The story of squaring the circle M. Laczkovich E otv os - - PowerPoint PPT Presentation

The story of squaring the circle

• M. Laczkovich

E¨

• tv¨
• s University, Budapest

Warwick, July 13, 2017

Theorem (G. Vitali 1905)

There is no translation invariant measure µ on P(R) satisfying µ([0, 1]) = 1.

Theorem (G. Vitali 1905)

There is no translation invariant measure µ on P(R) satisfying µ([0, 1]) = 1.

Theorem (F. Hausdorff 1914)

There is no rotation invariant finitely additive probability measure

• n the sphere S2 = {x ∈ R3 : |x| = 1}.

Theorem (G. Vitali 1905)

There is no translation invariant measure µ on P(R) satisfying µ([0, 1]) = 1.

Theorem (F. Hausdorff 1914)

There is no rotation invariant finitely additive probability measure

• n the sphere S2 = {x ∈ R3 : |x| = 1}.

Corollary

There is no isometry invariant finitely additive measure µ on P(R3) such that µ([0, 1]3) = 1.

Theorem (G. Vitali 1905)

There is no translation invariant measure µ on P(R) satisfying µ([0, 1]) = 1.

Theorem (F. Hausdorff 1914)

There is no rotation invariant finitely additive probability measure

• n the sphere S2 = {x ∈ R3 : |x| = 1}.

Corollary

There is no isometry invariant finitely additive measure µ on P(R3) such that µ([0, 1]3) = 1.

Theorem (S. Banach 1923)

There are isometry invariant finitely additive measures on P(R) and on P(R2) which are extensions of the Lebesgue measure.

• A. Tarski, J. von Neumann, A. Haar and others: measures on

groups, amenability, growth conditions etc.

• A. Tarski, J. von Neumann, A. Haar and others: measures on

groups, amenability, growth conditions etc.

Theorem (S. Banach and A. Tarski 1924)

If A, B ⊂ Rk (k ≥ 3) are bounded sets with nonempty interior, then A and B are equidecomposable.

• A. Tarski, J. von Neumann, A. Haar and others: measures on

groups, amenability, growth conditions etc.

Theorem (S. Banach and A. Tarski 1924)

If A, B ⊂ Rk (k ≥ 3) are bounded sets with nonempty interior, then A and B are equidecomposable. A and B are equidecomposable, if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn s.t. Ai is congruent to Bi (i = 1, . . . , n).

• A. Tarski, J. von Neumann, A. Haar and others: measures on

groups, amenability, growth conditions etc.

Theorem (S. Banach and A. Tarski 1924)

If A, B ⊂ Rk (k ≥ 3) are bounded sets with nonempty interior, then A and B are equidecomposable. A and B are equidecomposable, if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn s.t. Ai is congruent to Bi (i = 1, . . . , n).

• A. Tarski 1925 Is a disc in the plane equidecomposable with a

square of the same area?

• A. Tarski, J. von Neumann, A. Haar and others: measures on

groups, amenability, growth conditions etc.

Theorem (S. Banach and A. Tarski 1924)

If A, B ⊂ Rk (k ≥ 3) are bounded sets with nonempty interior, then A and B are equidecomposable. A and B are equidecomposable, if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn s.t. Ai is congruent to Bi (i = 1, . . . , n).

• A. Tarski 1925 Is a disc in the plane equidecomposable with a

square of the same area?

Theorem (M.L. 1990)

Yes.

Theorem (L. Grabowski, A. M´ ath´ e and O. Pikhurko 2015)

If A, B ⊂ Rk are bounded Borel sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A and B are equidecomposable using translations such that the pieces used in the decompositions are simultaneously Lebesgue measurable and have the Baire property.

Theorem (L. Grabowski, A. M´ ath´ e and O. Pikhurko 2015)

If A, B ⊂ Rk are bounded Borel sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A and B are equidecomposable using translations such that the pieces used in the decompositions are simultaneously Lebesgue measurable and have the Baire property.

