Book 11A

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Course 11Book 11A: The Ricci Flow: Existence and Maximum PrinciplesChapter 8

The Strong Maximum Principle and Splitting

Why a zero of nonnegative curvature spreads everywhere: flatness, local splittings into a surface times a line, and why cones cannot occur.

27 min read · Updated Oct 23, 2026

Read with Morgan and Tian's Ricci Flow and the Poincaré Conjecture, chapter 4, §4 (the strong maximum principle for curvature: Theorem 4.18, Corollaries 4.19–4.20 and Proposition 4.22), which this chapter follows, and with Hamilton's "Four-manifolds with positive curvature operator" (1986) for the general null-space theorem. The strong maximum principle for scalar heat equations is in 6A.5 Maximum Principles.

In this chapter · 7 sections
  1. 8.1The strong maximum principle for scalars
  2. 8.2Flatness from a single zero
  3. 8.3Splitting from a zero eigenvalue of Ricci
  4. 8.4Cones are excluded
  5. 8.5All dimensions, and compact manifolds
  6. 8.6History
  7. 8.7Exercises
  8. ↗Free resources

The maximum principles of 11A.7 Maximum Principles under Ricci Flow say that nonnegative curvature is preserved. The strong maximum principle says much more: if the curvature is nonnegative and vanishes somewhere, in some direction, at a positive time, then it vanishes in that direction everywhere, and the manifold splits off a flat factor. Positivity spreads instantly through a connected manifold under a heat equation; so does the structure of the zeros.

This is one of the main tools in Perelman's analysis of κ\kappa-solutions (12B.1 κ-Solutions, 12B.3 The Structure of κ-Solutions), and Morgan and Tian use it in the form proved here at least a dozen times. A three-dimensional κ\kappa-solution has nonnegative curvature. If its curvature operator has a zero eigenvalue anywhere, the strong maximum principle makes it split, locally a surface times a line, and globally (after passing to a cover) a product N2×RN^2\times\mathbb{R}; and then the two-dimensional theory shows that NN is a round sphere and the solution is a neck. If not, its curvature is positive, and the soul theorem makes it a ball or compact. The other application is negative: a nonflat cone cannot occur as a time slice of a flow with nonnegative curvature, which is how blow-down limits of κ\kappa-solutions are shown to be lower-dimensional.

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

  • state the strong maximum principle for the heat equation with Dirichlet boundary values, and the comparison trick that transfers it to curvature;
  • prove that a three-dimensional Ricci flow with nonnegative sectional curvature and R=0R = 0 at one point of a later time is flat;
  • prove that if, in addition, the Ricci curvature has a zero eigenvalue somewhere, the flow splits locally as a surface times a line, by showing that the null direction is parallel and constant in time;
  • describe the global versions on covers, and the four possible covers;
  • prove that a time slice of a flow with nonnegative curvature cannot contain an open piece of a nonflat cone;
  • state Hamilton's null-space theorem in all dimensions and the classification of compact 3-manifolds with nonnegative Ricci curvature.

The strong maximum principle for scalars

Theorem 8.1 The strong maximum principle for the heat equation (statement)

Let UU be a compact connected manifold with boundary and h(x,t)h(x, t), 0≤t≤T0 \leq t \leq T, a solution of ∂th=Δg(t)h\partial_th = \Delta_{g(t)}h (for a smooth family of metrics) with h=0h = 0 on ∂U×[0,T]\partial U\times[0, T] and h(⋅,0)≥0h(\cdot, 0) \geq 0. Then h≥0h \geq 0 on U×[0,T]U\times[0, T], and if h(y,0)>0h(y, 0) > 0 at some point yy, then h>0h > 0 on int⁡(U)×(0,T]\operatorname{int}(U)\times(0, T].

The first statement is the weak maximum principle. The second is the parabolic strong maximum principle of 6A.5 Maximum Principles (Nirenberg's), proved there with Hopf's barrier argument; equivalently, the Dirichlet heat kernel of UU is strictly positive in the interior for positive times. Heat started at one point is felt everywhere immediately.

