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On Mappings Preserving a Single Lorentz-Minkowski-Distance. I
Annals of Discrete Mathematics 18 (1983) 61-76 0 North-Holland Publishing Company
ON MAPPINGS PRESERVING A SINGLE
61
LORENTZ-MINKOWSKI-DISTANCE. I
Walter Benz
Consider a comnutative f i e l d K and denote by K2 the s e t o f a l l ordered p a i r s o f elements o f K .
distance
fl of
The elements o f K2 are c a l l e d points o f the plane K 2 ,
the p o i n t s P
fl:=
The f o l l o w i n g r e s u l t was proved by F. Rado, [ 3 1 :
(ql-pl)(q2-pz).
THEOREM: Assume char K
V
( p l , p 2 ) , Q = (ql , q 2 ) i s defined by
The
P,Q E Kz
#
2,3.
Given a bijection u
PQ = 1 i f a d only i f P'Q'
: K2
+
KZ such that
= 1.
Then u must be semilinear up t o a translation (considering K2 as a vector space over K). I n case K
R ! we have proved i n [ 1 ] the sharper r e s u l t
THEOREM: Given a mapping u of the plane lR2 i n t o i t s e l f such that
'P,Q
E
IR2
PQ = 1 implies P'Q'
= 1.
Then u must be semilinear up t o a translation and hence a Lorentz transformation. The main question we are i n t e r e s t e d i n t h i s series o f three papers I, 11, I11 i s PROBLEM A: Determine a l l mappings u of K2 i n t o i t s e l f such that
'P,Q
E K2
PQ = I implies P'Q'
=
1
W.Benz
62
We s h a l l present the s o l u t i o n o f t h i s problem f o r i n f i n i t e l y many f i e l d s K, e s p e c i a l l y f o r i n f i n i t e l y many G a l o i s f i e l d s by a p p l y i n g a r e s u l t o f G. T a l l i n i ,
An important step i n our c o n s i d e r a t i o n s i s t h e f o l l o w i n g g e n e r a l i z a t i o n ( f o r
141.
4
char K
5,7) o f Rado's theorem:
THEOREM 1: &=me c;i;tr K f 2,3,5,7.
Given an i n j e c t i v e mapping a K'
+
K'
such
L k t -
-
V
:'ha
i~
P,Q E K' w s t b,
PQ = 1 implies pupu = 1
K and not necessari-
s c m i i i w a r i u i t h respect t o a monomoq?iiisn: of
ly an c:L.:omorp;ifsn/ y' t o a t r m i s l a t i o n . The proof o f t h i s theorem w i l l be given i n p a r t I o f our s e r i e s .
We l i k e t o
f i n i s h t h i s i n t r o d u c t i o n w i t h a remark about t h e connection o f t h e d i s t a n c e w i t h the Lorentz-Minkowski-distance
The
question we are a c t u a l l y i n t e r e s t e d i n i s
PROBLEM B: Given a c o m t a t i v e f i e l d K o f c h a r a c t e r i s t i c f 2 and given a f i x e d e'emer~: k E K'' := K \ {O}.
'P,Q
E
K'
Determine a22 mappings
T
: K'
+
K2 such t h a t
d(P,Q) = k impZies d(PT,QT) = k.
We l i k e t o v e r i f y t h a t problem B i s solved f o r a l l f i e l d s K f o r which problem
A can be solved.
Define f o r t h i s purpose
Given now P,Q E K2 such t h a t for a
T
o f problem B .
= 1.
Then d(P',Qu)
= k and hence d(PuT,QUr) = k
~
A consequence o f t h e l a s t equation i s P'Q'
mapping according t o problem A and hence
= 1 f o r u := T
= u-' u U.
UTG-'.
Thus a i s a
63
On mappings preserving a single LMdistance I. -
Consider now a f i e l d K
A l l o c c u r i n g f i e l d s a r e assumed t o be commutative. o f arbitrary characteristic. LEMMA 1: Assume I K I
> 4.
Then every element of K2 can be w r i t t e n i n the form
,. . . ,xn
with f i n i t e l y many eZements x, ,x,
in K
.
(ActuaZly n can be choosen d 4.)
PROOF: We n o t i c e t h a t KZ i s an a b e l i a n group w i t h t h e a d d i t i o n x2 YY1 4- Y 2 1 1 1 Observe (0,O) = ( 1 t ( - l ) , - t -). - There e x i s t s an element 5 0,-1 i n K such 1 -1 1 t h a t E 2 t c t 1 f 0 because otherwise I K I S 4. Put - = c 2 t c t 1 and x, = t ( c Z + c ) , 1 1 1 x, = t ( 6 t l),x, = - t c . Hence (1,O) = (x, t x, t x,, - t - t - ) . x1 x2 x3 For a j 0 we have 1 1 1 t - t -). (a,o) = (ax, t ax, t ax3, ax, ax, ax, ( X I YX,
MY,
I n case b (a,b)
YY2
1
:= (XI
4
+
4
0 we observe 1 1 b). = ( a - by 0 ) +
(s,
For t h e r e p r e s e n t a t i o n o f (a,b) i n f o r m (x,
.
a t most f o u r elements x, ,x, ,x, ,x,
+ * . * + xn'
x,
t
... t -)1
we hence need
'n
REMARK: I n case I K I G 4 t h e r e does n o t e x i s t a r e p r e s e n t a t i o n as described i n lemma 1.
This w i l l p l a y a r o l e l a t e r on ( s . p a r t 111). 1 x), x
Denote by M t h e s e t o f a l l p o i n t s ( x , (IKI
E Ka.
Hence, by lemma 1
>4), M generates K 2 , i . e . K2 = < M > .
LEMMA 2: Assume char K
Given p o i n t s P j Q of K2.
{2,3,5,i3.
= 1'
onZy i f t h e r e e x i s t p o i n t s P.,B such t h a t AB = 12,.
PROOF: F o r P = ( p , t ) , A = ( p - - q-p, t 15
4 q,
Q = (q,t),
p
- y15- ) , B
= (p
9 P
-
-
, PB
= 14,
put 49 - (q-p), 15
t
-
60 -). q-p
,
Then = 4,,
= 0 i f and
@=
16'
,
W. Ben2
64
Q = ( t , q ) change t h e components i n t h e d e f i n i t i o n o f A,B. Consider v i c e versa p o i n t s P 4 Q and p o i n t s A,B such t h a t = l’, PB = 14’ , QA = 4 ’ , @ = 16’, AB = 1Z2. Without l o s s o f g e n e r a l i t y we assume A = (O,O),
I n case P = ( t , p ) ,
-
Then according t o t h e equations i t t u r n s o u t t h a t Q € I (1,16),(16,1)}
P = (1,l).
= 0.
and hence
i f and o n l y i f t h e r e e x i s t p o i n t s A,B,C
.-
@
QA = 4 ’ .
= 22,
qc =
-
4
Then @ = 0 = 1’ , AC = 1 2 , CP = l’,
REMARK: I n case char K = 7 t h e f o l l o w i n g can be proved: Given P
Q.
I
such t h a t
AB = 2’.
2’,
2. --
K be a f i e l d .
Let
0“ = 0; 0 := (O,O),
v P,Q Define
~p
-,K’
such t h a t
and
-
PQ = 1 i m p l i e s pago = 1 .
