Solution sheet 6. D-MATH Modular Forms HS 2015 Prof. Özlem Imamoglu. 1. Set Γ := SL 2 ( ) and let α GL + 2 (É).

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1 D-MATH Modular Forms HS 205 Prof. Özlem Imamoglu Solution sheet 6. Set Γ := SL 2 ( and let α GL + 2 (É. a Show that the subgroup Γ := α Γα Γis a congruence subgroup. It is enough to show that α Γα contains a principal subgroup. Choose a large enough positive integer N such that Nα,Nα M 2 (Z. We claim that Γ(N 2 α Γα. Anyγ Γ(N 2 is of the form id+n 2 g for someg M 2 ( and hence αγα = id+(nαg(nα has integer entries and determinant so thatαγα Γ. This shows thatαγ(n 2 α Γ and, equivalently, Γ(N 2 α Γα. b Show that for anyf,g S k (Γ we have that f k α,g k α S k (Γ and that f,g = f k α,g k α, where the latter inner product is with respect to Γ. Consider any τ Γ. Then there exists γ Γ such that τ = α γα and hence (f k α k τ = f k (ατ = f k (γα = (f k γ k α = f k α since f k γ = f by assumption. In order to see the second assertion let Γ 2 := Γ αγα so that α Γ 2 α = Γ. Let F 2 be a fundamental domain for Γ 2. Then α F 2 is a fundamental domain for Γ. Setting Γ := {±id}γ Γ and denoting by F the standard fundamental domain for Γ we thus get f k α,g k α := (f k αzg k αzy kdxdy SL 2 ( : Γ α F 2 = f(zg(zy kdxdy SL 2 ( : Γ F 2 = f(zg(zy kdxdy SL 2 ( : Γ 2 F 2 = f(zg(zy kdxdy F = f,g by the invariance of the inner product and since SL 2 ( : Γ = SL 2 ( : α Γ 2 α = SL 2 ( : Γ 2.

2 2. Let Γ = SL 2 (Z, fixα GL + 2 (É and set Γ := α Γα Γ. a Consider the double coset ΓαΓ ingl + 2 (Q. There is a natural group action byγonγαγ by left matrix multiplication. Show that the assignment Γ ΓαΓ, γ αγ yields a one-to-one correspondence between Γ\ΓαΓ and Γ \Γ. By Part a, it follows then that the orbit space Γ\ΓαΓ is finite. The assignment γ αγ induces a surjective map from Γ to Γ\ΓαΓ. Consider moreover any γ,γ 2 Γ such that Γαγ = Γαγ 2. Then we have as needed. γ γ 2 α Γα Γ = Γ Let the slash operator forgl + 2 (Q be defined as f k [γ](z = (detγ k j(γ,z k f(γz for any γ GL + 2 (Q. Let f M k(γ. Consider the double coset slash operator given by f k [ΓαΓ] := f k [γ]. [γ] Γ\ΓαΓ b Check that the above double coset operator is well-defined. Then show that the assignment f f k [ΓαΓ] defines an operator M k (Γ M k (Γ, resp. S k (Γ S k (Γ. Consider any complete sets of representatives R and R for Γ\ΓαΓ and any f M k (Γ. In order to see that the operator is well defined we need to show that f k γ = f k γ. γ R γ R For anyγ R there exist uniqueγ R andσ Γ such thatγ = σγ. Sincef M k (Γ, we have f k γ = f k (σγ = (f k σ k γ = f k γ from which the equality of the sums follows. Moreover, for anyg Γ the setrg is again a complete set of representatives forγ\γαγ. From this and the previously shown statement we conclude that (f k [ΓαΓ] k g = f k [ΓαΓ] for anyg Γ. Finally, it is immediately checked that any summand of f k [ΓαΓ] is holomorphic at all cusps and vanishes at all cusps if f does. From this the exercise follows. 2

3 3. We set Γ := SL 2 ( and let k be any integer. Let p be a prime number and consider ( 0 α p := SL 0 p 2 (. The p th Hecke operator of weight k is the operator T p : M k (Γ M k (Γ that sends f M k (Γ tot p (f := p k 2 f k [Γα p Γ]. a Show that Γα p Γ = {α GL 2 ( detα = p} = p Γ ( j 0 p Γ ( p 0 0 and conclude that T p (fτ = p f +p k f(pτ for any f M k (Γ. This shows that the definition of the Hecke operator T p given here agrees with the one given in class. By a Lemma shown in the lecture, a complete set of representatives for the action of SL 2 ( on {α M 2 ( detα = p} is given by {( {( {( } a b j p 0 R := ad = p, 0 b < d} = 0 j < p}. c d 0 p 0 Moreover, we have ( p 0 = S 0 and ( j = 0 p From this the first equality follows. Finally, (T p fτ = p k 2 γ Rf k γ(τ j= ( 0 S Γα 0 Γ ( 0 T j Γα 0 Γ. p ( ( = p k 2 j f k (τ+p k 2 p 0 f 0 p k (τ 0 p ( = p k 2 p k τ +j 2(0τ +p k f +p k 2 p k 2(0τ + k f (pτ p j= j= = p f +p k f (pτ. 3