Theorem (A.S. Marks and S.T. Unger 2016)

If A, B ⊂ Rk are bounded Borel sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A and B are equidecomposable with finitely many Borel pieces and only using translations.

Theorem (I. Dubins, M. Hirsch, J. Karush 1963)

The square and the disc are not “scissor-congruent”.

Theorem (I. Dubins, M. Hirsch, J. Karush 1963)

The square and the disc are not “scissor-congruent”.

Theorem (R.J. Gardner 1985)

The square and the disc are not equidecomposable if the pieces are moved by a locally discrete group of isometries.

Theorem (I. Dubins, M. Hirsch, J. Karush 1963)

The square and the disc are not “scissor-congruent”.

Theorem (R.J. Gardner 1985)

The square and the disc are not equidecomposable if the pieces are moved by a locally discrete group of isometries. A group G acting on Rk is locally discrete, if every finitely generated subgroup H ⊂ G is discrete; i.e., if C ∩ g(C) = ∅ for all but a finitely many g ∈ H for every bounded C.

Theorem (I. Dubins, M. Hirsch, J. Karush 1963)

The square and the disc are not “scissor-congruent”.

Theorem (R.J. Gardner 1985)

The square and the disc are not equidecomposable if the pieces are moved by a locally discrete group of isometries. A group G acting on Rk is locally discrete, if every finitely generated subgroup H ⊂ G is discrete; i.e., if C ∩ g(C) = ∅ for all but a finitely many g ∈ H for every bounded C. R.J. Gardner 1988 Conjecture: if a polytope and a convex body are equidecomposable under isometries of an amenable group, then they are equidecomposable under the same isometries with convex pieces.

u u 1−u 1−u 1 1

R

E = {(x, y): (x, y) ∈ R} is a bipartite graph between [0, 1] and [0, 1]. M ⊂ V is a matching if it is the graph of a bijection of [0, 1]

• nto itself.

u u 1−u 1−u 1 1

R

E = {(x, y): (x, y) ∈ R} is a bipartite graph between [0, 1] and [0, 1]. M ⊂ V is a matching if it is the graph of a bijection of [0, 1]

• nto itself.

Theorem (M.L. 1986)

If u ∈ [0, 1] \ Q then E contains a matching, but does not contain any matching which is Borel or measurable w.r.t. linear measure.

Let g1(x) = x + u, g2(x) = x − u, g3(x) = u − x, g4(x) = 1 − u − x. Then the gi are isometries of R, and they generate an amenable (in fact, solvable) group.

Let g1(x) = x + u, g2(x) = x − u, g3(x) = u − x, g4(x) = 1 − u − x. Then the gi are isometries of R, and they generate an amenable (in fact, solvable) group. R = [0, 1]2 ∩ 4

i=1 graph gi.

Let g1(x) = x + u, g2(x) = x − u, g3(x) = u − x, g4(x) = 1 − u − x. Then the gi are isometries of R, and they generate an amenable (in fact, solvable) group. R = [0, 1]2 ∩ 4

i=1 graph gi.

Let M ⊂ R be a matching. If Ai = π1(M ∩ graph gi), Bi = π2(M ∩ graph gi), then [0, 1] = A1 ∪ . . . ∪ A4, [0, 1] = B1 ∪ . . . ∪ B4 are decompositions and Bi = gi(Ai). So A = [0, 1] and B = [0, 1] are equidecomposable under isometries of an amenable group, but not equidecomposable under the same isometries with Borel pieces.

Let g1(x) = x + u, g2(x) = x − u, g3(x) = u − x, g4(x) = 1 − u − x. Then the gi are isometries of R, and they generate an amenable (in fact, solvable) group. R = [0, 1]2 ∩ 4

i=1 graph gi.

Let M ⊂ R be a matching. If Ai = π1(M ∩ graph gi), Bi = π2(M ∩ graph gi), then [0, 1] = A1 ∪ . . . ∪ A4, [0, 1] = B1 ∪ . . . ∪ B4 are decompositions and Bi = gi(Ai). So A = [0, 1] and B = [0, 1] are equidecomposable under isometries of an amenable group, but not equidecomposable under the same isometries with Borel pieces. R.J. Gardner 1988 “If my conjecture is false, then the circle can be squared.”

