transformation.
If g is parabolic, then for some h we have
hgh'1(z) =
z+ t
(t 0).t4.3. Fixed and ConjugacyClasses 73
Thus
= z
+ nt
and
= h1(hz + nt).
Observe that for each z in —÷ as thus in general, if g is parabolic then
—+
wherea is the fixed point of g.
If g is not parabolic, then g has two fixed points, say a and /1, and for some h we have
hgh'(z)_—tz and hence
=
These facts show that if g is loxodromic (equivalently, jt 1) and if z is not a or /3, then the images gfl(2) are distinct and accumulate at a and /3 only.
If —* a, say, as n +00, then a is called the attractive fixed point of g while is called the repulsivefixed point. Then for all zotherthan /3,
g g has invariant circles: indeed
each circle for which a and /3 are inverse points is a g-invariant circle and so each orbit under iterates of g is constrained to lie on such a circle. We collect these results together for future reference.
Theorem 4.3.10. (i) Let g be parabolic with fixed point a. Then for all z in C,
—+ aas n —+ +00, the convergence being uniform on compact subsets ofC—{a}.
(ii) Let g be loxodromic. Then the fixed points a and /3 of g can be labelled so that aas +00 (jfz /3), the convergence being uniform on compact subsets of C — {/3}.
(iii) Let ge be elliptic with fixed points a and /3. Then g leaves invariant each circle for which a and /3 are inverse points.
If a Möbius g is of finite order k (so gk, but no smaller power, is I) then g is necessarily elliptic. In this case we have
hgh
'(z) =
say,and so
0 =2itm/k,
74 4. Complex Möbius Transformations
where k and m are coprime. We deduce that tr2(g) =4cos2(9/2)
= 2[1 + cos(2itm/k)].
Notethat this can take different values depending on the prime factors of k.
If g is elliptic of order two, then k = 2 and necessarily, tr2(g) = 0:the con- verse is also true. Observe that among all g of order k, the largest value of tr2(g) occurs when m = 1 or k 1,
tr2(g) = 4cos2(ir/k)
and 9 = ±2ir/k. Again we record this for future reference.
Theorem 4.3.1l.Ler g be an elliptic transformation of order k. Then tr2(g) 4 cos2(ic/k),
with equality and only jfg is a rotation of angle
EXERCISE 4.3
1. Find Möbius transformations g and h such that
(i) tr[g, h] = —2;and —
(ii) g and h have no common fixed point in C.
2. Let g be any Möbius transformation which does not fix cia. Show that g = qjg2g3, whereg1 and g3 are parabolic elements fixing and where g2 is of order two, 3. An nth root of a Möbius transformation g is any Möbius transformation h satisfying
h" = g.Prove
(i) if g =Ithen g has infinitely many nth roots;
(ii) if g is parabolic then g has a unique nth root;
(iii) in all other cases, g has exactly n nth roots.
4. Show that if A and B are in SL(2, C) then
det(A — I) = 2— tr(A) and
det(AB — BA)=2 — tr[A,B]
([A, B] is the commutator of A and B). Deduce that if A and B viewed as Möbius transformations do not have a common fixed point in thenAB — BA is a non- singular matrix which represents a Möbius transformation or order two.
5. Let g(z) = z/(cz + 1). Verify (i) by induction and (ii) by considering a suitable hgh'' that
g"(z) = z ncz + 1
Findf wheref(z) = 6z/(z+ 3) and check your result by induction.
§44. CrossRatios 75
§4.4. Cross Ratios
Given four distinct points z2,z3, z4 of C we define the cross-ratio of these points as
(z1 — z3)(z2 — z4)
[z1,z2,z3,z4] =
(Zj — z2)(z3 — z4)
comparethis with (3.2.5) where division is not permitted. The definition is extended by continuity to include the case when one of the is cc so, for example,
z1 — z3
[z1,z2,z3,cxD] =
z1 —
Notethat in particular,
[0, 1, z, cc] = z. (4.4.1)
If
az + b
g(z) = (ad — bc 0),
cz + d
then
(z — w)(ad — bc) g(z)— g(w) =
(cz + d)(cw + d)
and it is immediate that the cross-ratio is invariant under Möbius trans- formations; that is,
[g(z1), g(z2), g(z3), g(z4)] = [z1,z2, z3, z4]. (4.4.2) This is a useful property which often leads to a considerable simplification.
