Curvature in a Moving Frame

11A.2 How Curvature Evolves derived the evolution of the scalar and Ricci curvatures exactly, and that of the full curvature tensor only in the schematic form $\partial_t\operatorname{Rm} = \Delta\operatorname{Rm} + Q(\operatorname{Rm})$. For much of the theory the schematic form is enough: it gives the doubling time and Shi's estimates. But the deepest results of Hamilton's theory, the preservation of nonnegative curvature, Hamilton–Ivey pinching, Hamilton's 1982 theorem and the strong maximum principle, all depend on the exact quadratic $Q$, written in the right frame. In that frame $Q$ becomes $\mathcal M^2 + \mathcal M^\#$, an expression built from the Lie algebra of rotations, and in dimension three it reduces to the three-dimensional ODE that the rest of Book 11A keeps returning to.

This chapter derives all of this carefully. It fixes the index conventions (which differ between this guide, Hamilton and Morgan–Tian, and are the most common source of sign errors in the subject), writes Hamilton's evolution equation for $R_{ijkl}$ in coordinates, explains Uhlenbeck's trick of evolving an orthonormal frame by the Ricci tensor, derives the curvature operator equation $\partial_t\mathcal M = \Delta\mathcal M + \mathcal M^2 + \mathcal M^\#$, and computes $\mathcal M^\#$ in dimension three. Every formula is checked on round spheres and cylinders, and the main identities were also checked numerically on random curvature tensors.

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

• translate between the guide's convention for $\operatorname{Rm}$ and Hamilton's indexing used in Morgan–Tian;
• state Hamilton's evolution equation for $R_{ijkl}$ with the tensor $B_{ijkl}$, and verify it on the shrinking sphere and cylinder;
• explain Uhlenbeck's trick: evolving frames, the bundle isometry $\Phi_t$, and why the Ricci terms disappear;
• derive $\partial_t\mathcal M = \Delta\mathcal M + \mathcal M^2 + \mathcal M^\#$ from the first Bianchi identity and the structure constants of $\mathfrak{so}(n)$;
• show that in dimension three $\mathcal M^\#$ is the adjugate of $\mathcal M$, and derive the curvature ODE and its basic invariant sets.

3.1Conventions

The guide writes $\operatorname{Rm}(X, Y, Z, W) = \langle R(X, Y)Z, W\rangle$ with $R(X, Y) = \nabla_X\nabla_Y - \nabla_Y\nabla_X - \nabla_{[X, Y]}$, so that the sectional curvature of the plane spanned by orthonormal $X, Y$ is $\operatorname{Rm}(X, Y, Y, X)$ (9A.4 Curvature and What It Means). Hamilton, and Morgan and Tian following him, index the curvature tensor so that $R_{ijij}$ is the sectional curvature. To match their formulas we use, in this chapter only,

the guide's tensor with the last two indices swapped. Then $R_{ijij} = K(e_i, e_j)$, the symmetries are the usual ones ($R_{ijkl} = -R_{jikl} = -R_{ijlk} = R_{klij}$, and the first Bianchi identity $R_{ijkl} + R_{iklj} + R_{iljk} = 0$), and the Ricci tensor is the contraction

The round sphere of sectional curvature $K$ has $R_{ijkl} = K(g_{ik}g_{jl} - g_{il}g_{jk})$, so $R_{1212} = K$. The round cylinder $S^2(\rho)\times\mathbb{R}$, with $e_1, e_2$ tangent to the sphere and $e_3$ along the line, has $R_{1212} = \frac{1}{\rho^2}$ and all components with an index $3$ equal to zero.

When reading Morgan and Tian, check which convention a formula uses by testing it on the sphere, as we do below. Their Theorem 3.13 is in Hamilton's indexing.

3.2Hamilton's evolution equation

Introduce the quadratic tensor

It has the symmetries $B_{ijkl} = B_{jilk} = B_{klij}$, but not those of a curvature tensor (Exercise 3.8).

