The Isoperimetric Problem in the Heisenberg group H

(1)

The Isoperimetric Problem in the Heisenberg group H

ⁿ

First Taiwan Geometry Symposium, NCTS South

November 20, 2010

First Taiwan Geometry Symposium, NCTS South ()The Isoperimetric Problem in the Heisenberg groupHn November 20, 2010 1 / 44

(2)

Talk outline

The Euclidean Isoperimetric Problem.

Relevant terminology, conepts, definitions in/of the Heisenberg group Hⁿ.

conjecture.

Existence of the Isoperimetric Profile in Carnot-Groups: Leonardi-Rigot.

Smooth Cylindrical Case and some properties of the Isoperimetric profile: Leonardi-Masnou, Ritoré-Rosales.

Partial Symmetry Case: Danielli-Garofalo-Nhieu.

The C²solution to the Isoperimetric problem inH¹: Ritore-Rosales. The non-smooth cases: Monti, Monti-Rickly.

An improvement of Danielli-Garofalo-Nhieu due to Ritoré: calibration argument.

(3)

Talk outline

The Isoperimetric Problem inHⁿ–Pansu’s Isoperimetric inequality and conjecture.

Existence of the Isoperimetric Profile in Carnot-Groups: Leonardi-Rigot.

(4)

Talk outline

Existence of the Isoperimetric Profile in Carnot-Groups:

Leonardi-Rigot.

profile: Leonardi-Masnou, Ritoré-Rosales. Partial Symmetry Case: Danielli-Garofalo-Nhieu.

(5)

Talk outline

Leonardi-Rigot.

(6)

Talk outline

Leonardi-Rigot.

The non-smooth cases: Monti, Monti-Rickly.

(7)

Talk outline

Leonardi-Rigot.

The C²solution to the Isoperimetric problem inH¹: Ritore-Rosales.

(8)

Talk outline

Leonardi-Rigot.

argument.

(9)

Talk outline

Leonardi-Rigot.

(10)

Leonardi-Rigot.

(11)

We begin with the following folklore which attributed the Isoperimetric Problem to Queen Dido, founder of the city of Carthage in North Africa.

Figure:Dido, Queen of Carthage. Engraving by Mathäus Merian the Elder 1630.

According to Virgil’s saga “Fleeing the vengeance of her brother, Dido (356-260 BC) lands on the coast of North Africa. For the bargain which Dido agrees to with a local potentate is this: she may have that portion of land which she is able to enclose with the hide of a bull. She then cut the hide into a seris of long thin strips and marked out a vast circumference. This area then eventually became the city of Carthage”.

(12)

Queen Dido’s problem/solution is a variant of what is now known as isoperimetric type problems. In more precise term, Dido’s problem is formulated as follows.

Among all bounded, connected open regions in the plane with a fixed perimeter, determine the one(s) that has the maximum volume.

connected open regions in the plane with a fixed volume, determine the one(s) that has the minimum perimeter.

Dido’s solution is correct: (although part of her region is bounded by a sea shore which we assume it to be a straight line when compared to the relative length of the bull’s hide): the extremal regions are precisely one half of the open circular planar discs.

(13)

The above problem is also equivalent to: Among all bounded, connected open regions in the plane with a fixed volume, determine the one(s) that has the minimum perimeter.

(14)

(15)

(16)

Over the centuries, the isoperimetric problem (in various forms) has stimulated substantial mathematical research in numerous areas:

Geometric measure theory: The precise setting for the study of classical questions in the calculus of variations and the proof of existence of an isoperimetric profile. The tools are compactness theorems for BV functions. Consequently, a priori solutions are only guaranteed within the class of sets of finite perimeter.

of constant mean curvature. The classification of such surfaces provides a characterization of isoperimetric profiles.

(17)

Differential Geometry: (Smooth) isoperimetric solutions are surfaces of constant mean curvature. The classification of such surfaces provides a characterization of isoperimetric profiles.

(18)

Differential Geometry: (Smooth) isoperimetric solutions are surfaces of constant mean curvature. The classification of such surfaces provides a characterization of isoperimetric profiles.

(19)

PDE: The introduction of dynamic altorihms of volume-constrained curvature flows which provides a way to smoothly deform a given region so that the isoperimetric ratio P(E)ⁿ⁻¹ⁿ /|E| cecreases monotonically. If the flow exists for all time, the deformed regions converge, in a suitable sense, to a solution of the isoperimetric problem.

