Helicoidal Surfaces in S2R (joint work with Brian White)Helicoidal Surfaces in S2 ×R (joint work with Brian White) David Hoﬀman, Stanford University Sevilla, 7 abril 2011 David

Helicoidal Surfaces in S2 × R(joint work with Brian White)

David Hoffman, Stanford University

Sevilla, 7 abril 2011

David Hoffman, Stanford University Helicoidal Surfaces in S2 × R (joint work with Brian White)

Helicoids in R3

The helicoid, H, is a minimal surface in R3 with the followingproperties.

! H contains a vertical line (the Axis, which we will denote byZ );

! H is ruled by horizontal lines;

! H is invariant under the screw motion σκt,t for some nonzeroreal number κ.


Helicoids in R3






Helicoids in R3






Helicoids in R3

! For some fixed real number κ,H is invariant under the screw motion σκt,t

which is rotation around Z by an angle κt, followed byvertical translation by t.

! The pitch of H is twice the distance between successive sheetsand equals 2π/κ.

! Pitch =∞ when κ = 0, and the surface is a vertical plane! Pitch → 0 as κ→∞ and the limit is a lamination by

horizontal planes with curvature blowing up on Z .


Helicoids in R3







Helicoids in R3







Helicoids in R3




! Pitch =∞ when κ = 0, and the surface is a vertical plane

! Pitch → 0 as κ→∞ and the limit is a lamination byhorizontal planes with curvature blowing up on Z .


Helicoids in R3





horizontal planes with curvature blowing up on Z .David Hoffman, Stanford University Helicoidal Surfaces in S2 × R (joint work with Brian White)

Helicoids H in S2 × R

! Let X be a great circle in S2 × 0.

! Choose antipodal points O and O∗ on X , and letZ = O × RandZ ∗ = O∗ × R.Z and Z ∗ will be the axes of a helicoid

! Fix a value of κ.

H = Hκ =⋃

t∈R

σκt,t(X )

where σκt,t is rotation by κt followed by vertical translationby t



! Let X be a great circle in S2 × 0.! Choose antipodal points O and O∗ on X , and let

Z = O × RandZ ∗ = O∗ × R.

Z and Z ∗ will be the axes of a helicoid


H = Hκ =⋃

t∈R

σκt,t(X )





Z = O × RandZ ∗ = O∗ × R.Z and Z ∗ will be the axes of a helicoid


H = Hκ =⋃

t∈R

σκt,t(X )





Z = O × RandZ ∗ = O∗ × R.Z and Z ∗ will be the axes of a helicoid


H = Hκ =⋃

t∈R

σκt,t(X )



Properties of Helicoids H in S2 × R

! They are properly embedded annuli

! They are minimal surfaces

! Helicoids with different pitch (i.e. different values of |κ|) arenot homothetically related as in R3. (There are nohomotheties in S2 × R.)

! When κ = 0 we get a flat cylinder X × R

! As κ→∞ we get a lamination of S2 × R by level spheresS2 × t with curvature blowing up on the axes Z ∪ Z ∗.






























More Properties of Helicoids H in S2 × R

Z Z*

X

0

0*Y+

Tc

T-c! Let Y be the great circle in S2 × O that passes through thepoints O and O∗ and is orthogonal to X .

180-rotation about Y is an isometry of H.

! This symmetry is referred to as a normal symmetry.

! H has the symmetry of reflection through a vertical cylinderC × 0, This symmetry does not exist in the Euclideansetting.



Z Z*

X

0

0*Y+

Tc







Z Z*

X

0

0*Y+

Tc






Properly embedded minimal surfaces in S2 × Rwith finite topology

Let M be a properly embedded minimal surface with finitetopology in S2 × R.

! If M is compact, then M = S2 × t.! (Rosenberg) All noncompact M have two ends, both annular,

one diverging upwards, the other downwards.

! (Meeks-Rosenberg) The Gauss curvature of M is bounded.

! (Hauswirth) Suppose E is an annulus (genus=0) fibred byhorizontal circles. Then E belongs to a two-parameter familyof examples on the boundary of which reside the helicoids. Itcontains examples constructed by Ritore and Pedrosa.




