Book 9B

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Course 9Book 9B: Comparison, Convergence and Heat on ManifoldsChapter 5

Splitting and Soul Theorems

Lines force splittings, souls, and asymptotic cones.

17 min read · Updated Oct 3, 2026

Read with Petersen's Riemannian Geometry, the chapters on Ricci curvature comparison (the splitting theorem) and on sectional curvature comparison (Toponogov's theorem and the soul theorem). Cheeger and Ebin's Comparison Theorems in Riemannian Geometry, chapters 8–9, is the classical account of splitting and souls.

In this chapter · 7 sections
  1. 5.1Lines and Busemann functions
  2. 5.2The splitting theorem
  3. 5.3Triangles: Toponogov's theorem
  4. 5.4Souls
  5. 5.5The view from infinity
  6. 5.6History
  7. 5.7Exercises

The compactness theorems of 9B.4 Convergence of Manifolds produce limits; this chapter says what limits with nonnegative curvature can look like. The answer comes from three structure theorems. The splitting theorem: nonnegative Ricci curvature plus a single straight line forces the manifold to be a product R×N\mathbb{R}\times N. The soul theorem: a complete noncompact manifold with nonnegative sectional curvature is a vector bundle over a compact totally geodesic submanifold, its soul. And the asymptotic cone: such a manifold, viewed from very far away, looks like a cone.

These theorems have few direct applications outside geometry, and this chapter does not invent any. Their use is internal. In Perelman's classification of the singularity models of the three-dimensional Ricci flow (12B.2 The Structure of κ-Solutions), a line in a limit forces a splitting, which produces the round cylinder S2×RS^2\times\mathbb{R}, the neck. The soul theorem says that every other noncompact model is diffeomorphic to R3\mathbb{R}^3, a cap. Necks and caps are the two pieces of the canonical neighbourhood theorem.

By the end of this chapter you will be able to:

  • define rays, lines and Busemann functions, and compute them in model spaces;
  • prove the splitting theorem from Laplacian comparison, the maximum principle and Bochner's formula;
  • state Toponogov's triangle comparison and check it on the sphere;
  • state the soul theorem and Perelman's solution of the soul conjecture, and find souls of examples;
  • describe asymptotic cones, and say what these results mean for the neck and the cap.

Lines and Busemann functions

A ray is a unit-speed geodesic γ:[0,∞)→M\gamma : [0, \infty) \to M that minimises between any two of its points; a line is such a geodesic defined on all of R\mathbb{R}. Every complete noncompact manifold has a ray from each point (a limit of minimising segments to points going to infinity). Lines are much rarer. A paraboloid has rays but no line, while a cylinder has many lines.

The Busemann function of a ray γ\gamma is

bγ(x)=lim⁡t→∞(d(x,γ(t))−t).b_\gamma(x) = \lim_{t\to\infty}\big(d(x, \gamma(t)) - t\big).

The limit exists because the expression is nonincreasing in tt (triangle inequality) and bounded below by −d(x,γ(0))-d(x, \gamma(0)). It is 11-Lipschitz. It measures "how far behind" xx is in a race to infinity along γ\gamma. In Rn\mathbb{R}^n, for γ(t)=tv\gamma(t) = tv, bγ(x)=−⟨x,v⟩b_\gamma(x) = -\langle x, v\rangle (Exercise 5.4): its level sets are the hyperplanes orthogonal to vv. In the hyperbolic plane they are horocycles.

The splitting theorem

Theorem 5.1 Cheeger–Gromoll splitting theorem

Let MM be complete with Ric⁡≥0\operatorname{Ric} \geq 0. If MM contains a line, then MM is isometric to a product R×N\mathbb{R}\times N, with NN complete with Ric⁡≥0\operatorname{Ric} \geq 0.

Proof. Let γ\gamma be the line, and b+b^+, b−b^- the Busemann functions of the rays t↦γ(t)t \mapsto \gamma(t) and t↦γ(−t)t \mapsto \gamma(-t), t≥0t \geq 0.

Step 1: b++b−≥0b^+ + b^- \geq 0, with equality on γ\gamma. By the triangle inequality, d(x,γ(t))+d(x,γ(−s))≥d(γ(−s),γ(t))=s+td(x, \gamma(t)) + d(x, \gamma(-s)) \geq d(\gamma(-s), \gamma(t)) = s + t; subtract t+st + s and let s,t→∞s, t \to \infty. On γ\gamma both sides are 00, because γ\gamma minimises.

