PRODUCTS of CONJUGACY CLASSES in SL2(R) S. Yu. Orevkov

PRODUCTS OF CONJUGACY CLASSES IN SL2(R)

S. Yu. Orevkov

Abstract. We compute the product of any n-tuple of conjugacy classes in SL2(R).

1. Introduction In this paper we compute the product of any n-tuple of conjugacy classes in the group SL2(R) (see Theorem 3.1 in §3). The computation is straightforward, the main difficulty being to find a suitable notation for the answer to be readable. All products of conjugacy classes are computed in [9] for simple finite groups of order less than million and for sporadic simple finite groups. In [12] the same is done for finite unitary groups GU3(Fq) and SU3(Fq) as well as for finite linear groups GL3(Fq) and SL3(Fq). Similar questions were studied by several authors, see [2, 5, 8, 10, 15] and references therein. My special interest in the computation of class products in any kind of linear or unitary groups is motivated by possible applications to plane real or complex algebraic curves, see [3, 11]. Perhaps, the most interesting and non-trivial case when the products of conjugacy classes are completely computed, is the case of the unitary groups SU(n), see [1, 4]. It seems that Belkale’s approach [4] could be extended (at least partially) to pseudo-unitary groups SU(p, q) using the techniques developed in [7]. We are going to do this in a subsequent paper. Some products of conjugacy classes in PU(n, 1) (especially for n = 2) are computed in [6, 13, 14]. Note that SU(1, 1) is isomorphic to SL2(R). Indeed,

1 i Φ: SL (R) → SU(1, 1), A 7→ Φ(A) = P −1AP, P = , (1) 2 i 1

∗ is an isomorphism; recall that SU(1, 1) = {A ∈ SL2(C) | A JA = J} where J = diag(1, −1). So, the main motivation for the computation of class products in SL2(R) was to get an idea what the answer for SU(p, q) could look like.

2. Conjugacy classes

Let G = SL2(R). The conjugacy classes in G are given in Table 1. This fact can be easily derived, e. g., from [5, §2]. A more geometric characterization of the conjugacy classes can be given as follows. For ~x =(x1,x2) and ~y =(y1,y2), we denote ~x ∧ ~y = x1y2 − x2y1. Proposition 2.1. Let A ∈ G \{I, −I} and 0 <α<π. Then: cε,δ R2 (a). A ∈ 2 , ε, δ = ±1, if and only if tr A =2ε and δ~x∧A~x ≥ 0 for any ~x ∈ ; cα R2 (b). A ∈ 2 if and only if tr A = 2 cos α and ~x ∧ A~x> 0 for any ~x ∈ .

Typeset by AMS-TEX 1 2 S. YU. OREVKOV

nessessary and suﬃcient class paramameters representative a b condition on A =( c d )

cε ε 0 1 {−1, 1} εI =( 0 ε ) A = εI cε,δ 2 ε 0 2 {−1, 1} ( δ ε ) tr A =2ε & δ(c − b) > 0 cα cos α − sin α 3 ]0, 2π[ \{π} sin α cos α tr A = 2 cos α & c sin α > 0 λ λ 0 c R −1 −1 4 \ [−1, 1] 0 λ tr A = λ + λ R cε,δ cα Table 1. Conjugacy classes in SL2( ). Note that for 2 (resp. 3 ) we have δ(c−b) > 0 ⇔ (δc > 0 or δb < 0) (resp. c sin α > 0 ⇔ b sin α < 0)

c... c We denote the union of all i by i (i = 1, 2, 3, 4). For X ⊂ G, we denote the complement G \ X by Xc. We set also cε,∗ cε,+ cε,− c+ cλ c− cλ 2 = 2 ∪ 2 , 4 = 4 , 4 = 4 , λ>[1 λ<[−1 c+ c+ c+∗ c− c− c−∗ 4 = 4 ∪{I} ∪ 2 , 4 = 4 ∪ {−I} ∪ 2 . If j is an interval (open or closed from either end) contained in ]0, 2π[\{π}, then cj cα we set 3 = α∈j 3 . If the left end of j is “]0 ... ” or “]π... ”, then the bracket “] ... ” can beS replaced by “[... ” or “h[... ”. Symmetrically, if the right end of j is “...π[” or “... 2π[”, then the bracket “... [” can be replaced by “... ]” or “... ]i”, and this means the following:

