Min-max theory and the Willmore conjecture Part II Fernando Codá Marques (IMPA, Brazil) & André Neves (Imperial College, UK) Sanya, China 2013
Min-max method C 1 is a critical point of the height function h with index(c 1 ) = 1.
Min-max method C 1 is a critical point of the height function h with index(c 1 ) = 1. Let π be the homotopy class of a loop γ 0 going into the hole. Then h(c 1 ) = inf γ π sup t [0,1] h(γ(t)).
Min-max method C 1 is a critical point of the height function h with index(c 1 ) = 1. Let π be the homotopy class of a loop γ 0 going into the hole. Then h(c 1 ) = inf γ π sup t [0,1] h(γ(t)). To find a critical point of index k, one needs to do min-max over families with at least k parameters.
Minimal surfaces (H = 0) are critical points for the area functional.
Minimal surfaces (H = 0) are critical points for the area functional. The equator (great sphere) is a minimal surface in S 3 with index 1.
Minimal surfaces (H = 0) are critical points for the area functional. The equator (great sphere) is a minimal surface in S 3 with index 1. It can be produced by doing min-max over 1-parameter families of surfaces.
Minimal surfaces (H = 0) are critical points for the area functional. The equator (great sphere) is a minimal surface in S 3 with index 1. It can be produced by doing min-max over 1-parameter families of surfaces. In fact, every compact Riemannian three-manifold contains an embedded closed minimal surface constructed in that way.
Almgren-Pitts Min-max Theory Φ : I n {closed surfaces in M}
Almgren-Pitts Min-max Theory Φ : I n {closed surfaces in M} Π homotopy class of Φ relative to I n
Almgren-Pitts Min-max Theory Φ : I n {closed surfaces in M} Π homotopy class of Φ relative to I n Φ Π, L(Φ ) = sup{area(φ (x)) : x I n }
Almgren-Pitts Min-max Theory Φ : I n {closed surfaces in M} Π homotopy class of Φ relative to I n Φ Π, L(Φ ) = sup{area(φ (x)) : x I n } L(Π) = inf{l(φ ) : Φ Π} width of Π
Almgren-Pitts Min-max Theory Φ : I n {closed surfaces in M} Π homotopy class of Φ relative to I n Φ Π, L(Φ ) = sup{area(φ (x)) : x I n } L(Π) = inf{l(φ ) : Φ Π} width of Π Min-max Theorem: If L(Π) > sup{area(φ(x)) : x I n }, then there exists a smooth embedded minimal surface (possibly disconnected, with integer multiplicities) Σ M such that area(σ) = L(Π).
Almgren-Pitts Min-max Theory Φ : I n {closed surfaces in M} Π homotopy class of Φ relative to I n Φ Π, L(Φ ) = sup{area(φ (x)) : x I n } L(Π) = inf{l(φ ) : Φ Π} width of Π Min-max Theorem: If L(Π) > sup{area(φ(x)) : x I n }, then there exists a smooth embedded minimal surface (possibly disconnected, with integer multiplicities) Σ M such that area(σ) = L(Π). Question: Can we produce the Clifford torus by min-max methods?
Let Σ S 3. We construct a five-parameter canonical family of surfaces Σ (v,t) S 3 such that Σ (0,0) = Σ and area(σ (v,t) ) W(Σ).
Let Σ S 3. We construct a five-parameter canonical family of surfaces Σ (v,t) S 3 such that Σ (0,0) = Σ and area(σ (v,t) ) W(Σ). Parameter space:
Let Σ S 3. We construct a five-parameter canonical family of surfaces Σ (v,t) S 3 such that Σ (0,0) = Σ and area(σ (v,t) ) W(Σ). Parameter space: For each v S 3 = B 4, {C(v, t)} is the standard family of round spheres centered along the axis passing through some Q(v) S 3.
Let Σ S 3. We construct a five-parameter canonical family of surfaces Σ (v,t) S 3 such that Σ (0,0) = Σ and area(σ (v,t) ) W(Σ). Parameter space: For each v S 3 = B 4, {C(v, t)} is the standard family of round spheres centered along the axis passing through some Q(v) S 3. The center map Q : S 3 S 3 has deg(q) = genus(σ)!
