Course 9 Book 9A: Metrics, Connections and CurvatureChapter 5

Computing Curvature

Model spaces, necks, Lie groups and the curvature operator.

29 min read · Updated Oct 3, 2026

Read with Lee, Introduction to Riemannian Manifolds (2nd edition), chapter 7 (the Ricci decomposition, the Weyl tensor, conformal changes) and the computations for the model spaces and warped products in chapters 3 and 8 and their problems. Petersen's Riemannian Geometry, chapter 4 (examples: rotationally symmetric metrics, doubly warped products, Berger spheres), is the second voice. Milnor's paper "Curvatures of left invariant metrics on Lie groups" (Advances in Mathematics, 1976) is the source for the Lie group section.

In this chapter · 8 sections

5.1Products
5.2Rotationally symmetric metrics
5.3Conformal change
5.4Left-invariant metrics on three-dimensional Lie groups
5.5The Ricci decomposition
5.6The curvature operator
5.7History
5.8Exercises

This chapter is a workshop. It computes the curvature of the examples that the Ricci flow needs, in a form you can reuse: the model spaces, products, rotationally symmetric metrics (necks, caps, cones and dumbbells), conformal changes, and left-invariant metrics on three-dimensional Lie groups. It then takes the curvature tensor apart into its Ricci decomposition. In dimension three that decomposition shows that the Ricci tensor determines the whole curvature tensor, which is why Hamilton's three-dimensional theory can work with Ricci curvature alone.

There are fewer real-world boxes here than elsewhere, deliberately. The payoff is the next three books: every curvature value quoted later for a neck, a cigar, a Berger sphere or a Bryant soliton comes from the formulas below. Every formula in this chapter has been checked numerically by the figures script for the chapter (in the guide's tools/ folder), on the examples given.

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

compute the curvature of products and of warped products $dr^2 + \varphi(r)^2g_{S^{n-1}}$ ;
read off the geometry of a neck from its profile;
transform curvature under a conformal change of metric;
compute the Ricci curvature of left-invariant metrics on three-dimensional Lie groups, including Berger spheres, Nil and Sol;
state the Ricci decomposition, explain why $W = 0$ in dimension $3$ , and describe the curvature operator.

Products

On a Riemannian product $(M_1\times M_2, g_1\oplus g_2)$ , the Levi-Civita connection and the curvature split: vectors tangent to the two factors do not interact. So $\operatorname{Rm} = \operatorname{Rm}_1\oplus\operatorname{Rm}_2$ , $\operatorname{Ric} = \operatorname{Ric}_1\oplus\operatorname{Ric}_2$ and $R = R_1 + R_2$ , and a mixed plane, spanned by $X_1 \in TM_1$ and $X_2 \in TM_2$ , has $K(X_1, X_2) = 0$ .

$S^2(\rho)\times\mathbb{R}$ , the round cylinder in dimension $3$ : $\operatorname{Ric}$ has eigenvalues $\frac{1}{\rho^2}, \frac{1}{\rho^2}, 0$ , the zero in the axis direction, and $R = \frac{2}{\rho^2}$ . Its sectional curvatures range over $[0, \frac{1}{\rho^2}]$ .
$\mathbb{H}^2\times\mathbb{R}$ : $\operatorname{Ric}$ has eigenvalues $-1, -1, 0$ .
$S^2\times S^2$ (equal radii $1$ ): Einstein, with $\operatorname{Ric} = g$ , but not of constant curvature; sectional curvatures range over $[0, 1]$ .

The cylinder is the most important of these. It is the model of a neck in a Ricci flow, and under the flow it shrinks by $\rho(t)^2 = \rho_0^2 - 2t$ (the last exercise), pinching off at $t = \frac{\rho_0^2}{2}$ everywhere at once.

Rotationally symmetric metrics

For the warped product $g = dr^2 + \varphi(r)^2g_{S^{n-1}}$ of 9A.1 Riemannian Metrics and Model Spaces, use an orthonormal frame $e_1 = \partial_r$ and $e_2, \dots, e_n$ tangent to the spheres. The curvature operator is diagonal in this frame, and there are only two kinds of planes:

K_{\mathrm{rad}} = K(\partial_r, e_i) = -\frac{\varphi''}{\varphi}, \qquad K_{\mathrm{tan}} = K(e_i, e_j) = \frac{1 - \varphi'^2}{\varphi^2}.

