how to solve falconer's distance problem using ergodic theory

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How to solve Falconer’s distance problem using ergodic theory Jonathan M. Fraser The University of Manchester joint with T. Sahlsten and A. Ferguson and with M. Pollicott Jonathan Fraser Falconer’s distance problem

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Page 1: How to solve Falconer's distance problem using ergodic theory

How to solve Falconer’s distance problemusing ergodic theory

Jonathan M. FraserThe University of Manchester

joint with T. Sahlsten and A. Ferguson and with M. Pollicott

Jonathan Fraser Falconer’s distance problem

Page 2: How to solve Falconer's distance problem using ergodic theory

Distance sets

Given a set E ⊂ Rd , the distance set of E is the set of all distancesrealised by pairs of points in E .

D(E ) = {|x − y | : x , y ∈ E}.

It is interesting to compare the ‘sizes’ of E and D(E ):

Question: Suppose E has cardinality n, what is the minimum possiblecardinality of D(E )?

Jonathan Fraser Falconer’s distance problem

Page 3: How to solve Falconer's distance problem using ergodic theory

Distance sets

Given a set E ⊂ Rd , the distance set of E is the set of all distancesrealised by pairs of points in E .

D(E ) = {|x − y | : x , y ∈ E}.

It is interesting to compare the ‘sizes’ of E and D(E ):

Question: Suppose E has cardinality n, what is the minimum possiblecardinality of D(E )?

Jonathan Fraser Falconer’s distance problem

Page 4: How to solve Falconer's distance problem using ergodic theory

Distance sets

Given a set E ⊂ Rd , the distance set of E is the set of all distancesrealised by pairs of points in E .

D(E ) = {|x − y | : x , y ∈ E}.

It is interesting to compare the ‘sizes’ of E and D(E ):

Question: Suppose E has cardinality n, what is the minimum possiblecardinality of D(E )?

Jonathan Fraser Falconer’s distance problem

Page 5: How to solve Falconer's distance problem using ergodic theory

Distance sets

Given a set E ⊂ Rd , the distance set of E is the set of all distancesrealised by pairs of points in E .

D(E ) = {|x − y | : x , y ∈ E}.

It is interesting to compare the ‘sizes’ of E and D(E ):

Question: Suppose E has cardinality n, what is the minimum possiblecardinality of D(E )?

Jonathan Fraser Falconer’s distance problem

Page 6: How to solve Falconer's distance problem using ergodic theory

Distance sets

Let g(n) = min{|D(E )| : |E | = n}.

Theorem (Erdos 1946)

For sets E in the plane√n − 3/4− 1/2 6 g(n) 6 cn/

√log n

Jonathan Fraser Falconer’s distance problem

Page 7: How to solve Falconer's distance problem using ergodic theory

Distance sets

Let g(n) = min{|D(E )| : |E | = n}.

Theorem (Erdos 1946)

For sets E in the plane√n − 3/4− 1/2 6 g(n) 6 cn/

√log n

Jonathan Fraser Falconer’s distance problem

Page 8: How to solve Falconer's distance problem using ergodic theory

Distance sets

The upper bound follows from the following simple example:

• Let E ⊂ Z2 be a√n ×√n square grid.

• Every distance realised by E is the square root of a number boundedabove by 2n which is the sum of two squares.

• But the Landau-Ramanujan Theorem says that the number of positiveintegers less than x that are the sum of two squares is bounded by aconstant times x/

√log x .

• So |D(E )| is bounded above by a constant times n/√

log n

Jonathan Fraser Falconer’s distance problem

Page 9: How to solve Falconer's distance problem using ergodic theory

Distance sets

The upper bound follows from the following simple example:

• Let E ⊂ Z2 be a√n ×√n square grid.

• Every distance realised by E is the square root of a number boundedabove by 2n which is the sum of two squares.

• But the Landau-Ramanujan Theorem says that the number of positiveintegers less than x that are the sum of two squares is bounded by aconstant times x/

√log x .

• So |D(E )| is bounded above by a constant times n/√

log n

Jonathan Fraser Falconer’s distance problem

Page 10: How to solve Falconer's distance problem using ergodic theory

Distance sets

The upper bound follows from the following simple example:

• Let E ⊂ Z2 be a√n ×√n square grid.

• Every distance realised by E is the square root of a number boundedabove by 2n which is the sum of two squares.

• But the Landau-Ramanujan Theorem says that the number of positiveintegers less than x that are the sum of two squares is bounded by aconstant times x/

√log x .

• So |D(E )| is bounded above by a constant times n/√

log n

Jonathan Fraser Falconer’s distance problem

Page 11: How to solve Falconer's distance problem using ergodic theory

Distance sets

The upper bound follows from the following simple example:

• Let E ⊂ Z2 be a√n ×√n square grid.

• Every distance realised by E is the square root of a number boundedabove by 2n which is the sum of two squares.

• But the Landau-Ramanujan Theorem says that the number of positiveintegers less than x that are the sum of two squares is bounded by aconstant times x/

√log x .

• So |D(E )| is bounded above by a constant times n/√

log n

Jonathan Fraser Falconer’s distance problem

Page 12: How to solve Falconer's distance problem using ergodic theory

Distance sets

The upper bound follows from the following simple example:

• Let E ⊂ Z2 be a√n ×√n square grid.