Step 2: b±b^\pm are superharmonic. By Laplacian comparison (9B.1 Laplacian Comparison), Δd(⋅,γ(t))≤n−1d(⋅,γ(t))\Delta d(\cdot, \gamma(t)) \leq \frac{n - 1}{d(\cdot, \gamma(t))}, in the barrier sense across cut loci. The right-hand side tends to 00 as t→∞t \to \infty, so in the limit Δb±≤0\Delta b^\pm \leq 0 in the barrier sense. So b++b−b^+ + b^- is superharmonic, nonnegative, and 00 on γ\gamma: it attains an interior minimum.

Step 3: harmonicity. The strong maximum principle for superharmonic functions in the barrier sense (Calabi, 1958; 6A.4 Maximum Principles for the smooth case) gives b++b−≡0b^+ + b^- \equiv 0. Then b+=−b−b^+ = -b^- is both superharmonic and subharmonic, hence harmonic, and by elliptic regularity smooth.

Step 4: splitting. Write b=b+b = b^+. It is 11-Lipschitz and along the rays asymptotic to γ\gamma it decreases at unit rate, so ∣∇b∣=1|\nabla b| = 1. Bochner's formula (9A.6 The Laplacian and the Bochner Formula) with Δb=0\Delta b = 0 and ∣∇b∣2=1|\nabla b|^2 = 1 gives 0=∣∇2b∣2+Ric⁡(∇b,∇b)≥∣∇2b∣20 = |\nabla^2b|^2 + \operatorname{Ric}(\nabla b, \nabla b) \geq |\nabla^2b|^2. So ∇2b=0\nabla^2b = 0: the field ∇b\nabla b is parallel. Its flow consists of isometries translating along unit-speed geodesics orthogonal to the level set N=b−1(0)N = b^{-1}(0), and the map R×N→M\mathbb{R}\times N \to M, (s,y)↦exp⁡y(−s∇b)(s, y) \mapsto \exp_y(-s\nabla b), is an isometry from the product metric.

Applying the theorem repeatedly, a complete manifold with Ric⁡≥0\operatorname{Ric} \geq 0 is isometric to Rk×N\mathbb{R}^k\times N, where NN contains no line. For a closed manifold with Ric⁡≥0\operatorname{Ric} \geq 0, the universal cover is Rk×N\mathbb{R}^k\times N with NN compact (Cheeger–Gromoll), which constrains the fundamental group: it has a finite normal subgroup whose quotient contains Zk\mathbb{Z}^k as a subgroup of finite index.

Figure 5.1. The cylinder S1×RS^1\times\mathbb{R} (and, one dimension up, the neck S2×RS^2\times\mathbb{R}) contains lines, the rulings {θ}×R\{\theta\}\times\mathbb{R}, and has Ric⁡≥0\operatorname{Ric} \geq 0. The splitting theorem says this is the only way: a line plus Ric⁡≥0\operatorname{Ric} \geq 0 forces a product.

Triangles: Toponogov's theorem

Sectional curvature bounds control triangles. A geodesic triangle has three minimising geodesic sides; its comparison triangle in the model space of curvature kk has the same side lengths.

Theorem 5.2 Toponogov's comparison theorem

Let MM be complete with K≥kK \geq k. Then every geodesic triangle in MM (with perimeter less than 2πk\frac{2\pi}{\sqrt k} if k>0k > 0) has angles at least as large as the corresponding angles of its comparison triangle in the model space of curvature kk. Equivalently, the distance from a vertex to any point of the opposite side is at least the corresponding distance in the comparison triangle.

With k=0k = 0: triangles in a manifold of nonnegative curvature are "fatter" than Euclidean ones, and their angle sums are at least π\pi, as on the sphere (Figure 5.2, Exercise 5.7). Victor Toponogov proved the theorem in 1959. It works with distances alone, so it makes sense on singular limits too. Alexandrov spaces, metric spaces in which Toponogov's conclusion holds, are the natural class of Gromov–Hausdorff limits of manifolds with K≥kK \geq k.

Figure 5.2. An equilateral geodesic triangle with sides 11 on the unit sphere (left) and its Euclidean comparison triangle with the same side lengths (right). The spherical angles are arccos⁡cos⁡11+cos⁡1≈69.5°\arccos\frac{\cos1}{1 + \cos1} \approx 69.5°, larger than 60°60°, as Toponogov's theorem with k=0k = 0 requires (computed).

Souls

Theorem 5.3 The soul theorem (Cheeger–Gromoll)

Let MM be complete and noncompact with K≥0K \geq 0. Then MM contains a compact, totally convex, totally geodesic submanifold SS, a soul, such that MM is diffeomorphic to the normal bundle of SS.