c[0,... c++ c]0,... c...,π] c...,π[ c−+ 3 = 2 ∪ 3 , 3 = 3 ∪ 2 , c[π,... c−− c]π,... c...,2π] c...,2π[ c+− 3 = 2 ∪ 3 , 3 = 3 ∪ 2 , ch[0,... c+ c[0,... c...,π]i c...,π] c− 3 = 4 ∪ 3 , 3 = 3 ∪ 4 , ch[π,... c− c[π,... c...,2π]i c...,2π] c+ 3 = 4 ∪ 3 , 3 = 3 ∪ 4 , ch[0,π] c+ c++ cα c−+ for example, 3 = 4 ∪ 2 ∪ 0<α<π 3 ∪ 2 . Let S + c+ c]0,π[ G = 4 ∪ 3 . 3. Statement of the main result Theorem 3.1. (a). For conjugacy classes contained in G+ (see Remark 3.2), their double and triple products are as shown in Tables 2 and 3; in Figure 1 we represent all triples of classes from c3 whose product contains I. (b). Let 0 <α,β,γ,δ<π. Then: c++c++c+−c+− c++c++c+−cα c 2 2 2 2 = 2 2 2 3 = {−I} , c++c+−c+−cα c++c+−cαcβ c 2 2 2 3 = 2 2 3 3 = {I} , c++c++cαcβ c 2 2 3 3 = {−I} if α + β ≥ π, c+−c+−cαcβ c 2 2 3 3 = {I} if α + β ≤ π, cαcβcγcδ c 3 3 3 3 = {−I} if π<α + β + γ + δ < 3π, PRODUCTS OF CONJUGACY CLASSES IN SL2(R) 3

cαcβcγc+± the products of the form 3 3 3 2 are as in Table 3, and for any other four non- scalar conjugacy classes contained in G+, their product is the whole G. (c). The product of any ﬁve non-scalar conjugacy classes is the whole G. Remark 3.2. We see in Table 1 that, for any conjugacy class c. either c ⊂ G+ or −c ⊂ G+. Thus any product of conjugacy classes can be immediately recovered if one knows all products of classes contained in G+. ˜ R c... ˜ c... Let G = PSL2( ). We denote the image of i in G by ˜i . cλcλ ˜ Corollary 3.3. (a). We have ˜4˜4 = G and c++c++ c+−c+− cλc ˜ ˜ c cλ ˜ ˜2 ˜2 = ˜2 ˜2 = ˜4˜ = G \{I}, where ˜ 6= ˜4 , I

(see Table 2 for the other double products of classes in G˜). (b). Let 0 <α,β,γ <π. The following triple products are equal to G˜ \{I˜}:

c++c+−cα c+−cαcβ c+±cαcβ ˜2 ˜2 ˜3 , ˜2 ˜3 ˜3 (if α + β<π), ˜2 ˜3 ˜3 (if α + β = π), c++cαcβ cαcβcγ ˜2 ˜3 ˜3 (if α + β>π), ˜3 ˜3˜3 (if π<α + β + γ < 2π).

The product of any other three non-trivial conjugacy classes is the whole G˜. (c). The product of any four non-trivial conjugacy classes is the whole G˜. Corollary 3.4. cn(G˜) = ecn(G˜)=4 (in the notation of [9]).

c++ c+− cγ cν I 2 2 3 4 c++c++ c[0,π]i c c[0,π]i c+ 2 2 3 {I} 3 ∪ 4 ւ ւ c++c+− c+ c[0,π]i c+ ch[π,2π] c+ 2 2 4 3 ∪ 4 3 ∪ 4 ւ ւ c+−c+− ch[π,2π] ch[π,2π] c+ c 2 2 3 3 ∪ 4 {I} ւ ւ c++cα c ]α,π]i c[0,α] c ch[0,π]i 2 3 3 {I} ∪ 3 3 ւ ւ c+−cα ch[0,α[ ch[0,π]i c[α,π] c 2 3 3 3 {−I} ∪ 3 ւ ւ c++cλ ch[0,π]i c c 2 4 3 {I} {−I} ւ ւ c+−cλ ch[π,2π]i c c 2 4 3 {−I} {I} ւ ւ c cαcβ, α + β<π c[α+β,π]i {I} ∪ c[0,α+β] ch[0,π]i ւ 3 3 3 3 3 See β h[π,2π]i h[0,π]i cαc , α + β = π {−I} ∪ c− c c ւ 3 3 4 3 3 Tbl. 3 cαcβ ch[π,α+β] ch[π,2π]i c[α+β,π] c 3 3 , α + β>π 3 3 {I} ∪ 3 ւ cαcλ ch[0,π]i c c c 3 4 3 {I} {−I} {I} G cλcλ c 4 4 {−I} G G G G cλcµ c 4 4 , λ 6= µ {I, −I} G G G G

