1 Second Facts About Spaces of Modular Forms

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1 April 30, :00 pm 1 Secon Facts About Spaces of Moular Forms We have repeately use facts about the imensions of the space of moular forms, mostly to give specific examples of an relations between moular forms of a given weight. We now prove these results in this section. Recall that a moular form f of weight 0 is a holomorphic function that is invariant uner the action of SL(, Z) an whose omain can be extene to inclue the point at infinity. Hence, we may regar f as a holomorphic function on the compact Riemann surface G\H, where H = H { }. If you haven t ha a course in complex analysis, you shouln t worry too much about this statement. We re really just saying that our funamental omain can be unerstoo as a compact manifol with complex structure. So locally, we look like C an there are compatibility conitions on overlapping local maps. By the maximum moulus principle from complex analysis (see section 3.4 of Ahlfors for example), any such holomorphic function must be constant. We can use this fact, together with the fact that given f 1, f M k, then f 1 /f is a meromorphic function invariant uner G (i.e. a weakly moular function of weight 0). Usually, one proves facts about the imensions of spaces of moular forms using founational algebraic geometry (the Riemann-Roch theorem, in particular) or the Selberg trace formula (an even more technical result relating ifferential geometry to harmonic analysis in a beautiful way). These apply to spaces of moular forms for other groups, but as we re only going to say a few wors about such spaces, we won t elve into either, but rather prove a much simpler result in the spirit of the Riemann-Roch theorem. Proposition 1 Let X be a compact Riemann surface. Fix a collection of points P 1,..., P n X an positive integers r 1,..., r n. Define V to be the vector space of meromorphic functions on X which are holomorphic except possibly at the P i, an at each P i, are either holomorphic or have a pole of orer at most r i. Then im(v ) r 1 + r + + r n + 1 Proof For each j with 1 j n, pick coorinate functions t = t j in a neighborhoo of P j so that (with respect to t j ) P j is the origin. (This is precisely what we re oing for moular forms when we take q = e πiz. This coorinate q takes the point at infinity i to the origin.) Then any meromorphic function φ in V has Laurent 1

2 expansion in t = t j of form φ(t) = a rj t r j + a rj +1t r j+1 + To each φ, we collect all the negative coefficients from the Laurent expansions for each coorinate t = t j. There are r := r r n of these, which we recor as a vector v(φ) C r. Given φ 1,..., φ N in V with N > r, then there exist coefficients c 1,..., c N C, not all 0, for which c 1 v(φ 1 ) + c v(φ ) + + c N v(φ N ) = 0. But then c 1 φ c N φ N has no poles (as we ve remove the poles at all possible P j ). Again, applying the maximum moulus principle, such a function must be constant. So any subspace of V with imension > r contains a constant function, hence im(v ) r + 1. Proposition The space M k is finite imensional. Proof Given a non-zero element g of M k, let X be the compact Riemann surface forme by G\H an let P 1,..., P n be the zeros of g, with orers r 1,..., r n. (Technicality: Some care nees to be taken in how we count these orers, as some points have non-trivial stabilizers, like i, ρ = e πi/3, ρ. If there are zeros at these points, their orers shoul be multiplie by the size of the stabilizer (which are or 3, resp.).) Let V be the vector space as in the previous proposition forme with the ata P j an r j s. If f M k, then f/g is a weakly moular function of weight 0. In particular, the map f f/g is an isomorphism of M k with V an hence has imension at most r + 1. We now prove a much more precise result about the imension of spaces of moular forms. Theorem 1 Write k = 1j + r with 0 r 10. Then { j + 1 if r = 0, 4, 6, 8, 10 im(m k ) = j if r = More succinctly, the ring k=0 M k is generate by G an G 3, the Eisenstein series of weights 4 an 6, respectively.

