Bounded Curvature at Bounded Distance

The outline of the canonical neighbourhood theorem in 12B.5 The Canonical Neighbourhood Theorem named one step as the genuinely new ingredient: bounded curvature at bounded distance. In a blow-up sequence normalised so that the base point has curvature $1$, the curvature must stay bounded on balls of each fixed radius, or no smooth limit can be taken. For $\kappa$-solutions this was the lemma of 12B.4 Compactness and Universal Estimates for κ-Solutions, proved with Hamilton's Harnack inequality and the asymptotic volume. For a general flow neither tool is available: there is no Harnack inequality without nonnegative curvature, and no asymptotic volume without completeness. What replaces them is a single hypothesis, that points of much higher curvature already have canonical neighbourhoods, together with the geometry of necks and an argument about cones.

Perelman disposed of this step in one paragraph (I, §12.1, Claim 2). Morgan and Tian devote a chapter to it, and state it as a stand-alone theorem about generalized Ricci flows, which is how it is used both in the proof of the canonical neighbourhood theorem and, later, for flows with surgery (12B.12 Canonical Neighbourhoods After Surgery). This chapter follows their proof.

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

• state the pinching condition and Morgan and Tian's Theorem 10.2 with every hypothesis, and explain why each is needed;
• construct, in a hypothetical counterexample, a tube of strong necks along a minimizing geodesic and an incomplete limit tube $U_\infty \cong S^2\times(0, 1)$ with curvature blowing up at one end;
• define the space of directions at that end, prove that its angle function is a metric, and prove that rescalings near the end converge to the cone over it;
• compare the two blow-up factors $d(x, E)^{-2}$ and $R(x)$, and deduce that a smooth Ricci flow limit is a piece of a non-flat cone;
• prove, from the strong maximum principle, that no Ricci flow with nonnegative curvature contains a piece of a non-flat cone.

6.1The setting

Generalized Ricci flows. Morgan and Tian prove the theorem for generalized Ricci flows: space-times $\mathcal M$ with a time function $t$, a vector field $\chi$ with $\chi(t) = 1$, and a metric $G$ on the horizontal distribution $\ker dt$ satisfying $\mathcal L_\chi G = -2\operatorname{Ric}(G)$. A Ricci flow $M\times[a, b]$ is one; a Ricci flow with surgery (12B.9 Surgery) is another, in which the time slices change topology at surgery times, and flow lines may begin or end. Every statement below is local in space-time, so the reader can keep an ordinary Ricci flow in mind; the generality is what lets the same theorem serve in 12B.12 Canonical Neighbourhoods After Surgery.

Pinching toward positive. For a point $x$ of the space-time let $\nu(x)$ be the smallest eigenvalue of the curvature operator and $X(x) = \max(0, -\nu(x))$, the size of its negative part. The flow has curvature pinched toward positive (Morgan–Tian, Definition 10.1) if at every point

This is the Hamilton–Ivey estimate of 11A.10 Hamilton–Ivey Pinching in the form preserved by the flow: if it holds at time $0$, in particular if $|\operatorname{Rm}| \leq 1$ at time $0$, it holds for all later times, and Morgan and Tian arrange that surgery preserves it too (12B.10 Surgery on a δ-Neck). Its content is that where $R$ is large the negative part of the curvature is small compared with $R$: $X \leq R/(2\log X - 6)$, roughly $X \lesssim R/\log R$. After rescaling by $Q \to \infty$, the negative part of the curvature tends to zero on regions of bounded rescaled curvature (12B.5 The Canonical Neighbourhood Theorem, exercise on pinching), so blow-up limits have nonnegative curvature.

Canonical neighbourhoods. Strong $(C, \varepsilon)$-canonical neighbourhoods are as defined in 12B.4 Compactness and Universal Estimates for κ-Solutions: strong $\varepsilon$-necks, cores of $(C, \varepsilon)$-caps, $C$-components and $\varepsilon$-round components. Fix $\varepsilon \leq \varepsilon_0$, a small universal constant for which the facts about necks below hold with $2\varepsilon$ in place of $\varepsilon$.