Table 2. Double and triple products of conjugacy classes in SL2(R) (“ւ” means “see some other cell(s) of this table”). The range of the parameters: 0 <α,β,γ <π and λ, µ, ν > 1. 4 S. YU. OREVKOV

cαcβcγ cαcβcγc++ cαcβcγc+− Condition on α,β,γ 3 3 3 3 3 3 2 3 3 3 2

c[0,α+β+γ[ c α + β + γ<π {I} ∪ 3 G G \{I} ch[π,2π]i α + β + γ = π {−I} ∪ 3 G \ {−I} G \{I} ch[π,2π]i π<α + β + γ < 2π 3 G \ {−I} G \{I} ch[π,2π]i α + β + γ =2π {I} ∪ 3 G \ {−I} G \{I} c]α+β+γ−2π,π] c α + β + γ > 2π {−I} ∪ 3 G \ {−I} G

Table 3. Triple and some quadruple products of conjugacy classes in R cαcβcγ SL2( ) involving 3 3 3 with 0 <α,β,γ <π.

γ γ

β β 2π π π α 0 α π 2π 0 π

3 cαcβcγ Figure 1. The sets (α,β,γ) ∈ ( ]0, 2π[ \ {π}) | I ∈ 3 3 3 3 cαcβcγ and (α,β,γ) ∈ (] − π,π [ \{0}) | I ∈ 3 3 3 . 4. Range of the trace on the product of two classes Let Φ be as in (1); see the introduction.

Lemma 4.1. SU , a b a,b C,aa b¯b a). (1 1) = { ¯b a | ∈ ¯ − =1}. ¯ cα λ 0 λ eαi b). Φ( 3 ) is the conjugacy class of λ¯ , = . 0 A a b ad bc A SU , A∗J JA−1 Proof. Let = c d , − = 1. Then ∈ (1 1) if and only if = . We have:

A∗J a¯ c¯ 1 0 a¯ −c¯ , JA−1 1 0 d −b d −b = ¯b d¯ 0 −1 = ¯b −d¯ = 0 −1 −c a = c −a c[0,π] c+ c[π,2π] Lemma 4.2. Let A ∈ 3 and B ∈ 4 . Then AB 6∈ 3 ∪{I, −I} unless A and B both belong to c2 and have common eigenvector. Proof. Let v be a real eigenvector of B that is not an eigenvector of A. The c+ condition B ∈ 4 implies that the corresponding eigenvalue λ is positive. We have PRODUCTS OF CONJUGACY CLASSES IN SL2(R) 5

v ∧Av ≥ 0 by Proposition 2.1. Moreover, v ∧Av 6= 0 since v is not an eigenvector of A. Hence v ∧ ABv = λv ∧ Av > 0 and the result follows from Proposition 2.1. Lemma 4.3. Let 0 <α,β<π. Then: cα cβ (a). {tr(AB) | A ∈ 3 ,B ∈ 3 } = ] −∞, 2 cos(α + β)]; cα cβ cα+β (b). If A ∈ 3 , B ∈ 3 , and tr(AB) = 2cos(α + β), then AB ∈ 3 . A cα B cβ A λ 0 Proof. Let ∈ 3 and ∈ 3 . We may assume by Lemma 4.1 that Φ( ) = λ¯ 0 and Φ(B) = Q µ 0 Q−1 with λ = eαi, µ = eβi, and Q = a b with aa¯ − b¯b = 1. 0µ ¯ ¯b a¯ Then we have

