A crash course. . . Day 2: Modular Forms Sharon Anne Garthwaite - PowerPoint PPT Presentation

A crash course. . . Day 2: Modular Forms Sharon Anne Garthwaite Bucknell University March 2008

Recall, p ( n ) q n = � � (1 − q n ) − 1 . � − 1 p ( n ) q 24 n − 1 = � � � (1 − q 24 n ) q = η − 1 (24 z ) , q := e 2 π iz . This is an example of a modular form.

Basic Definition f ( z ), a meromorphic function on the upper half plane H = { z := x + iy ∈ C : x , y ∈ R , y > 0 } , is a modular form of weight k if: ◮ f ( z + 1) = f ( z ). ◮ f ( − 1 / z ) = z k f ( z ) for some k ∈ Z . ◮ f ( z ) has a Fourier expansion ∞ � q := e 2 π iz . a f ( n ) q n ; f ( z ) := n ≥ n 0 “Behavior not too bad as z → i ∞ ”

Transformation Property � a b � SL 2 ( Z ) := { : a , b , c , d ∈ Z , ad − bc = 1 } c d � 0 − 1 ◮ Example: S := � , T := ( 1 1 0 1 ) ∈ SL 2 ( Z ). 1 0 ◮ S , T generate SL 2 ( Z ).

Transformation Property � a b � For V = ∈ SL 2 ( Z ), z ∈ H , c d Vz := az + b cz + d . Fractional Linear Transformations ◮ Inflations/Rotations z → az ◮ Translations z → z + b ◮ Inversions z → − 1 / z “Conformal maps” that map circles and lines to circles and lines.

Transformation Property � a b � For V = ∈ SL 2 ( Z ), z ∈ H , c d Vz := az + b cz + d . Examples: ◮ T := ( 1 1 0 1 ), T : z �→ z + 1 � 0 − 1 � S := , S : z �→ 1 / z . 1 0 ◮ ST : z �→ − 1 / ( z + 1), TS : z �→ − 1 / z + 1 = ( z − 1) / z . Check: If z ∈ H , Vz ∈ H . ◮ Fundamental domain F

More General Transformation Property � a b � For V = ∈ SL 2 ( Z ), z ∈ H , c d Vz := az + b cz + d . f ( Vz ) = ( cz + d ) k f ( z ) . Definition (“Slash” operator ) f | k V ( z ) := ( cz + d ) − k f ( Vz ) . Transformation property: f | k V ( z ) = f ( z )

Cusps ◮ Want to “complete” H = { z := x + iy ∈ C : x , y ∈ R , y > 0 } . ◮ Add point at infinity : ∞ . ◮ � a b � ∞ = a / c . c d ◮ Cusps: ∞ ∪ Q ◮ Same equivalency class in SL 2 ( Z ).

Simple Complex Analysis Let f ( z ) : C → C . ◮ z 0 is a zero of f if f ( z 0 ) = 0. ◮ z 0 is a singularity of f if f ( z 0 ) is undefined. ◮ f ( z ) is analytic if it is differentiable with respect to z . ◮ f ( z ) is meromorphic if it is analytic except an isloated singularities which are “poles.”

Singularities 1. Removable Example z 2 − 1 z − 1 . 2. Poles sin( z ) = 1 1 Example z + f ( z ), f ( z ) analytic. (Taylor series) 3. Essential Example e 1 / z .

Meromorphic vs. Holomorphic On SL 2 ( Z ), if f ( z ) has a Fourier expansion ∞ � q := e 2 π iz , a f ( n ) q n ; f ( z ) := n ≥ n 0 f ( z ) is holomorphic if n 0 ≥ 0. “ f ( z ) does not blow up at the cusps.” Let M k denote the space of weight k holomorphic modular forms on SL 2 ( Z ).

Cauchy’s Integral Formula 1 � f ( w ) f ( z ) = z − w dw . 2 π i ∂ D where ◮ D is a bounded domain with a smooth boundary. ◮ z ∈ D . ◮ f ( z ) analytic on D . ◮ f ( z ) extends smootly to ∂ D .

