Small unions of affine subspaces and skeletons via Baire category - - PowerPoint PPT Presentation

small unions of affine subspaces and skeletons via baire
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Small unions of affine subspaces and skeletons via Baire category - - PowerPoint PPT Presentation

Statement of results Proof outline for measure Small unions of affine subspaces and skeletons via Baire category Alan Chang (ac@math.uchicago.edu) http://math.uchicago.edu/~ac Joint work with Marianna Csrnyei, Kornlia Hra and Tams


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Statement of results Proof outline for measure

Small unions of affine subspaces and skeletons via Baire category

Alan Chang (ac@math.uchicago.edu)

http://math.uchicago.edu/~ac

Joint work with Marianna Csörnyei, Kornélia Héra and Tamás Keleti Workshop on Geometric Measure Theory 10-14 July 2017 The University of Warwick, Coventry, UK

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Statement of results Proof outline for measure

Alan Chang, Marianna Csörnyei, Kornélia Héra and Tamás Keleti

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Statement of results Proof outline for measure Previous results

Theorem (Stein 1976) Let n ≥ 3. Any subset of Rn which contains an (n − 1)-sphere centred at each point of Rn must have positive Lebesgue measure.

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Statement of results Proof outline for measure Previous results

Theorem (Stein 1976) Let n ≥ 3. Any subset of Rn which contains an (n − 1)-sphere centred at each point of Rn must have positive Lebesgue measure. Theorem (Bourgain 1986, Marstrand 1987) Same holds for n = 2.

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Statement of results Proof outline for measure Previous results

Theorem (Stein 1976) Let n ≥ 3. Any subset of Rn which contains an (n − 1)-sphere centred at each point of Rn must have positive Lebesgue measure. Theorem (Bourgain 1986, Marstrand 1987) Same holds for n = 2. What if we use something other than circles?

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Statement of results Proof outline for measure Previous results

Theorem (Stein 1976) Let n ≥ 3. Any subset of Rn which contains an (n − 1)-sphere centred at each point of Rn must have positive Lebesgue measure. Theorem (Bourgain 1986, Marstrand 1987) Same holds for n = 2. What if we use something other than circles? Theorem (Keleti, Nagy, and Shmerkin 2014) In R2, there exists a closed set of Hausdorff dimension 1 (and hence Lebesgue measure zero) that contains the boundary of an axis-parallel square around each point of R2.

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Statement of results Proof outline for measure The kinds of questions we ask

Let S ⊂ Rn be the k-skeleton of a polytope.

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Statement of results Proof outline for measure The kinds of questions we ask

Let S ⊂ Rn be the k-skeleton of a polytope. Suppose we “put a copy” of S around every point of Rn. What is the minimal dimension/measure of the resulting set?

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Statement of results Proof outline for measure The kinds of questions we ask

Let S ⊂ Rn be the k-skeleton of a polytope. Suppose we “put a copy” of S around every point of Rn. What is the minimal dimension/measure of the resulting set? Wait... what do we mean by “copy”?

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Statement of results Proof outline for measure The kinds of questions we ask

Let S ⊂ Rn be the k-skeleton of a polytope. Suppose we “put a copy” of S around every point of Rn. What is the minimal dimension/measure of the resulting set? Wait... what do we mean by “copy”? a scaled copy: x + rS (r > 0) a rotated copy: x + T(S) (T ∈ SO(n)) a scaled and rotated copy: x + rT(S)

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Statement of results Proof outline for measure (Corollaries of) our results

Theorem (C., Csörnyei, Héra, Keleti) For any integers 0 ≤ k < n, the minimal dimension of a Borel set A ⊂ Rn that contains the k-skeleton of

  • 1. a scaled copy of a cube around every point of Rn is n − 1

[Thornton 2015];

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Statement of results Proof outline for measure (Corollaries of) our results

Theorem (C., Csörnyei, Héra, Keleti) For any integers 0 ≤ k < n, the minimal dimension of a Borel set A ⊂ Rn that contains the k-skeleton of