• Every distance realised by E is the square root of a number boundedabove by 2n which is the sum of two squares.

• But the Landau-Ramanujan Theorem says that the number of positiveintegers less than x that are the sum of two squares is bounded by aconstant times x/

√log x .

• So |D(E )| is bounded above by a constant times n/√

log n

Jonathan Fraser Falconer’s distance problem

Page 13: How to solve Falconer's distance problem using ergodic theory

Distance sets

Landau and Ramanujan

Jonathan Fraser Falconer’s distance problem

Page 14: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.

This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 15: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 16: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2

(1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 17: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3

(2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 18: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7

(5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 19: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5

(4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 20: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7

(6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 21: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636...

(0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 22: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137...

(0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 23: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 24: How to solve Falconer's distance problem using ergodic theory

Distance sets

In 1946 Erdos conjectured that this upper bound (n1−o(1)) was sharp.This problem received a lot of attention over the years and the lowerbound was gradually improved:

• 1946 Erdos: n1/2 (1/2 ≈ 0.500)

• 1952 Moser: n2/3 (2/3 ≈ 0.667)

• 1984 Fan Chung: n5/7 (5/7 ≈ 0.714)

• 1993 Szekely: n4/5 (4/5 ≈ 0.800)

• 2001 Solymosi-Toth: n6/7 (6/7 ≈ 0.857)

• 2003 Tardos: n0.863636... (0.863636 · · · ≈ 0.864)

• 2004 Katz-Tardos: n0.864137... (0.864137 · · · ≈ 0.864)

• 2013 Guth-Katz: n/ log(n)

Jonathan Fraser Falconer’s distance problem

Page 25: How to solve Falconer's distance problem using ergodic theory

Distance sets

Larry Guth and Netz Hawk Katz

Jonathan Fraser Falconer’s distance problem

Page 26: How to solve Falconer's distance problem using ergodic theory

Distance sets

...but this talk is not about finite sets.

In 1985 Kenneth Falconer considered the same problem, but for Euncountable and ‘size’ interpreted as Hausdorff dimension.

Theorem (Falconer 1985)

For sets E ⊆ Rd , if dimH E ≥ d/2 + 1/2, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 27: How to solve Falconer's distance problem using ergodic theory

Distance sets

...but this talk is not about finite sets.

In 1985 Kenneth Falconer considered the same problem, but for Euncountable and ‘size’ interpreted as Hausdorff dimension.

Theorem (Falconer 1985)

For sets E ⊆ Rd , if dimH E ≥ d/2 + 1/2, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 28: How to solve Falconer's distance problem using ergodic theory

Distance sets

...but this talk is not about finite sets.

In 1985 Kenneth Falconer considered the same problem, but for Euncountable and ‘size’ interpreted as Hausdorff dimension.

Theorem (Falconer 1985)

For sets E ⊆ Rd , if dimH E ≥ d/2 + 1/2, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 29: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are several related conjectures on this problem.

Here is one

Conjecture

For sets E ⊆ Rd , if dimH E ≥ d/2, then dimHD(E ) = 1.

• 1985 Falconer: dimH E ≥ d/2 + 1/2⇒ dimHD(E ) = 1

• 1999 Wolff: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = 2)

• 2006 Erdogan: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = d)

• 2003 Bourgain: there exists a constant c > 1/2, such that for planar E

dimH E > 1⇒ dimHD(E ) > c

Jonathan Fraser Falconer’s distance problem

Page 30: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are several related conjectures on this problem. Here is one

Conjecture

For sets E ⊆ Rd , if dimH E ≥ d/2, then dimHD(E ) = 1.

• 1985 Falconer: dimH E ≥ d/2 + 1/2⇒ dimHD(E ) = 1

• 1999 Wolff: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = 2)

• 2006 Erdogan: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = d)

• 2003 Bourgain: there exists a constant c > 1/2, such that for planar E

dimH E > 1⇒ dimHD(E ) > c

Jonathan Fraser Falconer’s distance problem

Page 31: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are several related conjectures on this problem. Here is one

Conjecture

For sets E ⊆ Rd , if dimH E ≥ d/2, then dimHD(E ) = 1.

• 1985 Falconer: dimH E ≥ d/2 + 1/2⇒ dimHD(E ) = 1

• 1999 Wolff: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = 2)

• 2006 Erdogan: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = d)

• 2003 Bourgain: there exists a constant c > 1/2, such that for planar E

dimH E > 1⇒ dimHD(E ) > c

Jonathan Fraser Falconer’s distance problem

Page 32: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are several related conjectures on this problem. Here is one

Conjecture

For sets E ⊆ Rd , if dimH E ≥ d/2, then dimHD(E ) = 1.

• 1985 Falconer: dimH E ≥ d/2 + 1/2⇒ dimHD(E ) = 1

• 1999 Wolff: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = 2)

• 2006 Erdogan: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = d)

• 2003 Bourgain: there exists a constant c > 1/2, such that for planar E

dimH E > 1⇒ dimHD(E ) > c

Jonathan Fraser Falconer’s distance problem

Page 33: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are several related conjectures on this problem. Here is one

Conjecture

For sets E ⊆ Rd , if dimH E ≥ d/2, then dimHD(E ) = 1.

