Regularity of Lipschitz intrinsic - Purdue University · The other yields Lipschitz surfaces which arise as limit of Riemannian minimal surfaces (Pauls, 2004 and Cheng-Hwang-Yang,

The background geometry.Minimal graphs

Our resultsProofs

Regularity of Lipschitz intrinsicMinimal Graphs in Heisenberg groups Hn

Luca Capogna

May 27, 2009THE 4TH SYMPOSIUM ON ANALYSIS & PDE

Purdue University

Luca Capogna Regularity of Lipschitz intrinsic Minimal Graphs in Heisenberg groups Hn


Our resultsProofs

The Heisenberg group Hn

What this talk is about ...

One can define and construct minimal surfaces in thesub-Riemannian Heisenberg group (such surfaces arise for instancein the theory of perceptual completion, digital imagereconstruction and in the geometry of the visual cortex).

This notion was introduced by Garofalo in the 1990’s.



Our resultsProofs



Of the two ways known to construct such surfaces, one is purelyvariational and yields finite perimeter sets. (Garofalo-Nhieu, 1996).

The other yields Lipschitz surfaces which arise as limit ofRiemannian minimal surfaces (Pauls, 2004 andCheng-Hwang-Yang, 2007).

Both methods yield minimizers.



Our resultsProofs



Of the two ways known to construct such surfaces, one is purelyvariational and yields finite perimeter sets. (Garofalo-Nhieu, 1996).

The other yields Lipschitz surfaces which arise as limit ofRiemannian minimal surfaces (Pauls, 2004 andCheng-Hwang-Yang, 2007).

Both methods yield minimizers.



Our resultsProofs



Joint with Citti and Manfredini we prove sharp regularity of theRiemannian limits

Our work boils down to proving a-priori estimates for certainnon-linear, degenerate elliptic PDE.



Our resultsProofs


The group

Lie group with a Lie algebra hn = R2n+1 = V1 ⊕ V2, where V1 hasdimension 2n, and V2 = [V1,V1] has dimension 1 (pluscommutators relations)

Choose a basis Xs ,X1, ...,X2n−1 of theHorizontal space V1 and add a vectorX2n ∈ V2.Exponential coordinates. Denote (s, x) theelements of the group, where x = (x1, ..., x2n)and (x , s) ≈ exp(sXs) exp(x1X1 + ...+ x2nX2n).



Our resultsProofs


The group

Lie group with a Lie algebra hn = R2n+1 = V1 ⊕ V2, where V1 hasdimension 2n, and V2 = [V1,V1] has dimension 1 (pluscommutators relations)

Choose a basis Xs ,X1, ...,X2n−1 of theHorizontal space V1 and add a vectorX2n ∈ V2.Exponential coordinates. Denote (s, x) theelements of the group, where x = (x1, ..., x2n)and (x , s) ≈ exp(sXs) exp(x1X1 + ...+ x2nX2n).



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The group

Horizontal vector fields. By left translation we obtain

Xs = ∂s , Xi = ∂i , for i = 1, ..., n − 1,

Xi = ∂i + xi−n+1∂2n, for i = n, ..., 2n − 1.(1)

If n = 1 we have Xs = ∂s , X1 = ∂x1 + s∂x2 .

Results, Motivations and some HistoryMinimal graphs

Our resultsProofs


The group

Horizontal vector fields. Choose a basis of the Horizontal tangentspace V1 as follows:

Xs = !s ,Xi = !i , for i = 1, ..., n ! 1,

Xi = !i + xi!n+1!2n, for i = n, ..., 2n ! 1.(1)

For all p " Hn, the horizontal space HpM =span(Xs ,X1, ...,X2n!1)can be completed to be the tangent space by adding the vector!2n " V2.

Luca Capogna Regularity of Lipschitz intrinsic Minimal Graphs in Hn


Our resultsProofs


Hypersurfaces

M # Hn

Smooth hypersurface

! Characteristic points:

!(M) = p " M | HpM # TpM

! An intrinsic graph in Hn is an hypersurface parametrized by

x $ x exp("(x)V )

with x " U # A = exp(a) and such that a% span(V ) = hn,with V horizontal: With our choice of coordinatesM = (s, x) : s = u(x) where u : R2n $ R (alwaysnon-characteristic)


Horizontal bundle HpHn =span(Xs ,X1, ...,X2n−1)(p)



Our resultsProofs


The group


Xs = ∂s , Xi = ∂i , for i = 1, ..., n − 1,

Xi = ∂i + xi−n+1∂2n, for i = n, ..., 2n − 1.(1)



Our resultsProofs


The group


Xs = !s ,Xi = !i , for i = 1, ..., n ! 1,

Xi = !i + xi!n+1!2n, for i = n, ..., 2n ! 1.(1)




Our resultsProofs


Hypersurfaces

M # Hn

Smooth hypersurface


!(M) = p " M | HpM # TpM


x $ x exp("(x)V )






Our resultsProofs


The group


Xs = ∂s , Xi = ∂i , for i = 1, ..., n − 1,

Xi = ∂i + xi−n+1∂2n, for i = n, ..., 2n − 1.(1)



Our resultsProofs


The group


Xs = !s ,Xi = !i , for i = 1, ..., n ! 1,

Xi = !i + xi!n+1!2n, for i = n, ..., 2n ! 1.(1)




Our resultsProofs


Hypersurfaces

M # Hn

Smooth hypersurface


!(M) = p " M | HpM # TpM


x $ x exp("(x)V )






Our resultsProofs


How does the horizontal bundle arises in applications?

Hypersurfaces in Cn

If M ⊂ C2 is an hypersurface and p ∈ M, then TpM ≈ R3.

In this copy of R3 there lives a maximal 2-D plane which isinvariant by multiplication by the imaginary unit. We call suchplane HpM the horizontal plane at p. Clearly this gives rise to asub-bundle of the tangent bundle.

If M is given byIm(z2) = |z1|2

(biholomorphic to the sphere) we obtain exactly the Heisenberggroup with its horizontal bundle.



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Hypersurfaces in Cn

If M ⊂ C2 is an hypersurface and p ∈ M, then TpM ≈ R3.In this copy of R3 there lives a maximal 2-D plane which isinvariant by multiplication by the imaginary unit.