The soul is found by a convexity argument. Busemann functions are concave when K≥0K \geq 0 (by Toponogov). Their superlevel sets form an exhausting family of compact totally convex sets, and one shrinks such a set until no interior is left, repeating the process on its boundary structure if necessary. Detlef Gromoll and Wolfgang Meyer had proved in 1969 that K>0K > 0 everywhere forces MM to be diffeomorphic to Rn\mathbb{R}^n. Cheeger and Gromoll conjectured that K≥0K \geq 0 everywhere and K>0K > 0 at a single point should be enough. Perelman proved this soul conjecture in 1994: then the soul is a point, and MM is diffeomorphic to Rn\mathbb{R}^n.

Examples (Exercise 5.8): a paraboloid has a point as its soul and is diffeomorphic to R2\mathbb{R}^2; the cylinder S1×RS^1\times\mathbb{R} has any circle S1×{t}S^1\times\{t\} as a soul; the flat Möbius band has its central circle; S2×RS^2\times\mathbb{R} has S2×{t}S^2\times\{t\}; a capped half-cylinder, positively curved on its cap, has a point.

The view from infinity

If K≥0K \geq 0, the blow-downs (M,λ−2g,p)(M, \lambda^{-2}g, p), λ→∞\lambda \to \infty, converge in the pointed Gromov–Hausdorff sense to a metric cone C(X)C(X) over a compact space XX, the asymptotic cone of MM. The proof uses the monotonicity of angles that Toponogov's theorem provides. For a paraboloid, the asymptotic cone is a ray (Figure 5.3); for Rn\mathbb{R}^n it is Rn\mathbb{R}^n; for a cone, the cone itself. The asymptotic volume ratio of 9B.2 Volume Comparison is positive exactly when the asymptotic cone is nn-dimensional.

Figure 5.3. Blowing down the paraboloid z=x2+y2z = x^2 + y^2: the profile scaled by 1λ\frac1\lambda for λ=1,4,16\lambda = 1, 4, 16 (computed) narrows to a vertical ray, the asymptotic cone. The two-dimensional surface looks one-dimensional from far away, consistent with its asymptotic volume ratio 00.
Where this goes Necks and caps

Perelman's κ\kappa-solutions (12B.1 κ-Solutions) are ancient solutions of the three-dimensional Ricci flow with bounded nonnegative curvature that are κ\kappa-noncollapsed. 12B.2 The Structure of κ-Solutions classifies them with this chapter's tools. A κ\kappa-solution whose curvature operator has a zero eigenvalue splits, by a strong maximum principle for the flow in the spirit of the splitting theorem, and is a round cylinder S2×RS^2\times\mathbb{R} or a quotient of one: a neck. Otherwise the curvature is positive and the solution is compact (a quotient of a sphere) or, by the soul theorem, diffeomorphic to R3\mathbb{R}^3: a cap. Its asymptotic volume ratio is 00 (9B.2 Volume Comparison), and its blow-downs look like rays, which is how necks appear far out on a cap.

History

Herbert Busemann introduced his functions in The Geometry of Geodesics (1955). Toponogov proved his comparison theorem in 1959, and the splitting theorem for K≥0K \geq 0 in 1964. Jeff Cheeger and Detlef Gromoll proved the splitting theorem for Ric⁡≥0\operatorname{Ric} \geq 0 in 1971 and the soul theorem in 1972; Gromoll and Meyer's theorem on K>0K > 0 dates from 1969. Jost-Hinrich Eschenburg and Ernst Heintze gave the short proof of splitting followed above in 1984. Perelman's proof of the soul conjecture appeared in the Journal of Differential Geometry in 1994.

Recall Where we stand

Rays and lines are minimising geodesics to infinity; Busemann functions bγ=lim⁡(d(⋅,γ(t))−t)b_\gamma = \lim(d(\cdot, \gamma(t)) - t) are 11-Lipschitz, and with Ric⁡≥0\operatorname{Ric} \geq 0 they are superharmonic. If there is a line, b++b−≥0b^+ + b^- \geq 0 vanishes on it, the maximum principle makes b+b^+ harmonic, and Bochner makes ∇b+\nabla b^+ parallel: M=R×NM = \mathbb{R}\times N. Toponogov's theorem compares triangles with model triangles when K≥kK \geq k. With K≥0K \geq 0, a complete noncompact manifold is a vector bundle over a compact totally geodesic soul, and the soul is a point if K>0K > 0 somewhere (Perelman). Blow-downs converge to the asymptotic cone. In three-dimensional Ricci flow these facts give necks and caps. 9B.6 Scalar Curvature and Topology turns to the weakest curvature, scalar curvature, and what it can and cannot say about topology.

