Dimension Formulas
3.2 Automorphic forms
angle is 2π/3, again giving 2π; one with the two points where the angle is π, giving 2π; and two classes with one point each where the angle is 2π/3 as it is atµ3 inD, representing the unramified points. Determine whether each pair of boundary arcs is identified with orientation preserved or reversed. Show that under identification the figure is topologically a sphere.
(g) Carry out the same process forX0(11) andX0(17), convincing yourself that they are tori in accordance with part (e).
3.1.5.Letpbe prime and letX1(p) =Γ1(p)\H∗.
(a) Show that ifγ∈SL2(Z)swheres∈Q∪ {∞}(and as in Chapter 2 the group is its isotropy subgroup) thenγ has trace±2. Show that ifpis a prime then Γ0(p)s = {±I}Γ1(p)s. Deduce that natural projection f : X1(p) −→
X0(p) is unramified at the cusps. Combine this with part (b) of the previous problem to show thatX1(p) hasp−1 cusps ifp >2 and 2 cusps ifp= 2.
(b) Exercise 2.3.7 showed that X1(p) has no elliptic points ifp >3, that X1(3) has no elliptic points of period 2 and one elliptic point of period 3, and thatX1(2) has one elliptic point of period 2 and no elliptic points of period 3.
Combine this information with part (a) to compute the genus ofX1(p). (A hint for this exercise is at the end of the book.)
3.1.6.Letpbe prime and letX(p) =Γ(p)\H∗. Compute the genus ofX(p).
(A hint for this exercise is at the end of the book.)
(f[γ]k)(τ) =j(γ, τ)−kf(γ(τ)), j(γ, τ) =cτ+dforγ=a b
c d
.
As in Chapter 1, a meromorphic functionf :H −→C is calledweakly modular of weightkwith respect toΓ iff[γ]k=f for allγ∈Γ. To discuss meromorphy off at∞, lethbe the smallest positive integer such that [10 1h]∈Γ. Thus f has periodh. If alsof has no poles in some region{τ ∈ H: Im(τ)> c} then f has a Laurent series on the corresponding punctured disk about 0,
f(τ) = ∞ n=−∞
anqnh if Im(τ)> c, where qh=e2πiτ /h.
Thenf ismeromorphic at∞if this series truncates from the left, starting at somem∈Zwheream= 0, or iff = 0. Theorder off at∞, denotedν∞(f), is again defined as the starting indexm except when f = 0, in which case ν∞(f) = ∞. Now automorphic forms are defined the same way as modular forms except with meromorphy in place of holomorphy.
Definition 3.2.1.Let Γ be a congruence subgroup ofSL2(Z)and letk be an integer. A function f : H −→ C is an automorphic form of weight k with respect to Γ if
(1)f is meromorphic,
(2)f is weight-kinvariant under Γ,
(3)f[α]k is meromorphic at∞ for allα∈SL2(Z).
The set of automorphic forms of weightkwith respect toΓ is denotedAk(Γ).
Condition (3) makes sense since f[α]k is meromorphic on Hand weakly modular with respect toα−1Γ α, a congruence subgroup of SL2(Z). For any cusps∈Q∪ {∞}, the order off at sis defined as
νs(f) =ν∞(f[α]k) whereα(∞) =s,α∈SL2(Z).
This is independent of the choice ofαand well defined on the quotientX(Γ) (Exercise 3.2.1). Condition (3) implies beyond Condition (1) that the poles off inHcannot cluster at anys∈Q∪ {∞}.
Settingk= 0 in the definition shows thatA0(Γ) is the field of meromorphic functions onX(Γ), denotedC(X(Γ)). At elliptic points and cusps, where the local coordinate is many-to-one, functions f ∈ A0(Γ) are correspondingly periodic and thus meromorphic in the local variable of the Riemann surface structure. This statement will be elaborated later in the section.
