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.