Ricci Flow on Surfaces

On a surface, the Ricci flow is a scalar equation in disguise. Since $\operatorname{Ric} = Kg = \frac R2g$, the flow

moves the metric only by conformal factors, and it reduces to a nonlinear heat equation for one function. It was the first case Hamilton solved after 1982. The answer is the uniformization theorem, proved by a flow: every metric on a closed surface flows, after normalisation, to a metric of constant curvature. This chapter proves the easy cases, explains the hard one (the sphere), and introduces two tools that reappear in Perelman's work in more elaborate form: Hamilton's entropy and the Harnack inequality. It also meets the cigar, the first steady soliton.

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

• write the two-dimensional flow as an equation for the conformal factor, and derive $\partial_tR = \Delta R + R^2$;
• prove convergence of the normalised flow when $\chi(M) < 0$, and describe what happens on the sphere;
• define Hamilton's surface entropy and explain why it is monotone;
• state the two-dimensional Harnack inequality and check it on the cigar soliton;
• describe the ancient solutions on the sphere.

7.1Surfaces in the computer

7.2The conformal factor

Write $g(t) = e^{2u(t)}g_0$. Since the flow only rescales $g$ pointwise, the conformal class is preserved, and with $K = e^{-2u}(K_0 - \Delta_0u)$ (5A.5 Uniformization and the Two-Dimensional Ricci Flow),

This is a quasilinear heat equation for $u$, with diffusion coefficient $e^{-2u}$: it is strictly parabolic, so in two dimensions the DeTurck trick is unnecessary. The scalar curvature evolves by

since $2|\operatorname{Ric}|^2 = 2\cdot2\big(\frac R2\big)^2 = R^2$ (11A.2 How Curvature Evolves). The normalised flow, which preserves area $A$, is

where $r$ is constant in time by Gauss–Bonnet (8A.9 The Curvature of Surfaces). Under it, $\partial_tR = \Delta R + R(R - r)$.

7.3Convergence

When $\chi(M) < 0$. Here $r < 0$, and the reaction term $R(R - r)$ pushes $R$ towards $r$. Comparing with the ODE $\rho' = \rho(\rho - r)$, whose solutions converge to $r$ exponentially when $\rho$ starts between the roots, the maximum principle gives bounds on $R$, and a more careful argument (Hamilton's potential function $f$ with $\Delta f = R - r$) gives $|R - r| \leq Ce^{rt}$ (Exercise 7.4). The case $\chi = 0$ is similar, with polynomial convergence.

The sphere. Here $r > 0$, the reaction term is destabilising, and the proof is much harder. Hamilton proved convergence when $R > 0$ initially, using the entropy and Harnack inequality below. Bennett Chow (1991) showed that any metric on $S^2$ develops $R > 0$ in finite time. Hamilton's original argument used the uniformization theorem along the way, through the Kazdan–Warner identity, which rules out nontrivial solitons on $S^2$. Xiuxiong Chen, Peng Lu and Gang Tian (2006) removed that dependence, so the flow gives an independent proof of uniformization for the sphere.

Figure 7.1 shows an elongated sphere becoming round.

7.4Hamilton's entropy

The proof differentiates $N$ using $\partial_tR = \Delta R + R(R - r)$ and $\partial_t\,dA = (r - R)\,dA$, and finds $\frac{dN}{dt} = -\int\frac{|\nabla R|^2}{R}\,dA + \int(R - r)^2\,dA$. The second term has the wrong sign. Hamilton controls it with a potential function, rewriting $\frac{dN}{dt}$ as minus the integral of a square (Exercise 7.5 does the first part). Its form, an integral of $R\log R$ that decreases, is the ancestor of Perelman's $\mathcal F$- and $\mathcal W$-functionals (12A.2 Ricci Flow as a Gradient Flow, 12A.3 The 𝓦-Entropy), which do the same for the full Ricci flow in every dimension, and of the entropy of the heat equation (6A.10 Entropy, Information and Diffusion). For the round sphere of area $4\pi$, $R = 2$ and $N = 8\pi\log2 \approx 17.42$, its minimum value (Figure 7.2).

7.5The Harnack inequality and the cigar

For the unnormalised flow on a surface with $R > 0$, Hamilton proved the Harnack inequality

equivalently $\Delta\log R + R \geq -\frac1t$, since $\partial_t\log R = \Delta\log R + |\nabla\log R|^2 + R$. Like the Li–Yau inequality for the heat equation (9B.7 The Heat Equation on a Manifold), integrating it along paths compares curvature at different points and times. Its matrix version in all dimensions (11B.2 Ancient Solutions and the Harnack Inequality) is one of the main tools of Book 11B.