Consequently

\operatorname{Ric}(\partial_r, \partial_r) = -(n - 1)\frac{\varphi''}{\varphi}, \qquad \operatorname{Ric}(e_i, e_i) = -\frac{\varphi''}{\varphi} + (n - 2)\frac{1 - \varphi'^2}{\varphi^2},

R = -2(n - 1)\frac{\varphi''}{\varphi} + (n - 1)(n - 2)\frac{1 - \varphi'^2}{\varphi^2}.

Why. The radial planes: the surface swept out by the geodesics in the plane of $\partial_r$ and $e_i$ is $dr^2 + \varphi(r)^2d\theta^2$ , totally geodesic, and its Gauss curvature is $-\frac{\varphi''}{\varphi}$ by the formula of 8A.9 The Curvature of Surfaces (Exercise 5.1). The tangential planes: the sphere $\{r = r_0\}$ is a round sphere of radius $\varphi(r_0)$ , with intrinsic curvature $\frac{1}{\varphi^2}$ , sitting in $M$ with second fundamental form $\frac{\varphi'}{\varphi}$ times its metric; the Gauss equation (9A.8 Submanifolds and Minimal Surfaces) subtracts $(\frac{\varphi'}{\varphi})^2$ . Petersen works through the full computation for rotationally symmetric metrics.

Checks. The sphere ( $\varphi = \sin r$ ): $K_{\mathrm{rad}} = K_{\mathrm{tan}} = 1$ . Hyperbolic space ( $\varphi = \sinh r$ ): both $-1$ . The cylinder ( $\varphi = \rho$ ): $K_{\mathrm{rad}} = 0$ , $K_{\mathrm{tan}} = \frac{1}{\rho^2}$ . The cone ( $\varphi = ar$ , $0 < a < 1$ ): $K_{\mathrm{rad}} = 0$ and $K_{\mathrm{tan}} = \frac{1 - a^2}{a^2r^2} > 0$ , blowing up at the tip. So in dimension $n \geq 3$ a cone over a small sphere has positive curvature blowing up at the vertex, unlike the 2D cone, which is flat away from its tip. These formulas make the model spaces of 9A.1 Riemannian Metrics and Model Spaces honest: $\mathbb{R}^n$ , $S^n$ and $\mathbb{H}^n$ have constant curvature $0$ , $1$ , $-1$ .

A neck. Figure 5.1 shows the two curvatures along a dumbbell metric on $S^3$ whose neck has radius $0.2$ . In the middle of the neck, $K_{\mathrm{tan}} = 25$ and $K_{\mathrm{rad}} = 1$ , so the neck is nearly a round cylinder $S^2\times\mathbb{R}$ of radius $0.2$ , with scalar curvature $R = 2(2K_{\mathrm{rad}} + K_{\mathrm{tan}}) = 54$ , close to the cylinder's $\frac{2}{0.2^2} = 50$ . At the shoulders, where the neck widens quickly, $\varphi''$ is large and positive, and both curvatures become strongly negative, $K_{\mathrm{tan}}$ wherever $|\varphi'| > 1$ . That mix of large positive and negative curvature is what Angenent and Knopf had to control to prove that such necks pinch (11B.4 Singularities).

Figure 5.1. Sectional curvatures of a dumbbell metric

dr^2 + \varphi(r)^2g_{S^2}

S^3

with a long neck of radius

0.2

, computed from the formulas above. The dashed curve is the profile

\varphi

(scaled by

30

). Across the neck the tangential curvature is

\frac{1}{0.2^2} = 25

and the radial curvature about

1

: the neck is nearly a cylinder. At the shoulders both dip sharply negative.

