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

Combinatorial lines with few intervals

Nina Kamčev, joint work with David Conlon and Christoph Spiegel

The Hales-Jewett Theorem

NOTATION

• Ground set 𝑛 𝑜, usually 3 𝑜
• A root is a word 𝑥 of length 𝑜

in the symbols [𝑛] ∪ ∗ with at least one ∗.

• For 𝑗 ∈ [𝑛], 𝑥𝑗 is the word
• btained from 𝑥 by replacing

each ∗ by 𝑗.

• A (combinatorial) line is a set
• f words 𝑥𝑗: 𝑗 𝜗[𝑛] .

EXAMPLE: a combinatorial line in 3 12

𝑣 = 22 ∗ 32 ∗ 12 ∗∗∗ 22 𝑣 1 = 22𝟐32𝟐12𝟐𝟐𝟐22 𝑣 2 = 22𝟑32𝟑12𝟑𝟑𝟑22 𝑣 3 = 22𝟒32𝟒12𝟒𝟒𝟒22 Wildcard set

A line in 4 3 with wildcard set {1,2,3}

Theorem (Hales, Jewett, ‘63). Given 𝑛, 𝑠 ∈ ℕ, there is a natural number 𝑜 such that any 𝑠-colouring of [𝑛]𝑜 contains a monochromatic combinatorial line.

• 𝐼𝐾 𝑛, 𝑠 is the minimal 𝑜 for which the

conclusion holds

A warm-up…

• Claim. 𝐼𝐾 2, 𝑠 = 𝑠.
• Proof. Let 𝑑: [2]𝑠→ ℤ𝑠.

A warm-up…

• Claim. 𝐼𝐾 2, 𝑠 = 𝑠.
• Proof. Let 𝑑: [2]𝑠→ ℤ𝑠. 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

• 𝐼𝐾(𝑛, 𝑠) ≤ 22⋰

2

• 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? 2𝐼𝐾(𝑛 − 1, 𝑠) times

Lines with few intervals

• Definition. ℐ 𝑛, 𝑠 = min{𝑟: for large 𝑜, any 𝑠-colouring
• f [𝑛]𝑜 contains a monochromatic 𝑟-interval line}.

Observations.

• ℐ 2, 𝑠 = 1 for all 𝑠.
• ℐ 3, 𝑠 ≤ 𝑠, generally ℐ 𝑛, 𝑠 ≤ 𝐼𝐾(𝑛 − 1, 𝑠)

Theorem. (i) ℐ 3, 𝑠 = 𝑠 for odd 𝑠

(Conlon, K)

(ii) ℐ 3, 2 = 1

(Leader, Räty)

(iii) ℐ 3 𝑠 = 𝑠 − 1 for even 𝑠

(K, Spiegel).

1 2 3 4 5 6 7

1 3 5 7

ℐ(3, 𝑠) 𝑠

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

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

[3]𝑜

𝑑𝑝𝑜𝑢𝑠𝑏𝑑𝑢

[3]≤𝑜

𝑑𝑝𝑣𝑜𝑢

ℤ𝑠

3

𝑣1 = 2213211211122 ⟼ 𝑣1 = 2132 1212 ⟼ 𝜒 𝑣1 = 3, 4, 1 𝑣2 = 2223221222222 ⟼ 𝑣2 = 2 32 12 ⟼ 𝜒 𝑣2 = (1, 3, 1) 𝑣3 = 2233231233322 ⟼ 𝑣3 = 2 3231232 ⟼ 𝜒 𝑣3 = (1, 4, 3)

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

[3]𝑜

𝑑𝑝𝑜𝑢𝑠𝑏𝑑𝑢

[3]≤𝑜

𝑑𝑝𝑣𝑜𝑢

ℤ𝑠

3

𝑣1 = 22𝟐32𝟐12𝟐𝟐𝟐22 ⟼ 𝑣1 = 2𝟐32 12𝟐2 ⟼ 𝜒 𝑣1 = 3, 4, 1 𝑣2 = 22𝟑32𝟑12𝟑𝟑𝟑22 ⟼ 𝑣2 = 2 32 12 ⟼ 𝜒 𝑣2 = (1, 3, 1) 𝑣3 = 22𝟒32𝟒12𝟒𝟒𝟒22 ⟼ 𝑣3 = 2 32𝟒12𝟒2 ⟼ 𝜒 𝑣3 = (1, 4, 3) 𝑣 = 22 ∗ 32 ∗ 12 ∗∗∗ 22 ⟼ ഥ 𝑣 = 2 ∗ 32 ∗ 12 ∗ 2 ⟼ 𝜒 ഥ 𝑣 = (1, 4, 1)

𝜒 𝑣1 𝜒 𝑣 3 𝜒 𝑣 2 𝜒 ഥ 𝑣

ℤ𝑠

3

ℤ𝑠 ∙ (2, 2, −1)

(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→ ℤ𝑠 avoiding

𝑠 − 1 -interval lines with

𝑈 𝑣 = 𝑈′ 𝜒 ത 𝑣 for all 𝑣. That is, 𝑈 has the form [3]𝑜

ഥ

[3]≤𝑜

𝜒

ℤ𝑠

3 𝑈′

ℤ𝑠

3

• 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?