Moreover, the converse is also true: if
[w1,w2,w3,w4] = [z1,z2,z3,z4] (4.4.3)
holds then there is a Möbius transformation g with = To see this, letf and h be Möbius transformations which map z1, z2, z4 to 0, 1, cc and w1,w2,w4 to 0,1, cc respectively: these exist by Theorem 4.1.1. Then by (4.4.1), (4.4.2) and (4.4.3),
f(z3) =
[0, 1, f(z3), cc]= [f(zt),f(z2), f(z3), f(z4)]
= [z1,z2,z3,z4]
= [w1,w2,w3,w4]
= [h(w1),h(w2), h(w3), h(w4)]
= [0, 1, h(w3), co]
= h(w3).
It is now clear that g(z1) = where g =
h'
of.
76 4. Complex Möbius Transformations
We are now going to study how the cross ratio
= [zj, z2,23, 24] (4.4.4)
varies as we permute the With this in mind we let denote the permu- tation group of {1,.. .,n} and remark that (as with all functions) we regard permutations as acting on the left: for example, (12) (13) maps 3 to 2.
Each a inS4 induces a change in the value of the cross ratio by the formula
and it is essential to realize that the resulting value depends on a arid but
noton the individual values This is so because if
[zj,z2,z3, z4] =
[w1,w2,w3,w4], then there is some g with = and so2c3' =
= Wç3,
Because of this fact, we can introduce (a eS4) by the formula
= where A is given by (4.4.4). Because
=
[2 i,=
wehave the important relation
(44.5) Now suppose that a is the transposition (1, 2) and let g be the Möbius transformation which maps z1, z2, z4 to 0, 1, respectively. Then
=
[z3,z2,z3,z4]= [0,1,g(z3),cx]
=
g(z3) and so= [z3,z1,z3,z4]
=
=
I —A similar argument holds for all six transpositions in S4 and we find
(i) U = (1,2) or (3,4) then = 1 — A;
(ii) = (1,3)or(2,4) = — 1);
(iii)
a =
(1,4)or (2, 3) then = 1/2.§4.4. Cross Ratios 77
This information leads to a determination of As S4 is generated by transpositions, (i), (ii) and (iii) together with (4.4.5) suffice to give all
Note that for each transposition the function is actually a Möbius transformation which maps {O, 1, onto itself. Thus if we denote by the subgroup of Möbius transformations which map {O, 1, onto itself we find from (4.4.5) that the map
fe.,
is actually a homomorphism of 34 into A'0 (which is isomorphic to S3). In addition to this, it is clear from (i), (ii) and (iii) and (4.4.5) that the subgroup
K =
{1,(1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}of S4 is contained within the kernel of 8. We can now describe the situation completely.
Theorem4.4.1. The map 0: S4 A'0 is a homomorphism of 34 onto with kernel K.
PROOF. Theorem 4.1.1. implies that has exactly six elements: these are the functions
2, 1 A, 2/(1 — 1), 1/(1 — A),(2 — 1)/A
of 2. There are six permutations in S4 with i(4) = 4and a straightforward computation shows that the are precisely the six elements of This shows that 0 maps S4 onto A'0 and as this implies that the kernel of 0 has exactly four elements, the kernel must be K.
Four distinct points z1, z2, 23,z4 in are concyclic if and only if they lie on some circle. Let g be the Möbius transformation which maps z1, z2, z4 to 0, 1, respectively. Then the z1 are concyclic if and only if the are and this is so if and only if g(z3) is real. However,
g(z3) = [0, 1, g(z3),co]
= [z1,z2,z3,z4]:
thus z1,z2, z3, 24 are concyclic andonly jf[zt, Z2, 23, z4] is real.