E K?
: K”
+
K” by means o f
1 (Notice O ( x , -)” X
since
Consider an i n j e c t i v e mapping a : K2
0
1
1 = 1 because o f O(x, -) = 1 . )
= 0 ( x , -)“
X
X
Thus
(9
i s injective
i s injective.
G . Coizsider -
x1 ,xz
6
21.
K w i t h x’ f
X’Z
.
Then
1 1 0 1 1 Since (x, + x z , - t - ) i s o f PROOF: Denote (xi + x 2 , - t --) by (u,v). XI xz 1 1 = 1 and hence d i s t a n c e 1 from (xi, - ) we g e t (u,v)(ei, -) X i ’i u(cp1-101) = v q1 V Z ( P ~ - W ) , i . e . u = v cpI cpz because of 8 , # c p z . Now 1 (u W ) ( V - -) = 1 i m p l i e s v2 v1 = v(pl + 8 ’ ) - If v = 0 then u = v q1cpZ = 0 (91 1 1 l a 1 and hence (x, + XI, - + = 0, i . e . (x, t x z , - + -) = 0 because o f t h e i n j e c t i v i t y of o. Thus v
cp1cp2
XI
x,)
XI
T h i s c o n t r a d i c t s x12 = x2’.
= p, t W , i . e .
(u,v) = (cpl
+
(p2,
1 1 - + -). PI
82
X’
On mappings preserving a single LMdistance
b. Consider Pa =
P
x1 ,xz
(cpl
:= (XI
,. . . ,xn
+...t
cp
p.p = 1
-
y
*
K with xI2 ,xi
y . .
., x i p a i m i s e distinct.
Then
1 1 t...t -) with
'n
cp1
+...t xn, x,t...t
1 -)X and cp i := rp(xi). n
Consider n 2 3 and Pi :=P- (xi,
PROOF: By induction. Pa =: (u,v),
n
E
65
1 and P "= ( 1. i v+i
c,p
c -)1
v + i 'v
x). With I
i ( n o t i c e induction) we g e t
PiuPu = 1 and
(u - c
-
(v
)
v + i 'v
c
-)
I
= 1.
'v
v+i
Hence
c
v =
v+i
For i
l t u 'v
1
z
-
v + i 'v
+ j we g e t
<
=
' t
v+i
= z ' t
1
u - c
u+j
v + i 'v
'v
1
u - c
u + j 'u
i.e.
If u = c
cpv
1 then v = I - according t o (1) and we are ready.
Assume u
cpV
Hence
+
E cpv
and thus
(cpj
-
qi)
This implies u
(u
- "J
- ccpV
=
-
=
'pi
-
'pi2
-
'j
cpj2 *
for all i
+j
because o f
'pi
cp.
J
and hence
W. Benz
66 +
c'1
z
=
'02
-
Vb
u = WI +
q 3 ,
a contradiction.
char K $ I2,3,5}.
c. Assume -
iii
a:,
iiil
:il
,.. . ,a n
,.ii
+. . .+
a
2 ,
1 are p a i m i s e d i s t i n c t ,
1 =
=
n
Then there e x i s t elements al , a2
3 7, a2
-,1
=
1)
at
=
K"
with
1 1 +. ..+ . a
a1
n
PROOF: Case c h a r K $ I7,11,13,17,29,31,37,43,53,61,671 P u t at =
,... an 6
-2 -,
=
7
1 -,
6
as - 7 Y
a3
a6
:
1
= a5
.
Case c h a r K E (31,37,53,61,67} :
Case c h a r K E 129,431: Put
at
=
12 -,
ill
13
1 -,
=
a3
UI
-3 = -,
13
~ 1 4=
1 , a3
a5
=
4 -,
13
a6
1
= a5
Case c h a r K = 7 :
Case c h a r K = 11 :
Put
at
= -3,
a2
= -4,
Case c h a r K = 13 : Put a1 = 4,
a2
1
= a1
a3
=
2,
a4
= -5
.
.
Case c h a r K = 17 : Put a1 = 8,
J . Assume
a2
=-2,
a3
= 5, a4 = 7
char K $ {3,51.
Then cp(-x) =
PROOF: T h i s i s t r i v i a l f o r c h a r K g e t w i t h elements
=
-
(q(-x),
1 cp(-x)
-)
ai
.
= 2.
o f step c :
applying step b.
- cp(x)
f o r alZ x E K"
So assume c h a r K
f 2.
Given x
E
K" we
On mappings preserving a single L-Mdistance REMARK: d i s a l s o t r u e i n case c h a r K = 3.
e . Assume
char K
4
{3,5.1.
67
T h i s w i l l p l a y a r o l e i n p a r t 111.
Consider pairwise d i s t i n c t eZements x1 ,x2
,.. . ,xn
E
Then ( X l t...t x
with
‘pi
1 1 0 t . . . t -) = (cplt...tcpn, n’ x1 n
1
pl
+...t
1 -)
+
.
‘n
:= cp(xi).
PROOF: N o t h i n g i s t o prove f o r n = 1. Case n = 2: F o r xlz (Xl + xz
+ xi 9
Consider x12 = x z 2 and xl
see a.
1 1 u x, + x,’ = o0 =
a c c o r d i n g t o s t e p d. ( x , + x2
Now
1
Thus
x,’
Now a p p l y i n d u c t i o n .
Hence xz = -xl.
1 1 (030’ = (cp(x1 )+cp(-X1 1, -t cp(x1) cp(-x1)
1 1 0 xl + = (cp(X1 )fcp(X2
Y
xz
1
1
t-). 1, cp(x1) V(X2 1
L e t n be 2 3.
N o t h i n g i s t o prove i n case t h a t
= x2 x12 ,..., xz a r e p a i r w i s e d i s t i n c t because o f b. Otherwise suppose xz n n-1 n w i t h o u t l o s s o f g e n e r a l i t y and o f c o u r s e x ~ - xn.~ Hence xn = - x ~ - ~ Now .
+
(x, t...t xn
,... )“ =
(Xl t . . . +Xn-*
,... )“
= ( T I t . . . t cpn-2’...)
- . f Asswne char K
4
I2,3,5,7,11,13,17,19,23,31,37}.
Then there e x i s t
XI
,XZ
=
a
,x3 E K such that
1 1 1 xl t xz t x3 = 1, - t - t - = 0, x1 xz x3 (ii)I f k E 12,3,4,5} then the seven eZements 1 1 1 , - 1 are paimise d i s t i n c t . k x l y kxz 3 kx3 9 ~9 ~3
(il
PROOF: Case c h a r K $ {29,73,97,103,113,157,199,239,241,349,401} P u t x1 =
3 -, xz 7
=
2 - 7,
x3 =
6 7
‘
Case c h a r K E {29,73,97,113,157,199,241,349,401 10 15 6 P u t x1 = 19,xz = 19,x3 -19
I:
:
Kit
68
W.Benz
Case c h a r K E {103,2391:
Put
XI
=
12 -, 13
Asswne char K
(-I)
a
3 -13’
x* =
kl
4 13
x3 = -
$ {2,3,5,7}.
+..+a
Then f o r k E t2,3,4,5} then exist
1
=k=kn(k) akl
i i i j ak ,,... a k n ( k ) ,-1
.
