Hamilton–Ivey Pinching

The Ricci flow on a general 3-manifold has regions of negative curvature, and nothing prevents them from persisting. What the Hamilton–Ivey pinching estimate shows is that, where the curvature is large, negative curvature is small by comparison. Wherever the most negative sectional curvature is large in absolute value, the scalar curvature is larger still, by a logarithmic factor. Blow up the flow at a point of large curvature, rescaling so that the curvature there becomes $1$: the negative part of the curvature tends to zero. Every singularity model of the three-dimensional flow has nonnegative sectional curvature.

This is the estimate that makes three dimensions special. The singularity models of 11B.4 Singularities and Perelman's $\kappa$-solutions (12B.1 κ-Solutions) all have nonnegative curvature, and that is why they can be classified. The estimate is proved with the maximum principle of 11A.4 Maximum Principles under Ricci Flow: it identifies a set of curvature operators, convex and preserved by the ODE, and the principle does the rest.

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

• state the Hamilton–Ivey estimate in Hamilton's normalisation;
• explain why the pinching set is convex and preserved by the curvature ODE;
• deduce that rescaled limits at singularities have nonnegative sectional curvature;
• see from a computation how slowly the negative part becomes negligible.

5.1The estimate

Write the eigenvalues of the curvature operator in Hamilton's normalisation (11A.2 How Curvature Evolves) as $\lambda \geq \mu \geq \nu$, so that they are twice the sectional curvatures of the eigenplanes, $R = \lambda + \mu + \nu$, and $\nu$ is twice the smallest sectional curvature.

The normalisation $\nu \geq -1$ at $t = 0$ is not a restriction: any closed manifold has bounded curvature, and rescaling the metric by a constant (11A.1 The Equation and Its First Solutions) makes $\nu \geq -1$. The theorem is due to Hamilton (in his 1995 survey) and Thomas Ivey (1993), independently.

What it says. Only points where $|\nu|$ is large matter. There, $\frac{|\nu|}{R} \leq \frac{1}{\log|\nu| + \log(1 + t) - 3}$, which tends to $0$ as $|\nu| \to \infty$. If $|\nu|$ stays bounded while $R \to \infty$, the ratio tends to $0$ anyway. Either way, at points where $R$ is large, $\frac{\nu}{R}$ is close to $0$ or positive (Figure 5.1).

5.2Why it holds

The set of curvature operators satisfying the inequality (together with $\nu \geq -\frac{1}{1 + t}$) is a time-dependent set $K(t)$ in each fibre. Hamilton's maximum principle (11A.4 Maximum Principles under Ricci Flow), in a version for sets that depend on time, applies if three things hold.

1. Convexity. The function $x \mapsto x(\log x + \log(1 + t) - 3)$ is convex for $x > 0$ (its second derivative is $\frac1x$). The smallest eigenvalue $\nu$ is a concave function of $\mathcal M$, and $R$ is linear. Together these make $\{R \geq f(-\nu)\}$, for an increasing convex $f$, a convex set of operators (Exercise 5.3).
2. Parallel invariance. The conditions depend only on eigenvalues, which parallel transport preserves.
3. ODE invariance. Along the ODE $\lambda' = \lambda^2 + \mu\nu$ and cyclically (11A.2 How Curvature Evolves), a solution that starts in $K(0)$ stays in $K(t)$. This is the heart of the proof, a careful computation. On the boundary, where $R = x(\log x + \log(1 + t) - 3)$ with $x = -\nu > 0$, one shows that $R$ grows at least as fast as the right-hand side. The case $\mu \geq 0$ is the easier one (Exercise 5.4), and Chow–Knopf (chapter 9) carry out the case $\mu < 0$.

Figure 5.2 shows the set at $t = 0$ in the plane of $|\nu|$ and $R$, with ODE trajectories that start on its edge and move into it.

5.3Consequence: nonnegative curvature at singularities

This is how Hamilton and Perelman know that singularity models in dimension three, and the $\kappa$-solutions of 12B.1 κ-Solutions, have $\operatorname{Rm} \geq 0$. Without it, the tools of 9B.5 Splitting and Soul Theorems (splitting, souls), which need nonnegative curvature, could not be used.

