Surfaces in space

André Neves

Last timeGauss-Bonnet Thm. S genus g compact surface then

∫S KdA = 4π(1− g)

DefinitionThe Willmore energy of surface S is given by W (S) = 14

∫S |H|

2dA.

Examples

• S(r) = {p ∈ R3 : |p| = r}, W (S(r)) = 14∫

S(r)(2r−1)2dA = 4π

• A genus 0 surface with large Willmore energy.

Willmore energyExamples

•

Given a closed curve γ,consider T (r) tube of radius raround γ

With γ parametrized by arc-length, the curvature of γ is k = |γ′′(t)|

For r small, k1 ' k and k2 ' 1/r .

W (T (r)) =14

∫T (r)

H2dA '∫

T (r)r−2dA ' area (T (r))r−2 = r−1

•

A genus g surface with Willmoreenergy close to 8π

Willmore energyTheorem (Willmore)If S is compact surface then W (S) ≥ 4π with equality iff S is a round sphere.

1: With N : S → S2 ⊂ R3 the Gauss map, K = det (−dN) = det dN.

Area formula:∫

S|K |dA =

∫S|det dN|dA =

∫S2

#N−1(v)dv

2: Let S+ = {p ∈ S : K (p) ≥ 0} and S− = {p ∈ S : K (p) ≤ 0}.If S is a compact surface, the Gauss map restricted to S+ is surjective.

Indeed, given v ∈ S2 ⊂ R3, choose a plane P perpendicular to v and very far.Move it in the direction of v until the first time it touches S at a point p.At p we have K (p) ≥ 0 and N(p) = v .

3: One has K = k1k2 =(k1+k2)2−(k1−k2)2

4 =H24 −

(k1−k2)24 .

W (S) ≥ 14∫

S+ H2dA =

∫S+ KdA +

∫S+

(k1−k2)24 ≥

∫S+ KdA

=∫

S+ |K |dA =∫

S2 #(N−1(v) ∩ S+)dv ≥ 4π

Equality implies that S = S+ and S is totally umbilical.

Willmore energy

Theorem (Fenchel)If γ is a closed curve,

∫γ

kds ≥ 2π with equality iff γ is a plane convex curve.

Proof of inequality.

Consider T (r) tube of radius r around γ

Using coordinates (s, θ) (s is arc-length, 0 ≤ θ ≤ 2π) one computesKdA = −k cos θdtdθ.

4π ≤∫

T (r)+|K |dA =

∫T (r)+

KdA =∫γ

∫ 3π/2π/2

−k cos θdθds = 2∫γ

kds.

Theorem (Langevin-Rosenberg)

If S is a knotted torus, then W (S) ≥ 8π.

1: With v ∈ S2 ⊂ R3, set hv : S → R, hv (p) = p.v .

p ∈ S is critical point of hv ⇐⇒ v ⊥ TpS ⇐⇒ N(p) = ±v .

If Crt(hv ) are the critical points of hv , # Crt(hv ) = #N−1(v) + #N−1(−v).

Hence∫

S |K |dA =∫

S2 #N−1(v)dv =

∫S2

12# Crt(hv )dv .

2: They show that for almost all v ∈ S2, # Crt(hv ) ≥ 8.

Thus∫

S+ KdA−∫

S− KdA =∫

S |K |dA ≥ 16π.

3: From Gauss-Bonnet, 0 =∫

S KdA =∫

S+ KdA +∫

S− KdA.

Hence,∫

S+ KdA =12

∫S |K |dA ≥ 8π. This implies the result.

Willmore energy

Theorem (Milnor-Fary)

If γ is a knotted curved then∫γ

kds ≥ 4π.

Open questionAmong compact surfaces of genus g, which one has least Willmore energy?

When g = 0 it is the round sphere, when g = 1 is the Clifford torus, and forhigher genus should be the ones seen below.