We call suchplane HpM the horizontal plane at p. Clearly this gives rise to asub-bundle of the tangent bundle.





Our resultsProofs



Hypersurfaces in Cn

If M ⊂ C2 is an hypersurface and p ∈ M, then TpM ≈ R3.In this copy of R3 there lives a maximal 2-D plane which isinvariant by multiplication by the imaginary unit. We call suchplane HpM the horizontal plane at p.

Clearly this gives rise to asub-bundle of the tangent bundle.





Our resultsProofs



Hypersurfaces in Cn

If M ⊂ C2 is an hypersurface and p ∈ M, then TpM ≈ R3.In this copy of R3 there lives a maximal 2-D plane which isinvariant by multiplication by the imaginary unit. We call suchplane HpM the horizontal plane at p. Clearly this gives rise to asub-bundle of the tangent bundle.





Our resultsProofs



Hypersurfaces in Cn

If M ⊂ C2 is an hypersurface and p ∈ M, then TpM ≈ R3.In this copy of R3 there lives a maximal 2-D plane which isinvariant by multiplication by the imaginary unit. We call suchplane HpM the horizontal plane at p. Clearly this gives rise to asub-bundle of the tangent bundle.





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Constrained dynamics: The unicycle



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Constrained dynamics: The unicycleThe state space for the position of a unicycle is R2 × S1.

Allowed Directions X1 = cos θ∂x + sin θ∂y and X2 = ∂θ.Forbidden Direction X1 = − cos θ∂y + sin θ∂x



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Geometry of the visual cortex in mammals Hardwired visualprocessing in the first layer of the visual cortex:



Our resultsProofs



Geometry of the visual cortex in mammals Hypercolumnsstructure, with columns clusters which respond to the angle of the

image. Luca Capogna Regularity of Lipschitz intrinsic Minimal Graphs in Heisenberg groups Hn


Our resultsProofs



Geometry of the visual cortex in mammals Hypercolumns structure(pinwheel shaped) from experiments (cat’s brain).



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The sub-Riemannian structure and Riemannianapproximants

Assign a Riemannian metric g0 on HHn.

wlog we can assumeXs , ...,X2n−1 are g0−orthonormal.For each ε > 0 extend g0 to a Riemannian metric gε on R2n+1 byrequesting

Xs , ...,X2n−1, εX2n

is an orthonormal basis.• In the limit

gε →ε→0 g0

curvatures, volume, injectivity radius all blow up.• However length minimizing curves between two points converge.



Our resultsProofs



Assign a Riemannian metric g0 on HHn. wlog we can assumeXs , ...,X2n−1 are g0−orthonormal.

For each ε > 0 extend g0 to a Riemannian metric gε on R2n+1 byrequesting

Xs , ...,X2n−1, εX2n


gε →ε→0 g0




Our resultsProofs



Assign a Riemannian metric g0 on HHn. wlog we can assumeXs , ...,X2n−1 are g0−orthonormal.For each ε > 0 extend g0 to a Riemannian metric gε on R2n+1 byrequesting

Xs , ...,X2n−1, εX2n

is an orthonormal basis.

• In the limitgε →ε→0 g0




Our resultsProofs




Xs , ...,X2n−1, εX2n


gε →ε→0 g0

curvatures, volume, injectivity radius all blow up.

• However length minimizing curves between two points converge.



Our resultsProofs




Xs , ...,X2n−1, εX2n


gε →ε→0 g0




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Metric structure

Let x and y be points in H1. For δ > 0 we define the class C (δ) ofabsolutely continuous paths γ : [0, 1]→ R3 with endpointsγ(0) = x and γ(1) = y , so that

γ′(t) = a(t)X1|γ(t) + b(t)X2|γ(t) (2)

and a(t)2 + b(t)2 ≤ δ2 for a.e. t ∈ [0, 1]. Paths satisfying (2) arecalled horizontal or Legendrian paths.

Carnot-Caratheodory metric d(x , y) = infδ such that C (δ) 6= ∅.A dual formulation is

d(x , y) = inf

T :

∃γ : [0,T ]→ R3, γ(0) = x , γ(T ) = y ,

and γ′ = aX1|γ + bX2|γ with a2 + b2 ≤ 1 a.e.

,

that is d(x , y) is the shortest time that it takes to go from x to y ,traveling at unit speed along horizontal paths.



Our resultsProofs


Metric structure


γ′(t) = a(t)X1|γ(t) + b(t)X2|γ(t) (2)

and a(t)2 + b(t)2 ≤ δ2 for a.e. t ∈ [0, 1]. Paths satisfying (2) arecalled horizontal or Legendrian paths.Carnot-Caratheodory metric d(x , y) = infδ such that C (δ) 6= ∅.

A dual formulation is

d(x , y) = inf

T :

∃γ : [0,T ]→ R3, γ(0) = x , γ(T ) = y ,


,




Our resultsProofs


Metric structure


γ′(t) = a(t)X1|γ(t) + b(t)X2|γ(t) (2)

and a(t)2 + b(t)2 ≤ δ2 for a.e. t ∈ [0, 1]. Paths satisfying (2) arecalled horizontal or Legendrian paths.Carnot-Caratheodory metric d(x , y) = infδ such that C (δ) 6= ∅.A dual formulation is

d(x , y) = inf

T :

∃γ : [0,T ]→ R3, γ(0) = x , γ(T ) = y ,


,




Our resultsProofs


Metric structure


γ′(t) = a(t)X1|γ(t) + b(t)X2|γ(t) (2)

and a(t)2 + b(t)2 ≤ δ2 for a.e. t ∈ [0, 1]. Paths satisfying (2) arecalled horizontal or Legendrian paths.Carnot-Caratheodory metric d(x , y) = infδ such that C (δ) 6= ∅.A dual formulation is

d(x , y) = inf

T :

∃γ : [0,T ]→ R3, γ(0) = x , γ(T ) = y ,


,




Our resultsProofs


Metric structure as limit of collapsing Riemannian spaces

• The space (Hn, d) is a metric space of Hausdorff dimension2n + 2

• The gε geodesics converge as ε→ 0 to length minimizing curvesin (Hn, d). Indeed (R2n+1, dε)→ (Hn, d) as ε→ 0 in theGromov-Hausdorff sense.