The theorem is scale-free in its conclusion, $R(y)/R(x) \leq D$ on a ball of radius $A$ in the metric rescaled by $R(x)$, but it needs $R(x) \geq D_0$ so that pinching is effective. The second hypothesis is the "first bad point" structure of 12B.5 The Canonical Neighbourhood Theorem: in the proof of the canonical neighbourhood theorem the counterexample is chosen so that points of much higher curvature, at the same or earlier times, already have canonical neighbourhoods. Notice what the hypothesis does not say: nothing is assumed about points of curvature less than $4R(x)$, which may look like anything.

The shape of the proof. Suppose the theorem fails for some $A_0$. Then there are flows $(\mathcal M_n, G_n)$, pinched toward positive, and points $x_n$ satisfying hypothesis 2, and points $y_n$ at the same time with

Shift time so that $t(x_n) = 0$ and rescale by $Q_n = R(x_n)$; write everything in the rescaled metrics. So $R(x_n) = 1$, $d(x_n, y_n) < A_0$ and $R(y_n) \to \infty$: the curvature blows up at bounded distance. The plan, in four stages:

1. The path from $x_n$ to $y_n$ runs, beyond the point where $R = 4$, through points with canonical neighbourhoods; the curvature ratio forces these to be necks, and they line up into a tube.
2. A subsequence of the tubes converges to an incomplete limit $U_\infty \cong S^2\times(0, 1)$ of finite diameter, with nonnegative curvature, necks everywhere near one end $E$, and $R \to \infty$ at $E$. The limit carries a Ricci flow backward in time for a definite rescaled interval.
3. Seen from $E$, the limit is asymptotically a cone: rescalings near $E$ by $d(\cdot, E)^{-2}$ converge in the Gromov–Hausdorff sense to the cone over the space of directions at $E$, which is not a ray.
4. Rescalings at the same points by $R$ converge smoothly to a Ricci flow; the two scale factors are comparable, so the smooth limit is a piece of a non-flat cone that is a time slice of a Ricci flow. Hamilton's strong maximum principle forbids this.

Perelman's paragraph contains exactly these stages: "the radius of the cylinder goes to zero as we approach the (first) singular point", "the metric near $o$ must be cone-like", "we actually have the convergence of the solutions to the Ricci flow on some time interval", "a contradiction with the strong maximum principle".

6.2Stage 1: a tube of strong necks

Take a path $\gamma_n$ from $x_n$ to $y_n$ inside the ball $B(x_n, 0, A_0)$. Since $R(x_n) = 1$ and $R(y_n) \gg 1$, there is a last point $z_n$ on it with $R(z_n) = 4$; replace $\gamma_n$ by its piece from $z_n$ to $y_n$, along which $R \geq 4$. By hypothesis 2, every point of $\gamma_n$ now has a strong $(C, \varepsilon)$-canonical neighbourhood.

No compact components. If a point of $\gamma_n$ lay in an $\varepsilon$-round component or a $C$-component, the whole of the connected path $\gamma_n$ would lie in that component, on which $\sup R/\inf R$ is bounded by a constant depending on $C$ and $\varepsilon$. But $R(y_n)/R(z_n) \to \infty$. So for large $n$ every point of $\gamma_n$ is the centre of a strong $\varepsilon$-neck or in the core of a $(C, \varepsilon)$-cap.

Chains of necks and caps. Morgan and Tian's appendix on canonical neighbourhoods (chapter 19) shows how overlapping necks and caps fit together: the central spheres of two overlapping $\varepsilon$-necks are isotopic in their union, so a chain of necks whose centres march along a path builds an $\varepsilon$-tube, diffeomorphic to $S^2\times I$; a chain may also end in a cap. So $\gamma_n$ lies in an open submanifold $X_n$ which is an $\varepsilon$-tube, a $C$-capped $\varepsilon$-tube, a doubly capped tube, or the union of two caps (Proposition 19.21). The last is excluded for large $n$ by the curvature ratio again (on two overlapping caps the curvature ratio is bounded).

A minimizing geodesic (Claim 10.3). Inside $X_n$ there is a minimizing geodesic from $z_n$ to $y_n$. The point is not existence of geodesics in general, since $X_n$ is incomplete, but that minimizing sequences of paths stay away from the ends of $X_n$. This uses a basic fact about necks: in an $\varepsilon$-neck, a path that leaves through one end and comes back can be shortened (Lemma 19.1), because the neck is almost a product $S^2\times(-\varepsilon^{-1}, \varepsilon^{-1})$ and the shortest way between two points of a long cylinder does not wander out to its ends. So minimizing sequences live in a compact part of $X_n$ and converge. Replace $\gamma_n$ by this geodesic.