4 b For any any positive integer m and anyf = a nq n M k (Γ consider and Show that where b n := a pn +p k an and a n We have and V m (f := U m (f := a n q mn a m nq n. T p (f = U p (f+p k V p (f = f(mτ = b n q n, := 0 if p does not divide n. a n q mn = V m f(τ m f = m a n qme n 2πij n m m m m = a n q n m m m e 2πij n m = m na n q n m = Um f(τ, where we have used that { m e 2πij n if m n m = m 0 otherwise. By Part a we thus get (T p fτ = p f +p k f(pτ = U p f(τ+p k V p f(τ = a pn q n +p k n 0 p n from which the Excercise follows. c Show that an p qn, T p X +p k X 2 = ( U p X p k V p X 4

5 where both sides are regarded as polynomials in the variable X with coefficients in the Hecke algebra H which operates on the -subspace of [[q]] formed by the q- expansions of elements f M k. Prove that the following formal identities hold: ( T n n s = n k V n n s U n n s n= n= n= T n = d n It is immediately checked that U p V p = id. Using Part b we thus get T p X +p k X 2 = (U p +p k V p X +p k X 2 = U p X p k V p X +p k U p V p X 2 = ( U p X p k V p X. d k V d Un d Since the association n T n is multiplicative, we may use the first Exercise of Sheet 5 to get that formally n n n=t s = T p n p sn = ( T s +p k 2s, p P n p P where the second equality is checked via a straightforward computation which uses that T p r = T p T p r p k T p r 2 for any integer r 2. We moreover have that T s +p k 2s = ( p k V s ( U s by Part b. As is immediately checked, the associations n U n and n n k V n are completely multiplicative. Exercise of Sheet 5 thus implies that p P( p k V s ( ( U s = n k V n n s U n n s. p P n= n= We therefore get the desired formal identity ( T n n s = n k V n n s U n n s. n= n= Finally, we write the right hand side of the above equation in terms of one sum and get ( T n n s = n k V n n s U n n s = n k V n U m (nm s n= n= n= n,m >0 ( n k V = n m U mn s = d k V m d Un n s, d n= m n n= d n where in the second last equality we have replaced nm by n and the condition m > 0 by m n. This yields the second formal identity. n= 5

6 4. a Let k 4 be an even integer and let d be the -dimension of S k (SL 2 (. Choose any non-negative integers a,b such that 2 4a +6b 4 and 4a +6b = k mod 2. For each j d, define f j := j E 2(d j+b 6 E4 a = a (j n qn, n< where is the normalized discriminant function and E 4 and E 6 are the normalized Eisenstein series of weight 4 respectively 6. Verify that a (j n = 0 for n < j and a (j j =. Conclude that the f j form a basis for S k (SL 2 (. This basis is called the Miller basis. Show moreover, that a modular form in S k (SL 2 ( has integral Fourier coefficients if and only if it is a -linear combination of the Miller basis. Let = 0 n< a n q n and E 4 = 0 n< b n q n and E 6 = 0 n< c n q n denote the Fourier expansions of the respective forms. We know that a 0 = 0 and a = b 0 = c 0 =. From this we see that a (j n = 0 for any n < j and a (j j =. This implies that the f j s are -linearly independent. Moreover, the weight of any of the f j is 2j +2(d j+4a+6b = 2d+4a+6b which we claim is k so that indeed f j S k (Γ for any j d. By a theorem shown in the lecture, we haved = k 2 ifk = 2 mod2 andd = k 2 otherwise. Using that k 2 = k 4a 6b 2 and that 2 4a+6b 4, the claim thus follows. + 4a+6b 2 Let us turn to the second assertion of this exercise. By construction, any of the f j has integral Fourier coefficients. In order to see the converse, we use Gaussian elimination to find a basis g,...,g d of S k (Γ such that for any i,j d the i th Fourier coefficient of g j is δ i,j. Then the transformation matrix from the basis f,...,f d to the basis g,...,g d has integral coefficients and determinant. Consequently, any of the g j is a -linear combination of the f j. Consider now any f S k (Γ with integral coefficients. By construction of the g j, we have that f is a -linear combination of the g j. Hence f is also a -linear combination of the f j as desired. b Let f be a normalized Hecke eigenform for SL 2 (. Show that the Fourier coefficients of f are algebraic integers. Since f is a normalized Hecke eigenfunction, its Fourier coefficients are eigenvalues of Hecke operators. We thus show that the eigenvalues of the n th Hecke operator T n are algebraic integers for any integer n. Let k denote the weight of f and let f,...,f d denote the Miller basis fors k (Γ. By construction, any of thef j has integral coefficients. Thus also T n (f j has integral coefficients by the explicite formula shown in the class for 6

7 the coefficients oft n (f j in terms of the coefficients off j. By Part a,t n (f j is thus a - linear combination of thef,...,f d. Hence the matrix representing T n with respect to the Miller basis has entries in. Consequently, the characteristic polynomial of T n is monic with coefficients in. The eigenvalues of T n are therefore indeed algebraic integers. 7