A t ∼ B if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and vectors a1, . . . , an s.t. Bi = Ai + ai (i = 1, . . . , n).

A t ∼ B if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and vectors a1, . . . , an s.t. Bi = Ai + ai (i = 1, . . . , n).

Theorem (M.L. 1992)

If A, B ⊂ Rk are bounded measurable sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A t ∼ B.

A t ∼ B if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and vectors a1, . . . , an s.t. Bi = Ai + ai (i = 1, . . . , n).

Theorem (M.L. 1992)

If A, B ⊂ Rk are bounded measurable sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A t ∼ B. A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ A} is finite.

A t ∼ B if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and vectors a1, . . . , an s.t. Bi = Ai + ai (i = 1, . . . , n).

Theorem (M.L. 1992)

If A, B ⊂ Rk are bounded measurable sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A t ∼ B. A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ A} is finite. If A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and Bi = Ai + ai (i = 1, . . . , n),

A t ∼ B if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and vectors a1, . . . , an s.t. Bi = Ai + ai (i = 1, . . . , n).

Theorem (M.L. 1992)

If A, B ⊂ Rk are bounded measurable sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A t ∼ B. A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ A} is finite. If A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and Bi = Ai + ai (i = 1, . . . , n), then let φ(x) = x + ai (x ∈ Ai, i = 1, . . . , n).

A t ∼ B if there are decompositions A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and vectors a1, . . . , an s.t. Bi = Ai + ai (i = 1, . . . , n).

Theorem (M.L. 1992)

If A, B ⊂ Rk are bounded measurable sets with λ(A) = λ(B) > 0 and dimB ∂A < k, dimB ∂B < k, then A t ∼ B. A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ A} is finite. If A = A1 ∪ . . . ∪ An, B = B1 ∪ . . . ∪ Bn and Bi = Ai + ai (i = 1, . . . , n), then let φ(x) = x + ai (x ∈ Ai, i = 1, . . . , n). If {φ(x) − x : x ∈ A} = {a1, . . . , an}, then let Ai = {x ∈ A: φ(x) − x = ai} (i = 1, . . . , n).

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite.

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite. Let u1, . . . , ud ∈ Tk = [0, 1)k be linearly independent over Q, G = {n1u1 + . . . + ndud : ni ∈ Z}.

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite. Let u1, . . . , ud ∈ Tk = [0, 1)k be linearly independent over Q, G = {n1u1 + . . . + ndud : ni ∈ Z}. For E ⊂ Tk, u ∈ Tk let Eu = {(n1, . . . , nd) ∈ Zd : u + n1u1 + . . . + ndud ∈ E}.

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite. Let u1, . . . , ud ∈ Tk = [0, 1)k be linearly independent over Q, G = {n1u1 + . . . + ndud : ni ∈ Z}. For E ⊂ Tk, u ∈ Tk let Eu = {(n1, . . . , nd) ∈ Zd : u + n1u1 + . . . + ndud ∈ E}.

Proposition

Let A, B ⊂ Tk. Suppose there is a K > 0 s.t. ∀u ∃ a bijection φu : Au → Bu s.t. |φu(n) − n| ≤ K (n ∈ Au). Then A t ∼ B.

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite. Let u1, . . . , ud ∈ Tk = [0, 1)k be linearly independent over Q, G = {n1u1 + . . . + ndud : ni ∈ Z}. For E ⊂ Tk, u ∈ Tk let Eu = {(n1, . . . , nd) ∈ Zd : u + n1u1 + . . . + ndud ∈ E}.

Proposition

Let A, B ⊂ Tk. Suppose there is a K > 0 s.t. ∀u ∃ a bijection φu : Au → Bu s.t. |φu(n) − n| ≤ K (n ∈ Au). Then A t ∼ B. Proof (AC): Let U contain exactly one element of each coset of G.

