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 𝐼 𝑈 𝐼 = 𝑤𝐽 𝑦 𝑈 𝐼 𝑈 𝐼𝑦 = |𝜇| 2 𝑦 𝑦 𝑈 𝐼𝑦 = 𝜇𝑦 ⇒ 𝐼 𝑈 𝐼 = 𝑤𝐽 ⇒ 𝑦 𝑈 𝐼 𝑈 𝐼𝑦 = 𝑤𝑦 𝑦 𝑈 𝜇 ∈ ℂ eigenvalue of Hadamard matrix of order 𝑤 ⇓ |𝜇| 2 = 𝑤

Circulants 𝑏 0 𝑏 1 ⋯ 𝑏 𝑤−1 𝑏 𝑤−1 𝑏 0 ⋯ 𝑏 𝑤−2 𝐼 = ⋯ ⋯ ⋯ ⋯ 𝑏 1 𝑏 2 ⋯ 𝑏 0 Eigenvalues: 𝑤−1 𝑏 𝑗 𝜂 𝑗𝑙 , 𝑙 = 0, … , 𝑤 − 1, 𝜂 = exp( 2𝜌𝑗 𝑗=0 𝑤 ) 2 𝑤−1 𝑏 𝑗 𝜂 𝑗𝑙 ⇒ = 𝑤 𝐼 Hadamard matrix 𝑗=0 Note: Implies 𝑤 = 4𝑣 2

Basic Idea What does 2 𝑤−1 𝑏 𝑗 𝜂 𝑗𝑙 = 𝑤 𝑗=0 say about the 𝑏 𝑗 ’s?

Divisibility Method (Turyn) 15 𝑏 𝑗 𝜂 𝑗 2 = 16, 𝜂 = exp( 2𝜌𝑗 Suppose 𝑗=0 16 ) 1 − 𝜂 3 ⋯ (1 − 𝜂 15 ) Factorization: 2 = 1 − 𝜂 15 𝑏 𝑗 𝜂 𝑗 2 in ℤ[𝜂] Hence 1 − 𝜂 32 divides 𝑗=0 𝜂 = 1 − 𝜂 15 = (1 + ⋯ + 𝜂 14 )(1 − 𝜂) 1 − 1 − 𝜂 = (1 + ⋯ + 𝜂 14 )(1 − 𝜂) 15 𝑏 𝑗 𝜂 𝑗 Thus 1 − 𝜂 16 divides 𝑗=0

Divisibility Method (Turyn) 15 𝑏 𝑗 𝜂 𝑗 1 − 𝜂 16 divides 𝑗=0 15 𝑏 𝑗 𝜂 𝑗 4 divides 𝑗=0 15 𝑏 𝑗 𝜂 𝑗 = 𝑗=0 7 (𝑏 𝑗 −𝑏 𝑗+8 )𝜂 𝑗 Basis representation: 𝑗=0 4 divides 𝑏 𝑗 − 𝑏 𝑗+8 for all 𝑗 𝑏 𝑗 ’s cannot be ±1 No circulant Hadamard matrix of order 16

Turyn (1965) Suppose a circulant Hadamard matrix of order 𝑤 = 4𝑣 2 , 𝑣 ≥ 2, exists. Then: • 𝑣 is odd and 𝑣 ≥ 55 • If 𝑟 is a prime power dividing 𝑣 , then 𝑟 3 ≤ 2𝑣 2 • “Self - conjugacy” bound holds All results based on divisibility conditions for 𝑏 𝑗 ’s 𝑤−1 𝑏 𝑗 𝜂 𝑗𝑙 2 = 𝑤 coming from 𝑗=0

Further Known Results on Circulant Hadamard Conjecture Suppose a circulant Hadamard matrix of order 𝑤 = 4𝑣 2 , 𝑣 ≥ 2, exists. 2 𝑡−1 𝐺(4𝑣 2 , 𝑣) ≥ 𝑣 2 ( 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 4𝑣 2 , 𝑣 2 /(4𝜒(𝐺 4𝑣 2 , 𝑣 ) ≥ 𝑣 2 (S. 2001) • 𝐺 𝑣 2 , 𝑣 𝑣/𝜒(𝑣) ≥ 𝑣 2 (Leung, S. 2005) • • 𝑣 ≥ 11715 (Leung, S. 2005) • Improved F-bounds (Leung, S. 2012)