The trick for transferring this to curvature is comparison with a Dirichlet solution. Suppose a quantity s≥0s \geq 0 built from the curvature is a supersolution of the heat equation, (∂t−Δ)s≥0(\partial_t - \Delta)s \geq 0, at least in a weak sense. Put underneath it a solution hh of the heat equation with Dirichlet boundary values and with 0≤h(⋅,0)≤s(⋅,0)0 \leq h(\cdot, 0) \leq s(\cdot, 0), positive somewhere. The weak maximum principle applied to s−hs - h gives s≥hs \geq h, and the strong maximum principle for hh gives h>0h > 0 in the interior at positive times. So s>0s > 0 there.

For a scalar such as RR this is immediate. For quantities like "the smallest eigenvalue of Rm⁡\operatorname{Rm}", which are not smooth and satisfy no exact equation, Morgan and Tian use the maximum principle for systems (11A.7 Maximum Principles under Ricci Flow) on the bundle V⊕RV\oplus\mathbb{R} with the convex set {(T,h):s(T)≥h≥0}\{(T, h) : s(T) \geq h \geq 0\}: the conclusion is again s≥hs \geq h (their Proposition 4.17). The conditions on ss are those that make this set convex and invariant: ss is concave on each fibre (so {s≥h}\{s \geq h\} is convex), invariant under parallel transport, and the reaction term M2+M#\mathcal M^2 + \mathcal M^\# does not decrease it at points where it is nonnegative.

Flatness from a single zero

Theorem 8.2 A zero of scalar curvature forces flatness

Let (U,g(t))(U, g(t)), 0≤t≤T0 \leq t \leq T, T>0T > 0, be a three-dimensional Ricci flow with nonnegative sectional curvature, UU connected but not necessarily complete. If R(p,T)=0R(p, T) = 0 for some p∈Up \in U, then (U,g(t))(U, g(t)) is flat for every t∈[0,T]t \in [0, T].

Proof. Nonnegative sectional curvature makes R≥0R \geq 0, with R=0R = 0 at a point exactly when all sectional curvatures vanish there (the scalar curvature is a sum of sectional curvatures). Suppose the flow is not flat at some time; then R(q,t1)>0R(q, t_1) > 0 for some qq and some t1t_1, and by continuity we may take t1<Tt_1 < T. Shift time so that t1=0t_1 = 0. Choose a compact connected domain V⊂UV \subset U with smooth boundary, containing pp and qq in its interior, and a smooth function h0h_0 on VV with 0≤h0≤R(⋅,0)0 \leq h_0 \leq R(\cdot, 0), h0=0h_0 = 0 on ∂V\partial V and h0(q)>0h_0(q) > 0. Let hh solve ∂th=Δg(t)h\partial_th = \Delta_{g(t)}h on VV with h=0h = 0 on ∂V\partial V and h(⋅,0)=h0h(\cdot, 0) = h_0. Then

(∂t−Δ)(R−h)=2∣Ric⁡∣2≥0,(\partial_t - \Delta)(R - h) = 2|\operatorname{Ric}|^2 \geq 0,

and R−h≥0R - h \geq 0 on the parabolic boundary (t=0t = 0, where h0≤Rh_0 \leq R, and ∂V\partial V, where h=0≤Rh = 0 \leq R). By the weak maximum principle R≥hR \geq h on V×[0,T]V\times[0, T], and by the strong one h(p,T)>0h(p, T) > 0. So R(p,T)>0R(p, T) > 0: a contradiction.

Note what the theorem does not need: completeness, or any control at infinity. It is a local statement in space, which is why it applies to limits of flows that are only known on regions, as in Morgan and Tian's chapters 10 and 11.

Splitting from a zero eigenvalue of Ricci

Theorem 8.3 Local splitting

Let (U,g(t))(U, g(t)), 0≤t≤T0 \leq t \leq T, T>0T > 0, be a three-dimensional Ricci flow with nonnegative sectional curvature, UU connected, and (U,g(0))(U, g(0)) not flat. If the Ricci curvature at some (p,T)(p, T) has a zero eigenvalue, then for every t∈(0,T]t \in (0, T], (U,g(t))(U, g(t)) splits locally as the Riemannian product of a surface of positive curvature and a line, and the flow is locally the product of a Ricci flow on the surface and the static flow on the line.