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite. Let u1, . . . , ud ∈ Tk = [0, 1)k be linearly independent over Q, G = {n1u1 + . . . + ndud : ni ∈ Z}. For E ⊂ Tk, u ∈ Tk let Eu = {(n1, . . . , nd) ∈ Zd : u + n1u1 + . . . + ndud ∈ E}.

Proposition

Let A, B ⊂ Tk. Suppose there is a K > 0 s.t. ∀u ∃ a bijection φu : Au → Bu s.t. |φu(n) − n| ≤ K (n ∈ Au). Then A t ∼ B. Proof (AC): Let U contain exactly one element of each coset of

• G. If u ∈ U, x ∈ A ∩ (u + G) and φu(x) = (m1, . . . , md), then let

φ(x) = u + m1u1 + . . . + mdud ∈ B.

A t ∼ B ⇐ ⇒ ∃φ: A → B bijection s.t. {φ(x) − x : x ∈ Rk} is finite. Let u1, . . . , ud ∈ Tk = [0, 1)k be linearly independent over Q, G = {n1u1 + . . . + ndud : ni ∈ Z}. For E ⊂ Tk, u ∈ Tk let Eu = {(n1, . . . , nd) ∈ Zd : u + n1u1 + . . . + ndud ∈ E}.

Proposition

Let A, B ⊂ Tk. Suppose there is a K > 0 s.t. ∀u ∃ a bijection φu : Au → Bu s.t. |φu(n) − n| ≤ K (n ∈ Au). Then A t ∼ B. Proof (AC): Let U contain exactly one element of each coset of

• G. If u ∈ U, x ∈ A ∩ (u + G) and φu(x) = (m1, . . . , md), then let

φ(x) = u + m1u1 + . . . + mdud ∈ B. Then φ: A → B is a bijection, and φ(x) − x ∈ {n1u1 + . . . + ndud : |ni| ≤ K} for every x, so {φ(x) − x} is finite.

Theorem (M.L. 1992)

Suppose S1, S2 ⊂ Zd and there are α, ε, c > 0 s.t. ||Si ∩ Q| − αλd(Q)| ≤ c · s(Q)d−1−ε (i = 1, 2) for every lattice cube Q, where s(Q) is the side length of Q. Then then there is a bijection φ from S1 onto S2 s.t. supx∈S1 |φ(x) − x| < ∞.

Theorem (M.L. 1992)

Suppose S1, S2 ⊂ Zd and there are α, ε, c > 0 s.t. ||Si ∩ Q| − αλd(Q)| ≤ c · s(Q)d−1−ε (i = 1, 2) for every lattice cube Q, where s(Q) is the side length of Q. Then then there is a bijection φ from S1 onto S2 s.t. supx∈S1 |φ(x) − x| < ∞. If A ⊂ Tk, u ∈ Tk and Q = [0, N)d ∩ Zd, then |Au ∩ Q| − αλd(Q) = |A ∩ FN| − λ(A) · |FN|, where α = λ(A), FN = {u + n1u1 + . . . + ndud : 0 ≤ ni < N}.

Theorem (M.L. 1992)

Suppose S1, S2 ⊂ Zd and there are α, ε, c > 0 s.t. ||Si ∩ Q| − αλd(Q)| ≤ c · s(Q)d−1−ε (i = 1, 2) for every lattice cube Q, where s(Q) is the side length of Q. Then then there is a bijection φ from S1 onto S2 s.t. supx∈S1 |φ(x) − x| < ∞. If A ⊂ Tk, u ∈ Tk and Q = [0, N)d ∩ Zd, then |Au ∩ Q| − αλd(Q) = |A ∩ FN| − λ(A) · |FN|, where α = λ(A), FN = {u + n1u1 + . . . + ndud : 0 ≤ ni < N}. So we need that the discrepancy D(FN, A) =

• |FN ∩ A|

|FN| − λ(A)

• is small.

Theorem (M.L. 1992)

If A ⊂ Tk, λ(A) > 0, dimB(∂A) < k, then for every d large enough and for a.e. u1, . . . , ud ∈ Tk there are ε, M > 0 s.t. D(FN + u, A) ≤ M · N−1−ε for every u ∈ Tk and for every N.