For example, recall the modular invariant, defined as j= 1728g23
∆ :H −→C
where g2 and ∆ are the weight 4 Eisenstein series and the weight 12 cusp form from Chapter 1. Since the numerator and denominator are both weight- 12 invariants under SL2(Z) and holomorphic at ∞, the quotient is SL2(Z)- invariant and meromorphic at∞. Thusj is an automorphic form of weight 0 with respect to the full modular group SL2(Z) and is naturally viewed as a meromorphic function on the quotient,j : X(1)−→ C. Since ∆ is a cusp form,jhas a pole at∞. By Corollary 1.4.2, stating that∆is nonzero at every point ofH, this is its only pole. The leading 1728 normalizes its Laurent series to (Exercise 3.2.2)
j(τ) =1 q+
∞ n=0
anqn, an∈Z, q=e2πiτ. (3.1) Since its one pole is simple, j has degree 1 as a map j : X(1) −→ C of compact Riemann surfaces, cf. the beginning of Section 3.1. It follows thatjis a homeomorphism and conformal since it is locally analytic with nonvanishing derivative, and it takesY(1) toCand∞to ∞. In fact
A0(SL2(Z)) =C(j),
that is, the field of meromorphic functions on X(1) is the set of rational functions ofj. The containment C(j) ⊂ A0(SL2(Z)) is clear. For the other containment, letf be meromorphic and nonconstant onX(1) with finite zeros z1, . . . , zm and finite poles p1, . . . , pn, listed with multiplicity. Consider a functiong∈C(j),
g(τ) = m
i=1(j(τ)−j(zi)) n
i=1(j(τ)−j(pi)).
Then g has the same zeros and poles as f away from SL2(Z)∞, so g must vanish to the same order asf at SL2(Z)∞as well since for both functions the total number of zeros minus poles is 0. Thusf /g:X(1)−→C has no zeros or poles, making it constant and showing thatf ∈C(j).
Similarly, Chapter 7 will show that
A0(Γ0(N)) =C(j, jN), wherejN(τ) =j(N τ).
Chapter 7 will also describeA0(Γ0(N)),A0(Γ1(N)), andA0(Γ(N)) using the functionsf0,f0d, andf0v from the end of Section 1.5.
Returning to the general discussion, the sumA(Γ) =
k∈ZAk(Γ) forms a ring. The derivative j lies in A2(Γ), and it follows that Ak(Γ) contains nonzero elements for all even positive integers k (Exercise 3.2.3). For any k ∈ Z, if f is a nonzero element of Ak(Γ) then Ak(Γ) = C(X(Γ))f = {f0f :f0∈C(X(Γ))} (Exercise 3.2.4). These facts will be used to compute dimension formulas later in the chapter.
The transformation lawf(γ(τ)) =j(γ, τ)kf(τ) shows that an automorphic formf with respect to a congruence subgroupΓ is generally not well defined
on the quotient X(Γ). (The exceptional cases are when k = 0 or f = 0.) However, the order of vanishing of any automorphic formf does turn out to make sense on the quotient. Defining this notion appropriately requires taking the Riemann surface structure ofX(Γ) into account.
Letπ(τ)∈X(Γ) be a noncusp. Since the factor of automorphyj(γ, τ) has no zeros or poles onH, the order off is the same at all points ofΓ(τ). The local coordinate atπ(τ) is essentially q = (t−τ)h where τ has period has an elliptic point ofΓ. Excluding the zero function, sincef(t) =am(t−τ)m+
· · ·= amqm/h+· · · with am = 0 the natural definition of the order of f at π(τ)∈X(Γ) in terms of the ordermoff atτ ∈ Hism/h, that is
νπ(τ)(f) =ντ(f)
h whereτ has periodh. (3.2)
Thus νπ(τ)(f) = ντ(f) except at the elliptic points, where νπ(τ)(f) can be half-integral or third-integral. As remarked earlier, ifk= 0 then hdivides m andνπ(τ)(f) is integral.
Defining the order of f at cusps π(s) ∈ X(Γ) is trickier. First consider the cusp π(∞). Its local coordinate is qh = e2πiτ /h where h is the width, characterized by the conditions
{±I}Γ∞={±I} [10 1h], h∈Z+.
But since the negative identity matrix−Imight not lie inΓ this implies only Γ∞={±I} [10 1h] or Γ∞=[10 1h] or Γ∞=−[10 1h].
