The Anti-Field-Descent Method Bernhard Schmidt Nanyang - - PowerPoint PPT Presentation

The Anti-Field-Descent Method

Bernhard Schmidt Nanyang Technological University Joint work with Ka Hin Leung

Bordeaux, June 2017

Ryser (1963)

Eigenvalues

𝐼𝑈𝐼 = 𝑤𝐽 𝐼𝑦 = 𝜇𝑦 ⇒ 𝑦𝑈𝐼𝑈𝐼𝑦 = |𝜇|2𝑦 𝑦𝑈 𝐼𝑈𝐼 = 𝑤𝐽 ⇒ 𝑦𝑈𝐼𝑈𝐼𝑦 = 𝑤𝑦 𝑦𝑈 𝜇 ∈ ℂ eigenvalue of Hadamard matrix of order 𝑤 ⇓ |𝜇|2 = 𝑤

Circulants

𝐼 = 𝑏0 𝑏1 𝑏𝑤−1 𝑏0 ⋯ 𝑏𝑤−1 ⋯ 𝑏𝑤−2 ⋯ ⋯ 𝑏1 𝑏2 ⋯ ⋯ ⋯ 𝑏0 Eigenvalues: 𝑗=0

𝑤−1 𝑏𝑗𝜂𝑗𝑙, 𝑙 = 0, … , 𝑤 − 1, 𝜂 = exp( 2𝜌𝑗 𝑤 )

𝐼 Hadamard matrix Note: Implies 𝑤 = 4𝑣2 ⇒

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗𝑙

2

= 𝑤

Basic Idea

What does

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗𝑙

2

= 𝑤

say about the 𝑏𝑗’s?

Divisibility Method (Turyn)

Suppose 𝑗=0

15 𝑏𝑗𝜂𝑗 2 = 16, 𝜂 = exp( 2𝜌𝑗 16 )

Factorization: 2 = 1 − 𝜂 1 − 𝜂3 ⋯ (1 − 𝜂15) Hence 1 − 𝜂 32 divides 𝑗=0

15 𝑏𝑗𝜂𝑗 2 in ℤ[𝜂]

1 − 𝜂 = 1 − 𝜂15 = (1 + ⋯ + 𝜂14)(1 − 𝜂) 1 − 𝜂 = (1 + ⋯ + 𝜂14)(1 − 𝜂) Thus 1 − 𝜂 16 divides 𝑗=0

15 𝑏𝑗𝜂𝑗

Divisibility Method (Turyn)

1 − 𝜂 16 divides 𝑗=0

15 𝑏𝑗𝜂𝑗

4 divides 𝑗=0

15 𝑏𝑗𝜂𝑗

Basis representation: 𝑗=0

15 𝑏𝑗𝜂𝑗 = 𝑗=0 7

(𝑏𝑗−𝑏𝑗+8)𝜂𝑗 4 divides 𝑏𝑗 − 𝑏𝑗+8 for all 𝑗 𝑏𝑗’s cannot be ±1 No circulant Hadamard matrix of order 16

Turyn (1965)

Suppose a circulant Hadamard matrix of order 𝑤 = 4𝑣2, 𝑣 ≥ 2, exists. Then:

• 𝑣 is odd and 𝑣 ≥ 55
• If 𝑟 is a prime power dividing 𝑣, then 𝑟3 ≤ 2𝑣2
• “Self-conjugacy” bound holds

All results based on divisibility conditions for 𝑏𝑗’s coming from 𝑗=0

𝑤−1 𝑏𝑗𝜂𝑗𝑙 2 = 𝑤

Further Known Results on Circulant Hadamard Conjecture

Suppose a circulant Hadamard matrix of order 𝑤 = 4𝑣2, 𝑣 ≥ 2, exists.

• 2𝑡−1𝐺(4𝑣2, 𝑣) ≥ 𝑣2 (𝑡 = is number of distinct prime divisors of 𝑣)

(S. 1999)

• Only finitely many possible 𝑣 if the prime divisors of 𝑣 are bounded by a

constant (S. 1999)

• 𝐺 4𝑣2, 𝑣 2/(4𝜒(𝐺 4𝑣2, 𝑣 ) ≥ 𝑣2 (S. 2001)
• 𝐺 𝑣2, 𝑣 𝑣/𝜒(𝑣) ≥ 𝑣2 (Leung, S. 2005)
• 𝑣 ≥ 11715 (Leung, S. 2005)
• Improved F-bounds (Leung, S. 2012)

Parseval for Polynomials

Let 𝑔 𝑦 = 𝑗=0

𝑤−1 𝑏𝑗𝑦𝑗 and 𝜂 = exp( 2𝜌𝑗 𝑤 ) 𝑙=0 𝑤−1

𝑔 𝜂𝑙

2 = 𝑙=0 𝑤−1

𝑔 𝜂𝑙 𝑔 𝜂𝑙

=

𝑙=0 𝑤−1 𝑗,𝑘

𝑏𝑗𝑏𝑘 𝜂𝑙(𝑗−𝑘) =

𝑗,𝑘

𝑏𝑗𝑏𝑘

𝑙=0 𝑤−1

𝜂𝑙(𝑗−𝑘) = 𝑤 𝑏𝑗2

F-Bound (S.)