Conformal change

For $\tilde g = e^{2u}g$ on an $n$ -manifold,

\widetilde{\operatorname{Ric}} = \operatorname{Ric} - (n - 2)\big(\nabla^2u - du\otimes du\big) - \big(\Delta u + (n - 2)|\nabla u|^2\big)g,

\tilde R = e^{-2u}\big(R - 2(n - 1)\Delta u - (n - 1)(n - 2)|\nabla u|^2\big),

with $\Delta$ , $\nabla$ and norms taken in $g$ (Lee, chapter 7). In dimension $2$ this is $\tilde K = e^{-2u}(K - \Delta u)$ , the equation behind uniformization and the two-dimensional Ricci flow (5A.5 Uniformization and the Two-Dimensional Ricci Flow). In dimension $n \geq 3$ it is the starting point of the Yamabe problem: find a conformal metric of constant scalar curvature. Writing $e^{2u} = v^{4/(n - 2)}$ turns the scalar curvature equation into the semilinear equation $-\frac{4(n - 1)}{n - 2}\Delta v + Rv = \tilde Rv^{(n + 2)/(n - 2)}$ (the critical exponent of the Sobolev embedding, 4A.10 Sobolev Embeddings and Critical Exponents).

As a check, the Poincaré ball metric $\frac{4|dy|^2}{(1 - |y|^2)^2}$ is $e^{2u}$ times the flat metric with $u = \log2 - \log(1 - |y|^2)$ , and the formula gives $\tilde R = -n(n - 1)$ (Exercise 5.3).

Left-invariant metrics on three-dimensional Lie groups

A left-invariant metric on a Lie group $G$ is determined by an inner product on the Lie algebra $\mathfrak{g}$ (8A.5 Lie Groups and Group Actions). Its curvature is a computation in linear algebra, and its Ricci flow is a system of ODEs (11A.8 Homogeneous Flows). John Milnor (1976) showed that for a three-dimensional unimodular Lie group (one with $\operatorname{tr}\operatorname{ad}_X = 0$ for all $X$ , which includes every group with a compact quotient), any left-invariant metric has an orthonormal basis $e_1, e_2, e_3$ of $\mathfrak{g}$ with

[e_2, e_3] = \lambda_1e_1, \qquad [e_3, e_1] = \lambda_2e_2, \qquad [e_1, e_2] = \lambda_3e_3.

Set $\mu_i = \frac12(\lambda_1 + \lambda_2 + \lambda_3) - \lambda_i$ . Then the Ricci tensor is diagonal in this basis, with

\operatorname{Ric}(e_1, e_1) = 2\mu_2\mu_3, \qquad \operatorname{Ric}(e_2, e_2) = 2\mu_1\mu_3, \qquad \operatorname{Ric}(e_3, e_3) = 2\mu_1\mu_2.

The signs of $(\lambda_1, \lambda_2, \lambda_3)$ identify the group: $(+, +, +)$ is $SU(2)$ , so $S^3$ ; $(+, +, -)$ is $\widetilde{SL}(2, \mathbb{R})$ ; $(+, +, 0)$ is the universal cover of the Euclidean group of the plane; $(+, -, 0)$ is Sol; $(+, 0, 0)$ is the Heisenberg group, Nil; $(0, 0, 0)$ is $\mathbb{R}^3$ . These carry five of Thurston's eight geometries (10A.5 Thurston’s Eight Geometries).

The round $S^3$ . For $SU(2)$ , the unit quaternions, the left-invariant fields $X_1 = qi$ , $X_2 = qj$ , $X_3 = qk$ are orthonormal for the round metric of radius $1$ , and $[X_1, X_2] = 2X_3$ cyclically. So $\lambda_i = 2$ , $\mu_i = 1$ and $\operatorname{Ric} = 2g$ , as it must be.
Berger spheres. Rescale $X_1$ , the direction of the Hopf circles, to have length $\varepsilon$ : the orthonormal basis is $e_1 = X_1/\varepsilon$ , $e_2 = X_2$ , $e_3 = X_3$ , with $\lambda = (2\varepsilon, \frac2\varepsilon, \frac2\varepsilon)$ . Then $\mu_1 = \frac2\varepsilon - \varepsilon$ and $\mu_2 = \mu_3 = \varepsilon$ , so

\operatorname{Ric}(e_1, e_1) = 2\varepsilon^2, \qquad \operatorname{Ric}(e_2, e_2) = \operatorname{Ric}(e_3, e_3) = 4 - 2\varepsilon^2,

and the sectional curvatures of the coordinate planes are $K(e_1, e_2) = K(e_1, e_3) = \varepsilon^2$ and $K(e_2, e_3) = 4 - 3\varepsilon^2$ (Exercise 5.4). As $\varepsilon \to 0$ the curvatures stay between $0$ and $4$ while the Hopf circles shrink: collapse with bounded curvature (9B.3 Collapsing and Noncollapsing). Ricci curvature is positive exactly for $\varepsilon < \sqrt2$ (Figure 5.2).