Let Σ S 3. We construct a five-parameter canonical family of surfaces Σ (v,t) S 3 such that Σ (0,0) = Σ and area(σ (v,t) ) W(Σ). Parameter space: For each v S 3 = B 4, {C(v, t)} is the standard family of round spheres centered along the axis passing through some Q(v) S 3. The center map Q : S 3 S 3 has deg(q) = genus(σ)! If genus(σ) 1, the boundary of the cylinder is mapped onto the space of round spheres in a homotopically nontrivial way.
After a reparametrization, we have Φ : I 5 {closed surfaces in S 3 } that satisfies: (1) Φ is continuous (in the flat topology),
After a reparametrization, we have Φ : I 5 {closed surfaces in S 3 } that satisfies: (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4,
After a reparametrization, we have that satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3,
After a reparametrization, we have that satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x),
After a reparametrization, we have that satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x), (5) the center map Q : I 4 S 3 has degree 0.
After a reparametrization, we have that satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x), (5) the center map Q : I 4 S 3 has degree 0. Moreover, sup{area(φ(x)) : x I 5 } W(Σ).
Topological argument Theorem: genus(σ) 1 = L(Π) > 4π.
Topological argument Theorem: genus(σ) 1 = L(Π) > 4π. The proof is by contradiction. Suppose there exists Φ : I 5 {closed surfaces in S 3 }, Φ Π, with sup{area(φ (x)) : x I 5 } = 4π. Parameter space:
Every continuous path connecting the bottom to the top of I 5 must contain a great sphere.
Every continuous path connecting the bottom to the top of I 5 must contain a great sphere. We construct a 4-dimensional submanifold R 4 such that Φ (y) is a great sphere for every y R 4, R 4 = I 4 {1/2}.
Every continuous path connecting the bottom to the top of I 5 must contain a great sphere. We construct a 4-dimensional submanifold R 4 such that Φ (y) is a great sphere for every y R 4, R 4 = I 4 {1/2}. Hence Φ : R 4 S 3, and Φ R 4 = Q.
Every continuous path connecting the bottom to the top of I 5 must contain a great sphere. We construct a 4-dimensional submanifold R 4 such that Φ (y) is a great sphere for every y R 4, R 4 = I 4 {1/2}. Hence Φ : R 4 S 3, and Φ R = Q. In homology, 4 g S 3 = Φ ( I 4 {1/2}) = Φ ( R 4 ) = Φ (R 4 ) = 0.
Every continuous path connecting the bottom to the top of I 5 must contain a great sphere. We construct a 4-dimensional submanifold R 4 such that Φ (y) is a great sphere for every y R 4, R 4 = I 4 {1/2}. Hence Φ : R 4 S 3, and Φ R = Q. In homology, 4 g S 3 = Φ ( I 4 {1/2}) = Φ ( R 4 ) = Φ (R 4 ) = 0. Contradiction if g 1.
Statements Theorem A. (, Neves) Let Σ S 3 be an embedded closed surface of genus g 1. Then W(Σ) 2π 2, and the equality holds if and only if Σ is the Clifford torus up to conformal transformations of S 3.
Statements Theorem A. (, Neves) Let Σ S 3 be an embedded closed surface of genus g 1. Then W(Σ) 2π 2, and the equality holds if and only if Σ is the Clifford torus up to conformal transformations of S 3. Theorem A = Willmore conjecture is true!
Statements Theorem A. (, Neves) Let Σ S 3 be an embedded closed surface of genus g 1. Then W(Σ) 2π 2, and the equality holds if and only if Σ is the Clifford torus up to conformal transformations of S 3. Theorem A = Willmore conjecture is true! Theorem B. (, Neves) Let Σ S 3 be an embedded closed minimal surface of genus g 1. Then area(σ) 2π 2, and the equality holds if and only if Σ is the Clifford torus up to isometries of S 3.
Proof of Theorem B
Proof of Theorem B Let Σ be the minimal surface with lowest area among all minimal surfaces in S 3 with genus greater than or equal to 1. (The existence of Σ follows from standard arguments in Geometric Measure Theory.)