Theorem (M.L. 1992)

If A ⊂ Tk, λ(A) > 0, dimB(∂A) < k, then for every d large enough and for a.e. u1, . . . , ud ∈ Tk there are ε, M > 0 s.t. D(FN + u, A) ≤ M · N−1−ε for every u ∈ Tk and for every N. In the proof of A t ∼ B the bijection φ: A → B is constructed coset by coset.

Theorem (M.L. 1992)

If A ⊂ Tk, λ(A) > 0, dimB(∂A) < k, then for every d large enough and for a.e. u1, . . . , ud ∈ Tk there are ε, M > 0 s.t. D(FN + u, A) ≤ M · N−1−ε for every u ∈ Tk and for every N. In the proof of A t ∼ B the bijection φ: A → B is constructed coset by coset. If U contains exactly one element of each coset of G = {n1u1 + . . . + ndud : ni ∈ Z} then U is nonmeasurable.

Theorem (M.L. 1992)

If A ⊂ Tk, λ(A) > 0, dimB(∂A) < k, then for every d large enough and for a.e. u1, . . . , ud ∈ Tk there are ε, M > 0 s.t. D(FN + u, A) ≤ M · N−1−ε for every u ∈ Tk and for every N. In the proof of A t ∼ B the bijection φ: A → B is constructed coset by coset. If U contains exactly one element of each coset of G = {n1u1 + . . . + ndud : ni ∈ Z} then U is nonmeasurable. In the measurable circle squaring by Grabowski, M´ ath´ e, Pikhurko measurable sets U are constructed s.t. if x, y ∈ U ∩ (u + G) then x and y are far apart. Then local matchings are constructed in neighbourhoods of the points of U ∩ (u + G). A limit process using exhaustion produces the measurable matching between A and B.

Theorem (M.L. 1992)

If A ⊂ Tk, λ(A) > 0, dimB(∂A) < k, then for every d large enough and for a.e. u1, . . . , ud ∈ Tk there are ε, M > 0 s.t. D(FN + u, A) ≤ M · N−1−ε for every u ∈ Tk and for every N. In the proof of A t ∼ B the bijection φ: A → B is constructed coset by coset. If U contains exactly one element of each coset of G = {n1u1 + . . . + ndud : ni ∈ Z} then U is nonmeasurable. In the measurable circle squaring by Grabowski, M´ ath´ e, Pikhurko measurable sets U are constructed s.t. if x, y ∈ U ∩ (u + G) then x and y are far apart. Then local matchings are constructed in neighbourhoods of the points of U ∩ (u + G). A limit process using exhaustion produces the measurable matching between A and B. In the Borel circle squaring by Marks and Unger the Borel matching is obtained using Borel flows.

If (V , E) is a graph and f : V → R then an f -flow is φ: (E ∪ E ∗) → R s.t. φ(x, y) = −φ(y, x) ((x, y)) ∈ E) and f (x) =

• y, (x,y)∈V

φ(x, y) (x ∈ V ).

If (V , E) is a graph and f : V → R then an f -flow is φ: (E ∪ E ∗) → R s.t. φ(x, y) = −φ(y, x) ((x, y)) ∈ E) and f (x) =

• y, (x,y)∈V

φ(x, y) (x ∈ V ). Let E ⊂ A × B be a bipartite graph, M ⊂ E a matching. Then φ(x, y) =      1 if (x, y) ∈ M, −1 if (y, x) ∈ M, if (x, y) / ∈ M is an f -flow, where f = χA − χB.

u u 1−u 1−u 1 1

R

E = {(x, y): (x, y) ∈ R} is a 2-regular Borel bipartite graph between A and B, where A = B = [0, 1]. Let f = χA − χB.

u u 1−u 1−u 1 1

R

E = {(x, y): (x, y) ∈ R} is a 2-regular Borel bipartite graph between A and B, where A = B = [0, 1]. Let f = χA − χB. Then there is a Borel f -flow on V : φ(x, y) = 1/2, φ(y, x) = −1/2 ((x, y) ∈ E).