Exercises

Exercise 5.4 Busemann functions in the model spaces

(a) In Rn\mathbb{R}^n with γ(t)=tv\gamma(t) = tv, ∣v∣=1|v| = 1, show bγ(x)=−⟨x,v⟩b_\gamma(x) = -\langle x, v\rangle. (b) In the upper half-plane with the line γ(t)=iet\gamma(t) = ie^t, use cosh⁡d(z,w)=1+∣z−w∣22Im⁡zIm⁡w\cosh d(z, w) = 1 + \frac{|z - w|^2}{2\operatorname{Im}z\operatorname{Im}w} to show b+(z)=−log⁡Im⁡zb^+(z) = -\log\operatorname{Im}z and b−(z)=2log⁡∣z∣−log⁡Im⁡zb^-(z) = 2\log|z| - \log\operatorname{Im}z. (c) Show b++b−=2log⁡∣z∣Im⁡z≥0b^+ + b^- = 2\log\frac{|z|}{\operatorname{Im}z} \geq 0, vanishing exactly on γ\gamma, but not identically zero. Why doesn't this contradict the splitting theorem?

Solution

(a) ∣x−tv∣−t=t2−2t⟨x,v⟩+∣x∣2−t→−⟨x,v⟩|x - tv| - t = \sqrt{t^2 - 2t\langle x, v\rangle + |x|^2} - t \to -\langle x, v\rangle. (b) For w=ietw = ie^t, t→∞t \to \infty: ∣z−w∣2∼e2t|z - w|^2 \sim e^{2t}, so cosh⁡d∼et2y\cosh d \sim \frac{e^t}{2y} (y=Im⁡zy = \operatorname{Im}z) and d=t−log⁡y+o(1)d = t - \log y + o(1), giving b+=−log⁡yb^+ = -\log y. For w=ie−tw = ie^{-t}: ∣z−w∣2→∣z∣2|z - w|^2 \to |z|^2, so cosh⁡d∼∣z∣2et2y\cosh d \sim \frac{|z|^2e^t}{2y}, d=t+log⁡∣z∣2y+o(1)d = t + \log\frac{|z|^2}{y} + o(1), giving b−=2log⁡∣z∣−log⁡yb^- = 2\log|z| - \log y. (c) ∣z∣≥y|z| \geq y with equality on the imaginary axis. The hyperbolic plane has Ric⁡=−g<0\operatorname{Ric} = -g < 0, so the theorem does not apply; indeed b±b^\pm are not superharmonic there.

Exercise 5.5 Harmonic with unit gradient

Let bb be a smooth function with Δb=0\Delta b = 0 and ∣∇b∣=1|\nabla b| = 1 on a manifold with Ric⁡≥0\operatorname{Ric} \geq 0. Use Bochner's formula to show ∇2b=0\nabla^2b = 0 and Ric⁡(∇b,∇b)=0\operatorname{Ric}(\nabla b, \nabla b) = 0. Show that a parallel vector field XX has a flow by isometries (LXg=0\mathcal L_Xg = 0, 8A.6 Flows and the Lie Derivative).

Solution

12Δ∣∇b∣2=0=∣∇2b∣2+⟨∇b,∇Δb⟩+Ric⁡(∇b,∇b)=∣∇2b∣2+Ric⁡(∇b,∇b)\frac12\Delta|\nabla b|^2 = 0 = |\nabla^2b|^2 + \langle\nabla b, \nabla\Delta b\rangle + \operatorname{Ric}(\nabla b, \nabla b) = |\nabla^2b|^2 + \operatorname{Ric}(\nabla b, \nabla b), a sum of nonnegative terms. If ∇X=0\nabla X = 0, then (LXg)(Y,Z)=⟨∇YX,Z⟩+⟨Y,∇ZX⟩=0(\mathcal L_Xg)(Y, Z) = \langle\nabla_YX, Z\rangle + \langle Y, \nabla_ZX\rangle = 0.

Exercise 5.6 The neck has a line

Show that S2×RS^2\times\mathbb{R} with the product metric has Ric⁡≥0\operatorname{Ric} \geq 0 and that t↦(x,t)t \mapsto (x, t) is a line. Show that the round S3S^3 contains no line, and that the paraboloid contains no line.