Proof of Theorem B Let Σ be the minimal surface with lowest area among all minimal surfaces in S 3 with genus greater than or equal to 1. (The existence of Σ follows from standard arguments in Geometric Measure Theory.) Of course, area(σ) 2π 2.
Proof of Theorem B Let Σ be the minimal surface with lowest area among all minimal surfaces in S 3 with genus greater than or equal to 1. (The existence of Σ follows from standard arguments in Geometric Measure Theory.) Of course, area(σ) 2π 2. Recall Urbano s theorem: Σ S 3 minimal surface (H = 0), index( Σ) 5, g 0 = Σ is either the Clifford torus (index 5) or the great sphere (index 1).
Proof of Theorem B Let Σ be the minimal surface with lowest area among all minimal surfaces in S 3 with genus greater than or equal to 1. (The existence of Σ follows from standard arguments in Geometric Measure Theory.) Of course, area(σ) 2π 2. Recall Urbano s theorem: Σ S 3 minimal surface (H = 0), index( Σ) 5, g 0 = Σ is either the Clifford torus (index 5) or the great sphere (index 1). We claim that index(σ) 5.
Suppose, by contradiction, that index(σ) > 6.
Suppose, by contradiction, that index(σ) > 6. If {Σ (v,t) } (v,t) B 4 [ π,π] denotes the canonical family, then sup area(σ (v,t) ) W(Σ) = area(σ). (v,t) B 4 [ π,π]
Suppose, by contradiction, that index(σ) > 6. If {Σ (v,t) } (v,t) B 4 [ π,π] denotes the canonical family, then sup area(σ (v,t) ) W(Σ) = area(σ). (v,t) B 4 [ π,π] The function (v, t) area(σ (v,t) ) has a unique and nondegenerate global maximum point at (0, 0).
Suppose, by contradiction, that index(σ) > 6. If {Σ (v,t) } (v,t) B 4 [ π,π] denotes the canonical family, then sup area(σ (v,t) ) W(Σ) = area(σ). (v,t) B 4 [ π,π] The function (v, t) area(σ (v,t) ) has a unique and nondegenerate global maximum point at (0, 0). Jacobi operator of Σ : L = + A 2 + 2 Let {e 1, e 2, e 3, e 4 } be the standard orthonormal basis of R 4. For x Σ, define ψ i (x) = N(x), e i for each 1 i 4, and ψ 5 (x) = 1.
Suppose, by contradiction, that index(σ) > 6. If {Σ (v,t) } (v,t) B 4 [ π,π] denotes the canonical family, then sup area(σ (v,t) ) W(Σ) = area(σ). (v,t) B 4 [ π,π] The function (v, t) area(σ (v,t) ) has a unique and nondegenerate global maximum point at (0, 0). Jacobi operator of Σ : L = + A 2 + 2 Let {e 1, e 2, e 3, e 4 } be the standard orthonormal basis of R 4. For x Σ, define ψ i (x) = N(x), e i for each 1 i 4, and ψ 5 (x) = 1. Lemma. ψlψ dσ < 0 for every ψ = Σ 5 c i ψ i 0. i=1
Recall that area(σ (v,t) ) < W(Σ v ) = W(Σ) = area(σ), unless t = 0 and Σ v is a minimal surface.
Recall that area(σ (v,t) ) < W(Σ v ) = W(Σ) = area(σ), unless t = 0 and Σ v is a minimal surface. Last possibility cannot happen for v 0 since v, N(x) 2 area(σ v ) = area(σ) 4 Σ x v 4 dσ.
Recall that area(σ (v,t) ) < W(Σ v ) = W(Σ) = area(σ), unless t = 0 and Σ v is a minimal surface. Last possibility cannot happen for v 0 since v, N(x) 2 area(σ v ) = area(σ) 4 Σ x v 4 dσ. Because index(σ) 6, there exists ϕ C (Σ) such that ϕlϕ dσ < 0, ϕlψ i dσ = 0 for 1 i 5. Let X be any vector field such that X = ϕn along Σ, and let {Γ t } t 0 be the one parameter group of diffeomorphisms generated by X.
Define f (v, t, s) = area(γ s (Σ (v,t) )).
Define f (v, t, s) = area(γ s (Σ (v,t) )). We have f (0, 0, 0) = area(σ), Df (0, 0, 0) = 0 and D 2 f (0, 0, 0) < 0 (because of the choice of ϕ).
