Course 11 Book 11A: The Ricci Flow: Existence and Maximum PrinciplesChapter 4

Maximum Principles under Ricci Flow

Scalar and tensor maximum principles, and preserved curvature conditions.

14 min read · Updated Oct 3, 2026

Read with Topping's Lectures on the Ricci Flow, chapter 3 (the maximum principle and its first applications), and Chow and Knopf's The Ricci Flow: An Introduction, chapter 4 (maximum principles, including the tensor and vector bundle versions). Hamilton's "Four-manifolds with positive curvature operator" (1986) introduced the vector bundle version.

In this chapter · 6 sections

4.1Invariant regions, again
4.2The scalar maximum principle
4.3Hamilton's tensor maximum principle
4.4The ODE–PDE maximum principle
4.5History
4.6Exercises

The evolution equations of 11A.2 How Curvature Evolves are reaction–diffusion equations: heat equations with quadratic terms. For such equations the maximum principle is the main tool. It says, roughly, that diffusion cannot create new extremes, so whatever the reaction alone would preserve, the full equation preserves too. This chapter develops it in three forms of increasing strength:

for scalar quantities such as $R$ ;
for symmetric 2-tensors such as $\operatorname{Ric}$ , with Hamilton's null-eigenvector condition;
for the whole curvature operator, with Hamilton's ODE–PDE principle: a closed convex set of curvature operators that the curvature ODE preserves is preserved by the flow.

The third is the invariant-region principle of 6A.4 Maximum Principles, transplanted to a vector bundle. It turns statements about the three-dimensional ODE of 11A.2 How Curvature Evolves into theorems about the Ricci flow.

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

apply the scalar maximum principle and Hamilton's trick to $R$ , and derive the lower bound $R \geq -\frac{n}{2t}$ ;
bound the existence time of a flow with positive scalar curvature;
state and use Hamilton's maximum principle for symmetric 2-tensors;
state and sketch the proof of the ODE–PDE maximum principle for the curvature operator;
prove that nonnegative curvature in dimension three, and nonnegative Ricci curvature in dimension three, are preserved.

Invariant regions, again

In the world Model Concentrations stay where the chemistry keeps them

In 6A.4 Maximum Principles, a system of reaction–diffusion equations for chemical concentrations, $\partial_tu = \Delta u + F(u)$ with $u$ a vector, was shown to keep $u$ in a closed convex set $K$ whenever the reaction alone, $u' = F(u)$ , keeps it there. Concentrations that start nonnegative stay nonnegative because the reactions cannot make them negative, and diffusion, which averages, cannot leave a convex set. Hamilton's 1986 theorem is the same statement with the concentrations replaced by the curvature operator at each point, the diffusion by the Laplacian of the Ricci flow, and the reaction by $\mathcal M^2 + \mathcal M^\#$ . The convex sets are sets of curvature operators: nonnegative curvature, nonnegative Ricci curvature, pinching sets.

The scalar maximum principle

On a closed manifold, if a function satisfies $\partial_tu \leq \Delta_{g(t)}u + \langle X, \nabla u\rangle + F(u)$ for a time-dependent metric and vector field, then $u$ is bounded above by the solution of the ODE $\phi' = F(\phi)$ with $\phi(0) = \max u(\cdot, 0)$ (9B.7 The Heat Equation on a Manifold). At a spatial maximum, $\nabla u = 0$ and $\Delta u \leq 0$ , so Hamilton's trick gives $\frac{d}{dt}\max u \leq F(\max u)$ . The same holds for minima with the inequalities reversed.

For the scalar curvature, $\partial_tR = \Delta R + 2|\operatorname{Ric}|^2 \geq \Delta R + \frac2nR^2$ , and comparison with $\phi' = \frac2n\phi^2$ gives:

Proposition 4.1 Scalar curvature under the Ricci flow

On a closed manifold, $R_{\min}(t) = \min_MR(\cdot, t)$ is nondecreasing, and

R_{\min}(t) \geq \frac{R_{\min}(0)}{1 - \frac2nR_{\min}(0)t}.

Consequently $R \geq -\frac{n}{2t}$ for every $t > 0$ , whatever the initial metric, and if $R_{\min}(0) > 0$ , the flow becomes singular no later than $T = \frac{n}{2R_{\min}(0)}$ .