A tube with only necks (Claim 10.4). A cap must contain one end of $\gamma_n$: if the core $Y$ of a cap meets $\gamma_n$, then $Y$ together with half of its neck meets $\gamma_n$ in a subinterval containing $z_n$ or $y_n$ (Corollary 19.8). So a point $w$ in a cap whose core meets $\gamma_n$ has $R(w) < CR(z_n)$ or $R(w) > C^{-1}R(y_n)$, by the curvature ratio bound in the cap. Pass to a subsequence with $R(y_n)/R(z_n) > 4C^2$, and to the sub-geodesic $\gamma_n'$ from $z_n'$ to $y_n'$ where

Then no point of $\gamma_n'$ lies in the core of a cap, so every point of $\gamma_n'$ is the centre of a strong $\varepsilon$-neck. A balanced chain of these necks (consecutive necks overlapping in a controlled way, Proposition 19.19) has union a strong $\varepsilon$-tube $T_n \supset \gamma_n'$, beginning with a neck $N(z_n')$ centred at $z_n'$ and ending with a neck containing $y_n'$. A further topological argument, again with the isotopy of central spheres, shows that every point of $T_n$ is the centre of a strong $\varepsilon$-neck and has $R > 3$. Summary:

• the length of $\gamma_n'$ is at most $A_0$;
• $R(z_n') = 8C$ is bounded, $R(y_n') \to \infty$;
• every point of the tube $T_n \supset \gamma_n'$ is the centre of a strong $\varepsilon$-neck.

6.3Stage 2: the incomplete limit

The tubes cannot converge everywhere, since their curvature is unbounded. Morgan and Tian take the largest region on which it is bounded. Let $\mathcal A$ be the set of $A > 0$ for which the curvature on $B(z_n', A)\cap T_n$ is bounded independently of $n$, and $A_1 = \sup\mathcal A$; $A_1$ is finite, since $R(y_n') \to \infty$ at bounded distance, and positive, since the neck $N(z_n')$ has bounded curvature. Set $U_n = T_n\cap B(z_n', A_1)$.

The neck at $z_n'$ is inside (Claim 10.5). The curvature on $N(z_n')$ and on the next neck along is within $10\%$ of $R(z_n')$, so $U_n$ contains $N(z_n')$ once $A_1$ is compared with the neck's length $\approx 2\varepsilon^{-1}R(z_n')^{-1/2}$.

Curvature is bounded away from the frontier (Claim 10.6). For $\delta > 0$ let $\operatorname{Reg}_\delta(U_n)$ be the set of points whose $\delta$-ball lies in $U_n$. The claim is that $\operatorname{Reg}_\delta(U_n) \subset B(z_n', A_1 - \delta/4)$, so its curvature is bounded independently of $n$. The proof walks outward: from $w_n \in \operatorname{Reg}_\delta$, move along the $s$-direction of the neck centred at $w_n$, away from $z_n'$, then along the neck centred at the endpoint, and so on. In a neck the distance from a far-away point increases at a rate between $0.99$ and $1.01$ times the neck scale per unit of the neck coordinate (Lemma 19.9), so after walking a distance $\delta/2$ the distance from $z_n'$ has grown by more than $\delta/4$. The endpoint is still in $U_n$, at distance less than $A_1$ from $z_n'$; so $d(z_n', w_n) \leq A_1 - \delta/4$.

Compactness. On $\operatorname{Reg}_\delta(U_n)$ the curvature is bounded, and so are all its derivatives, by Shi's estimates on the strong necks (each comes with a backward parabolic neighbourhood on which the flow is close to a shrinking cylinder). Points at the centres of necks are noncollapsed on their own scale with universal constants. The tubes have bounded diameter and $R > 3$, hence bounded volume. Morgan and Tian's compactness theorem for incomplete manifolds (Theorem 5.6, the version of Cheeger–Gromov–Hamilton compactness of 11B.3 Compactness of Ricci Flows for manifolds exhausted by the regions $\operatorname{Reg}_\delta$) gives a subsequence converging to a smooth, incomplete limit $(U_\infty, g_\infty, z_\infty)$.

