Noncollapsing After Surgery

Perelman's noncollapsing theorem (12A.4 κ-Noncollapsing) was proved with the reduced volume: from a point $x$, find a point at an earlier time whose reduced length is at most $\frac n2$, take a ball of definite volume around it where the reduced length stays bounded, and push the volume lower bound forward to $x$ by monotonicity (12A.7 Monotonicity of the Reduced Volume, 12A.8 Local Noncollapsing from the Reduced Volume). Every step uses $\mathcal L$-geodesics from $x$ into the past. In a flow with surgery, the past of $x$ has holes: regions that were created by a surgery (the caps) have no past, and regions that disappeared at a surgery have no future. An $\mathcal L$-geodesic must avoid both, and the monotonicity argument needs minimizing $\mathcal L$-geodesics to "enough" of each earlier time slice.

Perelman put the difficulty candidly (II, §5.2): the argument would go through "if we knew that every barely admissible curve ... has reduced length at least $3/2 + \kappa'$", but "at the moment I don't see how to ensure that without imposing new restrictions on $\delta(t)$ for all $t$". His way around it, which Morgan and Tian develop into chapter 16, is to show that curves that come near a surgery cap have large $\mathcal L$-length, provided $\delta$ is small: approaching a cap from above means spending time in the evolving cap, where the curvature is large ($R \geq \frac{c}{1 - t}$ on the standard solution, 12B.8 The Standard Solution) or travelling a long way; approaching from below means crossing a neck in very little time. With that, minimizing $\mathcal L$-geodesics of moderate length stay in the smooth part of space-time, and the smooth proof runs.

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

• state the noncollapsing proposition for flows with surgery (Morgan–Tian, Proposition 16.1; Perelman II, Lemma 5.2) with its dependence of constants;
• prove noncollapsing at points of large curvature directly from canonical neighbourhoods;
• state and explain the evolution of surgery caps (Perelman's Lemma 4.5) and prove the length estimate for curves in an evolving cap (Corollary 4.6, Proposition 16.13);
• prove that curves of bounded $\mathcal L_+$-length avoid surgery caps and disappearing regions;
• assemble the proof: existence of minimizing $\mathcal L$-geodesics with $\min\ell \leq \frac32$, a point of controlled curvature on a short $\mathcal L$-geodesic, and the reduced-volume argument.

11.1The statement

Recall the inductive setting of 12B.10 Surgery on a δ-Neck: time steps $[T_i, T_{i+1}]$, $T_i = 2^it_0$, $t_0 = 2^{-5}$; parameters $r_i$ (canonical neighbourhoods), $\kappa_i$ (noncollapsing), $\delta_i$ (surgery control) on each step; and the surgery scale $h \leq \delta^2r$.

Two features of the quantifiers matter for the induction in 12B.13 Ricci Flow with Surgery for All Time. First, $\kappa$ depends only on the previous step's constants ($\kappa_i$, $r_i$, and the fixed $C$, $\varepsilon$, $t_0$), not on $r_{i+1}$; only the allowed surgery control $\delta$ depends on $r_{i+1}$. This is essential: in 12B.12 Canonical Neighbourhoods After Surgery the canonical neighbourhood parameter $r_{i+1}$ is found by a contradiction argument that needs $\kappa$ fixed in advance. Second, as always, "noncollapsed" has an escape clause: either the volume bound holds, or the point lies in a component of its time slice with positive sectional curvature (such components become extinct and are harmless).

11.2Points of large curvature

So the work is for points with $r_{i+1} < r \leq \varepsilon$, where noncollapsing is a genuine statement about the past.

11.3How surgery caps evolve

The standard solution enters through the following lemma, which is where the choice of $\delta$ small pays off.

In words: for rescaled time up to $\theta < 1$, a surgery cap evolves like the standard solution, unless a later surgery removes it, and then it is removed all at once. The reason for the second clause is Perelman's: while the cap is close to the standard solution it contains no point at the centre of a $\delta$-neck (the standard solution near its tip is far from a neck), and surgeries are only done along the central spheres of $\delta$-necks, so a later surgery cannot cut through the cap; it can only discard it whole.

Proof idea. By contradiction with $\delta_n \to 0$. The glued metrics near the tips are uniformly bounded in $C^\infty$ (the strong necks they were made from have Shi bounds, and the gluing is with fixed functions; Claim 16.6), so rescaled limits exist and, by conclusion 4 of the surgery theorem of 12B.10 Surgery on a δ-Neck, the limit initial metric is $g_0$. The flows from these initial data have curvature controlled on short time steps by the canonical neighbourhood derivative bound (Perelman steps forward in increments $\Delta t$ small compared with $Q^{-1}$, $Q$ the maximum curvature of the standard solution on $[0, \theta]$), so a subsequence converges to a Ricci flow with initial metric $g_0$ and bounded curvature, which by uniqueness (12B.8 The Standard Solution) is the standard solution. Iterating up to time $\theta$ gives closeness on growing balls, contradicting the choice of the sequence.

