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7—- _H,.- *UIDF 4... 4.

 

ABSTRACT
DISTANCE PRESERVING TRANSFORMATIONS
BY
Allen Jay Beadle

Let (M1,d1) and (M2,d2) be two metric spaces. An
injective map Tully-"PM.2 is called distance transforming if
for some pair of positive numbers a and b, d1(x,y) a a
implies that d2(Tx,Ty) = b. If a = b, we call T
distance preserving and say that a is preserved by T.

In 1953, F.S. Beckman and D.A. Quarles proved that if
T:En-*En, 2:5n<oo, is distance preserving, then T is an
isometry. In 1973, R. BishOp gave a different proof.

In this thesis, we study a number of generalizations of
the Beckman-Quarles result. We show that, up to a point, it
can be extended to the classical non-Euclidean spaces and to
most Minkowski planes. Beckman and Quarles noted that their
theorem did not extend to Hilbert space but the example they
gave was not continuous. We have been unable to find a
continuous distance preserving self map of a Hilbert space
which is not an isometry. In the cases of 5", El, and a
wide class of Banach spaces (not strictly convex) we have
examples of continuous distance preserving self maps which
are not isometries.

The thesis is organized into six sections. In Section
1, we develope a few basic lemmas in a general setting.

Using methods similar to those of BishOp, we show in

Allen Jay Beadle

Section 2 that if M and M2 are two Minkowski planes

1
and M has no flat spots of length greater than I on

2

its unit circle, then the only distance preserving maps
T:M1-*>M2 are the isometries. It is shown that in any
Banach space with a flat spot of length 2 on its unit
sphere, this theorem is not valid. However, the situation
in Munkowski spaces of dimension more than 2 is unknown.
In diMension 2, the only known execption is the space
with max norm.

In Section 3, further exploiting the methods of BishOp,
we show that for hyperbolic spaces H“, 2 s n<oo , if
T:Hn---5>Hn is distance transforming then T is an isometry.
While the methods are similar to those used to prove the
corresponding result in E“, the computational detail is
much more involved.

Spherical and elliptic spaces are analyzed in Section
4. An example is given of a map Tzsn—-—>Sn which has two
preserved distances, v and %, but is not an isometry.
This leads us to impose the condition that the transformed
distance a must be “small enough". Specific bounds are
found which force the mapping to be an isometry, but they
may not be the “best possible“.

In Section 5, we attempt to improve on some of the
results of D. Greenwell and P. Johnson who considered some
directional restrictions on the set of preserved distances.

More specifically, let T:E2———’B2 be a map such that

Allen Jay Beadle
there is a set ,0 of unit vectors such that $537605 implies
that d(Tx,Ty) = 1. If .0 = S , the set of all unit
vectors, then T preserves the distance 1 and is thus an
isometry. The question considered in this section is how
small 43 can be and still force T to be an isometry. we
show that 43 can have arbitrarily small measure and still
force T to be an isometry, but this requires that the
interior of $3, in the relative tOpology of S, to be non-
empty and requires 45' to contain certain vectors.

In Section 6, we suggest a number of directions for
further research, describe a few partial results and give a
few examples of a rather negative character showing bounds

on the type of theorems to be expected.

DISTANCE PRESERVING TRANSFORMATIONS

BY

Allen Jay Beadle

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1977

to my wife

Anne

ii

ACKNOWLEDGMENTS

I wish to express my gratitude to Professor Leroy M.
Kelly, not only for his advice and guidance during the
preparation of this theses, but also for the encouragement
he has given me and the influence he has been on me during
the last dozen years.

I also wish to thank Professor Fritz Herzog for the
interest he has shown and the help he has been to me,
especially during my undergraduate years.

Finally, I wish to thank Professors H. Davis, T. Yen,
J. Zaks, and E. Nordhaus for serving on my guidance

committee and for their suggestions concerning this theses.

iii

TABLE OF CONTENTS

List of Figures . . . . . . . . . . . . . V
List of Symbols . . . . . . . . . . . . . vii
Nomenclature . . . . . . . . . . . . . . viii
Introduction . . . . . . . . . . . . . . 1
Section
§l. Preliminaries . . . . . . . . . . . 3
§2. Minkowski Planes . . . . . . . . . . 8
53. Hyperbolic Space . . . . . . . . . . l6
§4. Spherical and Elliptic Spaces . . . . . 38
§5. Directional Restrictions . . . . . . . 49
§6. Counterexamples and Open Questions . . . . 55

Bibliography 0 O O O O O O O O O O O O O 65

iv

Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure
Figure

Figure

3-10
3-11

3-12

LIST

OF

FIGURES

10
1o
11
12
14
16
17
21
21
22
26
27
29
29
34
35
36
4o
48
so

52

Figure b-l

Figure 6-2

vi

LIST OF SYMBOLS

c (c): the distance between the centroid of a
c-equilateral n-simplex and any of its vertices.