Ifz1, z2, z3, 24 lie on a circle Q and are arranged in this order around Q,
then g(z3)> 1 and so
A = [z1,z2,z3,z4] > 1.
EXERCISE 4.4
1. Show that the unique Möbius transformation which maps 22, 24 tO 0, 1, co respectively is g where
g(z)= 22, Z, 2411.
2. Verify = 27(2— 1)when = (2,4).
4. ComplexMöbius Transformations
3. Let z2,z3, z4 bedistinct points in Show that the circle through z1, z2, z4 is orthogonal to the circle through z z3, z4 ifand only if
Re[z1, 22, 23, 24] = 0.
Generalize this to the case where the circles meet at an angle U (note that the are concyclic if and only if U = 0).
4. Let g be any Mobius transformation. Show that if g does not fix zthen [z,gz, g2z, g3z]
isindependent of z and evaluate this in terms of tr2(g).
§4.5. The Topology on
As described in Section 4.2, there is a homomorphism SL(2, C)
dl,
which associates to each g in dl exactly two matrices A and —A in SL(2, C).
The group SL(2, C) is a topological group with respect to the metric — BM
and the map induces the quotient topology on .11, namely the largest topology on dl with respect to which, is continuous. In addition, dl has a topology namely the topology of uniform convergence with respect to the chordal metric on C (see Section 3.7) and it is essential to know that
these topologies are the same. One method is to compare the action of
SL(2, C) through the action of dl on H3 (and then B3) to the matrix group3). However, a more direct approach is not without interest.
Theorem 4.5.1. The topology .9
induced on dl by
coincides with the topology of uniform convergence on C.PROOF. It is sufficient to show that the map
SL(2, C) (dl,37.*) (4.5.1)
is open and continuous: see Proposition 1.4.1.
Assuming that this has been established, observe that if X is in SL(2, C) then
— =
(see (x) of Section 2.2). This yields the next result.
Corollary 4.5.2. The restriction of radius inSL(2, C) is a homeomorphism: thus SL(2, C) is a two-sheeted covering space of if.
§4.5. TheTopology on 1/ 79
It remains to prove that the map (4.5.1) is open and continuous. Define
a(f,
g) = sup gz),where d is the chordal metric: thus is the metric topology induced by the metric a. We shall derive the continuity of 'l from the next result.
Proposition 4.5.3. If A in SL(2, C) represents g, then
a(g,I)
- 1(1.Explicitly, if B representsf, then
a(g,f) = a(gf1,I)
and so D is continuous at the general element B of SL(2, C).
PROOF OF PROPOSiTION 4.5.3. There is a unitary matrix B representing a Möbius map h such that hgh' fixes cc (h corresponds to a rotation of the sphere moving a selected fixed point of g to cc). By Theorems 2.5.2 and 4.2.2 we have
— 1 —
and
c(hgh',I) =
= a(g, I).
These remarks show that we may assume, without loss of generality, that g fixes cc. In addition, if g is loxodromic we may assume that the repulsive
fixed point of g is cc (we simply choose h appropriately).
Assume then that
czö=l:
the condition on the fixed point of g in the loxodromic case means that in all cases,
1
Now
d(z,gz) +
2!zIjl
(1 + 1z12)112(1 + +
4. Complex Möbius Transformations
the last line being an application of the Arithmetic—Geometric Mean inequality. This upper bound simplifies to a value independent of z and using = 1, we have
cr(g,I)
— + 21131(Ia — 112 ÷ II — + 1312)112(1 + 1 + 4)112
Finally, we must show that the map (4.5.1) is an open map and this will be derived from the next result.