Parseval for Polynomials 𝑤−1 𝑏 𝑗 𝑦 𝑗 and 𝜂 = exp( 2𝜌𝑗 Let 𝑔 𝑦 = 𝑗=0 𝑤 ) 𝑤−1 𝑤−1 2 = 𝑔 𝜂 𝑙 𝑔 𝜂 𝑙 𝑔 𝜂 𝑙 𝑙=0 𝑙=0 𝑤−1 𝑏 𝑗 𝑏 𝑘 𝜂 𝑙(𝑗−𝑘) = 𝑙=0 𝑗,𝑘 𝑤−1 𝜂 𝑙(𝑗−𝑘) = 𝑏 𝑗 𝑏 𝑘 𝑗,𝑘 𝑙=0 = 𝑤 𝑏 𝑗2

F-Bound (S.) Let’s try to prove 2 𝑤−1 𝑏 𝑗 𝜂 𝑗 < 𝑤 𝑗=0 2 = 𝑤 for all 𝑙 𝑤−1 𝑏 𝑗 𝑦 𝑗 . Then 𝑔 𝜂 𝑙 Write 𝑔 𝑦 = 𝑗=0 2 = 𝑤 2 𝑤−1 𝑔 𝜂 𝑙 Parseval ⇒ 𝑤 2 = 𝑤 𝑏 𝑗2 = 𝑙=0 2 = 𝜒 𝑤 𝑤 In particular, 𝑤 2 ≥ 𝑙:gcd 𝑙,𝑤 =1 𝑔 𝜂 𝑙

F-Bound (S.) Suppose there is a divisor 𝐺 of 𝑤 such that 𝑤−1 𝐺−1 𝑏 𝑗 𝜂 𝑗 = 𝑏 𝑗𝑤/𝐺 𝜂 𝑗𝑤/𝐺 𝑗=0 𝑗=0 By the same argument, 2 = 𝜒 𝐺 𝑤 𝐺 2 ≥ 𝑔 𝜂 𝑙𝑤/𝐺 𝑙:gcd 𝑙,𝐺 =1 𝐺 2 Thus 𝑤 ≤ (F-bound) 𝜒 𝐺

Field Descent (S. 1999) Goal: Find divisor 𝐺 of 𝑤 with 𝑤−1 𝐺−1 𝑏 𝑗 𝜂 𝑗 = 𝑏 𝑗𝑤/𝐺 𝜂 𝑗𝑤/𝐺 (∗) 𝑗=0 𝑗=0 Let 𝑞 is the largest prime divisor of 𝑤 = 4𝑣 2 If ord 𝑞 2 𝑟 ≡ 0 (mod 𝑞) for all prime divisors 𝑟 ≠ 𝑞 of 𝑣 , then (∗) holds with 𝐺 = 𝑤/𝑞 𝑤 𝑞 2 𝑤 Hence 𝑤 ≤ ⇒ 𝜒(𝑤) ≥ 𝑞 𝜒 𝑤 𝑞

What if Field Descent Fails? 𝑤 = 4𝑣 2 , 𝑣 odd, 𝑞 largest prime divisor of 𝑣 If field descent fails, then ord 𝑞 2 𝑟 ≢ 0 (mod 𝑞) for some prime divisor 𝑟 ≠ 𝑞 of 𝑣. In fact, 𝑤−1 𝑤 𝑞−1 𝑏 𝑗 𝜂 𝑗 ≠ 𝑏 𝑗𝑞 𝜂 𝑗𝑞 (∗∗) 𝑗=0 𝑗=0 Idea: Use (∗∗) to construct another cyclotomic integer with “strange” properties

“Twisted” Cyclotomic Integer 𝑤−1 𝑤 𝑞−1 𝑏 𝑗 𝜂 𝑗 ≠ 𝑏 𝑗𝑞 𝜂 𝑗𝑞 𝑌 = (∗∗) 𝑗=0 𝑗=0 ord 𝑞 2 𝑟 ≢ 0 (mod 𝑞) ord 𝑞 2 𝑠 ≡ 0 (mod 𝑞) for 𝑠|𝑣 , 𝑠 ≠ 𝑞, 𝑟 ℚ(ζ) ℚ) with Fix 𝜏 = ℚ(𝜂 𝑞 ) . Recall 𝑌 2 = 𝑤 Let 𝜏 ∈ Gal( 𝑌𝑌 𝜏 2 = 𝑤 2 and 𝑌𝑌 𝜏 ≡ 0 (mod 𝑤/𝑟 2 ) Thus

