A Beckman Quarles Type Theorem for Plane Lorentz Transformations

rvlathematische Zeitschrift

Math. Z. 177, 101-106 (1981)

9 Springer-Vertag t981

A Beckman Quarles Type Theorem for Plane Lorentz Transformations Walter Benz Mathematisches Seminar der Universitiit, Bundesstr. 55, D-2000 Hamburg, Federal Republic of Germany

1. The result of this note is

Theorem A. Given a fixed real number p ~=0 and given a mapping ~ of IR 2 into the set Po(IR2) of all non empty subsets of IR2 such that AB = p implies A'B' = p for all A, B M R 2, A ' ~ A ~, B'6BL Then a must be a Lorentz transformation. Here AB denotes the pseudo euclidean distance ( b l - a l ) 2 - ( b 2 - a 2 ) 2 for A = ( a l , a 2 ) , B =(bl,b2).

2. Our Theorem A corresponds to a result of F.S. Beckman and D.A. Quarles [1], concerning euclidean isometries. A similar characterization for hyperbolic isometries was established by A.D. Kuz'minyh [2]. The best known result before Theorem A for plane Lorentz transformations is due to E.M. Schr6der [4] (p. 144, Satz 6):

TheoremB. Given f i x e d real numbers a,b such that a~=O and b/a<=88 Given moreover a mapping o-: IR2--*Po(IR2) such that AB = p implies A ' B ' = p for all A, B M R 2, A' ~A ~, B' 6B ~ and pc{a, b}. Then a must be a Lorentz transformation.

We also refer to a paper of F. Rado [3], where the field case is treated for bijective mappings o- leaving invariant distance 1 in both directions. Since there exist no equilateral triangles in the pseudo euclidean geometry, the proof of Beckman, Quarles cannot be carried over to this case. - Schr6der's Theorem B will be used in our proof of Theorem A in Sect. 6. 3. Changing the coordinate system we shall replace the distance (b 1 - a l ) 2 - ( b 2 -az) 2 ofA, B b y AB := (b I - al) (b2 - a2)

(1)

and p by 1 without loss of generality. We also assume (0, 0)e(0, 0)L

0025-5874/81/0177/0101/$01.20

102

W. Benz

L e m m a l . Given x l , x 2 e l R * : = l R \ { 0 } such that xl+x24:0. Then there exist z, teN* satisfying z+t:~O and

(xl + x2 + z +t) ( i + l-+l+ q

x2 7 7I = 1

(2)

Proof. Put ct:--x, + x 2 + 0 , fl: = x I x 2 4:0. If [a[ > 0 is sufficiently large then fl a2 + c~fl a b:=fi_e2_aa is approximately equal to -fi-a. In the cases fi-> 0, - f i - > 0 choose respectively a > 0 , - a > 0 sufficiently large such that b<0. Hence a e > 4 b and there exist z, teN*, z + t ~ O , satisfying z + t = a , z t = b . Thus (2) holds. Lemma 2. Consider xl,xeelR* such that xl +xz~O. Then (0,0)r ~, where P: = ( x 1+x2, 1~+~1 t X1

X2]"

Proof Take z, t according to Lemma 1 and put 0 = (0, 0), 1

1

1\

2+2+1_+!t

Hence PA = 1, AB = 1, OB =1. Consider A ' e A ~ B'eB ~. Assume 0eP ~ Because of 0e0 ~ we get 0 A ' = 1, A ' B ' = 1, 0 B ' = 1. But there is no equilateral triangle in IRz with respect to pseudo euclidean distances. Lemma 3. Given x, yeIR* such that x ~=y. Then c~

(,7 Y

=0.

Proof. Put x l : = x, x2:= - y . Then take z, t according to Lemma 1. Put 1

1

Hence PA = 1, BA = 1, QB = 1. Now assume there would exist TeP~c~Q ~. Then TA' =1, B ' A ' = I , TB' =1, where A ' e A ~, B'eB ~. But there is no equilateral triangle in IR2. 4. Denote

x,

xelR*

}

by K. Because of 0e0 ~ we get

(1;

x=#0. Let q~: IR*--,IR* be a mapping such that ( q ~ ( x ) , ~ )

x,

c K for all

( x, 1)

~ for all

xelR*. Denote by 9 the set of all those mappings. Because of Lemma 3 all (peq) are injective. Let F: IR2~IR 2 be a mapping, F(x, y)=: (f(x, y), g(x, y)), such that F(x, y)e(x, y)~. Denote by A the set of all those mappings.