The ingredients are those already assembled:

• Finite diameter: every point is within $A_1$ of $z_\infty$.
• Necks (Claim 10.8): by the openness of the neck condition (12B.4 Compactness and Universal Estimates for κ-Solutions, Proposition 9.79), necks in $U_n$ with compact closure in the regions where convergence happens pass to $2\varepsilon$-necks in the limit.
• Topology: the side of the central sphere at $z_\infty$ toward $E$ is a $2\varepsilon$-tube, so $\cong S^2\times[0, 1)$; the other side is a limit of the negative halves of the necks $N(z_n')$, $\cong S^2\times(-1, 0]$; together $U_\infty \cong S^2\times(0, 1)$.
• $R \to \infty$ at $E$ (Claim 10.9): points $x_k \to E$ are at distance tending to zero from the end, yet each is the centre of a $2\varepsilon$-neck, whose length is about $2\varepsilon^{-1}R(x_k)^{-1/2}$; so $R(x_k) \to \infty$.
• The flow backward (Claim 10.10): each point $x$ of $U_n$ is the centre of a strong neck, so its flow line exists for rescaled time $R(x)^{-1}$ backward, with curvature at most $R(x)$ along it. On compact subsets $X$ of $U_\infty$ this gives bounded curvature on $X\times(-\inf_XR^{-1}, 0]$, and Hamilton's compactness for flows gives a limit flow there; a diagonal argument covers the whole region.
• Nonnegative curvature (Claim 10.11): the pinching condition, rescaled by $Q_n \to \infty$.

The limit is a solid object, a finite funnel $S^2\times(0, 1)$ closing up at $E$. Note that $E$ is an end, not a point of $U_\infty$: the funnel's tip is missing.

6.4Stage 3: the cone at the end

The next step is purely about Riemannian manifolds of nonnegative curvature, and uses Toponogov's comparison theorem (9B.5 Splitting and Soul Theorems) in its angle form: in nonnegative curvature, angles of geodesic triangles are at least those of the Euclidean triangle with the same sides, and the Euclidean comparison angle at a vertex is nonincreasing as the two sides are lengthened.

Distance to the end. For $x \in U_\infty$ let $d(x, E)$ be the infimum of lengths of rays from $x$ going out to $E$. It is finite. A minimizing geodesic ray from $x$ to $E$ is a ray every initial segment of which is minimizing.

• From every $x$ with $R(x) \geq 2R(z_\infty)$ there is a minimizing geodesic ray to $E$ (Claim 10.13): minimizing paths from $x$ to points $q_n \to E$ cannot exit the back end of the neck at $x$ or the front end of the neck at $q_n$, so they converge, and so do the geodesics as $q_n \to E$.
• Minimizing rays are exactly the shortest rays, and their length is $d(x, E)$ (Claim 10.14). The nontrivial direction: if a minimizing ray $\gamma$ were longer by $\delta$ than a shortest ray $\gamma'$, cut both at a central sphere $S_i$ of a neck near $E$, of diameter less than $\delta/3$ (spheres near $E$ have small extrinsic diameter since $R \to \infty$); the parts beyond $S_i$ have length less than $\delta/3$, and since the initial parts are minimizing they differ in length by less than $\delta/3$; adding up gives a contradiction.
• Subrays of minimizing rays are the unique shortest rays from their starting points (Corollary 10.15), and $d(x, y)$, $d(x, E)$, $d(y, E)$ satisfy the triangle inequality (Lemma 10.16), the last one since $d(x, E) = \lim d(x, q_n)$.

Directions. Two minimizing rays to $E$ are equivalent if one contains the other; an equivalence class is a direction at $E$ (Definition 10.17). There is more than one (Lemma 10.18): a point near $E$ off a given ray has its own ray.

Angles. For minimizing rays $\gamma$, $\mu$, parametrised by distance from $E$, and $0 < s, s'$, let $\theta(\gamma, s, \mu, s')$ be the angle at $e$ of the Euclidean triangle with sides $s$, $s'$ and $d(\gamma(s), \mu(s'))$, so that

The completion $\bar X(E)$ of the space of directions with this metric is compact, and has $\delta$-nets of realised directions for every $\delta$ (Lemma 10.27, Corollary 10.28): given $K$ directions with large $K$, follow them into a neck near $E$, where they cross its central sphere, a nearly round sphere of radius about the neck scale $r$; two crossing points are within $\frac\delta2r$ of each other, while both rays still have length at least $\frac{\varepsilon^{-1}}{2}r$ to go, so the comparison angle between them is less than $\delta$. Let $(C_E, g_E)$ be the cone on $\bar X(E)$, with metric

Since there are at least two directions, $C_E$ is not a ray (Proposition 10.24).