Proof. Step 1: positive scalar curvature. By Theorem 8.2, since the flow is not flat at time 00, R>0R > 0 on U×(0,T]U\times(0, T].

Step 2: the null space spreads. In Hamilton's normalisation, let λ≥μ≥ν≥0\lambda \geq \mu \geq \nu \geq 0 be the eigenvalues of M\mathcal M (11A.3 Curvature in a Moving Frame); the Ricci eigenvalues are μ+ν2≤λ+ν2≤λ+μ2\frac{\mu + \nu}{2} \leq \frac{\lambda + \nu}{2} \leq \frac{\lambda + \mu}{2}, so a zero Ricci eigenvalue means μ+ν=0\mu + \nu = 0, that is μ=ν=0\mu = \nu = 0. Let s=μ+νs = \mu + \nu, the sum of the two smallest eigenvalues. It is the minimum of the traces of M\mathcal M over 2-dimensional subspaces of Λ2\Lambda^2, so it is concave on each fibre and {s≥h}\{s \geq h\} is convex; it is invariant under parallel transport; and by 11A.3 Curvature in a Moving Frame the ODE does not decrease it where it is zero: (μ+ν)′=μ2+ν2+λ(μ+ν)(\mu + \nu)' = \mu^2 + \nu^2 + \lambda(\mu + \nu). The comparison trick, run as in the previous proof but with ss in place of RR and the maximum principle for systems in place of the scalar one, shows: if s(q,t1)>0s(q, t_1) > 0 for some (q,t1)(q, t_1) with t1<Tt_1 < T, then s(p,T)>0s(p, T) > 0. Since s(p,T)=0s(p, T) = 0, we conclude s≡0s \equiv 0 on U×[0,T)U\times[0, T) and, by continuity, on U×[0,T]U\times[0, T].

Step 3: the line field. So at every point and time, μ=ν=0\mu = \nu = 0 and, by Step 1, λ=R>0\lambda = R > 0 for t>0t > 0. The curvature operator has a 2-dimensional null space in Λ2TyU\Lambda^2T_yU and one positive eigenvalue λ\lambda, with eigenvector a 2-form e1∧e2e_1\wedge e_2 for some orthonormal e1,e2e_1, e_2. The planes on which the sectional curvature vanishes are exactly those containing the unit vector V=e3V = e_3 orthogonal to e1,e2e_1, e_2: the sectional curvatures are λ2\frac\lambda2 on e1∧e2e_1\wedge e_2 and 00 on e1∧e3e_1\wedge e_3, e2∧e3e_2\wedge e_3. So Rm⁡(V,⋅,⋅,⋅)=0\operatorname{Rm}(V, \cdot, \cdot, \cdot) = 0, and VV spans a smooth line field DD on U×(0,T]U\times(0, T] (smooth because λ\lambda is a simple eigenvalue and the null space is its orthogonal complement). Locally VV is a smooth unit vector field, defined up to sign.

Step 4: VV is parallel. Here the evolution equation does the work, as in Hamilton's original argument. Let ω=e1∧e2\omega = e_1\wedge e_2, a smooth unit section (up to sign) of the image of M\mathcal M, so that M=λ ω⊗ω\mathcal M = \lambda\,\omega\otimes\omega near any point, with λ>0\lambda > 0, and let σ\sigma be any smooth section of the null space ker⁡M\ker\mathcal M (which has constant rank 22, by Steps 2 and 3). Then ⟨Mσ,σ⟩≡0\langle\mathcal M\sigma, \sigma\rangle \equiv 0 in space and time. Differentiate twice in space: Mσ=0\mathcal M\sigma = 0 gives (∇iM)σ=−M∇iσ(\nabla_i\mathcal M)\sigma = -\mathcal M\nabla_i\sigma, and

0=Δ⟨Mσ,σ⟩=⟨(ΔM)σ,σ⟩+4⟨(∇iM)σ,∇iσ⟩+2⟨M∇iσ,∇iσ⟩=⟨(ΔM)σ,σ⟩−2⟨M∇iσ,∇iσ⟩.0 = \Delta\langle\mathcal M\sigma, \sigma\rangle = \langle(\Delta\mathcal M)\sigma, \sigma\rangle + 4\langle(\nabla_i\mathcal M)\sigma, \nabla_i\sigma\rangle + 2\langle\mathcal M\nabla_i\sigma, \nabla_i\sigma\rangle = \langle(\Delta\mathcal M)\sigma, \sigma\rangle - 2\langle\mathcal M\nabla_i\sigma, \nabla_i\sigma\rangle.