Proof. $R_{\min}$ satisfies $R_{\min}' \geq \frac2nR_{\min}^2 \geq 0$ in the sense of forward difference quotients, and the solution of $\phi' = \frac2n\phi^2$ , $\phi(0) = R_{\min}(0)$ , is the stated function. If $R_{\min}(0) < 0$ , the function is $\geq -\frac{n}{2t}$ (drop the $1$ in the denominator). If $R_{\min}(0) > 0$ , the comparison function tends to $+\infty$ at $\frac{n}{2R_{\min}(0)}$ , and $R$ is finite while the flow exists.

The round sphere shows the bound is sharp (Figure 4.1). The bound $R \geq -\frac{n}{2t}$ is used in 10A.8 Min–Max and Width (the width estimate) and throughout the long-time analysis (12C.4 Geometrization).

Figure 4.1. The comparison solutions

\phi(t) = \frac{\phi_0}{1 - \frac2n\phi_0t}

for

n = 3

(computed). With

\phi_0 = 3

the bound blows up at

t = \frac12

(dashed line): a flow with

R \geq 3

must become singular by then. With

\phi_0 = -3

the bound rises towards

0

and stays above

-\frac{3}{2t}

(dotted), the universal lower bound.

Hamilton's tensor maximum principle

Theorem 4.2 Maximum principle for symmetric 2-tensors (Hamilton 1982)

Let $g(t)$ be a Ricci flow on a closed manifold, and $M(t)$ a symmetric 2-tensor field satisfying

\partial_tM \geq \Delta M + \nabla_XM + N(M, g, t),

where $X$ is a vector field and $N$ is a symmetric 2-tensor depending smoothly on its arguments that satisfies the null-eigenvector condition: whenever $M \geq 0$ and $M(v, v) = 0$ , then $N(v, v) \geq 0$ . If $M(0) \geq 0$ , then $M(t) \geq 0$ for all $t$ .

The idea: if $M$ first fails to be nonnegative at a point, there is a null eigenvector $v$ there; extend $v$ to be parallel at that point, and the function $M(v, v)$ has a spatial minimum $0$ , so $\Delta M(v, v) \geq 0$ , while the null-eigenvector condition makes the reaction nonnegative. A perturbation by $\varepsilon(1 + t)g$ makes the inequalities strict and gives a contradiction (Chow–Knopf, chapter 4). Hamilton used this to show, for instance, that $\operatorname{Ric} \geq 0$ is preserved in dimension three (Exercise 4.5).

The ODE–PDE maximum principle

The curvature operator $\mathcal M$ is a section of the bundle of symmetric endomorphisms of $\Lambda^2TM$ . With Uhlenbeck's trick (11A.2 How Curvature Evolves), this bundle can be identified with a fixed one, and the evolution is $\partial_t\mathcal M = \Delta\mathcal M + F(\mathcal M)$ , $F(\mathcal M) = \mathcal M^2 + \mathcal M^\#$ .

Theorem 4.3 Hamilton's maximum principle for systems (1986)

Let $K$ be a closed subset of the bundle that is convex in each fibre and invariant under parallel transport. Suppose $K$ is preserved by the ODE $\frac{d}{dt}\mathcal M = F(\mathcal M)$ in each fibre. If $\mathcal M(x, 0) \in K$ for all $x$ , then $\mathcal M(x, t) \in K$ for all $x$ and $t$ .

The idea Why convexity and the ODE are enough

A closed convex set is the intersection of the half-spaces $\{\ell \leq c\}$ that contain it, where $\ell$ ranges over linear functionals. Fix a supporting functional $\ell$ at a point of the boundary, transported parallel. Then $f = \ell(\mathcal M) - c$ satisfies $\partial_tf = \Delta f + \ell(F(\mathcal M))$ , a scalar equation. Where $\mathcal M$ touches the boundary of $K$ , the ODE invariance says $F(\mathcal M)$ points into $K$ , so $\ell(F(\mathcal M)) \leq 0$ there. That is exactly what the scalar maximum principle needs to keep $f \leq 0$ . Parallel invariance of $K$ makes the choice of $\ell$ compatible with differentiation in space. Hamilton's proof makes this rigorous with the distance from $\mathcal M$ to $K$ , showing that its maximum satisfies a Gronwall inequality (2B.10 Ordinary Differential Equations) starting from $0$ .