A Beckman Quarles Type Theorem

103

Lemma 4. Given qo~gP, FcA and given xl, x2~IR such that O, x 2, x 2 are pairwise distinct. Then F (x I -~-X2, l + l t = ( g o ( x l ) + g o ( x 2 ) , x l x2I Proof. Put P:= x 1

1

+

~

1 ~ ~2)]"

t, er=:(f,g). Because of X 1

xi,7

P=I

for i

X 2]

= 1 , 2 we get (f-go(xl)) ( g- ~'i ))

=1 for i = 1 , 2 and hence (put goi:=go(xi))

(3)

f =g (ol go2

because of xl 4=xz and the injectivity of go. Thus

(4)

g2gol go2 = g (-;91q - g go 2 .

Assume g =0. Then f = 0 by means of (3). But 0~W according to Lemma 2. Thus g # 0 and Lemma 4 is established taking (4), (3) into account. Lemma 5. F(0, 0) = (0, 0) for all F~A. 1

4

~

Proof. We have to prove 0~ {0}. Assume 0 # Ae0 r and take goeqs. Since 0(x, ~) $

=1 for Mix 4=0 we get A (g0(x), ~--~x)) = 1. Because of the injecfivity of go we get AP = 1 for infinitely many PeK. But this implies A=0.

Lemma 6. Given go~q~, F~A and given xl, x2, x3EIR such that O, x 2, x 2, x~ are pairwise distinct. 7hen [

1

1

1\

=

t[go(X1)_I_go(X2)_i_go(X3),

1

1

+

1 \

9

Proof. Put 1

1

1

and

F (x i + x., 1 + 1 ] = (fu, gu) -

'xi

t

xj]

for i *j.

Hence ( f - f / j ) (g - gij) = 1 for i , j and thus (observe f - f o =~0) 1

1

1

for i#:j

(5)

f-goi f-cp~ = e j- ( f - ~ j - p,) p~. ( f - e ~ - p~)

(6)

g=--+--+ goi go~ f--qh-goj according to Lemma 4. This implies

104

W. Benz

for {i,j, k} = { 1, 2, 3}. Assume f - CO~= 0 = f injectivity of co. Thus (6) implies

COz- Then co1 = co2 contracting the

co~' ( f - co~- co,)= cok" ( f - cok- coO

(7)

for a certain i t ( l , 2} and (j, k} = {1, 2, 3 } \ {i}. But (7) leads to f=co~ +coz+co3. N o w observe (5). L e m m a 7 . Given (a,b)MR 2 such that (a,b)4:(0,0) and ab4= l. Then there exist infinitely many (x, y, z)6IR 3 such that (i)

O, x 2, y2, z 2 are pairwise distinct,

(ii)

a=x+y+z,

1

1

1

b =-+-+-. x y z

M o r e o v e r : I f (x, y, z), (x', y', z') are triples with these properties then either {x, y, z}

={x',y',z'} or { x , y , z } ~ { x ' , / , z ' } = O . Proof. There exist infinitely many z~IR* such that 1-4=b, a - z < 0 and b z 2 - 8 9 z 1 b-2

+ a b)z + a 4:0. Consider an infinite set F of those z. Take z e F and put

a-- z

u-

2

'

v=

]~

a-z

u2 - -

b--

1"

z

Then (u + v, u - v, z) has properties (i), (ii). Given now triples ( x , y , z ) , (x',y',z') fulfilling (i), (ii) such that {x, y, z} c~ {x', y', z'} 4= f). Assume z---z' without loss of generality. But then {x, y} = {x', y'} according to (ii), (i). L e m m a 8. CO(-x)= -CO(x)for all C O ~ and x M R * . P r o o f Given coeqo, F s A put F(x, y)=~ ( f ( x , y), g(x, y)). Given xMR* put x 1 = x ( 2 + 1/3), x 2 = x ( 2 - ] ~ ) . Hence 0, x 2, x22, ( - x ) 2 are pairwise distinct. Thus

4x = x 1 + x 2 ,

4 x

-

1 x~

1 4---, x2

3

3x=xl+x2+(-x),

x

-

1 x1

1 1 I---+-x 2 -x

imply

s(4x4)_ by L e m m a s 4, 6. Put y~ = 8 9 and 3x=yl+y2,

y2= 89 3 x

-

1 Yl

1 ~___, Y;

(3x3)=

x,

,8,

). Thus 0, yZ, y2, x ~ are pairwise distinct

4x=yl+y2+x,

4 x

-

1 Yl

1 1 +__+_ Y2 x

A Beckman Quarles Type Theorem

105

imply

again by Lemmas 4, 6. Hence c p ( - x ) = -qo(x). Lemma 9. The cardinality of P" is 1 for all PaIR 2.