• The Hausdorff dimension of horizontal curves is 1, the Hausdorffdimension of non-horizontal curves is 2.

• The Hausdorff dimension of hypersurfaces is 2n + 1. We callPHn(·) the 2n + 1 Hausdorff measure (perimeter measure).A less self-contained, but more natural approach, would involvedefining BV spaces and perimeter (Garofalo and collaborators, ... )



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• The Hausdorff dimension of horizontal curves is 1,

the Hausdorffdimension of non-horizontal curves is 2.




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• The Hausdorff dimension of hypersurfaces is 2n + 1.

We callPHn(·) the 2n + 1 Hausdorff measure (perimeter measure).A less self-contained, but more natural approach, would involvedefining BV spaces and perimeter (Garofalo and collaborators, ... )



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Our resultsProofs


Hypersurfaces geometry

M ⊂ Hn

Smooth hypersurface

Characteristic points:

Σ(M) = p ∈ M | HpM ⊂ TpM

A result of Balogh yields that the characteristic set has(Euclidean) surface measure zero if the surface is at least C 2.

However, below the C 2 regularity threshold there are examples(also due to Balogh) of C 1,α surfaces with large characteristic set.

Characteristic points are Bad!



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M ⊂ Hn

Smooth hypersurfaceCharacteristic points:







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M ⊂ Hn








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M ⊂ Hn








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M ⊂ Hn








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Intrinsic graphs

An intrinsic graph in Hn is an hypersurface parametrized by

x → x exp(φ(x)V )

with x ∈ U ⊂ A = exp(a) and such that a⊕ span(V ) = hn, with Vhorizontal:

With our choice of coordinates intrisic graphs are actual graphs:M = (s, x) : s = u(x) where u : R2n → R

Such graphs are always non-characteristic.



Our resultsProofs


Intrinsic graphs

An intrinsic graph in Hn is an hypersurface parametrized by

x → x exp(φ(x)V )

with x ∈ U ⊂ A = exp(a) and such that a⊕ span(V ) = hn, with Vhorizontal:

With our choice of coordinates intrisic graphs are actual graphs:M = (s, x) : s = u(x) where u : R2n → R

Such graphs are always non-characteristic.



Our resultsProofs


Implicit Function Theorem.

Intrinsic graphs are generic

Any sufficiently smooth hypersurface can be represented as anintrisic graph away from its characteristic set.



Our resultsProofs



Intrinsic graphs are generic

Any sufficiently smooth hypersurface can be represented as anintrisic graph away from its characteristic set.



Our resultsProofs



[Franchi-Serapioni-Serra Cassano, Citti-Manfredini]Any level surface f (s, x) = c ⊂ Hn of functions f : Hn → R withcontinuous derivatives along the directions Xs ,X1, ...,X2n−1,

canlocally (near non-characteristic points) be expressed as an intrinsicgraph of a function u : Ω→ R, Ω ⊂ R2n.Moreover the C 1

H smoothness of f implies that the function u isregular with respect to the projection on its domain of the vectorfields Xs ,X1, ...,X2n−1,

Since Xs has null projection of the domain of u, the regularity of thisfunction is described only in terms of the vector fields:Xi ,u = Xi for i ≤ 2n − 2,X2n−1,u = ∂2n−1 + u(x)∂2n

If n = 1 it is only X1,u = ∂1 + u(x1, x2)∂2.



Our resultsProofs



[Franchi-Serapioni-Serra Cassano, Citti-Manfredini]Any level surface f (s, x) = c ⊂ Hn of functions f : Hn → R withcontinuous derivatives along the directions Xs ,X1, ...,X2n−1,canlocally (near non-characteristic points) be expressed as an intrinsicgraph of a function u : Ω→ R, Ω ⊂ R2n.

Moreover the C 1H smoothness of f implies that the function u is

regular with respect to the projection on its domain of the vectorfields Xs ,X1, ...,X2n−1,





Our resultsProofs



[Franchi-Serapioni-Serra Cassano, Citti-Manfredini]Any level surface f (s, x) = c ⊂ Hn of functions f : Hn → R withcontinuous derivatives along the directions Xs ,X1, ...,X2n−1,canlocally (near non-characteristic points) be expressed as an intrinsicgraph of a function u : Ω→ R, Ω ⊂ R2n.Moreover the C 1






Our resultsProofs



[Franchi-Serapioni-Serra Cassano, Citti-Manfredini]Any level surface f (s, x) = c ⊂ Hn of functions f : Hn → R withcontinuous derivatives along the directions Xs ,X1, ...,X2n−1,canlocally (near non-characteristic points) be expressed as an intrinsicgraph of a function u : Ω→ R, Ω ⊂ R2n.Moreover the C 1






Our resultsProofs

What is a Minimal surface?

Minimal surfaces are critical points of the perimeter functional.

i.e. for any flow of Caccioppoli sets t → Et with E0 minimal onemust have

d

dtPH1(Et ,Ω)|t=0 = 0,

for some open set Ω ⊂ H1.

In particular, perimeter minimizers (among given classes ofcompetitors) are minimal surfaces.

Definition Let Ω ⊂ H1 be an open set. A set of finite perimeterE ⊂ H1 is a perimeter minimizer in Ω if for all competitors F (i.e.F is a set of finite perimeter, with F = E outside Ω) one has

PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,




PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs



i.e. for any flow of Caccioppoli sets t → Et

with E0 minimal onemust have

d

dtPH1(Et ,Ω)|t=0 = 0,




PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,




PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,


In particular,

perimeter minimizers (among given classes ofcompetitors) are minimal surfaces.


PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,




PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,



Definition Let Ω ⊂ H1 be an open set.

A set of finite perimeterE ⊂ H1 is a perimeter minimizer in Ω if for all competitors F (i.e.F is a set of finite perimeter, with F = E outside Ω) one has

PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,



Definition Let Ω ⊂ H1 be an open set. A set of finite perimeterE ⊂ H1 is a perimeter minimizer in Ω if

for all competitors F (i.e.F is a set of finite perimeter, with F = E outside Ω) one has

PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs




d

dtPH1(Et ,Ω)|t=0 = 0,




PH1(E ,Ω) ≤ PH1(F ,Ω).



