformation action consequently implies that the determinant map $\Gamma \to \{\pm 1\}$ has trivial kernel. Thus $|\Gamma |\le 2$.

If $\Gamma$ is nontrivial with generator $A$, then $A^2=I$. Its eigenvalues are $\pm 1$, and freeness excludes the eigenvalue $+1$. Hence $A=-I$, the antipodal involution. We obtainThis proves the even-dimensional part of the classification of complete positive constant-curvature manifolds.

The conclusion fails in odd dimensions. For $n=2m-1\ge 3$, identify $S^{2m-1}$ with the unit sphere in ${\mathbb{C}}^m$. Scalar multiplication by $\zeta =e^{2\pi i/r}$, with $r\ge 3$, generates a freely acting cyclic group of isometries: ${\zeta }^{j}z=z$ for nonzero $z$ forces ${\zeta }^j=1$. The quotient is complete with constant sectional curvature one and has fundamental group $C_r$, so it is neither the simply connected sphere nor Real projective space with fundamental group $C_2$. In particular $S^3/C_3$ is a lens space counterexample.

Let $\tau >0$ be the first conjugate time and choose a nonzero Jacobi field $J$ with $J(0)=J(\tau )=0$. It is normal by part 1(b), and $D_{t}J(\tau )\ne 0$, since zero value and derivative there would force $J=0$. Fix any $T>\tau$ and extend $J$ by zero to a continuous piecewise smooth vector field $V$ on $[0,T]$.

For endpoint-vanishing fields, the Riemannian index form isIntegration by parts on $[0,\tau ]$ and the Jacobi field equation give $I(V,V)=0$ andChoose a smooth normal field $W$, zero at $0,T$, with $W(\tau )=-D_{t}J(\tau )$. Thenfor sufficiently small positive $\epsilon$. The corner in $V$ causes no difficulty: smooth endpoint-vanishing fields approximate $V+\epsilon W$ in the Sobolev space $H^1$, and the Riemannian index form is continuous in this norm on the compact segment. A smooth approximant therefore still has negative index form.

Realize this smooth field as a fixed-endpoint curve variation using the Riemannian exponential map. The allowed second variation of geodesic energy gives $E^{\prime }(0)=0$ and $E^{\prime \prime }(0)<0$, so some nearby curve has energy less than $T/2$, the energy of the unit-speed reference geodesic. The Cauchy-Schwarz inequality bounds its length byIt is shorter than the reference segment. Thus a geodesic cannot minimize beyond its first conjugate time. This argument deliberately uses $T>\tau$; it does not rule out minimization up to the conjugate point itself.

Let $t_0$ be the finite cut time and put $q=\gamma (t_0)$. Continuity of Riemannian distance shows that the segment to $q$ still minimizes and $d(p,q)=t_0$. Choose $t_j\downarrow t_0$ with $t_j>t_0$. By the Hopf-Rinow theorem, there is a minimizing geodesic from $p$ to $q_j=\gamma (t_j)$ with length ${\ell }_j=d(p,q_j)<t_j$ and initial unit vector $v_j$. Continuity gives ${\ell }_j\to t_0$.

The unit tangent sphere is compact, so after a subsequence $v_j\to v$. Smooth dependence of the Riemannian exponential map gives ${\exp }_p(t_{0}v)=q$. The geodesic with initial vector $v$ has length $t_0=d(p,q)$ up to $q$, hence minimizes. If $v\ne \dot{\gamma }(0)$, this is the required second minimizing geodesic.

Suppose instead $v=\dot{\gamma }(0)$ and $q$ is not a conjugate point. Then the differential of ${\exp }_p$ at $t_0\dot{\gamma }(0)$ is nonsingular. The inverse function theorem makes ${\exp }_p$ injective on a neighborhood of this vector. For large $j$, both ${\ell }_j{v}_j$ and $t_j\dot{\gamma }(0)$ belong to that neighborhood and exponentiate to $q_j$. They must be equal, implying ${\ell }_j=t_j$, a contradiction. Thus either a distinct minimizing geodesic exists or $q$ is conjugate. In the latter case it is the first conjugate point, since part (a) excludes any earlier one on a minimizing segment. We have proved the cut-point dichotomy for complete Riemannian manifolds.

Here the relevant concept is a Riemannian cut point, not the topological cut point obtained by disconnecting a space through removal of a point.

Take unit-curvature Real projective space ${\mathbb{RP}}^n=S^n/\{\pm 1\}$ with $n\ge 2$, and any point $p=[v]$. It is complete with positive sectional curvature. A unit-speed geodesic through $p$ lifts toSince the two representatives of an endpoint are antipodal, Riemannian distance in the quotient is the smaller of the two spherical distances:This equals $t$ for $0\le t\le \pi /2$ and is strictly smaller just afterwards. Thus the Riemannian cut point occurs at time $\pi /2$ in every direction.

Along a curvature-one geodesic, every normal Jacobi field with initial value zero is $J(t)=\sin t{P}_{t}a$, where $P_t$ is parallel transport and $a\perp \dot{\gamma }(0)$. Its first further zero is $t=\pi$. The quotient map is a local isometry, so the Jacobi field equation and conjugacy times are unchanged. HenceThis is cut and conjugate times in round real projective space.

