Post on 30-Jan-2021
Min-max Theory, Willmore conjecture, and Energyof links
André Neves
(Joint with Fernando Marques)
Q: What is the best way of immersing a sphere in space?
A: The one that minimizes the bending energy∫Σ
H̄2 =∫
Σ
(k1 + k2
2
)2.
• bending energy is conformally invariant;
• every compact surface has∫
ΣH̄2 ≥ 4π with equality only for round
sphere.
Q: What is the best way of immersing a sphere in space?
A: The one that minimizes the bending energy∫Σ
H̄2 =∫
Σ
(k1 + k2
2
)2.
• bending energy is conformally invariant;
• every compact surface has∫
ΣH̄2 ≥ 4π with equality only for round
sphere.
What is the best way of immersing a torus in space?
Conjecture (Willmore, ’65)For every torus Σ immersed in R3∫
Σ
H̄2 ≥ 2π2.
What is the best way of immersing a torus in space?
Conjecture (Willmore, ’65)For every torus Σ immersed in R3∫
Σ
H̄2 ≥ 2π2.
Consider stereographic projection
π : S3 − {north pole} −→ R3.
• If Σ ⊂ S3 then ∫Σ
(1 + H2) =∫π(Σ)
H̄2.
• the Willmore energy of Σ ⊂ S3 is defined to be
W(Σ) =∫
Σ
(1 + H2).
Willmore Conjecture - ’65Every torus Σ immersed in S3 hasW(Σ) ≥ 2π2.
Consider stereographic projection
π : S3 − {north pole} −→ R3.
• If Σ ⊂ S3 then ∫Σ
(1 + H2) =∫π(Σ)
H̄2.
• the Willmore energy of Σ ⊂ S3 is defined to be
W(Σ) =∫
Σ
(1 + H2).
Willmore Conjecture - ’65Every torus Σ immersed in S3 hasW(Σ) ≥ 2π2.
Theorem (Marques-N., ’12)For every embedded surface Σ ⊂ S3 with genus g ≥ 1
W(Σ) ≥ 2π2
with equality if and only if Σ is conformal to S1(
1√2
)× S1
(1√2
).
Corollary AWillmore conjecture holds.
Corollary BThe only two minimal surfaces in S3 with area ≤ 2π2 are great spheres andClifford torus.
Links in space
• γ1 and γ2 two linked curves in R3.
• lk (γ1, γ2) is the linking number
lk = 1 lk = 3
Links in space
• γ1 and γ2 two linked curves in R3.
• lk (γ1, γ2) is the linking number
lk = 1 lk = 3
Links in space
• Möbius cross energy of a link (γ1, γ2) is
E(γ1, γ2) =∫
S1×S1
|γ′1(s)||γ′2(t)||γ1(s)− γ2(t)|2
ds dt .
• E(γ1, γ2) is conformally invariant and
E(γ1, γ2) ≥ 4π|lk (γ1, γ2)|
Links in space
Q: What is the best way of immersing a non-trivial link (γ1, γ2) in R3?
A: The one that minimizes the Möbius energy.
Conjecture (Freedman-He-Wang, ’94)
If lk (γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• If true, every non-trivial link has E(γ1, γ2) ≥ 2π2.
Links in space
Q: What is the best way of immersing a non-trivial link (γ1, γ2) in R3?
A: The one that minimizes the Möbius energy.
Conjecture (Freedman-He-Wang, ’94)
If lk (γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• If true, every non-trivial link has E(γ1, γ2) ≥ 2π2.
Theorem (Agol-Marques-N., ’12)
If lk (γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
If equality then (γ1, γ2) is conformal to the standard Hopf link in S3
β1(t) = (cos t , sin t ,0,0) and β2(s) = (0,0, cos s, sin s).
Theorem (Agol-Marques-N., ’12)
If lk (γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
If equality then (γ1, γ2) is conformal to the standard Hopf link in S3
β1(t) = (cos t , sin t ,0,0) and β2(s) = (0,0, cos s, sin s).
Partial results...• (Willmore ’71, Shioama-Takagi ’70): Conjecture is true for tubes of
constant radius around a space curve.
• (Langevin-Rosenberg ’76): If Σ ⊂ R3 is a knotted torus thenW(Σ) ≥ 8π.
• (Chen, ’83): Conjecture is true for flat tori in 3-sphere.