• 1985 Falconer: dimH E ≥ d/2 + 1/2⇒ dimHD(E ) = 1

• 1999 Wolff: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = 2)

• 2006 Erdogan: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = d)

• 2003 Bourgain: there exists a constant c > 1/2, such that for planar E

dimH E > 1⇒ dimHD(E ) > c

Jonathan Fraser Falconer’s distance problem

Page 34: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are several related conjectures on this problem. Here is one

Conjecture

For sets E ⊆ Rd , if dimH E ≥ d/2, then dimHD(E ) = 1.

• 1985 Falconer: dimH E ≥ d/2 + 1/2⇒ dimHD(E ) = 1

• 1999 Wolff: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = 2)

• 2006 Erdogan: dimH E ≥ d/2 + 1/3⇒ dimHD(E ) = 1 (for d = d)

• 2003 Bourgain: there exists a constant c > 1/2, such that for planar E

dimH E > 1⇒ dimHD(E ) > c

Jonathan Fraser Falconer’s distance problem

Page 35: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are also results which so far only apply to special classes of set.

Theorem (Orponen 2012)

Let E ⊆ R2 be a self-similar set with positive length. ThendimHD(E ) = 1.

• Orponen’s proof was split into two cases:

1) ‘Dense rotations’: recent pioneering work of Hochman and Shmerkin onfractal projection theorems gives the desired result.

2) ‘Discrete rotations’: This case can be reduced to the ‘no rotationscase’, and from there a delicate geometric argument yields the result.

Jonathan Fraser Falconer’s distance problem

Page 36: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are also results which so far only apply to special classes of set.

Theorem (Orponen 2012)

Let E ⊆ R2 be a self-similar set with positive length. ThendimHD(E ) = 1.

• Orponen’s proof was split into two cases:

1) ‘Dense rotations’: recent pioneering work of Hochman and Shmerkin onfractal projection theorems gives the desired result.

2) ‘Discrete rotations’: This case can be reduced to the ‘no rotationscase’, and from there a delicate geometric argument yields the result.

Jonathan Fraser Falconer’s distance problem

Page 37: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are also results which so far only apply to special classes of set.

Theorem (Orponen 2012)

Let E ⊆ R2 be a self-similar set with positive length. ThendimHD(E ) = 1.

• Orponen’s proof was split into two cases:

1) ‘Dense rotations’: recent pioneering work of Hochman and Shmerkin onfractal projection theorems gives the desired result.

2) ‘Discrete rotations’: This case can be reduced to the ‘no rotationscase’, and from there a delicate geometric argument yields the result.

Jonathan Fraser Falconer’s distance problem

Page 38: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are also results which so far only apply to special classes of set.

Theorem (Orponen 2012)

Let E ⊆ R2 be a self-similar set with positive length. ThendimHD(E ) = 1.

• Orponen’s proof was split into two cases:

1) ‘Dense rotations’: recent pioneering work of Hochman and Shmerkin onfractal projection theorems gives the desired result.

2) ‘Discrete rotations’: This case can be reduced to the ‘no rotationscase’, and from there a delicate geometric argument yields the result.

Jonathan Fraser Falconer’s distance problem

Page 39: How to solve Falconer's distance problem using ergodic theory

Distance sets

There are also results which so far only apply to special classes of set.

Theorem (Orponen 2012)

Let E ⊆ R2 be a self-similar set with positive length. ThendimHD(E ) = 1.

• Orponen’s proof was split into two cases:

1) ‘Dense rotations’: recent pioneering work of Hochman and Shmerkin onfractal projection theorems gives the desired result.

2) ‘Discrete rotations’: This case can be reduced to the ‘no rotationscase’, and from there a delicate geometric argument yields the result.

Jonathan Fraser Falconer’s distance problem

Page 40: How to solve Falconer's distance problem using ergodic theory

Distance sets

• Mike Hochman and Pablo Shmerkin published an important paper in2012 which studied the ergodic theory of the process of ‘blowing up’ ameasure.

Jonathan Fraser Falconer’s distance problem

Page 41: How to solve Falconer's distance problem using ergodic theory

What did Hochman-Shmerkin prove?

• Key idea: One can understand a set or measure by understanding itstangents.

• Refinement: One can understand a set or measure by understanding thedynamics of the process of zooming in to its tangents.

• Ideas date back to Hillel Furstenberg in the 60s-70s, but rediscoveredrecently by Furstenberg (2008), Gavish (2011), Hochman-Shmerkin (2012)and Hochman (2010/2013).

Jonathan Fraser Falconer’s distance problem

Page 42: How to solve Falconer's distance problem using ergodic theory

What did Hochman-Shmerkin prove?

• Key idea: One can understand a set or measure by understanding itstangents.

• Refinement: One can understand a set or measure by understanding thedynamics of the process of zooming in to its tangents.

• Ideas date back to Hillel Furstenberg in the 60s-70s, but rediscoveredrecently by Furstenberg (2008), Gavish (2011), Hochman-Shmerkin (2012)and Hochman (2010/2013).

Jonathan Fraser Falconer’s distance problem

Page 43: How to solve Falconer's distance problem using ergodic theory

What did Hochman-Shmerkin prove?

• Key idea: One can understand a set or measure by understanding itstangents.

• Refinement: One can understand a set or measure by understanding thedynamics of the process of zooming in to its tangents.

