Dimension Formulas
3.1 The genus
LetΓ be a congruence subgroup of SL2(Z). Topologically the compact Rie- mann surfaceX(Γ) =Γ\H∗ is a sphere withg handles for someg∈N. This g, thegenusofX(Γ), will figure in the pending dimension formulas for vector spaces of modular forms.
Let f : X −→ Y be a nonconstant holomorphic map between compact Riemann surfaces. As explained in Section 1.3,f surjects. For anyy∈Y the inverse imagef−1(y) is discrete inX sincef is nonconstant andX is closed, and thereforef−1(y) is finite sinceX is compact. In fact the mapf has a well defineddegreed∈Z+ such that|f−1(y)|=dfor all but finitely manyy ∈Y. More precisely, for each pointx∈X letex ∈Z+ be theramification degree of f at x, meaning the multiplicity with which f takes 0 to 0 as a map in local coordinates, makingf anex-to-1 map aboutx, cf. the Local Mapping Theorem of complex analysis. Then there exists a positive integerdsuch that
x∈f−1(y)
ex=d for ally∈Y .
To see this, letE ={x∈X :ex>1}denote theexceptional points inX, the finite set of points wheref is ramified. Let X =X\E and Y =Y\h(E) be the Riemann surfaces obtained by removing the exceptional points fromX
and their images fromY. SoX is all but finitely many points ofX, making it connected, and similarly forY. Let y be a point of Y. Each x∈f−1(y) has a neighborhood Ux where f is locally bijective. Shrinking these neigh- borhoods enough, they can be assumed disjoint and assumed to map to the same neighborhood ofy. Thusy has a neighborhood V in Y whose inverse image is a finite disjoint union of neighborhoodsUxinXwith each restriction fx:Ux−→V invertible. This makes the integer-valued functiony→ |f−1(y)| onY continuous and therefore constant since Y is connected. Let d be its value; so
x∈f−1(y)ex=d for all y ∈Y. The sum extends continuously to all ofY by definition of ramification degree: ify=f(x) then everyyin some neighborhood of y is the image of ex points x near x, each with ex = 1.
Thus
x∈f−1(y)ex=|f−1(y)|=das claimed, provided thaty is in a small enough neighborhood ofy to avoid ramification except possibly overy itself.
Again letf :X −→Y be a nonconstant holomorphic map between com- pact Riemann surfaces, of degreed. LetgX andgY be the genera ofX andY. To compute the genus of the modular curveX(Γ) we will use theRiemann–
Hurwitz formula,
2gX−2 =d(2gY −2) +
x∈X
(ex−1).
This shows, for instance, that there is no nonconstant map to a surface of higher genus, that a nonconstant map between surfaces of equal genus is unramified, and that there is no meromorphic function of degree d = 1 on any compact Riemann surface of positive genus. We sketch a proof of the Riemann–Hurwitz formula for the reader with some background in topology:
TriangulateY using VY vertices including all points of f(E),EY edges, and FY faces. This lifts underf−1 to a triangulation ofX withEX =dEY edges and FX = dFY faces but only VX =dVY −
x∈X(ex−1) vertices because of ramification. The Riemann–Hurwitz formula follows because 2−2gX = FX−EX+VX and similarly forY.
As a special case of all this, if Γ1 and Γ2 are congruence subgroups of SL2(Z) withΓ1⊂Γ2, then the natural projection of corresponding modular curves,
f :X(Γ1)−→X(Γ2), Γ1τ→Γ2τ,
is a nonconstant holomorphic map between compact Riemann surfaces, cf. Ex- ercise 2.4.7. Its degree is (Exercise 3.1.1)
deg(f) = [{±I}Γ2:{±I}Γ1] =
[Γ2:Γ1]/2 if−I∈Γ2 and−I /∈Γ1, [Γ2:Γ1] otherwise.
Recall from Figure 2.7 the local structure on the Riemann surfacesX(Γj) for j= 1,2. The commutative diagram
U id //
δ
zzvvvvvvvvvv
π1
U
π2
δ
$$H
HH HH HH HH H
δ(U)
ρ1
##H
HH HH HH HH
H π1(U) f //
ϕ1
π2(U)
ϕ2
δ(U)
ρ2
{{vvvvvvvvvv
V1
flocal // V2
shows that flocal◦ψ1 = ψ2, where ψj is the composition ρj◦δ. If U is a neighborhood ofτ∈ Hthenρ1(z) =zh1 andρ2(z) =zh2 so the local map is q→qh2/h1. Let Γj,τ denote the fixing subgroup ofτ in Γj forj = 1,2. Then the periods hj =|{±I}Γj,τ|/2 lie in {1,2,3} and since their ratio h2/h1 is integral this impliesh1= 1 orh1=h2. So the ramification degree is
eπ1(τ)=h2/h1=
h2 ifτ is an elliptic point forΓ2 but not forΓ1, 1 otherwise
= [{±I}Γ2,τ :{±I}Γ1,τ].