The cigar soliton is the complete metric on $\mathbb{R}^2$

which in geodesic polar coordinates is $ds^2 + \tanh^2s\,d\theta^2$ (9B.3 Collapsing and Noncollapsing). It is a steady soliton: under the Ricci flow it moves only by diffeomorphisms, flowing along a radial vector field (11B.1 Ricci Solitons). It satisfies the Harnack inequality with equality in the limit: $\Delta\log R + R = 0$ (Exercise 7.6).

Ancient solutions on the sphere. A solution is ancient if it exists for all $t \in (-\infty, T)$. Shrinking round spheres are ancient, and so is the King–Rosenau solution, a "sausage" that looks, far in the past, like two cigars joined end to end, and becomes round as it shrinks. John King (1993) and Philip Rosenau (1995) found it as a solution of the logarithmic fast diffusion equation, which the conformal factor satisfies. Physicists Fateev, Onofri and Zamolodchikov found it independently in 1993, as the "sausage" sigma model. Panagiota Daskalopoulos, Hamilton and Nataša Šešum (2012) proved that these are the only compact ancient solutions on $S^2$, apart from quotients.

7.6History

Hamilton's "The Ricci flow on surfaces" appeared in 1988, Chow's "The Ricci flow on the 2-sphere" in 1991, and Chen–Lu–Tian's note in 2006. Hamilton introduced the surface entropy and the Harnack inequality in the 1988 paper. The cigar appeared there as well, and in physics as Witten's two-dimensional black hole (1991).

7.7Exercises

Solution

$\partial_t(e^{2u}g_0) = 2(\partial_tu)e^{2u}g_0 = -Re^{2u}g_0$ gives $\partial_tu = -\frac R2 = -e^{-2u}(K_0 - \Delta_0u)$. For the normalised flow, $2\partial_tu = r - R$, so $\partial_tu = \frac r2 - e^{-2u}(K_0 - \Delta_0u)$.

Solution

The ODE $\rho' = \rho(\rho - r)$ with $r < 0$ and $r < \rho(0) < 0$ has the solution $\rho(t) = \frac{r}{1 - \big(1 - \frac{r}{\rho(0)}\big)e^{rt}}$, which tends to $r$ with $\rho - r = O(e^{rt})$ (since $r < 0$). The maximum principle gives $R_{\max}(t) \leq \rho(t)$ with $\rho(0) = R_{\max}(0)$, hence $R_{\max} - r \leq Ce^{rt}$. A symmetric argument bounds $R_{\min}$ from below once it is known to be greater than some value, and the potential function handles positive values of $R$.

Solution

$\frac{d}{dt}\int R\log R\,dA = \int(\log R + 1)\partial_tR\,dA + \int R\log R\,(r - R)\,dA$. Substituting, the terms $\int\log R\cdot R(R - r)$ cancel, leaving $\int(\log R + 1)\Delta R + \int R(R - r)$. Integrating by parts, $\int\log R\,\Delta R = -\int\frac{|\nabla R|^2}{R}$ and $\int\Delta R = 0$. Finally $\int R(R - r) = \int(R - r)^2 + r\int(R - r) = \int(R - r)^2$.

Solution

$e^{2u} = (1 + r^2)^{-1}$, $2u = -\log(1 + r^2)$, and $\Delta_E\log(1 + r^2) = f'' + \frac{f'}{r} = \frac{2 - 2r^2}{(1 + r^2)^2} + \frac{2}{1 + r^2} = \frac{4}{(1 + r^2)^2}$. So $K = \frac12(1 + r^2)\frac{4}{(1 + r^2)^2} = \frac{2}{1 + r^2}$ and $R = 2K$. Then $\log R = \log4 - \log(1 + r^2)$, $\Delta_g = (1 + r^2)\Delta_E$, and $\Delta_g\log R = -(1 + r^2)\frac{4}{(1 + r^2)^2} = -R$.

Solution

$R = \frac{2}{1 - 2t}$, $R' = \frac{4}{(1 - 2t)^2} = R^2$, with $\Delta R = 0$. The Harnack quantity is $\partial_t\log R = R > 0 > -\frac1t$: far from tight. For the cigar it is $0$, the value for an eternal (steady) solution, while the bound $-\frac1t$ is attained only asymptotically by expanding solutions that emerge from a point. The sphere is compact and shrinking; the inequality is designed to be sharp on solitons.