Differentiate once in time, in Uhlenbeck's frame (11A.3 Curvature in a Moving Frame): 0=∂t⟨Mσ,σ⟩=⟨(∂tM)σ,σ⟩+2⟨Mσ,∂tσ⟩=⟨(∂tM)σ,σ⟩0 = \partial_t\langle\mathcal M\sigma, \sigma\rangle = \langle(\partial_t\mathcal M)\sigma, \sigma\rangle + 2\langle\mathcal M\sigma, \partial_t\sigma\rangle = \langle(\partial_t\mathcal M)\sigma, \sigma\rangle. Now insert the evolution equation ∂tM=ΔM+M2+M#\partial_t\mathcal M = \Delta\mathcal M + \mathcal M^2 + \mathcal M^\#. Since M2σ=0\mathcal M^2\sigma = 0, and since in dimension three M#\mathcal M^\# is the adjugate (11A.3 Curvature in a Moving Frame), which for M=diag⁡(λ,0,0)\mathcal M = \operatorname{diag}(\lambda, 0, 0) is 00,

0=⟨(ΔM)σ,σ⟩+0+0=2⟨M∇iσ,∇iσ⟩.0 = \langle(\Delta\mathcal M)\sigma, \sigma\rangle + 0 + 0 = 2\langle\mathcal M\nabla_i\sigma, \nabla_i\sigma\rangle.

As M≥0\mathcal M \geq 0, this forces M∇iσ=0\mathcal M\nabla_i\sigma = 0: the derivative of a section of the null space stays in the null space. So ker⁡M\ker\mathcal M is a parallel subbundle of Λ2\Lambda^2, its orthogonal complement, spanned by ω\omega, is parallel too, and the unit section ω\omega is parallel (its derivative is orthogonal to it, because ∣ω∣=1|\omega| = 1, and lies in its span). Finally VV is the unit vector with ω=⋆V\omega = \star V, and the Hodge star commutes with covariant differentiation, so ∇V=0\nabla V = 0.

Step 5: VV is constant in time, and the flow splits. A parallel unit vector field splits the metric locally as a product Σ×R\Sigma\times\mathbb{R}, with VV the R\mathbb{R}-direction (de Rham's local splitting: the distribution V⊥V^\perp is integrable and totally geodesic because VV is parallel, and the flow of VV is by local isometries, as in the splitting theorem of 9B.5 Splitting and Soul Theorems). On a product the curvature is the pullback of the curvature of the surface factor, and Ric⁡(V,⋅)=0\operatorname{Ric}(V, \cdot) = 0. So ∂tg(V,⋅)=−2Ric⁡(V,⋅)=0\partial_tg(V, \cdot) = -2\operatorname{Ric}(V, \cdot) = 0: the metric does not change in the VV-direction or across it, and the evolution of Rm⁡\operatorname{Rm}, being the pullback of the evolution on Σ\Sigma, keeps VV in the null space. Hence the line field is independent of tt, and the flow is locally the product of a Ricci flow on Σ\Sigma, of curvature λ2>0\frac\lambda2 > 0, with the static line.

Step 4 is the heart of the matter. The null space persists in time (Step 2), so the quantity ⟨Mσ,σ⟩\langle\mathcal M\sigma, \sigma\rangle is identically zero; differentiating it once in time and twice in space, and comparing through the evolution equation, leaves a sum of two nonnegative terms that must vanish, 2⟨M∇σ,∇σ⟩2\langle\mathcal M\nabla\sigma, \nabla\sigma\rangle and ⟨M#σ,σ⟩\langle\mathcal M^\#\sigma, \sigma\rangle. The first says the null space cannot turn. Algebra alone (the curvature is nonnegative and vanishes on some planes) gives only first-order information and does not suffice; the parabolic equation supplies the second-order information.

