Let 4 be a Möbius transformation with = 0 and

= Then = xAfor some orthogonal matrix A.

PROOF. By Theorem 3.2.5, fixes and, as in the proof of Theorem 3.2.7, we see that is a Euclidean similarity. Because fixes the origin and leaves Sn-1 invariant, it is actually a Euclidean isometry. The result now follows from Theorem 3.1.4.

It is easy to see that the reflection in the plane P(a, t) leaves B" invariant if and only if t = 0. Better still, this reflection leaves B" invariant if and only if P(a, t) is orthogonal to S"' and in this form the statement is true for all reflections.

Theorem 3.4.2. Let be the reflection in S(a, r). Then the following are equiva- lent:

(i) S(a, r) and S" ¹ areorthogonal;

(ii) = 0(equivalently, = a*);

(iii) çb(B") = B".

PROOF. As

= a ^— r2a*

= (jaJ2 ^—

wesee that (i) and (ii) are equivalent. The assertion that (iii) implies (ii) is simply the fact that a and a* map to inverse points with respect to ¹

(Theorem 3.2.5).

Finally, (i) and (ii) together with (3.1.5) imply that q5(x)I = 4(x) —

—

— Jx ^— aI.Ia*

— x—aJ

§3.4. Self-mappings of the Unit Ball 39

1 - I4)(x)12 = (1 - 1x12)r2

(3.4.2)

—

andthis proves (iii).

E

Asanother application of (3.4.2) we observe that if 4)preserves then

— — yi2

. 343

(1

(.•)

this follows immediately from (3.1.5) and (3.4.2). In addition, (3.4.3) holds whenever 4) ^is the reflection in a plane P(a, 0) and hence for all Möbius 4) which preserve

The invariance expressed by (3.4.3) also yields j4)(y) — — 1 — I4)(x)12

andthis confirms once again the invariance of the hyperbolic metric in In two dimensions the complex conjugate 2 of z is available and in our notation this may be written as

= 1/2.

The familiar expression Ii — 2wI (where z and w are complex numbers) satisfies

Ii ^— = IziHz* — w(

andthis suggests the definition

[u,v] =

IuIIu* —vi Observe that

[u, v]2 = 1u121v12 —2(uv) + 1

= ^{lu — v12} + (1u12 - ^1)(1v12 - 1) (3.4.4) and this shows that

[u, v] = [v, u].

The identity (3.4.4) also shows that if I aI > 1 then Ix — a*I

[x, a*] — 1

ifand only if xl = ^1.Thus

= ^{lx — a*I 1}

I.

[x,a*]

40 3. Möbius Transformations on

and this is the n-dimensional version of the equation

of the unit circle in the complex plane.

Finally, we observe that (3.4.4) together with the invariance expressed by (3.4.3) yields the invariance

— [x,y]2

3

(1

-

^— - (1 ^1x12(1

-

^1y12)

EXERCISE 3.4 I. Show that for x in

p(0, x) =

±

H)

Deducethat if x and yare in then

!x — y12 sinh24p(x,y)

= ^{— X12)(l —} [Use(3.4.3).]

2. Let 4i and be reflections in the spheres S(a, r)aridS(b, t) respectively. Show that these spheres are orthogonal if and only if 4(b) =

3. Use Questions I and 2 to show that if S(a, r) isorthogonal to S(O, 1) and if 4 denotes reflection in S(ci, r)then

sinh 4p(0, = hr

and,for all x,

— al jx — = l/sinh2

§3.5. The General Form of a Möbius Transformation

We shall establish the following characterization of Möbius transformations.

Theorem 3.5.1. Let be a Möbius transformation.

(i) If

= then

= (cx)A,

where is a reflection in some sphere orthogonal to and A is an orthogonal matrix.

§3.5. The General Form of a Mhhius 41

(ii) If çb(oc) = then

= r(xA) + x0, where r > 0, x0 e and A is orthogonal.

(iii) then

4(x) = r(ax)A + xo for some r, x0, A and some reflection a.

Remark. a(x)A denotes a followed by A: the matrix A appears on the right as we are using row vectors.

PROOF. If preserves ^let a be the reflection in the sphere S(a, r) where

a =

= ¹ +

Bycomputation, a(0) = ^a*so

4(a(0)) = ^4(a*) = ^0,

(because 4 preserves inverse points): thus = xA. Replacing x by ax, we obtain (i).

