Introduction Method ASD congruences Congruences for sporadic sequences and modular forms for non-congruence subgroups Matija Kazalicki University of Zagreb Representation Theory XVI, Dubrovnik June 25, 2019
Introduction Method ASD congruences
Introduction Method ASD congruences Elementary congruences Denote by n k � 3 � n � � k � � ( − 1) k 8 n − k F ( n ) = . k j k =0 j =0
Introduction Method ASD congruences Elementary congruences Denote by n k � 3 � n � � k � � ( − 1) k 8 n − k F ( n ) = . k j k =0 j =0 Theorem (K.) For all primes p > 2 we have 2( a 2 − 6 b 2 ) (mod p ) if p = a 2 + 6 b 2 � � p − 1 � F ≡ 2 0 (mod p ) if p ≡ 5 , 11 , 13 , 17 , 19 , 23 (mod 24) .
Introduction Method ASD congruences Ap´ ery’s proof of the irrationality of ζ (3) In 1978 Roger Ap´ ery proved that ∞ 1 � ζ (3) = n 3 = 1 . 2020569031 . . . n =1 is irrational.
Introduction Method ASD congruences Ap´ ery’s proof of the irrationality of ζ (3) In 1978 Roger Ap´ ery proved that ∞ 1 � ζ (3) = n 3 = 1 . 2020569031 . . . n =1 is irrational. For that he defined sequences a n and a ′ n as a solutions of recursion ( n + 1) 3 u n +1 − (34 n 3 + 51 n 2 + 27 n + 5) u n + n 3 u n − 1 = 0 , with initial conditions ( a 0 , a 1 ) = (1 , 5) and ( a ′ 0 , a ′ 1 ) = (0 , 6).
Introduction Method ASD congruences Ap´ ery’s proof of the irrationality of ζ (3) In 1978 Roger Ap´ ery proved that ∞ 1 � ζ (3) = n 3 = 1 . 2020569031 . . . n =1 is irrational. For that he defined sequences a n and a ′ n as a solutions of recursion ( n + 1) 3 u n +1 − (34 n 3 + 51 n 2 + 27 n + 5) u n + n 3 u n − 1 = 0 , with initial conditions ( a 0 , a 1 ) = (1 , 5) and ( a ′ 0 , a ′ 1 ) = (0 , 6). He showed that for n sufficiently large relative to ǫ | ζ (3) − p n q n θ + ǫ , where p n 1 = a ′ n | < and θ = 1 . 080529 . . . , q n q n a n which implies that ζ (3) is irrational.
Introduction Method ASD congruences Ap´ ery’s proof of the irrationality of ζ (3) cont. One thing that is remarkable here is that a n ’s are integers, i.e. � 2 . � 2 � n + k a n = � n � n k =0 k k
Introduction Method ASD congruences Ap´ ery’s proof of the irrationality of ζ (3) cont. One thing that is remarkable here is that a n ’s are integers, i.e. � 2 . � 2 � n + k a n = � n � n k =0 k k Similarly for the proof of irrationality of ζ (2) he introduced � 2 � n + k numbers b n = � n � n � as a solutions of recursion k =0 k k ( n + 1) 2 u n +1 − (11 n 2 + 11 n + 3) u n − n 2 u n − 1 = 0 .
Introduction Method ASD congruences Zagier’s sporadic sequences Zagier performed a computer search on first 100 million triples ( A , B , C ) ∈ Z 3 and found that the recursive relation generalizing b n ( n + 1) 2 u n +1 − ( An 2 + An + B ) u n + Cn 2 u n − 1 = 0 , with the initial conditions u − 1 = 0 and u 0 = 1 has (non-degenerate i.e. C ( A 2 − 4 C ) � = 0) integral solution for only six more triples (whose solutions are so called sporadic sequences) (0 , 0 , − 16) , (7 , 2 , − 8) , (9 , 3 , 27) , (10 , 3 , 9) , (12 , 4 , 32) and (17 , 6 , 72) .
Introduction Method ASD congruences Zagier’s sporadic sequences Zagier performed a computer search on first 100 million triples ( A , B , C ) ∈ Z 3 and found that the recursive relation generalizing b n ( n + 1) 2 u n +1 − ( An 2 + An + B ) u n + Cn 2 u n − 1 = 0 , with the initial conditions u − 1 = 0 and u 0 = 1 has (non-degenerate i.e. C ( A 2 − 4 C ) � = 0) integral solution for only six more triples (whose solutions are so called sporadic sequences) (0 , 0 , − 16) , (7 , 2 , − 8) , (9 , 3 , 27) , (10 , 3 , 9) , (12 , 4 , 32) and (17 , 6 , 72) . The sequence F ( n ) corresponds to the triple (17 , 6 , 72).
Introduction Method ASD congruences The previous work Stienstra and Beukers proved congruences analogous to the one in the first slide for Apery numbers (and for two more sporadic sequences). Recently Osburn and Straub proved them for all sequences except for F ( n ) - for which they made a conjecture.
Introduction Method ASD congruences Connection with geometry Stienstra and Beukers showed that the generating functions of Ap´ ery’s numbers b n (and Zagier for other sporadic sequences) are holomorphic solutions of Picard-Fuchs differential equation of some elliptic surface.