  • 1. a scaled copy of a cube around every point of Rn is n − 1

[Thornton 2015];

  • 2. a scaled and rotated copy of a cube around every point of

Rn is k;

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Statement of results Proof outline for measure (Corollaries of) our results

Theorem (C., Csörnyei, Héra, Keleti) For any integers 0 ≤ k < n, the minimal dimension of a Borel set A ⊂ Rn that contains the k-skeleton of

  • 1. a scaled copy of a cube around every point of Rn is n − 1

[Thornton 2015];

  • 2. a scaled and rotated copy of a cube around every point of

Rn is k;

  • 3. a rotated copy of a cube around every point of Rn is k + 1;

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Statement of results Proof outline for measure (Corollaries of) our results

Theorem (C., Csörnyei, Héra, Keleti) For any integers 0 ≤ k < n, the minimal dimension of a Borel set A ⊂ Rn that contains the k-skeleton of

  • 1. a scaled copy of a cube around every point of Rn is n − 1

[Thornton 2015];

  • 2. a scaled and rotated copy of a cube around every point of

Rn is k;

  • 3. a rotated copy of a cube around every point of Rn is k + 1;
  • 4. a rotated cube of every size around every point of Rn is

k + 1.

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Statement of results Proof outline for measure (Corollaries of) our results

Theorem (C., Csörnyei, Héra, Keleti) For any integers 0 ≤ k < n, the minimal dimension of a Borel set A ⊂ Rn that contains the k-skeleton of

  • 1. a scaled copy of a cube around every point of Rn is n − 1

[Thornton 2015];

  • 2. a scaled and rotated copy of a cube around every point of

Rn is k;

  • 3. a rotated copy of a cube around every point of Rn is k + 1;
  • 4. a rotated cube of every size around every point of Rn is

k + 1. Theorem (C., Csörnyei, Héra, Keleti) Let n ≥ 2. Then there exists a set of Lebesgue measure zero that contains the boundary (i.e., (n − 1)-skeleton) of a rotated cube of every size around every point of Rn.

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Statement of results Proof outline for measure Strategy (Baire Category)

To prove these results, we show that in these cases, the minimal dimension/measure is also the typical one. (We also found situations where minimal = typical.) Recall that “typical” means “on a residual set”

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Statement of results Proof outline for measure Strategy (Baire Category)

To prove these results, we show that in these cases, the minimal dimension/measure is also the typical one. (We also found situations where minimal = typical.) Recall that “typical” means “on a residual set” General strategy: Prove residuality for a single affine k-plane not containing 0. Use Baire Category Theorem to obtain result for countable unions.

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Statement of results Proof outline for measure Statement of result

For simplicity we work in R2. Let P1 denote the set of all directions in R2. (We identify it with R/πZ.)

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Statement of results Proof outline for measure Statement of result

For simplicity we work in R2. Let P1 denote the set of all directions in R2. (We identify it with R/πZ.) For (x, θ) ∈ R2 × P1, we let ℓ(x, θ) ⊂ R2 be the line through x in direction θ (i.e., making angle θ with the horizontal axis).

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Statement of results Proof outline for measure Statement of result

For simplicity we work in R2. Let P1 denote the set of all directions in R2. (We identify it with R/πZ.) For (x, θ) ∈ R2 × P1, we let ℓ(x, θ) ⊂ R2 be the line through x in direction θ (i.e., making angle θ with the horizontal axis). Fix a nonempty compact set C ⊂ R2. Let P = {K ⊂ C × P1 : K compact, has full projection onto C}.

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Statement of results Proof outline for measure Statement of result

For simplicity we work in R2. Let P1 denote the set of all directions in R2. (We identify it with R/πZ.) For (x, θ) ∈ R2 × P1, we let ℓ(x, θ) ⊂ R2 be the line through x in direction θ (i.e., making angle θ with the horizontal axis). Fix a nonempty compact set C ⊂ R2. Let P = {K ⊂ C × P1 : K compact, has full projection onto C}. Fact: P is a complete metric space with respect to the Hausdorff metric d(A, B) = inf{ǫ ≥ 0 : A ⊂ Bǫ and B ⊂ Aǫ}.