! If M is compact, then M = S2 × t.

! (Rosenberg) All noncompact M have two ends, both annular,one diverging upwards, the other downwards.


























Theorem. (Hauswirth) Suppose E is an annulus (genus=0) fibredby horizontal circles. Then E belongs to a two-parameter family ofexamples on the boundary of which reside the helicoids.

QUESTION

! Is it true that each end of M asymptotic to one of theseannuli?

ANSWER (partial)

(H—,White) If M contains a vertical line, then the answer is ”Yes.”




QUESTION


ANSWER (partial)





QUESTION


ANSWER (partial)





QUESTION


ANSWER (partial)





QUESTION


ANSWER (partial)



Properly embedded minimal surfaces of finite topology inS2 × R of higher genus

Let κ and g ≥ 1 be given. Let µ be reflection in the verticalcylinder C × R.

Theorem (—–,White)

There exists a pair of properly embedded, genus = g, embeddedaxial minimal surfaces, M+1,M−1, each of whose ends isasymptotic to a helicoid with pitch 2π/κ.

If g is odd, µ(M+1) = M−1.

If g is even, µ(M+1) = M+1 and µ(M−1) = M+1, but M+1,M−1

are not congruent.

*Recent result of Pacard & Rosenberg: STATE IT


Axial minimal surfaces in S2 × R are helicoidal

Definition. A properly embedded minimal surface of finite topologyis Axial provided it contains a vertical line.

Axial Minimal Surfaces.

Theorem (H—,White) Suppose M is an axial minimal surface

i) If E is an annular end of M, then E is asymptotic to a helicoid.

ii.) If M is an annulus, then M is a helicoid

iii.) The annular ends of M are both asymptotic to helicoids withthe same pitch. (However, the two ends may not be ends of thesame helicoid Hκ.)



































Proposition. An axial minimal surface M contains two verticallines.

Proof. S2 × R \M consists of two components, M+, and M−.Suppose Z = O× R is contained in M. Define

Z ∗ = O′× R.

We claim that Z ∗ ⊂ M. Define

ρZ = rotation by π around Z .

The isometry ρZ fixes both Z and Z ∗. It also interchanges M+

and M−.Hence no point of M+ ∪M− can be on Z ∗.Therefore Z ∗ ⊂ M.



Proposition. An axial minimal surface M contains two verticallines.Proof. S2 × R \M consists of two components, M+, and M−.

Suppose Z = O× R is contained in M. Define

Z ∗ = O′× R.







Proposition. An axial minimal surface M contains two verticallines.Proof. S2 × R \M consists of two components, M+, and M−.Suppose Z = O× R is contained in M. Define

Z ∗ = O′× R.








Z ∗ = O′× R.








Z ∗ = O′× R.




and M−.

Hence no point of M+ ∪M− can be on Z ∗.Therefore Z ∗ ⊂ M.




Z ∗ = O′× R.






The angle function θ on S2 × R \ (Z ∪ Z ∗)

OnS2 × 0 \ O,O∗,

define θ by first identifying S2 × 0 \ O,O∗ with the complexplane via stereographic projection from O∗ , and then choosing θon this sphere to be the angle function in the plane.Extend θ to

S2 × R \ (Z ∪ Z ∗)

by vertical translation.

The function θ is defined up to an additive multiple of 2π


The angle function θ on S2 × R \ (Z ∪ Z ∗)

OnS2 × 0 \ O,O∗,

define θ by first identifying S2 × 0 \ O,O∗ with the complexplane via stereographic projection from O∗ , and then choosing θon this sphere to be the angle function in the plane.Extend θ to

S2 × R \ (Z ∪ Z ∗)

by vertical translation.

The function θ is defined up to an additive multiple of 2π


Axial Minimal surfaces: a single-valued angle function

Let I ⊂ R be a closed interval (possibly infinite).

Suppose that

E = M ∩ (S2 × I )

is an annulus. ThenE \ (Z ∪ Z ∗)

is the union of two simply connected domains congruent under theinvolution ρZ . Let D be one of them.