Let’s try to prove

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗

2

< 𝑤 Write 𝑔 𝑦 = 𝑗=0

𝑤−1 𝑏𝑗𝑦𝑗. Then 𝑔 𝜂𝑙 2 = 𝑤 for all 𝑙

Parseval ⇒ 𝑤2 = 𝑤 𝑏𝑗2 = 𝑙=0

𝑤−1 𝑔 𝜂𝑙 2 = 𝑤2

In particular, 𝑤2 ≥ 𝑙:gcd 𝑙,𝑤 =1 𝑔 𝜂𝑙

2 = 𝜒 𝑤 𝑤

F-Bound (S.)

Suppose there is a divisor 𝐺 of 𝑤 such that

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗 =

𝑗=0 𝐺−1

𝑏𝑗𝑤/𝐺𝜂𝑗𝑤/𝐺 By the same argument, 𝐺2 ≥

𝑙:gcd 𝑙,𝐺 =1

𝑔 𝜂𝑙𝑤/𝐺

2 = 𝜒 𝐺 𝑤

Thus 𝑤 ≤

𝐺2 𝜒 𝐺

(F-bound)

Field Descent (S. 1999)

Goal: Find divisor 𝐺 of 𝑤 with

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗 =

𝑗=0 𝐺−1

𝑏𝑗𝑤/𝐺𝜂𝑗𝑤/𝐺 (∗) Let 𝑞 is the largest prime divisor of 𝑤 = 4𝑣2 If ord𝑞2 𝑟 ≡ 0 (mod 𝑞) for all prime divisors 𝑟 ≠ 𝑞 of 𝑣, then (∗) holds with 𝐺 = 𝑤/𝑞 Hence 𝑤 ≤

𝑤 𝑞 2 𝜒 𝑤 𝑞

⇒

𝑤 𝜒(𝑤) ≥ 𝑞

What if Field Descent Fails?

𝑤 = 4𝑣2, 𝑣 odd, 𝑞 largest prime divisor of 𝑣 If field descent fails, then ord𝑞2 𝑟 ≢ 0 (mod 𝑞) for some prime divisor 𝑟 ≠ 𝑞 of 𝑣. In fact,

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗 ≠

𝑗=0 𝑤 𝑞−1

𝑏𝑗𝑞𝜂𝑗𝑞 (∗∗) Idea: Use (∗∗) to construct another cyclotomic integer with “strange” properties

“Twisted” Cyclotomic Integer

𝑌 =

𝑗=0 𝑤−1

𝑏𝑗𝜂𝑗 ≠

𝑗=0 𝑤 𝑞−1

𝑏𝑗𝑞𝜂𝑗𝑞 (∗∗)

• rd𝑞2 𝑟 ≢ 0 (mod 𝑞)
• rd𝑞2 𝑠 ≡ 0 (mod 𝑞) for 𝑠|𝑣, 𝑠 ≠ 𝑞, 𝑟

Let 𝜏 ∈ Gal( ℚ(ζ) ℚ) with Fix 𝜏 = ℚ(𝜂𝑞). Recall 𝑌 2 = 𝑤 Thus 𝑌𝑌𝜏 2 = 𝑤2 and 𝑌𝑌𝜏 ≡ 0 (mod 𝑤/𝑟2)

“Twisted” Cyclotomic Integer

𝑌𝑌𝜏 2 = 𝑤2 𝑌𝑌𝜏 ≡ 0 (mod 𝑤/𝑟2) Set 𝑍 = 𝑌𝑌𝜏/( 𝑤 𝑟2). Then 𝑍 is a cyclotomic integer with 𝑍 2 = 𝑟4 and 𝑍 ∉ ℚ(𝜂𝑞) Now write 𝑍 = 𝑐∈𝐶 𝑍

𝑐𝑐 with 𝑍 𝑐 ∈ ℤ[𝜂𝑞], where 𝐶 is an integral

basis of ℚ(𝜂)/ℚ(𝜂𝑞) Let 𝜐 ∈ Gal( ℚ 𝜂 ℚ), 𝜂𝜐 = 𝜂𝑟. Then 𝑍𝜐 = 𝜃𝑍 for some root of unity 𝜃