Figure 4.2. The convex set

\{\mu + \nu \geq 0\}

, which is

\operatorname{Ric} \geq 0

in dimension three, drawn in the

(\mu, \nu)

-plane for a fixed largest eigenvalue

\lambda = 1

(computed). Along its boundary

\mu + \nu = 0

, the ODE vector field

(\mu^2 + \lambda\nu, \nu^2 + \lambda\mu)

points inward, since

(\mu + \nu)' = \mu^2 + \nu^2 \geq 0

there. By the ODE–PDE principle, nonnegative Ricci curvature is preserved by the three-dimensional Ricci flow.

Consequences.

Nonnegative curvature operator $\{\mathcal M \geq 0\}$ is preserved in every dimension: it is convex and parallel-invariant, and Hamilton showed $\mathcal M^2 + \mathcal M^\#$ is nonnegative on the null space of a nonnegative $\mathcal M$ .
In dimension three, $\{\nu \geq 0\}$ (nonnegative sectional curvature) and $\{\mu + \nu \geq 0\}$ (nonnegative Ricci curvature) are preserved (Exercise 4.5, Exercise 4.6).
Pinching sets, which force curvature to become relatively nonnegative (11A.5 Hamilton–Ivey Pinching) or round (11A.6 Hamilton’s 1982 Theorem), are preserved, once they are shown to be convex and ODE-invariant.

Where this goes Where the principles are used

11A.5 Hamilton–Ivey Pinching builds the Hamilton–Ivey pinching set and 11A.6 Hamilton’s 1982 Theorem the pinching set for positive Ricci curvature. The same principle, applied to the Harnack quantity, proves Hamilton's Harnack inequality (11B.2 Ancient Solutions and the Harnack Inequality). Perelman's arguments use the scalar version constantly, with cut-off functions built from distance (9B.1 Laplacian Comparison).

History

Hamilton proved the tensor maximum principle in his 1982 paper and the vector bundle version in 1986. The invariant-region principle for reaction–diffusion systems is due to Chueh, Conley and Smoller (1977). Hamilton's trick for differentiating a maximum appears in his 1986 paper.

Recall Where we stand

On a closed manifold, maxima and minima of solutions of reaction–diffusion equations are controlled by the reaction ODE. For the scalar curvature: $R_{\min}$ is nondecreasing, $R \geq -\frac{n}{2t}$ always, and $R > 0$ forces a singularity by $\frac{n}{2R_{\min}(0)}$ . Symmetric 2-tensors stay nonnegative when the reaction satisfies the null-eigenvector condition. The curvature operator stays in any closed, convex, parallel set that the ODE $\mathcal M' = \mathcal M^2 + \mathcal M^\#$ preserves; so nonnegative curvature operator is preserved in all dimensions, and nonnegative sectional and Ricci curvature in dimension three. 11A.5 Hamilton–Ivey Pinching uses this to show that negative curvature becomes negligible at singularities.

Exercises

Exercise 4.4 The universal lower bound

From $R_{\min}' \geq \frac2nR_{\min}^2$ and $R_{\min}(0) = -a < 0$ , show $R_{\min}(t) \geq -\frac{a}{1 + \frac2nat} \geq -\frac{n}{2t}$ . Show that a hyperbolic metric attains the bound asymptotically: $R(t) = -\frac{n(n - 1)}{1 + 2(n - 1)t}$ , and compare with $-\frac{n}{2t}$ .

Solution

$\phi = \frac{-a}{1 + \frac2nat}$ solves $\phi' = \frac2n\phi^2$ , and $\frac{a}{1 + \frac2nat} < \frac{a}{\frac2nat} = \frac{n}{2t}$ . For the hyperbolic metric, $R \to -\frac{n(n - 1)}{2(n - 1)t} = -\frac{n}{2t}$ as $t \to \infty$ : asymptotically sharp. (Here $|\operatorname{Ric}|^2 = \frac{R^2}{n}$ exactly, so the comparison is an equality.)