u u 1−u 1−u 1 1

R

E = {(x, y): (x, y) ∈ R} is a 2-regular Borel bipartite graph between A and B, where A = B = [0, 1]. Let f = χA − χB. Then there is a Borel f -flow on V : φ(x, y) = 1/2, φ(y, x) = −1/2 ((x, y) ∈ E). There is no integer valued Borel f -flow:

u u 1−u 1−u 1 1

R

E = {(x, y): (x, y) ∈ R} is a 2-regular Borel bipartite graph between A and B, where A = B = [0, 1]. Let f = χA − χB. Then there is a Borel f -flow on V : φ(x, y) = 1/2, φ(y, x) = −1/2 ((x, y) ∈ E). There is no integer valued Borel f -flow: if φ was such a flow, then M = {(x, y): φ(x, y) > 0} would be a Borel matching.

V = Tk, E = {(x, y): y − x = n1u1 + . . . + ndud, n1 = 0, 1}. We need an integer valued bounded (χA − χB)-flow.

Theorem (Marks and Unger 2016)

Let f : Tk → R be s.t. for every x,

• y∈F2n+x

f (y)

• ≤ M · 2n(d−1−ε).

Then there exists a bounded f -flow on E.

V = Tk, E = {(x, y): y − x = n1u1 + . . . + ndud, n1 = 0, 1}. We need an integer valued bounded (χA − χB)-flow.

Theorem (Marks and Unger 2016)

Let f : Tk → R be s.t. for every x,

• y∈F2n+x

f (y)

• ≤ M · 2n(d−1−ε).

Then there exists a bounded f -flow on E. If X is a set, G is a group then G X (G acts on X) if g : X → X (g ∈ G) and gh(x) = g(hx) (g, h ∈ G).

V = Tk, E = {(x, y): y − x = n1u1 + . . . + ndud, n1 = 0, 1}. We need an integer valued bounded (χA − χB)-flow.

Theorem (Marks and Unger 2016)

Let f : Tk → R be s.t. for every x,

• y∈F2n+x

f (y)

• ≤ M · 2n(d−1−ε).

Then there exists a bounded f -flow on E. If X is a set, G is a group then G X (G acts on X) if g : X → X (g ∈ G) and gh(x) = g(hx) (g, h ∈ G). X standard Borel space, E ⊂ X × X a Borel equivalence relation.

V = Tk, E = {(x, y): y − x = n1u1 + . . . + ndud, n1 = 0, 1}. We need an integer valued bounded (χA − χB)-flow.

Theorem (Marks and Unger 2016)

Let f : Tk → R be s.t. for every x,

• y∈F2n+x

f (y)

• ≤ M · 2n(d−1−ε).

Then there exists a bounded f -flow on E. If X is a set, G is a group then G X (G acts on X) if g : X → X (g ∈ G) and gh(x) = g(hx) (g, h ∈ G). X standard Borel space, E ⊂ X × X a Borel equivalence relation. E is finite (countable) if each equivalence class is finite (countable).

V = Tk, E = {(x, y): y − x = n1u1 + . . . + ndud, n1 = 0, 1}. We need an integer valued bounded (χA − χB)-flow.

Theorem (Marks and Unger 2016)

Let f : Tk → R be s.t. for every x,

• y∈F2n+x

f (y)

• ≤ M · 2n(d−1−ε).

Then there exists a bounded f -flow on E. If X is a set, G is a group then G X (G acts on X) if g : X → X (g ∈ G) and gh(x) = g(hx) (g, h ∈ G). X standard Borel space, E ⊂ X × X a Borel equivalence relation. E is finite (countable) if each equivalence class is finite (countable). X = Tk is a standard Borel space. The group G = Zd acts on X: if g = (n1, . . . , nd), then let g(x) = x + n1u1 + . . . + ndud.

E is hyperfinite if E = ∞

n=1 En, where En is a finite Borel

equivalence relation for every n.