From this, the decisive estimate:

11.4Short curves avoid surgeries

The reduced length weights the integrand by $\sqrt\tau$, and $R$ may be negative (down to $-6$ by pinching), so Perelman and Morgan and Tian work with

which satisfies $\mathcal L \leq \mathcal L_+ \leq \mathcal L + 4\tau_0^{3/2}$ (Exercise 11.8).

The caps are approached "from above" (from later times). The disappearing regions are approached "from below": a curve going backward in time from a point that survives a surgery at time $t$ must not reach, just before $t$, the part that is about to disappear. Here the estimate is not universal.

The time $\tau_1$ depends on the spacing of surgeries in the particular flow, which is why Morgan and Tian construct, for each flow, a compact set $Y(\ell)$ of smooth space-time points by removing small open neighbourhoods $\nu_{\rm sing}(\ell, t)$ of each surgery (the evolving caps forward for rescaled time $\theta_0$, the disappearing regions backward for time $\tau_1$). The two lemmas say that curves from $x$ with bounded $\mathcal L$ stay in $Y(\ell)$, and since $Y(\ell)$ is compact and consists of smooth points, minimizing sequences of such curves converge (Proposition 16.21, Corollary 16.24).

11.5Minimizing $\mathcal L$-geodesics to enough points

This is the place where Perelman's worry about "barely admissible curves" is resolved: we never need to know the reduced length of curves on the boundary of the admissible set, only that curves of moderate length never get near it.

11.6The proof of noncollapsing

Proof of Proposition 11.1. Claim 15.1 handles the first step, so let $i \geq 1$, and let $x$ at time $T \in (T_i, T_{i+1}]$ satisfy the hypothesis of noncollapsing at scale $r$; by Proposition 11.2 assume $r_{i+1} < r \leq \varepsilon$. By Proposition 11.7 there is a minimizing $\mathcal L$-geodesic $\gamma$ from $x$ to the slice at time $T_{i-1}$ with $\mathcal L(\gamma) \leq 3\sqrt{T - T_{i-1}}$.

A point of moderate curvature on $\gamma$ (Claim 16.27). There is $\tau_0$ with $\max(\varepsilon^2, T - T_i) \leq \tau_0 \leq T - T_{i-1} - \varepsilon^2$ and $R(\gamma(\tau_0)) < r_i^{-2}$. Otherwise, with $T' = \max(\varepsilon^2, T - T_i)$ and $T'' = T - T_{i-1} - \varepsilon^2$, and using $R \geq -6$ outside $[T', T'']$,

by the elementary estimates of Claim 16.28 (Exercise 11.11), and since $r_i^{-2} \geq \varepsilon^{-2} \geq \frac{50}{t_0}$ and $T - T_{i-1} \geq t_0$, this is at least $4\sqrt{T - T_{i-1}}$, contradicting $\mathcal L(\gamma) \leq 3\sqrt{T - T_{i-1}}$. The point is that a curve through a long stretch of high curvature has large $\mathcal L$.

Let $y = \gamma(\tau_0)$ at time $t' = T - \tau_0 \leq T_i$, and $\gamma_1 = \gamma|_{[0, \tau_0]}$, with $\mathcal L(\gamma_1) \leq 3T_{i+1}^{1/2} + 4T_{i+1}^{3/2}$ (pinching again).

A parabolic neighbourhood of controlled size at $y$. By the canonical-neighbourhood lemma of 12B.7 The Singular Time: Horns and Strong Necks (Lemma 11.2), every point on backward flow lines from $B(y, t', \frac{r_i}{2C})$ for time $\Delta = \min(\frac{r_i^2}{16C}, \varepsilon^2)$ has $R \leq 2r_i^{-2}$. No such point lies in a surgery cap from a surgery in that time window, since caps have $R \geq D^{-1}\delta^{-4}r_i^{-2} \geq 1000r_i^{-2}$ (the scale is $h \leq \delta^2r_i$). So the parabolic neighbourhood exists, with $R \leq 2r_i^{-2}$, hence (pinching) $|\operatorname{Rm}| \leq C'(r_i)$. Shrinking to $r' = r'(r_i, C, \varepsilon)$ gives a parabolic neighbourhood $P(y, t'', r', -r'^2)$ with $|\operatorname{Rm}| \leq r'^{-2}$, at a time $t'' \leq T_i$ where the inductive hypothesis holds. So either $\operatorname{Vol}B(y, t'', r') \geq \kappa_ir'^3$, or $y$ is in a positively curved component, which then flows forward as a positively curved component until it disappears (Hamilton), and $x$, which is joined to $y$ by a curve, is in such a component.