If fixes then, for a suitable

x E—3.(x x0)/r,

the map fixes and and hence also the origin. Now (ii) follows from Theorem 3.4.1. Finally, (iii) follows by applying (ii) to 4a for a suitable reflection a mapping to

The characterization in (iii) leads to the notion of an isometric sphere.

Suppose that so that

= r(ax)A + x0,

where a is the reflection in some sphere S(a, t) and (necessarily) a = 1(cx).

By (3.1.5),

14(x) ^— = rlo(x)^—

—

-

^x

a Euclidean isometry on the sphere with equation Ix — aI

= ^t1 where t1 = Indeed,

lim ^—

y-.x Iy—xI

is greater than, equal to or less than one according as x is inside, on or outside S(a, t1). For this reason, S(a, t1) is called the isometric sphere of

42 3. MöbiusTransformationson

Notethat if a denotes reflection in the isometric sphere of 4) then 4)a fixes andalso acts as a Euclidean isometry on the isometric sphere. It follows that the expression in Theorem 3.5.1(u) must take the form

4)a(x) = xA

+ x0,

so in general, we see that

4)(x) =

wherea is the reflection in the isometric sphere and ^isa Euclidean isometry.

In the special case when 4)preserves thereflection a in Theorem 3.5.1(i) must be the reflection in the isometric sphere of 4)^asa and A act as Euclidean isometries on this sphere. We deduce that in this case, the isometric sphere is orthogonal to

EXERCISE 3.5

1. Show that if 4 preserves then the Euclidean radius of the isometric sphere of

isI /(Slflh 4i0).

2. Show that if is the isometric sphere of then is the isometric sphere of 4 -

§3.6.

Distortion Theorems

We prove two sharp distortion theorems for Möbius transformations.

Theorem 3.6.1. Let 4)bea Möbius transformation acting in and let p be the hyperbolic metric in ⁺ Then

d(4)x, çby))

sup _.11 = exp _+1'

4) satisfies a Lipschitz condition on with respect to the chordal metric d and actually exhibits the best Lipschitz constant in terms of 4) actingon the hyperbolic space ⁺

Thesecond result shows that if a family of Möbius transformations omits two values and in a domain D, then the family is equicontinuous on compact subsets of D: this enables one to develop, for example, the theory of normal families for

Theorem 3.6.2. Let D be a subdomain of fr and suppose that and are distinct points in _{If 4)}in does nor assume the values and in D, then for all x andy in D,

8d(x, y)

d(q5x,

aD)"2d(y, The constant 8 is best possible.

13.6. Distortion Theorems 43

PROOF OF THEOREM 3.6.1. By reflecting in ₊ = 0and applying stereographic projection, we may assume that 4) ^{preservesB" +} now we need to show that

—

sup = exp p(O,^4)0).

x,YGS" Ix —

ByTheorem 3.5.1(i), the Euclidean distortion under 4) isthe same as the distortion under the reflection a in the isometric sphere S(a, r) of 4). This is

maximal (as a limiting value) at the point of S" closest to the centre a of S(a, r). Thus from (3.1.5),

— r2

sup ₂

x,yeS" — — 1)

-

^{lal + 1}

- ^{- 1'}

becauseS(a, r) is orthogonal to S" (Section 3.5). Now

= ⁼

andso the supremum is

1 — expp(u, 4) ( ))

= exp p(4)O, 0).

E

PROOFOF THEOREM 3.6.2. Suppose that x andy are distinct points in D and that and flare distinct points outside of D. By Theorem 3.2.7, the product

[x, cz,y, 13]. [x, /3, y,

of cross-ratios is invariant under 4). Thus

[d(4)x,cby)12

[

^d(a,f3)

12[

¹⁶

[

^d(x,

y) ]

^— 4)13)] [d(x,n)d(x, f3)d(y, ^/3)

[

⁴ ]2[ _{1 1}

1[

^{1 1}

4)13)] [d(x, + d(x, /3)] [d(y, + d(y, /3) 64

4)/3)2d(x, öD)

The inequality follows by writing ₌ and /3 = 4) 1(c)

Toshow that the constant 8 cannot be improved, consider 4)(z) = ^z + 2m acting on C with D = ^{— {co,} —m}. Clearly, 4) omits the values X and

minDandifx =

—2m,we have

lim d(4)x, 4)y) ⁸

d(x,y) d(cc, m)d(x, 3D)' +cc.

