The Modular Group and the Fundamental Domain Seminar on Modular Forms Spring 2019

The Modular Group and the Fundamental Domain Seminar on Modular Forms Spring 2019

The modular group and the fundamental domain Seminar on Modular Forms Spring 2019 Johannes Hruza and Manuel Trachsler March 13, 2019 1 The Group SL2(Z) and the fundamental do- main Definition 1. For a commutative Ring R we define GL2(R) as the following set: a b GL (R) := A = for which det(A) = ad − bc 2 R∗ : (1) 2 c d We define SL2(R) to be the set of all B 2 GL2(R) for which det(B) = 1. Lemma 1. SL2(R) is a subgroup of GL2(R). Proof. Recall that the kernel of a group homomorphism is a subgroup. Observe ∗ that det is a group homomorphism det : GL2(R) ! R and thus SL2(R) is its kernel by definition. ¯ Let R = R. Then we can define an action of SL2(R) on C ( = C [ f1g ) by az + b a a b A:z := and A:1 := ;A = 2 SL (R); z 2 : (2) cz + d c c d 2 C Definition 2. The upper half-plane of C is given by H := fz 2 C j Im(z) > 0g. Restricting this action to H gives us another well defined action ":" : SL2(R)× H 7! H called the fractional linear transformation. Indeed, for any z 2 H the imaginary part of A:z is positive: az + b (az + b)(cz¯ + d) Im(z) Im(A:z) = Im = Im = > 0: (3) cz + d jcz + dj2 jcz + dj2 a b Lemma 2. For A = c d 2 SL2(R) the map µA : H ! H defined by z 7! A:z is the identity if and only if A = ±I. az+b 2 Proof. Assume z = cz+d for all z 2 H which reduces to cz + (d − a)z + b = 0. a b Having assumed A = c d is in SL2(R), it follows that the only solutions satisfying ad − bc = 1 are b = c = 0 and a = d so a = d = ±1 and hence ±I are the only elements of SL2(R) that act trivially on the upper half plane. 1 Definition 3. The full modular group is given by: Γ := SL2(Z) ≤ SL2(R): (4) In order to make the group act faithfully on H (recall that this means that there are no group elements A 2 ΓnfIdg such that A:z = z; 8z 2 H). We define the modular group Γ¯ := Γ={±Ig. The fact that Γ is indeed a subgroup of SL2(R) can be verified through check- a b −1 1 d −b ing that the inverse of an element in Γ lies in Γ too: c d = ad−bc −c a 1 and ad−bc = 1 2 Z by assumption. Also since Γ is a group it follows that Γ¯ = Γ={± Ig is a group, as {±Ig is normal in Γ. Definition 4. Let N > 0. Then we define the principle subgroup of level N by ( ) a b Γ(N) := 2 SL ( )ja ≡ d ≡ 1(mod N); b ≡ c ≡ 0(mod N) : (5) c d 2 Z First we show some basic properties of these subgroups: Lemma 3. Let N > 0. Then Γ(N) is a normal subgroup of Γ. Proof. We claim that Γ(N) is the kernel of a group homomorhism and hence normal. As a candidate we choose as follows: : SL2(Z) ! SL2(Z=NZ) (6) a b a (mod N) b (mod N) 7! : (7) c d c (mod N) d (mod N) This is indeed a group homomorphism, as it is one in every component. Its kernel is then exactly Γ(N). We are done. Remark: For all N > 2, the kernel Γ(¯ N) of the homomorphism : SL2(Z) ! SL2(Z=NZ) defined as in Lemma 3 is in fact equal to Γ(N). In- deed, for N > 2 we have −1 6≡ 1 mod(N) and thus −I 62 Γ(N). Definition 5. Let C subgroup of Γ. We call C a congruence subgroup of level N if C contains Γ(N). Similarly we may call a subgroup C0 of Γ¯ a congruence subgroup of level N if it contains Γ(¯ N). The most important examples of congruence subgroups, Γ0(N) and Γ1(N) will be defined here but are of little relevance for this lecture: ( ) a b Γ (N) := 2 Γjc ≡ 0 (mod N) (8) 0 c d ( ) a b Γ (N) := 2 Γ (N)ja ≡ 1 (mod N) : (9) 1 c d 0 The fact that these are indeed subgroups with Γ(N) ⊂ Γ1(N) ⊂ Γ0(N) ⊂ Γ will not be proven here. We now continue with the most important definition of the lecture, the fundamental domain. 