The global version. On a complete manifold the local splitting globalises after passing to a cover.

Corollary 8.4 Global splitting (statement)

Let (M,g(t))(M, g(t)), 0≤t≤T0 \leq t \leq T, T>0T > 0, be a Ricci flow of complete connected 3-manifolds with Rm⁡≥0\operatorname{Rm} \geq 0, with (M,g(0))(M, g(0)) not flat, and suppose Rm⁡(x,T)\operatorname{Rm}(x, T) has a zero eigenvalue for some xx. Then a cover M~\tilde M of MM splits, with its pulled-back flow, as (N,h(t))×(R,ds2)(N, h(t))\times(\mathbb{R}, ds^2), with (N,h(t))(N, h(t)) a Ricci flow of surfaces of positive curvature.

The cover is needed because the line field may not be orientable, or its orthogonal surfaces may close up. Morgan and Tian list the possibilities (their Remark 4.21): the cover can be trivial, an infinite cyclic cover, a double cover, or a cover with group the infinite dihedral group Z/2∗Z/2\mathbb{Z}/2 * \mathbb{Z}/2. In the first two cases there is a global parallel unit vector field spanning the null direction; in the last two only a nonorientable line field. Examples: S2×RS^2\times\mathbb{R} itself (trivial cover); S2×S1S^2\times S^1 (covered by S2×RS^2\times\mathbb{R} with group Z\mathbb{Z}); the cross-cap quotient (S2×R)/ι(S^2\times\mathbb{R})/\iota of 10A.6 Necks, Caps and Their Topology (double cover); and RP3#RP3\mathbb{RP}^3\#\mathbb{RP}^3 with a product-like metric (group Z/2∗Z/2\mathbb{Z}/2 * \mathbb{Z}/2).

Cones are excluded

The open cone over a Riemannian surface (N,h)(N, h) is N×(0,∞)N\times(0, \infty) with ds2+s2hds^2 + s^2h. Its curvature is computed from the warped product formulas of 9A.5 Computing Curvature with warping function φ(s)=s\varphi(s) = s: planes containing ∂s\partial_s have curvature −φ′′φ=0-\frac{\varphi''}{\varphi} = 0, and the plane tangent to N×{s}N\times\{s\} has curvature Kh−φ′2φ2=Kh−1s2\frac{K_h - \varphi'^2}{\varphi^2} = \frac{K_h - 1}{s^2}. So the curvature operator of a cone has a 2-dimensional null space at every point, and its third eigenvalue 2(Kh−1)s2\frac{2(K_h - 1)}{s^2} scales like s−2s^{-2} along the rays. The cone is flat exactly when Kh≡1K_h \equiv 1, that is when (N,h)(N, h) is locally the unit sphere and the cone is locally R3\mathbb{R}^3.

Proposition 8.5 No cones in nonnegatively curved flows

Let (U,g(t))(U, g(t)), 0≤t≤T0 \leq t \leq T, T>0T > 0, be a three-dimensional Ricci flow with nonnegative sectional curvature. If (U,g(T))(U, g(T)) is isometric to an open subset of a cone, then (U,g(t))(U, g(t)) is flat for every t∈[0,T]t \in [0, T].

Proof. If (U,g(T))(U, g(T)) is flat, Theorem 8.2 (applied at any point, where R=0R = 0) makes every (U,g(t))(U, g(t)) flat. Suppose (U,g(T))(U, g(T)) is not flat. Shrinking UU, we may assume the nonzero eigenvalue 2(Kh−1)s2\frac{2(K_h - 1)}{s^2} never vanishes on UU. At every point the curvature has a 2-dimensional null space, so the Ricci curvature has a zero eigenvalue, and by Theorem 8.3 (U,g(T))(U, g(T)) splits locally as a surface of positive curvature times a line, with the line in the null direction. In the cone, the null direction at each point is the radial direction ∂s\partial_s (it is the vector VV with Rm⁡(V,⋅,⋅,⋅)=0\operatorname{Rm}(V, \cdot, \cdot, \cdot) = 0). So the lines of the splitting are the rays of the cone. Along a line of a product Σ×R\Sigma\times\mathbb{R}, the curvature is constant (it is the curvature of Σ\Sigma at a fixed point). But along a ray of the cone the nonzero eigenvalue is proportional to s−2s^{-2}, which is not constant: a contradiction.