Define f (v, t, s) = area(γ s (Σ (v,t) )). We have f (0, 0, 0) = area(σ), Df (0, 0, 0) = 0 and D 2 f (0, 0, 0) < 0 (because of the choice of ϕ). This implies that area(γ s (Σ (v,t) )) < f (0, 0, 0) = area(σ) for every (v, t, s) (Bδ 4 (0) ( δ, δ) ( δ, δ)) \ {(0, 0, 0)}.
Define f (v, t, s) = area(γ s (Σ (v,t) )). We have f (0, 0, 0) = area(σ), Df (0, 0, 0) = 0 and D 2 f (0, 0, 0) < 0 (because of the choice of ϕ). This implies that area(γ s (Σ (v,t) )) < f (0, 0, 0) = area(σ) for every (v, t, s) (Bδ 4 (0) ( δ, δ) ( δ, δ)) \ {(0, 0, 0)}. Let β : R 5 R be a smooth function such that 0 β(y) δ 1 /2 for y R 5, β(y) = 0 if y δ 1 /2 and β(y) = δ 1 /2 if y δ 1 /4. Define Σ (v, t) = Γ β(v,t) (Σ (v,t) ) for (v, t) < δ 1.
Define f (v, t, s) = area(γ s (Σ (v,t) )). We have f (0, 0, 0) = area(σ), Df (0, 0, 0) = 0 and D 2 f (0, 0, 0) < 0 (because of the choice of ϕ). This implies that area(γ s (Σ (v,t) )) < f (0, 0, 0) = area(σ) for every (v, t, s) (Bδ 4 (0) ( δ, δ) ( δ, δ)) \ {(0, 0, 0)}. Let β : R 5 R be a smooth function such that 0 β(y) δ 1 /2 for y R 5, β(y) = 0 if y δ 1 /2 and β(y) = δ 1 /2 if y δ 1 /4. Define Σ (v, t) = Γ β(v,t) (Σ (v,t) ) for (v, t) < δ 1. Note that Σ (v, t) = Σ(v, t) if δ 1 /2 < (v, t) < δ 1.
Hence for every (v, t) B 4 [ π, π]. area(σ (v, t)) < area(σ)
Hence for every (v, t) B 4 [ π, π]. area(σ (v, t)) < area(σ) Let Φ be the min-max family produced out of {Σ (v,t) }, and let Π be the homotopy class of Φ.
Hence for every (v, t) B 4 [ π, π]. area(σ (v, t)) < area(σ) Let Φ be the min-max family produced out of {Σ (v,t) }, and let Π be the homotopy class of Φ. Since Φ agrees with Φ on I 5, and since g = genus(σ) 1, we can argue similarly as before to get L(Π ) > 4π.
Therefore we can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π ) = area( Σ) sup x I 5 area(φ (x)) < area(σ) 2π 2
Therefore we can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π ) = area( Σ) sup x I 5 area(φ (x)) < area(σ) 2π 2 Hence Σ has multiplicity one.
Therefore we can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π ) = area( Σ) sup x I 5 area(φ (x)) < area(σ) 2π 2 Hence Σ has multiplicity one. By Almgren, genus( Σ) 1.
Therefore we can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π ) = area( Σ) sup x I 5 area(φ (x)) < area(σ) 2π 2 Hence Σ has multiplicity one. By Almgren, genus( Σ) 1. Contradiction, since area( Σ) < area(σ).
Therefore we can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π ) = area( Σ) sup x I 5 area(φ (x)) < area(σ) 2π 2 Hence Σ has multiplicity one. By Almgren, genus( Σ) 1. Contradiction, since area( Σ) < area(σ). Therefore index(σ) 5, and Σ is the Clifford torus. This proves Theorem B.
Proof of Theorem A
Proof of Theorem A Let Σ be an embedded surface in S 3, not necessarily minimal, with genus g 1. Can assume W(Σ) < 8π.
Proof of Theorem A Let Σ be an embedded surface in S 3, not necessarily minimal, with genus g 1. Can assume W(Σ) < 8π. Let Φ be the min-max family associated with Σ, and let Π be its homotopy class. We have L(Π) > 4π.