Thus the width hhere is not necessarily the same as the period hfrom the beginning of the section. In all three cases the orbits Γ τ and Γ(τ+h) for τ∈ H are equal. In the first two cases,f has period hcompatibly with this.
But in the third case, which can arise only when−I /∈Γ (see Exercise 3.2.5 for an example), the factor of automorphyj(−[10 1h], τ) =−1 shows that if k is even then again f has period h, while if k is odd thenf(τ+h) =−f(τ) so that f has skew-period h but period 2h. In all cases set h = 2h, so h is twice the width and usually twice the period, and setqh =e2πi/h. Then f(τ) = g(qh) for Im(τ) > c where qh = e2πiτ /h and c is large enough to avoid the finite poles off, andg has Laurent expansion
g(qh) = ∞ n=m
anqhn, am= 0.
Sinceqh =q1/2h is the square root of the local variable, the order off atπ(∞) is naturally defined asm/2, half the lowest power ofqh, except whenf = 0 in which case νπ(∞)(f) = ∞. The relation f(τ+h) = g(−qh) shows that whenf in fact has periodh, the functiongis even andνπ(∞)(f) is the integer ν∞(f) from before. In the exceptional case where Γ∞ = −[10 1h] and k is
odd, the skew periodicityf(τ+h) =−f(τ) shows thatgis odd andνπ(∞)(f) isν∞(f)/2, possibly a half-integer.
For a general cusp π(s) ∈ X(Γ) where s ∈ Q∪ {∞}, take any matrix α∈SL2(Z) such thatα(∞) =s. Using αto conjugate the discussion above from ∞to sgives the order of f at π(s)∈X(Γ) in terms of the order of f ats∈Q∪ {∞},
νπ(s)(f) =
νs(f)/2 if (α−1Γ α)∞=−[10 1h]andkis odd,
νs(f) otherwise. (3.3)
This can be half-integral in the exceptional case, whenπ(s) orsitself is called anirregular cusp ofΓ. For example, whenk is odd 1/2 is an irregular cusp ofΓ1(4) by Exercise 3.2.5. Section 3.8 will show that this is the only example of an irregular cusp for the groupsΓ0(N), Γ1(N), andΓ(N). Again ifk= 0 thenνπ(s)(f) is integral.
The results so far in this chapter lead to more examples of modular forms.
Recall the Dedekind function η(τ) =q24
∞
n=1(1−qn) and the discriminant function ∆(τ) = (2π)12η(τ)24 = (2π)12q∞
n=1(1−qn)24 from Chapter 1, where as alwaysτ∈ Handq=e2πiτ, and q24=e2πiτ /24.
Proposition 3.2.2.LetkandN be positive integers such thatk(N+1) = 24.
Thus (k, N) is one of (1,23), (2,11), (3,7), (4,5), (6,3), (8,2), or (12,1).
Define a function
ϕk(τ) =η(τ)kη(N τ)k. (a)If Sk(Γ1(N))is nonzero then Sk(Γ1(N)) =Cϕk.
(b)If Sk(Γ0(N))is nonzero thenSk(Γ0(N)) =Sk(Γ1(N)) =Cϕk. Proof. Consider the functiong=ϕNk+1, i.e.,
g(τ) = (2π)−24∆(τ)∆(N τ) =qN+1 ∞ n=1
(1−qn)24(1−qN n)24.
Since ∆(τ) ∈ S12(SL2(Z)) it follows that ∆(N τ) ∈ S12(Γ0(N)) and so g ∈ S24(Γ0(N))⊂ S24(Γ1(N)). Let S =0−1
1 0
andqN =e2πiτ /N, and compute (Exercise 3.2.6)
(g[S]24)(τ) =N−12g(τ /N) =N−12qNN+1 ∞ n=1
(1−qnN)24(1−qN nN )24. Letπ0 :H∗ −→X0(N) andπ1 :H∗ −→X1(N) be the natural projections, e.g.,π0(τ) =Γ0(N)(τ) forτ ∈ H∗. All cusps ofΓ0(N) andΓ1(N) are regular sinceN = 4. If N = 1 then the only cusp of X0(N) is π0(∞), of width 1.