1
t . . . +-,
a
kn(k) are pairwise distinct.
PROOF: Case c h a r K $ {11,13,17,19,23,31,37}:
A p p l y i n g s t e p f we p u t
, ak2 = kx,, ak3 = k x 3 , 1 1 - _1 a -k4- kxl ’ ak5 - kx,’ ak6 - kx3 * Case c h a r K = 11 : P u t akl
= kxl
azl
= 1 , a,:
_-
= 2 , a 2 3 = -3, ,a.,
= -4,
a z s = -5
a 3 1 = -2, a s 2 = 5 ,
arl
= -3, aa2 = -4
aal
= 1, a
s2=
,
-3, a S 3 = -4
Case c h a r K = 13: Choose a
1 + (-3)
t
4 = 2 ,
V!J
. according t o
, ,
(-2) + 6 + 3 + ( - 4 b 3
(-2) t 6 = 4 1 + ( - 2 ) + 6 = 5 , Case c h a r K = 17 :
1
t
(-6)
t
(-3) +
2
t
9 = 2,
(-6)
t
(-3)
t
2
t
9
t
3
+
6 = 4 ,
t
( - 7 ) = 5,
7
t
5
(-5)
= 3,
Case char K = 19 :
1
t
(-3) 1+(-3 1
t
(-3)
t
, + 6 = 3 , )+6=4 , t 8 = 5 ,
(-7) t 8 = 2
6 + (-7)
,
On mappings preserving a single L-Mdistance
69
Case char K = 23 :
2+(-11)+
3
1+2+(-11)+
t 8 = 2 ,
3 t 8 = 3 ,
,
(-5) t 9 = 4
1 t(-5)
.
9 = 5
t
Case char K = 31 :
1
6 t ( - 5 ) = 2 ,
t
(-11) + 14 = 3 , 1
+
(-11) t 14 = 4 ,
1 t 6 + (-5)t Case char K
1
t
.
(-11) t 14 = 5
37 :
( - 3 ) t (-12) t
17 = 2
(-3) t (-12) t
17 = 3 ,
17
+ (-13)
(-4) t h. Assume char K -
4
= 4
9 = 5
.
.
I
{3,5,71.
Given x E K
k a
Then (kx, --)
k cp(x)
= (kcp(x), - f o r
k E {2,3,4,5]. So assume char K { 2.
PROOF: This i s t r i v i a l f o r char K = 2.
= (cp(X)Y
according t o e and d.
1
-1cp(x)
Hence (Zx,
2 0 X)
-
t vx,
T
+ -x -x
y
.
2 cp(x)
= (2cp(x), -)
The same procedure f o r (vx,
Applying g we g e t
3,4,5,
v
completes t h e p r o o f o f h.
i. Assume
char K
(Xl t . . . t xn,
4
{2,3,5,7}.
l a 1 x, t . . . + x) =
n
Given
(cpl
X,
+.. .t
,... ,xn cp n,
1
K
*.
Then
1
cpl +...+ -)
cpn
with
'pi
:= cp(Xi).
W.Benz
70
PROOF: This i s t r u e f o r n = 1 and a l s o f o r n = 2 according t o e and h. induction.
L e t n be
Now apply
3.
2
Case A : I t x l,..., x n l j
> 2.
x2 w i t h o u t l o s s o f g e n e r a l i t y . 1 := - -). Then, by i n d u c t i o n , ( w f i xv* w+i x
We assume x1 Put Pi
1
Pi0 = (vi;
‘? i j
1 Put P : = (zxw,z -)
, vf7 i -). qb
’+
5 ‘J
and Pa =: (u,v).
So
E.1
= 1 implies
1 u-
di
‘w
f o r i = 1 and i = 2 we g e t
u
1 0 implies v = i -according
zvW = - z~~
u .-
+ 0. -
((PI
We have
q1
Hence by ( 3 )
(PI
) (u
-
Tipw,)
=
(11
{ v2 because o f xI
U - E vW = -
i . e . u = v3 t
$1
q4+...+
-
t o ( 2 ) and we a r e ready.
IpL
cc2
1
#
-
$92
2
.
x2 and thus
9
p ‘.,
So ( 2 ) ( p u t i = 1) i m p l i e s
v =
1 1 +t...t 0:
P3
1 1 1 1 1 - + - = - t - +...+ v, (P2 cc3 $94 Wn
-
But according t o i n d u c t i o n we have
.
Assume now
On mappings preserving a single LMdistance (x3 t . . . t x
1 - t . . .+ n' x3
=
+)CJ
(cp3
-&1-
.
t . . -t cpn,
n
t...t
1 p' . n
This i mp l i e s, since u i s i n j e c t i v e , (x3
+...+
Hence
x
1 n ' x3 = 0
cpl+cpa
1 +...+x) P, i . e . =
n Thus
by d.
u
= I cp
x1 t xz = 0. by ( 4 ) , c o n t r a d i c t i n g the assumption
*o.
u - I l p
Case B : x1 = xz =
...
= x n = : x . n u n Here we have t o prove ( n x, X) = (n'p, -), cp
cp
:= cp(x).
Applying h we can assume n 2 6. Put y1 := 2x, y2 := 2x, y3 := Yl +y2 +y3 +y4 = 5 x
,
1 t -1+ - 1+ - = 1 - 5 Y1 Y2 Y3 Y, x
.
X -, y4 2
:=
X 7.
Hence
Now (n-6)x+yl+yzty3+y4+x, = ( x t . . . t x ty1 tyz 2
(n-6)-times c
ty3
ty4
-
n-6 t x
1 -t...t Y1
tx, 1 t . .. t 1 t 1 t . .. t 1 t 1-) X x Y1 Y4 x
0
2
(n-1) summands Similarly n-1 X)
CJ
((n-l)x,
= ( x t ...t x t y l
ty2 ty3 t y 4 ,
1 t...t X
1 1 x t y1 t . . .
1
y,'"
( n 1'6) -"tim i s Since i i s assumed t o be t r u e up t i l l n-1 summands we g e t
-
(nx,
((n-l)x, n o n Thus (nx, X) = (n'p, -). rp
j. Asswne
.
group 'K PROOF: K
char K $ {2,3,5,7).
CJ
IH .
-,Ka i s a monomorphism of the abelian
We have I. K .I > 4 because o f char Consider P,Q E K'. 1 1 +...+ -)1 Hence P = ( x l t txn, X, t . . . t X) , Q = ( y l t +Ym, n Ym
i s injective.
$ I2,31.
lemma 1.
Moreover M' c
Then u : K'
Now applying i we g e t
...
...
by
W. Benz
72
Since M is the set of all points of distance 1 from 0 we get h 'c
--k .
q
Assume char K
assumed t o be* 0.
'P,Q
E K'
Consider an element n = 1
t2,3,5,71.
:=
Fj
n' implies P'Q"
=
.
n
1 of K, which i s
Then
= n',
PROOF: Consider P = (pl,pl), Q = (ql,ql),
Put x
+...+
PI.
Thus x
*
%
Then (ql-pl)(ql-p2) = n' + 0.
= n'
0 and
n Q = P t ( n x , -). X
Hence
+ (nx, n
Q~ = P' by j and i.
U "
P Q
--1.
