Modular Curves as Riemann Surfaces
2.3 Elliptic points
q→α0
0β
(q1/h) h
= (α/β)hq and the map is clearly holomorphic.
So far the argument assumes that ϕ1(x) = 0. But it also covers the case ϕ2(x) = 0 since the inverse of a holomorphic bijection is again holomorphic.
And in general,ϕ2,1is a compositeϕ2,3◦ϕ3,1whereϕ3:π(U3)−→V3takesx to 0, so in fact the argument suffices for all cases.
Exercises
2.2.1.Let τ ∈ H be fixed only by the identity transformation in Γ. Use Proposition 2.1.1 to show that some neighborhood U of τ is homeomorphic to its image inY(Γ).
2.2.2.Compute 1−i
1 i 0−1 1 0
1−i 1 i
−1
and explain how this shows that the mapγ(τ) =−1/τ acts in the small as 180-degree rotation abouti.
2.2.3.Show that if τ ∈ H and γ ∈ SL2(Z) then the period of γ(τ) un- derγΓ γ−1 is the same as the period ofτ underΓ.
2.2.4.Prove Corollary 2.2.3.
2.2.5.Explain why ϕ:π(U)−→V as defined in the section is a homeomor- phism. (A hint for this exercise is at the end of the book.)
D
Figure 2.3.The fundamental domain for SL2(Z)
Proof. Given τ ∈ H it suffices to show that τ is SL2(Z)-equivalent to some point inD. Repeatedly apply one of1±1
0 1
:τ→τ±1 to translateτinto the vertical strip{|Re(τ)| ≤1/2}, and replaceτ by this transform. Now ifτ /∈ D then|τ|<1 and so Im(−1/τ) = Im(−τ /¯ |τ|2) = Im(τ /|τ|2)>Im(τ); replace τ by0−1
1 0
(τ) =−1/τ and repeat the process. Since there are only finitely many integer pairs (c, d) such that|cτ+d|<1 (because there are only finitely many lattice points inside a disk), the formula
Im(γτ) = Im(τ)
|cτ+d|2 forγ= a b
c d
∈SL2(Z)
shows that only finitely many transforms of τ have larger imaginary part.
Therefore the algorithm terminates with someτ∈ D. The surjection π : D −→ Y(1) isn’t injective. The translation [1 10 1] : τ → τ+ 1 identifies the two boundary half-lines, and the inversion0−1
1 0
: τ→ −1/τ identifies the two halves of the boundary arc. But these boundary identifications are the only ones:
Lemma 2.3.2.Supposeτ1 andτ2 are distinct points in Dand that τ2=γτ1
for someγ∈SL2(Z). Then either (1) Re(τ1) =±1/2 andτ2=τ1∓1, or (2)|τ1|= 1 andτ2=−1/τ1.
Proof. (Sketch—Exercise 2.3.1 asks for details.) Assume Im(τ2)≥Im(τ1) by symmetry. Let γ =a b
c d
. Then |cτ1+d|2 ≤ 1 since Im(τ2) ≥ Im(τ1), and Im(τ1)≥√
3/2 sinceτ1∈ D. So
|c|√
3/2≤ |c|Im(τ1) =|Im(cτ1+d)| ≤ |cτ1+d| ≤1, and sincec∈Zthis shows that|c| ∈ {0,1}.
Ifc= 0 thenγ=±[10 1b] and Re(τ2) = Re(τ1) +b, forcing|b|= 1 and (1) holds.
If |c| = 1 then the condition |cτ1+d|2 ≤ 1 becomes |τ1±d|2 ≤ 1, or (Re(τ1)±d)2+ (Im(τ1))2 ≤ 1, implying (Re(τ1)±d)2 ≤ 1−(Im(τ1))2 ≤ 1−3/4 = 1/4, so|Re(τ1)±d| ≤1/2, forcing|d| ≤1.
If |c| = 1 and |d| = 1 then in the preceding calculation all inequalities must be equalities. It follows that Im(τ1) =√
3/2 and|Re(τ1)±1|= 1/2, so Re(τ1) =±1/2 and both (1) and (2) hold.
