Dimension Formulas
3.5 Dimension formulas for even k
We can now express the results from Section 3.3 in the language of divisors.
Letkbe any even integer. Then using formulas (3.2) and (3.3), every nonzero automorphic form f ∈ Ak(Γ) has a well defined divisor
νx(f)x onX(Γ) even thoughf itself is not well defined onX(Γ). However, formula (3.2) shows that the coefficients in this divisor need not be integers at the elliptic points.
To allow for divisors with rational coefficients, make the definition
DivQ(X(Γ)) =,
nxx, nx∈Qfor allx,nx= 0 for almost allx -
. This is thefreeQ-module(orfree vector space)on the set of points ofX(Γ).
Thus div(f)∈DivQ(X(Γ)). The discussion of Div(X) as an Abelian group with a relation “≥” and a degree function extends to DivQ(X), but the Riemann–Roch Theorem requires integral divisors.
Now let the even integerksatisfyk≥2, and letf be any nonzero element ofAk(Γ) (Exercise 3.2.3 showed that suchf exists). Then (Exercise 3.5.1)
Ak(Γ) =C(X(Γ))f ={f0f :f0∈C(X(Γ))}.
The holomorphic subspaceMk(Γ) of modular forms can be described in terms of divisors,
Mk(Γ) ={f0f ∈ Ak(Γ) :f0f = 0 or div(f0f)≥0}
∼={f0∈C(X(Γ)) :f0= 0 or div(f0) + div(f)≥0},
where the isomorphism is a complex vector space isomorphism. If div(f) were integral then we could now compute dim(Mk(Γ)) with the Riemann–Roch Theorem, but it isn’t. To approximate div(f) by an integral divisor, define the greatest integer function on rational divisors by applying it to the coefficients,
.nxx /
=
nxx for
nxx∈DivQ(X(Γ)).
On the other hand, div(f0) is integral sincef0is meromorphic onX(Γ), giving the equivalence
div(f0) + div(f)≥0 ⇐⇒ div(f0) +div(f) ≥0.
This shows thatMk(Γ)∼=L(div(f)) and so dim(Mk(Γ)) =(div(f)).
To study the integral divisordiv(f)letω∈Ω⊗k/2(X(Γ)) be the mero- morphic differentialω(f) from Theorem 3.3.1 that pulls back tof(τ)(dτ)k/2 onH. Let{x2,i}be the period 2 elliptic points of X(Γ),{x3,i} the period 3 elliptic points, and{xi}the cusps. As in Section 3.1, letε2,ε3, andε∞be the sizes of these sets. Make the formal definition
div(dτ) =−
i
1
2x2,i−
i
2
3x3,i−
i
xi,
an element of DivQ(X(Γ)) even thoughdτ is not well defined onX(Γ). Then formulas (3.8) and (3.9), relating the vanishing orders ofω andf, show that compatibly with the identification ofωandf(τ)(dτ)k/2and with the rule that the divisor of a product is the sum of the divisors (Exercise 3.5.2(a)),
div(ω) = div(f) + (k/2)div(dτ). (3.10)
k Here the left side is integral, the right side a sum of rational expressions. It follows that (Exercise 3.5.2(b))
div(f)= div(ω) +
i
0k 4 1
x2,i+
i
0k 3 1
x3,i+
i
k
2xi. (3.11) Therefore by the discussion immediately after Corollary 3.4.2 showing that deg(div(ω)) =k(g−1),
deg(div(f)) =k(g−1) + 0k
4 1
ε2+ 0k
3 1
ε3+k 2ε∞
≥k
2(2g−2) +k−2
4 ε2+k−2 3 ε3+k
2ε∞ (Exercise 3.5.2(c))
= 2g−2 +k−2 2
2g−2 + ε2
2 +2ε3
3 +ε∞
+ε∞
≥2g−2 +ε∞ by Corollary 3.1.2, sincek≥2
>2g−2.
Sodiv(f)has large enough degree for Corollary 3.4.2(d) to apply and the simpler form of the Riemann–Roch Theorem says that for evenk≥2,
dim(Mk(Γ)) =(div(f)) = (k−1)(g−1) + 0k
4 1
ε2+ 0k
3 1
ε3+k 2ε∞. For the cusp forms Sk(Γ), a similar argument shows that Sk(Γ) ∼= L(div(f)−
ixi) and that (Exercise 3.5.2(d)) dim(Sk(Γ)) =(div(f)−
i
xi) =(div(f))−ε∞ fork≥4. (3.12) Fork= 2 the divisordiv(f)−
ixiworks out to a canonical divisor div(λ), cf. Exercise 3.4.2, and so its linear space has dimensiong.