This is the precise version of Perelman's "the curvature at $o$ is nonnegative in Aleksandrov's sense, thus the metric near $o$ must be cone-like": the tip of the funnel is a point of a nonnegatively curved length space, and the tangent cone at a point of such a space is the cone over its space of directions. Morgan and Tian prove the special case they need by hand rather than appeal to the theory of Aleksandrov spaces. (They remark that the space of directions is in fact homeomorphic to $S^2$, which is not needed.)

6.5Stage 4: two limits, one cone, and the contradiction

Now take the same points $x_n \to E$ and rescale differently, by $\lambda_n' = R(x_n)$. Each $x_n$ is the centre of an evolving $2\varepsilon$-neck defined for normalized time $\frac12$, so the balls of radius $\frac{\varepsilon^{-1}}{3}$ in $(U_\infty, \lambda_n'g_\infty, x_n)$ have bounded curvature, compact closure, and a flow on them for rescaled time $(-\frac12, 0]$ with bounded curvature. A subsequence converges smoothly, as a flow, to a limit on the time interval $(-\frac12, 0]$, with scalar curvature $1$ at the base point.

Pass to a subsequence with $\rho_n \to \rho_\infty \in (0, \infty)$. Then the Gromov–Hausdorff limit of $(U_\infty, \lambda_n'g_\infty, x_n)$ is the cone $C_E$ rescaled by $\rho_\infty$, still a cone. But on balls of radius less than $\frac{\varepsilon^{-1}}{2}$ about the base point the same sequence converges smoothly. Two limits of one sequence agree, so the smooth limit metric on a ball $B_\infty$ is isometric to an open subset of a cone. The cone is not flat, because the smooth limit has $R = 1$ at its base point. And this metric is the time-$0$ slice of a Ricci flow on $B_\infty\times(-\frac12, 0]$ with nonnegative curvature.

This contradicts Stage 4, so the sequence of counterexamples cannot exist, and Theorem 6.1 is proved. $\square$

6.6How the theorem is used

In the proof of the canonical neighbourhood theorem (12B.5 The Canonical Neighbourhood Theorem, Step 3) the theorem is applied to the counterexample sequence at the point $(\bar x, \bar t)$ chosen by point picking: hypothesis 2 is exactly the property of the chosen counterexample, that points of curvature at least $4$ times larger (at the same or earlier times, within the window) already have canonical neighbourhoods. With bounded curvature at bounded distance in hand, Steps 4–6 of 12B.5 The Canonical Neighbourhood Theorem produce a smooth limit, show it is ancient, and identify it as a $\kappa$-solution.

Perelman's own argument is organised slightly differently: he follows a geodesic out from $x_0$, and the alternative "either the curvature stays bounded at bounded distances along it, or not" is the alternative between the high-curvature point escaping to infinity in a cylindrical limit and the existence of a first singular point $o$ at finite distance. Morgan and Tian's tube $U_\infty$ is the limit "along $\gamma$" up to that singular point, and their end $E$ is Perelman's $o$.

12B.12 Canonical Neighbourhoods After Surgery applies the theorem again, unchanged, to flows with surgery. That is why Morgan and Tian state it for generalized Ricci flows: nothing in the proof uses that the flow is defined on a fixed manifold, only that points of high curvature have strong canonical neighbourhoods, which in a flow with surgery must be part of the inductive hypothesis.

6.7History

Perelman's Claim 2 in I, §12.1 (2002) gave this argument in a paragraph. The cone-limit step was expanded by Kleiner and Lott (2006, §52) and by Cao and Zhu (2006); Morgan and Tian (2007, chapter 10) isolated it as a theorem about generalized Ricci flows, so that it could be quoted verbatim in the surgery setting. The use of the strong maximum principle to rule out cones goes back to Hamilton's "Formation of singularities in the Ricci flow" (1995).