• Ideas date back to Hillel Furstenberg in the 60s-70s, but rediscoveredrecently by Furstenberg (2008), Gavish (2011), Hochman-Shmerkin (2012)and Hochman (2010/2013).

Jonathan Fraser Falconer’s distance problem

Page 44: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Let {Dk}k∈N be the filtration of Rd by half open dyadic cubes.

Let TD : Rd → Rd be the orientation preserving similitude that mapsa dyadic cube D onto [0, 1)d .

The magnification µD of a measure µ ∈ P([0, 1)d) to D withµ(D) > 0 is

µD =1

µ(D)TD(µ|D) ∈ P([0, 1)d)

If k ∈ N, let Dk(x) ∈ Dk be the cube with x ∈ Dk(x). Write

Ξ = {(x , µ) : µ ∈ P([0, 1)d) and µ(Dk(x)) > 0 for all k ∈ N}

and define the magnification operator M : Ξ→ Ξ by

M(x , µ) = (TD1(x)(x), µD1(x)).

Jonathan Fraser Falconer’s distance problem

Page 45: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Let {Dk}k∈N be the filtration of Rd by half open dyadic cubes.

Let TD : Rd → Rd be the orientation preserving similitude that mapsa dyadic cube D onto [0, 1)d .

The magnification µD of a measure µ ∈ P([0, 1)d) to D withµ(D) > 0 is

µD =1

µ(D)TD(µ|D) ∈ P([0, 1)d)

If k ∈ N, let Dk(x) ∈ Dk be the cube with x ∈ Dk(x). Write

Ξ = {(x , µ) : µ ∈ P([0, 1)d) and µ(Dk(x)) > 0 for all k ∈ N}

and define the magnification operator M : Ξ→ Ξ by

M(x , µ) = (TD1(x)(x), µD1(x)).

Jonathan Fraser Falconer’s distance problem

Page 46: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Let {Dk}k∈N be the filtration of Rd by half open dyadic cubes.

Let TD : Rd → Rd be the orientation preserving similitude that mapsa dyadic cube D onto [0, 1)d .

The magnification µD of a measure µ ∈ P([0, 1)d) to D withµ(D) > 0 is

µD =1

µ(D)TD(µ|D) ∈ P([0, 1)d)

If k ∈ N, let Dk(x) ∈ Dk be the cube with x ∈ Dk(x). Write

Ξ = {(x , µ) : µ ∈ P([0, 1)d) and µ(Dk(x)) > 0 for all k ∈ N}

and define the magnification operator M : Ξ→ Ξ by

M(x , µ) = (TD1(x)(x), µD1(x)).

Jonathan Fraser Falconer’s distance problem

Page 47: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Let {Dk}k∈N be the filtration of Rd by half open dyadic cubes.

Let TD : Rd → Rd be the orientation preserving similitude that mapsa dyadic cube D onto [0, 1)d .

The magnification µD of a measure µ ∈ P([0, 1)d) to D withµ(D) > 0 is

µD =1

µ(D)TD(µ|D) ∈ P([0, 1)d)

If k ∈ N, let Dk(x) ∈ Dk be the cube with x ∈ Dk(x). Write

Ξ = {(x , µ) : µ ∈ P([0, 1)d) and µ(Dk(x)) > 0 for all k ∈ N}

and define the magnification operator M : Ξ→ Ξ by

M(x , µ) = (TD1(x)(x), µD1(x)).

Jonathan Fraser Falconer’s distance problem

Page 48: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Let {Dk}k∈N be the filtration of Rd by half open dyadic cubes.

Let TD : Rd → Rd be the orientation preserving similitude that mapsa dyadic cube D onto [0, 1)d .

The magnification µD of a measure µ ∈ P([0, 1)d) to D withµ(D) > 0 is

µD =1

µ(D)TD(µ|D) ∈ P([0, 1)d)

If k ∈ N, let Dk(x) ∈ Dk be the cube with x ∈ Dk(x). Write

Ξ = {(x , µ) : µ ∈ P([0, 1)d) and µ(Dk(x)) > 0 for all k ∈ N}

and define the magnification operator M : Ξ→ Ξ by

M(x , µ) = (TD1(x)(x), µD1(x)).

Jonathan Fraser Falconer’s distance problem

Page 49: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Let {Dk}k∈N be the filtration of Rd by half open dyadic cubes.

Let TD : Rd → Rd be the orientation preserving similitude that mapsa dyadic cube D onto [0, 1)d .

The magnification µD of a measure µ ∈ P([0, 1)d) to D withµ(D) > 0 is

µD =1

µ(D)TD(µ|D) ∈ P([0, 1)d)

If k ∈ N, let Dk(x) ∈ Dk be the cube with x ∈ Dk(x). Write

Ξ = {(x , µ) : µ ∈ P([0, 1)d) and µ(Dk(x)) > 0 for all k ∈ N}

and define the magnification operator M : Ξ→ Ξ by

M(x , µ) = (TD1(x)(x), µD1(x)).

Jonathan Fraser Falconer’s distance problem

Page 50: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Iterating M, we see that

Mk(x , µ) = (TDk (x)(x), µDk (x)), (x , µ) ∈ Ξ.

Let (x , µ) ∈ Ξ and N ∈ N. The Nth scenery distribution of µ at x is

1

N

N−1∑k=0

δMk (x ,µ) ∈ P(Ξ).