= P
u
n
t ( n c p ( x ) , -1 V(X)
Thus = nw(x)
q
Assume char K
'P,Q
o
E K'
Fj
*
n -- n'. 9(x)
{2,3,5,71. =
o
Then
implies P ~ Q " = 0.
PROOF: Nothing is to prove for P =
Q. So assume
P
*
Q.
Then
existence of points A,B according to lemna 2 such thatfi = l',
F &
=
0 implies the
-= 14',
=
162 , @ = 12'. Since 14*, - 1,14,4,16,12 E K* we get PaAa = '1 , PUBu = QUA" = 4', QOBU = 16', A"B' = 12' by applying k. Now lemma 2 implies P'QU = 0
*
since P"
Q".
3. -
Consider a field mapping
U)
: K'
-+
K'
K
such that char
such that
K
$ {2,3,5,71.
Consider an injective
4')
On mappings preserving a single L-Mdbtance V
P u t (O,O)w
Ej
Ic
P,Q E
13
= 1 i m p l i e s pWqW= 1
=: (al , a z ) , ( l , l ) W =: (b, , b 2 ) .
Because o f (O,O)(l,l)
= 1 we have
(O,O)w(l,l)w
= 1, i . e .
(bl-al)(bz-az)
=
1.
Consider t h e d i s t a n c e p r e s e r v i n g mapping
1I
Y ' = (bl -al)(y
-
a,)
which i s b i j e c t i v e and l i n e a r up t o a t r a n s l a t i o n . Obviously, V
Fij
P,Q E K'
= 1 i m p l i e s pwagwa = 1
and ( O , O ) W a = (O,O), ( l , l ) W a = ( 1 , l ) .
So s e c t i o n 2 a p p l i e s t o t h e mapping w It i s I P
K'I
E
a.
Especially, w
~i
(0,O)P = 0 and ( 1 , l ) P = 01 = t ( l , O ) , ( O , l ) } .
I(l,O), (0,l)
Pa =
p r e s e r v e s d i s t a n c e 0. Hence
t ( 1 ,O) ,( 0 , l ) 3.
Define x' = y y' = x
which i s d i s t a n c e p r e s e r v i n g , b i j e c t i v e and l i n e a r . (
I
identity
1
6
( l , o ) W a = (1,O) for
8 = (
I
D e f i n e moreover
(l,O)Wa = ( 0 , l )
Then u := w a 6 i s i n j e c t i v e , p r e s e r v e s d i s t a n c e 1 and has (O,O),(l,l),(O,l),(l,O) as f i x e d p o i n t s . The l i n e y = 0 ( i . e .
{(x,o)/ x
4
{O,l},
x
E
K)) i s mapped under u i n t o t h e l i n e
fact!
Given (x,O),
(x,O)'
has a l s o d i s t a n c e 0 f r o m (0,O)'
y = 0.
t h i s p o i n t has d i s t a n c e 0 f r o m (O,O),(l,O). =
(O,O), (1,O)'
such t h a t xy
*
0.
So
= (1,O) and t h u s i s o f f o r m
l i n e x = 0 i s mapped under u i n t o t h e l i n e x = 0. ( ~ ' ~ 0 ) The . Consider a p o i n t (x,y)
In
The i n j e c t i v i t y o f u i m p l i e s
W. Benz
74
Hence u = xI
,v
*
= yl because o f (u,v)
(0,O).
We are now using t h e mapping 0 of s e c t i o n 2. (x,
l o
2)
1
(y,y)
0
= (:v(x), =
1 (v(y)' -j-' 0
f o r xy
*
($1
Thus (x,O)'
) for y
*
0.
By applying j o f s e c t i o n 2 we g e t
Now ( 5 ) i m p l i e s
Thus
and
= (cp(x),O).
*
0 it is
Also
Thus ( 0 , ~ ) ' = (0,
1 7) and hence V
0.
Consider xI ,x2 E K" such t h a t x1
For x
+
x2
*
$1
0 and consider x t K \ C0,lI.
Then
On mappings preserving a single L a d i s t a n c e A c c o r d i n g t o ( 6 ) we have v(x
(1-x)) = v(x)
t
v(1-x)
1 v ( 1 t ( - -1 ) = v ( 1 )
t
v(- -1
t
X
and
1
X
Thus ( 7 ) i m p l i e s 1 v(x)v(-) = X
But t h i s i s obviously a l s o t r u e f o r x = 1 Define
T
: K
r(X) =
I
+
L
.
K by means o f x * o for x = o
O
Thus -c i s i n j e c t i v e . r(x+y) = r(x)
+
For x t
1 0 we have T(X)T(T) = 1 .
T ( y ) f o r a l l x,y E K.
Because o f ( 6 ) t h i s i s t r u e i n case x,y,xty
Now assume x y
x = 0 o r y = 0.
We c l a i m moreover t h a t
+0
E
It i s a l s o g e n e r a l l y t r u e f o r
K".
and x t y = 0.
Thus x
+
0 and y = - x .
Now d i m p l i e s v ( y ) = - v ( x ) and hence T(X
+
y) =
T(0) =
0 = q(x)
A l t o g e t h e r we have t h a t 1 T(X)T(T) = 1 f o r a l l x
t
v ( y ) = T(X)
t
T(y).
i s a monomorphism o f t h e a d d i t i v e group o f K such t h a t
T
* 0.
By Hua's theorem (s.[ 21
, p. 324; t h e p r o o f 1 o c . c i t . a p p l i e s t o t h e p r e s e n t
s l i g h t l y more g e n e r a l s i t u a t i o n ) we t h u s g e t t h a t
T
We
i s a monomorphism o f K.
f i n a l l y p r o v e t h a t o i s a s e m i l i n e a r mapping o f t h e K-vectorspace K 2 . According t o j t h e mapping o i s a monomorphism o f K2.
(x,Y)'= f o r xy
* 0.
( T ( X ) , .(Y)) But t h i s e q u a t i o n i s a l s o t r u e i n case x y = 0.
( k x , ky)U Hence
w = u
Because o f ( 5 ) we have
c'
(r(kx),T(ky))
Thus f o r k E K
= T(~)(T(x)T , (Y)).
d' i s a s e m i l i n e a r mapping up t o a t r a n s l a t i o n .
t h e p r o o f o f theorem 1. theorem 1 a r e g i v e n by
T h i s completes
I t i s now t r i v i a l t o v e r i f y t h a t a l l t h e mappings
0
of
W. Benz
76 ( x . ~ ) + (a
T
(x)
(a
T
(Y) t b,
(x,Y)
+
where a,b,c E K , a
*
t
1 a 1
c)
b, -
T
(y)
a
T
(x) + c)
t
and
,
0.
BIBLIOGRAPHY 1. 2.
3. 4.
W . Benz, A Beckman Quarles Type Theorem f o r Plane Lorentz Transformations, b!ui,i?. z., 177 (1981), 101-106. W . Benz, V3rlesungen iiber Geometrie d e r AZgebren. Grundl. Bd. 197,
Springer-Verlag, New York 1973.
F. Rado, On the characterization of plane a f f i n e isometries, Resultate d. mat:^. , 3 (1980). 70-73. G . T a l l i n i , On a theorem by W. Benz characterizing plane Lorentz Transformations in J a e r n e f e l t ' s World. To appear in Jowlla2 of Geometry.