44 3. MObius Transformations on W

As an application of Theorem 3.6.2, we mention (briefly) the concept of a normal family. A family F of functions from one metric space (X, d) to another, say to (X', d'), is equicontinuous on X if and only if for every positive e there is a positive 6 such that for all x and y in X and all f in F,

d'(fx, fy) < e

whenever d(x, y) < 6.

Each function in an equicontinuous family is uniformly continuous on X and the uniformity is with respect to f as well as to the pair (x, y).

A family F (as above) is said to be normal in X if every sequencef1, f2, chosen from F has a subsequence that converges uniformly on each compact subset of X. There is a general result (the Arzela—Ascoli Theorem) which relates the concepts of equicontinuity and normal families. In the context in which we are primarily interested, it is sufficient to obtain the following special case.

Proposition 3.6.3. A family F of Möbius transformations of d) onto itself is normal in a subdomain D of it is equicontinuous on everycompact subset ofD.

PROOF.We only sketch the proof as the interested reader can find a proof of the Arzela—Ascoli Theorem elsewhere in the literature. Find a sequence x1, x2, .^..whichis dense in D. Given a sequence .. ^.in F we can find (because is compact) a subsequence which converges at x1, then a sub- sequence of this which converges at x2 and so on. By choosing a subsequence of the suitably, we can obtain a subsequence which is ultimately a sub- sequence of each of these chosen subsequences: thus we have constructed a subsequence which converges at each point

Now take any compact subset K of D and consider any positive c. We can cover K by a finite number of open balls (in the d-metric) of radius 6 (corresponding to a in the definition of equicontinuity). Select one point in each: let the selected points be (after relabelling) x1, x2 If y is in K then d(y, for some j and hence

d(&y, +

+ d(cbmxj,

2a + cbmxj).

For

n, m n0, say, the last term is at most a for all x1, .

.. , hence

3e on K.

We can now combine Theorem 3.6.2 and Proposition 3.6.3.

Theorem 3.6.4. Let D be a subdomain of W' and let F be a family of Möbius transformations. Suppose that for every 4 in F, there are two points in

which are not taken as values of 4) in D and suppose that also,

inD.

TheTopologicalGroup Structure 45

Remark. We can rewrite the inequality in Theorem 3.6.4 as inf [chordal diameter — D)] > 0.

PROOF. We simply apply Theorem 3.6.2 with = = and we find that F is equicontinuous (in fact, it satisfies a uniform Lipschitz condition) on

every compact subset of D. LII

Finally, this leads to the following result.

Theorem 3.6.5. Let be Möbius transformations and suppose that

—* for three distinct points x1, x2, x3 and three distinct points

Y2' ji3. Then ,...contains a subsequence which converges on li" to a Möbius transformation.

PROOF. By the deletion of a finite number of the (which clearly does not affect the result) we may assume that for each n, i and j(i j)^wehave

> 0.

It follows that the family {çb1, .} is normal in each of the sets —

(Theorem 3.6.4) and hence in their union, namely Thus there is a subsequence of the converging uniformly to some in 11" and by Theorem 3.2.7 (and its proof), is a Möbius transformation. LI EXERCISE 3.6

1. Show that a family F of Möbius transformations is normal in if and only if +

where (0 0, 1) in If +

2. Prove that if two Möbius transformations are equal on an open subset D of then they are the same transformation on Deduce that if the Möbius transformations converge uniformly to I on some open subset of then they converge uniformly to I on

§3.7. The Topological Group Structure

There are several ways to give the structure of a topological group.

The simplest construction is to observe that the elements of map the compact space onto itself so

D(çb, i/i) = qix): xc

(where d is the chordal metric on W') is a metric on Clearly, 4.

inthis metric if and only if -+ 4 uniformly on

46 3. Möbjus Transformations on

Theorem 3.7.1. is a topological group with respect to the topology induced by the metric D.

PROOF. From Theorem 3.6.1, we see that for each in there is a positive constant such that for all x and y we have

d(4x, çby) c(q5)d(x, y).

