Hamilton’s 1982 Theorem

In 1982 Hamilton proved the first theorem about topology by the Ricci flow:

The theorem says positive Ricci curvature is an open door to the round sphere: the flow walks through it. This chapter gives the proof in four steps, following Hamilton: positivity is preserved, the curvature pinches toward constant, its gradient is controlled, and the rescaled metrics converge. Each step uses tools already built: the maximum principles of 11A.4 Maximum Principles under Ricci Flow, the evolution equations of 11A.2 How Curvature Evolves, the extension criterion of 11A.3 Short-Time Existence and Uniqueness, and Myers' theorem (9A.7 Jacobi Fields and Curvature versus Topology). The chapter ends with what came after: the same strategy in higher dimensions, up to the differentiable sphere theorem.

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

• outline the four steps of Hamilton's proof and say which tool each uses;
• explain the pinching estimate $|\mathring{\operatorname{Ric}}|^2 \leq CR^{2 - \delta}$ and why it forces constant curvature;
• describe how the gradient estimate and Myers' theorem control the shape at the singular time;
• compute the flow of Berger spheres and watch them round out;
• say what the theorem implies for the Poincaré conjecture, and what it does not.

6.1A computed example: Berger spheres become round

6.2The four steps

Step 1: positive Ricci curvature is preserved. In dimension three, $\operatorname{Ric} \geq 0$ is a convex, ODE-invariant set of curvature operators (11A.4 Maximum Principles under Ricci Flow). Hamilton proved more: $\operatorname{Ric} \geq \varepsilon Rg$ is preserved for some $\varepsilon > 0$ depending on $g_0$. On a compact manifold with $\operatorname{Ric} > 0$, such an $\varepsilon$ exists at $t = 0$. By 11A.4 Maximum Principles under Ricci Flow, the flow becomes singular at a finite time $T$, and $R_{\min} \to \infty$ as $t \to T$.

Step 2: pinching toward constant curvature. Let $\mathring{\operatorname{Ric}} = \operatorname{Ric} - \frac R3g$, the trace-free part. Hamilton's key estimate:

for constants $C, \delta > 0$ depending on $g_0$. It follows from the maximum principle applied to $f = \frac{|\mathring{\operatorname{Ric}}|^2}{R^{2 - \delta}}$, whose evolution equation has a reaction term that is nonpositive when $\delta$ is small, thanks to $\operatorname{Ric} \geq \varepsilon Rg$. Dividing by $R^2$: $\frac{|\mathring{\operatorname{Ric}}|^2}{R^2} \leq CR^{-\delta} \to 0$ where $R \to \infty$. In dimension three the Weyl tensor vanishes (9A.5 Computing Curvature), so the curvature tensor is determined by $\operatorname{Ric}$, and $\operatorname{Ric}$ close to $\frac R3g$ means all sectional curvatures are close to one another, relative to their size (Figure 6.2).

Step 3: the gradient estimate. Hamilton proved that for every $\eta > 0$, $\frac{|\nabla R|^2}{R^3} \leq \eta$ wherever $R$ is large enough, up to an error that decays like a negative power of $R$: as curvature becomes large, its gradient becomes small relative to it. Integrating along geodesics, $R$ is nearly constant on balls of radius $\frac{c}{\sqrt{R_{\max}}}$.

Step 4: convergence. As $t \to T$, Step 3 gives $\frac{R_{\min}}{R_{\max}} \to 1$. By Step 2, $\operatorname{Ric} \geq \frac{R}{3}(1 - o(1))g$, and Myers' theorem (9A.7 Jacobi Fields and Curvature versus Topology) bounds the diameter by $\frac{C}{\sqrt{R_{\min}}} \to 0$: the manifold shrinks to a point while becoming round. Rescale to fixed volume: the normalised flow (11A.1 The Equation and Its First Solutions) exists for all time, and Hamilton showed it converges exponentially fast, in every $C^k$ norm, to a metric with $\mathring{\operatorname{Ric}} = 0$. In dimension three, an Einstein metric has constant sectional curvature, here positive. Its universal cover is the round $S^3$ (9A.7 Jacobi Fields and Curvature versus Topology and 10A.5 Thurston’s Eight Geometries), so $M = S^3/\Gamma$.