Mathematisches Seminar Bundesstrasse 55 2000 Hamburg 13 Federal Republic of Germany
ON MAPPINGS PRESERVING A SINGLE
61
LORENTZ-MINKOWSKI-DISTANCE. I
Walter Benz
Consider a comnutative f i e l d K and denote by K2 the s e t o f a l l ordered p a i r s o f elements o f K .
distance
fl of
The elements o f K2 are c a l l e d points o f the plane K 2 ,
the p o i n t s P
fl:=
The f o l l o w i n g r e s u l t was proved by F. Rado, [ 3 1 :
(ql-pl)(q2-pz).
THEOREM: Assume char K
V
( p l , p 2 ) , Q = (ql , q 2 ) i s defined by
The
P,Q E Kz
#
2,3.
Given a bijection u
PQ = 1 i f a d only i f P'Q'
: K2
+
KZ such that
= 1.
Then u must be semilinear up t o a translation (considering K2 as a vector space over K). I n case K
R ! we have proved i n [ 1 ] the sharper r e s u l t
THEOREM: Given a mapping u of the plane lR2 i n t o i t s e l f such that
'P,Q
E
IR2
PQ = 1 implies P'Q'
= 1.
Then u must be semilinear up t o a translation and hence a Lorentz transformation. The main question we are i n t e r e s t e d i n t h i s series o f three papers I, 11, I11 i s PROBLEM A: Determine a l l mappings u of K2 i n t o i t s e l f such that
'P,Q
E K2
PQ = I implies P'Q'
=
1
W.Benz
62
We s h a l l present the s o l u t i o n o f t h i s problem f o r i n f i n i t e l y many f i e l d s K, e s p e c i a l l y f o r i n f i n i t e l y many G a l o i s f i e l d s by a p p l y i n g a r e s u l t o f G. T a l l i n i ,
An important step i n our c o n s i d e r a t i o n s i s t h e f o l l o w i n g g e n e r a l i z a t i o n ( f o r
141.
4
char K
5,7) o f Rado's theorem:
THEOREM 1: &=me c;i;tr K f 2,3,5,7.
Given an i n j e c t i v e mapping a K'
+
K'
such
L k t -
-
V
:'ha
i~
P,Q E K' w s t b,
PQ = 1 implies pupu = 1
K and not necessari-
s c m i i i w a r i u i t h respect t o a monomoq?iiisn: of
ly an c:L.:omorp;ifsn/ y' t o a t r m i s l a t i o n . The proof o f t h i s theorem w i l l be given i n p a r t I o f our s e r i e s .
We l i k e t o
f i n i s h t h i s i n t r o d u c t i o n w i t h a remark about t h e connection o f t h e d i s t a n c e w i t h the Lorentz-Minkowski-distance
The
question we are a c t u a l l y i n t e r e s t e d i n i s
PROBLEM B: Given a c o m t a t i v e f i e l d K o f c h a r a c t e r i s t i c f 2 and given a f i x e d e'emer~: k E K'' := K \ {O}.
'P,Q
E
K'
Determine a22 mappings
T
: K'
+
K2 such t h a t
d(P,Q) = k impZies d(PT,QT) = k.
We l i k e t o v e r i f y t h a t problem B i s solved f o r a l l f i e l d s K f o r which problem
A can be solved.
Define f o r t h i s purpose
Given now P,Q E K2 such t h a t for a
T
o f problem B .
= 1.
Then d(P',Qu)
= k and hence d(PuT,QUr) = k
~
A consequence o f t h e l a s t equation i s P'Q'
mapping according t o problem A and hence
= 1 f o r u := T
= u-' u U.
UTG-'.
Thus a i s a
63
On mappings preserving a single LMdistance I. -
Consider now a f i e l d K
A l l o c c u r i n g f i e l d s a r e assumed t o be commutative. o f arbitrary characteristic. LEMMA 1: Assume I K I
> 4.
Then every element of K2 can be w r i t t e n i n the form
,. . . ,xn
with f i n i t e l y many eZements x, ,x,
in K
.
(ActuaZly n can be choosen d 4.)
PROOF: We n o t i c e t h a t KZ i s an a b e l i a n group w i t h t h e a d d i t i o n x2 YY1 4- Y 2 1 1 1 Observe (0,O) = ( 1 t ( - l ) , - t -). - There e x i s t s an element 5 0,-1 i n K such 1 -1 1 t h a t E 2 t c t 1 f 0 because otherwise I K I S 4. Put - = c 2 t c t 1 and x, = t ( c Z + c ) , 1 1 1 x, = t ( 6 t l),x, = - t c . Hence (1,O) = (x, t x, t x,, - t - t - ) . x1 x2 x3 For a j 0 we have 1 1 1 t - t -). (a,o) = (ax, t ax, t ax3, ax, ax, ax, ( X I YX,
MY,
I n case b (a,b)
YY2
1
:= (XI
4
+
4
0 we observe 1 1 b). = ( a - by 0 ) +
(s,
For t h e r e p r e s e n t a t i o n o f (a,b) i n f o r m (x,
.
a t most f o u r elements x, ,x, ,x, ,x,
+ * . * + xn'
x,
t
... t -)1
we hence need
'n
REMARK: I n case I K I G 4 t h e r e does n o t e x i s t a r e p r e s e n t a t i o n as described i n lemma 1.
This w i l l p l a y a r o l e l a t e r on ( s . p a r t 111). 1 x), x
Denote by M t h e s e t o f a l l p o i n t s ( x , (IKI
E Ka.
Hence, by lemma 1
>4), M generates K 2 , i . e . K2 = < M > .
LEMMA 2: Assume char K
Given p o i n t s P j Q of K2.
{2,3,5,i3.
= 1'
onZy i f t h e r e e x i s t p o i n t s P.,B such t h a t AB = 12,.
PROOF: F o r P = ( p , t ) , A = ( p - - q-p, t 15
4 q,
Q = (q,t),
p
- y15- ) , B
= (p
9 P
-
-
, PB
= 14,
put 49 - (q-p), 15
t
-
60 -). q-p
,
Then = 4,,
= 0 i f and
@=
16'
,
W. Ben2
64
Q = ( t , q ) change t h e components i n t h e d e f i n i t i o n o f A,B. Consider v i c e versa p o i n t s P 4 Q and p o i n t s A,B such t h a t = l’, PB = 14’ , QA = 4 ’ , @ = 16’, AB = 1Z2. Without l o s s o f g e n e r a l i t y we assume A = (O,O),
I n case P = ( t , p ) ,
-
Then according t o t h e equations i t t u r n s o u t t h a t Q € I (1,16),(16,1)}
P = (1,l).
= 0.
and hence
i f and o n l y i f t h e r e e x i s t p o i n t s A,B,C
.-
@
QA = 4 ’ .
= 22,
qc =
-
4
Then @ = 0 = 1’ , AC = 1 2 , CP = l’,
REMARK: I n case char K = 7 t h e f o l l o w i n g can be proved: Given P
Q.
I
such t h a t
AB = 2’.
2’,
2. --
K be a f i e l d .
Let
0“ = 0; 0 := (O,O),
v P,Q Define
~p
-,K’
such t h a t
and
-
PQ = 1 i m p l i e s pago = 1 .