3 Proof First we show im(m k ) = 1 for k = 4, 6, 8 or 10. For such k, suppose f is not in the one-imensional space generate by G k. By subtracting an appropriate multiple of G k, we may assume that f has constant term 0 in its q-expansion (i.e. is a cusp form). Consier the function G h (f/ ) 6, where h = 6(1 k) an is the cusp form of weight 1 given by the Ramanujan iscriminant function. This is a moular form of weight 0 with no poles (the prouct form of shows that it is never 0 except at, where f too is 0) an hence is constant. Thus G h = c( /f) 6 for some constant c, an hence G h can have no zeros on H. But then m /G h with m = 6h is a moular form of weight 0 with no poles an a zero of orer m at, a contraiction (the zeros an poles must be equal, counting with multiplicity). Hence, im(m k ) = 1 for k = 4, 6, 8 or 10. To show M is 0, suppose f is a non-zero moular form of weight. Then from what we ve just shown, fg = cg 3 for some constant c. But then since G (e πi/3 ) = 0, we woul have G 3 (e πi/3 ) = 0 an hence (e πi/3 ) = 0 since is a linear combination of G 3 an G 3, a contraiction since is never 0. So we re one for k < 1 (noting that im(m 0 ) = 1, containing the constants). For k 1, we recall that im(s k ) = im(m k ) 1. Moreover, multiplication by is an isomorphism between M k 1 an S k. (It s clearly injective, an if f S k, then f/ again has no poles, so is in M k 1. To show that G an G 3 generate, first note that M 8 an M 10 are one imensional, so G 4 an G 5 must be linear combinations of G an G G 3, respectively. This finishes the proof for k 10. But is generate by G an G 3 as note above, so bootstrapping from k 10, we have M k 1 = S k generate by them as well. Finally, noting that G r G s 3 with 4r + 6s = k is not cuspial an in M k finishes the claim. Moular Forms for Other Groups We ve been ealing with moular forms for the group SL(, Z), or more properly the moular group G. This group arose naturally for us, relating to the equivalence on basis vectors for lattices. But suppose we want to generalize this notion of moular forms to groups other than G. We ll nee to take stock of which properties of G were necessary in formulating reasonable efinitions for moular forms. Recall that the action of matrices γ on points in the upper half plane is wellefine for any γ SL(, R). In fact SL(, R) acts transitively on H because the group of upper triangular matrices B acts transitively (i.e. takes any element in H 3

4 to any other element by an upper-triangular matrix): ( ) y 1/ xy 1/ 0 y 1/ : i x + iy so every element in H is in the orbit of i. The stabilizer of i is (check this for yourself) {( ) } a b SO() = a + b = 1 b a whose entries are often represente by sin θ an cos θ for a single parameter θ. So H may be ientifie with cosets SL(, R)/SO(). Further, since B acts transitively on H, it can be mae to act transitively on SL(, R)/SO(). Equivalently, we may write SL(, R) = B SO(); this is known as the Iwasawa ecomposition for SL(, R). This is the precise sense in which we mean that H is a moel for SL(, R). To efine moular forms, we consiere a particular iscrete subgroup SL(, Z) in SL(, R). We then ientifie SL(, Z)\H with a funamental omain D whose bounary was forme by the intersection of geoesics in the hyperbolic plane. (Remember, these geoesics for H are either vertical lines starting at the real line in C, or they re semicircles intersecting R at right angles. This comes from computing the shortest istance between two points with the metric x y/y where z = x + iy.) Moular forms f(z) for SL(, Z) of weight k coul then be unerstoo as invariant ifferential forms f(z)z k on the compact Riemann surface built from the funamental omain D. In choosing a ifferent subgroup of SL(, R), we like to retain the same properties, most importantly having a funamental omain realize as a hyperbolic polygon (i.e. boune by geoesics. In Eucliean geometry, geoesics are of course just straight lines, so hyperbolic polygon generalizes the notion of usual polygon.) In a moment, we ll give some explicit examples an eal almost exclusively with them, but first a wor about the general problem. On a first reaing, one may initially want to skip these generalities. The group SL(, R) can be embee in M (R), the group of all real matrices. This has an inner prouct g, h = Trace(gh t ) for g, h M (R) so that ( ) a b g, g = g = a + b + c + for g =. c This efines a topology on M (R) an SL(, R) inherits a topology inuce from the embeing. 4

