Homogeneous Flows

A left-invariant metric on a Lie group looks the same at every point, so its Ricci flow stays left-invariant (8A.6 Flows and the Lie Derivative): the flow reduces from a PDE to an ODE for finitely many numbers. For the three-dimensional unimodular groups of 9A.5 Computing Curvature, those numbers are the lengths of a Milnor frame, and Milnor's curvature formulas give the ODE explicitly. This chapter derives it once, in general, and solves it for each geometry: $SU(2)$, Nil, Sol, $\widetilde{SL}(2, \mathbb{R})$ and the Euclidean group. It is a laboratory in which the long-time behaviours predicted in 10A.7 Geometrization and Ricci Flow can be computed exactly: rounding and extinction, convergence to flat, and collapse with bounded or decaying curvature.

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

• derive the Ricci flow ODE for diagonal left-invariant metrics on a three-dimensional unimodular Lie group;
• solve it exactly for Nil and describe its solutions for Sol, $\widetilde{SL}(2, \mathbb{R})$, $SU(2)$ and the Euclidean group;
• read off which directions shrink and which expand, and what the rescaled limits are;
• check the general estimates of 11A.4 Maximum Principles under Ricci Flow on these examples;
• connect the behaviours to the roles of the geometries in geometrization.

8.1The ODE

Let $X_1, X_2, X_3$ be a basis of left-invariant fields with $[X_2, X_3] = c_1X_1$, $[X_3, X_1] = c_2X_2$, $[X_1, X_2] = c_3X_3$ (Milnor's form, 9A.5 Computing Curvature), and consider metrics that are diagonal in this basis, $g = \operatorname{diag}(A, B, C)$. The orthonormal frame $e_1 = X_1/\sqrt A$, and so on, has structure constants $\lambda_1 = \frac{c_1A}{\sqrt{ABC}}$, $\lambda_2 = \frac{c_2B}{\sqrt{ABC}}$, $\lambda_3 = \frac{c_3C}{\sqrt{ABC}}$. Milnor's formula $\operatorname{Ric}(e_1, e_1) = 2\mu_2\mu_3$ and $A' = -2A\operatorname{Ric}(e_1, e_1)$ give (Exercise 8.1)

The flow keeps the metric diagonal, because the Ricci tensor is diagonal in Milnor's frame.

8.2The geometries one by one

$SU(2)$, $c = (2, 2, 2)$: the sphere. With $B = C$, this is the Berger flow of 11A.6 Hamilton’s 1982 Theorem: $A' = -\frac{4A^2}{B^2}$, $B' = -4(2 - \frac AB)$. The metric becomes round as it shrinks to a point in finite time.

The Euclidean group, $c = (1, 1, 0)$. The metric with $A = B$ is flat. From any diagonal metric the flow converges to a flat one, with $A - B \to 0$, while the curvature decays.

Nil, $c = (1, 0, 0)$. The equations are $A' = -\frac{A^2}{BC}$, $B' = \frac AC$, $C' = \frac AB$. With $B = C$, the product $AB$ is constant (Exercise 8.2), and

The fibre direction $X_1$ shrinks like $t^{-1/3}$ and the base directions grow like $t^{1/3}$. The curvature decays like $\frac1t$.

Sol, $c = (1, -1, 0)$. The equations are $A' = -\frac{A^2 - B^2}{BC}$, $B' = \frac{A^2 - B^2}{AC}$, $C' = \frac{(A + B)^2}{AB}$. The difference $A - B$ decays (Exercise 8.3), and once $A = B$, $C = C_0 + 4t$: the two directions of the torus fibre stay bounded while the circle direction grows linearly.

$\widetilde{SL}(2, \mathbb{R})$, $c = (-2, 2, 2)$. With $B = C$: $A' = -\frac{4A^2}{B^2}$, $B' = 8 + \frac{4A}{B}$. The base directions grow like $8t$, while the fibre direction $A$ converges to a positive constant.