Our resultsProofs

Existence results

Garofalo-Nhieu (1996): Existence of BV minimizers (withprescribed boundary data).

Pauls (2004): Existence of Sobolev regular minimal t−graphs,which are limit of Riemannian minimal surfaces.

Cheng-Hwang-Yang (2007): Existence of Lipschitz minimizers(with prescribed, suitably smooth, boundary data) which are limitof Riemannian minimal surfaces. In the contest of t−graphs.

...More and more individual examples built out of Legendrianfoliations.



Our resultsProofs

What is the regularity of minimal surfaces?

A priori we only know they are Caccioppoli sets.

- Ritore builds examples of minimizers in H1 which are onlyLipschitz (in the Euclidean sense).

- Pauls builds a family of shears x3 = 1/2(x1x2)− g(x2) which isminimal and depending on the regularity of g is C k but not C k+1,k ≥ 2.

-Cheng, Hwang, Malchiodi and Yang, build an example of C 1,1

minimizer with smooth boundary data.



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Minimal surfaces and horizontal mean curvature

Let M ⊂ H1 be an oriented C 2 immersed surface withg1-Riemannian normal ν1 and horizontal normal νH = ProjH(ν1).

Suppose that U is a C 1 vector field with compact support onM \Σ(M), let φt(p) = expp(tU) and let Mt be the surface φt(M).

The first variation formula tells us the rate of change of theperimeter of Mt as t varies.

d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)

[u(divSνH) dσ, (3)

where u = 〈U, ν1〉1, (normal component of U)divMνH = H0 the horizontal mean curvature,dσ denotes the Riemannian surface area element on M.



Our resultsProofs





d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)





Our resultsProofs





d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)





Our resultsProofs





d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)


where

u = 〈U, ν1〉1, (normal component of U)divMνH = H0 the horizontal mean curvature,dσ denotes the Riemannian surface area element on M.



Our resultsProofs





d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)


where u = 〈U, ν1〉1, (normal component of U)

divMνH = H0 the horizontal mean curvature,dσ denotes the Riemannian surface area element on M.



Our resultsProofs





d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)


where u = 〈U, ν1〉1, (normal component of U)divMνH = H0 the horizontal mean curvature,

dσ denotes the Riemannian surface area element on M.



Our resultsProofs





d

dtPH1(Mt)

∣∣∣∣t=0

=

∫M\Σ(S)





Our resultsProofs


If M ⊂ H1 is an oriented C 2 immersed minimal surface then thefirst variation formula implies that its mean curvature vanishesuniformly on the complement of the characteristic set

divMνH = H = 0 in M \ Σ(M)

Note that a priori, the expression νH and consequently H0 is notdefined at Σ(M).



Our resultsProofs


If M ⊂ H1 is an oriented C 2 immersed minimal surface then thefirst variation formula implies that its mean curvature vanishesuniformly on the complement of the characteristic set

divMνH = H = 0 in M \ Σ(M)

Note that a priori, the expression νH and consequently H0 is notdefined at Σ(M).



Our resultsProofs

Horizontal mean curvature and the Legendrian foliation

Consider a manifold M and a point x ∈ M.



Our resultsProofs


The tangent space TxM at x to M is a plane in R3.



Our resultsProofs


The horizontal space Hx at x is a plane in R3. If x isnon-characteristic then TxM 6= Hx .



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The intersection of TxM and Hx at non-characteristic pointsx 6 ∈Σ(M) form a line bundle HM over M.

Horizontal Tangent Bundle



Our resultsProofs


The flow lines of the horizontal tangent bundle yield a foliation ofM \ Σ(M) with Legendrian curves.



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The projection on the horizontal space of thecurves γ = (γ1, γ2, γ3) in the Legendrianfoliation are plane curves whose curvaturecoincides with the horizontal mean curvatureof M at the point γ.



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C 2 minimal surfaces are foliated by legendrian lifts of linesegments

Since a C 2 minimal surfaces have zero horizontal mean curvaturethen

it is foliated by Legendrian curves whose horizontalprojections are plane curves with zero curvature i.e. segments.

C 2 minimal surfaces are foliated (outside the characteristic set) bylegendrian lifts of line segments

How these segments are allowed to cross at Σ(M) is the heart ofthe study of smooth minimal surfaces in Hn

(Danielli-Garofalo-Nhieu-Pauls, Cheng-Hwang-Malchiodi-Yang,Ritore’-Rosales, ... ).



Our resultsProofs


Since a C 2 minimal surfaces have zero horizontal mean curvaturethen it is foliated by Legendrian curves whose horizontalprojections are plane curves with zero curvature

i.e. segments.






Our resultsProofs


Since a C 2 minimal surfaces have zero horizontal mean curvaturethen it is foliated by Legendrian curves whose horizontalprojections are plane curves with zero curvature i.e. segments.






Our resultsProofs


Since a C 2 minimal surfaces have zero horizontal mean curvaturethen it is foliated by Legendrian curves whose horizontalprojections are plane curves with zero curvature i.e. segments.






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The minimal surface PDE for C 2 intrinsic graphs

2n−1∑i=1

Xi ,u

(Xi ,uu√

1 + |∇uu|2

)= 0.

whereXi ,u = ∂xi if i 6= 2n − 1 and X2n−1,u = ∂x2n−1 + u(x)∂x2n .∇uf = (X1,uf , ...,X2n−1,uf ).

This is a Non-Linear, Degenerate Elliptic PDE.

If n = 1 it reads (∂x1 + u(x)∂x2)

(∂x1u(x)+u(x)∂x2u(x)√

1+(∂x1u(x)+u(x)∂x2u(x))2

)= 0.

Holds also in a much weaker form without the C 2 assumption(Ambrosio, Serra Cassano, Vittone).



Our resultsProofs


2n−1∑i=1

Xi ,u

(Xi ,uu√

1 + |∇uu|2

)= 0.

whereXi ,u = ∂xi if i 6= 2n − 1 and X2n−1,u = ∂x2n−1 + u(x)∂x2n .∇uf = (X1,uf , ...,X2n−1,uf ).This is a Non-Linear, Degenerate Elliptic PDE.