If|c|= 1 andd= 0 then the condition|cτ1+d| ≤1 becomes|τ1| ≤1, so in fact|τ1|= 1 (sinceτ1∈ D) and Im(τ2) = Im(τ1). So also|τ2|= 1 by symmetry since nowτ1andτ2have the same conditions on their imaginary parts and on thec-entries of the matrices transforming each to the other. Thus τ1 and τ2
have the same absolute value and the same imaginary part but are distinct, forcing their real parts to be opposites and (2) holds.
So with suitable boundary identification the setDis a model forY(1) = SL2(Z)\H, also called a fundamental domain for SL2(Z). Topologically, D modulo the identification is a plane. Figure 2.4 shows some SL2(Z)-translates ofD. The whole configuration repeats horizontally with period 1 under iter- ations of1±1
0 1
:τ →τ±1. The figure shows how the SL2(Z)-translates of a pointτ ∈ H can cluster only down toward the real axis, perhaps giving a more intuitive understanding of Proposition 2.1.1.
Figure 2.4. Some SL2(Z)-translates ofD
Returning to elliptic points, supposeτ∈ H is fixed by a nontrivial trans- formationa b
c d
∈ SL2(Z). Then aτ +b = cτ2+dτ; solving forτ with the quadratic equation (c = 0 is impossible since τ /∈Q) and remembering that τ∈ Hshows that|a+d|<2 (Exercise 2.3.2). Thus the characteristic polyno- mial ofγisx2+ 1 orx2±x+ 1. Sinceγsatisfies its characteristic polynomial, one ofγ4 =I, γ3 =I, γ6 =I holds, and γ has order 1, 2, 3, 4, or 6 as a matrix. Orders 1 and 2 give the identity transformation (Exercise 2.3.3). So the following proposition describes all nontrivial fixing transformations.
Proposition 2.3.3.Let γ∈SL2(Z).
(a)If γ has order 3 thenγ is conjugate to 0 1
−1−1
±1
inSL2(Z).
(b)If γ has order 4 thenγ is conjugate to0−1
1 0
±1
inSL2(Z).
(c)If γ has order 6 thenγ is conjugate to0−1
1 1
±1
inSL2(Z).
Proof. (c) Since γ6 = I the lattice L = Z2 of integral column vectors is a module over the ringZ[µ6] where µ6 =e2πi/6, defining the scalar-by-vector product (a+bµ6)·v fora, b∈Zandv∈Las the matrix-by-vector product (aI+bγ)v.
The ring Z[µ6] is known to be a principal ideal domain and L is finitely generated over it. The structure theorem for modules over a principal ideal domain therefore says thatLisZ[µ6]-isomorphic to a sum+
kZ[µ6]/Ik where theIk are ideals. As an Abelian group,Lis free of rank 2. Every nonzero Ik
of Z[µ6] has rank 2 as an Abelian group, making the quotient Z[µ6]/Ik a torsion group, so no such terms appear in the sum. Only one free summand appears, for otherwise the sum would be too big as an Abelian group. Thus there is aZ[µ6]-module isomorphismφγ :Z[µ6]−→L.
Letu=φγ(1) andv=φγ(µ6) and let [u v] denote the matrix with columns uandv. ThenL=Zu+Zv so det[u v]∈ {±1}(Exercise 2.3.4(a)). Compute thatγu=µ6·φγ(1) =φγ(µ6) =v, and similarlyγv=µ6·φγ(µ6) =φγ(µ26) = φγ(−1 +µ6) =−u+v. Thus
γ[u v] = [v −u+v] = [u v]0−1
1 1
, soγ= [u v]0−1
1 1
[u v]−1, and
γ[v u] = [−u+v v] = [v u] 1 1
−1 0
, soγ= [v u]0−1
1 1
−1
[v u]−1. One of [u v], [v u] is in SL2(Z), proving (1).
Parts (a) and (b) are Exercise 2.3.4(b).