Proposition 4.5.4. Let g1, g2,... be Möbius transformations and suppose that
-+ wforw = 0, 1, Then:
(i) there exist matrices representing which converge to I; and
(ii) -÷
I
on C.PROOF. Choose matrices
=
in SL(2, C) representing where is 1 or —1 and is to be chosen later.
In the following argument, trivial modifications are required if =
weignore these cases.
As
— 1
— —
—*1,
we can select so that —+ 1.Next,
=
cc)
—
so I also. As
c,, =
=
we see that and tend to zero: thus —+ I.This proves (i). Observe that
(ii) follows from (i) and Proposition 4.5.3. D
Finally, we can complete the proof of Theorem 4.5.1. Let be an open subset of SL(2, C) and suppose that is not an open subset of .A' (with
§4.5. TheTopology on .11
respect to the metric topology 3*)• Then there is some g in and some
g2,... not in
withg) —*0.
As
g) = I),
we see from Proposition 4.5.4 that there are matrices representing with -+I. If B (in represents g, then B so is in for all
large n. It follows that is in for these n and this is a contradiction.
A subgroup G of is discrete if and only if the topology described by Theorem 4.5.1 induces the discrete topology on G. It is clear from Corollary 4.5.2 that if G is discrete, then '(G) is a discrete subgroup of SL(2, C).
Conversely, if F is a discrete subgroup of SL(2, C), then cb(F) is a discrete subgroup of .4'.
Of course, if G is a discrete subgroup of .11, then G is countable (see Section 2.3), say G = {g1,g2,...},and
as n —* + In view of this, the next result is of interest.
Theorem 4.5.5. Suppose that K is a compact subset of a domain D in C and that g omits the values 0 and in D. Then for some positive m depending only on D and K, we have
md(z, w) d(gz,gw)
for all z and w in K.
PROOF. Define m1 by
2m1 = inf{d(z,w): z e K, D}
and suppose that
az + b
ad—bc=1.
cz + d As g '(cc) D, we see that for z in K,
2m, d(z,g'co)
2Icz + dl
(1 + 1z12)"2(1c12 + !d12)"2
82 4. Complex Möbius Transformations
A similar inequality holds for g 10 so
(I + 1z12)
az + hi2 + jcz
+ dj2.As
d(gz,gw) ( 1 + 1z12
\i/2f
1 + \1/2d(z, w) + b12 + icz
+d12) bj2 + icw +dV)
the result follows.
The implication of this is that if G is discrete, then under the assumptions in Theorem 4.5.5, the chordal diameters of the sets tend to zero.
EXERCISE 4.5
1, Prove that if ad —bc= 1 then for all z
(1a12 + c12)(iaz + hi2 + icz +di2) 1
with equality if and only if z = —(lIb +icV). Show that if g(z) =
(ab + b)(cz + d) 'then for all z,
1
Iaz+b12+icz+d12
< <ugh2.
hIglh2 — I +
2. Let G be a group of Möbius transformations preserving H2. Show that each g in G can be written uniquely in the form g =fh wheref(z) =az+ b (a > 0, be and h(i) = i.Deduce that G is homeomorphic to R2 x S'.
3. Show that a sequence of loxodromic transformations can converge to an elliptic element but if this is so, then is strictly loxodromic for almost all n. Show that a sequence of elliptic elements cannot converge to a loxodromic element.
§4.6. Notes
For a discussion of quaternions and Möbius transformations see [1], [5]
and [26]. The problem of obtaining a subgroup of SL(2, C) isomorphic to a given subgroup of .A' has been considered in [2] and [74]. For general information on Möbius transformations see [30] (especially for isometric circles), [51] and [52]. See [53] for Theorems 4.2.2 and 4.3.7.
CHAPTER 5
Discontinuous Groups
§5.1. The Elementary Groups
In this section we shall define and describe a class of subgroups of which have a particularly simple structure. This class contains all finite subgroups
of each point in
Definition 5.1.1. A subgroup G of is said to be elementary if and only if there exists a finite G-orbit in
Of course, the emphasis here is on the word finite. Also, note that this definition makes no reference to discreteness. The group acts as the group of directly conformal isometries of H3 and G is elementary if there is a finite G-orbit in the closure of hyperbolic space.