tr(AB)=(λµ + λ¯µ¯)aa¯ − (λµ¯ + λµ¯ )b¯b = 2(1+ b¯b) cos(α + β) − 2b¯b cos(α − β) = 2 cos(α + β) − 4b¯b sin α sin β and the result easily follows. cα c++ Lemma 4.4. Let 0 <α<π. Then {tr(AB) | A ∈ 3 ,B ∈ 2 } = ] −∞, 2 cos α[ cα c+− and {tr(AB) | A ∈ 3 ,B ∈ 2 } = ]2 cos α, ∞[. cα c+± Proof. Let A ∈ 3 and B ∈ 2 . Let us fix a quadratic form invariant under A and choose a positively oriented orthonormal base (e1, e2) such that e2 is an eigenvector of B. In this base, the matrices of the corresponding operators are: A′ = cos α − sin α B′ 1 0 p > A′B′ α p α α α and = ±p 1 with 0, thus tr = 2 cos ∓ sin . sin cos c++ c+− Lemma 4.5. {tr(AB) | A,B ∈ 2 } = {tr(AB) | A,B ∈ 2 } = ] −∞, 2] and c++ c+− {tr(AB) | A ∈ 2 ,B ∈ 2 } = [2, ∞[. c++ Proof. We consider only the case A,B ∈ 2 (the other two cases are similar). A 1 0 B a b c++ b AB Let = 1 1 and = c d ∈ 2 . Then ≤ 0 (see Table 1) and tr = a + b + d = b +tr B = b + 2. Moreover, b can attain any non-positive value. Indeed, 1 b consider the matrices B1 = (b < 0) and B2 = A. 0 1 Lemma 4.6. Let A ∈ c4. Then for any t1,t2 ∈ R there exist matrices B, C ∈ G such that tr B = t1, tr C = t2, and AB = C. Proof. Without loss of generality we may assume that A = diag(λ, λ−1), |λ| > 1. B a b B a d AB λa λ−1d Let = c d . Then we have tr = + and tr = + . Thus, it is −1 enough to find a and d from the simultaneous equations a + d = t1, λa + λ d = t2 and then to find b and c such that bc = ad − 1.

5. Double products of conjugacy classes cαcβ Lemma 5.1. Let 0 <α,β<π. Then 3 3 is as shown in Table 2. cα cβ Proof. Let A ∈ 3 , B ∈ 3 , and let C = AB. It follows from Lemma 4.3 that the range of tr C is as required. So, it remains to show that the conjugacy class of C is uniquely determined by the trace. This is evidently so when C ∈ c4. Let us consider all the other cases. Case 1. α + β<π. It is clear that C 6∈ {±I}. 6 S. YU. OREVKOV

cε,δ Case 1.1. C ∈ 2 , ε, δ = ±1. We have ε = −1 by Lemma 4.3, so it remains to c−− −1 show that δ 6= −1. Suppose that δ = −1, i.e., C ∈ 2 . We have B =(−A )(−C) −1 cπ−α c++ with −A ∈ 3 and −C ∈ 2 . Hence Lemma 4.4 implies 2 cos β = tr B < tr(−A−1) = 2cos(π − α) which contradicts the assumption that α + β<π. c cγ Case 1.2. C ∈ 3. Let C ∈ 3 , γ ∈ ]0, 2π[ \{π}. Then, by Lemma 4.3, we have 1 cos γ = 2 tr C ≤ cos(α + β), hence

γ ≥ α + β. (2)