This proposition is what makes asymptotic cones of κ\kappa-solutions useful. Rescaling a nonnegatively curved κ\kappa-solution down (a blow-down) produces, in the limit, a cone (9B.5 Splitting and Soul Theorems); if the limit were a smooth flow with nonnegative curvature through a nonflat cone, the proposition would force flatness. So the blow-down is either flat or singular, and in three dimensions Perelman uses this to show that the asymptotic volume of a κ\kappa-solution vanishes (12B.3 The Structure of κ-Solutions; Morgan and Tian, chapter 9, §6).

All dimensions, and compact manifolds

Hamilton proved the strong maximum principle for the full curvature operator in every dimension (1986; statement here).

Theorem 8.6 Hamilton's null-space theorem (statement)

Let g(t)g(t), 0≤t≤T0 \leq t \leq T, be a Ricci flow with M≥0\mathcal M \geq 0 on a connected manifold. For t∈(0,T]t \in (0, T], the null space of M\mathcal M has constant rank and is invariant under parallel transport, and the image of M\mathcal M is a Lie subalgebra of Λ2≅so(n)\Lambda^2 \cong \mathfrak{so}(n), invariant under parallel transport; by the Ambrose–Singer theorem it is the Lie algebra of the restricted holonomy group.

The proof is the argument of Step 4 above in general dimension: on the null space, 2⟨M∇iσ,∇iσ⟩+⟨M#σ,σ⟩=02\langle\mathcal M\nabla_i\sigma, \nabla_i\sigma\rangle + \langle\mathcal M^\#\sigma, \sigma\rangle = 0, both terms are nonnegative, and the vanishing of the second, which is built from the Lie bracket, is what makes the image closed under brackets. In dimension three the possible null spaces are 00 (positive curvature), a 2-dimensional subspace {V∧W}\{V\wedge W\} (the splitting above), or all of Λ2\Lambda^2 (flat).

Combining with Hamilton's 1982 theorem gives the classification of compact 3-manifolds with nonnegative Ricci curvature (Hamilton 1986; statement): if MM is closed and has a metric with Ric⁡≥0\operatorname{Ric} \geq 0, then under the Ricci flow either Ric⁡\operatorname{Ric} becomes positive at once, and MM is a spherical space form (11A.11 Hamilton’s 1982 Theorem); or the flow splits locally, and MM is a quotient of S2×RS^2\times\mathbb{R}; or Ric⁡≡0\operatorname{Ric} \equiv 0, which in three dimensions means flat, and MM is a quotient of R3\mathbb{R}^3. The middle case uses the analogue of Theorem 8.3 for Ricci curvature, whose proof is the same with s=μ+νs = \mu + \nu.

Where this goes Where the strong maximum principle is used

12B.1 κ-Solutions and 12B.3 The Structure of κ-Solutions: a κ\kappa-solution whose curvature operator has a zero eigenvalue is a quotient of the round cylinder; asymptotic cones of κ\kappa-solutions are not smooth nonflat cones; the asymptotic volume vanishes. Morgan and Tian also use Theorem 8.2 and Theorem 8.3 on incomplete limits in their chapters 10 and 11, where completeness is not available.

History

Hamilton proved the strong maximum principle for the curvature operator, and the classification of compact 3-manifolds with nonnegative Ricci curvature and 4-manifolds with nonnegative curvature operator, in "Four-manifolds with positive curvature operator" (1986). Louis Nirenberg proved the strong maximum principle for parabolic equations in 1953. Perelman used the splitting consequences throughout his study of κ\kappa-solutions (2002), and Morgan and Tian's chapter 4, §4 gives the local versions used here.