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Statement of results Proof outline for measure Statement of result

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ)

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Statement of results Proof outline for measure Statement of result

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ) We’ll sketch the proof of the following Nikodym-type result. Theorem (C., Csörnyei, Héra, Keleti) For a typical K ∈ P, the set

  • (x,θ)∈K

ℓ(x, θ) \ {x} has measure zero.

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Statement of results Proof outline for measure Statement of result

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ) We’ll sketch the proof of the following Nikodym-type result. Theorem (C., Csörnyei, Héra, Keleti) For a typical K ∈ P, the set

  • (x,θ)∈K

ℓ(x, θ) \ {x} has measure zero. (The same ideas are used in the previously stated theorem about boundaries of cubes.)

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Statement of results Proof outline for measure Turning it into a geometric problem

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ)

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Statement of results Proof outline for measure Turning it into a geometric problem

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ) Fix a ball B ⊂ R2. Given any K ∈ P, can we find an L ∈ P close to K, such that AL ∩ B is small? (B is “bad.”)

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Statement of results Proof outline for measure Turning it into a geometric problem

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ) Fix a ball B ⊂ R2. Given any K ∈ P, can we find an L ∈ P close to K, such that AL ∩ B is small? (B is “bad.”) Local version: Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a L ⊂ R2 × P1 s.t.

  • 1. L is in a small neighbourhood of (x, θ).
  • 2. AL ∩ B has small measure.
  • 3. The projection of L onto R2 contains some nbhd of x.

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Statement of results Proof outline for measure Turning it into a geometric problem

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ) Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a L ⊂ R2 × P1 s.t.

  • 1. L is in a small neighbourhood of (x, θ).
  • 2. AL ∩ B has small measure.
  • 3. The projection of L onto R2 contains some nbhd of x.

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Statement of results Proof outline for measure Turning it into a geometric problem

P = {K ⊂ C × P1 : K compact, has full projection onto C} AK =

  • (x,θ)∈K

ℓ(x, θ) Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a L ⊂ R2 × P1 s.t.

  • 1. L is in a small neighbourhood of (x, θ).
  • 2. AL ∩ B has small measure.
  • 3. The projection of L onto R2 contains some nbhd of x.

Goal (restated, more geometric): Find a set of lines s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

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Statement of results Proof outline for measure The main geometric construction

Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a “set of lines” s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

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Statement of results Proof outline for measure The main geometric construction

Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a “set of lines” s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

If B is disjoint from ℓ(x, θ), then this is easy: Form a small double cone by rotating ℓ(x, θ) around some point on the line

  • ther than x. Then the double cone is disjoint from B and

contains a neighbourhood of x.

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Statement of results Proof outline for measure The main geometric construction

Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a set of lines s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

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Statement of results Proof outline for measure The main geometric construction

Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a set of lines s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

If B intersects ℓ(x, θ) the same construction does not work... what do we do instead?

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Statement of results Proof outline for measure The main geometric construction

Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a set of lines s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

If B intersects ℓ(x, θ) the same construction does not work... what do we do instead? We take the double cone as before. We cut it into many smaller double cones and translate them so that they overlap with each

  • ther a lot in B.

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Statement of results Proof outline for measure The main geometric construction

Fix a ball B ⊂ R2 and a (x, θ) ∈ R2 × P1. Goal: Find a set of lines s.t.

  • 1. Each line is “close” to ℓ(x, θ).
  • 2. (union of these lines) ∩ B has small measure.
  • 3. The union of these lines contains some nbhd of x.

If B intersects ℓ(x, θ) the same construction does not work... what do we do instead? We take the double cone as before. We cut it into many smaller double cones and translate them so that they overlap with each

  • ther a lot in B.

Note: this doesn’t work if x ∈ B.