Solution

Ric⁡=gS2⊕0≥0\operatorname{Ric} = g_{S^2}\oplus0 \geq 0 (9A.5 Computing Curvature). d((x,s),(x,t))=∣s−t∣d((x, s), (x, t)) = |s - t| because the distance in a product is dS22+∣s−t∣2\sqrt{d_{S^2}^2 + |s - t|^2}. S3S^3 is compact, so it has no geodesic minimising on all of R\mathbb{R}. On the paraboloid (K>0K > 0, Ric⁡≥0\operatorname{Ric} \geq 0), a line would split it as R×N\mathbb{R}\times N, forcing K=0K = 0 in some plane at every point, which is false.

Exercise 5.7 Toponogov on the sphere

For an equilateral geodesic triangle with side a<2π3a < \frac{2\pi}{3} on the unit sphere, use the spherical law of cosines cos⁡a=cos⁡2a+sin⁡2acos⁡α\cos a = \cos^2a + \sin^2a\cos\alpha to show cos⁡α=cos⁡a1+cos⁡a\cos\alpha = \frac{\cos a}{1 + \cos a}. Check that α>π3\alpha > \frac{\pi}{3} for 0<a<2π30 < a < \frac{2\pi}{3}, and compute the angle sum for a=π2a = \frac{\pi}{2}.

Solution

cos⁡α=cos⁡a−cos⁡2asin⁡2a=cos⁡a(1−cos⁡a)(1−cos⁡a)(1+cos⁡a)=cos⁡a1+cos⁡a\cos\alpha = \frac{\cos a - \cos^2a}{\sin^2a} = \frac{\cos a(1 - \cos a)}{(1 - \cos a)(1 + \cos a)} = \frac{\cos a}{1 + \cos a}. This is <12< \frac12 iff 2cos⁡a<1+cos⁡a2\cos a < 1 + \cos a iff cos⁡a<1\cos a < 1, true for a>0a > 0; and α\alpha is defined while cos⁡a>−12\cos a > -\frac12. For a=π2a = \frac{\pi}{2}: cos⁡α=0\cos\alpha = 0, three right angles, sum 3π2\frac{3\pi}{2} (the octant of 9A.2 Connections).

Exercise 5.8 Finding souls

Find a soul of: (a) the paraboloid; (b) the cylinder S1×RS^1\times\mathbb{R}; (c) S2×RS^2\times\mathbb{R}; (d) a capped half-cylinder, a hemisphere glued smoothly to S1×[0,∞)S^1\times[0, \infty) along a transition region with K≥0K \geq 0, and positively curved on the cap. Check in each case that MM is diffeomorphic to the normal bundle of the soul.

Solution

(a) The vertex; M≅R2M \cong \mathbb{R}^2, the normal bundle of a point. (b) Any circle S1×{t}S^1\times\{t\}; MM is its trivial line bundle. (c) S2×{t}S^2\times\{t\}; M=S2×RM = S^2\times\mathbb{R}, the trivial line bundle. (d) K>0K > 0 on the cap, so by Perelman's theorem the soul is a point (the tip of the cap, by symmetry); M≅R2M \cong \mathbb{R}^2.

Exercise 5.9 Rehearsal: three-manifolds with a line

Let M3M^3 be complete, orientable, with K≥0K \geq 0, containing a line. (a) Show that M=R×NM = \mathbb{R}\times N with NN a complete orientable surface with K≥0K \geq 0. (b) If NN is compact and has K>0K > 0 somewhere, use Gauss–Bonnet (8A.9 The Curvature of Surfaces) to show NN is a sphere, so MM is a cylinder S2×RS^2\times\mathbb{R} with some metric of positive curvature on the S2S^2. (c) In 12B.2 The Structure of κ-Solutions, the analogous statement for κ\kappa-solutions shows that the round cylinder is the only one with a line. Explain informally why a κ\kappa-solution that splits must have the round metric on its S2S^2 factor (think of the Ricci flow on the two-sphere, 5A.5 Uniformization and the Two-Dimensional Ricci Flow).

Solution

(a) K≥0K \geq 0 implies Ric⁡≥0\operatorname{Ric} \geq 0, so the splitting theorem applies; the curvature of NN is a sectional curvature of MM, so KN≥0K_N \geq 0, and NN is orientable because MM is. (b) ∫NK dA=2πχ(N)>0\int_NK\,dA = 2\pi\chi(N) > 0, so χ(N)=2\chi(N) = 2 and NN is a sphere. (c) A split κ\kappa-solution is R×N\mathbb{R}\times N with NN an ancient solution of the two-dimensional Ricci flow with positive curvature on S2S^2, defined for all negative times, and noncollapsed. On S2S^2, the only such solutions that remain noncollapsed are the round shrinking spheres (the classification of ancient solutions on S2S^2, 11B.2 Ancient Solutions and the Harnack Inequality), so the factor is round.

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