(∂x1u(x)+u(x)∂x2u(x)√

1+(∂x1u(x)+u(x)∂x2u(x))2

)= 0.




Our resultsProofs


2n−1∑i=1

Xi ,u

(Xi ,uu√

1 + |∇uu|2

)= 0.



(∂x1u(x)+u(x)∂x2u(x)√

1+(∂x1u(x)+u(x)∂x2u(x))2

)= 0.




Our resultsProofs


2n−1∑i=1

Xi ,u

(Xi ,uu√

1 + |∇uu|2

)= 0.



(∂x1u(x)+u(x)∂x2u(x)√

1+(∂x1u(x)+u(x)∂x2u(x))2

)= 0.




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Riemannian approximations

We will consider Lipschitz solutions that arise as limits of solutionsof elliptic/Riemannian regularizations.

i.e. Consider uε, ε > 0smooth solution of the Elliptic nonlinear PDE

Lε,uεuε =2n∑i=1

X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,

withX ε

i ,u = Xi ,u if i = 1, ..., 2n − 1 and X ε2n,u = ε∂x2n .

This is the minimal surface PDE for the graph s = uε(x) in theApproximating Riemannian metrics gε, i.e. this is the Riemannianmetric on R2n+1 s.t. the frame

(∂s ,X1, ...,X2n−1, ε∂x2n)

is orthonormal.



Our resultsProofs


We will consider Lipschitz solutions that arise as limits of solutionsof elliptic/Riemannian regularizations. i.e. Consider uε, ε > 0smooth solution of the Elliptic nonlinear PDE


X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,

withX ε



(∂s ,X1, ...,X2n−1, ε∂x2n)

is orthonormal.



Our resultsProofs




X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,

withX ε


This is the minimal surface PDE for the graph s = uε(x) in theApproximating Riemannian metrics gε,

i.e. this is the Riemannianmetric on R2n+1 s.t. the frame

(∂s ,X1, ...,X2n−1, ε∂x2n)

is orthonormal.



Our resultsProofs




X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,

withX ε



(∂s ,X1, ...,X2n−1, ε∂x2n)

is orthonormal.Luca Capogna Regularity of Lipschitz intrinsic Minimal Graphs in Heisenberg groups Hn


Our resultsProofs

Vanishing viscosity solutions

Are Lipschitz functions u : Ω→ R s.t. there existsuεj sequence of smooth solutions of theapproximating PDE such that

uεj → u

uniformly on compacts and for some C > 0

||uεj ||LIP(Ω) ≤ C .



Our resultsProofs


• A priori, the only link between u and the original equation is viathe approximating PDE.

• The existence of particular vanishing viscosity solutions is due toCheng-Hwang-Yang, but see also the work of Pauls.

Question How regular are vanishing viscositysolutions? What relation do they have withminimal surfaces?



Our resultsProofs







Our resultsProofs




Question

How regular are vanishing viscositysolutions? What relation do they have withminimal surfaces?



Our resultsProofs







Our resultsProofs

Main Theorem (C.- Citti-Manfredini)

Let u be a vanishing viscosity minimal graph.

(i) u ∈ C 1,αE (Ω).

(ii) If n > 1 then u is smooth.(iii) If n = 1 then X k

u u ∈W 1,ploc (Ω) for all k ∈ N and p ≥ 1.

Moreover the PDE holds pointwise and Ω is foliated by quadricswhich lift (via u) to the Legendrian foliation of the graph.

Related Results: Cheng-Hwang-Yang: C 1 t−graphs, Hcontinuous prescribed, they prove the Legendrian foliation is C 2.Bigolin-Serra Cassano: Deal with weak (broad∗) solutions, theyprove that If X1,uu is Lipschitz, then u is Lipschitz.Vittone: Boundedness of perimeter minimizers.



Our resultsProofs



(i) u ∈ C 1,αE (Ω).







Our resultsProofs



(i) u ∈ C 1,αE (Ω).

(ii) If n > 1 then u is smooth.

(iii) If n = 1 then X ku u ∈W 1,p

loc (Ω) for all k ∈ N and p ≥ 1.Moreover the PDE holds pointwise and Ω is foliated by quadricswhich lift (via u) to the Legendrian foliation of the graph.




Our resultsProofs



(i) u ∈ C 1,αE (Ω).







Our resultsProofs



(i) u ∈ C 1,αE (Ω).







Our resultsProofs



(i) u ∈ C 1,αE (Ω).




Related Results:

Cheng-Hwang-Yang: C 1 t−graphs, Hcontinuous prescribed, they prove the Legendrian foliation is C 2.Bigolin-Serra Cassano: Deal with weak (broad∗) solutions, theyprove that If X1,uu is Lipschitz, then u is Lipschitz.Vittone: Boundedness of perimeter minimizers.



Our resultsProofs



(i) u ∈ C 1,αE (Ω).




Related Results: Cheng-Hwang-Yang: C 1 t−graphs, Hcontinuous prescribed, they prove the Legendrian foliation is C 2.

Bigolin-Serra Cassano: Deal with weak (broad∗) solutions, theyprove that If X1,uu is Lipschitz, then u is Lipschitz.Vittone: Boundedness of perimeter minimizers.



Our resultsProofs



(i) u ∈ C 1,αE (Ω).




Related Results: Cheng-Hwang-Yang: C 1 t−graphs, Hcontinuous prescribed, they prove the Legendrian foliation is C 2.Bigolin-Serra Cassano: Deal with weak (broad∗) solutions, theyprove that If X1,uu is Lipschitz, then u is Lipschitz.

Vittone: Boundedness of perimeter minimizers.



Our resultsProofs



(i) u ∈ C 1,αE (Ω).







Our resultsProofs

Intrinsic and Euclidean Regularity

In the case n = 1 we obtain that the solution u is infinitely manytimes differentiable along the vector Xu = ∂x1 − u(x1, x2)∂x2 .