8.3Rescaled limits: collapse

Divide each metric by $t$, as Perelman does for long-lived flows (10A.7 Geometrization and Ricci Flow):

geometry	behaviour of $(A, B, C)$	$\frac{g(t)}{t}$ as $t \to \infty$	role in geometrization
$SU(2)$	rounds, extinct at finite $T$	(extinct)	spherical pieces disappear
Euclidean group	converges to flat	collapses to a point	flat pieces are thin
Nil	$t^{-1/3}$, $t^{1/3}$, $t^{1/3}$	collapses to a point	graph-manifold (thin) pieces
Sol	$1$, $1$, $t$	collapses to a line	thin, Sol torus bundles
$\widetilde{SL}(2, \mathbb{R})$	$1$, $t$, $t$	fibre collapses; base $\to$ hyperbolic plane	Seifert (thin) pieces

Every geometry except $SU(2)$ collapses after rescaling: some directions become negligible compared with $\sqrt t$, the volume of $\frac{g(t)}{t}$-balls of unit radius tends to zero, and the curvature of $\frac{g(t)}{t}$ stays bounded. This is the thin part of 10A.7 Geometrization and Ricci Flow and 12C.4 Geometrization, computed exactly. The hyperbolic geometry, which is not a Lie group geometry, is the one that does not collapse: $\frac{g(t)}{t} \to 4g_0$ (10A.6 Hyperbolic Three-Manifolds). Isenberg and Jackson classified these behaviours in 1992; Knopf and McLeod (2001) described the convergence of the normalised flows in detail, including on non-unimodular groups.

8.4History

Milnor's curvature formulas date from 1976. James Isenberg and Martin Jackson studied the Ricci flow of locally homogeneous metrics on closed 3-manifolds in 1992, and Dan Knopf and Kevin McLeod refined the picture in 2001. Lott later studied the long-time limits of these flows as geometric objects ("On the long-time behavior of type-III Ricci flow solutions", 2007).

8.5Exercises

Solution

$2\mu_2\mu_3 = \frac12(\lambda_1 - (\lambda_2 - \lambda_3))(\lambda_1 + (\lambda_2 - \lambda_3)) = \frac12(\lambda_1^2 - (\lambda_2 - \lambda_3)^2) = \frac{(c_1A)^2 - (c_2B - c_3C)^2}{2ABC}$. Then $A' = -2\operatorname{Ric}(X_1, X_1) = -2A\cdot\frac{(c_1A)^2 - (c_2B - c_3C)^2}{2ABC}$. Round: $A' = -\frac{4A^2}{A^2} = -4$.

Solution

$(AB)' = A'B + AB' = -\frac{A^2}{B} + \frac{A^2}{B} = 0$, and $(B^3)' = 3B^2\cdot\frac AB = 3AB$, constant, so $B^3 = B_0^3 + 3A_0B_0t$. Ricci in the orthonormal frame is $(\frac{A}{2B^2}, -\frac{A}{2B^2}, -\frac{A}{2B^2})$, so $R = -\frac{A}{2B^2} = -\frac{A_0B_0}{2B^3} \sim -\frac{A_0B_0}{6A_0B_0t} = -\frac{1}{6t}$.

Solution

$A' - B' = -\frac{A^2 - B^2}{BC} - \frac{A^2 - B^2}{AC} = -\frac{(A^2 - B^2)(A + B)}{ABC}$. With $A = B$: $C' = \frac{4A^2}{A^2} = 4$. Ricci: $\operatorname{Ric}(e_1) = \operatorname{Ric}(e_2) = 0$ and $\operatorname{Ric}(e_3) = -\frac{(A + B)^2}{2ABC} = -\frac2C$, so $R = -\frac{2}{C_0 + 4t} \geq -\frac{2}{4t} = -\frac{1}{2t} > -\frac{3}{2t}$.

Solution

$B' = 8 + \frac{4A}{B} \geq 8$. Then $\frac{d}{dt}\big(\frac1A\big) = -\frac{A'}{A^2} = \frac{4}{B^2} \leq \frac{4}{(B_0 + 8t)^2}$, which is integrable, so $\frac1A$ stays bounded and $A$ is bounded below. The base metric $\frac{B}{t}(\omega_2^2 + \omega_3^2) \to 8(\omega_2^2 + \omega_3^2)$, a hyperbolic metric on the base, while $\frac At \to 0$: the circle fibres collapse.

Solution

Nil: $R = -\frac{A_0B_0}{2B^3} \sim -\frac{1}{6t}$. Sol: $R = -\frac{2}{C_0 + 4t} \sim -\frac{1}{2t}$. $\widetilde{SL}(2, \mathbb{R})$: with $B = C$, Milnor's formulas give $R = -\frac{8}{B} - \frac{2A}{B^2}$, which is $\sim -\frac{1}{t}$ since $B \sim 8t$ and $A$ is bounded. Each is $\geq -\frac{3}{2t}$ for large $t$, as the bound requires; the full curvature tensor is of the same order, so the flows are Type III.