Geometric Structure of Normal and Integral Currents · Overview of the main results Some...
Transcript of 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
Overview of the main resultsSome unavoidable notation
Key identities and proofs
1 Overview of the main results
2 Some unavoidable notation
3 Key identities and proofs
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
Section 1
Overview of the main results
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Proof
Remark. There is no integral current tangent to the horizontal distribution in H1.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Proof
Remark. There is no integral current tangent to the horizontal distribution in H1.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Proof
Remark. There is no integral current tangent to the horizontal distribution in H1.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
COMFORTING FACTS ABOUT INTEGRAL CURRENTS
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))
holds at almost every point of the boundary.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
COMFORTING FACTS ABOUT INTEGRAL CURRENTS
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))
holds at almost every point of the boundary.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Proof
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Proof
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Proof
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
(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.
Proof
Example. Any current T = (X ∧ Y)µ in H1 has µ� L3. (Optimality of Zworski’scounterexample)
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
(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.
Proof
Example. Any current T = (X ∧ Y)µ in H1 has µ� L3. (Optimality of Zworski’scounterexample)
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
(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.
Proof
Example. Any current T = (X ∧ Y)µ in H1 has µ� L3. (Optimality of Zworski’scounterexample)
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
Section 2
Some unavoidable notation
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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 .
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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 .
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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 .
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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
vj .
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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
vj .
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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
vj .
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
Section 3
Key identities and proofs
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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〉 .
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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〉 .
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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).
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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.
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
THANKS FOR YOUR ATTENTION!
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
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 .
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS
Overview of the main resultsSome unavoidable notation
Key identities and proofs
THE INTEGRALGEOMETRIC MEASURE
Definition. The d-dimensional integralgeometric measure of a Borel setB ⊂ Rn is defined as
Id(B) :=
∫Gr(d,n)
∫W
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.
Back
Annalisa Massaccesi GEOMETRIC STRUCTURE OF NORMAL AND INTEGRAL CURRENTS