What does this mean in terms of Euclidean regularity? Not much:Let u(x1, x2) = |x2| then

Xuu = ∂x1 |x2|+ |x2|∂x2 |x2| = x2

X 2u u = ∂x1x2 + |x2|∂x2x2 = |x2|

.... and so on.



Our resultsProofs


In the case n = 1 we obtain that the solution u is infinitely manytimes differentiable along the vector Xu = ∂x1 − u(x1, x2)∂x2 .What does this mean in terms of Euclidean regularity?

Not much:Let u(x1, x2) = |x2| then

Xuu = ∂x1 |x2|+ |x2|∂x2 |x2| = x2

X 2u u = ∂x1x2 + |x2|∂x2x2 = |x2|

.... and so on.



Our resultsProofs


In the case n = 1 we obtain that the solution u is infinitely manytimes differentiable along the vector Xu = ∂x1 − u(x1, x2)∂x2 .What does this mean in terms of Euclidean regularity? Not much:Let u(x1, x2) = |x2| then

Xuu = ∂x1 |x2|+ |x2|∂x2 |x2| = x2

X 2u u = ∂x1x2 + |x2|∂x2x2 = |x2|

.... and so on.



Our resultsProofs


In the case n = 1 we obtain that the solution u is infinitely manytimes differentiable along the vector Xu = ∂x1 − u(x1, x2)∂x2 .What does this mean in terms of Euclidean regularity? Not much:Let u(x1, x2) = |x2| then

Xuu = ∂x1 |x2|+ |x2|∂x2 |x2| = x2

X 2u u = ∂x1x2 + |x2|∂x2x2 = |x2|

.... and so on.



Our resultsProofs

Strategy of the proof

A priori estimates:

||X kuεuε||W 1,p ≤ Ck , for n = 1, p > 1, k ∈ N.

||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).

uniform in ε as ε→ 0.

I C 1,α regularity. (new idea)

I Linearization and Freezing. (bulk of the work, a new take onideas of Citti and collaborators, and related to work ofSawyer-Wheeden)

I Caccioppoli inequalities. (technically very involved, but notsurprising)



Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).







Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).







Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).







Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).


I C 1,α regularity.

(new idea)





Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).







Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).



I Linearization and Freezing.

(bulk of the work, a new take onideas of Citti and collaborators, and related to work ofSawyer-Wheeden)




Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).







Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).




I Caccioppoli inequalities.

(technically very involved, but notsurprising)



Our resultsProofs


A priori estimates:


||uε||C k,α ≤ Ck , for n > 1, k ∈ N, α ∈ (0, 1).







Our resultsProofs

The Euclidean setting

div

(∇u√

1 + |∇u|2

)= 0, in Ω.

With u Lipschitz in Ω, weak solution.

First we use difference quotients + Caccioppoli inequalities toshow u ∈W 2,2

loc (Ω) Moreover w = ∂xlu solves∑

i ,j ∂i

(Aij(∇u)∂jw

)= 0, with Aij(ξ) =

δij−ξi ξj√1+|ξ|2√

1+|ξ|2.

De Giorgi-Nash-Moser => w ∈ Cα => Aij(∇u) ∈ Cα (viaSchauder esitmates)=> w ∈ C 1,α => ... => u smooth.



Our resultsProofs


div

(∇u√

1 + |∇u|2

)= 0, in Ω.

With u Lipschitz in Ω, weak solution.First we use difference quotients + Caccioppoli inequalities toshow u ∈W 2,2

loc (Ω)

Moreover w = ∂xlu solves∑

i ,j ∂i

(Aij(∇u)∂jw



1+|ξ|2.




Our resultsProofs


div

(∇u√

1 + |∇u|2

)= 0, in Ω.



i ,j ∂i

(Aij(∇u)∂jw



1+|ξ|2.




Our resultsProofs


div

(∇u√

1 + |∇u|2

)= 0, in Ω.



i ,j ∂i

(Aij(∇u)∂jw



1+|ξ|2.

De Giorgi-Nash-Moser => w ∈ Cα

=> Aij(∇u) ∈ Cα (viaSchauder esitmates)=> w ∈ C 1,α => ... => u smooth.



Our resultsProofs


div

(∇u√

1 + |∇u|2

)= 0, in Ω.



i ,j ∂i

(Aij(∇u)∂jw



1+|ξ|2.

De Giorgi-Nash-Moser => w ∈ Cα => Aij(∇u) ∈ Cα

(viaSchauder esitmates)=> w ∈ C 1,α => ... => u smooth.



Our resultsProofs


div

(∇u√

1 + |∇u|2

)= 0, in Ω.



i ,j ∂i

(Aij(∇u)∂jw



1+|ξ|2.

De Giorgi-Nash-Moser => w ∈ Cα => Aij(∇u) ∈ Cα (viaSchauder esitmates)=> w ∈ C 1,α

=> ... => u smooth.



Our resultsProofs


div

(∇u√

1 + |∇u|2

)= 0, in Ω.



i ,j ∂i

(Aij(∇u)∂jw



1+|ξ|2.




Our resultsProofs

From Lipschitz to C 1,αE

Differentiate the PDE

2n∑i=1

X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,

with respect to X εk,u and ∂x2n

Compute all commutators and setz = X ε

k,uu + 2||u||Lip > 0, w = ∂x2nu + 2||u||Lip > 0 which solve

(X εi ,u)∗

[Aij(∇εu)(X ε

j ,u)∗w

]= 0, and X ε

i ,u

[Aij(∇εu)X ε

j ,uz

]= −

∑i

[X εk,u,X

εi ,u]

(X ε

i ,u√1 + |∇εuu|2

)+ higher order commutators

These PDE yield Caccioppoli inequalities for w and z



Our resultsProofs



2n∑i=1

X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,

with respect to X εk,u and ∂x2n Compute all commutators and set

z = X εk,uu + 2||u||Lip > 0, w = ∂x2nu + 2||u||Lip > 0 which solve

(X εi ,u)∗

[Aij(∇εu)(X ε

j ,u)∗w

]= 0, and X ε

i ,u

[Aij(∇εu)X ε

j ,uz

]= −

∑i

[X εk,u,X

εi ,u]