• (Langer-Singer ’84): Conjecture is true for tori of revolution.
Partial results...• (Willmore ’71, Shioama-Takagi ’70): Conjecture is true for tubes of
constant radius around a space curve.
• (Langevin-Rosenberg ’76): If Σ ⊂ R3 is a knotted torus thenW(Σ) ≥ 8π.
• (Chen, ’83): Conjecture is true for flat tori in 3-sphere.
• (Langer-Singer ’84): Conjecture is true for tori of revolution.
Partial results...• (Willmore ’71, Shioama-Takagi ’70): Conjecture is true for tubes of
constant radius around a space curve.
• (Langevin-Rosenberg ’76): If Σ ⊂ R3 is a knotted torus thenW(Σ) ≥ 8π.
• (Chen, ’83): Conjecture is true for flat tori in 3-sphere.
• (Langer-Singer ’84): Conjecture is true for tori of revolution.
Partial results...• (Willmore ’71, Shioama-Takagi ’70): Conjecture is true for tubes of
constant radius around a space curve.
• (Langevin-Rosenberg ’76): If Σ ⊂ R3 is a knotted torus thenW(Σ) ≥ 8π.
• (Chen, ’83): Conjecture is true for flat tori in 3-sphere.
• (Langer-Singer ’84): Conjecture is true for tori of revolution.
Partial results...
• (Li-Yau ’82): If Σ ⊂ S3 covers a point k times thenW(Σ) ≥ 4πk . Thus
Σ not embedded =⇒ W(Σ) ≥ 8π.
• (Li-Yau ’82): Conjecture is true for a set of conformal classes whichcontains the conformal class of the Clifford torus.
• (Montiel-Ros ’86):Conjecture is true for a larger set of conformal classesthan the set given by Li-Yau.
Partial results...
• (Li-Yau ’82): If Σ ⊂ S3 covers a point k times thenW(Σ) ≥ 4πk . Thus
Σ not embedded =⇒ W(Σ) ≥ 8π.
• (Li-Yau ’82): Conjecture is true for a set of conformal classes whichcontains the conformal class of the Clifford torus.
• (Montiel-Ros ’86):Conjecture is true for a larger set of conformal classesthan the set given by Li-Yau.
Partial results...
• (Li-Yau ’82): If Σ ⊂ S3 covers a point k times thenW(Σ) ≥ 4πk . Thus
Σ not embedded =⇒ W(Σ) ≥ 8π.
• (Li-Yau ’82): Conjecture is true for a set of conformal classes whichcontains the conformal class of the Clifford torus.
• (Montiel-Ros ’86):Conjecture is true for a larger set of conformal classesthan the set given by Li-Yau.
Partial results...
(Benisom-Mutz ’91)Conjecture is true for observed toroidal vesicles in some cells.
Partial results...
• (Simon ’93): There is a torus which has least Willmore energy among alltori.
• (Ros ’99, Topping ’00): Conjecture is true for tori symmetric under theantipodal map.
• (Ros ’00): Conjecture is true for tori symmetric with respect to a point.
Partial results...
• (Simon ’93): There is a torus which has least Willmore energy among alltori.
• (Ros ’99, Topping ’00): Conjecture is true for tori symmetric under theantipodal map.
• (Ros ’00): Conjecture is true for tori symmetric with respect to a point.
Partial results...
• (Simon ’93): There is a torus which has least Willmore energy among alltori.
• (Ros ’99, Topping ’00): Conjecture is true for tori symmetric under theantipodal map.
• (Ros ’00): Conjecture is true for tori symmetric with respect to a point.
Other results...
Theorem (Bryant ’84)
Classified immersions of Willmore spheres.
Other results...
Theorem (Pinkall ’85)
Found many embedded Willmore tori which are not conformal to minimalsurface.
Min-max Theory
How to find critical points for area functional?
• Z2(S3) = oriented surfaces in S3.
• φ : Ik → Z2(S3) continuous, where Ik = k -cube.• [φ] = homotopy class relative to the boundary (i.e. φ|∂Ik is fixed).• L([φ]) = infψ∈[φ] maxx∈Ik area (ψ(x)).
Min-max Theory
How to find critical points for area functional?
• Z2(S3) = oriented surfaces in S3.• φ : Ik → Z2(S3) continuous, where Ik = k -cube.
