Combinatorial lines Nina Kam ev , joint work with David Conlon and - PowerPoint PPT Presentation

Combinatorial lines Nina Kam čev , joint work with David Conlon and Christoph with few intervals Spiegel

The Hales-Jewett Theorem N OTATION E XAMPLE : a combinatorial line in 3 12 • Ground set 𝑛 𝑜 , usually 3 𝑜 𝑣 = 22 ∗ 32 ∗ 12 ∗∗∗ 22 Wildcard • A root is a word 𝑥 of length 𝑜 𝑣 1 = 22𝟐32𝟐12𝟐𝟐𝟐22 set 𝑣 2 = 22𝟑32𝟑12𝟑𝟑𝟑22 in the symbols [𝑛] ∪ ∗ with 𝑣 3 = 22𝟒32𝟒12𝟒𝟒𝟒22 at least one ∗ . • For 𝑗 ∈ [𝑛] , 𝑥 𝑗 is the word Theorem (Hales, Jewett, ‘63). Given obtained from 𝑥 by replacing 𝑛, 𝑠 ∈ ℕ , there is a natural number 𝑜 each ∗ by 𝑗 . such that any 𝑠 -colouring of [𝑛] 𝑜 • A ( combinatorial ) line is a set contains a monochromatic of words 𝑥 𝑗 : 𝑗 𝜗[𝑛] . combinatorial line. A line in 4 3 with • 𝐼𝐾 𝑛, 𝑠 is the minimal 𝑜 for which the wildcard set {1,2,3} conclusion holds

A warm- up… Claim. 𝐼𝐾 2, 𝑠 = 𝑠. Proof. Let 𝑑: [2] 𝑠 → ℤ 𝑠 .

A warm- up… Claim. 𝐼𝐾 2, 𝑠 = 𝑠. Proof. Let 𝑑: [2] 𝑠 → ℤ 𝑠 . Consider the words 1111111 … 1 11 … 111 … 2 11 … 122 … 2 ⋮ 1222222 … 2 2222222 … 2 . Among 𝑠 + 1 words, two have the same colour ⇒ monochromatic line.

Shelah’s proof of HJT Consequences 2 • 𝐼𝐾(𝑛, 𝑠) ≤ 2 2 ⋰ 2𝐼𝐾(𝑛 − 1, 𝑠) times • For 𝑛 = 3 , there exists an 𝑠 - interval line (a line whose wildcard set consists of at most 𝑠 intervals), e.g. 𝑣 = 33 ∗2∗21 ∗∗∗ 33 ≤ 𝑠 intervals Is this the best possible?

Lines with few intervals Definition. ℐ 𝑛, 𝑠 = min { 𝑟 : for large 𝑜 , any 𝑠 -colouring of [𝑛] 𝑜 contains a monochromatic 𝑟 -interval line}. 7 Observations. ℐ(3, 𝑠) 6 • ℐ 2, 𝑠 = 1 for all 𝑠 . 5 • ℐ 3, 𝑠 ≤ 𝑠, generally ℐ 𝑛, 𝑠 ≤ 𝐼𝐾(𝑛 − 1, 𝑠) 4 3 Theorem. 2 (i) ℐ 3, 𝑠 = 𝑠 for odd 𝑠 (Conlon, K) 1 ℐ 3, 2 = 1 (ii) (Leader, Räty) 1 3 5 7 𝑠 (iii) ℐ 3 𝑠 = 𝑠 − 1 for even 𝑠 (K, Spiegel).

(i) The colouring avoiding 𝑠 − 1 -interval lines

(i) The colouring avoiding 𝑠 − 1 -interval lines 𝑑𝑝𝑜𝑢𝑠𝑏𝑑𝑢 𝑑𝑝𝑣𝑜𝑢 [3] 𝑜 [3] ≤𝑜 3 ℤ 𝑠 𝑣 1 = 2213211211122 ⟼ 𝑣 1 = 2132 1212 ⟼ 𝜒 𝑣 1 = 3, 4, 1 𝑣 2 = 2223221222222 ⟼ 𝑣 2 = 2 32 12 ⟼ 𝜒 𝑣 2 = (1, 3, 1) 𝑣 3 = 2233231233322 ⟼ 𝑣 3 = 2 3231232 ⟼ 𝜒 𝑣 3 = (1, 4, 3)

(i) The colouring avoiding 𝑠 − 1 -interval lines 𝑑𝑝𝑜𝑢𝑠𝑏𝑑𝑢 𝑑𝑝𝑣𝑜𝑢 [3] 𝑜 [3] ≤𝑜 3 ℤ 𝑠 𝑣 1 = 22𝟐32𝟐12𝟐𝟐𝟐22 ⟼ 𝑣 1 = 2𝟐32 12𝟐2 ⟼ 𝜒 𝑣 1 = 3, 4, 1 𝑣 2 = 22𝟑32𝟑12𝟑𝟑𝟑22 ⟼ 𝑣 2 = 2 32 12 ⟼ 𝜒 𝑣 2 = (1, 3, 1) 𝑣 3 = 22𝟒32𝟒12𝟒𝟒𝟒22 ⟼ 𝑣 3 = 2 32𝟒12𝟒2 ⟼ 𝜒 𝑣 3 = (1, 4, 3) 𝑣 = 22 ∗ 32 ∗ 12 ∗∗∗ 22 ⟼ ഥ 𝑣 = 2 ∗ 32 ∗ 12 ∗ 2 ⟼ 𝜒 ഥ 𝑣 = (1, 4, 1) 𝜒 𝑣 3 ∙ (2, 2, −1) 𝜒 ഥ 𝑣 𝜒 𝑣 1 ℤ 𝑠 𝜒 𝑣 2 3 ℤ 𝑠

(iii) The upper bound Theorem (K, Spiegel). ℐ 3, 𝑠 = 𝑠 − 1 for even 𝑠 . • Idea: reduce the problem to the ‘linear’ colourings Proposition. For any 𝑠 , if ℐ 3 𝑠 > 𝑠 − 1 , then there is a colouring 𝑈: [3] 𝑜 → ℤ 𝑠 avoiding 𝑠 − 1 -interval lines with 𝑈 𝑣 = 𝑈′ 𝜒 ത 𝑣 for all 𝑣 . That is, 𝑈 has the form 𝑈 ′ ഥ 𝜒 [3] 𝑜 [3] ≤𝑜 3 3 ℤ 𝑠 ℤ 𝑠 • Odd 𝑠 – there is a colouring • Even 𝑠 – contradiction.

Onwards and upwards Improve the bounds 𝑠 ≤ ℐ 𝑛, 𝑠 ≤ 𝐼𝐾 𝑛 − 1, 𝑠 . New proofs of the Hales-Jewett theorem?