E K?
: K”
+
K” by means o f
1 (Notice O ( x , -)” X
since
Consider an i n j e c t i v e mapping a : K2
0
1
1 = 1 because o f O(x, -) = 1 . )
= 0 ( x , -)“
X
X
Thus
(9
i s injective
i s injective.
G . Coizsider -
x1 ,xz
6
21.
K w i t h x’ f
X’Z
.
Then
1 1 0 1 1 Since (x, + x z , - t - ) i s o f PROOF: Denote (xi + x 2 , - t --) by (u,v). XI xz 1 1 = 1 and hence d i s t a n c e 1 from (xi, - ) we g e t (u,v)(ei, -) X i ’i u(cp1-101) = v q1 V Z ( P ~ - W ) , i . e . u = v cpI cpz because of 8 , # c p z . Now 1 (u W ) ( V - -) = 1 i m p l i e s v2 v1 = v(pl + 8 ’ ) - If v = 0 then u = v q1cpZ = 0 (91 1 1 l a 1 and hence (x, + XI, - + = 0, i . e . (x, t x z , - + -) = 0 because o f t h e i n j e c t i v i t y of o. Thus v
cp1cp2
XI
x,)
XI
T h i s c o n t r a d i c t s x12 = x2’.
= p, t W , i . e .
(u,v) = (cpl
+
(p2,
1 1 - + -). PI
82
X’
On mappings preserving a single LMdistance
b. Consider Pa =
P
x1 ,xz
(cpl
:= (XI
,. . . ,xn
+...t
cp
p.p = 1
-
y
*
K with xI2 ,xi
y . .
., x i p a i m i s e distinct.
Then
1 1 t...t -) with
'n
cp1
+...t xn, x,t...t
1 -)X and cp i := rp(xi). n
Consider n 2 3 and Pi :=P- (xi,
PROOF: By induction. Pa =: (u,v),
n
E
65
1 and P "= ( 1. i v+i
c,p
c -)1
v + i 'v
x). With I
i ( n o t i c e induction) we g e t
PiuPu = 1 and
(u - c
-
(v
)
v + i 'v
c
-)
I
= 1.
'v
v+i
Hence
c
v =
v+i
For i
l t u 'v
1
z
-
v + i 'v
+ j we g e t
<
=
' t
v+i
= z ' t
1
u - c
u+j
v + i 'v
'v
1
u - c
u + j 'u
i.e.
If u = c
cpv
1 then v = I - according t o (1) and we are ready.
Assume u
cpV
Hence
+
E cpv
and thus
(cpj
-
qi)
This implies u
(u
- "J
- ccpV
=
-
=
'pi
-
'pi2
-
'j
cpj2 *
for all i
+j
because o f
'pi
cp.
J
and hence
W. Benz
66 +
c'1
z
=
'02
-
Vb
u = WI +
q 3 ,
a contradiction.
char K $ I2,3,5}.
c. Assume -
iii
a:,
iiil
:il
,.. . ,a n
,.ii
+. . .+
a
2 ,
1 are p a i m i s e d i s t i n c t ,
1 =
=
n
Then there e x i s t elements al , a2
3 7, a2
-,1
=
1)
at
=
K"
with
1 1 +. ..+ . a
a1
n
PROOF: Case c h a r K $ I7,11,13,17,29,31,37,43,53,61,671 P u t at =
,... an 6
-2 -,
=
7
1 -,
6
as - 7 Y
a3
a6
:
1
= a5
.
Case c h a r K E (31,37,53,61,67} :
Case c h a r K E 129,431: Put
at
=
12 -,
ill
13
1 -,
=
a3
UI
-3 = -,
13
~ 1 4=
1 , a3
a5
=
4 -,
13
a6
1
= a5
Case c h a r K = 7 :
Case c h a r K = 11 :
Put
at
= -3,
a2
= -4,
Case c h a r K = 13 : Put a1 = 4,
a2
1
= a1
a3
=
2,
a4
= -5
.
.
Case c h a r K = 17 : Put a1 = 8,
J . Assume
a2
=-2,
a3
= 5, a4 = 7
char K $ {3,51.
Then cp(-x) =
PROOF: T h i s i s t r i v i a l f o r c h a r K g e t w i t h elements
=
-
(q(-x),
1 cp(-x)
-)
ai
.
= 2.
o f step c :
applying step b.
- cp(x)
f o r alZ x E K"
So assume c h a r K
f 2.
Given x
E
K" we
On mappings preserving a single L-Mdistance REMARK: d i s a l s o t r u e i n case c h a r K = 3.
e . Assume
char K
4
{3,5.1.
67
T h i s w i l l p l a y a r o l e i n p a r t 111.
Consider pairwise d i s t i n c t eZements x1 ,x2
,.. . ,xn
E
Then ( X l t...t x
with
‘pi
1 1 0 t . . . t -) = (cplt...tcpn, n’ x1 n
1
pl
+...t
1 -)
+
.
‘n
:= cp(xi).
PROOF: N o t h i n g i s t o prove f o r n = 1. Case n = 2: F o r xlz (Xl + xz
+ xi 9
Consider x12 = x z 2 and xl
see a.
1 1 u x, + x,’ = o0 =
a c c o r d i n g t o s t e p d. ( x , + x2
Now
1
Thus
x,’
Now a p p l y i n d u c t i o n .
Hence xz = -xl.
1 1 (030’ = (cp(x1 )+cp(-X1 1, -t cp(x1) cp(-x1)
1 1 0 xl + = (cp(X1 )fcp(X2
Y
xz
1
1
t-). 1, cp(x1) V(X2 1
L e t n be 2 3.
N o t h i n g i s t o prove i n case t h a t
= x2 x12 ,..., xz a r e p a i r w i s e d i s t i n c t because o f b. Otherwise suppose xz n n-1 n w i t h o u t l o s s o f g e n e r a l i t y and o f c o u r s e x ~ - xn.~ Hence xn = - x ~ - ~ Now .
+
(x, t...t xn
,... )“ =
(Xl t . . . +Xn-*
,... )“
= ( T I t . . . t cpn-2’...)
- . f Asswne char K
4
I2,3,5,7,11,13,17,19,23,31,37}.
Then there e x i s t
XI
,XZ
=
a
,x3 E K such that
1 1 1 xl t xz t x3 = 1, - t - t - = 0, x1 xz x3 (ii)I f k E 12,3,4,5} then the seven eZements 1 1 1 , - 1 are paimise d i s t i n c t . k x l y kxz 3 kx3 9 ~9 ~3
(il
PROOF: Case c h a r K $ {29,73,97,103,113,157,199,239,241,349,401} P u t x1 =
3 -, xz 7
=
2 - 7,
x3 =
6 7
‘
Case c h a r K E {29,73,97,113,157,199,241,349,401 10 15 6 P u t x1 = 19,xz = 19,x3 -19
I:
:
Kit
68
W.Benz
Case c h a r K E {103,2391:
Put
XI
=
12 -, 13
Asswne char K
(-I)
a
3 -13’
x* =
kl
4 13
x3 = -
$ {2,3,5,7}.
+..+a
Then f o r k E t2,3,4,5} then exist
1
=k=kn(k) akl
i i i j ak ,,... a k n ( k ) ,-1
.