For a connected complete $n$-dimensional Riemannian manifold, $n\ge 2$, the Bonnet-Myers theorem states thatThe fundamental group conclusion follows by applying the same diameter and compactness result to the complete universal cover: its covering fiber is closed and discrete in a compact space, hence finite.

There is an important Ricci curvature convention in the printed question. With the standard trace convention used in this wiki, the literal hypothesis $\mathrm{Ric}\ge kg$ yields instead $\mathrm{diam}M\le \pi \sqrt{(n-1)/k}$. The sphere constants in parts (b) and (c) require normalized Ricci curvature $\overline{\mathrm{Ric}}=\mathrm{Ric}/(n-1)\ge kg$, equivalently the stronger trace bound $\mathrm{Ric}\ge (n-1)kg$. We prove those conclusions with this intended normalization below and exhibit counterexamples to the literal trace interpretation. Thus no factor of $n-1$ is silently discarded.

Use the intended trace bound $\mathrm{Ric}\ge (n-1)kg$ and write $D=\pi /\sqrt{k}$. We state precisely the allowed Bishop volume comparison with a positive-curvature model. Setwhere ${\omega }_{n-1}$ is the area of the unit $(n-1)$-sphere. For $V_p(r)=\mathrm{Vol}B(p,r)$, the ratio $V_p(r)/v_k(r)$ is nonincreasing on $0<r<D$, tends to one at zero, and is at most one. Before the cut time, the radial volume density $J(r,\theta )$ in geodesic polar coordinates is at most $s_k(r{)}^{n-1}$.

By the Bonnet-Myers theorem, $\mathrm{diam}M\le D$. Therefore $V_p(D)=\mathrm{Vol}M$. If distance-$D$ endpoints exist, they do not affect this volume equality: the distance sphere is contained in the smooth image ${\exp }_p(D{S}_p^{n-1})$, which has zero $n$-dimensional Riemannian volume. Taking limits in the comparison ratio at $D$ is legitimate. The assumed equality of total volume gives $V_p(D)/v_k(D)=1$, so monotonicity forcesfor every $p$.

We now extract the equality geometry, rather than merely naming a rigidity theorem. On a small normal neighborhood of $p$, the continuous nonnegative difference $s_k(r{)}^{n-1}-J(r,\theta )$ has integral zero. It is therefore identically zero. Let $A(X)={\nabla }_X{\partial }_r$ on a distance sphere and $h=\mathrm{tr}A={\partial }_r\log J$. Equality of the density gives $h=(n-1)s_k^{\prime }/s_k$. The traced radial Riccati equation for distance spheres readsSince $s_k^{\prime \prime }=-k{s}_k$, substituting $h$ makes the sum of the following two nonnegative quantities zero:Equality in the Cauchy-Schwarz inequality for the eigenvalues of the self-adjoint operator $A$ forces $A=(s_k^{\prime }/s_k)I$. The angular metric $g_r$ in geodesic polar coordinates consequently satisfies ${\partial }_r{g}_r=2(s_k^{\prime }/s_k)g_r$. Smoothness at the center fixes its integration constant, since $r^{-2}{g}_r\to g_{S^{n-1}}$. Henceon the normal ball: the Riemannian metric is locally the round curvature-$k$ metric. As $p$ was arbitrary, $M$ has constant sectional curvature $k$.

The classification of complete positive constant-curvature manifolds now gives $M=S_k^n/\Gamma$ with $\Gamma$ a finite freely acting group of isometries. Covering degree gives $\mathrm{Vol}M=\mathrm{Vol}{S}_k^n/|\Gamma |$. Volume equality forces $|\Gamma |=1$, provingThis establishes spherical rigidity of maximal total volume under the correct normalization.

Under the literal trace bound, the assertion is false in dimension four. Take the Riemannian product $M=S^2(r)\times S^2(r)$ with $r^2=1/(\sqrt{6}k)$. Its Ricci curvature is $r^{-2}g=\sqrt{6}kg\ge kg$, andBut a mixed tangent two-plane has sectional curvature zero, whereas a factor plane has curvature $\sqrt{6}k$. Thus this manifold is not the round sphere despite having the literal hypotheses.

Again assume the intended trace Ricci curvature bound $\mathrm{Ric}\ge (n-1)kg$, and put $D=\pi /\sqrt{k}$. Compactness from the Bonnet-Myers theorem ensures that the diameter is attained: choose $p,q$ with $d(p,q)=D$. For $0<r<D$, the open balls $B(p,r)$ and $B(q,D-r)$ are disjoint, because a point in both would violate the triangle inequality.

