Geometric Structure of Normal and Integral Currents · Overview of the main results Some...

Overview of the main resultsSome unavoidable notation

Key identities and proofs

GEOMETRIC STRUCTURE OF NORMAL AND

INTEGRAL CURRENTS

Annalisa Massaccesi

Venice, 06 - 07 - 2017

Joint work with G. Alberti and E. Stepanov

Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS

1 Overview of the main results

2 Some unavoidable notation

3 Key identities and proofs

Section 1

Overview of the main results

LEXICON

The Lie bracket of a pair of vectorfields vi, vj ∈ C1(U; Λ1(Rn)) is

[vi, vj] := 〈∇vi; vj〉 − 〈∇vj; vi〉 .

A smooth distribution of k-planes V = span(v1, . . . , vk) (resp. a simplevectorfield v = v1 ∧ . . . ∧ vk) is involutive at a point x0 ∈ U if

[vi, vj](x0) = 0 for every i, j .

Example. The horizonal distribution of the Heisenberg group H1 is spanned byX := (1, 0,−y/2), Y := (0, 1, x/2). Since [X, Y] = (0, 0, 1), then X ∧ Y is nowhereinvolutive.

LEXICON

[vi, vj] := 〈∇vi; vj〉 − 〈∇vj; vi〉 .

LEXICON

[vi, vj] := 〈∇vi; vj〉 − 〈∇vj; vi〉 .

LEXICON

A k-current T is a linear and continuous functional on C∞c (U; Λk(Rn)),its boundary is a (k − 1)-current defined through the relation∂T(α) := T(dα) and its mass is the operator norm

M(T) := sup{T(ω) : ‖ω‖ ≤ 1, ω ∈ C∞c (U; Λk(Rn))} .

A normal current can be represented as T = vµ, with boundary∂T = v′µ′, where µ, µ′ are positive measures and v, v′ are vectorfields.An integral current is a normal current with a representation asT(ω) =

∫Σ〈v(x);ω(x)〉θ(x) dH k(x), with Σ ⊂ Rn k-rectifiable, v

tangent to Σ a.e. and θ ∈ L1(Σ;Z).

We assume:T is oriented by a unit, simple k-vectorfield v = v1 ∧ . . . ∧ vk;T is tangent to the distribution of k-planes V ∈ C1(U; Gr(k,Rn)), i.e.,span(v(x)) ⊂ V(x) for µ-a.e. x ∈ U.

LEXICON

COMFORTING FACTS ABOUT INTEGRAL CURRENTS

How much does an integral current generalize a surface?

Theorem (Frobenius). Take a distribution of planes V ∈ C1.(i) If S ⊂ Rn is a k-dimensional surface tangent to V (that is, the tangent

space Tan(S, x) = V(x) at every x ∈ S), then V must be involutive atevery point of S.

(ii) If V is everywhere involutive, then Rn can be locally foliated withk-dimensional surfaces which are tangent to V .

In our case, (i) turns into

Theorem 1. If T = vH k xΣ is an integral current tangent to V ∈ C1,then V is involutive at almost every point in Σ.

Remark. There is no integral current tangent to the horizontal distribution in H1.

Theorem 2. If T is an integral current tangent to V ∈ C1, then

span(v′(x)) ⊂ V(x)

holds at almost every point of the boundary.

T = vHkxΣv′(x)

v(x)To compare with

Theorem (Alberti- M., 2014). If T is an integral current oriented byv ∈ C0(U; Λk(Rn)), then

span(v′(x)) ⊂ span(v(x))

Theorem 2. If T is an integral current tangent to V ∈ C1, then

span(v′(x)) ⊂ V(x)

T = vHkxΣv′(x)

v(x)To compare with

Theorem (Alberti- M., 2014). If T is an integral current oriented byv ∈ C0(U; Λk(Rn)), then

span(v′(x)) ⊂ span(v(x))

GEOMETRIC PROPERTY OF THE BOUNDARY

Definition. A normal current T = vµ satisfies the geometric property ofthe boundary at some point x0 if

span(v′(x0)) ⊂ span(v(x0)) . (GPB)

Write µ′ = µ′a + µ′s, where µ′a � µ and µ′s ⊥ µ.