Clearly, for any and i,l, we ^{also have}

= D(q51, so

+

D(çb, +

Thisshows that the composition map _p4') is continuous at Similarly,the map cli t—+ -1 is continuous at 4 ^as

= I)

c(cb

Fora different construction of the same topology we proceed as follows.

The group is conjugate in ¹⁾to the group ^of

all Möbius transformations preserving If ^tj in corresponds to in ¹⁾then (by definition of the chordal metric)

= sup{!41x —

Thuswe may consider insteadof GM(W') with the metric (which wecontinue to denote by D)ofuniform convergence in Euclideanterms on S"

and the conjugation is then an isometry between and ^1).

For each non-zero a in ¹ let be the reflection in the sphere with

centre a* that is orthogonal to

thus preserves ¹ and aa(a) = 0.

Also, let Tadenotethe reflection in the plane x^.a = 0. Then, defining Ta to

be the composition ;aa, we find that the isometry Ta of

leaves the Euclidean diameter through a ^invariantand T0(a) = 0. We call any isometry

Ta constructedin this way a pure translation: if a = 0 we define Ta to be the identity.

Lemma 3.7.2. (i) Themap 4''—p 4(0) of ^1)onto is continuous

(ii) ^The map a Ta isa homeomorphism of onto the set of pure transla- tions.

PROOF. To ^{prove (i)}

we suppose first that I) <a. Each Euclidean

diameter of ¹ is mapped by &to a circular arc (orthogonal to in whose end-points are at most a distance a from those of We

§3.7. The Topological Group Structure 47

deduce that the Euclidean cylinder with axis L3 and radius of cross-section r contains Thus

&(O) =

fl c fl

= < e}.

Thisshows that if 4.,, —+ I uniformly on Sn, then -+0:in fact, I).

Suppose now that tb,, (as ¹⁾ and are metric spaces,

it is sufficient to consider sequential convergence). From Theorem 3.7.1 we have - ^-# I:thus (from above) -+0and hence

This proves (i).

To prove (ii) observe first that the map ^{F—* 1} iscontinuous (Theorem 3.7.1). By (i), the composite map

TQ1 namely, a,is continuous.

It remains to prove that the map a

7, is continuous: explicitly, as b -+aso Tb uniformly on Sn. We have explicit formulae for

and;

and the continuity follows from straightforward (if tedious) estimates: we

omit the details.

E

We know from Theorem 3.5.1 that every element of ^I) canbe expressed uniquely as

=

where a = '(0) and A is an orthogonal matrix (A acts after it appears on the right because we are using row vectors). It follows that we can also write (uniquely)

=

where A4, (namely, ; followed by A) is also an orthogonal matrix and this description establishes a natural bijection between ¹⁾andO(n + 1)

x by the correspondence

a =

Now the group O(n + 1) of orthogonal matrices is itself a metric space.

First, there is the natural metric

-

1/2

=

-

48 3. Möbius Transformations on

and second, there is the metric D induced by regarding O(n + 1) as a subset of In fact, these metrics yield the same topology because if A =

B =

C =

A — Band x is on S", then D(A,B)2 = sup xA —

= SUp + +

1x11 j1

sup

x11

j1 i1 ⁱ¹

= A ^— i=1 j=1

= ^{— e1Bl2}

nD(A, B)2.

The space O(n + 1) x now inherits a natural product topology and we have the following result.

Theorem 3.7.3. The bijection çb t—' (Ag,, a) is a homeomorphism of ontoO(n + 1) x

PROOF. The proof consists of repeated applications of Theorem 3.7.1 and Lemma 3.7.2. First, a t—÷ is continuous, hence so is the map (Ag,, a) t—+

Ta). Also the map of Ti,) into their composition, namely 4), is

continuous thus so is the map (A a) 4).

Next,

4

T'

is continuous. We deduce that the composition

4) (A a).

Remark. Theorem 3.7.3 simply means that the topology on ¹⁾ induced by the bijection from O(n + 1) x coincides with the topology induced by the metric D. As has been identified isometrically with 1), this result providesa new construction for the topology induced on by the metric D.

For our third and final construction of the topology we need another model of hyperbolic space.