6.3What the theorem does and does not say about Poincaré

If $M$ is simply connected and carries some metric with $\operatorname{Ric} > 0$, the theorem gives $M = S^3/\Gamma$ with $\Gamma = \pi_1(M) = 1$: $M = S^3$. So the Poincaré conjecture holds for such manifolds. But a simply connected 3-manifold comes with no metric of positive Ricci curvature, and finding one is as hard as the conjecture itself. For an arbitrary initial metric, the flow develops regions of negative curvature, necks that pinch, and other singularities. Hamilton spent the next two decades building tools for that case (11B.5 Hamilton’s Program in 2002), and Perelman finished the job.

6.4History

Hamilton's paper was submitted in 1982 and introduced the Ricci flow, its short-time existence, the evolution equations and the tensor maximum principle, all to prove this theorem. Hamilton's 1986 paper treated positive curvature operator in dimension four; Böhm–Wilking's paper appeared in the Annals of Mathematics in 2008, and Brendle–Schoen's in the Journal of the AMS in 2009.

6.5Exercises

Solution

With $R_{ij} = \frac R3g_{ij}$: $R_{il}g_{jk} + R_{jk}g_{il} - R_{ik}g_{jl} - R_{jl}g_{ik} = \frac{2R}{3}(g_{il}g_{jk} - g_{ik}g_{jl})$, and subtracting $\frac R2(g_{il}g_{jk} - g_{ik}g_{jl})$ leaves $\frac R6(\dots)$. Schur's lemma makes $R$ constant in dimension $3$.

Solution

(a) $A' = -2\operatorname{Ric}(X_1, X_1) = -2A\cdot\frac{2A}{B^2}$; $B' = -2B\cdot\frac{2(2B - A)}{B^2} = -\frac{4(2B - A)}{B}$. (b) $x' = \frac{A'B - AB'}{B^2} = \frac{-\frac{4A^2}{B} + \frac{4A(2B - A)}{B}}{B^2} = \frac{4A(2B - 2A)}{B^3} = \frac{8x(1 - x)}{B}$. (c) $A' = -4 = -4\cdot\frac{(1 - 4t)^2}{(1 - 4t)^2}$ and $B' = -4(2 - 1) = -4$.

Solution

The function $\min_{|v| = 1}\frac{\operatorname{Ric}(v, v)}{R}$ is continuous and positive on the compact unit sphere bundle, so it has a positive minimum. For the Berger sphere, the Ricci eigenvalues are $\frac{2A}{B^2}$ and $\frac{2(2B - A)}{B^2}$ (twice), and $R = \frac{8B - 2A}{B^2}$; with $A = 0.3$, $B = 1$: eigenvalues $0.6$ and $3.4$, $R = 7.4$, so $\varepsilon = \frac{0.6}{7.4} \approx 0.081$.

Solution

Myers with $(n - 1)k = 2k = \frac{R_{\min}(1 - \eta)}{3}$ gives $\operatorname{diam} \leq \frac{\pi}{\sqrt k} = \pi\sqrt{\frac{6}{(1 - \eta)R_{\min}}}$. As $t \to T$, $R_{\min} \to \infty$ (Step 1 with the gradient estimate), so the diameter tends to zero.

Solution

By the theorem $M = S^3/\Gamma$ with $\Gamma \cong \pi_1(M) = 1$, so $M = S^3$. The conjecture concerns every closed simply connected 3-manifold, and nothing guarantees that such a manifold admits a metric with $\operatorname{Ric} > 0$; proving that it does would itself prove the conjecture. What is needed is a flow that works from an arbitrary metric, handling the singularities that then form: Ricci flow with surgery (12B.5 Ricci Flow with Surgery for All Time, 12C.3 The Poincaré Conjecture, Assembled).