2 Definition 6. Let G be a group acting on a topological space X. A fundamental domain F for the action of G is a (closed) subset, such that exactly one point of each orbit is contained in (the interior of) F . Applying this to our situation we call a closed subset F ⊂ H a fundamental domain for a subgroup G of Γ if every z 2 HnF is G-equivalent to one point in ◦ ◦ the interior of F (denoted F ) and no two points z1; z2 2 F are G-equivalent. We will prove and describe the fundamental domain of the whole group Γ itself before turning to subgroups. Theorem 1. A fundamental domain of Γ under fractional linear transforma- tions (2) is given by 1 1 F := fz 2 j − ≤ Re(z) ≤ and jzj ≥ 1g: H 2 2 Figure 1: Fundamental domain of Γ and approximating γ = T ◦ S ◦ T −1 such that γ:x 2 F . Proof. The idea is to fix any point z 2 H and then to find a combination of translations and inversions until z is mapped to F . This process is demonstrated for some x in Figure 1. Fix z 2 H. Define Γ0 to be the subgroup generated by T and S, where T and S are: 1 1 T := so that T:z = z + 1; (10) 0 1 3 0 −1 1 S := so that S:z = − : (11) 1 0 z a b 0 −2 Now let γ = c d 2 Γ . Then by (3) we have Im(γ:z) = Im(z)jcz+dj and thus Im(γ:z) > 0. By choosing the factors c and d so that jcz + dj is as small as possible but not zero. This is possible as cz +d must be on the lattice generated by 1 and z. Hence jcz + dj must be bigger or equal than one of the following: fj ± z + 0j; j ± z ± 1j; j0 + ±1jg by construction. But this set is finite hence has a minimum. So we are able to find a γ such that Im(γ:z) is maximal. In a next j 0 1 1 step find a suitable j 2 N [ f0g so that (T γ):z 2 fz 2 Hj − 2 ≤ Re(z) ≤ 2 g. Define T jγ =: γ0 and claim that γ0:z lies in F . If this was not the case, so jγ0:zj < 1, then again by (3) we would have Im((Sγ0):z) = Im(γ0:z)jγ0:zj−2 > Im(γ0:z) contradicting the maximality re- quirement we set above. Hence 9γ 2 Γ0 such that γ:z 2 F; 8z 2 H. We now have to prove that no two points in the interior of F share the same Orbit. For this assume z1; z2 2 F and wlog that Im(z2) ≥ Im(z1) and lastly that there exists an A 2 Γ so that z2 = A:z1. Having assumed that it follows that −2 Im(z2) = Im(A:z1) = Im(z1)jcz1 + dj ≥ Im(z1): by (3) and hence jcz + dj2 ≤ 1. As all entries in A must lie in and p 1 Z 3 Im(z1) ≥ 2 , it follows that the absolute value of c must be less than 2 oth- 1 erwise 1 ≤ Im(cz1) = Im(cz1 + d) ≤ jcz1 + dj. Moreover, we have jdj − 2 ≤ jd + Re(z)cj ≤ jcz + dj ≤ 1. This slims down the possibilities for A and we can deal with them case by case. Case 1: c = 0, d = ±1. By enforcing the ad − bc = 1 constraint we see that j 1 j then A or −A must be equal to T = 0 1 . But for A:z1 to lie in F again only j 2 {−1; 0; 1g come in question. For j = 0, we have z1 = z2. For j = ±1 recall the definition of T and we see that z2 and z1 must lie on the boundary lines 1 Re(z) = ± 2 of F and hence not in the interior of F . Case 2: c = ±1, d = 0. Then z1 lies on the unit circle, as we have jcz1j ≤ 1 a −1 a and jz1j ≥ 1. To satisfy ad−bc = 1, A has to be of the form A = ± 1 0 = T S a for some a 2 Z. For jaj > 2 we see that T S:z1 2= F , thus a 2 {−1; 0; 1g. 1 For a = 0 we get A = ±S and hence that z1; z2 lie in S \ F and are symmetric with respect to i as jz j = 1. R 1 p 1 −3 For a = 1 we have A = ±TS, but then z1 = + . The case a = −1 is p 2 2 1 −3 similar: We get A = ±ST and z1 = − 2 + 2 . p 1 −3 Case 3: Suppose c = d = ±1. Since jcz+dj ≤ 1, we see that z1 = − 2 + 2 , as for any other z1 2 F we find that j ± z1 ± 1j > 1.

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