Definition 3.7.4. Let Qbethe hyperboloid model defined by

Q = 1,x0 >O},

§3.7. The Topological Group Structure 49

where

q(x, y) = XoYo ^— (x1y1 + +

Observe that Q is one sheet of a hyperboloid of two sheets and that if xe Q then

so, in fact, x0 1.

Now let y = (Yo y,,) be any smooth curve on Q. Thus for all t, y0(t)2 = y1(t)2 + + + 1,

so differentiating,

= +

+

(more briefly, q(y, y) = ¹ so q(y, = 0).We deduce that

= + +

+

+

-

(L

=

0,

the summations being overj = ^1,

...,

^n.Observe also that a strict inequality holds unless = = = 0in which case,

=

⁰also. It follows that we can construct a metric on Q in the usual way by the line element

ds2 =

+ ...

+ ^{— (3.7.1)}

the distance between two points on Q being the infimum of dt

over all curves joining the two points. The associated metric topology is the Euclidean topology on Q. We shall now compare Q and this metric with the model and the metric

4 dx2

ds2 (1 — ¹²⁾² (3.7.2)

Theorem 3.7.5. The map

( ^x1

1+x0

is an isometry of Q with the metric (3.7.1) onto with the metric (3.7.2).

50 3. Möbius Transformations on PROOF. For brevity, we write

I

^{xi xn}

(Ys,..., = ( ^{, .}

l+x0

and denote the vectors by x and yin the obvious way. As x e Q,acomputation yields

2 = ^—_1,

(33)

x0 + 1

so 0

^< 1 and F maps Q into

By direct computation we find that the map

±

₁

1y12

(3.7.4) is indeed the inverse of F and so F is a bijection of Qonto

To verify that F is an isometry, we observe that

d ^{— —}

1 + x0

(1 +

Thus, using this and (3.7.3) we have

+ ...

+

— 1 2

(

^{xj dx0}

\

²

—( +x0) +x0(1+x0)2)

— V d 2 V ²

— + _/1 ^Xf

r X0)

j1

^{m X0}

=

+

^{(xo — 1)d4 — dx0 — 1)}

j=1

(o+

¹

1+x0

—

Itis now clear that the group G(Q) of isometries of Qandthe group of isometries of are isomorphic by virtue of the relation

=

F(G(Q))F1

Our aim now is to prove an alternative characterization of G(Q) and hence of

Theorem 3.7.6. The isometries of Q are

precisely the (n + 1) x (n + 1)

matrices which preserve both the quadratic form q(x, x) and the half-space given by x0 > 0.

PROOF. First, let A be any matrix with the prescribed properties. As x0 > 0 is preserved and as

q(xA, xA) = q(x,

x) =

^1,

§3.7. The Topological Group Structure 51

when x e Q we see that A preserves Q. Moreover, for any curve on Q, let

r =

^yA. Then F = ^so

q(t, t)

=

and this simply expresses the fact that y and yA have the same length. Thus each such A is an isometry of Q onto itself.

It remains to show that every 4) in is of the form for some such matrix A and to do this, we simply compute the action of F(A)F -¹

on Suppose then that A = where i,j = 0, 1 ii. Withthe obvious notation, we write

(Yi (u0,

u1,..,

(vs,^i.'1

F- (w1 wa).

Now

(v0 = (u0

V3 = 4- +

Using (3.7.4), this yields

(1 — = (1 + y12)a01 ^{+ 2(y1a13} +

... +

Thus

w.=

_' V.

1+v0

-

^{(1 —}

-

^{(1 — + (1 —}

— (1 + y12)a03 + 2(y1a13 + -• +

(375

— 1y12(aoo — 1) + 2(y1a10 + ... + + ¹⁾ and this is the explicit expression for the map F(A)F '.

If A0 is an orthogonal ii x it matrix(viewed as an isometry of B"), then

A=(0

0

0)

preserves q and the condition x0 > 0. In this case, (3.7.5) yields w = yA0

and so every isometry of B" which fixes the origin does arise in the form

-

52 3. Mäbius Transformations on

It is only necessary to show now that the reflection in the sphere r)

orthogonal to ¹ isof the form F(A)F'. Because orthogonal transforma- tions are of this form, we need only consider the case when is of the form (s, 0 0). It is actually more convenient to introduce another positive parameter t with

cosh t

= (c(t),0 0), c(t)

= sinh t

and

r =

1/sinht,

so the orthogonality requirement = 1 + r2 is satisfied.