Since the periodshj are defined on the curvesXj, the ramification degree is indeed defined onX1. IfU is a neighborhood of s∈Q∪ {∞}thenρ1(z) = e2πiz/h1 andρ2(z) =e2πiz/h2 so the local map isq→qh1/h2. Here the widths arehj = [SL2(Z)∞:{±I}Γj,s] forj= 1,2 and the ramification degree is the reciprocal of their ratio,
eπ1(s)=h1/h2= [{±I}Γ2,s:{±I}Γ1,s].
Again this is well defined onX1. IfΓ1is normal inΓ2then all points ofX(Γ1) lying over a given point of X(Γ2) have the same ramification degree (Exer- cise 3.1.2).
To compute the genus of X(Γ), further specialize to Γ1 = Γ and Γ2 = SL2(Z). Lety2= SL2(Z)i,y3= SL2(Z)µ3, andy∞= SL2(Z)∞be the elliptic point of period 2, the elliptic point of period 3, and the cusp of X(1) = SL2(Z)\H∗. Let ε2 and ε3 be the number of elliptic points of Γ in f−1(y2) andf−1(y3), i.e., the number of elliptic points of period 2 and 3 inX(Γ), and letε∞ be the number of cusps of X(Γ). Then recalling that d= deg(f) and lettingh= 2 or h= 3, the formula fordat the beginning of the section and then the formula for eπ1(τ) at the nonelliptic points and the elliptic points over SL2(Z)yh show that (Exercise 3.1.3(a))
d=
x∈f−1(yh)
ex=h·(|f−1(yh)| −εh) + 1·εh, and using these equalities twice gives
x∈f−1(yh)
(ex−1) = (h−1)(|f−1(yh)| −εh) =h−1
h (d−εh).
Also,
x∈f−1(y∞)
(ex−1) =d−ε∞.
SinceX(1) has genus 0, the Riemann–Hurwitz formula now shows
Theorem 3.1.1.LetΓ be a congruence subgroup ofSL2(Z). Letf :X(Γ)−→
X(1) be natural projection, and letd denote its degree. Letε2 andε3 denote the number of elliptic points of period 2 and 3 inX(Γ), andε∞ the number of cusps ofX(Γ). Then the genus ofX(Γ)is
g= 1 + d 12−ε2
4 −ε3
3 −ε∞ 2 .
Proof. Exercise 3.1.3(b).
The next result will be used in Sections 3.5 and 3.6.
Corollary 3.1.2.LetΓ,g,ε2,ε3, andε∞ be as above. Then 2g−2 +ε2
2 +2ε3
3 +ε∞>0.
Proof. Exercise 3.1.3(c).
Exercises 3.1.4, 3.1.5, and 3.1.6 compute the genus ofX(Γ) forΓ =Γ0(p), Γ =Γ1(p), and Γ =Γ(p) where pis prime. The results for general N will take considerably more work.
A fundamental domain forΓ0(13) is shown in Figure 3.1, including repre- sentatives of the two elliptic points of order 2 (the light dots) and the two of order 3 (the dark ones—Exercise 3.1.4(f) explains how the figure was gener- ated and justifies the description to follow in this paragraph and the next).
Each boundary arc with an order-2 elliptic point folds together at the elliptic point, identifying its two halves with each other, and each pair of boundary arcs joined at an order-3 elliptic point folds together at the elliptic point, iden- tifying with each other. The remaining two curved boundary arcs in the left half that don’t include the cusp 0 are identified with each other, and similarly in the right half. The vertical boundary segments in the left half identify with those in the right half, and so the remaining two arcs are identified with each other as well. Gluing boundary arcs together under these identifications gives a sphere per the formula in Exercise 3.1.4(e).
The 13 points of SL2(Z)(i) in the figure identify to eight points ofX0(13), and the 14 points of SL2(Z)(µ3) in the figure identify to six points ofX0(13).