A micromeasure distribution of µ at x is an accumulation point ofthe scenery distributions in P(Ξ) w.r.t. the weak topology.

The measure component of a micromeasure distribution is supportedon the micromeasures of µ at x (i.e. accumulation points of the‘minimeasures’ µDk (x), as k →∞)

Jonathan Fraser Falconer’s distance problem

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Magnification dynamics

Iterating M, we see that

Mk(x , µ) = (TDk (x)(x), µDk (x)), (x , µ) ∈ Ξ.

Let (x , µ) ∈ Ξ and N ∈ N. The Nth scenery distribution of µ at x is

1

N

N−1∑k=0

δMk (x ,µ) ∈ P(Ξ).

A micromeasure distribution of µ at x is an accumulation point ofthe scenery distributions in P(Ξ) w.r.t. the weak topology.

The measure component of a micromeasure distribution is supportedon the micromeasures of µ at x (i.e. accumulation points of the‘minimeasures’ µDk (x), as k →∞)

Jonathan Fraser Falconer’s distance problem

Page 52: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Iterating M, we see that

Mk(x , µ) = (TDk (x)(x), µDk (x)), (x , µ) ∈ Ξ.

Let (x , µ) ∈ Ξ and N ∈ N. The Nth scenery distribution of µ at x is

1

N

N−1∑k=0

δMk (x ,µ) ∈ P(Ξ).

A micromeasure distribution of µ at x is an accumulation point ofthe scenery distributions in P(Ξ) w.r.t. the weak topology.

The measure component of a micromeasure distribution is supportedon the micromeasures of µ at x (i.e. accumulation points of the‘minimeasures’ µDk (x), as k →∞)

Jonathan Fraser Falconer’s distance problem

Page 53: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Iterating M, we see that

Mk(x , µ) = (TDk (x)(x), µDk (x)), (x , µ) ∈ Ξ.

Let (x , µ) ∈ Ξ and N ∈ N. The Nth scenery distribution of µ at x is

1

N

N−1∑k=0

δMk (x ,µ) ∈ P(Ξ).

A micromeasure distribution of µ at x is an accumulation point ofthe scenery distributions in P(Ξ) w.r.t. the weak topology.

The measure component of a micromeasure distribution is supportedon the micromeasures of µ at x (i.e. accumulation points of the‘minimeasures’ µDk (x), as k →∞)

Jonathan Fraser Falconer’s distance problem

Page 54: How to solve Falconer's distance problem using ergodic theory

Magnification dynamics

Iterating M, we see that

Mk(x , µ) = (TDk (x)(x), µDk (x)), (x , µ) ∈ Ξ.

Let (x , µ) ∈ Ξ and N ∈ N. The Nth scenery distribution of µ at x is

1

N

N−1∑k=0

δMk (x ,µ) ∈ P(Ξ).

A micromeasure distribution of µ at x is an accumulation point ofthe scenery distributions in P(Ξ) w.r.t. the weak topology.

The measure component of a micromeasure distribution is supportedon the micromeasures of µ at x (i.e. accumulation points of the‘minimeasures’ µDk (x), as k →∞)

Jonathan Fraser Falconer’s distance problem

Page 55: How to solve Falconer's distance problem using ergodic theory

CP-distributions

• A CP-distribution Q is a special type of micromeasure distribution.

• We say µ generates a CP-distribution Q if

(1) Q is the only micromeasure distribution of µ at µ almost every x ;

(2) and at µ almost every x the q-sparse scenery distributions

1

N

N−1∑k=0

δMqk (x ,µ) ∈ P(Ξ)

converge to some distribution Qq for any q ∈ N, where each Qq maybe different from Q.

Condition (2) seems strange at first sight, but is essential to carrygeometric information from the micromeasure back to µ.

In ‘nice’ situations, (2) does not cause any problems in the proofs andoften Qq = Q for all q ∈ N.

Jonathan Fraser Falconer’s distance problem

Page 56: How to solve Falconer's distance problem using ergodic theory

CP-distributions

• A CP-distribution Q is a special type of micromeasure distribution.

• We say µ generates a CP-distribution Q if

(1) Q is the only micromeasure distribution of µ at µ almost every x ;

(2) and at µ almost every x the q-sparse scenery distributions

1

N

N−1∑k=0

δMqk (x ,µ) ∈ P(Ξ)

converge to some distribution Qq for any q ∈ N, where each Qq maybe different from Q.

Condition (2) seems strange at first sight, but is essential to carrygeometric information from the micromeasure back to µ.

In ‘nice’ situations, (2) does not cause any problems in the proofs andoften Qq = Q for all q ∈ N.

Jonathan Fraser Falconer’s distance problem

Page 57: How to solve Falconer's distance problem using ergodic theory

CP-distributions

• A CP-distribution Q is a special type of micromeasure distribution.

• We say µ generates a CP-distribution Q if

(1) Q is the only micromeasure distribution of µ at µ almost every x ;

(2) and at µ almost every x the q-sparse scenery distributions

1

N

N−1∑k=0

δMqk (x ,µ) ∈ P(Ξ)

converge to some distribution Qq for any q ∈ N, where each Qq maybe different from Q.

Condition (2) seems strange at first sight, but is essential to carrygeometric information from the micromeasure back to µ.

