Symmetric monochromatic subsets in colorings of the Lobachevsky plane

Abstract

Combinatorics International audience We prove that for each partition of the Lobachevsky plane into finitely many Borel pieces one of the cells of the partition contains an unbounded centrally symmetric subset.

Symmetry and colorings: some results and open problems, II

“…The answer to this problem is affirmative for r = 2. This follows from the subsequent theorem that can be proved by analogy with Theorem 3.1 of [10]: Theorem 9.4. The family S = {s c : c ∈ H 2 } of central symmetries of the hyperbolic plane H 2 contains a 3-element subfamily S ′ such that for the any 2-coloring of H 2 there is a monochromatic S ′ -subset A ⊂ H 2 of asymptotic dimension asdim(A) = 2.…”

“…The following unexpected result proved in [10] shows that by its Ramsey-theoretic properties, the hyperbolic plane substantially differs from its Euclidean counterpart. Theorem 9.2.…”

Centerpole sets for colorings of abelian groups

“…We claim that the set H ∩ C is not k-centerpole for Borel colorings of the topological group H. By (the proof) of Proposition 1 in [3], c B 2 (H) = 3. Since H is a support hyperplane for C and |C \ H| ≤ 2, we can apply Lemma 6 and conclude that C is not k-centerpole for Borel colorings of H ⊕ R = A.…”

“…and is affinely equivalent to any triangle ∆ = {a, b, c} in R 2 ; • Ξ 2 0 has cardinality c 3 (R 3 ) = 6 and is affinely equivalent to ∆ ∪ (x − ∆) where ∆ ⊂ R 3 is a triangle centered at zero and x ∈ R 3 does not belong to the linear span of ∆; • Ξ 3 1 has cardinality c 4 (R 4 ) = 12 and is affinely equivalent to (x − ∆) ∪ ∆ ∪ (−x − ∆) where ∆ ⊂ R 4 is a tetrahedron centered at zero and x ∈ R 4 does not belong to the linear span of ∆. To see that Ξ 3 1 is of this form, observe that c = ( 1 4 , 1 2 , 1 2 , 1 2 ) is the barycenter of Ξ 3 1 and Ξ 3 1 − c = (x− ∆)∪∆∪(−x− ∆) for the tetrahedron ∆ = {(0, 0, 0, 1), (0, 0, 1, 0), (0, 1, 0, 0), (…”