Thus it suﬃces to show that γ cannot be >π. Suppose that γ>π. Without loss of generality we may assume that α ≥ β. We −1 −1 cπ−α cγ−π have (−A )(−C) = B with −A ∈ 3 , −C ∈ 3 . The both angles π − α and γ − π are in ]0,π[. Thus, by Lemma 4.3 applied to the matrices −A−1, −C, 1 and B, we have cos β = 2 tr B ≤ cos((π − α)+(γ − π)) = cos(γ − α). Combining this inequality with (2), we obtain γ − α ≥ 2π − β. Thus γ ≥ 2π + α − β which contradicts our assumptions α ≥ β and γ < 2π. Case 2. α + β = π. Follows immediately from Lemma 4.3. −1 −1 −1 −1 cπ−β Case 3. α + β >π. We have C = (−B )(−A ) with −B ∈ 3 , −1 cπ−α −A ∈ 3 , and (π −β)+(π −α) ∈ ]0,π[, thus we reduce this case to Case 1. cαc+± Lemma 5.2. Let 0 <α<π. Then 3 2 is as shown in Table 2. Proof. Combine Lemma 4.4 with Lemma 4.2. c++c++ c++c+− c+−c+− Lemma 5.3. 2 2 , 2 2 , and 2 2 are as shown in Table 2. c++c++ Proof. 2 2 is as required due to Lemma 4.5 combined with Lemma 4.2. We c+−c+− c++c++ obtain 2 2 from 2 2 by passing to the inverse matrices. The computa- c++c+− tion of 2 2 is immediate from Lemma 4.5 combined with the observation that 1 0 1 0 c++ c+− −t 1 belongs to 2 , {I}, or 2 according to the sign of 1 − t. 1 1 cλ Lemma 5.4. The product of 4 with any other conjugacy class is as in Table 2. Proof. Let λ > 1, and c be any conjugacy class. Let X be the set that should cλc cλc coincide with 4 according to Table 2. The fact that the product 4 is contained in X, is either evident or follows from Lemma 4.2. Let us prove the inverse inclusion. cλ c Let A ∈ 4 , B0 ∈ , and C0 ∈ X. We assume that B0, C0 6= ±I (otherwise everything is evident). We set t1 = tr B0 and t2 = tr C0. By Lemma 4.6, we can choose matrices B and C such that tr B = t1, tr C = t2, and AB = C. By passing −1 to inverse matrices if necessary, we may assume that B ∼ B0 (note that A ∼ A). We have tr C = tr C0 whence C belongs either to the class of C0 or to the class of −1 C0 . However only one of these two classes may be contained in X (see Table 2) which completes the proof. All double products of conjugacy classes are computed in Lemmas 5.1–5.4. Using them, one easily computes triple and quadruple products as well; see the subsequent sections.

6. Triple products of conjugacy classes Due to previous computations we have: PRODUCTS OF CONJUGACY CLASSES IN SL2(R) 7

c++c++c++ c[0,π]ic++ c]0,π[ c++ c+− c+ c++ 2 2 2 = 3 2 = 3 ∪ 2 ∪ − 2 ∪ − 4 2 c]0,π]i c[0,π]i c+ ch[0,π]i c = 3 ∪ 3 ∪ − 4 ∪ − 3 = {I} ,

c++c++c+− c[0,π]ic+− c]0,π[ c++ c+− c+ c+− 2 2 2 = 3 2 = 3 ∪ 2 ∪ − 2 ∪ − 4 2 ch[0,π[ c+ ch[π,2π] ch[π,2π]i c[π,2π[ c = 3 ∪ 4 ∪ − 3 ∪ − 3 = {−I} ∪ 3 ,

cα c++c++ c]α,π]ic++ c]α,π[ c+− c+ c++ 3 2 2 = 3 2 = 3 ∪ − 2 ∪ − 4 2 c]α,π]i c+ ch[0,π]i c[0,α] c = 3 ∪ − 4 ∪ − 3 = {I} ∪ 3 ,

cα c++c+− c]α,π]ic+− c]α,π[ c+− c+ c+− 3 2 2 = 3 2 = 3 ∪ − 2 ∪ − 4 2 ch[0,π[ ch[π,2π] ch[π,2π]i ch[0,π]i = 3 ∪ − 3 ∪ − 3 = 3 ,

cλ c++c++ cλc[0,π]i cλ c[0,π] c ch[0,π]i c c 4 2 2 = 4 3 = 4 3 ∪ − 4 = 3 ∪ −{−I} = {I} , cλ c++c+− cλc+ cλ c+ c c c 4 2 2 = 4 4 = 4 4 ∪ha subset of 4 i = {−I} , cλ cα c++ cλc]α,π]i cλ c+ c c c 4 3 2 = 4 3 = 4 − 4 ∪ha subset of 4 i = {I} , cλ cα cβ cλ c+ c c c 4 3 3 = 4 − 4 ∪ha subset of 4 i = {I} , cλ cµ c c 4 4 = G for any non-scalar class . If α + β<π, then

cα cβ c++ c[α+β,π]ic++ c[α+β,π[ c+− c+ c++ 3 3 2 = 3 2 = 3 ∪ − 2 ∪ − 4 2 c]α+β,π]i c+ ch[0,π]i c[0,α+β] c = 3 ∪ − 4 ∪ − 3 = {I} ∪ 3 . If α + β = π, then

cα cβ c++ c+ c++ c++ ch[0,π]i ch[π,2π]i 3 3 2 = {−I} ∪ − 4 2 = − 2 ∪ − 3 = 3 . If α + β>π, then

cα cβ c++ ch[π,α+β]c++ c]0,α+β−π[ c++ c+ c++ 3 3 2 = 3 2 = − 3 ∪ − 2 ∪ − 4 2 c]0,π]i c[0,π]i ch[0,π]i ch[π,2π]i = − 3 ∪ − 3 ∪ − 3 = 3 .