5.4History

Thomas Ivey proved the estimate in "Ricci solitons on compact three-manifolds" (Differential Geometry and its Applications, 1993), using it to show that compact three-dimensional Ricci solitons have constant curvature. Hamilton proved it independently and stated it in the form above in "The formation of singularities in the Ricci flow" (1995). Perelman used it in the definition of the canonical neighbourhoods and of Ricci flow with surgery, where the estimate must be shown to survive each surgery (12B.4 Surgery).

5.5Exercises

Solution

$-\nu(\mathcal M) = \max_{|\omega| = 1}\langle-\mathcal M\omega, \omega\rangle$ is convex, and a nondecreasing convex function of a convex function is convex. So $\mathcal M \mapsto f(-\nu(\mathcal M)) - R(\mathcal M)$ is convex, and its sublevel set $\{\leq 0\}$ is convex. Here $f''(x) = \frac1x > 0$ and $f'(x) = \log(x(1 + t)) - 2 \geq 0$ exactly when $x(1 + t) \geq e^2$. Where $x(1 + t) < e^3$, $f(x) < 0$, and the proof uses instead the lower bound $R \geq -\frac{3}{1 + 2t}$, which follows from 11A.4 Maximum Principles under Ricci Flow because $R = \lambda + \mu + \nu \geq 3\nu \geq -3$ initially.

Solution

(a) Expanding, $\frac12[(\lambda + \mu)^2 + (\lambda + \nu)^2 + (\mu + \nu)^2] = \lambda^2 + \mu^2 + \nu^2 + \lambda\mu + \lambda\nu + \mu\nu$, which is the sum of the three ODE right-hand sides. (b) $x' = -(\nu^2 + \lambda\mu) \leq -x^2$ since $\lambda\mu \geq 0$. With $\nu = -x$: $R' = \frac12[(\lambda + \mu)^2 + (\lambda - x)^2 + (\mu - x)^2]$; for fixed $x$, minimising over $\lambda = \mu = s$ gives $2s^2 + (s - x)^2$, smallest at $s = \frac x3$, where it equals $\frac{2x^2}{3}$; so $R' \geq \frac{x^2}{3}$. (c) $\frac{d}{dt}f = f_x\,x' + \frac{x}{1 + t}$ with $f_x = \log(x(1 + t)) - 2 \geq 0$ in the region, and $x' \leq 0$, so $\frac{d}{dt}f \leq \frac{x}{1 + t}$. Hence $\frac{d}{dt}(R - f) \geq \frac{x^2}{3} - \frac{x}{1 + t} = x\big(\frac x3 - \frac{1}{1 + t}\big) \geq 0$, because $x(1 + t) \geq e^2 > 3$.

Solution

Solve $x(\log x - 3) = 10^6$: $x \approx 9.9\times10^4$ gives $x(\log x - 3) \approx 9.9\times10^4\times8.5 \approx 8.4\times10^5$; $x \approx 1.15\times10^5$ gives about $1.0\times10^6$. So $x \approx 1.2\times10^5$ and $\frac xR \approx 0.12$. For $10^{12}$: $x \approx 4.7\times10^{10}$ ($\log x \approx 24.6$), ratio about $0.05$. The negative part shrinks relative to $R$, but only logarithmically.

Solution

For a shrinking soliton, $g(t) = (T - t)\phi_t^*g_0/T$, so the curvature scales like $\frac{1}{T - t}$ and $\frac{\nu}{R}$ at corresponding points is constant in time. As $t \to T$, $R \to \infty$ everywhere, so Hamilton–Ivey forces $\frac{\nu}{R} \to$ a limit $\geq 0$. Since the ratio is constant along the flow, $\frac\nu R \geq 0$ from the start: $\operatorname{Rm} \geq 0$.

Solution

Rescaling the metric by $Q_k$ divides the curvature operator by $Q_k$, so $\nu_k \geq -\frac{A}{Q_k}$. Without a uniform bound, $\nu$ could go to $-\infty$ along with $R$; Hamilton–Ivey says that if $|\nu|$ is large then $R \geq |\nu|\log|\nu|$ roughly, so after dividing by $R$ the negative part is at most $\frac{1}{\log|\nu|}$, still tending to $0$.