Obviously, if a single point is G-invariant then G is elementary. If G is abelian, then either G contains only elliptic elements and I or G contains some parabolic or loxodromic element g. In the first case (whether G is abelian or not), G is elementary by virtue of Theorem 4.3.7: in the second case, G is elementary by Theorem 4.3.6(iii). Thus every abelian subgroup of is elementary.
Remark. Elementary groups are sometimes defined by the condition that for every g and h in G which are of infinite order, we have trace[q, h] = 2:
equivalently, g and h have a common fixed point in C (Theorem 4.3.5).
However, with this definition, the stabilizer of a point in H3 is not necessarily elementary.
5. Discontinuous Groups
Let
us now assume that G is an elementary group and examine the
possibilities. Suppose that the finite orbit is {x1 If g is in G then the points m = 0, 1, 2 cannot all be distinct so there is an integerwith the property that gm fixes x3. If rn is now the product of the then
gm fixes each xj. With this available we can now classify the elementary groups into three types.
Type 1: suppose that n 3or that {x1,.. .,x,,} is not in
Ifthe points are not in C then each g in G has some power gm fixing and so gm, and hence g itself, is elliptic (or 1). If n 3 and the are in C, then gtm has at least three fixed points and so is the identity: thus again, each non-trivial element of G is elliptic. This shows that if G is of Type 1, then G contains only elliptic elements and I. By Theorem 4.3.7, there is some x in H3 which is fixed by every element of G arid by mapping H3 onto B3 and x to 0 we see that G is conjugate in to a subgroup of the Special Orthogonal group SO(3) (see Theorem 3.4.1).
Type 2: suppose that n = I and x1 is in C.
In this case, G is conjugate to a subgroup of .A, every element of which fixes and so is of the form z az + b. Thus G is conjugate to a group of Euclidean similarities of C.
Type 3: suppose that n = 2and that x1, x2 are in
Inthis case, G is conjugate to a subgroup of .A1, every element of which leaves {0, invariant and is therefore of the form
a
0,s2 =
1.Note that G is then conjugate to a group of isometries of the space C — {0}
with the metric derived from dz I/ z
We shall now describe all discrete elementary groups. If G is a discrete elementary group of Type I we may assume that every element of G fixes the point] in H3. Thus by Theorem 4.2.1, = 2 for every gin G and (as G is discrete) G is necessarily finite. Thus G is conjugate to a finite sub- group of SO(3) and hence to one of the symmetry groups of the regular solids.
We can use the fact that G is finite to obtain the possible structures of G without reference to the regular solids. We say that v in C is a vertex if v is fixed by some g I) in G and we denote the set of vertices by V. Now consider the number IE of elements of the finite set
E =
{(g,v):geG,g V,g(v)= v}.As each g in G (g I) is elliptic it fixes exactly two vertices and we have El = 2(IGI — 1).
§5.1. The Elementary Groups
The stabilizer of a vertex v is so we also have
= - 1).
The
set V is partitioned by G into disjoint orbits V1,.
.,and as the
stabilizers of each v in have the same number, say of elements we havej=1
=
-1).
j=I
Finally, each orbit G(v) is in 1—1 correspondence with the class of cosets
G/GVsofor v in we have = G(v)and
Eliminating we obtain
2(1
—k).
Weshall exclude the trivial group, so IGI 2 and
I
2(1—
<2.
By definition,
2 so
—
Theseinequalities together with (5.1.1) show that s = 2 or s = 3.
Case 1: s =2.
In this case, (5.1.1) becomes
nl n2
and hence (as
G),
IGI=n1=n2,
V11=1V21=1.In this case there are only two vertices and each is fixed by every element of G. By conjugation, we may take the vertices to be 0 and and G is then a finite, cyclic group of rotations of C.