The equation has three kinds of terms, and it is worth seeing where each comes from.

1. The derivative of the connection. By 11A.2 How Curvature Evolves, $\partial_tR_{ijk}{}^l$ is a combination of $\nabla(\partial_t\Gamma)$, and $\partial_t\Gamma$ is a combination of first derivatives of $\partial_tg = -2\operatorname{Ric}$. So the first contribution is a combination of second derivatives of the Ricci tensor: terms $\nabla_i\nabla_kR_{jl}$ and their index permutations.
2. The Laplacian of the curvature. The second Bianchi identity says $\nabla_mR_{ijkl}$ is cyclic in three of its indices (9A.4 Curvature and What It Means). Taking a divergence and using the contracted Bianchi identity expresses $\Delta R_{ijkl}$ as the same combination of $\nabla^2\operatorname{Ric}$, plus quadratic terms that come from commuting covariant derivatives (the Ricci identity, 9A.6 The Laplacian and the Bochner Formula). Those quadratic terms are exactly $-2(B_{ijkl} - B_{ijlk} - B_{iljk} + B_{ikjl})$ plus Ricci-times-curvature terms. This identity, the Laplacian of the curvature tensor, is proved in Topping's §2.4; it is pure Riemannian geometry, with no flow.
3. Lowering the index. $R_{ijkl}$ has its last index lowered with the changing metric $g(t)$, and $\partial_tg_{lm} = -2R_{lm}$ contributes Ricci-times-curvature terms.

Subtracting (2) from (1), the second derivatives of the Ricci tensor cancel, and what remains is the stated equation. The structural content is that the second derivatives of $\operatorname{Ric}$ assemble into a Laplacian of $\operatorname{Rm}$: the evolution of curvature is a heat equation, not merely a second-order equation, and this rests entirely on the second Bianchi identity.

3.3Uhlenbeck's trick

The Ricci-times-curvature terms in Theorem 3.1 are an artefact of measuring the curvature against a metric that is itself moving. Karen Uhlenbeck's suggestion, used by Hamilton in 1986, is to measure it instead in a frame that moves with the metric so as to stay orthonormal.

The evolution is linear in the frame and does not depend on which initial frame was chosen: if $F'_a = A_a^bF_b$ at $t = 0$ with constant coefficients, then the same relation holds for all $t$. So the construction defines a bundle map $\Phi_t : TM \to TM$, covering the identity of $M$, with $\Phi_0 = \operatorname{id}$ and

an evolving isometry from the fixed bundle $(TM, g(0))$ to the moving one $(TM, g(t))$. Pulling the curvature tensor back by $\Phi_t$ means writing it in the evolving frame: $R_{abcd} = R_{ijkl}F_a^iF_b^jF_c^kF_d^l$.

$\Phi_t$ is a map of bundles, not the differential of a diffeomorphism, so the pulled-back curvature is not constant even though the pulled-back metric is. Its evolution is what we want.

Check on the sphere. In a frame, $R_{abcd} = K(\delta_{ac}\delta_{bd} - \delta_{ad}\delta_{bc})$ and $B_{abcd} = K^2(\delta_{ac}\delta_{bd} + (n - 2)\delta_{ab}\delta_{cd})$, so the bracket is $(n - 1)K^2(\delta_{ac}\delta_{bd} - \delta_{ad}\delta_{bc})$ and $\partial_tR_{abcd} = 2(n - 1)K^2(\ldots) = 2(n - 1)K\,R_{abcd}$. In the frame, $R_{1212} = K(t)$ itself, and indeed $K(t) = \frac{K_0}{1 - 2(n - 1)K_0t}$ has $K' = 2(n - 1)K^2$. The sign has changed compared with the coordinate computation: in coordinates the components shrink with the metric; in the orthonormal frame the curvature grows.