The function θ has a single-valued branch that is well defined on D.This θ extends continuously to D.

DEFINE

α(t) = maxθ(p, t) : (p, t) ∈ D ∩ S2 × t,β(t) = minθ(p, t) : (p, t) ∈ D ∩ S2 × t,

φ(t) = α(t)− β(t).



Let I ⊂ R be a closed interval (possibly infinite).Suppose that

E = M ∩ (S2 × I )

is an annulus.

ThenE \ (Z ∪ Z ∗)



DEFINE


φ(t) = α(t)− β(t).




E = M ∩ (S2 × I )


is the union of two simply connected domains congruent under theinvolution ρZ .

Let D be one of them.


DEFINE


φ(t) = α(t)− β(t).




E = M ∩ (S2 × I )



The function θ has a single-valued branch that is well defined on D.

This θ extends continuously to D.

DEFINE


φ(t) = α(t)− β(t).




E = M ∩ (S2 × I )




DEFINE


φ(t) = α(t)− β(t).




E = M ∩ (S2 × I )




DEFINE


φ(t) = α(t)− β(t).


Axial Minimal surfaces: the angle functions

α(t) = maxθ(p, t) : (p, t) ∈ D ∩ S2 × t,

β(t) = minθ(p, t) : (p, t) ∈ D ∩ S2 × t,

φ(t) = α(t)− β(t).

Observe that for a fixed value of t,

α(t) = β(t) ⇐⇒ D ∩ (S2 × t) is a great semicircle.

Observe that E is a part of a helicoid if and only if

α(t) ≡ β(t) ≡ κt + b,

for some real constants κ and b.


Axial Minimal surfaces: the angle functions

α(t) = maxθ(p, t) : (p, t) ∈ D ∩ S2 × t,

β(t) = minθ(p, t) : (p, t) ∈ D ∩ S2 × t,

φ(t) = α(t)− β(t).

Observe that for a fixed value of t,

α(t) = β(t) ⇐⇒ D ∩ (S2 × t) is a great semicircle.

Observe that E is a part of a helicoid if and only if

α(t) ≡ β(t) ≡ κt + b,

for some real constants κ and b.


Convexity Lemma for the angle functions

α(t) = maxθ(p, t) : (p, t) ∈ D ∩ S2 × tβ(t) = minθ(p, t) : (p, t) ∈ D ∩ S2 × t,

————————–

Lemma. i.) α(t) and −β(t) are both strictly convex functionsunless D is part of a half-helicoid (in which case they are linear).

ii.)φ(t) = α(t)− β(t) is either strictly convex or D is part of ahalf-helicoid, in which case φ(t) ≡ 0.

Proof. We prove this for α(t). The same proof works for −β(t).φ(t) is the sum of convex functions, and so is strictly convex if oneof the functions is strictly convex. If they are not both strictlyconvex, then D lies on a half-helicoid and therefore

φ(t) ≡ 0.




————————–




φ(t) ≡ 0.




————————–




φ(t) ≡ 0.


Convexity Lemma for the the angle functions

α(t) = maxθ(p, t) : (p, t) ∈ D ∩ S2 × t

————————–

If α(t) is not strictly convex, there exists values t1 < t2, such thatfor some value t1 ≤ t3 ≤ t2,

α(t3) ≥ κ(t3 − t1) + α(t1),

where κ is the slope of the line connecting (t1,α(t1)) to(t2,α(t2)). It follows that there is a parallel line of the form

y = κt + b,

that touches the graph at a point (t∗,α(t∗)) for some t∗ ∈ (t1, t2)and is above the graph of α(t) over some interval (t∗ − ε, t∗ + ε).