The reduced-volume argument. For $z \in B(y, t'', r')$ join $y$ to $z$ by a $g(t')$-geodesic run backward over $[t'', t']$; with the metric bounds its $\mathcal L$-length is at most $\sqrt T$. So every $z$ is reached from $x$ by a curve of length at most $4T_{i+1}^{1/2} + 4T_{i+1}^{3/2} \leq \frac L2$, and by Proposition 11.7 by a minimizing $\mathcal L$-geodesic in the smooth part. The set of points of $B(y, t'', r')$ in the $\mathcal L$-injectivity domain has full measure, so it has volume at least $\kappa_ir'^3$, with $\ell_x \leq \frac{L}{2}$ (after normalising by $2\sqrt{T - t''}$) on it. Morgan and Tian's Theorem 8.1 (12A.8 Local Noncollapsing from the Reduced Volume) now gives $\kappa$-noncollapsing at $x$ on scales $\leq \varepsilon$, with $\kappa$ depending on $\kappa_i$, $r'$, $\varepsilon$ and $L$, hence only on $\kappa_i$ and $r_i$. $\square$

11.7History

Perelman's Lemma 5.2 (II, 2003) proves noncollapsing for flows with cutoff with the remark about barely admissible curves quoted above, and Lemmas 4.5 and 5.3 and Corollary 4.6 supply the length estimates. Kleiner and Lott (§§78–79) and Morgan and Tian (chapter 16, 2007) gave complete proofs, including the existence of minimizing $\mathcal L$-geodesics in the presence of surgery; Cao and Zhu did the same.

11.8Exercises

Solution

$R_+ \geq R$, so $\mathcal L_+ \geq \mathcal L$; and $R_+ - R = \max(0, -R) \leq 6$, so $\mathcal L_+ - \mathcal L \leq \int_0^{\tau_0}6\sqrt\tau\,d\tau = 4\tau_0^{3/2}$. Lower bounds for $\mathcal L_+$ over a sub-curve add up: the integrand is nonnegative, so a lower bound on any piece is a lower bound on the whole. For $\mathcal L$, negative curvature elsewhere along the curve could cancel the gain.

Solution

$\int_{t'}^\theta\frac{c}{1 - t}dt = c\log\frac{1 - t'}{1 - \theta} \geq c\log\frac{1/2}{1 - \theta}$. It exceeds $\ell$ when $1 - \theta < \frac12e^{-\ell/c}$. So $1 - \theta_0(\ell)$ decreases exponentially in $\ell$: to force very long curves, one must follow the cap almost to the extinction time of the standard solution, which in turn requires closeness to the standard solution over almost all of $[0, 1)$, hence a smaller $\delta$.

Solution

$\frac{A}{5} \leq \int|X|_{g(T')}dt \leq \big(\int|X|^2_{g(T')}\big)^{1/2}(\Delta t)^{1/2} \leq \big(3\int|X|^2_{g(t)}\big)^{1/2}$, so $\int|X|^2_{g(t)} \geq \frac{A^2}{75}$. In Euclidean space the cheapest crossing is at constant speed, with energy $\frac{D^2}{\Delta t}$: crossing quickly is expensive, and the cost scales like distance squared over time, which is the parabolic scaling of $\mathcal L$.

Solution

(a) $\varepsilon^2 \leq \frac{t_0}{50} \leq \frac{\Delta}{50}$ gives $T'' \geq 0.98\Delta$. If $T' = T - T_i$: $T - T_i = \Delta - (T_i - T_{i-1}) = \Delta - T_{i-1}$, and $\Delta = T - T_{i-1} < 3T_{i-1}$, so $T' < \Delta - \frac\Delta3 = \frac23\Delta$; if $T' = \varepsilon^2$, $T' \leq \frac\Delta{50}$. (b) $0.9^{3/2} > 0.85$ and $(\frac23)^{3/2} < 0.6$, so the difference exceeds $0.25\Delta^{3/2}$. (c) Since $\Delta \geq t_0$, $\Delta^{3/2} = \Delta\sqrt\Delta \geq t_0\sqrt\Delta$, so $(\frac16r_i^{-2} - 4)\Delta^{3/2} \geq (\frac{50}{6t_0} - 4)t_0\sqrt\Delta = (\frac{50}{6} - 4t_0)\sqrt\Delta$. Subtracting $\frac{2t_0}{25}\sqrt\Delta$ leaves at least $(8 - 5t_0)\sqrt\Delta \geq 4\sqrt\Delta$ for $t_0 = 2^{-5}$.

Solution

$R \geq D^{-1}h^{-2} \geq D^{-1}\delta^{-4}r_i^{-2} \geq \delta^{-3}r_i^{-2} \geq 1000r_i^{-2}$, using $D^{-1} \geq \delta$. But along backward flow lines from $B(y, t', \frac{r_i}{2C})$ over time $\Delta$, $R \leq 2r_i^{-2}$ by Lemma 11.2. A point cannot have both. So no flow line in the region runs into a cap created in the window, which is what "the backward parabolic neighbourhood exists" requires.