(X ε

i ,u√1 + |∇εuu|2





Our resultsProofs



2n∑i=1

X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,



(X εi ,u)∗

[Aij(∇εu)(X ε

j ,u)∗w

]= 0, and X ε

i ,u

[Aij(∇εu)X ε

j ,uz

]= −

∑i

[X εk,u,X

εi ,u]

(X ε

i ,u√1 + |∇εuu|2





Our resultsProofs



2n∑i=1

X εi ,uε

(X ε

i ,uεuε√

1 + |∇εuεuε|2

)= 0,



(X εi ,u)∗

[Aij(∇εu)(X ε

j ,u)∗w

]= 0, and X ε

i ,u

[Aij(∇εu)X ε

j ,uz

]= −

∑i

[X εk,u,X

εi ,u]

(X ε

i ,u√1 + |∇εuu|2





Our resultsProofs


For p 6= 2∫|∂2nz |2zp−2φ2 ≤ C

(∫|∇εuz |2zp−2φ2 +

∫zp(φ2 + |∇εuφ|2)

)∫|∇εuz |zp−2φ2 ≤ C

(∫zp(φ2 + |∇εuφ|2)

).

Since |∇E z |2 ≤∑

k<2n |X εk,uz |2 + ∂x2nz |2,∫

|∇E z |2zp−2φ2 ≤ C

∫zp(φ2 + |∇Eφ|2).

(Euclidean Caccioppoli inequality)Ready made to start Moser iteration and obtain C 1,α andX ε

i ,uXεj ,uuε ∈ L2 regularity uniform in ε.



Our resultsProofs


For p 6= 2∫|∂2nz |2zp−2φ2 ≤ C

(∫|∇εuz |2zp−2φ2 +

∫zp(φ2 + |∇εuφ|2)

)∫|∇εuz |zp−2φ2 ≤ C

(∫zp(φ2 + |∇εuφ|2)

).


k<2n |X εk,uz |2 + ∂x2nz |2,∫

|∇E z |2zp−2φ2 ≤ C

∫zp(φ2 + |∇Eφ|2).

(Euclidean Caccioppoli inequality)

Ready made to start Moser iteration and obtain C 1,α andX ε




Our resultsProofs


For p 6= 2∫|∂2nz |2zp−2φ2 ≤ C

(∫|∇εuz |2zp−2φ2 +

∫zp(φ2 + |∇εuφ|2)

)∫|∇εuz |zp−2φ2 ≤ C

(∫zp(φ2 + |∇εuφ|2)

).


k<2n |X εk,uz |2 + ∂x2nz |2,∫

|∇E z |2zp−2φ2 ≤ C

∫zp(φ2 + |∇Eφ|2).

(Euclidean Caccioppoli inequality)Ready made to start Moser iteration and obtain C 1,α andX ε




Our resultsProofs

The non-divergence form PDE

Since we have now two derivatives in L2 we can ”differentiate” thePDE and obtain

Lε,uu =2n∑i ,j

aεij(∇εuu)X εi ,uX

εj ,uu = 0

Which we ”linearize” as

Lε,uz =2n∑i ,j


εj ,uz = 0

Note that the coefficients are good (i.e. Cα) but the vector fieldsare not smooth, there is no ready-made Schauder Theory.What do we know for Schauder theory in this setting?



Our resultsProofs



Lε,uu =2n∑i ,j


εj ,uu = 0


Lε,uz =2n∑i ,j


εj ,uz = 0




Our resultsProofs



Lε,uu =2n∑i ,j


εj ,uu = 0


Lε,uz =2n∑i ,j


εj ,uz = 0

Note that the coefficients are good (i.e. Cα) but the vector fieldsare not smooth, there is no ready-made Schauder Theory.

What do we know for Schauder theory in this setting?



Our resultsProofs



Lε,uu =2n∑i ,j


εj ,uu = 0


Lε,uz =2n∑i ,j


εj ,uz = 0




Our resultsProofs

Schauder Theory in 1 min.

Key Idea: We know how to obtain C 2,α estimates for∆u = f ∈ Cα.

Hence we can deal with∑

ij aij∂i∂j , constant coefficients, elliptic.If aij(x) ∈ Cα then we write

aij(x)∂i∂ju(x) = aij(x0)∂i∂ju(x) + (aij(x)− aij(x0))∂i∂ju(x)

This is what we call freezing the coefficients.



Our resultsProofs









Our resultsProofs




ij aij∂i∂j , constant coefficients, elliptic.

If aij(x) ∈ Cα then we write





Our resultsProofs









Our resultsProofs

Schauder Theory in 1 min. (Apologies to the experts!)

In our setting the role of the Laplacian is played by Hormandersum of squares of vector fields (sub-Laplacians)

L =∑

i

X 2i ,

with X1, ...,Xm smooth vector fields, whose Lie brackets up to stepr ∈ N, [Xi1 , [Xi2 , ...,Xir ]...] generate all of Rn.

Here the analogue of the Schauder estimates are Holder estimatesfor second order horizontal derivatives, e.g. ||XiXju||α ≤ C ||Lu||α.

• (const. coeff.) aijXiXj is ok, • aij(x)XiXj is ok if aij ∈ Cα



Our resultsProofs



L =∑

i

X 2i ,

with X1, ...,Xm smooth vector fields, whose Lie brackets up to stepr ∈ N, [Xi1 , [Xi2 , ...,Xir ]...] generate all of Rn.Here the analogue of the Schauder estimates are Holder estimatesfor second order horizontal derivatives, e.g. ||XiXju||α ≤ C ||Lu||α.




Our resultsProofs



L =∑

i

X 2i ,


• (const. coeff.) aijXiXj is ok,

• aij(x)XiXj is ok if aij ∈ Cα



Our resultsProofs



L =∑

i

X 2i ,





Our resultsProofs

Rule of Thumb

The higher is the step of the commutator needed to span Rn, theworse is the regularity loss: e.g.

||f ||H2/r ≤ C ||Lf ||L2

(subelliptic estimates)



Our resultsProofs

The Linearization scheme: Freezing.

• Our operator is

Lε,uz =2n∑i ,j


εj ,uz

• Freezing yields

2n∑i ,j

aεij(∇εuu(x0))X εi ,uX

εj ,uz

• This is still not good enough because X εi ,u although C∞ do not

satisfy estimates uniform in ε.