Consider now the matrix

/

^{cosh2t sinh 2t 0}

^...

/ —sinh 2t —cosh 2t 0

...

⁰

0 0

\o

⁰ In—I

observe that det(P) = — 1

and that P preserves both the quadratic form

q(x, x) and the half-space x0 > 0. The effect y w

of F(A)F1 on

^is given by (3.7.5) and the denominator of this expression can be simplified as follows:

1y12(aoo

— l)+

2(y1a10 +

...

+(a00 + 1)

= 2Jy12 ^sinh2t — 2Yi sinh(2t) + 2 cosh2 t

=

^{2jy — sinh2 t}

=

Nowfor] = 2,

...,

ⁿ the formula (3.7.5) yields

—

ryj

2 WJ

—

Also,

(1 + 1y12)sinh(2t)_{— 2Yi} cosh(2t)

1 21y—Cl2sinh2t

— sinh(2t)[fy^—

+ 1— + 2(y.ç)] —

2y1[2cosh2t —1]

= c(t) +

r2

— c(t)).

§3.7. The Topological Group Structure 53

This proves that F(P)F ¹ is

+ ^{r2(y —}

thatis, the reflection in r). U

In view of Theorem 3.7.6, we examine briefly the group 0(1, n) of matrices which preserve the quadratic form q(x, x). If A n 0(1, n), then

q(x, x) = q(xA,xA),

so

I4JA = J, (3.7.6)

where

(10. ^0).

We deduce that det(A)2 = 1: the subgroup of 0(1, 11) withdeterminant 1 is S0(l, ii).

Next,we show that the set of matrices A in 0(1, n) with a00 > 0 is also a subgroup. We denote this subgroup by n) with

n) = S0(1,

n) n 01(1,

Supposethat the matrices A, B and C satisfy a00 > 0, b00 > 0 and C =AB:

then

c00 = a00b00

+ ...

+

a00boo Iao1b1o + ... +

ao0b00 —

+ ...

+ + +

Because of (3.7.6), we have

(a00, —a01,...,

a01,...,

= 1,

so

Taking the transpose of both sides of (3.7.6) after replacing A with B yields

so C00 > 0.

Finally, the inverse of A (= is (JAJ)t because A(JAJ)' = AJAtJ

j2

= I.

54 3. Mdbjus Transformations _on

Thus A preserves the condition a00 >0 and so

n) IS indeed

a group. Observe that an element A of 0(1, n) leaves the hyperbolojd of two sheets {x: q(x, x) = 1}invariant:the component Q is A-invariant if

and only

if a00 > 0.

We have proved that the isometries of Q are precisely the elements of

Ot(1,n) and that in the isomorphism A

of

n) onto

the subgroup n) corresponds exactly to the directly

con-

formal elements of (in the proof of Theorem 3.7.6, each reflection corresponds to a matrix of determinant —1).We can now induce a topology on by transferring the natural topology from n) to

and it is not hard to see that convergence of matrices in 0 ÷ (1, n) corresponds exactly to uniform convergence on

thus this topology agrees with

those previously constructed. Reverting back to

we have proved

the following result.

Theorem 3.7.7. with the topology of

convergence in the

chordal metric is isomorphic as a topological group to the group (1, n ±

1)

ofmatrices.

In particular, if we identify with the extended complex plane, then is (as we shall see) the class of complex Möbius transformations

cz + d

and this is isomorphic to the Lorentz group of matrices preserving both the quadratic form + + ^— t2 and the inequality t > 0.

EXERCISE 3.7

1. Show that if the Möbius transformations _preserve ⁺ _andif —. 1 uniformly on some relatively open subset of then ^—* I uniformly on and on S's..

[Identify with and consider convergence on first.]

2. Suppose that n =2so that Q in Definition 37.4 lies in Show that the geodesics in B2 through the origin correspond via F and F to the intersections of Q with certain planes through the origin in

§3.8.

Notes

For recent treatments of Möbius transformations in see

[5], [101] and

[1101: for shorter works see (for example) [3], [33] and [108]. A

more

algebraic treatment based on quadratic forms is

given in [19]. Theorem

3.1.5 is well documented: see, for instance, [36], _{p. 133.}

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