5 Definition 1 A subgroup Γ SL(, R) is iscrete if the inuce topology is iscrete, i.e. {γ Γ γ < C} is a finite set for any constant C > 0. Definition Let X be a topological space with an action of a group Γ giving homeomorphisms of X. Then Γ acts iscontinuously on X if the orbit Γ x of any x X has no limit point in X. The following ol result of Poincaré connects these two efinitions in our case. Proposition 3 (Poincaré) A subgroup of SL(, R) is iscrete if an only if, when consiere as a subgroup of P SL(, R) = SL(, R)/{±1}, it acts iscontinuously on H. Such a subgroup of P SL(, R) is calle a Fuchsian group. Any Fuchsian group Γ can be shown to have a funamental omain D. Recall, a funamental omain for Γ has three properties: D is a omain (connecte, open set) in H. Distinct points of D are not equivalent uner Γ. The orbit of any point z H contains a point in the closure of D. These omains D are not unique, but one can show they all have the same volume (possibly infinite). We like the volume to be finite so that we can integrate over D AND we like to be able to choose D to be a (hyperbolic) polygon. For this, we nee a final assumption: Definition 3 A Fuchsian group Γ is of the first kin if every point on the bounary of H, H = R { }, is a limit point of the orbit Γ z for some z H. One can show that SL(, Z) is a Fuchsian group of the first kin (see Bump for etails). Moreover, any finite inex subgroup of a Fuchsian group of the first kin is again such a group. There are many nice results for such groups, but here s just one (whose proof is ue to Siegel): Theorem Every Fuchsian group of the first kin Γ has a funamental omain D which is a hyperbolic polygon with an even number of sies. The sies of D can be arrange in pairs, whose points are equivalent uner Γ an the elements of Γ that pair each sie generate Γ (hence a finite number of generators). Moreover, D has finite volume. 5

6 3 Moular Forms for Congruence Subgroups As we lai out in sketch in the previous section, Fuchsian groups of the first kin have all the right properties for efining moular forms. We briefly note there that SL(, Z) an its subgroups of finite inex are Fuchsian groups of the first kin. There is a nice family of groups of finite inex, efine by { ( ) ( ) a b a b Γ(n) = γ = SL(, Z) c c ( ) } (mo n). Sometimes we write SL(, Z) = Γ(1). Any subgroup of finite inex Γ with Γ(n) Γ Γ(1) = SL(, Z) for some n is calle a congruence subgroup of SL(, Z). If Γ has inex m, then one can make a funamental omain D Γ from the funamental omain D for SL(, Z) by D Γ = m γ j (D) for certain γ j Γ(1). j=1 One shoul pick the γ j to be istinct coset representatives for Γ in Γ(1) for which D Γ is a omain. There is a java applet online (very easy to play aroun with, an great for getting intuition about the shape of these funamental omains) at for rawing these funamental omains for Γ(n) as well as the congruence subgroups { ( ) } a b Γ 0 (n) = γ = SL(, Z) c 0 (mo n). c Of particular importance are the images of { } uner γ j. Recall that ( ) a b γ = : z az + b an γ( ) = a c cz + c Q. This will give a set of inequivalent rational numbers p 1,..., p r calle cusps (where we take p 1 = ). We are now reay to efine moular forms for a congruence subgroup. Definition 4 A moular form of weight k for Γ, a congruence subgroup of Γ(1), is a holomorphic function f(z) on H satisfying 6

