A Beckman-Quarles type theorem for finite desarguesian planes

0047-2468/82/010089-0551.50+0.20/0 9 Birkh~user Verlag, Basel

J o u r n a l of G e o m e t r y V o l . 1 9 (1982)

A BECKMA_N-QUARLES

Walter

TYPE

THEOREM

FOR FINITE

DESARGUESIAN

PLANES

Benz

Dedicated

to G.

Pickert

on the

occasion

of h i s

65. b i r t h d a y .

Abstract: The characterization t h e o r e m g i v e n in [2] for L o r e n t z transformations of t h e ~ a is c a r r i e d o v e r to t h e c a s e of f i n i t e planes. Consider

the plane

F 2 over

Lorentz-Minkowski-distance Q=

(ql,q~)

the G a l o i s between

o f F ~ is d e f i n e d d(P,Q)

:=

field

the

points

_

s e t of all n o n - e m p t y

result

of t h i s

is

M

: Given

fixed

elements

Then

~PT

plane

in

subsets

of F 2. T h e m a i n

Given

a mapping

S(F 2)

that

Vp,Q s F 2

The

P = (pl,pa),

a,b:s F ~ := F ~ {O}.

T : F2+ such

The

(q~ _ p ~ ) 2

b y S(F 2) t h e

Theorem

q = pn.

by

(ql _ p i ) 2

Denote

note

F = GF(q),

a | for a l l

F a if o n e

P s

of the

p 9 2,3,5,7

b)

p6

c)

q=

d)

p = 7, n o d d a n d

[8],

{5,7}

2 and

= a

implies

T is a b i j e c t i v e conditions

d(P',Q')

= b.

collineation

of the

holds:

n even

7

GF(7),

who was

proved

a square

and

d(P,Q)

following

a)

cases

M are

Vp, s pT,Q, 6 QT

in

GF(11)

3/n

.

of this

theorem

applying

a computer.

[3] w i t h

the i n f i n i t e l y

in F, b u t n o t

-11.

Our proof

were

treated

The other

in

many [3]

by H . - J . S a m a g a

results

exceptions in c a s e

of t h e o r e m that

that

-3 is

-3 is n o t

Benz

90

a square

applies

noticed der),

that

q= 5

Different

an i m p o r t a n t

theorem

(H.-J.Samaga,[8]), authors

have

Beckman,

D.A.Quarles

minyh

[5]

in

J.Lester

for the

1.

For

P,Q s

We

like

real

PQ

Then

Given

:=

ping

Beckman-Quarles

[I] for the

real

hyperbolic

Type

euclidean

case,

It m i g h t

case,

Theorems: case,

W.Benz

in

be

E.M.Schr~-

F.S.

A.V.Kuz'-

[2],[4]

B.Farrahi

and

(unpub-

case.

theorem

~ : F2§ Q,s Qa

Consider

M is a c o n s e q u e n c e

2) s u c h

that

PQ = 1

implies

P 6F 2 and

conditions

a),

bijection

c),

d)

a

:

a ~ xl

- x2)

a by b in t h i s ~ = a T 8 -I . H e n c e

F 2

§

collineation

of F 2

e of F ~ ,

a ( ~ xl + x2,

M define

= 1

holds.

8, r e p l a c i n g

T of t h e o r e m

of

P'Q'

~ is a b i j e c t i v e

b),

the b i j e c t i o n

(x1,x~) ~ := and the

([3]).

[9].

(H.Schaeffer,

(ql - Pl ) (q~ - P~)

that

I for all

if one of t h e In fact!

q= 9

elliptic

~ Vp, 6 pO,

~PO=

for q = 3

2 define

to v e r i f y

Vp,Q s

in

of G . T a l l i n i ,

true

[6] for the r e a l m i n k o w s k i a n

lished)

A:

proved

for the r e a l

in

Theorem

result

M is not

,

formula.

Given

a map-

S(F 2)

2

Consider

P,Q s F

with

d ( P ~ , Q ~) = a and d((P') 8 (Q')8) = b By'means

PQ = I a n d m o r e o v e r

(P')86 since

of t h e o r e m

P' 6 P~,

Q' 6 Qo.

(P~)~ , ( Q ), ~ s (Q~) T . This

implies

T is a m a p p i n g

A now

of t h e o r e m

~ is a c o l l i n e a t i o n

and

M.