Convexity of the angle functions

α(t) = maxθ(p, t) : (p, t) ∈ D ∩ S2 × tIt follows that there is a parallel line of the form

y = κt + b,


————————–

Let (p∗, t∗) ∈ S2 × t∗ be the point where θ(p∗, t∗) = α(t∗).It follows from the definition of α(t) that in some neighborhood of(p∗, t∗) in D, the half-helicoid

H represented by y = κt + b

lies on one side of D and touches D at (p∗, t∗), the point whereθ(p∗, t∗) = α(t∗). By the maximum principle (or the boundarymaximum principle) D ⊂ H.




y = κt + b,


————————–

Let (p∗, t∗) ∈ S2 × t∗ be the point where θ(p∗, t∗) = α(t∗).

It follows from the definition of α(t) that in some neighborhood of(p∗, t∗) in D, the half-helicoid






y = κt + b,


————————–



lies on one side of D and touches D at (p∗, t∗), the point whereθ(p∗, t∗) = α(t∗).

By the maximum principle (or the boundarymaximum principle) D ⊂ H.




y = κt + b,


————————–





Proof of the first statment of the theorem


Assume E ⊂ S2 × [a,∞). As usual let D be one of the simplyconnected regions defined by removing the axes from E . Define

c = lim supt→∞

φ(t).

Let tn be a diverging sequence with c = limn→∞ φ(tn).

Pull down E and D by a translation of −tn to produce En and Dn.We get subsequential convergence

En → E Dn → D with φ(t + tn) =: φn(t)→ φ(t).

where φn(t) and φ(t) are the angle-difference functions associatedwith Dn and D.



i) If E is an annular end of M, then E is asymptotic to a helicoid.Assume E ⊂ S2 × [a,∞). As usual let D be one of the simplyconnected regions defined by removing the axes from E .

Define

c = lim supt→∞

φ(t).







i) If E is an annular end of M, then E is asymptotic to a helicoid.Assume E ⊂ S2 × [a,∞). As usual let D be one of the simplyconnected regions defined by removing the axes from E . Define

c = lim supt→∞

φ(t).







i) If E is an annular end of M, then E is asymptotic to a helicoid.Assume E ⊂ S2 × [a,∞). As usual let D be one of the simplyconnected regions defined by removing the axes from E . Define

c = lim supt→∞

φ(t).






Proof of the first statment of the theorem continued


c = lim supt→∞

φ(t) ≥ limn→∞

φ(t + tn) =: limn→∞

φn(t)→ φ(t),

where φn(t) and φ(t) are the angle-difference functions associatedwith Dn and D. The equality above implies: φ(0) = c . Apply the

lemma to D, noticing that since 0 ≤ φ(t) ≤ c , we must haveφ ≡ 0. In particular,

c = 0.


Proof of the first statment of the theorem concluded


c = lim supt→∞

φ(t) = 0

Let A+ be the open region in the plane above the graph of α.Let B− be the open region in the plane below the graph of β.These are disjoint convex regions and therefore can be separatedby a line

y = κt + b.

Since α(t)− β(t)→ c = 0 as t →∞, the graphs of α and β areasymptotic to the line y = κt + b.

Therefore, D is C 0-asymptotic to the half-helicoid given byθ = κt + b.Because D has bounded curvature, it is smoothly asymptotic tothis helicoid.




c = lim supt→∞

φ(t) = 0


y = κt + b.

Since α(t)− β(t)→ c = 0 as t →∞, the graphs of α and β areasymptotic to the line y = κt + b.Therefore, D is C 0-asymptotic to the half-helicoid given byθ = κt + b.

Because D has bounded curvature, it is smoothly asymptotic tothis helicoid.




c = lim supt→∞

φ(t) = 0


y = κt + b.

Since α(t)− β(t)→ c = 0 as t →∞, the graphs of α and β areasymptotic to the line y = κt + b.Therefore, D is C 0-asymptotic to the half-helicoid given byθ = κt + b.Because D has bounded curvature, it is smoothly asymptotic tothis helicoid.


Proof of the second statment of the theorem

ii) If M is an annulus, then M is a helicoid.Let D be one of the simply connected components M \ (Z ∪ Z ∗)The angle-difference function φ is defined on all of R and isnonnegative.By the Lemma, either φ is strictly convex, D must lie on ahalf-helicoid. Apply the first statement of the theorem to concludethat

limt→±∞

φ(t) = 0.