Theorem 3. A normal current T = vµ tangent to V ∈ C1 satisfies (GPB)for µ′s-a.e. x and µ′a-a.e. x at which v is involutive.

OPTIMALITY OF THEOREM 3

The following counterexample (Zworski, 1988) shows that the geometricproperty of the boundary cannot hold in general µ′-a.e.

Counterexample. Take the (locally) normal current T := (X ∧ Y)L3 inR3, where X,Y are the horizontal vectorfields of H1. Then ∂T has acomponent (0, 0, 1)L3 and (0, 0, 1) /∈ span(X,Y).

In this casethe singular part of µ′ with respect to µ is 0;the orientation X ∧ Y is nowhere involutive.

(NON-)INVOLUTIVITY OF V AND DIFFUSION OF µ

Notation.We set V(x) := span(v(x)) + span({[vi, vj](x) : 1 ≤ i < j ≤ k});We “measure” the (non-)involutivity of a vectorfield through thestratification

Cd = {x ∈ U : dim(V(x)

)= d} .

N :=⋃

d>k Cd, hence Ck \ N is the set of points where v is involutive.

Theorem 4. If T = vµ is a normal k-current tangent to V ∈ C1, then themeasure µ x Cd is absolutely continuous with respect to the d-dimensionalintegralgeometric measure, k < d ≤ n. In particular, if µ is concentratedon Cn, then µ� Ln.

Example. Any current T = (X ∧ Y)µ in H1 has µ� L3. (Optimality of Zworski’scounterexample)

)= d} .

N :=⋃

)= d} .

N :=⋃

Section 2

Some unavoidable notation

HODGE OPERATOR AND DIVERGENCE

We recall the following definitions for an h-covector υ, a k-vector u and ak-vectorfield v:

if h ≤ k, the interior product u x υ gives 〈u x υ; υ〉 := 〈u; υ ∧ υ〉 forevery test υ ∈ Λk−h(Rn);if h ≥ k, the interior product u y υ gives 〈u; u y υ〉 := 〈u ∧ u; υ〉 forevery test u ∈ Λh−k(Rn);the Hodge star operator is ?u := u y dx1 ∧ . . . ∧ dxn ∈ Λn−k(Rn);the divergence of v is div(v) := (−1)n−k ? (d(?v)). In coordinates

div v =∑

1≤i1<...<ik≤n

n∑j=1

∂vi1...ik

∂xjei1 ∧ . . . ∧ eik x dxj .

div v =∑

1≤i1<...<ik≤n

n∑j=1

∂vi1...ik

∂xjei1 ∧ . . . ∧ eik x dxj .

div v =∑

1≤i1<...<ik≤n

n∑j=1

∂vi1...ik

∂xjei1 ∧ . . . ∧ eik x dxj .

INVOLUTIVITY THROUGH THE DIVERGENCE

Proposition (Cartan’s formula). Consider a simple vectorfieldv = v1 ∧ . . . ∧ vk, with vi ∈ C1, then

div v =

k∑i=1

(−1)i−1div vi

∧j 6=i

vj +∑

1≤i1<i2≤k

(−1)i1+i2−1[vi1 , vi2 ] ∧∧

j6=i1,i2

Example. If k = 2, then div(X ∧ Y) = (div X)Y − (div Y)X + [X, Y].

We can now rewrite

V(x) = span(v(x)) + span(div v(x)) .

In particular V is involutive at a point x0 if and only if V(x0) = V(x0).

div v =

k∑i=1

(−1)i−1div vi

∧j 6=i

vj +∑

1≤i1<i2≤k

(−1)i1+i2−1[vi1 , vi2 ] ∧∧

j6=i1,i2

We can now rewrite

div v =

k∑i=1

(−1)i−1div vi

∧j 6=i

vj +∑

1≤i1<i2≤k

(−1)i1+i2−1[vi1 , vi2 ] ∧∧

j6=i1,i2

We can now rewrite

Section 3

A SPECIAL FAMILY OF TESTS

Given a simple vectorfield v = v1 ∧ . . . ∧ vk, we extract a simple2-vectorfield v tangent to span(v) (for instance a pair vi, vj). Chooseu ∈ Λn−k−1(Rn), we set

α := ?(v ∧ u) ∈ C1(U; Λk−1(Rn)) .