Nil, $\lambda = (1, 0, 0)$ : $\mu = (-\frac12, \frac12, \frac12)$ and $\operatorname{Ric}$ has eigenvalues $\frac12, -\frac12, -\frac12$ ; $R = -\frac12$ .
Sol, $\lambda = (1, -1, 0)$ : $\mu = (-1, 1, 0)$ and $\operatorname{Ric}$ has eigenvalues $0, 0, -2$ .

Figure 5.2. Curvatures of the Berger spheres, the round

S^3

with the Hopf circles rescaled to length

2\pi\varepsilon

, from Milnor's formulas. At

\varepsilon = 1

everything equals the round values (

K = 1

\operatorname{Ric} = 2

). Ricci curvature is positive for

\varepsilon < \sqrt2

; as

\varepsilon \to 0

the curvature stays bounded while the volume goes to zero.

The Ricci decomposition

The Kulkarni–Nomizu product of two symmetric 2-tensors is the 4-tensor

(h \mathbin{\bigcirc\mkern-15mu\wedge} k)_{ijkl} = h_{il}k_{jk} + h_{jk}k_{il} - h_{ik}k_{jl} - h_{jl}k_{ik},

which has the symmetries of a curvature tensor. In this notation a constant curvature tensor is $\frac{\kappa}{2}g \mathbin{\bigcirc\mkern-15mu\wedge} g$ . Every curvature tensor splits orthogonally as

\operatorname{Rm} = W + \frac{1}{n - 2}\mathring{\operatorname{Ric}} \mathbin{\bigcirc\mkern-15mu\wedge} g + \frac{R}{2n(n - 1)}g \mathbin{\bigcirc\mkern-15mu\wedge} g,

where $\mathring{\operatorname{Ric}} = \operatorname{Ric} - \frac Rng$ is the trace-free Ricci tensor and the Weyl tensor $W$ is the part with all traces zero (Lee, chapter 7). The three pieces are the scalar part (what constant curvature has), the trace-free Ricci part (what Einstein metrics lack), and the Weyl part, the curvature that Ricci curvature does not see. The Weyl tensor is conformally invariant: $W(e^{2u}g) = e^{2u}W(g)$ as a $(0, 4)$ -tensor. In general relativity it is the curvature of empty space, the tidal shape change of 9A.4 Curvature and What It Means.

The idea Why three dimensions is special

The Weyl tensor has the symmetries of a curvature tensor and all its traces vanish. In dimension $3$ , a curvature tensor has $6$ independent components and the Ricci tensor also has $6$ , so no room is left: $W = 0$ in dimension $3$ , and the Ricci tensor determines the whole curvature tensor,

R_{ijkl} = R_{il}g_{jk} + R_{jk}g_{il} - R_{ik}g_{jl} - R_{jl}g_{ik} - \frac R2(g_{il}g_{jk} - g_{ik}g_{jl}).

This is why Hamilton's 1982 theorem (11A.6 Hamilton’s 1982 Theorem) can be stated and proved with Ricci curvature: in dimension $3$ , controlling $\operatorname{Ric}$ controls everything. It is also why the three-dimensional Ricci flow is so much more tractable than the four-dimensional one, where the Weyl tensor has $10$ components of its own.

The curvature operator

Since $\operatorname{Rm}$ is antisymmetric in each pair of slots, it defines a symmetric linear map on 2-vectors, the curvature operator $\mathcal{R} : \Lambda^2T_pM \to \Lambda^2T_pM$ , by

\langle\mathcal{R}(X\wedge Y), Z\wedge W\rangle = \operatorname{Rm}(X, Y, W, Z),

where $\Lambda^2$ has the inner product that makes $e_i\wedge e_j$ ( $i < j$ ) orthonormal. With this normalisation $\langle\mathcal{R}(X\wedge Y), X\wedge Y\rangle = K(X, Y)|X\wedge Y|^2$ , and the unit sphere has $\mathcal{R} = \operatorname{id}$ . (Other books scale $\mathcal{R}$ by a factor of $2$ or change its sign; test on the unit sphere.) Positive curvature operator, $\mathcal{R} > 0$ , implies positive sectional curvature, but is stronger in dimension $4$ and above, because not every 2-vector is a plane $X\wedge Y$ .