Proof of Theorem A Let Σ be an embedded surface in S 3, not necessarily minimal, with genus g 1. Can assume W(Σ) < 8π. Let Φ be the min-max family associated with Σ, and let Π be its homotopy class. We have L(Π) > 4π. We can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π) = area( Σ) sup x I 5 area(φ(x)) W(Σ) < 8π.
Proof of Theorem A Let Σ be an embedded surface in S 3, not necessarily minimal, with genus g 1. Can assume W(Σ) < 8π. Let Φ be the min-max family associated with Σ, and let Π be its homotopy class. We have L(Π) > 4π. We can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π) = area( Σ) sup x I 5 area(φ(x)) W(Σ) < 8π. Hence genus( Σ) 1.
Proof of Theorem A Let Σ be an embedded surface in S 3, not necessarily minimal, with genus g 1. Can assume W(Σ) < 8π. Let Φ be the min-max family associated with Σ, and let Π be its homotopy class. We have L(Π) > 4π. We can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π) = area( Σ) sup x I 5 area(φ(x)) W(Σ) < 8π. Hence genus( Σ) 1. By Theorem B, area( Σ) 2π 2.
Proof of Theorem A Let Σ be an embedded surface in S 3, not necessarily minimal, with genus g 1. Can assume W(Σ) < 8π. Let Φ be the min-max family associated with Σ, and let Π be its homotopy class. We have L(Π) > 4π. We can apply the Min-Max Theorem to Π in order to find an embedded minimal surface Σ (with possible multiplicities) in S 3 such that 4π < L(Π) = area( Σ) sup x I 5 area(φ(x)) W(Σ) < 8π. Hence genus( Σ) 1. By Theorem B, area( Σ) 2π 2. Therefore W(Σ) 2π 2 (Willmore conjecture).
Suppose W(Σ) = 2π 2.
Suppose W(Σ) = 2π 2. Then area(σ (v,t) ) 2π 2 for every (v, t) B 4 [ π, π].
Suppose W(Σ) = 2π 2. Then area(σ (v,t) ) 2π 2 for every (v, t) B 4 [ π, π]. There must be (v, t) with area(σ (v,t) ) = 2π 2.
Suppose W(Σ) = 2π 2. Then area(σ (v,t) ) 2π 2 for every (v, t) B 4 [ π, π]. There must be (v, t) with area(σ (v,t) ) = 2π 2. Since W(Σ v ) = W(Σ) = 2π 2, we get that t = 0 and Σ v is a minimal surface.
Suppose W(Σ) = 2π 2. Then area(σ (v,t) ) 2π 2 for every (v, t) B 4 [ π, π]. There must be (v, t) with area(σ (v,t) ) = 2π 2. Since W(Σ v ) = W(Σ) = 2π 2, we get that t = 0 and Σ v is a minimal surface. Since area(σ v ) = 2π 2, Σ v is the Clifford torus by Theorem B.
Suppose W(Σ) = 2π 2. Then area(σ (v,t) ) 2π 2 for every (v, t) B 4 [ π, π]. There must be (v, t) with area(σ (v,t) ) = 2π 2. Since W(Σ v ) = W(Σ) = 2π 2, we get that t = 0 and Σ v is a minimal surface. Since area(σ v ) = 2π 2, Σ v is the Clifford torus by Theorem B. Therefore Σ is a conformal image of the Clifford torus. This finishes the proof of Theorem A.
This theory implies that if satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x), (5) the center map Q : I 4 S 3 has degree 0,
This theory implies that if satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x), (5) the center map Q : I 4 S 3 has degree 0, then there exists y I 5 with area(φ(y)) 2π 2.
A problem in Topology γ 1 and γ 2 two linked curves in R 3.
A problem in Topology γ 1 and γ 2 two linked curves in R 3. lk (γ 1, γ 2 ) is the linking number lk = 1 lk = 3
Let γ i : S 1 R 3, i = 1, 2, be a 2-component link, i.e., a pair of closed curves in Euclidean three-space with γ 1 (S 1 ) γ 2 (S 1 ) =.