If α + β + γ<π, then

cα cβ cγ c[α+β,π]icγ c[α+β,π−γ[ cπ−γ c]π−γ,π[ c+− c+ cγ 3 3 3 = 3 3 = 3 ∪ 3 ∪ 3 ∪ − 2 ∪ − 4 3 c[α+β+γ]i c+ ch[π,π+γ[ ch[0,γ[ ch[0,π]i = 3 ∪ ({−I} ∪ − 4 ) ∪ 3 ∪ − 3 ∪ − 3 c[0,α+β+γ[ c = {I} ∪ 3 . 8 S. YU. OREVKOV

If α + β + γ = π, then cα cβ cγ c[α+β,π]icγ cα+β c]α+β,π[ c+− c+ cγ 3 3 3 = 3 3 = 3 ∪ 3 ∪ − 2 ∪ − 4 3 c− ch[π,π+γ[ ch[0,γ[ ch[0,π]i ch[π,2π]i =({−I} ∪ 4 ) ∪ 3 ∪ − 3 ∪ − 3 = {−I} ∪ 3 . If π<α + β + γ < 2π and α + β<π, then cα cβ cγ c[α+β,π]icγ c[α+β,π[ c+− c+ cγ 3 3 3 = 3 3 = 3 ∪ − 2 ∪ − 4 3 ch[π,π+γ[ ch[0,γ[ ch[0,π]i ch[π,2π]i = 3 ∪ − 3 ∪ − 3 = 3 . If π<α + β + γ < 2π and α + β = π, then cα cβ cγ c+ cγ cγ ch[0,π]i ch[π,2π]i 3 3 3 = {−I} ∪ − 4 3 = − 3 ∪ − 3 = 3 . All other triple products reduce to these by passing to the inverse matrices (and maybe changing the sign). For example, if π<α + β + γ < 2π and α + β>π, then π < (π − α)+(π − β)+(π − γ) < 2π, (π − α)+(π − β) < π, hence cα cβ cγ cπ−α −1 cπ−β −1 cπ−γ −1 ch[π,2π]i −1 ch[π,2π]i 3 3 3 =(− 3 ) (− 3 ) (− 3 ) = − 3 = 3 . 7. Quadruple products. End of proof of Theorem 3.1 We see in Table 2 that any triple product of non-scalar conjugacy classes contains c c4, and the product of c4 with any other non-scalar class contains {I, −I} . Thus, to compute the product X of four non-scalar classes, it is enough to check if I and −I belongs to it. In its turn, to decide whether or not ±I ∈ X, it is enough to check if the inverse of one of these four classes multiplied by ±1 belongs to the product of the others. cαcβcγc++ c For example, let us check that X := 3 3 3 2 = {−I} when α + β + γ = π (see Table 3). Indeed,

c++ −1 c+− cαcβcγ ch[π,2π]i 2 = 2 ∈ 3 3 3 = {−I} ∪ 3 , hence I ∈ X, c++−1 c−+ cαcβcγ ch[π,2π]i − 2 = 2 6∈ 3 3 3 = {−I} ∪ 3 , hence −I 6∈ X. cαcβcγcδ As another example, let us compute the product 3 3 3 3 which we denote by X. We consider only the case when α+β +γ +δ ≤ 2π because the other case, when this sum is in the range [2π, 4π[, is reduced to this one by passing to the inverse cαcβcγ matrices. Let Y = 3 3 3 . We have c c[0,α+β+γ[ {I} ∪ 3 , α + β + γ <π,  Y =  {−I} ∪ ch[π,2π]i, α + β + γ = π,  3 ch[π,2π]i, α + β + γ >π.  3  cδ −1 cπ−δ cδ −1 cπ−δ Therefore 3 = − 3 ∈ Y whence I ∈ X, and since − 3 = 3 , we have cπ−δ −I ∈ X ⇔ 3 ∈ Y ⇔ π − δ 6∈ ]0,α + β + γ[ ⇔ π − δ ≥ α + β + γ whence X = G if α + β + γ + δ ≤ π, and X = {−I}c if π<α + β + γ + δ ≤ 2π. PRODUCTS OF CONJUGACY CLASSES IN SL2(R) 9

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