In dimension $3$ , every 2-vector is decomposable, $\Lambda^2 \cong TM$ (by $e_i\wedge e_j \mapsto e_k$ for $(i, j, k)$ cyclic), and if $\operatorname{Ric}$ has eigenvalues $r_1, r_2, r_3$ in an orthonormal eigenbasis, then $\mathcal{R}$ is diagonal in the basis $e_2\wedge e_3, e_3\wedge e_1, e_1\wedge e_2$ with eigenvalues

\frac{r_2 + r_3 - r_1}{2}, \qquad \frac{r_3 + r_1 - r_2}{2}, \qquad \frac{r_1 + r_2 - r_3}{2},

which are the sectional curvatures of the three coordinate planes (Exercise 5.5). So in dimension $3$ , $\mathcal{R} \geq 0$ if and only if all sectional curvatures are $\geq 0$ , and the Hamilton–Ivey pinching estimate (11A.5 Hamilton–Ivey Pinching) is a statement about the smallest eigenvalue of $\mathcal{R}$ .

Where this goes Curvature as a moving target

Under the Ricci flow, the curvature operator evolves by a heat equation with a quadratic reaction term. In Hamilton's normalisation $\mathcal M = 2\mathcal R$ , it reads $\partial_t\mathcal M = \Delta\mathcal M + \mathcal M^2 + \mathcal M^\#$ , where $\mathcal M^\#$ is a quadratic expression in $\mathcal M$ (11A.2 How Curvature Evolves). Hamilton's maximum principle for systems (11A.4 Maximum Principles under Ricci Flow) reduces many questions about the flow to the ODE $\frac{d}{dt}\mathcal M = \mathcal M^2 + \mathcal M^\#$ , which in dimension $3$ is an ODE for the three eigenvalues computed above (doubled).

History

Rotationally symmetric metrics have been the first examples since Riemann. Milnor's paper on left-invariant metrics appeared in 1976. The Weyl tensor was introduced by Hermann Weyl in 1918; the Kulkarni–Nomizu product is named after Ravindra Kulkarni and Katsumi Nomizu. The Yamabe problem was posed by Hidehiko Yamabe in 1960 and solved by the combined work of Neil Trudinger (1968), Thierry Aubin (1976) and Richard Schoen (1984). Berger spheres are named after Marcel Berger.

Recall Where we stand

Products split curvature, with mixed planes flat; the cylinder $S^2\times\mathbb{R}$ has Ricci eigenvalues $(\frac{1}{\rho^2}, \frac{1}{\rho^2}, 0)$ . Warped products $dr^2 + \varphi^2g_{S^{n-1}}$ have only two sectional curvatures, $-\frac{\varphi''}{\varphi}$ and $\frac{1 - \varphi'^2}{\varphi^2}$ ; a thin neck has a large tangential and small radial curvature, nearly a cylinder, and its shoulders have negative curvature. Conformal change transforms $\operatorname{Ric}$ and $R$ by explicit formulas generalising $\tilde K = e^{-2u}(K - \Delta u)$ . Milnor's formulas give the Ricci curvature of left-invariant metrics on unimodular 3D Lie groups: Berger spheres have Ricci $> 0$ for $\varepsilon < \sqrt2$ , Nil has $(\frac12, -\frac12, -\frac12)$ , Sol $(0, 0, -2)$ . Curvature splits into scalar, trace-free Ricci and Weyl parts; in dimension $3$ , $W = 0$ and Ricci determines everything, and the curvature operator has eigenvalues $\frac12(r_j + r_k - r_i)$ . 9A.6 The Laplacian and the Bochner Formula builds the calculus (Laplacian, Bochner formula, commuting derivatives) in which the flow's evolution equations are written.

Exercises

Exercise 5.1 The radial curvature

For the surface $dr^2 + \varphi(r)^2d\theta^2$ , use the formula $K = -\frac{(\sqrt G)_{rr}}{\sqrt G}$ for a metric $dr^2 + G\,d\theta^2$ (8A.9 The Curvature of Surfaces) to show $K = -\frac{\varphi''}{\varphi}$ . Check the sphere, the hyperbolic plane and the cone.