In ‘nice’ situations, (2) does not cause any problems in the proofs andoften Qq = Q for all q ∈ N.

Jonathan Fraser Falconer’s distance problem

Page 58: How to solve Falconer's distance problem using ergodic theory

CP-distributions

• A CP-distribution Q is a special type of micromeasure distribution.

• We say µ generates a CP-distribution Q if

(1) Q is the only micromeasure distribution of µ at µ almost every x ;

(2) and at µ almost every x the q-sparse scenery distributions

1

N

N−1∑k=0

δMqk (x ,µ) ∈ P(Ξ)

converge to some distribution Qq for any q ∈ N, where each Qq maybe different from Q.

Condition (2) seems strange at first sight, but is essential to carrygeometric information from the micromeasure back to µ.

In ‘nice’ situations, (2) does not cause any problems in the proofs andoften Qq = Q for all q ∈ N.

Jonathan Fraser Falconer’s distance problem

Page 59: How to solve Falconer's distance problem using ergodic theory

CP-distributions

• A CP-distribution Q is a special type of micromeasure distribution.

• We say µ generates a CP-distribution Q if

(1) Q is the only micromeasure distribution of µ at µ almost every x ;

(2) and at µ almost every x the q-sparse scenery distributions

1

N

N−1∑k=0

δMqk (x ,µ) ∈ P(Ξ)

converge to some distribution Qq for any q ∈ N, where each Qq maybe different from Q.

Condition (2) seems strange at first sight, but is essential to carrygeometric information from the micromeasure back to µ.

In ‘nice’ situations, (2) does not cause any problems in the proofs andoften Qq = Q for all q ∈ N.

Jonathan Fraser Falconer’s distance problem

Page 60: How to solve Falconer's distance problem using ergodic theory

Example: CP distributions for self-similar measures

When one zooms in on a set or measure with a self-similar structure,roughly speaking, one expects the tangent objects to be the same as theoriginal object.

Proposition (Hochman-Shmerkin 2012)

Let µ be a self-similar measure in Rd satisfying the strong separationcondition. Then there exists a Borel-set B with µ(B) > 0 and a similitudeS of Rd , such that

ν = µ(B)−1S(µ|B)

generates an ergodic CP distribution.

Jonathan Fraser Falconer’s distance problem

Page 61: How to solve Falconer's distance problem using ergodic theory

Example: CP distributions for self-similar measures

When one zooms in on a set or measure with a self-similar structure,roughly speaking, one expects the tangent objects to be the same as theoriginal object.

Proposition (Hochman-Shmerkin 2012)

Let µ be a self-similar measure in Rd satisfying the strong separationcondition. Then there exists a Borel-set B with µ(B) > 0 and a similitudeS of Rd , such that

ν = µ(B)−1S(µ|B)

generates an ergodic CP distribution.

Jonathan Fraser Falconer’s distance problem

Page 62: How to solve Falconer's distance problem using ergodic theory

Example: dimensions of projections

Let Πd ,k be the set of all orthogonal projections Rd → Rk , k < d .

Theorem (Hochman-Shmerkin 2012)

Suppose µ generates an ergodic CP distribution Qand let k ∈ N andε > 0. Then there exists an open dense set Uε ⊂ Πd ,k (which is also of fullmeasure) such that for all π ∈ Uε

dimH π(µ) > min{k , dimH µ} − ε.

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Page 63: How to solve Falconer's distance problem using ergodic theory

Example: dimensions of projections

Let Πd ,k be the set of all orthogonal projections Rd → Rk , k < d .

Theorem (Hochman-Shmerkin 2012)

Suppose µ generates an ergodic CP distribution Qand let k ∈ N andε > 0. Then there exists an open dense set Uε ⊂ Πd ,k (which is also of fullmeasure) such that for all π ∈ Uε

dimH π(µ) > min{k , dimH µ} − ε.

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Example: dimensions of C 1 images

Theorem (Hochman-Shmerkin 2012)

Suppose that a measure µ on [0, 1]d generates an ergodic CP-chain Q.Let π ∈ Πd ,k and ε > 0. Then there exists a δ > 0 such that for all C 1

maps g : [0, 1]d → Rk such that the maximal norm

supx∈supp(µ)

‖Dxg − π‖ < δ,

we havedimH gµ > dimH πµ− ε.

Jonathan Fraser Falconer’s distance problem

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Tuomas Sahlsten

Jonathan Fraser Falconer’s distance problem

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Tuomas Sahlsten and Andy Ferguson

Jonathan Fraser Falconer’s distance problem

Page 67: How to solve Falconer's distance problem using ergodic theory

An application to the distance problem

Theorem (Ferguson, F, Sahlsten, 2013)

If µ on R2 generates an ergodic CP distribution and H1(sptµ) > 0, then

dimHD(sptµ) ≥ min{1, dimH µ}.

Theorem (Ferguson, F, Sahlsten, 2013)

If µ is a self-affine measure supported on a Bedford-McMullen carpet, thenµ generates an ergodic CP-distribution.

Corollary (Ferguson, F, Sahlsten, 2013)

If E is a Bedford-McMullen carpet with dimH E ≥ 1, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

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An application to the distance problem

Theorem (Ferguson, F, Sahlsten, 2013)

If µ on R2 generates an ergodic CP distribution and H1(sptµ) > 0, then

dimHD(sptµ) ≥ min{1, dimH µ}.