Hence φ cannot be strictly convex, so D is a half-helicoid, and

M = D ∪ ρz(D)

is a helicoid.


Proof of the third statment of the theorem

ii) The helicoids to which the two ends of M are asymptotic differby a rotationLet ∂/∂θ be the Killiing field of rotation around Z ∪ Z ∗.Let Ma = M ∩ (S2 × a). The ROTATIONAL FLUX of M

∫

Ma

< ∂/∂θ, ν > ds,

where ν is the outward-pointing normal to M ∩ (S2 × t ≤ a), isindependent of a.The helicoids asymptotic to the top and bottom ends of M, havethe same rotational flux (equal to the rotational flux of M).

Two helicoids θ = κ1 + b1(mod π) and θ = κ2 + b2(mod π) havethe same rotational flux if and only if κ1 = κ2.


The vertical flux of a helicoid is determined by itspitch=2π/κ

Two helicoids θ = κ1 + b1(mod π) and θ = κ2 + b2(mod π) havethe same rotational flux if and only if κ1 = κ2.The vertical flux of the helicoid H given by θ = κt + b (mod π) is

∫

C< ∂/∂θ, ν > ds,

where C is great circle H ∩ (S2 × 0)∂/∂θ is tangent to S2 × 0 (and orthogonal to C ).At each point of C \ O,O∗ the angle that the ν makes withS2 × 0 is a monotonic increasing function of the|pitch| = 2π/|κ|. Hence the integrand is a monotonic decreasingfunction of the absolute value of the pitch. The vertical fluxes ofhelicoids with different values κ are different.


Why do we use rotational flux and not vertical flux?

Why Rotational Flux and not Vertical Flux∫

Ma

< ∂/∂t, ν > ds ?

The vertical flux will not distinguish between helicoids Hκ and H−κ.The rotational flux does make that distinction.


Why do we use rotational flux and not vertical flux?

Why Rotational Flux and not Vertical Flux∫

Ma

< ∂/∂t, ν > ds ?

The vertical flux will not distinguish between helicoids Hκ and H−κ.The rotational flux does make that distinction.


Why aren’t the two ends of M asymptotic to the samehelicoid?

We are not claiming that the helicoids asymptotic to the ends ofM differ by a rotation.

After all this is not true if M is a helicoid!However there is good reason to believe that in general they dodiffer by a rotation.A rotation by γ is equivalent to a vertical translation by b = γ/κ.The examples of higher genus we mentioned at the beginning ofthe lecture (and will discuss soon) can be thought of asdesingularizing the intersection of S2 × 0 and a helicoid H. .Such desingularization might well cause a vertical separation of thetop and bottom of ends of H in order to “make room” for handles.



We are not claiming that the helicoids asymptotic to the ends ofM differ by a rotation. After all this is not true if M is a helicoid!

However there is good reason to believe that in general they dodiffer by a rotation.A rotation by γ is equivalent to a vertical translation by b = γ/κ.The examples of higher genus we mentioned at the beginning ofthe lecture (and will discuss soon) can be thought of asdesingularizing the intersection of S2 × 0 and a helicoid H. .Such desingularization might well cause a vertical separation of thetop and bottom of ends of H in order to “make room” for handles.



We are not claiming that the helicoids asymptotic to the ends ofM differ by a rotation. After all this is not true if M is a helicoid!However there is good reason to believe that in general they dodiffer by a rotation.A rotation by γ is equivalent to a vertical translation by b = γ/κ.

The examples of higher genus we mentioned at the beginning ofthe lecture (and will discuss soon) can be thought of asdesingularizing the intersection of S2 × 0 and a helicoid H. .Such desingularization might well cause a vertical separation of thetop and bottom of ends of H in order to “make room” for handles.



We are not claiming that the helicoids asymptotic to the ends ofM differ by a rotation. After all this is not true if M is a helicoid!However there is good reason to believe that in general they dodiffer by a rotation.A rotation by γ is equivalent to a vertical translation by b = γ/κ.The examples of higher genus we mentioned at the beginning ofthe lecture (and will discuss soon) can be thought of asdesingularizing the intersection of S2 × 0 and a helicoid H.