1
t . . . +-,
a
kn(k) are pairwise distinct.
PROOF: Case c h a r K $ {11,13,17,19,23,31,37}:
A p p l y i n g s t e p f we p u t
, ak2 = kx,, ak3 = k x 3 , 1 1 - _1 a -k4- kxl ’ ak5 - kx,’ ak6 - kx3 * Case c h a r K = 11 : P u t akl
= kxl
azl
= 1 , a,:
_-
= 2 , a 2 3 = -3, ,a.,
= -4,
a z s = -5
a 3 1 = -2, a s 2 = 5 ,
arl
= -3, aa2 = -4
aal
= 1, a
s2=
,
-3, a S 3 = -4
Case c h a r K = 13: Choose a
1 + (-3)
t
4 = 2 ,
V!J
. according t o
, ,
(-2) + 6 + 3 + ( - 4 b 3
(-2) t 6 = 4 1 + ( - 2 ) + 6 = 5 , Case c h a r K = 17 :
1
t
(-6)
t
(-3) +
2
t
9 = 2,
(-6)
t
(-3)
t
2
t
9
t
3
+
6 = 4 ,
t
( - 7 ) = 5,
7
t
5
(-5)
= 3,
Case char K = 19 :
1
t
(-3) 1+(-3 1
t
(-3)
t
, + 6 = 3 , )+6=4 , t 8 = 5 ,
(-7) t 8 = 2
6 + (-7)
,
On mappings preserving a single L-Mdistance
69
Case char K = 23 :
2+(-11)+
3
1+2+(-11)+
t 8 = 2 ,
3 t 8 = 3 ,
,
(-5) t 9 = 4
1 t(-5)
.
9 = 5
t
Case char K = 31 :
1
6 t ( - 5 ) = 2 ,
t
(-11) + 14 = 3 , 1
+
(-11) t 14 = 4 ,
1 t 6 + (-5)t Case char K
1
t
.
(-11) t 14 = 5
37 :
( - 3 ) t (-12) t
17 = 2
(-3) t (-12) t
17 = 3 ,
17
+ (-13)
(-4) t h. Assume char K -
4
= 4
9 = 5
.
.
I
{3,5,71.
Given x E K
k a
Then (kx, --)
k cp(x)
= (kcp(x), - f o r
k E {2,3,4,5]. So assume char K { 2.
PROOF: This i s t r i v i a l f o r char K = 2.
= (cp(X)Y
according t o e and d.
1
-1cp(x)
Hence (Zx,
2 0 X)
-
t vx,
T
+ -x -x
y
.
2 cp(x)
= (2cp(x), -)
The same procedure f o r (vx,
Applying g we g e t
3,4,5,
v
completes t h e p r o o f o f h.
i. Assume
char K
(Xl t . . . t xn,
4
{2,3,5,7}.
l a 1 x, t . . . + x) =
n
Given
(cpl
X,
+.. .t
,... ,xn cp n,
1
K
*.
Then
1
cpl +...+ -)
cpn
with
'pi
:= cp(Xi).
W.Benz
70
PROOF: This i s t r u e f o r n = 1 and a l s o f o r n = 2 according t o e and h. induction.
L e t n be
Now apply
3.
2
Case A : I t x l,..., x n l j
> 2.
x2 w i t h o u t l o s s o f g e n e r a l i t y . 1 := - -). Then, by i n d u c t i o n , ( w f i xv* w+i x
We assume x1 Put Pi
1
Pi0 = (vi;
‘? i j
1 Put P : = (zxw,z -)
, vf7 i -). qb
’+
5 ‘J
and Pa =: (u,v).
So
E.1
= 1 implies
1 u-
di
‘w
f o r i = 1 and i = 2 we g e t
u
1 0 implies v = i -according
zvW = - z~~
u .-
+ 0. -
((PI
We have
q1
Hence by ( 3 )
(PI
) (u
-
Tipw,)
=
(11
{ v2 because o f xI
U - E vW = -
i . e . u = v3 t
$1
q4+...+
-
t o ( 2 ) and we a r e ready.
IpL
cc2
1
#
-
$92
2
.
x2 and thus
9
p ‘.,
So ( 2 ) ( p u t i = 1) i m p l i e s
v =
1 1 +t...t 0:
P3
1 1 1 1 1 - + - = - t - +...+ v, (P2 cc3 $94 Wn
-
But according t o i n d u c t i o n we have
.
Assume now
On mappings preserving a single LMdistance (x3 t . . . t x
1 - t . . .+ n' x3
=
+)CJ
(cp3
-&1-
.
t . . -t cpn,
n
t...t
1 p' . n
This i mp l i e s, since u i s i n j e c t i v e , (x3
+...+
Hence
x
1 n ' x3 = 0
cpl+cpa
1 +...+x) P, i . e . =
n Thus
by d.
u
= I cp
x1 t xz = 0. by ( 4 ) , c o n t r a d i c t i n g the assumption
*o.
u - I l p
Case B : x1 = xz =
...
= x n = : x . n u n Here we have t o prove ( n x, X) = (n'p, -), cp
cp
:= cp(x).
Applying h we can assume n 2 6. Put y1 := 2x, y2 := 2x, y3 := Yl +y2 +y3 +y4 = 5 x
,
1 t -1+ - 1+ - = 1 - 5 Y1 Y2 Y3 Y, x
.
X -, y4 2
:=
X 7.
Hence
Now (n-6)x+yl+yzty3+y4+x, = ( x t . . . t x ty1 tyz 2
(n-6)-times c
ty3
ty4
-
n-6 t x
1 -t...t Y1
tx, 1 t . .. t 1 t 1 t . .. t 1 t 1-) X x Y1 Y4 x
0
2
(n-1) summands Similarly n-1 X)
CJ
((n-l)x,
= ( x t ...t x t y l
ty2 ty3 t y 4 ,
1 t...t X
1 1 x t y1 t . . .
1
y,'"
( n 1'6) -"tim i s Since i i s assumed t o be t r u e up t i l l n-1 summands we g e t
-
(nx,
((n-l)x, n o n Thus (nx, X) = (n'p, -). rp
j. Asswne
.
group 'K PROOF: K
char K $ {2,3,5,7).
CJ
IH .
-,Ka i s a monomorphism of the abelian
We have I. K .I > 4 because o f char Consider P,Q E K'. 1 1 +...+ -)1 Hence P = ( x l t txn, X, t . . . t X) , Q = ( y l t +Ym, n Ym
i s injective.
$ I2,31.
lemma 1.
Moreover M' c
Then u : K'
Now applying i we g e t
...
...
by
W. Benz
72
Since M is the set of all points of distance 1 from 0 we get h 'c
--k .
q
Assume char K
assumed t o be* 0.
'P,Q
E K'
Consider an element n = 1
t2,3,5,71.
:=
Fj
n' implies P'Q"
=
.
n
1 of K, which i s
Then
= n',
PROOF: Consider P = (pl,pl), Q = (ql,ql),
Put x
+...+
PI.
Thus x
*
%
Then (ql-pl)(ql-p2) = n' + 0.
= n'
0 and
n Q = P t ( n x , -). X
Hence
+ (nx, n
Q~ = P' by j and i.
U "
P Q
--1.