Theorem (Ferguson, F, Sahlsten, 2013)

If µ is a self-affine measure supported on a Bedford-McMullen carpet, thenµ generates an ergodic CP-distribution.

Corollary (Ferguson, F, Sahlsten, 2013)

If E is a Bedford-McMullen carpet with dimH E ≥ 1, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 69: How to solve Falconer's distance problem using ergodic theory

An application to the distance problem

Theorem (Ferguson, F, Sahlsten, 2013)

If µ on R2 generates an ergodic CP distribution and H1(sptµ) > 0, then

dimHD(sptµ) ≥ min{1, dimH µ}.

Theorem (Ferguson, F, Sahlsten, 2013)

If µ is a self-affine measure supported on a Bedford-McMullen carpet, thenµ generates an ergodic CP-distribution.

Corollary (Ferguson, F, Sahlsten, 2013)

If E is a Bedford-McMullen carpet with dimH E ≥ 1, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 70: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 71: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 72: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 73: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1.

This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 74: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.

Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 75: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 76: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1.

Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 77: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let µ be a probability measure with support K ⊂ R2 satisfyingH1(K ) > 0 and suppose µ generates an ergodic CP-distribution.

• Define the direction set of K by

Dir(K ) =

{x − y

|x − y |: x , y ∈ K , x 6= y

}

• Case 1: Dir(K ) is not dense in S1. This means K is 1-rectifiable.Combined with the fact that K has positive length, a result of Besicovitchand Miller gives that D(K ) contains an interval.

• Case 2: Dir(K ) is dense in S1. Let ε > 0 and choose π ∈ Dir(K ) suchthat

dimH π(µ) > min{1, dimH µ} − ε.

Jonathan Fraser Falconer’s distance problem

Page 78: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let g be the pinned distance map at x , g(z) = |x − z |

• Choose r > 0 small enough to guarantee that

supz∈B(y ,r)

‖Dzg − π‖ < δ.

Jonathan Fraser Falconer’s distance problem

Page 79: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let g be the pinned distance map at x , g(z) = |x − z |

• Choose r > 0 small enough to guarantee that

supz∈B(y ,r)

‖Dzg − π‖ < δ.

Jonathan Fraser Falconer’s distance problem

Page 80: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• Let g be the pinned distance map at x , g(z) = |x − z |

• Choose r > 0 small enough to guarantee that

supz∈B(y ,r)

‖Dzg − π‖ < δ.

Jonathan Fraser Falconer’s distance problem

Page 81: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• It follows form the Besicovitch density point theorem that the measure νdefined to be the restriction of µ to B(y , r) generates exactly the sameCP-chain as µ.

• Therefore

dimHD(K ) ≥ dimH g(ν) > dimH πµ− ε > min{1, dimH µ} − 2ε

completing the proof.

Jonathan Fraser Falconer’s distance problem

Page 82: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• It follows form the Besicovitch density point theorem that the measure νdefined to be the restriction of µ to B(y , r) generates exactly the sameCP-chain as µ.

• Therefore

dimHD(K ) ≥ dimH g(ν) > dimH πµ− ε > min{1, dimH µ} − 2ε

completing the proof.

Jonathan Fraser Falconer’s distance problem

Page 83: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• It follows form the Besicovitch density point theorem that the measure νdefined to be the restriction of µ to B(y , r) generates exactly the sameCP-chain as µ.

• Therefore

dimHD(K ) ≥

dimH g(ν) > dimH πµ− ε > min{1, dimH µ} − 2ε

completing the proof.

Jonathan Fraser Falconer’s distance problem

Page 84: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• It follows form the Besicovitch density point theorem that the measure νdefined to be the restriction of µ to B(y , r) generates exactly the sameCP-chain as µ.

• Therefore

dimHD(K ) ≥ dimH g(ν) >

dimH πµ− ε > min{1, dimH µ} − 2ε

completing the proof.

Jonathan Fraser Falconer’s distance problem

Page 85: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• It follows form the Besicovitch density point theorem that the measure νdefined to be the restriction of µ to B(y , r) generates exactly the sameCP-chain as µ.

• Therefore

dimHD(K ) ≥ dimH g(ν) > dimH πµ− ε >

min{1, dimH µ} − 2ε

completing the proof.

Jonathan Fraser Falconer’s distance problem

Page 86: How to solve Falconer's distance problem using ergodic theory

An outline of the proof

• It follows form the Besicovitch density point theorem that the measure νdefined to be the restriction of µ to B(y , r) generates exactly the sameCP-chain as µ.

• Therefore

dimHD(K ) ≥ dimH g(ν) > dimH πµ− ε > min{1, dimH µ} − 2ε

completing the proof.

Jonathan Fraser Falconer’s distance problem

Page 87: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

In recent joint work with Mark Pollicott, we’ve been trying to apply thisresult to prove the distance set conjecture for conformally generatedfractals.

• self-conformal sets

• Totally disconnected Hyperbolic Julia sets are self-conformal

• Limit sets of Schottky groups are subsets of self-conformal setscorresponding to subshifts of finite type.

Jonathan Fraser Falconer’s distance problem

Page 88: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

In recent joint work with Mark Pollicott, we’ve been trying to apply thisresult to prove the distance set conjecture for conformally generatedfractals.