Our resultsProofs


• Our operator is

Lε,uz =2n∑i ,j


εj ,uz

• Freezing yields

2n∑i ,j


εj ,uz





Our resultsProofs


• Our operator is

Lε,uz =2n∑i ,j


εj ,uz

• Freezing yields

2n∑i ,j


εj ,uz





Our resultsProofs

The Linearization scheme: Freezing

• We need to approximate X2n−1,u = ∂2n−1 + u(x)∂2n with

X2n−1,x0 = ∂2n−1 + [P1u ](x)∂2n

where [P1u ] is a sort of Taylor Polynomial of order one of u. (in the

same spirit as the homogenous polynomials used by Folland,Folland-Stein, Rothschild-Stein, Nagel-Stein-Weinger).



Our resultsProofs

Here is a fine point: On the one hand it would be nice to havebetter approximation of u (something like higher order Taylorpolynomials).

On the other hand (1) u is not sufficent smooth at the beginningof the bootstrap to allow for higher order Taylor polynomials(smoothness uniform in ε as usual).

(2) Higher order Taylor polynomials would yield more and morecomplicated Lie algebra structures. i.e. the number ofcommutators needed to generate Rn would increase with thedegree of the polynomial, thus leading to loss of regularity in theestimates.



Our resultsProofs






Our resultsProofs






Our resultsProofs


• Setting Lε,x0z =∑2n

i ,j aεij(∇εuu(x0))X εi ,x0

X εj ,x0

z we can approximate

Lε,uz =2n∑i ,j


εj ,uz = Lε,x0 + Error Term

The analysis of the error term is based in part on precise estimateson kernels in representation formula.For instance, we need estimates on the fundamental solution forthe ”regularized sub-Laplacians”

Lε =2n−1∑i=1

X 2i ,x0

+ εX 22n,x0

which are well-behaved in ε (this is a result of Citti-Manfredini).



Our resultsProofs




X εj ,x0


Lε,uz =2n∑i ,j




Lε =2n−1∑i=1

X 2i ,x0

+ εX 22n,x0




Our resultsProofs




X εj ,x0


Lε,uz =2n∑i ,j




Lε =2n−1∑i=1

X 2i ,x0

+ εX 22n,x0




Our resultsProofs

Breakdown at n = 1. Why?

• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n

i=1 aεi ,2nXεi ,x0∂x2nz .

= ”PRINCIPAL PART” + REMAINDER.• If n > 1 the ”principal part” is a sub-Laplacian (so we haveSchauder, etc. etc.)• if n = 1 the principal part is composed of one term, it is not asum of squares Hormander operator. (no theory, no previousresults)



Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n





Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n


= ”PRINCIPAL PART”

+ REMAINDER.• If n > 1 the ”principal part” is a sub-Laplacian (so we haveSchauder, etc. etc.)• if n = 1 the principal part is composed of one term, it is not asum of squares Hormander operator. (no theory, no previousresults)



Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n


= ”PRINCIPAL PART” + REMAINDER.

• If n > 1 the ”principal part” is a sub-Laplacian (so we haveSchauder, etc. etc.)• if n = 1 the principal part is composed of one term, it is not asum of squares Hormander operator. (no theory, no previousresults)



Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n


= ”PRINCIPAL PART” + REMAINDER.• If n > 1 the ”principal part” is a sub-Laplacian

(so we haveSchauder, etc. etc.)• if n = 1 the principal part is composed of one term, it is not asum of squares Hormander operator. (no theory, no previousresults)



Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n


= ”PRINCIPAL PART” + REMAINDER.• If n > 1 the ”principal part” is a sub-Laplacian (so we haveSchauder, etc. etc.)

• if n = 1 the principal part is composed of one term, it is not asum of squares Hormander operator. (no theory, no previousresults)



Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n


= ”PRINCIPAL PART” + REMAINDER.• If n > 1 the ”principal part” is a sub-Laplacian (so we haveSchauder, etc. etc.)• if n = 1 the principal part is composed of one term, it is not asum of squares Hormander operator.

(no theory, no previousresults)



Our resultsProofs


• Lε,x0z =∑2n


X εj ,x0

z

=∑2n−1


X εj ,x0

z + ε∑2n





Our resultsProofs

The new lifting technique in n = 1

In n = 1 there is no background sub-Laplacian structure. This isdue to the fact that the intersection of the horizontal plane andthe tangent plane is generically one-dimensional, so we have a sortof one-dimensional surface laplacian.

Bottom line: There is not enough room in n = 1.Our method Lift to a step three group by adding a new variable τSo that Ω→ Ω× (−1, 1), the vector fields are lifted to

X1 = ∂x1 + (P1u(x) + τ2)∂x2 , X2 = ε∂x2 , and X3 = ∂τ

If z(x) is a solution of Lε,x0z(x) = 0 then it also solves∑3i ,j=1 aij(∇εuu(x0))Xi Xjz = 0 for τ = 0

Which has a ’background sub-Laplacian’∑3i ,j 6=2;i ,j=1 aij(∇εuu(x0))Xi Xjz



Our resultsProofs


In n = 1 there is no background sub-Laplacian structure. This isdue to the fact that the intersection of the horizontal plane andthe tangent plane is generically one-dimensional, so we have a sortof one-dimensional surface laplacian.Bottom line: There is not enough room in n = 1.

Our method Lift to a step three group by adding a new variable τSo that Ω→ Ω× (−1, 1), the vector fields are lifted to






Our resultsProofs


In n = 1 there is no background sub-Laplacian structure. This isdue to the fact that the intersection of the horizontal plane andthe tangent plane is generically one-dimensional, so we have a sortof one-dimensional surface laplacian.Bottom line: There is not enough room in n = 1.Our method Lift to a step three group by adding a new variable τSo that Ω→ Ω× (−1, 1), the vector fields are lifted to






Our resultsProofs








Our resultsProofs







Documents

Regularity of Lipschitz intrinsic - Purdue University · The other yields Lipschitz surfaces which arise as limit of Riemannian minimal surfaces (Pauls, 2004 and Cheng-Hwang-Yang,