The clearly stated Bishop volume comparison with a positive-curvature model from part (b), applied between radius $r$ and radius $D$, givesHere both radius-$D$ ball volumes equal the total volume, as shown in part (b). The symmetry $s_k(D-t)=s_k(t)$ givesThe two lower bounds therefore sum to $\mathrm{Vol}M$, while disjointness bounds that same sum above by $\mathrm{Vol}M$. Both inequalities must be equalities. In particularLetting $r\downarrow 0$, the local Euclidean volume asymptotic gives one on the left. Thus $\mathrm{Vol}M=v_k(D)$, and the detailed equality proof in part (b) applies. We obtainThis proves maximal diameter rigidity from disjoint comparison balls directly, including the required equality argument.

The literal trace interpretation again has a counterexample. On $M=S^2(r)\times S^2(r)$ with $r^2=1/(2k)$, the Ricci curvature equals $2kg\ge kg$. For the product Riemannian metric, distances satisfy $d^2=d_1^2+d_2^2$: every product curve has length at least $\sqrt{d_1^2+d_2^2}$ by the integral triangle inequality, and a pair of constant-speed minimizing factor geodesics attains it. ConsequentlyThe mixed sectional curvature is zero, so this manifold is not $S_k^4$. The stronger bound, or normalized Ricci curvature convention, is essential.

Choose a finite symmetric generating set $S$ for $\Gamma$, a base point $p$, and put $L={\max }_{s\in S}d(p,sp)$. If the word length of $\gamma$ is at most $m$, the triangle inequality and invariance of Riemannian distance under isometries give $d(p,\gamma p)\le Lm$.

A properly discontinuous group action has finite point stabilizer $F={\Gamma }_p$ and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move $p$ into any bounded ball, since closed bounded balls are compact by the Hopf-Rinow theorem. Thus there is $\rho >0$ such that distinct orbit points have distance more than $2\rho$: choose it using the finitely many orbit points within a fixed small ball of $p$, and translate the resulting separation by isometries. If the whole orbit is one point, the group itself is finite and the conclusion is immediate.

Let $N_m$ be the number of distinct orbit points represented by the word metric ball of radius $m$. Their radius-$\rho$ balls are disjoint, have equal positive volume $v_0=\mathrm{Vol}B(p,\rho )$, and lie inside $B(p,Lm+\rho )$. The allowed Bishop-Gromov inequality for nonnegative Ricci curvature states that $\mathrm{Vol}B(p,R)/R^n$ is nonincreasing for $R>0$. ThereforeEach orbit point has at most $|F|$ preimages inside the word ball, since its full fiber is a coset of $F$. For the growth function of a finitely generated group, this givesHence $\Gamma$ has polynomial growth of a group of degree at most $n$. The finite stabilizer factor makes the proof valid even when the properly discontinuous action is not free. This is polynomial growth of properly discontinuous isometry groups.

Let $\Gamma ={\pi }_1(M)$. A compact connected smooth manifold has finite CW complex homotopy type, hence a finitely generated fundamental group. Its universal cover with the lifted Riemannian metric is complete by the lifted-geodesic argument in part 2(b) and has nonnegative Ricci curvature. The deck transformations act freely and properly discontinuously by isometries. Part (a) givesThe degree-one Hurewicz theorem identifies $H_1(M;\mathbb{Z})$ with the abelianization of $\Gamma$. This finitely generated abelian group has free rank $b=b_1(M)$, so there is a surjection $\Gamma \to {\mathbb{Z}}^b$. Choose lifts ${\gamma }_1,\dots ,{\gamma }_b$ of a basis and let $A\ge 1$ bound their word lengths. For every $(z_1,\dots ,z_b)$ with $\sum |z_i|\le m$, the product ${\gamma }_1^{z_1}\cdots {\gamma }_b^{z_b}$ has word length at most $Am$ and maps to that vector. Distinct vectors therefore give distinct group elements. The integer cube $|z_i|\le \lfloor m/b\rfloor$ shows, for $b>0$, thatfor some $c>0$ and all sufficiently large $m$. Comparing with the degree-$n$ upper bound forces $b\le n$; the case $b=0$ is immediate. ThusThis is the first Betti-number bound from polynomial fundamental-group growth.

The standard connectedness convention for a Riemannian manifold is needed here. If disconnected manifolds are allowed, the bound applies to each connected component; a disjoint union of two flat $n$-tori has first Betti number $2n$ and contradicts a global bound by $n$.

Consider the orientation-preserving affine group of the real line, realized asThe coordinates $(\log a,b)$ identify it smoothly with ${\mathbb{R}}^2$, so this Lie group is connected. Choose the Lie algebra element $E$ and the group element $D$:Direct matrix multiplication gives ${\mathrm{Ad}}_{D}E=DE{D}^{-1}=2E$.

If a bi-invariant Riemannian metric existed, every conjugation map $x\mapsto Dx{D}^{-1}$ would be an isometry fixing the identity. Its differential there, the Adjoint representation of a Lie group, would preserve the positive-definite inner product at the identity. Thusa contradiction since $E\ne 0$. Therefore this connected Lie group admits no bi-invariant Riemannian metric. This is the adjoint dilation obstruction to a bi-invariant Riemannian metric, an elementary alternative permitted by the question. The positive definite Riemannian metric interpretation matters; no claim about indefinite metrics is needed.