Check on the cylinder. For $S^2(\rho)\times\mathbb{R}$, with $K = \frac{1}{\rho^2}$, only $R_{1212} = K$ and its symmetric images are nonzero. Then $B_{1212} = \sum_{e,f}R_{1e2f}R_{1e2f} = R_{1221}R_{1221} = K^2$... computing all four terms, $\partial_tR_{1212} = 2K^2$, and $\rho^2 = \rho_0^2 - 2t$ gives $K' = \frac{2}{\rho^4} = 2K^2$. The components with an index $3$ have zero right-hand side and stay zero: the line factor stays flat. (Exercise 3.9 fills in the computation.)

3.4The curvature operator

The quadratic terms simplify further if the curvature tensor is viewed as an operator on 2-forms. Choose an orthonormal basis $\varphi^1, \ldots, \varphi^N$ of $\Lambda^2$, $N = \frac{n(n - 1)}{2}$, where 2-forms are identified with antisymmetric matrices $\varphi_{ab}$ and the inner product is $\langle\varphi, \psi\rangle = \sum_{a,b}\varphi_{ab}\psi_{ab}$. (For instance $\varphi^{(pq)}_{ab} = \frac{1}{\sqrt2}(\delta_{pa}\delta_{qb} - \delta_{pb}\delta_{qa})$ for $p < q$.) Write

The symmetric matrix $\mathcal M = (\mathcal M_{\alpha\beta})$ is Hamilton's curvature operator. With this normalisation the round sphere of curvature $K$ has $\mathcal M = 2K\cdot\operatorname{id}$: indeed $\sum_\alpha\varphi^\alpha_{ab}\varphi^\alpha_{cd} = \frac12(\delta_{ac}\delta_{bd} - \delta_{ad}\delta_{bc})$, so $R_{abcd} = K(\delta_{ac}\delta_{bd} - \delta_{ad}\delta_{bc})$ needs $\mathcal M = 2K$. This is the factor $2$ relating Hamilton's $\mathcal M$ to the operator $\mathcal R$ of 9A.5 Computing Curvature; it is why, in dimension three, the eigenvalues of $\mathcal M$ are twice the sectional curvatures of the eigenplanes.

The space $\Lambda^2$ is the Lie algebra $\mathfrak{so}(n)$ of antisymmetric matrices, with the commutator bracket $[\varphi, \psi] = \varphi\psi - \psi\varphi$. Its structure constants in the basis $\varphi^\alpha$ are $c_{\alpha\beta\gamma} = \langle[\varphi^\alpha, \varphi^\beta], \varphi^\gamma\rangle$, and they are antisymmetric in all three indices. Define

$\mathcal M^2$ is the square of the operator; $\mathcal M^\#$ is the Lie algebra square, which uses the bracket.

Two remarks. The pieces $\mathcal M^2$ and $\mathcal M^\#$ separately need not satisfy the first Bianchi identity in dimension four and higher, while their sum does, as it must (it is the time derivative of a curvature operator minus a Laplacian of one). In dimension three every symmetric operator on $\Lambda^2$ comes from an algebraic curvature tensor, and the question does not arise. Second, the reaction term $\mathcal M^2 + \mathcal M^\#$ is intrinsic: it is defined by the inner product and Lie bracket on $\Lambda^2$, both of which are preserved by parallel transport. That is what makes Hamilton's maximum principle for systems (11A.7 Maximum Principles under Ricci Flow) applicable to sets defined by conditions on $\mathcal M$ that are invariant under rotations.

3.5Three dimensions

In dimension three, $\Lambda^2$ is three-dimensional and $\mathfrak{so}(3)$ is the Lie algebra of $\mathbb{R}^3$ with the cross product. With the basis $\varphi^1 = \varphi^{(23)}$, $\varphi^2 = \varphi^{(31)}$, $\varphi^3 = \varphi^{(12)}$, a computation of commutators gives $[\varphi^\alpha, \varphi^\beta] = \pm\frac{1}{\sqrt2}\varphi^\gamma$ for $(\alpha, \beta, \gamma)$ a permutation of $(1, 2, 3)$, that is $c_{\alpha\beta\gamma} = s\,\epsilon_{\alpha\beta\gamma}$ with $s^2 = \frac12$, $\epsilon$ the alternating symbol.