“Twisted” Cyclotomic Integer 𝑌𝑌 𝜏 2 = 𝑤 2 𝑌𝑌 𝜏 ≡ 0 (mod 𝑤/𝑟 2 ) Set 𝑍 = 𝑌𝑌 𝜏 /( 𝑤 𝑟 2 ) . Then 𝑍 is a cyclotomic integer with 𝑍 2 = 𝑟 4 and 𝑍 ∉ ℚ(𝜂 𝑞 ) 𝑐 ∈ ℤ[𝜂 𝑞 ] , where 𝐶 is an integral Now write 𝑍 = 𝑐∈𝐶 𝑍 𝑐 𝑐 with 𝑍 basis of ℚ(𝜂)/ℚ(𝜂 𝑞 ) ℚ), 𝜂 𝜐 = 𝜂 𝑟 . Then 𝑍 𝜐 = 𝜃𝑍 for some root of Let 𝜐 ∈ Gal( ℚ 𝜂 unity 𝜃

“Twisted” Cyclotomic Integer 𝑍 2 = 𝑟 4 and 𝑍 ∉ ℚ(𝜂 𝑞 ) 𝑐 ∈ ℤ[𝜂 𝑞 ] 𝑍 = 𝑐∈𝐶 𝑍 𝑐 𝑐 with 𝑍 𝑍 𝜐 = 𝜃𝑍 𝑍 𝑐 ≠ 0 for some 𝑐 ≠ 1 ⇒ 𝑍 𝑐 ≠ 0 for at least ord 𝑞 (𝑟) elements 𝑐 𝑍 𝛽 2 ≥ ord 𝑞 (𝑟) Gal( ⇒ ℚ 𝜂 ℚ) 𝛽∈Gal( ℚ 𝜂 ℚ)

“Twisted” Cyclotomic Integer 𝑍 𝛽 2 ≥ ord 𝑞 (𝑟) Gal( ℚ 𝜂 ℚ) 𝛽∈Gal( ℚ 𝜂 ℚ) 𝑍 2 = 𝑟 4 ⇒ ord 𝑞 (𝑟) ≤ 𝑟 4

Result (Leung, S. 2016), Simplified Suppose a circulant Hadamard matrix of order 4𝑣 2 exists Write 𝑣 = 𝑞 𝑏 𝑥 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 𝑞 2𝑏 → 𝑞 2𝑏−𝑦 2 𝑡−1 Then ord 𝑞 𝑟 ≤ 𝑟 4𝑐 max 𝑟 2 ∪ 𝑡 ) : 𝑡 ∈ 𝑇 𝑔 𝑡 (𝑡−𝑔

Circulant Hadamard Open Cases Order 4𝑣 2 • Smallest open cases: 𝑣 = 11715 , 82005, 550605, 3854235 . • 1371 open cases with 𝑣 ≤ 10 13 (computation by Borwein, Mossinghoff 2014) • 423 of the 1371 cases ruled out by twisted cyclotomic integer result (Leung, S. 2016)

Barker Sequences 𝑏 0 , … , 𝑏 𝑤−1 with 𝑏 𝑗 = ±1 such that 𝑜−𝑙−1 𝑏 𝑗 𝑏 𝑗+𝑙 ≤ 1, 𝑙 = 1, … , 𝑤 − 1 𝑗=0 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 ∙ 10 12 (“field descent”, S . 1999) • 𝑤 > 10 22 (Leung, S. 2005) • 𝑤 > 2 ∙ 10 30 (Leung, S. 2012, Mossinghoff 2009)

Open Cases with Length ≤ 10 50 𝒗 Factorization (computation by Borwein, Mossinghoff 2014)

Result 2 𝑡 − 1 ord 𝑞 𝑟 ≤ 𝑟 4𝑐 max 𝑟 2 ∪ 𝑡 ) : 𝑡 ∈ 𝑇 𝑔 𝑡 (𝑡 − 𝑔

Application to Length ≤ 10 50

Consequences for Barker Sequences • There is no Barker sequence of length 𝑤 with 13 < 𝑤 ≤ 4 ∙ 10 33 All known open cases with 𝑤 ≤ 10 50 ruled out • 229,305 out of 237,807 open cases with 𝑤 ≤ 10 100 ruled out • • Smallest case known not to be ruled out: 𝑤 = 4 ∙ 30109 2 ∙ 1128713 2 ∙ 167849 2 ∙ 268813277 2 ≈ 1.57 ∙ 10 51

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