7 1. f(γz) = (cz + ) k f(z) for all γ Γ. f(z) is holomorphic at EACH cusp p i. This secon item requires a bit more explanation. Just as q = e πiz mappe the point to the origin in the coorinate q for SL(, Z), for our subgroup Γ, there are coorinate functions which take each cusp p i to the origin, an for which we again have a Fourier expansion. This is usually achieve by mapping each p i to by a matrix γ i an then composing with q = e πiz/m where γ i Γγ 1 i ( ) = ( 1 M 0 1 (as we can no longer guarantee that the function is perioic with perio 1, since the matrix T may not be in Γ, but the fact that Γ is of finite inex guarantees that some translation by an integer M will be a member of γ i Γγ 1 i, an hence any moular form f for Γ will satisfy f(z + M) = f(z) an be expressible in a Fourier expansion.) ) 4 Half-integral weight moular forms Recall that we previously efine (for the lattice Z R) θ(t) = m= e πm t t > 0 which we can exten to a efinition for any z H by θ(z) = m= e iπm z z H, (which will still efine an absolutely convergent function, accoring to the arguments we presente earlier for the convergence of the theta function.) Then we have the following transformations of θ(z): θ(z + ) = θ(z), θ( 1/z) = iz θ(z) where the first equality is clear from the efinition, an the secon follows from Poisson summation along the imaginary axis, an analytic continuation. Let us choose the usual branch for the square root function, so that it is positive on positive reals. While these transformation laws implicitly suggest how θ transforms uner 7

8 ( ) ( ) any matrix in the group generate by an, they on t give an explicit transformation in terms of an arbitrary matrix. We want such an explicit transformation in orer to suggest the appropriate efinition of a half-integral weight moular form. ( ) b a To this en, consier SL(, Z) with b 0 (), c 0 () (this c somewhat strange choice will be justifie later in the computation). Further take > 0 (since < 0 follows similarly). Then ( ) ( ) bz a b θ = θ z c 1 (z c) = z c m (mo ) e iπm b/ t= = (i) 1/ (z c) 1/ e iπ( m +t) m (mo ) e iπm b/ u= mu πi e +iu π(z c ) where the last equality is obtaine by applying Poisson summation to the inner sum in the secon line. Interchanging the orers of summation, the inner sum over m (mo ) becomes ) m (mo ) e iπm b/+πi mu = m (mo ) ( αm + mu e with e(x) = e πix an b = β, β Z. Since bc = (β)c 1 (mo ) an o (as a bc = 1) we may write m = r β mo where β β 1 (). Then ( ) βm + mu ( ) β(r + ru) e = e m (mo ) r (mo ) ( β(r + u) = e β 4u ) r (mo ) ( ) β 4u ( ) βr = e e r (mo ) where we ve continue to write x for the inverse of x mo. Substituting back into 8

9 the sum over u above, we have ( ) bz a θ = (i) 1/ (z c) 1/ z c r (mo ) ( ) βr e θ(z). Substituting 1/z for z in our equation for θ, we have ( ) az + b θ = (i) 1/ (( 1/z) c) 1/ ( ) βr e θ( 1/z) cz + r (mo ) = i(cz + ) 1/ 1/ ( ) βr e θ(z) r (mo ) where we use the transformation θ( 1/z) = θ(z) iz in the last step. Finally, recall that the sum in brackets is closely relate to the Gauss sum 1 ( ) { m 1/ if 1 (mo 4) e = i 1/ if 3 (mo 4). m=0 This evaluation is worke out carefully in many number theory textbooks, incluing Davenport s Multiplicative Number Theory. Using this evaluation in the above, we have ( ) c θ(γ(z)) = ε 1 (cz + )1/ θ(z) for any γ SL(, Z) with b, c 0 (), ε = 1 or i accoring to 1 or 3 mo 4, respectively, an ( ) c is the Legenre symbol appropriately extene to inclue = 1 (there are slight annoyances in this case, like ( ) 0 = 1.) Setting θ(z) = θ(z) we have just shown: Theorem 3 For any γ Γ 0 (4), θ(γ(z)) = j(γ, z) θ(z) where as efine above. ( c j(γ, z) = ) ε 1 (cz + )1/ Any such function that transforms by j(γ, z) k for k a fixe half-integer, an all γ Γ 0 (N) an is holomorphic at each cusp, is calle a half-integral weight moular form. Note this efinition is consistent with our earlier efinitions for integral weight moular forms if k is a proper even integer, j(γ, z) reuces to the usual efinition. 9

10 References [1] D. W. Bump, Automorphic Forms an Representations, Cambrige Stuies in Avance Mathematics, vol. 55, Cambrige Univ. Press (1996). [] J.-P. Serre, A Course in Arithmetic, Grauate Texts in Mathematics, vol. 7, Springer-Verlag (1973). 10