Since the reaction term of a diagonal $\mathcal M$ is diagonal, the ODE $\frac{d}{dt}\mathcal M = \mathcal M^2 + \mathcal M^\#$ preserves the eigenframe, and reduces to the system of 11A.2 How Curvature Evolves:

With $\lambda \geq \mu \geq \nu$ these are twice the sectional curvatures of the eigenplanes, $R = \lambda + \mu + \nu$, and the Ricci eigenvalues are $\frac{\mu + \nu}{2} \leq \frac{\lambda + \nu}{2} \leq \frac{\lambda + \mu}{2}$.

These four facts, with the maximum principle for systems (11A.7 Maximum Principles under Ricci Flow), give the preservation of nonnegative sectional and Ricci curvature in dimension three; the pinching estimates of 11A.10 Hamilton–Ivey Pinching and 11A.11 Hamilton’s 1982 Theorem are more refined invariant sets of the same ODE.

3.6History

Hamilton derived the evolution of $R_{ijkl}$ in his 1982 paper. Karen Uhlenbeck suggested the evolving frame, and Hamilton introduced the curvature operator form $\mathcal M^2 + \mathcal M^\#$ in "Four-manifolds with positive curvature operator" (1986), proving there that nonnegative curvature operator is preserved in all dimensions. The Lie algebra square has since been the starting point of every new preserved curvature condition, including Böhm and Wilking's (2008) and Brendle and Schoen's (2009).

3.7Exercises

Solution

$B_{jilk} = \sum R_{jpiq}R_{lpkq}$. Using $R_{jpiq} = R_{iqjp}$ (pair symmetry) and $R_{lpkq} = R_{kqlp}$, this is $\sum R_{iqjp}R_{kqlp}$, which is $B_{ijkl}$ after renaming $p \leftrightarrow q$. And $B_{klij} = \sum R_{kplq}R_{ipjq} = B_{ijkl}$ (swap the factors). On the round $S^3$, $B_{ijkl} = K^2(\delta_{ik}\delta_{jl} + \delta_{ij}\delta_{kl})$, so $B_{1111} = 2K^2 \neq 0$, while a curvature tensor has $R_{1111} = 0$; also $B_{1212} = 0$ but $B_{1122} = K^2$, while $-B_{2112} = 0$.

Solution

The nonzero components are $R_{1212} = R_{2121} = K$, $R_{1221} = R_{2112} = -K$. $B_{abcd} = \sum_{e,f}R_{aebf}R_{cedf}$ is nonzero only when $a, b, c, d \in \{1, 2\}$; for instance $B_{1111} = \sum R_{1e1f}R_{1e1f} = R_{1212}^2 = K^2$, $B_{1122} = \sum R_{1e1f}R_{2e2f} = R_{1212}R_{2121} = K^2$, $B_{1212} = \sum R_{1e2f}R_{1e2f} = R_{1221}^2 = K^2$, $B_{1221} = \sum R_{1e2f}R_{2e1f} = R_{1221}R_{2112} = K^2$, while $B_{1112} = 0$. Then for $(a, b, c, d) = (1, 2, 1, 2)$ the bracket is $B_{1212} + B_{1122} - B_{1221} - B_{1121} = K^2 + K^2 - K^2 - 0 = K^2$, so $\partial_tR_{1212} = 2K^2$, matching $K' = 2K^2$ from $\rho^2 = \rho_0^2 - 2t$. Any component with an index $3$ has every $B$ term zero, so it stays zero. For the operator: $R_{1212} = K$ means $\mathcal M_{33}\varphi^3_{12}\varphi^3_{12}\cdot(\text{four sign-matched terms}) = K$, i.e. $\mathcal M_{33}\cdot\frac12 = K$, $\mathcal M_{33} = 2K$; the others vanish. Then $\mathcal M^2 = \operatorname{diag}(0, 0, 4K^2)$ and $\mathcal M^\# = \operatorname{diag}(0\cdot2K, 0\cdot2K, 0\cdot0) = 0$, so $\mathcal M' = \operatorname{diag}(0, 0, 4K^2)$, that is $(2K)' = 4K^2$, again $K' = 2K^2$.