Lemma. The (k − 1)-form α above fulfills(i) v xα = 0;

(ii) v x dα = 〈v ∧ div v ∧ u; dx1 ∧ . . . ∧ dxn〉, in particular, ifv ∧ div v ∧ u 6= 0, then v x dα 6= 0.

A SPECIAL FAMILY OF TESTS

Given a simple vectorfield v = v1 ∧ . . . ∧ vk, we extract a simple2-vectorfield v tangent to span(v) (for instance a pair vi, vj). Chooseu ∈ Λn−k−1(Rn), we set

α := ?(v ∧ u) ∈ C1(U; Λk−1(Rn)) .

Lemma. The (k − 1)-form α above fulfills(i) v xα = 0;

(ii) v x dα = 〈v ∧ div v ∧ u; dx1 ∧ . . . ∧ dxn〉, in particular, ifv ∧ div v ∧ u 6= 0, then v x dα 6= 0.

PROOF OF THE LEMMA

(i) Take a 1-covector ϕ, then

(−1)k−1〈v xα;ϕ〉=〈v xϕ;α〉=〈(v xϕ) ∧ v︸︷︷︸∈span(v)

∧u; dx1 ∧ . . . ∧ dxn〉 = 0 ;

(ii) By definition dα = ?(div(v ∧ u)). By Cartan’s formula

div(v1 ∧ v2 ∧ u) = [v1, v2] ∧ u + w1 + w2 , (C)

where wi is a sum of simple vectorfields having vi as a factor. Thus

v x dα = 〈v ∧ div(v ∧ u); dx1 ∧ . . . ∧ dxn〉(C)= 〈v ∧ divv ∧ u; dx1 ∧ . . . ∧ dxn〉 .

PROOF OF THE LEMMA

(i) Take a 1-covector ϕ, then

(−1)k−1〈v xα;ϕ〉=〈v xϕ;α〉=〈(v xϕ) ∧ v︸︷︷︸∈span(v)

∧u; dx1 ∧ . . . ∧ dxn〉 = 0 ;

(ii) By definition dα = ?(div(v ∧ u)). By Cartan’s formula

div(v1 ∧ v2 ∧ u) = [v1, v2] ∧ u + w1 + w2 , (C)

where wi is a sum of simple vectorfields having vi as a factor. Thus

v x dα = 〈v ∧ div(v ∧ u); dx1 ∧ . . . ∧ dxn〉(C)= 〈v ∧ divv ∧ u; dx1 ∧ . . . ∧ dxn〉 .

THE KEY IDENTITY

Observe that T xα = (v xα)µ(i)= 0, thus

0 = ∂(T xα) = ∂T xα+ T x dα = (v′ xα)µ′ + (v x dα)µ

(ii)= 〈v′ ∧ v ∧ u; dx1 ∧ . . . ∧ dxn〉+ 〈v ∧ div v ∧ u; dx1 ∧ . . . ∧ dxn〉 .

By the arbitrariness of u ∈ Λn−k−1(Rn) we conclude

(v′ ∧ v)µ′ = −(v ∧ div v)µ (KI1)

for every choice of v tangent to span(v).

PROOFS OF THM. 1 AND THM. 3

(v′ ∧ v)µ′ = −(v ∧ div v)µ

Focus on µ, which is singular with respect to µ′ if the current isintegral: for µ-a.e. x we have that v ∧ div v = 0 and, thanks to Cartan’sformula, this proves Theorem 1.Focus on µ′s (which is singular with respect to µ): for µ′s-a.e. x we havethat v′(x) ∧ v(x) = 0, which proves the first part of Theorem 3.Moreover µ xN � µ′a.