• self-conformal sets

• Totally disconnected Hyperbolic Julia sets are self-conformal

• Limit sets of Schottky groups are subsets of self-conformal setscorresponding to subshifts of finite type.

Jonathan Fraser Falconer’s distance problem

Page 89: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

In recent joint work with Mark Pollicott, we’ve been trying to apply thisresult to prove the distance set conjecture for conformally generatedfractals.

• self-conformal sets

• Totally disconnected Hyperbolic Julia sets are self-conformal

• Limit sets of Schottky groups are subsets of self-conformal setscorresponding to subshifts of finite type.

Jonathan Fraser Falconer’s distance problem

Page 90: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

In recent joint work with Mark Pollicott, we’ve been trying to apply thisresult to prove the distance set conjecture for conformally generatedfractals.

• self-conformal sets

• Totally disconnected Hyperbolic Julia sets are self-conformal

• Limit sets of Schottky groups are subsets of self-conformal setscorresponding to subshifts of finite type.

Jonathan Fraser Falconer’s distance problem

Page 91: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

In recent joint work with Mark Pollicott, we’ve been trying to apply thisresult to prove the distance set conjecture for conformally generatedfractals.

• self-conformal sets

• Totally disconnected Hyperbolic Julia sets are self-conformal

• Limit sets of Schottky groups are subsets of self-conformal setscorresponding to subshifts of finite type.

Jonathan Fraser Falconer’s distance problem

Page 92: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

Proposition (Hochman-Shmerkin 2012)

Let µ be a self-similar measure in Rd satisfying the strong separationcondition. Then there exists a Borel-set B with µ(B) > 0 and a similitudeS of Rd , such that

ν = µ(B)−1S(µ|B)

generates an ergodic CP distribution.

Proposition (F, Pollicott 2014)

Let µ be a Gibbs measure supported on a self-conformal set satisfying thestrong separation condition. Then there exists a Borel-set B withµ(B) > 0, a conformal map S of Rd and a measure µ′ ≡ µ, such that

ν = µ′(B)−1S(µ′|B)

generates an ergodic CP distribution.

Jonathan Fraser Falconer’s distance problem

Page 93: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

Proposition (Hochman-Shmerkin 2012)

Let µ be a self-similar measure in Rd satisfying the strong separationcondition. Then there exists a Borel-set B with µ(B) > 0 and a similitudeS of Rd , such that

ν = µ(B)−1S(µ|B)

generates an ergodic CP distribution.

Proposition (F, Pollicott 2014)

Let µ be a Gibbs measure supported on a self-conformal set satisfying thestrong separation condition. Then there exists a Borel-set B withµ(B) > 0, a conformal map S of Rd and a measure µ′ ≡ µ, such that

ν = µ′(B)−1S(µ′|B)

generates an ergodic CP distribution.

Jonathan Fraser Falconer’s distance problem

Page 94: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E ) - which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 95: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E ) - which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 96: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E )

- which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 97: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E ) - which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 98: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E ) - which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 99: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E ) - which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 100: How to solve Falconer's distance problem using ergodic theory

Conformally generated fractals

• The fact that only an image of a little piece of the original measure isshown to generate an ergodic CP-distribution is not a problem in theself-similar case.

• It is a problem in the self-conformal case, however, especially if the mapS is not a similarity.

• You can only obtain the distance set conjecture for S(E ) - which is veryunsatisfying!

Jonathan Fraser Falconer’s distance problem

Page 101: How to solve Falconer's distance problem using ergodic theory

A further application to the distance problem

Theorem (F, Pollicott 2014)

If µ on R2 generates an ergodic CP distribution and H1(sptµ) > 0, then

dimHD(S(sptµ)) ≥ min{1, dimH µ}

for any conformal map S .

Corollary (F, Pollicott 2014)

If E is a self-conformal set with dimH E > 1, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 102: How to solve Falconer's distance problem using ergodic theory

A further application to the distance problem

Theorem (F, Pollicott 2014)

If µ on R2 generates an ergodic CP distribution and H1(sptµ) > 0, then

dimHD(S(sptµ)) ≥ min{1, dimH µ}

for any conformal map S .

Corollary (F, Pollicott 2014)

If E is a self-conformal set with dimH E > 1, then dimHD(E ) = 1.

Jonathan Fraser Falconer’s distance problem

Page 103: How to solve Falconer's distance problem using ergodic theory

Thank you!

Jonathan Fraser Falconer’s distance problem

Page 104: How to solve Falconer's distance problem using ergodic theory

Some References

K. J. Falconer. On the Hausdorff dimensions of distance sets, Mathematika, 32,(1985), 206–212.

A. Ferguson, J. M. Fraser & T. Sahlsten. Scaling scenery of (×m,×n)invariant measures, Advances in Mathematics, 268, (2015), 564–602.

J. M. Fraser & M. Pollicott. Micromesure distributions and applications forconformally generated fractals, preprint, (2014).

M. Hochman & P. Shmerkin. Local entropy averages and projections of fractalmeasures, Annals of Mathematics, 175, (2012), 1001–1059

T. Orponen. On the distance sets of self-similar sets, Nonlinearity, 25, (2012),1919–1929.

Jonathan Fraser Falconer’s distance problem