6.8Exercises

Solution

Write $g = dr^2 + f(r)^2h$ with $f = r$. For a warped product the radial curvatures are $-\frac{f''}{f} = 0$, the mixed terms vanish, and the tangential curvature is $\frac{K_h - f'^2}{f^2} = \frac{K_h - 1}{r^2}$. (Check with $S$ the unit sphere: the cone is $\mathbb{R}^3$, flat, and $K_h = 1$.) The curvature operator is diagonal in the basis $\partial_r\wedge e_1$, $\partial_r\wedge e_2$, $e_1\wedge e_2$, with eigenvalues $0, 0, \frac{K_h - 1}{r^2}$. Scalar curvature is twice the sum of sectional curvatures of three orthogonal planes: $2\frac{K_h - 1}{r^2}$. Nonnegative curvature means $K_h \geq 1$: the cross-section is more curved than the unit sphere, so it is "smaller", as for the cone over a small round sphere, which is a cone of angle less than $2\pi$.

Solution

(a) The law of cosines in the plane holds with the true angle. (b) By the spherical law of cosines, $\cos d = \cos^2s + \sin^2s\cos\theta$ for $s = s'$; the Euclidean comparison angle $\tilde\theta$ satisfies $d^2 = 2s^2(1 - \cos\tilde\theta)$. For small $s$, $d \approx s\sqrt{2(1 - \cos\theta)}$ and $\tilde\theta \to \theta$; as $s$ grows, $d$ is smaller than the Euclidean value for the same hinge (positive curvature pulls geodesics together), so $\tilde\theta < \theta$, and Toponogov's hinge theorem gives the monotonicity. (c) The lemma is the same statement for rays coming out of the end $E$, with $E$ playing the role of the vertex: the angle is defined as the limit of comparison angles as the sides shrink to zero, and monotonicity makes the limit exist.

Solution

Pick $\ell_1 \neq \ell_0$. The path $s \mapsto (\ell_1, s)$, $\frac12 \leq s \leq 2$, joins the two regions and avoids $p$ (its points have direction $\ell_1$, and the only point identified across directions is the cone point, at $s = 0$). If $X$ is a point, $C(X)$ is a ray $[0, \infty)$, and removing $1$ does separate it. In the lemma, $X$ is the space of directions at $E$, which has more than one point (Lemma 10.18), so this case does not arise; that is why Morgan and Tian prove Lemma 10.18 first.

Solution

(a) In a product the distance from $(\sigma_0, s_0)$ to $(\sigma, s)$ is $\sqrt{d_{S^2}(\sigma_0, \sigma)^2 + (s - s_0)^2}$, with $d_{S^2} \leq \pi r$; its derivative in $s$ is $\frac{s - s_0}{\sqrt{d_{S^2}^2 + (s - s_0)^2}}$, between $\big(1 + \frac{\pi^2r^2}{(s - s_0)^2}\big)^{-1/2}$ and $1$. Far points have $s - s_0 \gg r$, so the rate is close to $1$. Morgan and Tian's Lemma 19.9 is the version for $\varepsilon$-close metrics, with rates between $0.99$ and $1.01$ times the neck scale. (b) $R = \frac{2}{r^2}$ for $S^2(r)\times\mathbb{R}$, so $R = 1$ gives $r = \sqrt2$. (c) Two points of the middle half are joined inside the neck by a path of length at most $\pi r + L$, while any path leaving the neck must travel at least $\frac L2$ to the end and $\frac L2$ back; for small $\varepsilon$ the second exceeds the first.

Solution

After rescaling by $R(x)$ the neck has half-length about $\frac{\sqrt2}{2\varepsilon}$ and lies inside $U_\infty$, so $\sqrt{R(x)}d(x, E) \geq \frac{\sqrt2}{2\varepsilon}(1 - O(\varepsilon))$, giving $c$ about $\frac12$ (Morgan and Tian use the cruder $\frac14$). A neck is long compared with its radius; if the funnel narrows to its tip within a distance comparable to the radius, there is no room for the neck. So near $E$ the funnel's radius $\sim R^{-1/2}$ is at most a small multiple $\sim\varepsilon$ of the distance to the tip: Perelman's "the radius of the cylinder is at most $\varepsilon$ times the distance from $o$". This is what makes the tip cone-like rather than cusp-like.