Recall Where we stand

Heat started at one point is felt everywhere at once: a Dirichlet solution with nonnegative, not identically zero initial data is positive in the interior. Comparing a curvature quantity with such a solution transfers this to curvature. So in a three-dimensional flow with K≥0K \geq 0, R=0R = 0 at one point of a later time forces flatness; and a zero Ricci eigenvalue at one point forces μ=ν=0\mu = \nu = 0 everywhere, a parallel null vector field (its derivative vanishes because nonnegative curvature has a minimum on its planes), and a local splitting Σ×R\Sigma\times\mathbb{R} preserved by the flow, global on a cover. Cones have curvature scaling like s−2s^{-2} along rays, incompatible with splitting, so nonflat cones never occur in such flows. Hamilton's null-space theorem gives the analogue in all dimensions. 11A.10 Hamilton–Ivey Pinching proves Hamilton–Ivey pinching.

Exercises

Exercise 8.7 Positivity of heat

On the interval U=[0,π]U = [0, \pi] with the flat metric, solve the heat equation with Dirichlet conditions and initial data h0h_0 a small nonnegative bump near x=1x = 1, by Fourier series, and explain why h(x,t)>0h(x, t) > 0 for every x∈(0,π)x \in (0, \pi) and t>0t > 0, even far from the bump.

Solution

h(x,t)=∑k≥1bke−k2tsin⁡kxh(x, t) = \sum_{k \geq 1}b_ke^{-k^2t}\sin kx with bk=2π∫0πh0sin⁡kx dxb_k = \frac2\pi\int_0^\pi h_0\sin kx\,dx. Positivity for all t>0t > 0 follows from the representation h(x,t)=∫0πpt(x,y)h0(y) dyh(x, t) = \int_0^\pi p_t(x, y)h_0(y)\,dy with the Dirichlet heat kernel pt(x,y)>0p_t(x, y) > 0 for x,y∈(0,π)x, y \in (0, \pi) and t>0t > 0 (a consequence of the strong maximum principle, or of the method of images, which writes ptp_t as an alternating sum of Gaussians that is positive inside the interval). Since h0≥0h_0 \geq 0 and is positive on a set of positive measure, h(x,t)>0h(x, t) > 0 everywhere in the interior. For large tt, h≈b1e−tsin⁡xh \approx b_1e^{-t}\sin x with b1>0b_1 > 0: positive everywhere, and with the shape of the first eigenfunction.

Exercise 8.8 No partial flattening

Explain why a three-dimensional Ricci flow cannot start from a metric that is round near one point and flat near another, with K≥0K \geq 0 everywhere, and remain flat near the second point for a positive time.

Solution

If the flow stayed flat near the second point pp up to some time T>0T > 0, then R(p,T)=0R(p, T) = 0. The flow has nonnegative sectional curvature for all time (preserved, 11A.7 Maximum Principles under Ricci Flow), so Theorem 8.2 makes it flat everywhere on [0,T][0, T], including at t=0t = 0 near the first point, where it is round: a contradiction. The positive curvature near the first point immediately makes R>0R > 0 everywhere.

Exercise 8.9 The curvature of a cone

For the cone ds2+s2hds^2 + s^2h over a surface (N,h)(N, h) of Gauss curvature KhK_h, verify with the warped product formulas that the sectional curvatures are 00 on planes containing ∂s\partial_s and Kh−1s2\frac{K_h - 1}{s^2} on planes tangent to N×{s}N\times\{s\}. For NN a round sphere of radius a<1a < 1, show the cone has positive curvature, and describe it near its vertex.

Solution

For dr2+φ(r)2hdr^2 + \varphi(r)^2h, 9A.5 Computing Curvature gives K(∂r,X)=−φ′′φK(\partial_r, X) = -\frac{\varphi''}{\varphi} and K(X,Y)=Kh−φ′2φ2K(X, Y) = \frac{K_h - \varphi'^2}{\varphi^2} for X,YX, Y tangent to NN. With φ=s\varphi = s: φ′′=0\varphi'' = 0 and φ′=1\varphi' = 1, giving 00 and Kh−1s2\frac{K_h - 1}{s^2}. A round sphere of radius aa has Kh=1a2>1K_h = \frac{1}{a^2} > 1 when a<1a < 1, so the tangential curvature a−2−1s2\frac{a^{-2} - 1}{s^2} is positive and blows up at the vertex s=0s = 0, where the cone is singular: it is a cone of total angle less than that of R3\mathbb{R}^3, like a sharpened pencil tip in one dimension higher. By Proposition 8.5 no smooth flow with nonnegative curvature can pass through a time slice containing such a piece, away from the vertex.