Solution

With $R_{abef} = -R_{aefb} - R_{afbe}$ and $R_{cdef} = -R_{cefd} - R_{cfde}$, the product is $R_{aefb}R_{cefd} + R_{aefb}R_{cfde} + R_{afbe}R_{cefd} + R_{afbe}R_{cfde}$. In the first, $R_{aefb}R_{cefd} = R_{aebf}R_{cedf}$ (one antisymmetry in each factor), whose sum over $e, f$ is $B_{abcd}$; the fourth, after swapping the names $e \leftrightarrow f$, is the same: $B_{abcd}$. In the second, $R_{aefb} = -R_{aebf}$ and $R_{cfde} = R_{decf}$ (pair symmetry), so its sum is $-\sum R_{aebf}R_{decf} = -B_{abdc}$, since $B_{abdc} = \sum_{e,f}R_{aebf}R_{decf}$. In the third, rename $e \leftrightarrow f$ to get $\sum R_{aebf}R_{cfed}$, and $R_{cfed} = R_{edcf} = -R_{decf}$, so it is also $-B_{abdc}$. So the total is $2B_{abcd} - 2B_{abdc}$. (A numerical check on random algebraic curvature tensors in dimensions three and four confirms this form of the identity; some printed sources have the indices of the second $B$ permuted.)

Solution

For $g(t) = (1 - 2\lambda t)g_0$, $\operatorname{Ric}(g(t)) = \lambda g_0 = \frac{\lambda}{1 - 2\lambda t}g(t)$, so as an endomorphism (index raised with $g(t)$) $\operatorname{Ric} = \frac{\lambda}{1 - 2\lambda t}\operatorname{id}$. The ODE $\partial_tF = \frac{\lambda}{1 - 2\lambda t}F$ has solution $F(t) = (1 - 2\lambda t)^{-1/2}F(0)$. Then $g(t)(F_a, F_b) = (1 - 2\lambda t)\cdot(1 - 2\lambda t)^{-1}g_0(F_a, F_b) = \delta_{ab}$. A metric that shrinks ($\lambda > 0$) needs longer vectors to have unit length.

Solution

(a) On $\{\nu \geq 0\}$, $\mu \geq \nu \geq 0$ and $\lambda \geq 0$, so at a boundary point $\nu' = \lambda\mu \geq 0$; and the ordering is preserved, so $\nu$ stays the smallest eigenvalue. On $\{\mu + \nu \geq 0\}$, at a boundary point $(\mu + \nu)' = \mu^2 + \nu^2 \geq 0$. To conclude invariance rigorously one perturbs: the sets $\{\nu \geq -\varepsilon e^{Ct}\}$ are entered strictly, and $\varepsilon \to 0$. (b) $\nu(\mathcal M) = \min_{|v| = 1}\langle\mathcal Mv, v\rangle$ is a minimum of linear functions, hence concave, and $\{\nu \geq 0\}$ is a superlevel set of a concave function, convex; it is a cone because $\nu(c\mathcal M) = c\nu(\mathcal M)$ for $c \geq 0$. Similarly $\mu + \nu = \min_{\dim P = 2}\operatorname{tr}(\mathcal M|_P)$ is concave and homogeneous. (c) The sets are closed, convex, invariant under parallel transport (they depend only on eigenvalues), and preserved by the ODE; Hamilton's maximum principle for systems keeps $\mathcal M(x, t)$ in them if it starts there.