Applications For f ( z ) with Fourier expansion (about infinity) ∞ a f ( n ) q n = g ( q ); � q := e 2 π iz , f ( z ) := n ≥ n 0 1 g ( q ) � a f ( n ) = q n +1 dq . 2 π i γ Residue Theorem � � f ( z ) dz = 2 π i Res ( f ) | z = a . ∂ D a ∈ D

Examples of Modular Forms: Eisenstein series Let 1 � G k ( z ) := ( mz + n ) k ( m , n ) ∈ Z 2 ( m , n ) � =(0 , 0) This sum is absolutely convergent for k > 2.

Normalized Eisenstein series For even k ≥ 4, ∞ E k ( z ) := 1 − 2 k � σ k − 1 ( n ) q n , B k n =1 where the rational (Bernoulli) numbers B k are ∞ B n · t n e t − 1 = 1 − 1 2 t + 1 t 12 t 2 + · · · , � n ! = n =0 and � d k − 1 . σ k − 1 ( n ) = 1 ≤ d | n This is G k ( z ) normalized to have constant coefficient 1.

Eisenstein Series ∞ � σ 3 ( n ) q n ◮ E 4 ( z ) := 1 + 240 n =1 ∞ � ◮ E 6 ( z ) := 1 − 504 σ 5 ( n ) q n n =1 ∞ � ◮ E 8 ( z ) := 1 + 480 σ 7 ( n ) q n n =1 ∞ � σ 9 ( n ) q n ◮ E 10 ( z ) := 1 − 264 n =1 ∞ ◮ E 12 ( z ) := 1 + 65520 � σ 11 ( n ) q n 691 n =1 ∞ � σ 13 ( n ) q n ◮ E 14 ( z ) := 1 − 24 n =1 If ℓ ≥ 5 is prime then E ℓ − 1 ( z ) ≡ 1 (mod ℓ ).

Eisenstein Series ∞ � σ 1 ( n ) q n ? What about E 2 ( z ) := 1 − 24 n =1 ◮ E 2 ( z + 1) = E 2 ( z ). ◮ E 2 ( − 1 / z ) = z 2 E 2 ( z ) + 12 z 2 π i “Quasi-modular form”. Note: For p ≥ 3 prime, E 2 ( z ) ≡ E p +1 ( z ) (mod p ).

More Modular Forms How can we make new modular forms? ◮ Add and subtract? Only if the same weight. ◮ Multiplication? Yes Example: E 4 ( z ) E 6 ( z ) ∈ M 10 . Can we get all modular forms from Eisenstein series?

How ‘big’ is M k ? Finite dimensional C vector space. For k ≥ 4 even, dimension of M k is � ⌊ k 12 ⌋ + 1 if k �≡ 2 (mod 12) , d ( k ) := ⌊ k 12 ⌋ if k ≡ 2 (mod 12) . If k odd, d ( k ) = 0. Dimension formula consequence of the Valence formula.

Valence Formula If f ( z ) ∈ M k , 12 = v ∞ ( f ) + 1 k 2 v i ( f ) + 1 � 3 v ω ( f ) + v τ ( f ) . τ ∈F τ �∈{ i ,ω } ◮ ω := e 2 π i / 3 . ◮ F fundamental domain of H . ◮ v τ ( f ) the order of vanishing of f at τ

The Valence formula can be computed with complex integration.

� ⌊ k 12 ⌋ + 1 if k �≡ 2 (mod 12) , d ( k ) := ⌊ k 12 ⌋ if k ≡ 2 (mod 12) . ◮ For which weights is M k one dimensional? ◮ What is the first k with d ( k ) = 2?

Delta Function E 4 ( z ) 3 , E 6 ( z ) 2 ∈ M 12 . Check: E 4 ( z ) 3 � = cE 6 ( z ) 2 for any c ∈ C . Define ∆( z ) : = E 2 4 ( z ) − E 3 6 ( z ) 1728 ∞ � (1 − q n ) 24 = q n =1 = q − 24 q 2 + 252 q 3 − · · · ∈ M 12 ∩ Z [[ q ]] . ∞ ◮ ∆( z ) := η 24 ( z ) where η ( z ) := q 1 / 24 � (1 − q n ). n =1 ◮ First example of a cusp form.

Cusp Forms If f ( z ) ∈ M k has Fourier expansion: ∞ � a f ( n ) q n , f ( z ) := n ≥ n 0 f ( z ) is a cusp form if a f (0) = 0. Let S k denote the space of cusp forms of weight k .