Let γ i : S 1 R 3, i = 1, 2, be a 2-component link, i.e., a pair of closed curves in Euclidean three-space with γ 1 (S 1 ) γ 2 (S 1 ) =. The Möbius cross energy of the link (γ 1, γ 2 ) is defined to be γ 1 E(γ 1, γ 2 ) = (s) γ 2 (t) ds dt. γ 1 (s) γ 2 (t) 2 S 1 S 1 The Möbius energy is invariant under conformal transformations of R 3.
Let γ i : S 1 R 3, i = 1, 2, be a 2-component link, i.e., a pair of closed curves in Euclidean three-space with γ 1 (S 1 ) γ 2 (S 1 ) =. The Möbius cross energy of the link (γ 1, γ 2 ) is defined to be γ 1 E(γ 1, γ 2 ) = (s) γ 2 (t) ds dt. γ 1 (s) γ 2 (t) 2 S 1 S 1 The Möbius energy is invariant under conformal transformations of R 3. The Gauss formula lk(γ 1, γ 2 ) = 1 4π S 1 S 1 det(γ 1 (s), γ 2 (t), γ 1(s) γ 2 (t)) γ 1 (s) γ 2 (t) 3 ds dt implies E(γ 1, γ 2 ) 4π lk (γ 1, γ 2 ). It is natural to search for the optimal non-trivial link in R 3.
Conjecture (Freedman-He-Wang, 94) The Möbius cross energy should be minimized, among the class of all non-trivial links in R 3, by the stereographic projection of the standard Hopf link. The standard Hopf link (ˆγ 1, ˆγ 2 ) is described by ˆγ 1 (s) = (cos s, sin s, 0, 0) S 3 and ˆγ 2 (t) = (0, 0, cos t, sin t) S 3. Easy to check that E(ˆγ 1, ˆγ 2 ) = 2π 2.
Conjecture (Freedman-He-Wang, 94) The Möbius cross energy should be minimized, among the class of all non-trivial links in R 3, by the stereographic projection of the standard Hopf link. The standard Hopf link (ˆγ 1, ˆγ 2 ) is described by ˆγ 1 (s) = (cos s, sin s, 0, 0) S 3 and ˆγ 2 (t) = (0, 0, cos t, sin t) S 3. Easy to check that E(ˆγ 1, ˆγ 2 ) = 2π 2. We can assume lk(γ 1, γ 2 ) = 1, since He proved that the minimal Möbius cross energy among nontrivial 2-component links is achieved by a link isotopic to the Hopf link.
Theorem (Agol,, Neves) The conjecture is true, i.e., if lk((γ 1, γ 2 ) = 1 then E(γ 1, γ 2 ) 2π 2. If equality then (γ 1, γ 2 ) is conformal to the standard Hopf link in S 3.
Theorem (Agol,, Neves) The conjecture is true, i.e., if lk((γ 1, γ 2 ) = 1 then E(γ 1, γ 2 ) 2π 2. If equality then (γ 1, γ 2 ) is conformal to the standard Hopf link in S 3.
The Gauss map of the link (γ 1, γ 2 ) R 4 is defined by for (s, t) S 1 S 1. G(γ 1, γ 2 )(s, t) = γ 1(s) γ 2 (t) γ 1 (s) γ 2 (t) S 3
The Gauss map of the link (γ 1, γ 2 ) R 4 is defined by for (s, t) S 1 S 1. G(γ 1, γ 2 )(s, t) = γ 1(s) γ 2 (t) γ 1 (s) γ 2 (t) S 3 It satisfies area(g(s 1 S 1 )) E(γ 1, γ 2 ).
The Gauss map of the link (γ 1, γ 2 ) R 4 is defined by for (s, t) S 1 S 1. G(γ 1, γ 2 )(s, t) = γ 1(s) γ 2 (t) γ 1 (s) γ 2 (t) S 3 It satisfies area(g(s 1 S 1 )) E(γ 1, γ 2 ). If equality holds, then γ 1(s), γ 2(t) = γ 1(s), γ 1 (s) γ 2 (t) = γ 2(t), γ 1 (s) γ 2 (t) = 0.
