How Curvature Evolves

Under the Ricci flow the metric changes, so its curvature changes, and the curvature is what drives the flow. This chapter computes how. The result is a family of reaction–diffusion equations: each curvature quantity satisfies a heat equation (diffusion) with a quadratic term in the curvature (reaction). The scalar curvature satisfies

the Ricci tensor evolves by the Lichnerowicz Laplacian, and the full curvature tensor by $\Delta\operatorname{Rm} + Q(\operatorname{Rm})$. In dimension three, the reaction term reduces to an ODE for three eigenvalues. The ODE describes what curvature does when diffusion is ignored, and the maximum principles of 11A.4 Maximum Principles under Ricci Flow make that precise.

The chapter is pure computation, by design. The derivations are done once, in full, for a general variation $\partial_tg = h$, and then specialised to $h = -2\operatorname{Ric}$. Every formula is checked on the exact solutions of 11A.1 The Equation and Its First Solutions.

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

• derive the variations of the Christoffel symbols, the Ricci tensor, the scalar curvature and the volume form under $\partial_tg = h$;
• derive the evolution of $R$, $\operatorname{Ric}$ and $dV$ under the Ricci flow;
• state the evolution of the curvature tensor and explain Uhlenbeck's trick;
• write the three-dimensional curvature ODE and describe its phase portrait;
• check each formula on shrinking spheres and cylinders.

2.1The variation formulas

Let $g(t)$ be any family of metrics with $\partial_tg_{ij} = h_{ij}$. All the formulas below are tensorial, so they can be checked at a point in coordinates in which the Christoffel symbols vanish (9A.3 Geodesics and the Exponential Map); this is how each is derived.

The connection (9A.2 Connections):

The curvature. From $R_{ijk}{}^l = \partial_i\Gamma_{jk}^l - \partial_j\Gamma_{ik}^l + \Gamma\Gamma$-terms (9A.4 Curvature and What It Means), at a point where $\Gamma = 0$,

The Ricci tensor. Contracting $i$ with $l$ (since $R_{jk} = R_{ljk}{}^l$),

using $\partial_t\Gamma_{lk}^l = \frac12\nabla_k\operatorname{tr}h$ (Exercise 2.2).

The scalar curvature. Since $R = g^{jk}R_{jk}$ and $\partial_tg^{jk} = -h^{jk}$,

where $\operatorname{div}\operatorname{div}h = \nabla^j\nabla^kh_{jk}$.

The volume form (9B.7 The Heat Equation on a Manifold): $\partial_t\,dV = \frac12\operatorname{tr}h\,dV$.

2.2Under the Ricci flow

Now put $h = -2\operatorname{Ric}$, so $\operatorname{tr}h = -2R$.

Checks on the exact solutions (Exercise 2.3): on the shrinking unit sphere in dimension $n$, $R = \frac{n(n - 1)}{1 - 2(n - 1)t}$, and $R' = 2|\operatorname{Ric}|^2 = \frac{2R^2}{n}$, since $\operatorname{Ric}$ is a multiple of $g$ and $R$ is constant in space. On the shrinking cylinder $S^2\times\mathbb{R}$, $R = \frac{2}{\rho^2}$ and $|\operatorname{Ric}|^2 = \frac{2}{\rho^4} = \frac{R^2}{2}$, so $R' = R^2$, which matches $\rho^2 = \rho_0^2 - 2t$.

Two consequences used constantly:

• Since $|\operatorname{Ric}|^2 \geq \frac{R^2}{n}$ (8A.7 Tensors and Index Notation), $\partial_tR \geq \Delta R + \frac2nR^2$. By the maximum principle (11A.4 Maximum Principles under Ricci Flow), the minimum of $R$ cannot decrease, and positive scalar curvature blows up in finite time (the last exercise of 9B.6 Scalar Curvature and Topology).
• $\frac{d}{dt}\operatorname{Vol} = -\int R\,dV$: volume decreases where scalar curvature is positive.

2.3Uhlenbeck's trick and the curvature operator

The quadratic term $Q(\operatorname{Rm})$ is clean only in the right frame. Uhlenbeck's trick evolves an orthonormal frame $\{e_a\}$ by $\partial_te_a = \operatorname{Ric}(e_a)$ (raising an index with $g$), which keeps it orthonormal for $g(t)$ (Exercise 2.4). Written in this moving frame, the curvature tensor evolves by a heat equation whose reaction term involves no Ricci-times-Riemann terms, only an intrinsic quadratic in the curvature.

Hamilton (1986) wrote this reaction term through the curvature operator. Here the literature's normalisation differs from the one in 9A.5 Computing Curvature. Hamilton's operator $\mathcal M$ is twice the guide's $\mathcal R$: on the unit sphere, $\mathcal M = 2\operatorname{id}$, and in dimension three its eigenvalues are twice the sectional curvatures of the eigenplanes, with $R = \operatorname{tr}\mathcal M$. With that normalisation,

where $\mathcal M^\#$ is a quadratic expression built from the Lie algebra structure of $\Lambda^2 \cong \mathfrak{so}(n)$. In dimension three, $\mathcal M^\#$ is the adjugate: if $\mathcal M = \operatorname{diag}(\lambda, \mu, \nu)$, then $\mathcal M^\# = \operatorname{diag}(\mu\nu, \lambda\nu, \lambda\mu)$. Ignoring the Laplacian gives the three-dimensional curvature ODE

which preserves the ordering $\lambda \geq \mu \geq \nu$. It is the system previewed in 2B.10 Ordinary Differential Equations, and 11A.4 Maximum Principles under Ricci Flow shows that its invariant sets are invariant for the PDE.