Finally we dispense with the cases where the weight k is nonpositive. A modular form of weight 0 with respect toΓis a meromorphic function onX(Γ) with no poles. Since any holomorphic map between compact Riemann surfaces either surjects or maps to a single point (this was discussed in Section 1.3), this shows that M0(Γ) = C (the constant functions) and S0(Γ) = {0}. It follows thatMk(Γ) ={0} fork <0, for iff ∈ Mk(Γ) then f12∆−k (where
∆ is the weight 12 cusp form from Chapter 1) lies in S0(Γ) = {0}. This argument works whetherk is even or odd.
Assembling the results of this section gives
Theorem 3.5.1.Let k be an even integer. Let Γ be a congruence subgroup ofSL2(Z),gthe genus ofX(Γ),ε2the number of elliptic points with period 2, ε3 the number of elliptic points with period 3, and ε∞ the number of cusps.
Then
dim(Mk(Γ)) =
⎧⎪
⎨
⎪⎩
(k−1)(g−1) +2k
4
3ε2+2k
3
3ε3+k2ε∞ if k≥2,
1 if k= 0,
0 if k <0,
and
dim(Sk(Γ)) =
⎧⎪
⎨
⎪⎩
(k−1)(g−1) +2k
4
3ε2+2k
3
3ε3+ (k2−1)ε∞ ifk≥4,
g ifk= 2,
0 ifk≤0.
In concert with the results from Exercises 3.1.4 through 3.1.6 this gives dimension formulas forΓ0(p),Γ1(p), andΓ(p). When Γ is the full modular group SL2(Z), the theorem is
Theorem 3.5.2.The modular forms of weight0 areM0(SL2(Z)) =C. For any nonzero even integer k < 4, Mk(SL2(Z)) = {0}. For any even integer k <4,Sk(SL2(Z)) ={0}. For any even integerk≥4,
Mk(SL2(Z)) =Sk(SL2(Z))⊕CEk
whereEk is the normalized weightk Eisenstein series from Chapter 1, and dim(Sk(SL2(Z))) =
12k −1 if k≡2 (mod 12), 12k otherwise.
The ring of modular forms M(SL2(Z)) =
k∈ZMk(SL2(Z)) and the ideal of cusp formsS(SL2(Z)) =
k∈ZSk(SL2(Z)) are a polynomial ring in two variables and a principal ideal,
M(SL2(Z)) =C[E4, E6], S(SL2(Z)) =∆· M(SL2(Z)) where∆ is the discriminant function from Chapter 1.
The proof is Exercise 3.5.3. Exercises 3.5.4 and 3.5.5 use the results of this section to describe some spaces of modular forms explicitly.
Exercises
3.5.1.Show thatAk(Γ) =C(X(Γ))f for any nonzerof ∈ Ak(Γ).
3.5.2.(a) Prove formula (3.10).
(b) Prove formula (3.11).
(c) Prove that fork≥2 even,k/4 ≥(k−2)/4 andk/3 ≥(k−2)/3.
(d) Prove formula (3.12). Where does the conditionk≥4 arise?
k 3.5.3.Prove Theorem 3.5.2. Write bases of Mk(SL2(Z)) for allk ≤24, ex- pressing the basis elements as polynomials in E4 and E6. (A hint for this exercise is at the end of the book.).
3.5.4.(a) Let p be prime. Exercise 3.1.4(b) showed that X0(p) has exactly two cusps. Use this to show that
M2(Γ0(p)) =S2(Γ0(p))⊕CG2,p
whereG2,pwas defined in Section 1.2.
(b) LetkandNbe positive withk(N+ 1) = 24 andkeven. Thus (k, N) is one of (2,11), (4,5), (6,3), (8,2), or (12,1). Show that dim(Sk(Γ0(N))) = 1. It follows from Proposition 3.2.2(b) thatSk(Γ0(N)) =Sk(Γ1(N)) =Cϕk where ϕk(τ) =η(τ)kη(N τ)k.
(c) What do (a) and (b) combine to say aboutM2(Γ0(11))?
3.5.5.Combine Exercise 3.1.4 and Theorem 3.5.2 to show that letting k = p+1 forpan odd prime,Sk(SL2(Z)) andS2(Γ0(p)) are isomorphic as complex vector spaces. What is the isomorphism whenp= 11? (A hint for this exercise is at the end of the book.)