• [φ] = homotopy class relative to the boundary (i.e. φ|∂Ik is fixed).• L([φ]) = infψ∈[φ] maxx∈Ik area (ψ(x)).
Min-max Theory
How to find critical points for area functional?
• Z2(S3) = oriented surfaces in S3.• φ : Ik → Z2(S3) continuous, where Ik = k -cube.• [φ] = homotopy class relative to the boundary (i.e. φ|∂Ik is fixed).
• L([φ]) = infψ∈[φ] maxx∈Ik area (ψ(x)).
Min-max Theory
How to find critical points for area functional?
• Z2(S3) = oriented surfaces in S3.• φ : Ik → Z2(S3) continuous, where Ik = k -cube.• [φ] = homotopy class relative to the boundary (i.e. φ|∂Ik is fixed).• L([φ]) = infψ∈[φ] maxx∈Ik area (ψ(x)).
Min-max Theory
Theorem (Pitts, ’81)Assume
L([φ]) = infψ∈[φ]
maxx∈Ik
area (ψ(x)) > maxx∈∂Ik
area (φ(x)),
(this means [φ] 6= 0 ∈ πk (Z2(S3), φ|∂Ik ).)
There is Σ ⊂ S3 smooth embedded minimal surface (with multiplicities) so that
L([φ]) = area(Σ).
• Conjecture: index(Σ) ≤ k .
Min-max Theory
Theorem (Pitts, ’81)Assume
L([φ]) = infψ∈[φ]
maxx∈Ik
area (ψ(x)) > maxx∈∂Ik
area (φ(x)),
(this means [φ] 6= 0 ∈ πk (Z2(S3), φ|∂Ik ).)
There is Σ ⊂ S3 smooth embedded minimal surface (with multiplicities) so that
L([φ]) = area(Σ).
• Conjecture: index(Σ) ≤ k .
Min-max Theory
Theorem (Urbano, ’90)Σ ⊂ S3 compact embedded minimal surface with index(Σ) ≤ 5. Then either• Σ is a great sphere (index(Σ) = 1 and area(Σ) = 4π), or• Σ is the Clifford torus (index(Σ) = 5 and area(Σ) = 2π2).
(Almgren, ’62) The sweepout
φ0(t) = {x4 = 2t − 1} ∩ S3, 0 ≤ t ≤ 1,
has L([φ0]) > 0 and so L([φ0]) = 4π.
Min-max Theory
Theorem (Urbano, ’90)Σ ⊂ S3 compact embedded minimal surface with index(Σ) ≤ 5. Then either• Σ is a great sphere (index(Σ) = 1 and area(Σ) = 4π), or• Σ is the Clifford torus (index(Σ) = 5 and area(Σ) = 2π2).
(Almgren, ’62) The sweepout
φ0(t) = {x4 = 2t − 1} ∩ S3, 0 ≤ t ≤ 1,
has L([φ0]) > 0 and so L([φ0]) = 4π.
Min-max Theory
• T = {oriented great spheres} ≈ S3
• R = {oriented round spheres} ≈ S3 × [0, π]
Theorem (Marques-N.)Consider φ : I5 → Z2(S3) continuous with
1 φ(I4 × {0}) = φ(I4 × {1}) = 0;2 φ(∂I4 × I) ⊂ R and for x ∈ ∂I4, φ(x , t) ∈ T ⇐⇒ t = 1/2;3 the sweepout t 7→ φ(1/2,1/2,1/2,1/2,1/2, t) is non-trivial.
If φ : ∂I4 × {1/2} ≈ S3 → T ≈ S3 has non-zero degree then
L([φ]) > 4π = maxx∈∂I5
area(φ(x)).
Min-max Theory• T = {oriented great spheres} ≈ S3
• R = {oriented round spheres} ≈ S3 × [0, π]
Theorem (Marques-N.)Consider φ : I5 → Z2(S3) continuous with
1 φ(I4 × {0}) = φ(I4 × {1}) = 0;2 φ(∂I4 × I) ⊂ R and for x ∈ ∂I4, φ(x , t) ∈ T ⇐⇒ t = 1/2;3 the sweepout t 7→ φ(1/2,1/2,1/2,1/2,1/2, t) is non-trivial.