(v′ ∧ v)µ′ = −(v ∧ div v)µ

A NEW KEY IDENTITY

(v′ ∧ v)µ′ = −(v ∧ div v)µ

Consider the key identity respect to µ′a (which is absolutely continuous withrespect to µ), we easily get the following

Theorem 5. If T = vµ is normal current tangent to a distribution V and∂T = v′µ′ is its boundary, then µ′a almost every point we have

span(v) + span(v′) = V . (KI2)

In particular, since µ xN � µ′a, the same (KI2) holds at µ-a.e. x ∈ N .

Proof of Thm. 4

CONSEQUENCES OF THM. 5

span(v) + span(v′) = V µ′a − a.e.

µ′a-a.e. x at which V is involutive, the geometric property of theboundary holds (and this completes the proof of Theorem 3);a weak geometric property of the boundary holds µ′a-a.e. (henceµ′-a.e.), namely

span(v′) ⊂ span(v) + span(div v) ;

on the other hand, we have a “control” on the Lie brackets of avectorfields through

span(div v) ⊂ span(v) + span(v′) µ′a − a.e.

PROOF OF THM. 4

We give the following (operational) definition of the decomposabilitybundle V(µ, ·) of the measure µ:

Definition. Take a measure λ� µ which can be decomposed asλ =

∫λt dt with λt 1-rectifiable measure (i.e., λt � H1 xEt and Et

rectifiable). Then the tangent τt to Et belongs to the decomposabilitybundle of µ, that is

τt(x) ⊂ V(µ, x) for a.e. t andH1 xEt − a.e. x .

Theorem (Alberti-Marchese, 2016). If T = vµ is a normal current, thenspan(v(x)) is contained in V(µ, x) for µ-a.e. x.

We exploit this theorem for both T and ∂T (because µ′ � µ in this case).We get that span(v′(x)) and span(v(x)) are both in V(µ, x).

PROOF OF THM. 4

Thanks to Theorem 5 we have that

dim (span(v′(x)) + span(v(x))) = n µ− a.e. x ∈ N ,

thusV(µ, x) = Rn .

So we can apply

Theorem (De Philippis-Rindler, 2016). If the decomposability bundle ofµ is of full dimension, then µ� Ln.

This theorem is the combination of Lemma 3.1 and Corollary 1.12 by DePhilippis-Rindler.

PROOF OF THM. 4

Thanks to Theorem 5 we have that

dim (span(v′(x)) + span(v(x))) = n µ− a.e. x ∈ N ,

thusV(µ, x) = Rn .

So we can apply

Theorem (De Philippis-Rindler, 2016). If the decomposability bundle ofµ is of full dimension, then µ� Ln.

This theorem is the combination of Lemma 3.1 and Corollary 1.12 by DePhilippis-Rindler.

THANKS FOR YOUR ATTENTION!

STRUCTURE OF A -FREE MEASURES AND APPLICATIONS

The abovementioned theorem is a consequence of this

Theorem (De Philippis-Rindler, 2016). If Ti = viµi ∈ Nki(U) fori = 1, . . . , r and λ� µi for every i, then λs-a.e. x ∈ U there exists acommon ωx ∈ Λ1(Rn) such that

vi xωx = 0 .

This theorem is an application of the following

Theorem (De Philippis-Rindler, 2016). Let U ⊂ Rn be an open set, letA be a k’th-order linear constant-coefficient differential operator, and letµ be a vector valued Radon measure with A µ = 0. Then

dµd|µ|

(x) ∈ ΛA for |µ|s − a.e. x ∈ U .

THE INTEGRALGEOMETRIC MEASURE

Definition. The d-dimensional integralgeometric measure of a Borel setB ⊂ Rn is defined as

Id(B) :=

∫Gr(d,n)

H 0 (B ∩ p−1W ({w})

)dH d(w) dγd,n(W) ,

where γd,n is the measure induced on the Grassmannian Gr(d, n) by theHaar measure on the orthogonal group O(n).

Remark. The integralgeometric measure Id coincides with H d onrectifiable sets of (Hausdorff) dimension d.

Geometric Structure of Normal and Integral Currents · Overview of the main results Some...

Documents

Transcript of Geometric Structure of Normal and Integral Currents · Overview of the main results Some...