The Gauss map of the link (γ 1, γ 2 ) R 4 is defined by for (s, t) S 1 S 1. G(γ 1, γ 2 )(s, t) = γ 1(s) γ 2 (t) γ 1 (s) γ 2 (t) S 3 It satisfies area(g(s 1 S 1 )) E(γ 1, γ 2 ). If equality holds, then γ 1(s), γ 2(t) = γ 1(s), γ 1 (s) γ 2 (t) = γ 2(t), γ 1 (s) γ 2 (t) = 0. If the link (γ 1, γ 2 ) is contained in an oriented affine hyperplane with unit normal vector p S 3 compatible with the orientation, then G(S 1 S 1 ) = lk(γ 1, γ 2 ) B π/2 ( p).
Let (γ 1, γ 2 ) S 3. Given v B 4, we define the conformal map F v : R 4 \ {v} R 4, F v (x) = x v x v 2. We have that F v (S 3 1 (0)) = S 3 1 1 v 2 (c(v)) where c(v) = v 1 v 2.
Let (γ 1, γ 2 ) S 3. Given v B 4, we define the conformal map F v : R 4 \ {v} R 4, F v (x) = x v x v 2. We have that F v (S 3 1 (0)) = S 3 1 1 v 2 (c(v)) where c(v) = v 1 v 2. Let g (v,λ) = G(F v γ 1, λ(f v γ 2 c(v)) + c(v)), λ (0, ).
Let (γ 1, γ 2 ) S 3. Given v B 4, we define the conformal map F v : R 4 \ {v} R 4, F v (x) = x v x v 2. We have that F v (S 3 1 (0)) = S 3 1 1 v 2 (c(v)) where c(v) = v 1 v 2. Let g (v,λ) = G(F v γ 1, λ(f v γ 2 c(v)) + c(v)), λ (0, ). We have area(g (v,λ) (S 1 S 1 )) E(γ 1, γ 2 ).
Let (γ 1, γ 2 ) S 3. Given v B 4, we define the conformal map F v : R 4 \ {v} R 4, F v (x) = x v x v 2. We have that F v (S 3 1 (0)) = S 3 1 1 v 2 (c(v)) where c(v) = v 1 v 2. Let g (v,λ) = G(F v γ 1, λ(f v γ 2 c(v)) + c(v)), λ (0, ). We have area(g (v,λ) (S 1 S 1 )) E(γ 1, γ 2 ). If v S 3, then F v (S1 3 (0) \ {v}) is the hyperplane P v = {x R 4 : x, v = 1/2}.
Let (γ 1, γ 2 ) S 3. Given v B 4, we define the conformal map F v : R 4 \ {v} R 4, F v (x) = x v x v 2. We have that F v (S 3 1 (0)) = S 3 1 1 v 2 (c(v)) where c(v) = v 1 v 2. Let g (v,λ) = G(F v γ 1, λ(f v γ 2 c(v)) + c(v)), λ (0, ). We have area(g (v,λ) (S 1 S 1 )) E(γ 1, γ 2 ). If v S 3, then F v (S1 3 (0) \ {v}) is the hyperplane P v = {x R 4 : x, v = 1/2}. Hence g (v,1) (S 1 S 1 ) = lk(γ 1, γ 2 ) B π/2 (v) for every v S 3.
Suppose lk(γ 1, γ 2 ) = 1.
Suppose lk(γ 1, γ 2 ) = 1. These Gauss maps can be used to construct a family that satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x), (5) the center map Q : I 4 S 3 has degree 1. Moreover, sup{area(φ(x)) : x I 5 } E(γ 1, γ 2 ).
Suppose lk(γ 1, γ 2 ) = 1. These Gauss maps can be used to construct a family that satisfies: Φ : I 5 {closed surfaces in S 3 } (1) Φ is continuous (in the flat topology), (2) Φ(x, 0) = Φ(x, 1) = 0 for any x I 4, (3) x I 4 {Φ(x, t)} t [0,1] is the standard foliation of S 3 by round spheres centered along the axis passing through some Q(x) S 3, (4) Φ(x, 1/2) is the great sphere orthogonal to Q(x), (5) the center map Q : I 4 S 3 has degree 1. Moreover, sup{area(φ(x)) : x I 5 } E(γ 1, γ 2 ). Previous theory = E(γ 1, γ 2 ) 2π 2.