Functional Analysis: An equivalent way of formulating the isoperimetric problem consists in viewing it as a best constant problem for a Sobolev inequality, relating mean values of a given smooth function with those of its derivatives.

Geometric function theory: Symmetrization procedures that replace a given mathematical object or region with one admitting a larger symmetry group while retaining certain properties.

(20)

(21)

(22)

We recall the classical isoperimetric inequality in the Euclidean space.

Theorem 1

For every Borel setΩ ⊂ Rⁿ, n ≥ 2, with finite perimeter P(Ω),

min{|Ω|,|Rⁿ\Ω|} ≤ Ciso(Rⁿ)P(Ω)ⁿ⁻¹ⁿ , (1)

where

C_iso(Rⁿ) = 1 nω_n−1ⁿ⁻¹¹

,

Here,ωkis the surface measure of the unit sphere S^kinR^k+1. Equality holds in (1) if and only if almost everywhereΩ = B(x,R) (i.e. a ball) for some x ∈ Rⁿand R > 0. In the case where ∂Ω is smooth say C¹then P(Ω) coincide with surface measure of∂Ω. In the non-smooth case P(Ω) = Var(χΩ,Rⁿ) whereχΩis the indicator function ofΩ and

Var(u) = sup (Z

Rⁿu(x)

n

X

i=1

∂xiGidx

¯

¯Gi∈ Co^∞(Rⁿ), G²₁+ · · · + C²n≤ 1 )

is the variation of u inRⁿ. Note thatΩ need not to be bounded in a priori.

(23)

We recall the classical isoperimetric inequality in the Euclidean space.

Theorem 1

For every Borel setΩ ⊂ Rⁿ, n ≥ 2, with finite perimeter P(Ω),

min{|Ω|,|Rⁿ\Ω|} ≤ Ciso(Rⁿ)P(Ω)ⁿ⁻¹ⁿ , (1)

where

C_iso(Rⁿ) = 1 nω_n−1ⁿ⁻¹¹

,

Here,ωkis the surface measure of the unit sphere S^kinR^k+1. Equality holds in (1) if and only if almost everywhereΩ = B(x,R) (i.e. a ball) for some x ∈ Rⁿand R > 0. In the case where ∂Ω is smooth say C¹then P(Ω) coincide with surface measure of∂Ω. In the non-smooth case P(Ω) = Var(χΩ,Rⁿ) whereχΩis the indicator function ofΩ and

Var(u) = sup (Z

Rⁿu(x)

n

X

i=1

∂xiGidx

¯

¯Gi∈ Co^∞(Rⁿ), G²₁+ · · · + C²n≤ 1 )

is the variation of u inRⁿ. Note thatΩ need not to be bounded in a priori.

(24)

Roughtly speaking, the isoperimetric problem consists in finding the smallest constant Ciso(Rⁿ) and classifying setsΩ such that inequality (1) becomes an equality. This problem is equivalent to the two following formulations:

Among all bounded, connected open sets of fixed perimeter L, find one with largest volume V .

Among all bounded, connected open sets with fixed volume V , find one with smallest perimeter L.

However, it was not until 1841 that Jacob Steiner gave the first proof (which contain gaps) but later on completed by many

mathematicians. Steiner proved that if such a region exists in the plane, it must be a circle.

(25)

Of course, the solution (inR²anyway) was known long long time ago.

(26)

Of course, the solution (inR²anyway) was known long long time ago.

(27)

The idea of his proof can be outlined in the following three steps. Assume therefore that there is a regionG in the plane such that among all other regions with the same perimeter ofG , then G must be a disc.

Step I: The regionG must be convex. For if not, using reflection, we can construct another region with the same perimeter but enclose a larger area, this contradict our assumption onG .

Figure:Steiner’s proof, step I.

(28)

(29)

(30)

Step II: Any straight line that divides the perimeter ofG in half must also divide the area ofG in half. Since G is convex, each half of the bounding curve lies entirely on one side of the line through A and B (see the figures). Suppose the line AB does not divide the area ofG in half, reflect the larger area across AB to obtain another region having the same perimeter ofG but with a larger area.