“Twisted” Cyclotomic Integer

𝑍 2 = 𝑟4 and 𝑍 ∉ ℚ(𝜂𝑞) 𝑍 = 𝑐∈𝐶 𝑍

𝑐𝑐 with 𝑍 𝑐 ∈ ℤ[𝜂𝑞]

𝑍𝜐 = 𝜃𝑍 𝑍

𝑐 ≠ 0 for some 𝑐 ≠ 1 ⇒ 𝑍 𝑐 ≠ 0 for at least ord𝑞(𝑟) elements 𝑐

⇒

𝛽∈Gal( ℚ 𝜂 ℚ)

𝑍𝛽 2 ≥ ord𝑞(𝑟) Gal( ℚ 𝜂 ℚ)

“Twisted” Cyclotomic Integer

𝛽∈Gal( ℚ 𝜂 ℚ)

𝑍𝛽 2 ≥ ord𝑞(𝑟) Gal( ℚ 𝜂 ℚ) 𝑍 2 = 𝑟4 ⇒ ord𝑞(𝑟) ≤ 𝑟4

Result (Leung, S. 2016), Simplified

Suppose a circulant Hadamard matrix of order 4𝑣2 exists Write 𝑣 = 𝑞𝑏𝑥 where 𝑞 is the largest prime divisor of 𝑣 and gcd 𝑞, 𝑥 = 1 Let 𝑟 be the prime divisor of 𝑥 which “prevents” the field descent 𝑞2𝑏 → 𝑞2𝑏−𝑦 Then ord𝑞 𝑟 ≤ 𝑟4𝑐max

2 𝑟2 ∪ 𝑡−1 𝑔

𝑡(𝑡−𝑔 𝑡) : 𝑡 ∈ 𝑇

Circulant Hadamard Open Cases

Order 4𝑣2

• Smallest open cases: 𝑣 = 11715, 82005, 550605, 3854235.
• 1371 open cases with 𝑣 ≤ 1013 (computation by Borwein,

Mossinghoff 2014)

• 423 of the 1371 cases ruled out by twisted cyclotomic integer

result (Leung, S. 2016)

Barker Sequences

𝑏0, … , 𝑏𝑤−1with 𝑏𝑗 = ±1 such that

𝑗=0 𝑜−𝑙−1

𝑏𝑗𝑏𝑗+𝑙 ≤ 1, 𝑙 = 1, … , 𝑤 − 1 Exist for: 𝑤 = 1, 2, 3, 4, 5, 7, 11, 13 Conjecture: Barker sequences of length 𝑤 > 13 do not exist

Known Results

Suppose a Barker sequence of length 𝑤 exists.

• 𝑤 even ⇒ ∃ circulant Hadamard matrix of order 𝑤
• 𝑤 odd ⇒ 𝑤 ≤ 13 (Storer, Turyn 1961)
• 𝑤 > 13 ⇒ 𝑤 ≥ 12,100 (Turyn 1965)
• 𝑤 > 13 ⇒ 𝑤 ≥ 1,898,884 and 𝑞 ≡ 1 (mod 4) for all odd

primes 𝑞 dividing 𝑤 (Eliahou, Kervaire, Saffari 1990)

Known Results (continued)

Suppose a Barker sequence of length 𝑤 > 13 exists.

• 𝑤 > 4 ∙ 1012 (“field descent”, S. 1999)
• 𝑤 > 1022 (Leung, S. 2005)
• 𝑤 > 2 ∙ 1030 (Leung, S. 2012, Mossinghoff 2009)

Open Cases with Length ≤ 1050

𝒗 Factorization

(computation by Borwein, Mossinghoff 2014)

Result

• rd𝑞 𝑟 ≤ 𝑟4𝑐max

2 𝑟2 ∪ 𝑡 − 1 𝑔

𝑡(𝑡 − 𝑔 𝑡) : 𝑡 ∈ 𝑇

Application to Length ≤ 1050

Consequences for Barker Sequences

• There is no Barker sequence of length 𝑤 with 13 < 𝑤 ≤ 4 ∙ 1033
• All known open cases with 𝑤 ≤ 1050 ruled out
• 229,305 out of 237,807 open cases with 𝑤 ≤ 10100 ruled out
• Smallest case known not to be ruled out:

𝑤 = 4 ∙ 301092 ∙ 11287132 ∙ 1678492 ∙ 2688132772 ≈ 1.57 ∙ 1051

Thank you!