Now we can understand the elliptic points and their isotropy subgroups.
Corollary 2.3.4.The elliptic points for SL2(Z) are SL2(Z)i and SL2(Z)µ3
where µ3 = e2πi/3. The modular curve Y(1) = SL2(Z)\H has two elliptic points. The isotropy subgroups ofi andµ3 are
SL2(Z)i=0−1
1 0
and SL2(Z)µ3 =0−1
1 1
.
For each elliptic point τ of SL2(Z) the isotropy subgroup SL2(Z)τ is finite cyclic.
Proof. The fixed points inHof the matrices in Proposition 2.3.3 areiandµ3. The first statement follows (Exercise 2.3.5(a)). The second statement follows since i and µ3 are not equivalent under SL2(Z). The third statement can be verified directly (Exercise 2.3.5(a) again), and the fourth statement fol- lows since all other isotropy subgroups of order greater than 2 are conjugates of SL2(Z)i and SL2(Z)µ3. See Exercise 2.3.5(b) for a more conceptual proof
of the fourth statement.
Proposition 2.2.2 and a bit more now follow.
Corollary 2.3.5.Let Γ be a congruence subgroup of SL2(Z). The modular curveY(Γ)has finitely many elliptic points. For each elliptic pointτofΓ the isotropy subgroupΓτ is finite cyclic.
Proof. If SL2(Z) =d
j=1Γ γj then the elliptic points ofY(Γ) are a subset of EΓ ={Γ γj(i), Γ γj(µ3) : 1≤j ≤d}. The second statement is clear since Γτ
is a subgroup of SL2(Z)τ for allτ∈ H.
For a congruence subgroupΓ, if SL2(Z) =
j{±I}Γ γjthen the set
jγjD surjects toY(Γ) (Exercise 2.3.8), and with suitable boundary identification this is a bijection as before. The set need not be a fundamental domain, however, since a fundamental domain is also required to be connected.
Exercises
2.3.1.Fill in any details as necessary in the proof of Lemma 2.3.2.
2.3.2.Ifa b
c d
∈SL2(Z) fixesτ∈ H, show that|a+d|<2.
2.3.3.Show that ifγ∈SL2(Z) has order 2 thenγ=−I.
2.3.4.(a) In the proof of Proposition 2.3.3, why does the condition L = Zu+Zv imply det[u v] =±1?
(b) Prove the other two parts of Proposition 2.3.3. (Hints for this exercise are at the end of the book.)
2.3.5.(a) Complete the proof of Corollary 2.3.4. (A hint for this exercise is at the end of the book.)
(b) Give a more conceptual proof of the third statement in Corollary 2.3.4 as follows: The results from Exercise 2.1.3(c), that the isotropy subgroup ofi in SL2(R) is SO2(R) and thatH ∼= SL2(R)/SO2(R), show that an element s(τ) of SL2(R) movesτ to iand the isotropy subgroup ofτ correspondingly conjugates to a discrete subgroup of SO2(R). But since SO2(R) is the rota- tions of the circle, any such subgroup is cyclic.
2.3.6.In the proof of Corollary 2.3.5, need the 2d points in Y(Γ) listed in the description ofEΓ be distinct? Need the points ofEΓ all be elliptic points ofΓ?
2.3.7.Prove that there are no elliptic points for the following groups: (a) Γ(N) forN >1, (b)Γ1(N) forN >3 (also, find the elliptic points forΓ1(2) andΓ1(3) given that each group has one), (c) Γ0(N) for N divisible by any primep≡ −1 (mod 12). (A hint for this exercise is at the end of the book.) In the next chapter we will extend the technique of proving Proposition 2.3.3 to count the elliptic points ofΓ0(N) for allN.
Also, show that if the congruence subgroupΓ does not contain the negative identity matrix−I thenΓ has no elliptic points of period 2.
2.3.8.Suppose SL2(Z) =
j{±I}Γ γj where Γ is a congruence subgroup.
Show that
jγjDsurjects toY(Γ). (A hint for this exercise is at the end of the book.)