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Statement of results Proof outline for measure The main geometric construction

Key ideas: (WLOG ℓ(x, θ) is the vertical line {(y1, y2) : y2 = 0}.)

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Statement of results Proof outline for measure The main geometric construction

Key ideas: (WLOG ℓ(x, θ) is the vertical line {(y1, y2) : y2 = 0}.)

  • 1. Take a (one-sided) cone D with vertex at the origin,
  • pening vertically upwards. If we translate the cone

downwards to ˜ D, the translated copy contains D.

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Statement of results Proof outline for measure The main geometric construction

Key ideas: (WLOG ℓ(x, θ) is the vertical line {(y1, y2) : y2 = 0}.)

  • 1. Take a (one-sided) cone D with vertex at the origin,
  • pening vertically upwards. If we translate the cone

downwards to ˜ D, the translated copy contains D.

  • 2. Furthermore, suppose D is very thin. If we look above the

x1-axis, ˜ D is contained in a small neighborhood D.

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Statement of results Proof outline for measure The main geometric construction

Key ideas: (WLOG ℓ(x, θ) is the vertical line {(y1, y2) : y2 = 0}.)

  • 1. Take a (one-sided) cone D with vertex at the origin,
  • pening vertically upwards. If we translate the cone

downwards to ˜ D, the translated copy contains D.

  • 2. Furthermore, suppose D is very thin. If we look above the

x1-axis, ˜ D is contained in a small neighborhood D. i.e., D ∩ {y1 > 0} ⊂ ˜ D ∩ {y1 > 0} ⊂ (small nbhd of D) ∩ {y1 > 0}

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Statement of results Proof outline for measure The main geometric construction

Key ideas: (WLOG ℓ(x, θ) is the vertical line {(y1, y2) : y2 = 0}.)

  • 1. Take a (one-sided) cone D with vertex at the origin,
  • pening vertically upwards. If we translate the cone

downwards to ˜ D, the translated copy contains D.

  • 2. Furthermore, suppose D is very thin. If we look above the

x1-axis, ˜ D is contained in a small neighborhood D. i.e., D ∩ {y1 > 0} ⊂ ˜ D ∩ {y1 > 0} ⊂ (small nbhd of D) ∩ {y1 > 0}

  • 3. The tip of a cone has small area.

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Statement of results Proof outline for measure The main geometric construction

For B ⊂ R2 a ball and ǫ > 0, let A(B, ǫ) be the set {K ∈ P : B ∩

  • (x,θ)∈K

x∈2B

ℓ(x, θ) has measure < ǫ}

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Statement of results Proof outline for measure The main geometric construction

For B ⊂ R2 a ball and ǫ > 0, let A(B, ǫ) be the set {K ∈ P : B ∩

  • (x,θ)∈K

x∈2B

ℓ(x, θ) has measure < ǫ} The argument we just presented shows that A(B, ǫ) is dense in P.

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Statement of results Proof outline for measure The main geometric construction

For B ⊂ R2 a ball and ǫ > 0, let A(B, ǫ) be the set {K ∈ P : B ∩

  • (x,θ)∈K

x∈2B

ℓ(x, θ) has measure < ǫ} The argument we just presented shows that A(B, ǫ) is dense in

  • P. With a little more work, we can deduce the theorem:

Theorem For a typical K ∈ P, the set

  • (x,θ)∈K

ℓ(x, θ) \ {x} has measure zero.

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Statement of results Proof outline for measure The main geometric construction

For B ⊂ R2 a ball and ǫ > 0, let A(B, ǫ) be the set {K ∈ P : B ∩

  • (x,θ)∈K

x∈2B

ℓ(x, θ) has measure < ǫ} The argument we just presented shows that A(B, ǫ) is dense in

  • P. With a little more work, we can deduce the theorem:

Theorem For a typical K ∈ P, the set

  • (x,θ)∈K

ℓ(x, θ) \ {x} has measure zero. Thank you!

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