For a complete connected -dimensional Riemannian manifold with , , put andHere is the area of the unit -sphere, and is the Riemannian volume of a radius- ball in the round curvature- sphere. The Bishop-Gromov inequality asserts that is nonincreasing for , tends to one at zero, and is at most one. In geodesic polar coordinates, the radial volume density before the cut time is at most . The bound uses trace Ricci curvature; with normalized Ricci curvature it reads .
The Bishop-Gromov inequality forces every volume ratio to equal one if the asymptotic ratio is one. Equality in the radial Jacobian comparison forces the radial operator of each distance sphere to be (it is the negative of the outward shape operator in the convention ), by the traced radial Riccati equation for distance spheres and equality in Cauchy-Schwarz inequality. Any finite cut time would remove a positive cone of radial volume at a later radius, contradicting equality. Thus the Riemannian exponential map is global and one-to-one, and the polar metric is , the Euclidean metric.
Past exam of the mathematics course of the University of Cambridge 2003 iii Paper 16 5 a Solution Created 2026-10-03 Updated 2026-10-14
Choose a finite symmetric generating set for , a base point , and put . If the word length of is at most , the triangle inequality and invariance of Riemannian distance under isometries give .
A properly discontinuous group action has finite point stabilizer and a discrete orbit. More explicitly, properness on compact sets shows that only finitely many elements move into any bounded ball, since closed bounded balls are compact by the Hopf-Rinow theorem. Thus there is such that distinct orbit points have distance more than : choose it using the finitely many orbit points within a fixed small ball of , 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 be the number of distinct orbit points represented by the word metric ball of radius . Their radius- balls are disjoint, have equal positive volume , and lie inside . The allowed Bishop-Gromov inequality for nonnegative Ricci curvature states that is nonincreasing for . ThereforeEach orbit point has at most preimages inside the word ball, since its full fiber is a coset of . For the growth function of a finitely generated group, this givesHence has polynomial growth of a group of degree at most . 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.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 19 5 a Solution Created 2026-10-03 Updated 2026-10-14
Choose a finite symmetric generating set for , let be its word metric ball, and fix . Set . The triangle inequality and invariance of distance implyThe properly discontinuous group action has a finite point stabilizer . It also gives uniform separation of distinct orbit points. To see this, only finitely many group elements can send into a fixed compact ball about , 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 . Thus whenever the centers differ.
Take and write . Balls of radius about distinct orbit points are disjoint and all have volume , because the action is by isometries. Those corresponding to words of length at most lie in . Each orbit point arises from at most group elements in , soThe final inequality is the allowed Bishop-Gromov inequality for nonnegative Ricci curvature. If the entire group fixes , 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 , proved by polynomial group growth from orbit packing. Neither freeness of the action nor compactness of the quotient was assumed.
Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 19 5 b Solution Created 2026-10-03 Updated 2026-10-14
Put . The Bishop-Gromov inequality says that is nonincreasing and at most one, while smoothness gives as . The prescribed limit at infinity is also one. Hence for every .
Here is how equality gives the global Euclidean rigidity of maximal asymptotic volume ratio. Assume first . In geodesic polar coordinates centered at , let be the cut time along unit direction , and the positive radial volume density before that time. The comparison giving the Bishop-Gromov inequality says , andNo cut time can be finite. If , then the radial segment at time is not minimizing, so . This strict inequality persists for nearby directions by continuity, and their cut times are also at most . At any radius , that open set of directions omits all radial volume from to . Even the maximal density cannot then give Euclidean volume, contradicting . Thus for all directions. Equality of all ball volumes and continuity now force everywhere.
Let on the distance sphere. This is the negative of the outward shape operator in the convention . Put . The traced radial Riccati equation for distance spheres isSince , it givesBut by the Cauchy-Schwarz inequality, and the Ricci term is nonnegative. Equality therefore holds in both, so . Writing the polar metric as , its evolution is . 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 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.
Polynomial group growth from orbit packing 2026-10-14
A properly discontinuous group action by isometries of a finitely generated group on a complete -dimensional Riemannian manifold of nonnegative Ricci curvature yields the displayed group growth bound. Distinct orbit points are uniformly separated, while a word of length moves a basepoint by at most . Pack disjoint equal-volume small balls about the orbit points into and apply the Bishop-Gromov inequality. A finite point stabilizer contributes its order to the bound; freeness is unnecessary.
A finitely generated group of isometries acting properly discontinuously on a complete connected -dimensional Riemannian manifold with nonnegative Ricci curvature has polynomial growth of a group of degree at most . Choose so distinct points of an orbit are separated by more than . Let be the largest displacement of the base point by a generator. The radius- word metric ball maps into the metric ball of radius , with fibers of size the finite stabilizer . Disjoint radius- balls around these orbit points have equal volume and fit inside the ball of radius . The Bishop-Gromov inequality bounds their number by , giving the displayed estimate.
Radial Riccati equation for distance spheres 2026-10-14
Before the cut locus, define on the tangent space of a distance sphere. This is the negative of the shape operator with outward normal under the convention . A nonsingular matrix of radial Jacobi fields satisfies and , yielding the displayed matrix Riccati equation. Taking its trace gives , where . This is the local differential identity behind the Bishop-Gromov inequality.