If φ : ∂I4 × {1/2} ≈ S3 → T ≈ S3 has non-zero degree then
L([φ]) > 4π = maxx∈∂I5
area(φ(x)).
Min-max Theory• T = {oriented great spheres} ≈ S3
• R = {oriented round spheres} ≈ S3 × [0, π]
Theorem (Marques-N.)Consider φ : I5 → Z2(S3) continuous with
1 φ(I4 × {0}) = φ(I4 × {1}) = 0;
2 φ(∂I4 × I) ⊂ R and for x ∈ ∂I4, φ(x , t) ∈ T ⇐⇒ t = 1/2;3 the sweepout t 7→ φ(1/2,1/2,1/2,1/2,1/2, t) is non-trivial.
If φ : ∂I4 × {1/2} ≈ S3 → T ≈ S3 has non-zero degree then
L([φ]) > 4π = maxx∈∂I5
area(φ(x)).
Min-max Theory• T = {oriented great spheres} ≈ S3
• R = {oriented round spheres} ≈ S3 × [0, π]
Theorem (Marques-N.)Consider φ : I5 → Z2(S3) continuous with
1 φ(I4 × {0}) = φ(I4 × {1}) = 0;2 φ(∂I4 × I) ⊂ R and for x ∈ ∂I4, φ(x , t) ∈ T ⇐⇒ t = 1/2;
3 the sweepout t 7→ φ(1/2,1/2,1/2,1/2,1/2, t) is non-trivial.
If φ : ∂I4 × {1/2} ≈ S3 → T ≈ S3 has non-zero degree then
L([φ]) > 4π = maxx∈∂I5
area(φ(x)).
Min-max Theory• T = {oriented great spheres} ≈ S3
• R = {oriented round spheres} ≈ S3 × [0, π]
Theorem (Marques-N.)Consider φ : I5 → Z2(S3) continuous with
1 φ(I4 × {0}) = φ(I4 × {1}) = 0;2 φ(∂I4 × I) ⊂ R and for x ∈ ∂I4, φ(x , t) ∈ T ⇐⇒ t = 1/2;3 the sweepout t 7→ φ(1/2,1/2,1/2,1/2,1/2, t) is non-trivial.
If φ : ∂I4 × {1/2} ≈ S3 → T ≈ S3 has non-zero degree then
L([φ]) > 4π = maxx∈∂I5
area(φ(x)).
Min-max Theory• T = {oriented great spheres} ≈ S3
• R = {oriented round spheres} ≈ S3 × [0, π]
Theorem (Marques-N.)Consider φ : I5 → Z2(S3) continuous with
1 φ(I4 × {0}) = φ(I4 × {1}) = 0;2 φ(∂I4 × I) ⊂ R and for x ∈ ∂I4, φ(x , t) ∈ T ⇐⇒ t = 1/2;3 the sweepout t 7→ φ(1/2,1/2,1/2,1/2,1/2, t) is non-trivial.
If φ : ∂I4 × {1/2} ≈ S3 → T ≈ S3 has non-zero degree then
L([φ]) > 4π = maxx∈∂I5
area(φ(x)).
Min-max Theory
CorollaryFor φ as in previous theorem, maxx∈I5 area(φ(x)) ≥ 2π2.
Idea:L([φ]) > max
x∈∂I5area(φ(x)) = 4π
(Pitts’ Thm) =⇒ ∃Σ minimal: L([φ]) = area(Σ) > 4π and index(Σ) ≤ 5
(Urbano’s Thm) =⇒ Σ = Clifford torus =⇒ maxx∈I5
area(φ(x)) ≥ L([φ]) = 2π2.
Min-max Theory
CorollaryFor φ as in previous theorem, maxx∈I5 area(φ(x)) ≥ 2π2.
Idea:L([φ]) > max
x∈∂I5area(φ(x)) = 4π
(Pitts’ Thm) =⇒ ∃Σ minimal: L([φ]) = area(Σ) > 4π and index(Σ) ≤ 5
(Urbano’s Thm) =⇒ Σ = Clifford torus =⇒ maxx∈I5
area(φ(x)) ≥ L([φ]) = 2π2.
Min-max Theory
CorollaryFor φ as in previous theorem, maxx∈I5 area(φ(x)) ≥ 2π2.