.Such desingularization might well cause a vertical separation of thetop and bottom of ends of H in order to “make room” for handles.



We are not claiming that the helicoids asymptotic to the ends ofM differ by a rotation. After all this is not true if M is a helicoid!However there is good reason to believe that in general they dodiffer by a rotation.A rotation by γ is equivalent to a vertical translation by b = γ/κ.The examples of higher genus we mentioned at the beginning ofthe lecture (and will discuss soon) can be thought of asdesingularizing the intersection of S2 × 0 and a helicoid H. .Such desingularization might well cause a vertical separation of thetop and bottom of ends of H in order to “make room” for handles.


Higher genus helicoidal minimal surfaces in S2 × R

Z Z*

X

0

0*Y+

Tc

T-c

THECONSRUCTION PROCEDURE

! Y -Surfaces: Symmetry

! Helicoidal Barrier: A Y -symmetric, mean-convex region

! Double the surface by rotating about the axes

! Degree Theory with symmetries.

! Curvature Estimates in order to get limit surface.



Z Z*

X

0

0*Y+

Tc

T-c









Z Z*

X

0

0*Y+

Tc

T-c









Z Z*

X

0

0*Y+

Tc

T-c









Z Z*

X

0

0*Y+

Tc

T-c









Z Z*

X

0

0*Y+

Tc

T-c








Y -surfaces

Consider a connected and open surface Σ, (possibly the interior ofa surface with boundary) in a three-manifold N.Suppose that Σ is invariant under an isometric involution ρ of Nthat preserves orientation.

! Here are three possible properties of ρ:

! 1. ρ∗ acts on the homology of Σ by multiplication by −1.

! 2. The quotient surface Σ/ρ is a disk.

! 3. If k is the number of fixed points of ρ in Σ, then

k = 2− χ(Σ).

! Proposition. The three conditions are equivalent.


Y -surfaces






k = 2− χ(Σ).



Y -surfaces






k = 2− χ(Σ).



Y -surfaces






k = 2− χ(Σ).



Y -surfaces






k = 2− χ(Σ).



Y -surfaces






k = 2− χ(Σ).



Why the name Y -surface?

Consider a connected and open minimal surface Σ, (possibly theinterior of a surface with boundary) in a three-manifold N.Suppose that Σ is invariant under the isometric involution, ρY , ofN, consisting of 180-rotation about a geodesic Y .

! The fixed points of ρY acting on Σ are the points in Y ∩ Σ.

! Definition. Σ is a Y -surface if it satisfies one of the equivalentconditions 1., 2. or 3.

! Example. The helicoid in R3 (or S2 × R), with Y the normalline (or great circle).














“Proof” by example.

Y

! By “inspection,” ρY acts by multiplication by −1 onhomology.

! |Y ∩ Σ| = 82− χ(Σ) = 2− (2− 2genus(Σ)− ends(Σ)) =2 genus(Σ) + ends(Σ) = 6 + 2.



Y Y





Y Y





Y Y


! |Y ∩ Σ| = 8

2− χ(Σ) = 2− (2− 2genus(Σ)− ends(Σ)) =2 genus(Σ) + ends(Σ) = 6 + 2.



Y Y


! |Y ∩ Σ| = 82− χ(Σ)

= 2− (2− 2genus(Σ)− ends(Σ)) =2 genus(Σ) + ends(Σ) = 6 + 2.



Y Y


! |Y ∩ Σ| = 82− χ(Σ) = 2− (2− 2genus(Σ)− ends(Σ))

=2 genus(Σ) + ends(Σ) = 6 + 2.



Y Y


! |Y ∩ Σ| = 82− χ(Σ) = 2− (2− 2genus(Σ)− ends(Σ)) =2 genus(Σ) + ends(Σ)

= 6 + 2.



Y Y





Y

The quotient Σ/σY is a disk.



Y Y




Y Y



Helicoidal barrier: a Y -symmetric mean convex region

! Let H ⊂ S2 × R be a helicoid.