Hence

Hence

thus

P'Q' = 1

also

--I

= ~

a 8

9

Furthermore Theorem

we

like

B: G i v e n

to v e r i f y

~ : F2 §

2 such

V p , Q E F 2 P-~= I Then a),

~ is a b i j e c t i v e b),

In fact! point

c),

P such

mappings

d)

Assume

that

theorem

A is a c o n s e q u e n c e

of

that

implies

collineation

P~QO

=

I

of F 2 if one

of the

conditions

holds. there

that ~P

would

exist

a mapping

a~ 2 . Consider

points

a of t h e o r e m VsW

A and a

of pa a n d d e f i n e

Benz

91

oI,~ 2 : F2+ F 2 as

follows: X

gl

Hence,

by

X~2

=

P~

= V,

X~

s

for all X~P,

pS2 = W

theorem

B,

.

the mappings

F 2 . But there do not exist oI o 1 P * P~= and X = X ~= So w h a t

one actually

theorem

B.

It m i g h t

for Galois Theorem # 2,3

fields

Vp,Q s Then

c a s e d) square

and

being

2

the

Now

-11

being

exclude Lemma

following

to

theorem

show

theorem

M is

B is a g e n e r a l i z a t i o n

([7])

if p g Q o =

I

of F 2

of

[3] it r e m a i n s

fields

F = GF(pn),

to p r o v e p >7,

theorem

such

that

B in -3 is a

-11.

proof

of theorem

-3,

-11

is u s e d

are

only

2 P-~ = 3

in o r d e r

implies

not a square

B in

[3],

squares

part

in F we

II,(section

realize

that

to e x c l u d e . P ~ = Q ~

3), -11

in t h e

p ~ Q a = 3 o r P ~ = QO

we have

to f i n d

some

other

argument

to

pa = Q~

I:

Let

a square

Proof:

F be a finite

in F. T h e n

AB = AC =

3

and

there

Now

It m i g h t

B-~ =

may

assume

of t r a n s i t i v i t y I

be noticed

c h a r F % 2,3.

or

infinite

do n o t

exist

field

such that

points

-11

is n o t

A,B,C 6 F 2 with

BC = I .

We otherwise

x y # O, b e c a u s e tries.

not

that

implication YP,Q6F

in o r d e r

that

P-Q = I if a n d o n l y

in F, b u t

a square

that

a c o m m u t a t i v e f i e l d F of c h a r a c t e r i s t i c a bijection g : F 2 ~ F 2 such that

to r e s u l t s

for t h e c a s e

oI,~ a such

of

Given

for Galois

If w e c h e c k

be collineations

for all X s

to p r o v e

o is a c o l l i n e a t i o n

2. A c c o r d i n g

must

two collineations

be noticed

of t h e

of F.Rado:

and given

9

has

~I,~2

implies that

A =

(O,0),

properties

B =

of the group

the contradiction

-1!

being

(x, ~i,), C =

(5 - 6 ~ ) 2

not a square

3 (y, ~ ) ,

of i s o m e = -11.

in F i m p l i e s

92

Benz

Lemma with

2: L e t

F be a finite

or infinite

field

of characteristic

~2,3

the property

(~) T h e r e

exists

square Given

5

t s

3,4}

9 t2+t_-T~ is a

such that

in F.

then

such that

an element

points PA =

P,Q 6F 2 with

~

= 3 there

exist

points

A,B 6F 2

I a n d A B = BQ = 3.

Proof:

B e c a u s e of 9 2 ta + ~-~_4 = s .

(~) t h e r e

exist

t,s 6 F

with

s ( t-3

I)

t #~,

3,4

t E {3,

~_}

and

Define t-4 x = ~

t_s3 t-4 ( _ + I), y = 2 t - 5

It is x y @ 0 b e c a u s e o t h e r w i s e s2 9 a I = O, i.e. (t-3) 2 = t 2 + ~ (t-3) Furthermore (1 + y - x)(I Because P =

of

3)= x

3.

(O,O) (x, ~ ) = 3 w e m a y

(O,O),

Q =

Lemma

3: L e t

lemma

2 and

a mapping

+ 3 y

(x, x ) .

Now

such

that

g : F2

-3

assume

put A

F be a f i n i t e

or

, i.e.

:=

without

(1,1)

infinite

is a s q u a r e

and B

loss :=

field with

in F,

of g e n e r a l i t y

(1+y,

1+--~). Y

property

but not

-11.

(~) o f

Consider

F = with

vp,Qs F 2

P-Q = I

~

p~Qs =

I .

Vp,Qs

P-Q = 3

~

poQs = 3

Then

Proof:

Consider

we h a v e

char

F~2,3

P,Q 6F 2 with of F.

Put V

e 6 F with and

e2=-3.

thus

P-~ = 3. H e n c e := P +

Q = P+' (z, ~ ) , z % O a s u i t a b l e

(ez,-~'),

W

= QV= QW = I .

This

implies

If pS # Q S

the

last

equations

[3],

II.)

part

Vp,Q s Now assume with

We t h u s

:= P +

(Bz, ~ ) .

