A tutorial on maximum principles

SCALAR MAXIMUM PRINCIPLE We assume:

• M^{n} with g (t) metrics

• X (t) vector fields

• v : M^{n} × [0, T) → ℝ satisfies

Theorem 1 (Scalar MP) Suppose v (x, 0) ≥ 0 for all x ∈ M^{n} and

\frac{∂v}{∂t} ≥ Δ_{g(t)}v + ⟨X, ∇v⟩

at any (x, t) ∈ M^{n}×[0, T) such that v (x, t) < 0. Then v (x, t) ≥ 0 for all x ∈ M^{n} and t ∈ [0, T).

Proof. First and second derivative tests in calculus: If u : M^{n} × [0, T) → ℝ and (x_{0}, t_{0}) is a point and time where u attains its minimum among all points and earlier times, namely

u (x_{0}, t_{0}) = min_{M^{n}×[0,t_{0}]} u,

then

\frac{∂u}{∂t} (x_{0}, t_{0}) ≤ 0,
Δu (x_{0}, t_{0}) ≥ 0,
∇u (x_{0}, t_{0}) = 0.

Let

u (x, t) ≑ v (x, t) + εt + ε,

where ε is a positive number. Note that u ≥ ε > 0 at t = 0. We have

\frac{∂u}{∂t} ≥ Δu + ⟨X, ∇u⟩ + ε  (2)

at any point where u ≤ 0 (actually where v < 0.) The theorem follows from

Claim: u (x, t) > 0 for all x, t.

Proof of claim: Suppose that u ≤ 0 at some (x_{1}, t_{1}) ∈ M^{n} × [0, T). Then since M^{n} is compact and u > 0 at t = 0, there is a first time t_{0} ∈ (0, t_{1}] such that there exists a point x_{0} ∈ M^{n} such that u (x_{0}, t_{0}) = 0.

Now (2) implies that at (x_{0}, t_{0})

0 ≥ \frac{∂u}{∂t} ≥ Δu + ⟨X, ∇u⟩ + ε ≥ ε > 0,

a contradiction.

MAXIMUM PRINCIPLE FOR SYSTEMS

For a solution g (t) to the RF the curvature Rm : ∧^{2}M → ∧^{2}M

satisfies a PDE:

\frac{∂}{∂t}Rm = ΔRm + Q (Rm)

where

Q ( \begin{matrix}a \\ & b \\ & & c \end{matrix} ) = ( \begin{matrix}a^{2} + bc \\ & b^{2} + ac \\ & & c^{2} + ab \end{matrix} ) .

Consider the ODE:

\frac{d}{dt}M = Q (M) ,

that is,

\frac{d}{dt}a = a^{2} + bc
\frac{d}{dt}b = b^{2} + ac
\frac{d}{dt}c = c^{2} + ab.

Remarks:

• At a point ∧^{2}M ≅ so (3) since the components of 2-forms are skew-symmetric 3 × 3 matrices. We diagonalize to get ℝ^{3}.

• Rm is a section of the vector bundle E = ∧^{2}M ⊗_{S} ∧^{2}M.

• ∧^{2}M ≅ ∧^{1}M so E ≅ ∧^{1}M ⊗_{S} ∧^{1}M the bundle of symmetric 2-tensors.

• The ODE \frac{d}{dt}M = Q (M) is defined on each fiber E_{x}.

• We say that the PDE preserves a subset K ⊂ E if Rm (0) ⊂ K implies Rm (t) ⊂ K for all t ≥ 0.

• We say that the ODE preserves a subset K_{x} ⊂ E_{x} if M (0) ∈ K_{x} impliesM(t) ∈ K_{x} for all t ≥ 0.

Theorem 2 (Systems MP) Let K ⊂ E be a set such that K is invariant under parallel translation and K_{x} = K ∩ E_{x} is convex for each x. If the ODE preserves K_{x} for each x, then the PDE preserves K.

Example: Let K = {M : M ≥ 0} (nonnegative operators). Rm ∈ K iff Rm ≥ 0. The ODE preserves K_{x} = {M : M ≥ 0} (nonnegative matrices) since if M ≥ 0, then \frac{d}{dt}M ≥ 0. Hence by the MP,

Rm ≥ 0 is preserved under the RF.

STRONG MAX. PRIN. (SYSTEMS)

Theorem 3 (Strong MP) Suppose g (t) is a solution to RF on U ×(α, ω) with Rm (t) ≥ 0. Then either:

1. Rm (t) > 0 everywhere,

2. Rm (0) = 0 everywhere, or

3. locally U is isometric to Σ^{2} ×ℝ the product metric of a surface with positive curvature with the line.

This result applies to limits of dilations.

IVEY-HAMILTON PINCHING EST.

Assume a ≥ b ≥ c. To motivate we study the ODE under the assumption a > b = c < 0. Let z = −c > 0 so that z + a > 0. We then have the system of two ODEs:

\frac{da}{dt} = a^{2} + z^{2}
\frac{dz}{dt} = az − z^{2}.

• z (0) ≤ 1 (assumption).

• Since \frac{da}{dt} ≥ 0 we have a ≥ −C is preserved under the ODE.

• z > 0 and z + a > 0 are preserved under the ODE.

We have

\frac{dz}{da} = \frac{az − z^{2}}{a^{2} + z^{2}}

which is a homogeneous equation whose solution is given by:

a = z log (\frac{(z + a)^{2}}{z}) − Cz.

If z (t) is sufficiently large and positive, this leads to a contradiction unless

a (t) > z (t) log z (t) .

In particular, a is positive and much larger than z. On the other hand, if z is sufficiently small and positive, we get a contradiction unless −z < a < 0.

Theorem 4 Let (M^{3}, g (t)) be any solution of the RF on a closed 3-manifold for 0 ≤ t < T. If min_{x∈M^{3}} c (x, 0) ≥ −1, then at any point (x, t) ∈ M^{3} × [0, T) where c (x, t) < 0,

R ≥ |c| (log |c| + log (1 + t) − 3) .

Corollary 5 At points where c < 0 R ≥ |c| (log |c| − 3)

•

Rm ≥ −ϕ (R) R − C

where ϕ (R) → 0 as R → ∞. That is, |c| ≤ ϕ (R) R + C. (Since if |c| ≥ e^{6}, then R ≥ \frac{1}{2} |c| log |c| .)

• To control |Rm| in dimension 3, it suffices to bound R from above. I.e., if R ≤ C, then |Rm| ≤ C. (Since R ≤ C implies |c| ≤ C which implies |Rm| ≤ C.)

Sketch of Proof. Let K (t) ⊂ E be the set of all M such that tr M ≥ −3 (1 + t) and tr M ≥ |c (M)| (log |c (M)| + log (1 + t) − 3) if c (M) ≤ −1/ (1 + t) . Then K satisfies all of the hypotheses of the MP and Rm (0) ⊂ K (0) . By analyzing the system of ODEs one can show that K_{x} (t) is preserved by the ODE. Hence by the MP,

Rm (t) ⊂ K (t) for all t ≥ 0. That is

R ≥ −3 (1 + t)

and

R ≥ |c| (log |c| + log (1 + t) − 3)

if c ≤ −1/ (1 + t) . On the other hand, if 0 > c > −1/ (1 + t) , then

\frac{R}{|c|} − log |c| ≥ \frac{\frac{−3}{1+t}}{\frac{1}{1+t}} − log (\frac{1}{1 + t})
= log (1 + t) − 3
≥ |c| (log |c| + log (1 + t) − 3) .