Surgery

A Ricci flow on a closed 3-manifold may reach a time $T$ at which the curvature blows up somewhere. By 12B.3 The Canonical Neighbourhood Theorem, the regions of high curvature just before $T$ are made of necks and caps. Perelman's second preprint turns this into a procedure. Look at the limit of the metric as $t \to T$ on the part of the manifold where it survives; find the long thin horns of necks running into the singularity; cut each one along the central sphere of a neck of a small, fixed radius $h$; throw away the horn; glue in a standard cap; and restart the flow. This chapter describes the cap that is glued in (the standard solution), the structure of the flow at the singular time, the cutting and gluing, and what each surgery does to the topology: it undoes a connected sum. That is how the Ricci flow finds the prime decomposition (10A.3 The Prime Decomposition).

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

• describe the standard solution and list Perelman's five claims about it;
• describe the limit metric at the first singular time and its pieces: tubes, caps, horns, capped horns and double horns;
• state the surgery procedure with its parameters $\varepsilon$, $r$, $\delta$, $\rho$ and $h$;
• explain why each surgery removes volume at least $h^3$;
• show that undoing a surgery is a connected sum, and read off the topology of the manifold before surgery.

4.1Cutting and sealing

4.2The standard solution

The cap that is glued in comes with its own future: Perelman needs to know how the flow behaves near a fresh cap, and he gets it from one model flow.

The cap is a choice, made once. The obvious choice, a round hemisphere glued to the cylinder, is not smooth enough: its curvature jumps where the hemisphere meets the cylinder (Figure 4.1). Any smooth choice close to it will do. Perelman's claims about the standard solution, with his sketches of proof, are:

1. Symmetry. The solution stays rotationally symmetric, because Killing fields remain Killing under the flow (by the maximum principle applied to their evolution).
2. Behaviour at infinity. It converges at infinity to the shrinking round cylinder of scalar curvature $1$ at time $0$, which becomes singular at time $1$; so the solution's maximal time is at most $1$ (Exercise 4.5).
3. Uniqueness, by reducing the linearised flow, using the symmetry, to a system of two equations on the line.
4. Existence on $[0, 1)$, as a limit of flows on $S^3$ from long capped cylinders, with pseudolocality (12A.6 Pseudolocality) ruling out an earlier singularity.
5. Canonical neighbourhoods. Every point has a canonical neighbourhood as in 12B.2 The Structure of κ-Solutions, except that a neck need not be strong at times before $\frac34$; and $R_{\min}(t) \geq c(1 - t)^{-1}$.

Perelman's proof of Claim 3 is a sketch. Bing-Long Chen and Xi-Ping Zhu later proved that Ricci flows with bounded curvature on complete noncompact manifolds are unique (Journal of Differential Geometry 74, 2006), a result which covers the standard solution.

4.3The flow at the first singular time

Let $g(t)$, $t \in [0, T)$, be a Ricci flow on a closed oriented 3-manifold whose curvature blows up as $t \to T$. By 12B.3 The Canonical Neighbourhood Theorem there is $r > 0$ such that every point with $R \geq r^{-2}$ has a canonical neighbourhood, and the derivative estimates of 12B.2 The Structure of κ-Solutions hold there. If some canonical neighbourhood is a closed positively curved component, that component becomes extinct at time $T$. Otherwise, let

The derivative estimates show that $\Omega$ is open and that $R \to \infty$ on $M\setminus\Omega$. On $\Omega$ the metrics converge smoothly to a limit metric $\bar g$ (Shi's derivative estimates). For $\rho < r$ let $\Omega_\rho = \{x \in \Omega : \bar R(x) \leq \rho^{-2}\}$; it is compact.

The argument walks along necks. Start at an $\varepsilon$-neck in $\Omega$ and look at a point on one of its boundary spheres. If the point has curvature above $\rho^{-2}$, its canonical neighbourhood is another neck or a cap, attached to the first. Continuing gives a chain of necks that ends in $\Omega_\rho$, ends in a cap, or goes on forever into the singularity, forming a horn. The pieces of types (tube, cap, horn) touching $\Omega_\rho$ have volume bounded below in terms of $\rho$, so there are finitely many of them. Looking at $g(t)$ just before $T$ shows how $M$ is reassembled from the $\Omega_j$, by gluing tubes and caps along the horns.

4.4The surgery

Two scales are now in play: the canonical neighbourhood scale $r$, and the much smaller surgery radius $h$, the radius of the necks along which one cuts. Perelman explains that the point is to make $h$ arbitrarily small while keeping $r$ bounded away from zero. The surgery is done deep inside horns, where the geometry is not merely an $\varepsilon$-neck but a strong $\delta$-neck for a much smaller $\delta$.

The proof is by contradiction: a sequence of bad points with $h(x) \to 0$ has rescaled limits that are complete, nonnegatively curved and have two ends, so they split as $S^2\times\mathbb{R}$ (Perelman cites Toponogov; this is the splitting theorem of 9B.5 Splitting and Soul Theorems). Repeating backwards in time gives an ancient limit that is the round cylinder, contradicting badness.

Gluing in the cap. Perelman takes the gluing from Hamilton's four-dimensional work. Near the middle of the neck, multiply the metric by a conformal factor $e^{-f}$, where $f = f(z)$ depends only on the coordinate $z$ along the neck, is positive on the part to be removed and zero on the part to be kept. Where $f$ is small, the dominant terms in the change of curvature are positive multiples of $f''$, so the pinching improves, and all curvatures become positive where $f > \delta'$.