! H divides S2 × R into two connected regions. Label one ofthem H+.

! Fix a value c > 0. Then

N = Nc = (p, t) ∈ S2 × R | p ∈ H+, |t| < c..

! ∂N consists of part of H and totally geodesic hemispheres atthe levels ±c . Therefore N is mean convex.



! Let H ⊂ S2 × R be a helicoid.! H divides S2 × R into two connected regions. Label one of

them H+.

! Fix a value c > 0. Then

N = Nc = (p, t) ∈ S2 × R | p ∈ H+, |t| < c..





them H+.! Fix a value c > 0. Then

N = Nc = (p, t) ∈ S2 × R | p ∈ H+, |t| < c.

.! ∂N consists of part of H and totally geodesic hemispheres at

the levels ±c . Therefore N is mean convex.




them H+.! Fix a value c > 0. Then

N = Nc = (p, t) ∈ S2 × R | p ∈ H+, |t| < c..


Z Z*

X

0

0*Y+

Tc

T-c


Constructing the desired surface of desired genus =k

Z Z*

X

0

0*Y+

Tc

T-c

Suppose S ⊂ N is a Y -surface with boundary Γ.Suppose |Y ∩ S | = k(We have not yet indicated how to do this...)Extend S by rotation about Z to produce M.M has two boundary components, they arefull circles at levels ±c .

M is also a Y -surface, and|Y ∩M| = |Y ∩ S |+ |Y ∩ ρZ (S)|+ |O,O∗|

= k + k + 2 = 2k + 2

|Y ∩M| = 2− χ(M) = 2− (2− 2genus(M)− ends(M))

or2k + 2 = 2genus(M)) + 2.



Z Z*

X

0

0*Y+

Tc

T-c

Suppose S ⊂ N is a Y -surface with boundary Γ.Suppose |Y ∩ S | = k(We have not yet indicated how to do this...)

Extend S by rotation about Z to produce M.M has two boundary components, they arefull circles at levels ±c .


= k + k + 2 = 2k + 2


or2k + 2 = 2genus(M)) + 2.



Z Z*

X

0

0*Y+

Tc

T-c



= k + k + 2 = 2k + 2


or2k + 2 = 2genus(M)) + 2.



Z Z*

X

0

0*Y+

Tc

T-c



= k + k + 2 = 2k + 2


or2k + 2 = 2genus(M)) + 2.



Z Z*

X

0

0*Y+

Tc

T-c



= k + k + 2 = 2k + 2

|Y ∩M| = 2− χ(M)

= 2− (2− 2genus(M)− ends(M))

or2k + 2 = 2genus(M)) + 2.



Z Z*

X

0

0*Y+

Tc

T-c



= k + k + 2 = 2k + 2


or2k + 2 = 2genus(M)) + 2.



Z Z*

X

0

0*Y+

Tc

T-c



= k + k + 2 = 2k + 2


or2k + 2 = 2genus(M)) + 2.


Producing the desired example of genus = k by taking alimit as c →∞

We have produced a genus = k surface in S2 × (−c , c) thatcontains the axes Z ∪ Z ∗ and has Y -symmetry. We prove uniformcurvature estimates for

M = Mc ⊂ S2 × (−c , c)

and take subsequential limits as cn →∞.

! The limit surface M contains the axes, so each end must beasymptotic to helicoid—in fact the helicoid H used as thebarrier, or a rotation of H.

! M ∩H = Z ∪ Z ∗ ∪ X .

! Each Mc is a Y -surface, so M is a Y -surface

! The intersection of each Mc with Y contains 2k + 2 points,so the same is true for M. Hence genus(M) = k


Counting minimal surfaces.

Let Σ be a fixed compact 2-manifold with boundary. andLet N be a compact three-manifold with boundary.

If Γ is an embedded curve in ∂N diffeomorphic to ∂Σ, we make thefollowing definitions:

Definition. M∗(N, Γ) denotes the set of embedded minimalsurfaces in N that are diffeomorphic to Σ and that have boundaryΓ.