Checks. The unit sphere: $\lambda = \mu = \nu = 2K$ with $K = \frac{1}{1 - 4t}$, and $(2K)' = 8K^2 = (2K)^2 + (2K)^2$. The cylinder $S^2\times\mathbb{R}$: $\lambda = \frac{2}{\rho^2}$, $\mu = \nu = 0$, and $\lambda' = \lambda^2$, while $\mu$ and $\nu$ stay $0$.

2.4History

Hamilton derived the evolution equations in his 1982 paper, introduced the curvature operator form with $\mathcal M^2 + \mathcal M^\#$ in "Four-manifolds with positive curvature operator" (1986), and used the moving frames that Karen Uhlenbeck had suggested. The Lichnerowicz Laplacian dates from Lichnerowicz's 1961 work on Einstein deformations.

2.5Exercises

Solution

$\nabla_l\partial_t\Gamma_{jk}^l = \frac12(\nabla^m\nabla_jh_{km} + \nabla^m\nabla_kh_{jm} - \Delta h_{jk})$; $\partial_t\Gamma_{lk}^l = \frac12g^{lm}(\nabla_lh_{km} + \nabla_kh_{lm} - \nabla_mh_{lk}) = \frac12\nabla_k\operatorname{tr}h$, the first and third terms cancelling. Contracting: $g^{jk}\partial_tR_{jk} = \frac12(2\operatorname{div}\operatorname{div}h - \Delta\operatorname{tr}h - \Delta\operatorname{tr}h) = \operatorname{div}\operatorname{div}h - \Delta\operatorname{tr}h$, and $\partial_tR = (\partial_tg^{jk})R_{jk} + g^{jk}\partial_tR_{jk}$ with $\partial_tg^{jk} = -h^{jk}$.

Solution

(a) $R = \frac{n(n - 1)}{r^2}$, $R' = \frac{n(n - 1)\cdot2(n - 1)}{r^4}$; $|\operatorname{Ric}|^2 = n\frac{(n - 1)^2}{r^4}$. (b) $R = \frac{2}{\rho^2}$, $R' = \frac{4}{\rho^4}$ ($\rho^{2\prime} = -2$); $\operatorname{Ric} = \frac{1}{\rho^2}g_{S^2(\rho)}$ restricted, eigenvalues $\frac{1}{\rho^2}, \frac{1}{\rho^2}, 0$, so $2|\operatorname{Ric}|^2 = \frac{4}{\rho^4}$. For Ricci: as a tensor, $\operatorname{Ric} = g_{S^2(1)}\oplus0$ is constant in time, so $\partial_t\operatorname{Ric} = 0$; and $2R_{pjkq}R^{pq} - 2R_{jp}R^p{}_k$: on the $S^2$ factor, with $\operatorname{Ric} = \frac{1}{\rho^2}g$ there and constant curvature $\frac{1}{\rho^2}$, the first term is $2\cdot\frac{1}{\rho^2}\cdot\frac{1}{\rho^2}(2 - 1)g_{jk} = \frac{2}{\rho^4}g_{jk}$ and the second is $\frac{2}{\rho^4}g_{jk}$, so they cancel; on the line factor both vanish. (c) $\operatorname{Vol} = |S^n|r^n$, $\frac{d}{dt}\operatorname{Vol} = |S^n|nr^{n-1}r' = |S^n|nr^{n-2}\cdot(-(n - 1)) = -\frac{n(n - 1)}{r^2}|S^n|r^n = -R\operatorname{Vol}$.

Solution

$\partial_t(g_{ij}e_a^ie_b^j) = -2R_{ij}e_a^ie_b^j + g_{ij}R^i{}_ke_a^ke_b^j + g_{ij}e_a^iR^j{}_ke_b^k = -2\operatorname{Ric}(e_a, e_b) + \operatorname{Ric}(e_a, e_b) + \operatorname{Ric}(e_a, e_b) = 0$.

Solution

$(\lambda - \mu)' = \lambda^2 - \mu^2 + \nu(\mu - \lambda) = (\lambda - \mu)(\lambda + \mu - \nu)$; similarly for $\mu - \nu$. A linear ODE $x' = a(t)x$ keeps the sign of $x$. For $\nu$: $\nu' = \nu^2 + \lambda\mu$, and while $\nu \geq 0$ we have $\mu \geq \nu \geq 0$ and $\lambda \geq 0$, so $\nu' \geq 0$ wherever $\nu = 0$; more carefully, at a first time where $\nu$ reaches $0$, $\nu' = \lambda\mu \geq 0$, so $\nu$ cannot become negative.

Solution

$\frac{d}{dt}\frac\nu\lambda = \frac{\nu'\lambda - \nu\lambda'}{\lambda^2} = \frac{(\nu^2 + \lambda\mu)\lambda - \nu(\lambda^2 + \mu\nu)}{\lambda^2} = \frac{\lambda^2(\mu - \nu) + \nu^2(\lambda - \mu)}{\lambda^2}$, a sum of nonnegative terms, positive since $\mu > \nu$. So $\frac\nu\lambda$, which is negative, increases towards $0$: the negative eigenvalue becomes small compared with the largest. Along a blow-up, this is the ODE form of pinching toward nonnegative curvature.