Again, we obtain a contradiction.

Figure:The argument only works for a convex region.

(31)

Figure:Steiner’s proof, step II.

(32)

Figure:Steiner’s proof, step II.

(33)

Step III: Now we concentrate on “half of the figure”. Pick any point C on this half curve and join it to A and B to obtain two

“lunes” AMC and BNC and a triangle ACB. The angle at C can be either < π/2,

= π/2 or > π/2. Imagine that the lunes are made of non-deformable material and they are hinged at C. Now the area of the region is the area of the two lunes plus the area of the triangle. The area of the triangle is computed by the base AC and heightBD. By adjusting the angle at C by moving the lunes, we see that the largest height is obtained when B = D, that is the angle at C isπ/2.

Hence, from elementary geometry, the angle ACB is a right angle if and only if C lies on a semi-circle with diameter AB.

(34)

= π/2 or > π/2. Imagine that the lunes are made of non-deformable material and they are hinged at C. Now the area of the region is the area of the two lunes plus the area of the triangle. The area of the triangle is computed by the base AC and height BD. By adjusting the angle at C by moving the lunes, we see that the largest height is obtained when B = D, that is the angle at C isπ/2.

(35)

= π/2 or > π/2. Imagine that the lunes are made of non-deformable material and they are hinged at C. Now the area of the region is the area of the two lunes plus the area of the triangle. The area of the triangle is computed by the base AC and height BD. By adjusting the angle at C by moving the lunes, we see that the largest height is obtained when B = D, that is the angle at C isπ/2.

(36)

Since Steiner’s proof there are many many proofs for the isoperimetric inequality. We present a few for theR²case.

Proof by complex function theory. Let z = x + iy and

dA = dx ∧ dy =¹₂idz ∧ dz. Using the fact that winding number of ∂Ω is one, Green and Fubini’s theorem we find

4πA = Z

Ω2πidz ∧ dz = Z

Ω

Z

∂Ω

dξ

ξ − zdz ∧ dz = Z

∂Ω

Z

∂Ω

ξ − z

ξ − zdz dξ ≤ L². The case of equality is easy to analyze in the above. The interplay between geometric extremal problems (e.g. isoperimetric problem) and sharp analytic inequalities is witnessed in the following analytic proof of the planar isoperimetric inequality. First, let’s recall Wirtinger’s inequality. If f is in the Sobolev space W^1,2([0, 2π]) satisfying

R_2π

0 f (t)dt = 0 then Z 2π

0 |f (t)|²dt ≤ Z 2π

0 |f⁰(t)|²dt , (2)

with equality holds only when f (t) = Acos(t) + B sin(t). The proof of Wirtinger’s inequality is an easy exercise in Fourier series.

(37)

Let ds denote the element of arc length and assume that∂Ω is a Lipschitz curve which is the boundary of a domainΩ ⊂ R². Denote by x = (x1, x2) the position vector. By translating the regionΩ which preserves area and perimeter, we may assume thatR

∂Ωxds = 0. The divergence theorem and Wirtinger’s inequality applied to x1, x2then gives

2A = Z

Ωdiv(x) dA = Z

∂Ω< x,~n > ds ≤ Z

∂Ω|x|ds ≤p L

µZ

∂Ω|x|²ds

¶¹₂

≤p L

·µ L 2π

¶2Z

∂Ω

¯

¯ dx ds

¯

2

ds

¸

1 2

≤L² 2π. Equality holds if and only ifΩ is a disc.

(38)

The proof for higher dimension Euclidean spaces are more technical and we will not recall them here. However a few remarks should be made.

The case ofR³was proved by Schwarz in 1884. His argument can be described in two steps:

symmetrization) that reduce the problem to finding the solution among a class of rotationally symmetric objects.

A geometric construction to rule out rotationally symmetric candidates different from the sphere.

The full generality of Theorem 1 was established by de Giorgi in 1958. There are many proofs of the Euclidean isoperimetric problem. There are also analogues of the same type of problem an in different settings, i.e., Riemannian manifolds. Our goal is to investigate the

Sub-Riemannian counterpart of this problem and we start with the Heisenberg group.

