On equipartitions of measures

Sini ša Vrećica

University of Belgrade

Applied topology - Bedlewo 2013.

July 21 - 27, 2013.

EQUIPARTITIONS – P. 1/1

Equipartition by hyperplanes

B. Grünbaum: For which d, j, k, any collectionof j (sufficiently nice) measures in Rd couldbe partitioned by k hyperplanes into 2k conescontaining the same amount of everymeasure?

EQUIPARTITIONS – P. 2/1

Equipartition by hyperplanes

B. Grünbaum: For which d, j, k, any collectionof j (sufficiently nice) measures in Rd couldbe partitioned by k hyperplanes into 2k conescontaining the same amount of everymeasure?

The general case still open, as it is theimportant special case d = 4, j = 1, k = 4.

EQUIPARTITIONS – P. 2/1

Equipartition by hyperplanes

If some hyperplanes are allowed to beparallel, then the number of cones is smallerthan 2k.

EQUIPARTITIONS – P. 3/1

Equipartition by hyperplanes

If some hyperplanes are allowed to beparallel, then the number of cones is smallerthan 2k.

For every continuous measure in R4, thereare 4 hyperplanes, two of which are parallel,partitioning a measure into 12 cones of thesame measure.

EQUIPARTITIONS – P. 3/1

Equipartition in convex sets

For every collection of d "nice" measures inR

d, and k ∈ N, there is a partition of Rd in kconvex sets containing the same amount ofeach of the measures.

EQUIPARTITIONS – P. 4/1

Equipartition in convex sets

For every collection of d "nice" measures inR

d, and k ∈ N, there is a partition of Rd in kconvex sets containing the same amount ofeach of the measures.

Proof uses power diagrams and the V.Vassiliev’s result on the nonexistence of theequivariant map from the configuration spaceFn(R

d) to S(n−1)(d−1)−1.

EQUIPARTITIONS – P. 4/1

Equipartition of necklaces

If q thieves steal a necklace with beads of kdifferent colors, they could divide thenecklace using at most (q − 1)k cuts andpartition the obtained intervals in q collectionswith equal number of beads of each color.

EQUIPARTITIONS – P. 5/1

Equipartition of necklaces

If q thieves steal a necklace with beads of kdifferent colors, they could divide thenecklace using at most (q − 1)k cuts andpartition the obtained intervals in q collectionswith equal number of beads of each color.

Prove the continuous version first.

EQUIPARTITIONS – P. 5/1

Equipartition of necklaces

If q thieves steal a necklace with beads of kdifferent colors, they could divide thenecklace using at most (q − 1)k cuts andpartition the obtained intervals in q collectionswith equal number of beads of each color.

Prove the continuous version first.

The result extends to higher dimensionalnecklaces (cubes).

EQUIPARTITIONS – P. 5/1

Equipartition of necklaces

For any k continuous measures on [0, 1]d, thiscube could be cut by at most k(q − 1)hyperplanes orthogonal to the edges, suchthat it is possible to partition the obtainedcuboids into q collections containing the sameamount of each of the measures.

EQUIPARTITIONS – P. 6/1

Equipartition of necklaces

For any k continuous measures on [0, 1]d, thiscube could be cut by at most k(q − 1)hyperplanes orthogonal to the edges, suchthat it is possible to partition the obtainedcuboids into q collections containing the sameamount of each of the measures.

The number of cuts orthogonal to any edgecould be prescribed in advance.

EQUIPARTITIONS – P. 6/1

Pattern avoidance

P. Erdös (1961.): Is there a coloring ofintegers by 4 colors such that no two adjacentintervals contain the same number of integersof each color?

EQUIPARTITIONS – P. 7/1

Pattern avoidance

P. Erdös (1961.): Is there a coloring ofintegers by 4 colors such that no two adjacentintervals contain the same number of integersof each color?

V. Keränen (1991.): Yes!

EQUIPARTITIONS – P. 7/1

A link

N. Alon, J. Grytczuk, M. Lason, and M.Michalek established a connection betweenthese two problems.

EQUIPARTITIONS – P. 8/1

A link

N. Alon, J. Grytczuk, M. Lason, and M.Michalek established a connection betweenthese two problems.

They considered the colorings of the real line(partitions into disjoint measurable subsets).

EQUIPARTITIONS – P. 8/1

A link

N. Alon, J. Grytczuk, M. Lason, and M.Michalek established a connection betweenthese two problems.

They considered the colorings of the real line(partitions into disjoint measurable subsets).

There is a measurable coloring of the real lineby 4 colors such that no two adjacent intervalscontain the same amount of each measure.