Idea:L([φ]) > max
x∈∂I5area(φ(x)) = 4π
(Pitts’ Thm) =⇒ ∃Σ minimal: L([φ]) = area(Σ) > 4π and index(Σ) ≤ 5
(Urbano’s Thm) =⇒ Σ = Clifford torus =⇒ maxx∈I5
area(φ(x)) ≥ L([φ]) = 2π2.
Min-max Theory
CorollaryFor φ as in previous theorem, maxx∈I5 area(φ(x)) ≥ 2π2.
Idea:L([φ]) > max
x∈∂I5area(φ(x)) = 4π
(Pitts’ Thm) =⇒ ∃Σ minimal: L([φ]) = area(Σ) > 4π and index(Σ) ≤ 5
(Urbano’s Thm) =⇒ Σ = Clifford torus =⇒ maxx∈I5
area(φ(x)) ≥ L([φ]) = 2π2.
Sketch of proof of Theorem:
• Assume L([φ]) = 4π = maxx∈Ik area (ψ(x)) for some ψ ∈ [φ].
• Set K = ψ−1(T ), where ∂K ⊂ ∂I4 × {1/2}.
(A) [∂K ] = [∂I4 × {1/2}] in H3(∂I4 × {1/2},Z)
φ = ψ on ∂K and ψ : K → T ≈ S3
=⇒ degree(φ|∂I4×{1/2}) = degree(ψ|∂K ) = 0
(B) [∂K ] = 0 in H3(∂I4 × {1/2},Z)ψ ◦ c non-trivial sweepout with max
t∈Iarea (ψ ◦ c(t)) ≤ max
x∈Ikarea (ψ(x)) = 4π
=⇒ ψ ◦ c(I) ∩ T 6= ∅ =⇒ c(I) ∩ K 6= ∅.
Sketch of proof of Theorem:
• Assume L([φ]) = 4π = maxx∈Ik area (ψ(x)) for some ψ ∈ [φ].• Set K = ψ−1(T ), where ∂K ⊂ ∂I4 × {1/2}.
(A) [∂K ] = [∂I4 × {1/2}] in H3(∂I4 × {1/2},Z)
φ = ψ on ∂K and ψ : K → T ≈ S3
=⇒ degree(φ|∂I4×{1/2}) = degree(ψ|∂K ) = 0
(B) [∂K ] = 0 in H3(∂I4 × {1/2},Z)ψ ◦ c non-trivial sweepout with max
t∈Iarea (ψ ◦ c(t)) ≤ max
x∈Ikarea (ψ(x)) = 4π
=⇒ ψ ◦ c(I) ∩ T 6= ∅ =⇒ c(I) ∩ K 6= ∅.
Sketch of proof of Theorem:
• Assume L([φ]) = 4π = maxx∈Ik area (ψ(x)) for some ψ ∈ [φ].• Set K = ψ−1(T ), where ∂K ⊂ ∂I4 × {1/2}.
(A) [∂K ] = [∂I4 × {1/2}] in H3(∂I4 × {1/2},Z)
φ = ψ on ∂K and ψ : K → T ≈ S3
=⇒ degree(φ|∂I4×{1/2}) = degree(ψ|∂K ) = 0
(B) [∂K ] = 0 in H3(∂I4 × {1/2},Z)ψ ◦ c non-trivial sweepout with max
t∈Iarea (ψ ◦ c(t)) ≤ max
x∈Ikarea (ψ(x)) = 4π
=⇒ ψ ◦ c(I) ∩ T 6= ∅ =⇒ c(I) ∩ K 6= ∅.
Sketch of proof of Theorem:
• Assume L([φ]) = 4π = maxx∈Ik area (ψ(x)) for some ψ ∈ [φ].• Set K = ψ−1(T ), where ∂K ⊂ ∂I4 × {1/2}.
(A) [∂K ] = [∂I4 × {1/2}] in H3(∂I4 × {1/2},Z)
φ = ψ on ∂K and ψ : K → T ≈ S3
=⇒ degree(φ|∂I4×{1/2}) = degree(ψ|∂K ) = 0
(B) [∂K ] = 0 in H3(∂I4 × {1/2},Z)ψ ◦ c non-trivial sweepout with max
t∈Iarea (ψ ◦ c(t)) ≤ max
x∈Ikarea (ψ(x)) = 4π
=⇒ ψ ◦ c(I) ∩ T 6= ∅ =⇒ c(I) ∩ K 6= ∅.