(39)

A symmetrization process (known nowadays as Schwarz

The full generality of Theorem 1 was established by de Giorgi in 1958. There are many proofs of the Euclidean isoperimetric problem. There are also analogues of the same type of problem an in different settings, i.e., Riemannian manifolds. Our goal is to investigate the

(40)

There are many proofs of the Euclidean isoperimetric problem. There are also analogues of the same type of problem an in different settings, i.e., Riemannian manifolds. Our goal is to investigate the

(41)

The full generality of Theorem 1 was established by de Giorgi in 1958.

(42)

(43)

(44)

The Heisenberg groupHⁿand relevant concept/quantities.

The Heisenberg groupHⁿis a Lie group onR²ⁿ⁺¹with the following group law:

(x, y, t) ◦ (x⁰, y⁰, t⁰) = (x + x⁰, y + y⁰, t + t⁰+1

2(x⁰· y − y⁰· x)) . where x = (x1, ..., xn), y = (y1, ..., yn), t ∈ R and the dot product is the standard dot product on Euclidean spaces.

Associated to this group law we work with the following standard left-invariant vector fields: (i = 1..n)

Xi= ∂

∂xi−y_i 2

∂

∂t, Yi= ∂

∂yi+x_i 2

∂

∂t, T = [Xi, Yi] = ∂

∂t, together with an inner product <, > with respect to which these 2n + 1 vector fields form an orthonormal system.Hⁿequiped with <, > is then a Riemannian manifold.

measure and therefore a Haar measure onHⁿ. For sets E ⊂ Hⁿ, the volume of E is the Lebesgue measure of E and will be denoted by |E|.

(45)

(x, y, t) ◦ (x⁰, y⁰, t⁰) = (x + x⁰, y + y⁰, t + t⁰+1

Xi= ∂

∂xi−y_i 2

∂

∂t, Yi= ∂

∂yi+x_i 2

∂

∂t, T = [Xi, Yi] = ∂

The Lebesgue measure onR²ⁿ⁺¹is both left and right translation measure and therefore a Haar measure onHⁿ. For sets E ⊂ Hⁿ, the volume of E is the Lebesgue measure of E and will be denoted by |E|.

(46)

(x, y, t) ◦ (x⁰, y⁰, t⁰) = (x + x⁰, y + y⁰, t + t⁰+1

Xi= ∂

∂xi−y_i 2

∂

∂t, Yi= ∂

∂yi+x_i 2

∂

∂t, T = [Xi, Yi] = ∂

The Lebesgue measure onR²ⁿ⁺¹is both left and right translation measure and therefore a Haar measure onHⁿ. For sets E ⊂ Hⁿ, the volume of E is the Lebesgue measure of E and will be denoted by |E|.

(47)

Given an oriented C²embedded hypersurfaceS ⊂ Hⁿ(and after an orientation is chosen) we let N be the Riemannian unit normal toS and we write N =Pn

i=1(p_iX_i+ qiY_i) + ωT

The projection of N onto the horizontal plane span{X_i, Y_i| i = 1, .., n} at each point g ∈ S is called the horizontal normal and is denoted by

N_H=Pn

i=1p_iX_i+ qiY_i.

The set

ΣSdef

= {g ∈ S | NH(g) = 0}

is called the characteristic set (singular set by some authors) of the hypersurfaceS . For our purpose, it suffices to know that for any C² hypersurfaceS we have σ(ΣS) = 0 where dσ is the Riemannian volume onS .

For points g ∉ ΣSwe define the horizontal Gauss map (the horizontal unit normal) by setting

νH(g) = N_H(g)

|NH(g)|= Xn i=1

p_iXi+ q_iYi.

The vector fieldνHplays an important role in the analysis of hypersurfaces in this setting.

(48)

N_H=Pn

i=1p_iX_i+ qiY_i. The set

ΣSdef

= {g ∈ S | NH(g) = 0}

normal) by setting

νH(g) = N_H(g)

|NH(g)|= Xn i=1

p_iXi+ q_iYi.

(49)

N_H=Pn

ΣSdef

= {g ∈ S | NH(g) = 0}

νH(g) = N_H(g)

|NH(g)|= Xn i=1

p_iX_i+ q_iY_i.

(50)

N_H=Pn

ΣSdef

= {g ∈ S | NH(g) = 0}

νH(g) = N_H(g)

|NH(g)|= Xn i=1

p_iX_i+ q_iY_i.