E is hyperfinite if E = ∞

n=1 En, where En is a finite Borel

equivalence relation for every n. Vitali’s relation x − y ∈ Q is hyperfinite.

E is hyperfinite if E = ∞

n=1 En, where En is a finite Borel

equivalence relation for every n. Vitali’s relation x − y ∈ Q is hyperfinite.

Theorem (Feldman, Moore 1977)

E is a countable Borel equivalence relation on X ⇐ ⇒ E is Borel and E = {(x, g(x)): x ∈ X, g ∈ G}, where G is a countable group G acting on X.

E is hyperfinite if E = ∞

n=1 En, where En is a finite Borel

equivalence relation for every n. Vitali’s relation x − y ∈ Q is hyperfinite.

Theorem (Feldman, Moore 1977)

E is a countable Borel equivalence relation on X ⇐ ⇒ E is Borel and E = {(x, g(x)): x ∈ X, g ∈ G}, where G is a countable group G acting on X.

Theorem (B. Weiss 1981)

If G = Zd and G X is a Borel action, then E = {(x, g(x)): x ∈ X, g ∈ G} is hyperfinite.

E is hyperfinite if E = ∞

n=1 En, where En is a finite Borel

equivalence relation for every n. Vitali’s relation x − y ∈ Q is hyperfinite.

Theorem (Feldman, Moore 1977)

E is a countable Borel equivalence relation on X ⇐ ⇒ E is Borel and E = {(x, g(x)): x ∈ X, g ∈ G}, where G is a countable group G acting on X.

Theorem (B. Weiss 1981)

If G = Zd and G X is a Borel action, then E = {(x, g(x)): x ∈ X, g ∈ G} is hyperfinite.

Theorem (S. Gao, S. Jackson 2015)

If G is countable and Abelian and G X is a Borel action, then E = {(x, g(x)): x ∈ X, g ∈ G} is hyperfinite.

E is hyperfinite if E = ∞

n=1 En, where En is a finite Borel

equivalence relation for every n. Vitali’s relation x − y ∈ Q is hyperfinite.

Theorem (Feldman, Moore 1977)

E is a countable Borel equivalence relation on X ⇐ ⇒ E is Borel and E = {(x, g(x)): x ∈ X, g ∈ G}, where G is a countable group G acting on X.

Theorem (B. Weiss 1981)

If G = Zd and G X is a Borel action, then E = {(x, g(x)): x ∈ X, g ∈ G} is hyperfinite.

Theorem (S. Gao, S. Jackson 2015)

If G is countable and Abelian and G X is a Borel action, then E = {(x, g(x)): x ∈ X, g ∈ G} is hyperfinite. Thus X = R+, x/y ∈ Q is hyperfinite.

Lemma

[Marks and Unger 2016] Suppose G = Zd, G X is a free Borel action (if g = id. then g has no fixed points), and E = {(x, y): y = gn1...nd(x), ni = 0, 1}. If f : X → Z is Borel, φ is an f -flow, then there is an integer valued f -flow ψ s.t. |φ − ψ| ≤ 3d.

Lemma

[Marks and Unger 2016] Suppose G = Zd, G X is a free Borel action (if g = id. then g has no fixed points), and E = {(x, y): y = gn1...nd(x), ni = 0, 1}. If f : X → Z is Borel, φ is an f -flow, then there is an integer valued f -flow ψ s.t. |φ − ψ| ≤ 3d.

Lemma

[S. Gao, S. Jackson 2015] Suppose G = Zd, and G X is a free Borel action. Then for every n there is a partition C of X into sets

• f the form {gn1...nd(x): 0 ≤ ni ≤ n or n + 1}.

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow.

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow. If R ∈ C, then let N(R) = {S ∈ C : ∂S ∩ ∂R = ∅}.

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow. If R ∈ C, then let N(R) = {S ∈ C : ∂S ∩ ∂R = ∅}. Ψ(R, S) =

• (x,y)∈R×S

φ(x, y).

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow. If R ∈ C, then let N(R) = {S ∈ C : ∂S ∩ ∂R = ∅}. Ψ(R, S) =

• (x,y)∈R×S

φ(x, y).

