Frank Morgan

Jian Ge’s recent ArXiv post on “Comparison theorems for manifolds with mean convex boundary,” Theorem 0.1, has a generalization to manifolds with density, here within a factor of 2 of sharp for constant density:

Proposition. Consider a smooth complete connected n-dimensional Riemannian manifold with nonempty boundary with smooth (positive) density e^ψ. Suppose that the (n-1) Bakry Emery Ricci curvature

Ric –  Hess ψ – (1/(n-1))dψ@dψ

is at least (n-1)δ and that the generalized (inward) mean curvature (sum of principal curvatures) H = H₀ – ∂ψ/∂n of the boundary satisfies

H/(n-1) ≥ h,

with h > √(2|δ|) if δ ≤ 0. Then the distance to the boundary is at most

 2/h                                       if δ = 0,

(√(2/δ))arccot(h/√2δ)          if δ > 0,

 (√(2/|δ|))arccoth(h/√2|δ|)   if δ < 0.

Proof (cf. Chapter 18 of my Geometric Measure Theory book, 4th ed.). Let dA = e^fdA₀ be the element of weighted surface area of the boundary as it flows inward. Then f′ = –H and

f′′ = -II² + ∂²f/∂n² – Ric ≤ –H₀²/(n-1) + ∂²ψ/∂n² – Ric.

 Note that H² ≤ 2H₀² + 2(∂ψ/∂n)²  [with strict inequality unless |H₀| = |∂ψ/∂n|], so that

f′′  = –H²/2(n-1) + (∂ψ/∂n)²/(n-1) + ∂²ψ/∂n² – Ric ≤ -f′²/2(n-1) – (n-1)δ.

Let a = √(1/2(n-1)), b = √((n-1)|δ|). Since by hypothesis initially af′ < –b when δ ≤ 0, by the lemma below f′  = –H goes to -∞ and e^f goes to 0 as asserted.

Example. For small ε > 0, consider a disk of radius 2 about a point O, with Gaussian curvature 1 for 0 ≤ r < ε, Gaussian curvature -1/2 for ε < r < 2ε, and flat for 2ε < r ≤ 2. Smooth or relax the smoothness hypotheses slightly. Note that Ge’s Theorem 0.1 does not apply, because the curvature of the boundary, 1/2, is not greater than 1. Consider a function ψ with ψ(O) = 0 and

 ψ′(r) = .9r           for 0 < r < ε ,

ψ′(r)  = -.9r         for ε  < r < 2ε ,

ψ′(r) = 0              for 2ε < r < 2.

 For 0 < r < ε , Ric – Hess ψ – dψ@dψ ~ 1 – .9 > 0. For ε  < r < 2ε , Ric – Hess ψ – dψ@dψ ~ -1/2 + .9 > 0. For 2ε < r < 2, Ric – Hess ψ – dψ@dψ = 0. Therefore the Proposition applies and implies that the distance to the boundary is always less than 2/(1/2) = 4, twice as large as the sharp value of 2.

Remark. There is no such proposition for density alone. The hypotheses

 Hess ψ + (1/N))dψ@dψ ≤ –Nδ

 and on the boundary

 ∂δ/∂n < –Nh

 with h > √(|δ|)) if δ ≤ 0 are incompatible. Given any point, consider the shortest geodesic to the boundary. Along that geodesic, starting from the boundary,

ψ” ≤ -(1/N)ψ’²  – Nδ.

Since by hypothesis initially ψ’ < –Nh, by the lemma below ψ’ would go to -∞ and the density would go to 0 unless it hits the boundary first, where ∂ψ/∂n = -ψ’ would now be greater than Nh.

Here’s a more trivial version for density alone, unrelated to generalized Ricci curvature.

Proposition (trivial density case). Consider a smooth complete connected n-dimensional Riemannian manifold with nonempty boundary with smooth (positive) density f. Suppose that

 Hess f  ≤ -δ < 0

and that on the boundary

 ∂f/∂n ≤ h > 0.

 Then the distance to the boundary is at most h/δ.

Proof. Suppose there is a point at distance greater h/δ from the boundary. Consider the shortest geodesic to the boundary. By the time the geodesic reaches the point, df/ds < 0. If the geodesic never reaches the boundary again, f would eventually go negative, a contradiction. By the time the geodesic reaches the boundary again, df/ds < –h, so ∂f/∂n > h, a contradiction of the hypotheses.

Remark. To see that this is sharp, consider density 2-.5x² on [-1,1].

Lemma. Suppose y’≤ -a²y² +/- b² with a > 0, b ≥ 0 and ay(0) < -b except for the – sign case. Then y goes to -∞ within:

-1/a²y₀                                 if b = 0,

(1/ab)arccot(-ay₀ /b)       if b > 0,

(1/ab)arccoth(-ay₀/b)     if b < 0.

 Proof. By comparison with the equality cases y = 1/a²(x₀+x), y = (b/a)cot(ab(x₀+x)), y = (b/a)coth(ab(x₀+x)). Note that x₀ is negative; the three functions blow up as x approaches 0.