Dimension formula ◮ Can use the valence formula to prove d ( k ) = 1 for k = 4 , 6 , 8 , 10 , 14. ◮ Can check by valence formula that only zero of ∆( z ) is at ∞ . ◮ Can set up a correspondence between M k − 12 and S k using multiplication/division by ∆( z ). ◮ S k = ∆ M k − 12 , k ≥ 14. ◮ M k = C E k ⊕ S k , k ≥ 4. 4 ( z ) E j ◮ It is true that M k is generated by { E i 6 ( z ) : 4 i + 6 j = k } .

Ramanujan’s τ function ∞ � (1 − q n ) 24 ∆( z ) = q n =1 = q − 24 q 2 + 252 q 4 − 1472 q 4 + 4830 q 5 − 6048 q 6 − 16744 q 7 + · · · Define τ ( n ) by ∞ (1 − q n ) 24 = � � τ ( n ) q n . q n =1 Questions? ◮ For which n is τ ( n ) = 0? ◮ How big are the values of τ ( n )?

Ramanujan’s τ function ◮ For which n is τ ( n ) = 0? Lehmer’s Conjecture (1947): τ ( n ) � = 0 for any n ≥ 1. ◮ How big are the values of τ ( n )? Deligne (1974) | τ ( p ) | ≤ 2 p 11 / 2 if p is prime.

How quickly does p ( n ) grow? Hardy-Ramanujan (1918): � P k ( n ) + O ( n − 1 / 4 ) p ( n ) = k <α √ n α a constant, P k ( n ) an exponential function. ◮ Infinite sum diverges for all n (Lehmer, 1937) ◮ Gives e π √ 1 2 n / 3 √ p ( n ) ∼ 4 n 3 ◮ Get exact value if n large enough to make error less than 1 / 2. ◮ Marks the start of the circle method.

Circle Method Idea 1. Let P ( q ) := � (1 − q n ) − 1 be generating function for p ( n ). For each n ≥ 0, ∞ p ( k ) q k P ( q ) � q n +1 = 0 < | q | < 1 q n +1 k =0 2. Cauchy’s residue theorem: 1 � P ( q ) p ( n ) = q n +1 dq 2 π i C C closed counter-clockwise contour around origin inside unit disk. 3. Want to use information about singularities of P ( q ). These occur at each root of unity ( x k = 1)

Circle Method Idea 1. Choose circular contour C with radius close to 1. 2. Divide C into disjoint arcs C h , k about roots of unity e 2 π ih / k for 1 ≤ h < k ≤ N , ( h , k ) = 1. 3. On each arc C h , k replace P ( x ) that has the same behavior near e 2 π ih / k . (This is where modular transformation properties of P ( x ) comes into play.) 4. Evaluate these integrals along each arc. Make distinction between major/minor arcs for main terms and error.

Rademacher, 1937: Change the contour C : Convergent formula (exact formula) for p ( n ): � π √ 24 n − 1 ∞ A k ( n ) � p ( n ) = 2 π (24 n − 1) − 3 � · I 3 , 4 k 6 k 2 k =1 � 2 z � sinh z � d ◮ I 3 / 2 ( z ) = π dz z ◮ A k ( n ) is a “Kloosterman-type” sum. � x � A k ( n ) := 1 k � 12 � � � · e , 2 12 x 12 k (mod 24 k ) x x 2 ≡− 24 n +1 (mod 24 k ) where e ( x ) := e 2 π ix .

Other types of Modular Forms Recall that ∞ η ( z ) := q 1 / 24 � (1 − q n ) n =1 (1 − q n ) 24 = ∆( z ) ∈ S 12 . � η 24 ( z ) = q What about η (24 z ) = q � (1 − q 24 n )?

By Euler’s identity, � (1 − q 24 n ) η (24 z ) = q ( − 1) k q (6 k +1) 2 � = k ∈ Z Can write this as χ 12 ( n ) q n 2 � n ∈ Z Where  1 n ≡ 1 , 11 (mod 12)   χ 12 ( n ) := − 1 n ≡ 5 , 7 (mod 12)  0 else .  χ 12 ( n ) is a called a Dirichlet character.

Dirichlet characters Definition A function ψ : Z → C is a Dirichlet character modulo m if ◮ ψ ( n ) = ψ ( n + m ). ◮ ψ (1) = 1. ◮ ψ ( n 1 n 2 ) = ψ ( n 1 ) ψ ( n 2 ). ◮ ψ ( n ) = 0 if gcd( n , m ) = 1.