Exercise 8.10 Why the null vector is parallel

In Step 4 of the proof of Theorem 8.3, the key claim is that if f≥0f \geq 0 is a smooth function along a curve with f(0)=0f(0) = 0, then f′(0)=0f'(0) = 0. Use this to show: if A(τ)A(\tau) is a smooth family of nonnegative symmetric matrices with A(0)v=0A(0)v = 0, then A′(0)vA'(0)v is orthogonal to vv, in the sense that ⟨A′(0)v,v⟩=0\langle A'(0)v, v\rangle = 0; and that A′(0)v∈im⁡A(0)A'(0)v \in \operatorname{im}A(0). (Hint for the last part: apply the first part to v+cwv + cw for w∈ker⁡A(0)w \in \ker A(0) and every real cc.)

Solution

f(τ)=⟨A(τ)v,v⟩≥0f(\tau) = \langle A(\tau)v, v\rangle \geq 0 with f(0)=0f(0) = 0, so f′(0)=⟨A′(0)v,v⟩=0f'(0) = \langle A'(0)v, v\rangle = 0. For the last part, let w∈ker⁡A(0)w \in \ker A(0) and apply the first part to u=v+cwu = v + cw, which is in ker⁡A(0)\ker A(0) for every real cc: 0=⟨A′(0)u,u⟩=⟨A′(0)v,v⟩+2c⟨A′(0)v,w⟩+c2⟨A′(0)w,w⟩0 = \langle A'(0)u, u\rangle = \langle A'(0)v, v\rangle + 2c\langle A'(0)v, w\rangle + c^2\langle A'(0)w, w\rangle for all cc, so all three coefficients vanish, in particular ⟨A′(0)v,w⟩=0\langle A'(0)v, w\rangle = 0. So A′(0)vA'(0)v is orthogonal to ker⁡A(0)\ker A(0), which for a symmetric matrix means A′(0)v∈im⁡A(0)A'(0)v \in \operatorname{im}A(0). This kind of first-order information is what a minimum provides; in the proof of Theorem 8.3 the second-order information from the evolution equation is what forces the null space to be parallel.

Exercise 8.11 Rehearsal: the neck as the splitting case

Let (M,g(t))(M, g(t)), t∈(−∞,0]t \in (-\infty, 0], be a complete ancient solution of the three-dimensional Ricci flow with bounded nonnegative curvature operator, not flat, and suppose the curvature operator has a zero eigenvalue at some point. Using Corollary 8.4, show that a cover of MM is (N,h(t))×R(N, h(t))\times\mathbb{R} with (N,h(t))(N, h(t)) an ancient solution on a surface with positive curvature. Using the classification of compact ancient solutions on S2S^2 (Daskalopoulos–Hamilton–Šešum, 11A.12 Ricci Flow on Surfaces), explain why, if NN is compact and the flow is κ\kappa-noncollapsed, the cover is the round shrinking cylinder. What input from Perelman's theory handles a noncompact NN?

Solution

By the global splitting, a cover is (N,h(t))×R(N, h(t))\times\mathbb{R} with (N,h(t))(N, h(t)) an ancient solution of the Ricci flow on a surface, with positive curvature (and complete, since MM is). If NN is compact it is S2S^2 (or RP2\mathbb{RP}^2, covered by S2S^2), and compact ancient solutions on S2S^2 are the round shrinking spheres or the King–Rosenau sausages. The sausage is collapsed in the far past (it looks like two cigars), so if one knows that the factor is κ\kappa-noncollapsed it is round, and the cover is the round shrinking cylinder. If NN were noncompact, it would be a complete ancient surface solution with positive curvature, and the cigar is the model; it is collapsed at large scales. Perelman's κ\kappa-noncollapsing (and his classification of two-dimensional κ\kappa-solutions as round spheres, 12B.1 κ-Solutions) is the extra input: it excludes the sausage and the cigar, leaving the round cylinder and its quotients.

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