86 5. Discontinuous Groups
Case 2: s = 3.
In this case, (5.1.1) becomes
1 1 1 2
—+—+—= 1
and we may assume that n1
n3. Clearly n1 3 leads to a contra-
diction:thusn1 = 2and1 1 1 2
n2 n3 2
If n3 n2 4 we again obtain a contradiction, so n2 =
2 or 3. The case= 2 leads to
(IGI,ni,n2,n3) =
(2n,2,2,n)(n 2)
and this is isomorphic to the group of orientation preserving symmetries of a regular plane n-gon (the dihedral group Dr).
The remaining cases are those with s = 3,n1 = 2,n2 = 3 and
1 1 2
n33,
and the (integer) solutions of this are
(i)
"i,
n2, n3) = (12,2,3,3);(ii)
G n1,
n3) = (24, 2, 3, 4);(iii) (IGI,n1, n2, n3) = (60, 2, 3, 5).
These groups are isomorphic to A4, S4 and A5 respectively and they corre- spond to the symmetry groups of the tetrahedron, the octahedron and the icosahedron respectively. For more details, see the references in Section 5.5.
We continue with our discussion of discrete, elementary groups. The next result essentially distinguishes between groups of Types 2 and 3.
Theorem 5.1.2. Let g be loxodromic and suppose that f and g have exactly one fixed point in common. Then g> is not discrete.
PROOF. As discreteness is preserved under conjugation we may assume that the common fixed point is co and, say,
g(z) = > 1),
f(z)
= az b(if necessary, we may replace g by g').
Then
= az +
§5.1. The Elementary Groups 87
Asf and g have only one common fixed point, we see that b 0. As > 1, we find that the sequence
n = 1,2,...
is a convergent sequence of distinct terms: thus g> is not discrete. For a much more illuminating proof, the reader need only draw a diagram and locate (for large n) the points 2,g'z, and
Suppose now that G is elementary, discrete but not of Type 1. Then G must contain parabolic or loxodromic elements. If G contains a parabolic element g, fixing say, then every element of G fixes (because all other orbits are infinite) and by Theorem 5.1.2, G has no loxodromic elements.
Such a group is of Type 2. If G contains a loxodromic element g, fixing 0 and say, then every element of G must leave the set {0, invariant. This implies that G cannot contain parabolic elements and such a group is of Type 2 or 3.
Let us now examine the structure of a discrete group of Type 2 with parabolic elements. Thus G contains only I, parabolic elements and possibly some elliptic elements.
By conjugation, we may assume that every element of G fixes oo and so is of the form z H+ + /3. As this is either elliptic or parabolic, we see that
= 1: thus G is conjugate to a group of Euclidean isometries of C.
We call the multiplier of the map z —÷ + /3 and in general, we denote
the multiplier of g by;. Note that ; =
1 if and only if g is parabolic or I.It is a trivial matter to check that the set S of multipliers of g in G is a (multi- plicative) subgroup of = 1} and that the map 6: G —+ S defined by 0(g) =
;
is a homomorphism of G into S. The statement that; 1 if and only if g is parabolic or I is precisely the statement that the kernel, T, ofo is the subgroup of translations in G. As G/T is isomorphic to S (=0(G)), we can describe G by giving explicit descriptions of S and T: this effectively separates the parabolic and elliptic elements.
First, we show that S is a finite cyclic group. Now G contains a trans- lation, say f(z) = z + and if g(z) = cz + /3 is in G, then so is
gfg'(z) =
z +We deduce that G contains z i—+z + st for every s in S and as G is discrete, S cannot accumulate in C. Thus S is a finite subgroup of = 1} and (as is easily seen) it is necessarily cyclic.
We can obtain even more information about S. With f and g as above,
f1(gfg')(z) =
z + — 1)Aand so if —
fl
< 1, then there is a translation + in G with= — < L21.