Definition. |M∗(N, Γ)| denote the number of surfaces inM∗(N, Γ).




















Counting G-invariant minimal surfaces, mod 2.

Fix Σ.

Theorem (White)

Let N be a smooth, compact, mean convex riemannian 3-manifoldthat is homeomorphic to a ball, that has piecewise smoothboundary, and that contains no closed minimal surfaces. Let G bea group of isometries of N. Let Γ ⊂ ∂N be a smooth curve that isG-invariant and G-bumpy.Suppose that no two contiguous connected components of(∂N) \ Γ form a smooth, G-invariant minimal surface.Suppose also that Γ = ∂Ω for some G-invariant region Ω ⊂ ∂N. ∗∗

Then |M∗G (N, Γ)| is even unless Σ is a union of disks, in which

case |M∗G (N, Γ)| is odd.

The subscript “G” in M∗G (N, Γ) means we are looking only at

G -invariant surfaces.

∗∗ We need this condition on Γ in order to be able to deform it toa point in a G -invariant fashion.


Counting G-invariant minimal surfaces, mod 2.

Fix Σ.

Theorem (White)

Let N be a smooth, compact, mean convex riemannian 3-manifoldthat is homeomorphic to a ball, that has piecewise smoothboundary, and that contains no closed minimal surfaces. Let G bea group of isometries of N. Let Γ ⊂ ∂N be a smooth curve that isG-invariant and G-bumpy.Suppose that no two contiguous connected components of(∂N) \ Γ form a smooth, G-invariant minimal surface.Suppose also that Γ = ∂Ω for some G-invariant region Ω ⊂ ∂N. ∗∗

Then |M∗G (N, Γ)| is even unless Σ is a union of disks, in which

case |M∗G (N, Γ)| is odd.

The subscript “G” in M∗G (N, Γ) means we are looking only at

G -invariant surfaces.∗∗ We need this condition on Γ in order to be able to deform it toa point in a G -invariant fashion.


Higher genus helicoids in S2 × R: Orienting Γ.

Z Z*

X

0

0*Y+

Tc

T-c

Near the point O, the axis Z is oriented.

We may orient Y (not on Γ) by the choice of H+. This allows usto orient X . We can now talk about positive and negativequadrants at O and at O∗.



Z Z*

X

0

0*Y+

Tc

T-c

Near the point O, the axis Z is oriented.We may orient Y (not on Γ) by the choice of H+.

This allows usto orient X . We can now talk about positive and negativequadrants at O and at O∗.



Z Z*

X

0

0*Y+

Tc

T-c

Near the point O, the axis Z is oriented.We may orient Y (not on Γ) by the choice of H+. This allows usto orient X .

We can now talk about positive and negativequadrants at O and at O∗.



Z Z*

X

0

0*Y+

Tc

T-c

Near the point O, the axis Z is oriented.We may orient Y (not on Γ) by the choice of H+. This allows usto orient X . We can now talk about positive and negativequadrants at O and at O∗.


Higher genus helicoids in S2 × R: Positive Roundings of Γ

0

0

0*

0*

! When k is even (left) we connect positively at O∗ to producetwo components.

! When k is odd (right), we connect negatively at O∗ toproduce one component.



0

0

0*

0*





0

0

0*

0*




Higher genus helicoids in S2 × R: The fundamentalrelationship

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

Suppose k = 0 or k = 1.This is the case when Σ is either two disks or one disk.

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ 0

By Counting Mod Two Theorem:

1 ∼= |M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|



|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|


|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ 0


1 ∼= |M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|



|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|


|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ 0


1 ∼= |M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|



|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|


|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ 0


1 ∼= |M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|



|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|


|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ 0


1 ∼= |M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|



|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|


|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ |M∗Y (Γ, k − 2,−)|

|M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|+ 0


1 ∼= |M∗Y (Γt , k)| = |M∗

Y (Γ, k,+)|


Documents

Helicoidal Surfaces in S2R (joint work with Brian White)Helicoidal Surfaces in S2 ×R (joint work with Brian White) David Hoﬀman, Stanford University Sevilla, 7 abril 2011 David