= P
u
n
t ( n c p ( x ) , -1 V(X)
Thus = nw(x)
q
Assume char K
'P,Q
o
E K'
Fj
*
n -- n'. 9(x)
{2,3,5,71. =
o
Then
implies P ~ Q " = 0.
PROOF: Nothing is to prove for P =
Q. So assume
P
*
Q.
Then
existence of points A,B according to lemna 2 such thatfi = l',
F &
=
0 implies the
-= 14',
=
162 , @ = 12'. Since 14*, - 1,14,4,16,12 E K* we get PaAa = '1 , PUBu = QUA" = 4', QOBU = 16', A"B' = 12' by applying k. Now lemma 2 implies P'QU = 0
*
since P"
Q".
3. -
Consider a field mapping
U)
: K'
-+
K'
K
such that char
such that
K
$ {2,3,5,71.
Consider an injective
4')
On mappings preserving a single L-Mdbtance V
P u t (O,O)w
Ej
Ic
P,Q E
13
= 1 i m p l i e s pWqW= 1
=: (al , a z ) , ( l , l ) W =: (b, , b 2 ) .
Because o f (O,O)(l,l)
= 1 we have
(O,O)w(l,l)w
= 1, i . e .
(bl-al)(bz-az)
=
1.
Consider t h e d i s t a n c e p r e s e r v i n g mapping
1I
Y ' = (bl -al)(y
-
a,)
which i s b i j e c t i v e and l i n e a r up t o a t r a n s l a t i o n . Obviously, V
Fij
P,Q E K'
= 1 i m p l i e s pwagwa = 1
and ( O , O ) W a = (O,O), ( l , l ) W a = ( 1 , l ) .
So s e c t i o n 2 a p p l i e s t o t h e mapping w It i s I P
K'I
E
a.
Especially, w
~i
(0,O)P = 0 and ( 1 , l ) P = 01 = t ( l , O ) , ( O , l ) } .
I(l,O), (0,l)
Pa =
p r e s e r v e s d i s t a n c e 0. Hence
t ( 1 ,O) ,( 0 , l ) 3.
Define x' = y y' = x
which i s d i s t a n c e p r e s e r v i n g , b i j e c t i v e and l i n e a r . (
I
identity
1
6
( l , o ) W a = (1,O) for
8 = (
I
D e f i n e moreover
(l,O)Wa = ( 0 , l )
Then u := w a 6 i s i n j e c t i v e , p r e s e r v e s d i s t a n c e 1 and has (O,O),(l,l),(O,l),(l,O) as f i x e d p o i n t s . The l i n e y = 0 ( i . e .
{(x,o)/ x
4
{O,l},
x
E
K)) i s mapped under u i n t o t h e l i n e
fact!
Given (x,O),
(x,O)'
has a l s o d i s t a n c e 0 f r o m (0,O)'
y = 0.
t h i s p o i n t has d i s t a n c e 0 f r o m (O,O),(l,O). =
(O,O), (1,O)'
such t h a t xy
*
0.
So
= (1,O) and t h u s i s o f f o r m
l i n e x = 0 i s mapped under u i n t o t h e l i n e x = 0. ( ~ ' ~ 0 ) The . Consider a p o i n t (x,y)
In
The i n j e c t i v i t y o f u i m p l i e s
W. Benz
74
Hence u = xI
,v
*
= yl because o f (u,v)
(0,O).
We are now using t h e mapping 0 of s e c t i o n 2. (x,
l o
2)
1
(y,y)
0
= (:v(x), =
1 (v(y)' -j-' 0
f o r xy
*
($1
Thus (x,O)'
) for y
*
0.
By applying j o f s e c t i o n 2 we g e t
Now ( 5 ) i m p l i e s
Thus
and
= (cp(x),O).
*
0 it is
Also
Thus ( 0 , ~ ) ' = (0,
1 7) and hence V
0.
Consider xI ,x2 E K" such t h a t x1
For x
+
x2
*
$1
0 and consider x t K \ C0,lI.
Then
On mappings preserving a single L a d i s t a n c e A c c o r d i n g t o ( 6 ) we have v(x
(1-x)) = v(x)
t
v(1-x)
1 v ( 1 t ( - -1 ) = v ( 1 )
t
v(- -1
t
X
and
1
X
Thus ( 7 ) i m p l i e s 1 v(x)v(-) = X
But t h i s i s obviously a l s o t r u e f o r x = 1 Define
T
: K
r(X) =
I
+
L
.
K by means o f x * o for x = o
O
Thus -c i s i n j e c t i v e . r(x+y) = r(x)
+
For x t
1 0 we have T(X)T(T) = 1 .
T ( y ) f o r a l l x,y E K.
Because o f ( 6 ) t h i s i s t r u e i n case x,y,xty
Now assume x y
x = 0 o r y = 0.
We c l a i m moreover t h a t
+0
E
It i s a l s o g e n e r a l l y t r u e f o r
K".
and x t y = 0.
Thus x
+
0 and y = - x .
Now d i m p l i e s v ( y ) = - v ( x ) and hence T(X
+
y) =
T(0) =
0 = q(x)
A l t o g e t h e r we have t h a t 1 T(X)T(T) = 1 f o r a l l x
t
v ( y ) = T(X)
t
T(y).
i s a monomorphism o f t h e a d d i t i v e group o f K such t h a t
T
* 0.
By Hua's theorem (s.[ 21
, p. 324; t h e p r o o f 1 o c . c i t . a p p l i e s t o t h e p r e s e n t
s l i g h t l y more g e n e r a l s i t u a t i o n ) we t h u s g e t t h a t
T
We
i s a monomorphism o f K.
f i n a l l y p r o v e t h a t o i s a s e m i l i n e a r mapping o f t h e K-vectorspace K 2 . According t o j t h e mapping o i s a monomorphism o f K2.
(x,Y)'= f o r xy
* 0.
( T ( X ) , .(Y)) But t h i s e q u a t i o n i s a l s o t r u e i n case x y = 0.
( k x , ky)U Hence
w = u
Because o f ( 5 ) we have
c'
(r(kx),T(ky))
Thus f o r k E K
= T(~)(T(x)T , (Y)).
d' i s a s e m i l i n e a r mapping up t o a t r a n s l a t i o n .
t h e p r o o f o f theorem 1. theorem 1 a r e g i v e n by
T h i s completes
I t i s now t r i v i a l t o v e r i f y t h a t a l l t h e mappings
0
of
W. Benz
76 ( x . ~ ) + (a
T
(x)
(a
T
(Y) t b,
(x,Y)
+
where a,b,c E K , a
*
t
1 a 1
c)
b, -
T
(y)
a
T
(x) + c)
t
and
,
0.
BIBLIOGRAPHY 1. 2.
3. 4.
W . Benz, A Beckman Quarles Type Theorem f o r Plane Lorentz Transformations, b!ui,i?. z., 177 (1981), 101-106. W . Benz, V3rlesungen iiber Geometrie d e r AZgebren. Grundl. Bd. 197,
Springer-Verlag, New York 1973.
F. Rado, On the characterization of plane a f f i n e isometries, Resultate d. mat:^. , 3 (1980). 70-73. G . T a l l i n i , On a theorem by W. Benz characterizing plane Lorentz Transformations in J a e r n e f e l t ' s World. To appear in Jowlla2 of Geometry.
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