• S∈N(R)

Ψ(R, S) = |R ∩ A| − |R ∩ B|.

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow. If R ∈ C, then let N(R) = {S ∈ C : ∂S ∩ ∂R = ∅}. Ψ(R, S) =

• (x,y)∈R×S

φ(x, y).

• S∈N(R)

Ψ(R, S) = |R ∩ A| − |R ∩ B|. We connect Ψ(R, S) elements of A ∩ R with the same number of elements of B ∩ S. Then the cardinality of the rest of A ∩ R is the same as the cardinality of the rest of B ∩ R.

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow. If R ∈ C, then let N(R) = {S ∈ C : ∂S ∩ ∂R = ∅}. Ψ(R, S) =

• (x,y)∈R×S

φ(x, y).

• S∈N(R)

Ψ(R, S) = |R ∩ A| − |R ∩ B|. We connect Ψ(R, S) elements of A ∩ R with the same number of elements of B ∩ S. Then the cardinality of the rest of A ∩ R is the same as the cardinality of the rest of B ∩ R. We construct a matching between them.

Proof of A

t

∼ B with Borel pieces:

17

Let φ be an integer valued bounded (χA − χB)-flow. If R ∈ C, then let N(R) = {S ∈ C : ∂S ∩ ∂R = ∅}. Ψ(R, S) =

• (x,y)∈R×S

φ(x, y).

• S∈N(R)

Ψ(R, S) = |R ∩ A| − |R ∩ B|. We connect Ψ(R, S) elements of A ∩ R with the same number of elements of B ∩ S. Then the cardinality of the rest of A ∩ R is the same as the cardinality of the rest of B ∩ R. We construct a matching between them. All these matchings can be done constructively, resulting a Borel matching between A and B.

Can the condition on the boundaries dimB(∂A), dimB(∂A) < k be relaxed?

Can the condition on the boundaries dimB(∂A), dimB(∂A) < k be relaxed? No, if k = 2 or if the group of isometries used is amenable.

Can the condition on the boundaries dimB(∂A), dimB(∂A) < k be relaxed? No, if k = 2 or if the group of isometries used is amenable.

Theorem (M.L. 2003)

If G is an amenable group of isometries of Rk, then there are bounded sets A, B ⊂ Rk with differentiable boundaries s.t. λ(A) = λ(B) > 0 and A, B are not equidecomposable under G.

Can the condition on the boundaries dimB(∂A), dimB(∂A) < k be relaxed? No, if k = 2 or if the group of isometries used is amenable.

Theorem (M.L. 2003)

If G is an amenable group of isometries of Rk, then there are bounded sets A, B ⊂ Rk with differentiable boundaries s.t. λ(A) = λ(B) > 0 and A, B are not equidecomposable under G. In particular, there are Jordan domains A, B ⊂ R2 with differentiable boundaries s.t. λ(A) = λ(B) > 0 and A, B are not equidecomposable using arbitrary isometries.

Is it true that if A, B are bounded Borel sets in R3 with nonempty interior and of the same measure, then they are equidecomposable with Borel pieces?

Is it true that if A, B are bounded Borel sets in R3 with nonempty interior and of the same measure, then they are equidecomposable with Borel pieces? A set A ⊂ Rk is Jordan measurable if it is bounded and its boundary has measure zero.

Is it true that if A, B are bounded Borel sets in R3 with nonempty interior and of the same measure, then they are equidecomposable with Borel pieces? A set A ⊂ Rk is Jordan measurable if it is bounded and its boundary has measure zero. Is it true that if A, B are Jordan measurable sets in R2 of the same measure, then they are equidecomposable with Jordan measurable pieces?

Is it true that if A, B are bounded Borel sets in R3 with nonempty interior and of the same measure, then they are equidecomposable with Borel pieces? A set A ⊂ Rk is Jordan measurable if it is bounded and its boundary has measure zero. Is it true that if A, B are Jordan measurable sets in R2 of the same measure, then they are equidecomposable with Jordan measurable pieces? Thank you for your attention.