Proof (9) is valid for all FeA, (pe~b, xelR*. Working with the same F in (9) we get ~o(x)=0(x) for all q0, 0 ~ b and x~IR*. Thus the cardinality of ~b is 1. This implies card X,x

=1 f o r x + 0 s i n c e

X,x

c K . By Lemma 5 we have card 0"

=1. Given now (a,b)ElR 2 such that (a, b) + (0, 0) and a b + l (x, y, z)EIR 3 such that (i)

0, x 2, y2, z 2 are pairwise distinct.

(ii)

a=x+y+z,

there exists

b = l - + - l + -1, x y z

according to Lemma7. Since L e m m a 6 is valid for all F~A we get card(a, b)~

=card(x+y+z,l+ts x

y

z/

5. Because of L e m m a 9 we know that our mapping a : IR2--~Po(IRz) can be written as a mapping a : ]R2----~]R 2. SO we shall write (x,y)~ /

4

~

=(f(x, y), g(x, y))in the sequel. Observe (0, 0)~ =(01 0 ) a n d F (x, xt-)= (cp(x), (o~--~) T for

\

/-

x+0. A similar proof as that of Lemma 6 leads to

Lemma 10. Given x 1, x2, x3, x4~lR such that O, x 2, x22, x 23, x2 are pairwise distinct.

Then F i Xi, i= 1 =

=

i=1

~o(xi),

i=1

Lemma 11. Given x, y, zelR such that z ~=O. Then

f (x + z, y + ~ ) - f (x, y) = cp(z),

(10)

1

g (x + z, y + z ) - g ( x , y)=qol(z).

(11)

/

t\

\

z]

Proof We only need to prove (10)since (x,y)Ix + z , y + ~ ) =1. Observe that (10) is valid for x = y = 0. Case x y = 1. Obviously, Lemma 4 yields (10) in case x z +z 2. For z = - x we make use of Lemma8. Consider now z=x. According to L e m m a 7 we find (xl,x2, x3)elR 3

106

W. Benz

such that (i)

0, x 2, x 2, x~ are pairwise distinct

(ii)

2X =X1 q-X2 q-X3,

2

1

--

X

X1

1 1 "k-- q - - - . X2

X3

Observing also the second statement of L e m m a 7 we can assume

x2

2

Hence f (2x+(-x),

by L e m m a s 10, 6, 8. and x y = 1.

2 l x ) - f (2x, 2)=~o(- x)= -qffx) x+

1 2\ Whusf{2x, c.)=2~o(x)and(lO)isestablishedincasez=x \ A/

Case (x, y) ,t= (0, 0) and

xy+l.

Again working with L e m m a 7,

X=XI @X2-I-X3,

1

1

1

X1

X2

X3

y=--+--+--,

z2 {x ,

2 X 32} , N2~

we get (10) by applying L e m m a s 10, 6. 6. We are now in the position to finish the p r o o f of T h e o r e m A . Given a

AB = n 2 implies A~B~=n; for all A, BMR2. In order to put A=:(a,b), B=:(p,q), x::~(p-a), Then B=(a+nx, b ft \

positive integer n then

4

prove this statement +x

because of

AB = n 2. By (10) and

/

induction

n f (a+nx, b+ n) :f(a, b)+n~o(x), g (a+nx, b+ n) = g ( a , b)+qg~x ).

A~B~ 2.

Hence N o w according to T h e o r e m B of E.M. Schr6der a must be a Lorentz transformation.

References

1. Beckman, F.S., Quarles, D.A.: On isometries of Euciidean spaces. Proc. Am. Math. Soc. 4, 810815 (1953) 2. Kuz'minyh, A.V.: On a characteristic property of isometric mappings. Dokl. Akad. Nauk. SSSR, Tom 226, Nr. 1 (1976) 3. Rado, F.: On the characterization of plane anne isometries. Resultate d. Math. 3, 70-73 (1980) 4. Schr6der, E.M.: Zur Kennzeichnung der Lorentz-Transformationen. Aequationes Math. 19, 134144 (1979) Received July 21, 1980; in final form October 15, 1980