Solution

$\sqrt G = \varphi$ , so $K = -\frac{\varphi''}{\varphi}$ . Sphere: $-\frac{-\sin r}{\sin r} = 1$ ; hyperbolic: $-\frac{\sinh r}{\sinh r} = -1$ ; cone $\varphi = ar$ : $\varphi'' = 0$ , so $K = 0$ .

Exercise 5.2 Products and cylinders

Compute $\operatorname{Ric}$ and $R$ for $S^2(\rho)\times\mathbb{R}$ , $\mathbb{H}^2\times\mathbb{R}$ , $S^2\times S^1$ and $S^2\times S^2$ . Which are Einstein? Which have $\operatorname{Ric} \geq 0$ ?

Solution

$S^2(\rho)\times\mathbb{R}$ and $S^2(\rho)\times S^1$ : $\operatorname{Ric} = \frac{1}{\rho^2}g_{S^2(\rho)}\oplus0$ , $R = \frac{2}{\rho^2}$ , $\operatorname{Ric} \geq 0$ , not Einstein. $\mathbb{H}^2\times\mathbb{R}$ : $\operatorname{Ric} = -g_{\mathbb{H}^2}\oplus0$ , $R = -2$ . $S^2\times S^2$ with equal radii $1$ : $\operatorname{Ric} = g$ , Einstein, $R = 4$ . (In 9A.7 Jacobi Fields and Curvature versus Topology you prove that a compact manifold with $\operatorname{Ric} > 0$ has finite fundamental group, so $S^2\times S^1$ , with infinite $\pi_1$ , has no metric with $\operatorname{Ric} > 0$ , though it has one with $\operatorname{Ric} \geq 0$ .)

Exercise 5.3 Hyperbolic space by conformal change

With $u = \log2 - \log(1 - |y|^2)$ on the unit ball in $\mathbb{R}^n$ and the flat background ( $\operatorname{Ric} = 0$ ), compute $\nabla u$ , $|\nabla u|^2$ and $\Delta u$ , and show $\tilde R = -n(n - 1)$ .

Solution

With $\rho = |y|$ : $\nabla u = \frac{2y}{1 - \rho^2}$ , $|\nabla u|^2 = \frac{4\rho^2}{(1 - \rho^2)^2}$ , $\Delta u = \frac{2n}{1 - \rho^2} + \frac{4\rho^2}{(1 - \rho^2)^2}$ . Then $\tilde R = \frac{(1 - \rho^2)^2}{4}(n - 1)\Big(-\frac{4n}{1 - \rho^2} - \frac{8\rho^2}{(1 - \rho^2)^2} - \frac{4(n - 2)\rho^2}{(1 - \rho^2)^2}\Big) = \frac{n - 1}{4}\big(-4n(1 - \rho^2) - 4n\rho^2\big) = -n(n - 1)$ .

Exercise 5.4 Berger spheres

Verify the brackets $\lambda = (2\varepsilon, \frac2\varepsilon, \frac2\varepsilon)$ for the Berger basis, and the Ricci curvatures. Using $\operatorname{Ric}(e_i, e_i) = \sum_{j \neq i}K(e_i, e_j)$ (the curvature operator is diagonal in Milnor's frame), find the three coordinate sectional curvatures. For which $\varepsilon$ is the scalar curvature positive?

Solution

$[e_2, e_3] = [X_2, X_3] = 2X_1 = 2\varepsilon e_1$ ; $[e_3, e_1] = \frac1\varepsilon[X_3, X_1] = \frac2\varepsilon X_2 = \frac2\varepsilon e_2$ ; similarly $[e_1, e_2] = \frac2\varepsilon e_3$ . Then $\frac12\sum\lambda = \varepsilon + \frac2\varepsilon$ , so $\mu_1 = \frac2\varepsilon - \varepsilon$ , $\mu_2 = \mu_3 = \varepsilon$ , and $\operatorname{Ric}$ is $(2\varepsilon^2, 4 - 2\varepsilon^2, 4 - 2\varepsilon^2)$ . Writing $K_{ij}$ for the coordinate planes: $K_{12} + K_{13} = 2\varepsilon^2$ , $K_{12} + K_{23} = 4 - 2\varepsilon^2 = K_{13} + K_{23}$ , so $K_{12} = K_{13} = \varepsilon^2$ and $K_{23} = 4 - 3\varepsilon^2$ . $R = 8 - 2\varepsilon^2 > 0$ for $\varepsilon < 2$ .