EQUIPARTITIONS – P. 8/1

Higher-dimensional extension

M. Lason extended this results to higherdimensions.

EQUIPARTITIONS – P. 9/1

Higher-dimensional extension

M. Lason extended this results to higherdimensions.

There is a measurable coloring of Rd with2d + 3 colors such that no two nontrivialcubes have the same measure of each color.

EQUIPARTITIONS – P. 9/1

Higher-dimensional extension

M. Lason extended this results to higherdimensions.

There is a measurable coloring of Rd with2d + 3 colors such that no two nontrivialcubes have the same measure of each color.

The same result for cuboids is true forcolorings with 4d + 1 colors.

EQUIPARTITIONS – P. 9/1

Questions

Are these estimates the best possible?

EQUIPARTITIONS – P. 10/1

Questions

Are these estimates the best possible?

What could be said if we replace thepartitions of the Euclidean space intomeasurable subsets, with the family ofcontinuous measures on Rd?

EQUIPARTITIONS – P. 10/1

Questions

Are these estimates the best possible?

What could be said if we replace thepartitions of the Euclidean space intomeasurable subsets, with the family ofcontinuous measures on Rd?

S. V., R. Živaljević, Measurable necklaces andsets indiscernible by measure,arXiv:1305.7474 [math.CO]

EQUIPARTITIONS – P. 10/1

Cubes

Theorem. For every collection µ1, ..., µd of dcontinuous finite measures on Rd and n ∈ N,there are n pairwise distinct nontrivialaxis-aligned cubes C1, ..., Cn such thatµi(Cj) = µi(Ck) for all i, j, k.

EQUIPARTITIONS – P. 11/1

Cubes

Theorem. For every collection µ1, ..., µd of dcontinuous finite measures on Rd and n ∈ N,there are n pairwise distinct nontrivialaxis-aligned cubes C1, ..., Cn such thatµi(Cj) = µi(Ck) for all i, j, k.

Example. There are d + 1 continuous finitemeasures on Rd so that no two distinctnontrivial axis-aligned cubes contain thesame amount of each of these measures.

EQUIPARTITIONS – P. 11/1

Cuboids

Theorem. For every collection µ1, ..., µ2d−1 of2d − 1 continuous finite measures on Rd andn ∈ N, there are n pairwise distinct nontrivialaxis-aligned cuboids C1, ..., Cn such thatµi(Cj) = µi(Ck) for all i, j, k.

EQUIPARTITIONS – P. 12/1

Cuboids

Theorem. For every collection µ1, ..., µ2d−1 of2d − 1 continuous finite measures on Rd andn ∈ N, there are n pairwise distinct nontrivialaxis-aligned cuboids C1, ..., Cn such thatµi(Cj) = µi(Ck) for all i, j, k.

Example. There are 2d continuous finitemeasures on Rd so that no two distinctnontrivial axis-aligned cuboids contain thesame amount of each of these measures.

EQUIPARTITIONS – P. 12/1

The proof again depends on the V. Vassiliev’sresult on the nonexistence of the equivariantmap from the configuration space Fn(Rd) toS(n−1)(d−1)−1.

EQUIPARTITIONS – P. 13/1

The proof again depends on the V. Vassiliev’sresult on the nonexistence of the equivariantmap from the configuration space Fn(Rd) toS(n−1)(d−1)−1.

The result for cubes is true also for balls, andthe result for cuboids is also true forellipsoids.

EQUIPARTITIONS – P. 13/1

More general families of sets

Let GDT be the group of all transformations ofR

d generated by positive-axis aligneddilatations and translations, and let K be acentrally symmetric convex body in Rd.

EQUIPARTITIONS – P. 14/1

More general families of sets

Let GDT be the group of all transformations ofR

d generated by positive-axis aligneddilatations and translations, and let K be acentrally symmetric convex body in Rd.

Then the same result as for cuboids is alsotrue for the family of setsFK = {L(K) | L ∈ GDT}.

EQUIPARTITIONS – P. 14/1

THANK YOU

FOR YOUR

ATTENTION!

EQUIPARTITIONS – P. 15/1

Equipartition by hyperplanesEquipartition by hyperplanes

Equipartition by hyperplanesEquipartition by hyperplanes

Equipartition in convex setsEquipartition in convex sets

Equipartition of necklacesEquipartition of necklacesEquipartition of necklaces

Equipartition of necklacesEquipartition of necklaces

Pattern avoidancePattern avoidance

A linkA linkA link

Higher-dimensional extensionHigher-dimensional extensionHigher-dimensional extension

QuestionsQuestionsQuestions

CubesCubes

CuboidsCuboids

More general families of setsMore general families of sets