Exercise 4.5 Nonnegative Ricci curvature in dimension three

In Hamilton's normalisation, the Ricci eigenvalues in dimension three are $\frac{\lambda + \mu}{2}$ , $\frac{\lambda + \nu}{2}$ , $\frac{\mu + \nu}{2}$ (each direction lies in two eigenplanes). (a) Show $\operatorname{Ric} \geq 0$ iff $\mu + \nu \geq 0$ . (b) Show the set $\{\mu + \nu \geq 0\}$ is convex, using that $\mu + \nu$ is the minimum of $\langle\mathcal Mu, u\rangle + \langle\mathcal Mv, v\rangle$ over orthonormal pairs. (c) Show $(\mu + \nu)' = \mu^2 + \nu^2 + \lambda(\mu + \nu)$ , which is $\geq 0$ on the boundary. Conclude that $\operatorname{Ric} \geq 0$ is preserved.

Solution

(a) The smallest Ricci eigenvalue is $\frac{\mu + \nu}{2}$ . (b) The minimum over a family of linear functions of $\mathcal M$ is concave, so its superlevel set $\{\geq 0\}$ is convex; it is also invariant under parallel transport, which acts by isometries of $\Lambda^2$ . (c) $(\mu + \nu)' = \mu^2 + \lambda\nu + \nu^2 + \lambda\mu$ , and on $\mu + \nu = 0$ this is $\mu^2 + \nu^2 \geq 0$ , so the ODE preserves the set; the ODE–PDE principle finishes.

Exercise 4.6 Nonnegative sectional curvature in dimension three

Show that $\{\nu \geq 0\}$ , the set of nonnegative curvature operators, is convex and that the ODE preserves it (the last exercise of 11A.2 How Curvature Evolves). Why is this the same as nonnegative sectional curvature in dimension three (9A.5 Computing Curvature)?

Solution

$\nu = \min_{|\omega| = 1}\langle\mathcal M\omega, \omega\rangle$ is concave, so $\{\nu \geq 0\}$ is convex. On $\nu = 0$ , $\nu' = \lambda\mu \geq 0$ since $\lambda \geq \mu \geq 0$ . In dimension three every 2-vector is decomposable, so the eigenvalues of $\mathcal M$ are (twice) sectional curvatures and $\mathcal M \geq 0$ iff $K \geq 0$ .

Exercise 4.7 The maximum of a family

Let $u(x, t)$ be smooth on $M\times[0, T]$ , $M$ compact. Show by example ( $M$ two points, $u = \sin t$ and $u = \cos t$ ) that $\max_xu(x, t)$ need not be differentiable, and explain why Hamilton's trick uses one-sided derivatives.

Solution

$\max(\sin t, \cos t)$ has a corner at $t = \frac\pi4$ , where the left derivative is $-\sin\frac\pi4$ (from $\cos$ ) and the right derivative is $\cos\frac\pi4$ (from $\sin$ ). The maximum of smooth functions is only Lipschitz, so one works with the upper right derivative, which at each time is the largest $\partial_tu$ over the points achieving the maximum.

Exercise 4.8 Rehearsal: volume growth from the lower bound

For a Ricci flow on a closed $n$ -manifold, use $R \geq -\frac{n}{2t}$ and $\frac{d}{dt}\operatorname{Vol} = -\int R\,dV$ to show that $t^{-n/2}\operatorname{Vol}(g(t))$ is nonincreasing. Check that it is constant to leading order for a hyperbolic metric as $t \to \infty$ . Perelman uses this monotone quantity in the long-time analysis (12C.4 Geometrization).

Solution

$\frac{d}{dt}\log\operatorname{Vol} = -\frac{\int R}{\operatorname{Vol}} \leq \frac{n}{2t}$ , so $\frac{d}{dt}\big(\log\operatorname{Vol} - \frac n2\log t\big) \leq 0$ . For a hyperbolic metric $\operatorname{Vol}(t) = (1 + 2(n - 1)t)^{n/2}\operatorname{Vol}(0)$ , and $t^{-n/2}\operatorname{Vol} \to (2(n - 1))^{n/2}\operatorname{Vol}(0)$ , a constant.

© 2026 NeckPinch (www.neckpinch.com). All content in the guidebook (text, mathematics, figures and exercises) is protected by copyright. All rights reserved. No part may be copied, republished or redistributed without written permission.

4.1Invariant regions, again

4.2The scalar maximum principle

4.3Hamilton's tensor maximum principle

4.4The ODE–PDE maximum principle

4.5History

4.6Exercises

Invariant regions, again

The scalar maximum principle

Hamilton's tensor maximum principle

The ODE–PDE maximum principle

History

Exercises