Introduction Method ASD congruences Picard-Fuchs differential equations for the Legendre family of elliptic curves For t ∈ C let y 2 = x ( x − 1)( x − t ) , E t : be Legendre’s family of elliptic curve with period integrals � 1 � ∞ dx dx Ω 1 ( t ) = , Ω 2 ( t ) = . � � x ( x − 1)( x − t ) x ( x − 1)( x − t ) 1 t
Introduction Method ASD congruences Picard-Fuchs differential equations for the Legendre family of elliptic curves For t ∈ C let y 2 = x ( x − 1)( x − t ) , E t : be Legendre’s family of elliptic curve with period integrals � 1 � ∞ dx dx Ω 1 ( t ) = , Ω 2 ( t ) = . � � x ( x − 1)( x − t ) x ( x − 1)( x − t ) 1 t They satisfy Picard-Fuchs differential equation t ( t − 1)Ω ′′ ( t ) + (2 t − 1)Ω ′ ( t ) + 1 4Ω( t ) = 0 , whose unique holomorphic solution at t = 0 is hyperelliptic function ∞ ((1 / 2) n ) 2 � t n . − π Ω 2 ( t ) = n ! n =0
Introduction Method ASD congruences Modular elliptic surface and sequence F ( n ) Consider modular rational elliptic surface for Γ 1 (6) W : ( x + y )( x + z )( y + z ) − 8 xyz = 1 t xyz , with fibration φ : W → P 1 , ( x , y , z , t ) �→ t .
Introduction Method ASD congruences Modular elliptic surface and sequence F ( n ) Consider modular rational elliptic surface for Γ 1 (6) W : ( x + y )( x + z )( y + z ) − 8 xyz = 1 t xyz , with fibration φ : W → P 1 , ( x , y , z , t ) �→ t . Picard-Fuchs differential equation associated to this elliptic surface (8 t + 1)(9 t + 1) P ( t ) ′′ + t (144 t + 17) P ( t ) ′ + 6 t (12 t + 1) P ( t ) = 0 , has a holomorphic solution around t = 0 ∞ � ( − 1) n F ( n ) t n . P ( t ) = n =0
Introduction Method ASD congruences Modular forms We can identify t with a modular function (for Γ 0 (6)) t ( τ ) = η (2 τ ) η (6 τ ) 5 η ( τ ) 5 η (3 τ ) , τ ∈ H n =0 ( − 1) n F ( n ) t ( τ ) n is a weight one modular form then P ( τ ) := � ∞ for Γ 1 (6).
Introduction Method ASD congruences The main idea Proposition (Beukers) n =1 b n t n − 1 dt a differential form Let p be a prime and ω ( t ) = � ∞ with b n ∈ Z p . Let t ( q ) = � ∞ n =1 A n q n ,A n ∈ Z p , and suppose ∞ � c n q n − 1 dq . ω ( t ( q )) = n =1 Suppose there exist α p , β p ∈ Z p with p | β p such that (mod p r ) , b mp r − α p b mp r − 1 + β p b mp r − 2 ≡ 0 ∀ m , r ∈ N . Then c mp r − α p c mp r − 1 + β p c mp r − 2 ≡ 0 (mod p r ) , ∀ m , r ∈ N . Moreover, if A 1 is p-adic unit then the second congruence implies the first, and we have that b p ≡ α p b 1 (mod p ) .
Introduction Method ASD congruences Congruences for F ( n )
Introduction Method ASD congruences Congruences for F ( n ) Now consider a two cover S of W , a K3-surface given by the equation S : ( x + y )( x + z )( y + z ) − 8 xyz = 1 s 2 xyz , � η (2 τ ) η (6 τ ) 5 where t = s 2 . Then s ( τ ) = η ( τ ) 5 η (3 τ ) is a corresponding modular function for index two genus zero subgroup Γ 2 ⊂ Γ 1 (6)
Introduction Method ASD congruences Congruences for F ( n ) Now consider a two cover S of W , a K3-surface given by the equation S : ( x + y )( x + z )( y + z ) − 8 xyz = 1 s 2 xyz , � η (2 τ ) η (6 τ ) 5 where t = s 2 . Then s ( τ ) = η ( τ ) 5 η (3 τ ) is a corresponding modular function for index two genus zero subgroup Γ 2 ⊂ Γ 1 (6) Given prime p > 2, we apply the previous proposition to the differential form ∞ � ( − 1) n F ( n ) s 2 n ds , ω ( s ) = n =1 and s ( q ) - the q -expansion of modular function s ( τ ) (where q = e π i τ ).
Introduction Method ASD congruences Congruences for F ( n ) cont. We obtain that ∞ � c n q n − 1 dq , ω ( s ( q )) = n =0 where c n are Fourier coefficients of weight 3 cusp form g ( τ ) ∈ S 3 (Γ 2 ) ∞ dq s ( q ) = q +3 2 q 3 − 9 8 q 5 − 85 16 q 7 − 981 g ( q ) = P ( q ) q d 128 q 9 + · · · = � c n q n . n =1
Introduction Method ASD congruences It is enought to prove that g ( τ ) satisfy three term Atkin and Swinnerton-Dyer (ASD) congruence relation.
Introduction Method ASD congruences It is enought to prove that g ( τ ) satisfy three term Atkin and Swinnerton-Dyer (ASD) congruence relation. Proposition (K.) Let p > 3 be a prime. Then for all m , r ∈ N , we have that � − 1 � � − 6 � p 2 c mp r − 2 ≡ 0 (mod p 2 r ) , c mp r − γ ( p ) c mp r − 1 + p p where 2( a 2 − 6 b 2 ) (mod p ) if p = a 2 + 6 b 2 � γ ( p ) = . 0 (mod p ) if p ≡ 5 , 11 , 13 , 17 , 19 , 23 (mod 24) .
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