The boundary arc for each order-2 elliptic pointxfolds together like the local coordinate aboutiin X(1) so that the ramification degree of the projection f :X0(13)−→X(1) at these points isex = 1, whileex= 2 at the other six pointsxthat project to SL2(Z)(i) and the ramification degrees sum to 14 = [SL2(Z) : Γ0(13)] as they should. Each pair of boundary arcs joined at an
Figure 3.1.Fundamental domain forΓ0(13)
order-3 elliptic point x folds together so that ex = 1, while ex = 3 at the other four pointsxthat project to SL2(Z)(µ3) and again the sum is 14. The ramification degrees at the cuspsΓ0(p)(∞) andΓ0(13)(0) are visibly 1 and 13.
To generate pictures of fundamental domains, see Helena Verrill’s web page [Ver].
Exercises
3.1.1.Show that natural projection f : X(Γ1) −→ X(Γ2) has the degree claimed in the text. (A hint for this exercise is at the end of the book.) 3.1.2.Show that if Γ1 is normal in Γ2 then all points of X(Γ1) lying over a given point ofX(Γ2) have the same ramification degree. (A hint for this exercise is at the end of the book.)
3.1.3.(a) Fill in details as necessary in the computations preceding Theo- rem 3.1.1.
(b) Prove Theorem 3.1.1.
(c) Prove Corollary 3.1.2.
3.1.4.Letpbe prime, letX0(p) =Γ0(p)\H∗, letαj =1 0
j1
forj= 0, . . . , p− 1, and letα∞=1−1
1 0
. (a) Show that SL2(Z) =
jΓ0(p)αj, a disjoint union.
(b) Show thatX0(p) has exactly two cusps. (Hints for this exercise are at the end of the book.)
(c) Show thatγαj(i) =αj(i) for someγ∈Γ0(p) of order 4 if and only if j2+ 1≡0 (modp). Thus the number of elliptic points of period 2 inX0(p) is the number of solutions of the congruencex2+ 1≡0 (modp). This number is 2 ifp≡1 (mod 4), 0 if p≡3 (mod 4), and 1 ifp= 2.
(d) Show thatγαj(µ3) =αj(µ3) for someγ∈Γ0(p) of order 6 if and only ifj2−j+ 1≡0 (modp). Thus the number of elliptic points of period 3 in X0(p) is the number of solutions ofx2−x+ 1≡0 (modp). This number is 2 ifp≡1 (mod 3), 0 ifp≡2 (mod 3), and 1 if p= 3. Along with part (c), this shows that the number of elliptic points is determined by p(mod 12).
The examplep= 13 in the text is the smallest where all four possible elliptic points exist.
(e) Letg be the genus ofX0(p) and let k=p+ 1. Show that g=
12k −1 ifk≡2 (mod 12), 12k otherwise.
(The symbol k is introduced here because the same formula with k as the weight will recur later in the chapter.)
(f) Figure 3.1 was generated using the coset representatives βj=1 0
j1 0−1 1 0
, j=−6, . . . ,6,
andβ∞ =1−1
0 1
α∞0−1
1 0
. Find each translate βj(D) in the figure, where D is the fundamental domain from Chapter 2. Show that the 13 points of SL2(Z)(i) in the figure are βj(i) for j = −6, . . . ,6. Since 0−1
1 0
fixes i, the elliptic points of order 2 areβj(i) whenj2+ 1≡0 (mod 13). Show that γβj(i) = βj(i) for some γ ∈ Γ0(13) of order 4 if and only if jj + 1 ≡ 0 (mod 13). Use this to partition the 13 points of SL2(Z)(i) in the figure into eight equivalence classes underΓ0(13), five with two points each where the an- gle isπ, giving a total of 2π; one with one point where the angle is 2π; and two with one point where the angle isπas it is atiinD, representing the unrami- fied points. Identify the boundary arcs pairwise except for two arcs that fold in on themselves. Note that SL2(Z)(µ3) = SL2(Z)(µ6). Since0−1
1 0
takesµ6to µ3, the elliptic points of order 3 areβj(µ6) whenj2−j+1≡0 (mod 13). Show that the 14 points of SL2(Z)(µ3) in the figure areβj(µ6) forj =−6, . . . ,7.
Show thatγβj(µ6) =βj(µ6) for someγ∈Γ0(13) of order 3 or 6 if and only ifjj−j+ 1≡0 (mod 13) orjj−j+ 1≡0 (mod 13). Use this to partition the 14 points of SL2(Z)(µ3) in the figure into six equivalence classes under Γ0(13), one with the four points where two of the angles are 2π/3 and two of the angles areπ/3, giving a total of 2π; two with with three points where the
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.)