Each surgery removes volume. The removed horn contains, at least, a half-neck of radius $h$ and length comparable to $\delta^{-1}h$, while the cap glued in has volume comparable to $h^3$. Perelman records that every surgery reduces the volume by at least $h^3$. Between surgeries, $\frac{d}{dt}\operatorname{Vol} = -\int R\,dV$, and $R$ is bounded below (11A.4 Maximum Principles under Ricci Flow), so the volume grows at most at a controlled rate. So surgery times cannot accumulate: on any finite time interval there are finitely many (11B.5 Hamilton’s Program in 2002 rehearsed this). This is what lets the flow with surgery continue, provided the hypotheses of 12B.3 The Canonical Neighbourhood Theorem survive the surgeries, which is 12B.5 Ricci Flow with Surgery for All Time.

4.5The topology of surgery

Surgery changes the manifold, but in a controlled way. Cutting along a 2-sphere and capping both sides with balls is undone by a connected sum (10A.3 The Prime Decomposition, Exercise 4.7). There are two cases.

• The sphere separates the component. Cutting and capping produces two closed manifolds $M_1$ and $M_2$ with $M \cong M_1\#M_2$.
• The sphere does not separate. Cutting and capping produces one manifold $M'$ with $M \cong M'\#(S^2\times S^1)$ (Exercise 4.8).

Discarded components are covered by necks and caps, or are nearly round. They are among $S^3$, $\mathbb{RP}^3$, $S^2\times S^1$, $\mathbb{RP}^3\#\mathbb{RP}^3$ and spherical space forms $S^3/\Gamma$. So, keeping track through all the surgeries, the original manifold is

where $M_{\text{final}}$ is whatever remains (Figure 4.3). The Ricci flow with surgery is carrying out the prime decomposition geometrically: necks are where the essential spheres of 10A.3 The Prime Decomposition become small. 12C.1 Reading Off the Topology makes this bookkeeping precise, and 12C.2 Finite Extinction shows that for simply connected manifolds nothing remains.

4.6History

Hamilton introduced surgery for the Ricci flow in his 1997 paper on four-manifolds with positive isotropic curvature, including the conformal capping construction that Perelman uses. Perelman wrote at the start of his second preprint (2003) that Hamilton's argument, as written, contained an unjustified statement that he could not fix, and that his own approach was different, with two scales, $h$ and $r$. Perelman's longer-term aim, which he said he had not yet achieved, was a canonical Ricci flow defined on as large a part of space-time as possible. Bing-Long Chen and Xi-Ping Zhu proved uniqueness for complete noncompact flows with bounded curvature in 2006. Perelman's canonical flow was realised much later: Bruce Kleiner and John Lott constructed singular Ricci flows through singularities in dimension three (Acta Mathematica 219, 2017), and Richard Bamler and Kleiner proved their uniqueness (Acta Mathematica 228, 2022) (12C.5 After Perelman).

4.7Exercises

Solution

$\operatorname{Ric}$ of $S^2(r)$ is $\frac{1}{r^2}g$, so $(r^2)' = -2$ and $r^2 = 2 - 2t$; $R = \frac{2}{r^2} = \frac{1}{1 - t}$. The standard solution converges at infinity to this cylinder (Claim 2), so its curvature at infinity becomes unbounded as $t \to 1$, and it cannot be continued with bounded curvature past $t = 1$.

Solution

(a) Both numerators are then nonnegative and $\psi > 0$. (b) Hemisphere: $\psi'' = -\frac{1}{\sqrt2}\sin(s/\sqrt2)$, so $K_{\text{rad}} = \frac12$; $K_{\text{tan}} = \frac{1 - \cos^2(s/\sqrt2)}{2\sin^2(s/\sqrt2)} = \frac12$; $R = 2 + 1 = 3$. Cylinder: $K_{\text{rad}} = 0$, $K_{\text{tan}} = \frac12$, $R = 1$. (c) At $s = \pi/\sqrt2$ both pieces have $\psi = \sqrt2$ and $\psi' = 0$, but $\psi''$ jumps from $-\frac{1}{\sqrt2}$ to $0$, so the curvature jumps from $3$ to $1$. A smooth admissible profile has $\psi'$ decreasing smoothly from $1$ at the tip to $0$, with all derivatives of $\psi'$ vanishing where it reaches $0$, and total rise $\sqrt2$; the solid profile in Figure 4.1 is one.

Solution

The connected sum $M_1\#M_2$ removes the interior of the glued-in ball $B^3$ from each $M_i$, which leaves $N_1$ and $N_2$, and glues them along their boundary spheres. That is exactly $M$, cut along $\Sigma$ and reglued. (Up to diffeomorphism the gluing map does not matter, since orientation-preserving diffeomorphisms of $S^2$ are isotopic to the identity.)

Solution

Remove two balls from $M'$ (the caps) to recover $N$. Gluing the two boundary spheres of $N$ together gives $M$. On the other hand, $M'\#(S^2\times S^1)$ is $M'$ minus a ball, glued to $S^2\times S^1$ minus a ball, which is $S^2\times[0, 1]$ with a ball removed; equivalently, it is $M'$ with two balls removed and the two boundary spheres joined by a tube $S^2\times[0, 1]$. Joining two boundary spheres by a tube is the same as gluing them directly. So both are $N$ with its two boundary spheres identified, which is $M$.

Solution

If the chain of necks closes up, the manifold is an $S^2$-bundle over $S^1$, and orientability makes it $S^2\times S^1$. Otherwise the tube $S^2\times I$ has a cap at each end. Two balls give $B^3\cup_{S^2}B^3 = S^3$. A ball and a punctured $\mathbb{RP}^3$ give $\mathbb{RP}^3$. Two punctured copies of $\mathbb{RP}^3$, joined by a tube, give $\mathbb{RP}^3\#\mathbb{RP}^3$, by the definition of the connected sum.