Sketch of proof of Theorem:
• Assume L([φ]) = 4π = maxx∈Ik area (ψ(x)) for some ψ ∈ [φ].• Set K = ψ−1(T ), where ∂K ⊂ ∂I4 × {1/2}.
(A) [∂K ] = [∂I4 × {1/2}] in H3(∂I4 × {1/2},Z)
φ = ψ on ∂K and ψ : K → T ≈ S3
=⇒ degree(φ|∂I4×{1/2}) = degree(ψ|∂K ) = 0
(B) [∂K ] = 0 in H3(∂I4 × {1/2},Z)
ψ ◦ c non-trivial sweepout with maxt∈I
area (ψ ◦ c(t)) ≤ maxx∈Ik
area (ψ(x)) = 4π
=⇒ ψ ◦ c(I) ∩ T 6= ∅ =⇒ c(I) ∩ K 6= ∅.
Sketch of proof of Theorem:
• Assume L([φ]) = 4π = maxx∈Ik area (ψ(x)) for some ψ ∈ [φ].• Set K = ψ−1(T ), where ∂K ⊂ ∂I4 × {1/2}.
(A) [∂K ] = [∂I4 × {1/2}] in H3(∂I4 × {1/2},Z)
φ = ψ on ∂K and ψ : K → T ≈ S3
=⇒ degree(φ|∂I4×{1/2}) = degree(ψ|∂K ) = 0
(B) [∂K ] = 0 in H3(∂I4 × {1/2},Z)ψ ◦ c non-trivial sweepout with max
t∈Iarea (ψ ◦ c(t)) ≤ max
x∈Ikarea (ψ(x)) = 4π
=⇒ ψ ◦ c(I) ∩ T 6= ∅ =⇒ c(I) ∩ K 6= ∅.
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)
3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!
6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Marques-N.)If Σ ⊂ S3 has positive genus thenW(Σ) ≥ 2π2.
1 Given v ∈ B4 consider Fv ∈ Conf (S3), x 7→ 1−|v |2
|x−v |2 (x − v)− v
2 C : B4 × [−π, π]→ Z2(S3), (v , t) 7→ surface at distance t from Fv (Σ)3 (Ros, 99) area(C(v , t)) ≤ W(C(v ,0)) =W(Fv (Σ)) =W(Σ)
4 Reparametrize C to obtain φ : B4 × [−π, π]→ Z2(S3) continuous with
• image(φ) = image(C);• φ(S3 × [−π, π]) ⊂ R and φ(x , t) ∈ T ⇐⇒ t = 0;• t 7→ φ(0, t), π ≤ t ≤ π, is non-trivial sweepout
5 degree(φ|S3×{0}) = genus of Σ!6 Corollary to our Min-max Thm can be applied and so
2π2 ≤ max(v ,t)∈B4×[−π,π]
area(φ(x)) = max(v ,t)∈B4×[−π,π]
area(C(v , t)) ≤ W(Σ)
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).
• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).
3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).
Theorem (Agol-Marques-N.)If (γ1, γ2) link in S3 has lk(γ1, γ2) = ±1, then E(γ1, γ2) ≥ 2π2.
• Given (σ1, σ2) ⊂ R4 consider G(σ1, σ2) : S1 × S1 → S3
G(γ1, γ2)(s, t) =(σ1(s)− σ2(t))|σ1(s)− σ2(t)|
.
• area(G(σ1, σ2)) ≤ E(γ1, γ2).• Given v ∈ B4, Fv (x) = x−v|x−v |2 ∈ Conf (R
4). Fv (B1(0)) = Br(v)(c(v)).
1 φ : B4 × [0,+∞]→ Z2(S3),
φ(v , λ) = G(Fv (γ1), λ(Fv (γ2)− c(v)) + c(v)).
2 area(φ(v , λ)) ≤ E(φ(v ,1)) = E(Fv (γ1),Fv (γ2)) = E(γ1, γ2).3 Given v ∈ S3,
φ(v ,1) = lk(γ1, γ2)∂Bπ/2(−v) =⇒ degree(φ|S3×{1}) = lk(γ1, γ2) 6= 0
4 Can apply our Min-max Thm to conclude
2π2 ≤ max(v ,t)∈B4×[0,+∞] area(φ(v , t)) ≤ E(γ1, γ2).