Exercise 5.5 Dimension three

In an orthonormal basis diagonalising $\operatorname{Ric}$ with eigenvalues $r_1, r_2, r_3$ , use the formula for $R_{ijkl}$ in the box above to show $K(e_2, e_3) = R_{2332} = \frac{r_2 + r_3 - r_1}{2}$ , and that the components $R_{ijkl}$ with $\{i, j\} \neq \{k, l\}$ vanish. Check the round $S^3$ and $S^2\times\mathbb{R}$ .

Solution

$R_{2332} = R_{22}g_{33} + R_{33}g_{22} - R_{23}g_{32} - R_{32}g_{23} - \frac R2(g_{22}g_{33} - g_{23}g_{32}) = r_2 + r_3 - \frac{r_1 + r_2 + r_3}{2}$ . If $\{i, j\} \neq \{k, l\}$ , every term of the formula contains an off-diagonal $g$ or an off-diagonal $R$ , which vanish. Round $S^3$ : $r_i = 2$ gives $K = 1$ . $S^2\times\mathbb{R}$ with $r = (1, 1, 0)$ ( $e_3$ the axis): $K(e_1, e_2) = \frac{1 + 1 - 0}{2} = 1$ , $K(e_2, e_3) = \frac{1 + 0 - 1}{2} = 0$ .

Exercise 5.6 Nil

For the Heisenberg group with $[e_2, e_3] = e_1$ and the other brackets of the basis zero, compute $\mu$ , $\operatorname{Ric}$ and $R$ , and the sectional curvatures of the coordinate planes. Why can no left-invariant metric on Nil have constant curvature?

Solution

$\lambda = (1, 0, 0)$ , $\frac12\sum\lambda = \frac12$ , $\mu = (-\frac12, \frac12, \frac12)$ ; $\operatorname{Ric} = (2\cdot\frac14, 2\cdot(-\frac14), 2\cdot(-\frac14)) = (\frac12, -\frac12, -\frac12)$ , $R = -\frac12$ . Coordinate planes: $K_{23} = \frac{r_2 + r_3 - r_1}{2} = -\frac34$ , $K_{12} = K_{13} = \frac14$ . Any left-invariant metric on Nil can be put in Milnor's form with $\lambda = (\lambda_1, 0, 0)$ , $\lambda_1 > 0$ , and then $\operatorname{Ric}$ has eigenvalues of both signs, so it is not a multiple of $g$ .

Exercise 5.7 Rehearsal: the shrinking neck

(a) For the cylinder $S^{n-1}(\rho)\times\mathbb{R}$ , show that $g(t) = \rho(t)^2g_{S^{n-1}} + dz^2$ is a Ricci flow when $\rho(t)^2 = \rho_0^2 - 2(n - 2)t$ , so it becomes singular at $T = \frac{\rho_0^2}{2(n - 2)}$ with $R(t) = \frac{(n - 1)(n - 2)}{\rho_0^2 - 2(n - 2)t}$ . (b) Check that $(T - t)R(t)$ is constant: curvature blows up like $\frac{1}{T - t}$ , a Type I singularity (11B.4 Singularities). (c) Why does the $\mathbb{R}$ factor not move?

Solution

(a) $\operatorname{Ric}(g(t)) = (n - 2)g_{S^{n-1}}\oplus0$ (scale invariance, 9A.4 Curvature and What It Means), so $\partial_t(\rho^2) = -2(n - 2)$ . $R = \frac{(n - 1)(n - 2)}{\rho^2}$ . (b) $\rho(t)^2 = 2(n - 2)(T - t)$ , so $(T - t)R = \frac{n - 1}{2}$ . (c) $\operatorname{Ric}$ vanishes in the axis direction, so $\partial_tg(\partial_z, \partial_z) = 0$ .

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5.1Products

5.2Rotationally symmetric metrics

5.3Conformal change

5.4Left-invariant metrics on three-dimensional Lie groups

5.5The Ricci decomposition

5.6The curvature operator

5.7History

5.8Exercises

Products

Rotationally symmetric metrics

Conformal change

Left-invariant metrics on three-dimensional Lie groups

The Ricci decomposition

The curvature operator

History

Exercises