OtherwiseN is prime and Exercise 3.1.4 shows that the cusps ofX0(N) are π0(∞), of width 1, and π0(0), of width N. Exercise 3.1.5 shows that the
natural projectionπ:X1(N)−→X0(N) is unramified at the cusps when N is prime, and the same assertion holds trivially whenN = 1. Thus ifπ1(s) is a cusp ofX1(N) overπ0(∞) then it has width 1,s=α(∞) for someα∈Γ0(N), and the identityg[α]24=g combines with the displayed product form ofgto show that νπ1(s)(g) = N+ 1 by (3.3) since the cusp is regular, i.e., g has a zero of order N+ 1 at π1(s). Similarly, ifπ1(s) is a cusp overπ0(0) then it has widthN+ 1, s =α(0) =αS(∞) for someα∈Γ0(N), and the identity g[αS]24=g[S]24combines with the displayed product form ofg[S]24 to show that alsoνπ1(s)(g) =N+ 1 and againg has a zero of orderN+ 1 atπ1(s).
Supposef ∈ Sk(Γ1(N)). ThenfN+1∈ S24(Γ1(N)). The quotientfN+1/g lies in A0(Γ1(N)) = C(X1(N)), it is holomorphic on Y1(N) because ∆ is nonzero on H, and it is holomorphic at the cusps of X1(N) since fN+1 has zeros of order at least N+ 1 at the cusps. The only holomorphic functions on the compact Riemann surfaceX1(N) are the constants, sofN+1=cgfor somec∈C. Taking (N+ 1)st roots givesf(τ) =cµe(τ)N ϕk(τ) for eachτ ∈ H where µN = e2πi/N and e(τ) is a continuous integer-valued function of τ, making it constant, andcdenotes different constants in different places. That is,f =cϕk, proving part (a). Part (b) follows immediately sinceSk(Γ0(N))⊂
Sk(Γ1(N)).
The proposition does not show that ϕk is a modular form at all until we know by some other means that a nonzero modular form of the desired type exists; the only possible f in the proof could be f = 0, in which case the relation f = cϕk does not imply ϕk = cf. When (k, N) = (12,1) the proposition shows thatS12(SL2(Z)) =C∆since∆is a nonzero cusp form of weight 12. The dimension-counting formulas to be obtained in this chapter will show thatSk(Γ0(N)) is 1-dimensional when (k, N) is any of (2,11), (4,5), (6,3), (8,2), so part (b) of the proposition applies in these cases. Although Sk(Γ0(N)) ={0}whenkis odd since−I∈Γ0(N), the dimension formulas will also show thatS3(Γ1(7)) is 1-dimensional, and part (a) applies. The formulas will not help with the remaining caseS1(Γ1(23)), but here one can still verify directly thatϕ1lies in the space (though we omit this) and so the proposition says that it spans the space. Thusϕk spansSk(Γ1(N)) wheneverk(N+ 1) = 24, and it also spansSk(Γ0(N)) whenkis even.
Exercises
3.2.1.Let f ∈ Ak(Γ) be an automorphic form and let s ∈ Q∪ {∞} be a cusp. Show that the definition ofνs(f) given in the text is independent of the choice ofαand well defined on the quotientX(Γ).
3.2.2.Show that the modular invariantjhas Laurent series (3.1). (A hint for this exercise is at the end of the book.)
3.2.3.LetΓ be a congruence subgroup of SL2(Z). Show that the derivative of the modular invariantjlies inA2(Γ). Show that consequentlyAk(Γ) contains
nonzero elements for all even positive integers k. (A hint for this exercise is at the end of the book.)
3.2.4.Show that for any k ∈ Z, if f is a nonzero element of Ak(Γ) then Ak(Γ) =C(X(Γ))f.
3.2.5.Letα=1−1
2−1
and letΓ =α−1Γ1(4)α. Show thatΓ∞=−[1 10 1]. (A hint for this exercise is at the end of the book.)
3.2.6.Verify the expression forg[S]24 in the proof of Proposition 3.2.2.