P~Vd = PSWS

have

2 P-Q = 3

p S = QS

S i n c e -11 is n o t a s q u a r e 3+~ ~ 3 e e := --6--30, 8 := ~ - - ~ O " Given

= VaW ~ = QSVO = QSWS

implies

2 there

It is A S #

otherwise

B ~ because

exist

points

( o b s e r v e pS = QS)

B ~ PC; = 3 or B s = pS

1

= I. 3,b,

in

p s Q s = 3 or pS = QS

of l e m m a

P-~ = I, A-B = B-Q = 3. T h i s

=

proved

A S B ~ = 3 or A s = B ~,

PSAs =

element

PV=PW=VW

l e a d to P S Q S = 3. (S. s e c t i o n generally

implies

Because

Hence

in F

A,B s F a

poAs =

.

~AsP G = 3 or A s = pS c o n t r a d i c t i n g

I,

Benz

93

It is B ~ # p a Thus

because

otherwise

A--dP~ =

3 or A ~ = P ~

P ~ A ~ = 1, A O B ~ = 3, BOP ~ = 3, w h i c h

Lemma

4:

Given

F = GF (pn)

contradicts

lemma

1.

with

1)

p >7

2)

p = 7, n o d d and 3/n

Then there

.

or

exists

a t ,~

.

,3,4 in F such that t 2 +

is a s q u a r e

in F. Proof:

I)

If 2 is a s q u a r e

If -2 is a square

observe

observe 5 I #~,3,4

6 ,5,3,4 and

and 6 2 + ~ - C ~ 9 12 + ~ = -2.

If 2 and -2 are n o n - s q u a r e s then 2. (-2) m u s t be & square 5 02 9 32 a l s o -1 Now observe O #~,3,4 and + = -( ) 0-4

= 2.(

)2.

and h e n c e

"

2)

In this

case GF(73)

is a s u b f i e l d

of F. The p o l y n o m i a l

t3 + 3 t2 - t - I

is i r r e d u c i b l e o v e r GF(7) and hence has a zero t 9 = 1 is a square in F. in GF(73) . Thus t ~ 5 ,3,4 and t 2 + t-4

Remark: square

In all

GF(7 n) , n odd and 3/n, we h a v e t h a t

-3 is a

b u t not -11.

The proof section

fields

of t h e o r e m

2, part

III,

B in case

d) m a k e s

use of c o n s i d e r a t i o n s

in

[3]

References [1]

Beckman, F.S., Quarles, D.A.: On i s o m e t r i e s Proc. Am. Math. Soc. 4, 810-815 (1953).

of E u c l i d e a n

[2]

Benz, W.: A B e c k m a n - Q u a r l e s T y p e T h e o r e m for Plane T r a n s f o r m a t i o n s . Math. Z. 177, 101-106 (1981).

[3]

Benz, W.: On m a p p i n g s p r e s e r v i n g a single L o r e n t z - M i n k o w s k i distance. I,II,III. I (Proc. Conf. in m e m o r i a m B e n i a m i n o Segre, R o m e 1981), II,III (To a p p e a r in J o u r n a l of Geometry).

[4]

Benz, W.: Eine B e c k m a n ~ Q ~ a r l e s - C h a r a k t e r i s i e r u n g der Lorentzt r a n s f o r m a t i o n e n des ~ . A r c h i v d.Math. 34, 550-559 (1980).

[5]

Kuz'minyh, A.V.: On a c h a r a c t e r i s t i c p r o p e r t y of i s o m e t r i c m a p p i n g s . Dokl. Akad. Nauk. SSSR, T o m 226, Nr.1 (1976).

~6]

Lester, J.: The B e c k m a n - Q u a r l e s T h e o r e m in M i n k o w s k i Space a S p a c e l i k e S q u a r e , d i s t a n c e . To a p p e a r A r c h i v d. Math.

[7]

Rado, F.: On the c h a r a c t e r i z a t i o n of plane R e s u l t a t e d. Math. 3, 70-73 (1980).

[8]

S ~ m a g a , H . - J . : Zur K e n n z e i c h n u n g von L o r e n t z t r a n s f o r m a t i o n e n e n d l i c h e n Ebenen. To appear J o u r n a l of Geometry.

[9]

Tallini, G.: On a t h e o r e m by W.Benz c h a r a c t e r i z i n g plane L o r e n t z T r a n s f o r m a t i o n s in J ~ r n e f e l t ' s World. To appear J o u r n . o f Geometry.

affine

Lorentz

13

!Eingegangen

am 2.Oktober

for

isometries.

Mathematisches Seminar der Universit~t Hamburg B u n d e s s t r . 55 0-2000 Hamburg

spaces.

1981)

in