Cn(c): = cos(cn(c)) in elliptic or spherical spaces,

cosh(cn(c)) in hyperbolic spaces.
- l -
dn(c). - 2hn(c) cn(c).
D (c): = cos(dn(c)) in elliptic or spherical spaces,

= cosh(dn(c)) in hyperbolic spaces.

3“: n-dimensional Euclidean space.

8?: n-dimensional elliptic space.

H“: n-dimensional hyperbolic space.

hn(c): the distance between the two remaining points of
two c-equilateral n-simplices having n points
in common.

Hn(c): = cos(hn(c)) in elliptic or spherical spaces,
= cosh(hn(c)) in hyperbolic spaces.

5“: n-dimensional spherical space.

xn(c): the angle between two adjacent (n-l)-simplices
which are the faces of a c—equilateral n-simplex.

[xyz]: y is between x and 2.

vii

NOMENCLATURE

between: A point y in a metric space (M,d) is between
points x and y if d(x,y)+d(y,z) = d(x,z), and

this relation will be denoted by [xyz].

centroid: For n 21, the centroid of an n—dimensional
equilateral simplex in an elliptic, spherical,
hyperbolic, or Euclidean space is the intersection
point of the n+1 segments which have one end at a
vertex of the simplex and the other end at the centroid
of the (n-l)-dimension simplex which is the face
Opposite that vertex. For n = O, the simplex is just

one point which is its own centroid.

convex metric space: A metric space (M,d) is convex if
for every pair of points x,y in M, there is a point

z in M such that [xZY].

diameter of M: For a metric space (M,d), we will say

that the diameter of M is sup {d(x,yl}.
x,yeM

l—"’M2 between two metric

spaces (M1,d1) and (M2,d2) is distance preserving

distance preserving: A map T:M

if there is some distance a such that x,y in M1

and d1(x,y) = a implies d2(Tx,Ty) = b.

viii

distance transforming: A map T:M1-—-!>M2 between two metric
spaces (M1,d1) and (M2,d2) is distance transforming
if there are two distances a and b such that for

any x,y in M, d1(x,y) = a implies d2(Tx,Ty) = b.

r—equilateral n-simplex: A set of n+1 points in a metric
space is an r-equilateral n-simplex if the distance

between any two of them is equal to r.

equilateral unit lattice: A set of points in E2 is an
equilateral unit lattice if it consists of the points
_,

{31+nv2+m33} where n and m range over all integers,

61 is any vector, ani V? and 33 are two unit

vectors at an angle of 60° to each other.

extension property: A metric Space (M,d) has the exten-
sion pr0perty if for each two points x,y of M and
each distance r such that d(x,y)s rsédiameter of M,
there exists a point z in M such that [xyz] and

d(x,z) = r.

externally convex metric space: A metric space (M,d) is
externally convex if for any x,y in M, there is

some 2 in M such that d(x,y)+d(y,z) = d(x,z).

flat spot: A Banach space is said to have a flat spot on

its unit sphere if there is a segment of length greater

than 0 on any unit sphere.

ix

isometry: An injective map T:M1---a--M2 between two metric
spaces (M1,d1) and (M2,d2) is an isometry if for

all x,y in M1, d1(x,y) = d2(TX,TY).

length of a segment: If a segment in a metric space has

endpoints x and y, then the length of the segment

is d(x,y) .

local isometry: A map T:M1—erM2 between two metric spaces
(M1,d1) and (M2,d2) is a local isometry if for every
x in M1, there is a distance r(x)> 0 such that

y,z in M dl(x,y)<r(x) and d1(x,z)( r(x)

1,
implies that d2(Ty,Tz) = d1(y,z).

locgl:isosceles prOperty: A metric space has the local
isosceles prOperty if there is a p(M) > 0 such that
for any positive :1 less than p and any x,y in M,
with d(x,y)$'2A, there is a point 2 such that

d<x,z) = «2.1» = A.

Minkowski space: A Minkowski space is a finite dimensional

Banach space.

T :preserves a: A map T:M1—--I>M2 between two metric
spaces preserves a if for all x,y in M,

d1(x,y) = a implies d2(Tx,Ty) = a.

segment: A set of points in a metric space (M,d) is a
segment if it is the isometric image of a finite

closed interval of the real line.

segmentally connected: A metric space is segmentally
connected if every two points in it are the endpoints

of some segment.

I, transforms a into b: A map T:Ml~-—a>M2 between two
metric spaces (M1,d1) and (M2,d2) transforms
distance a into distance b if for any x,y in M

l
with d1(x,y) = a, it follows that d2(Tx,Ty) = b.

xi

DISTANCE PRESERVING TRANSFORMATIONS

maonucn ON

In 1953, F.S. Beckman and D.A. Quarles [1] proved that
if T is a mapping of E“ into En for 25 n<0° which
preserves one distance, then T is an isometry. In 1973,
R. BishOp [3] gave a different proof.

The aim of this thesis is to extend this result to
other geometries. It will be shown that this theorem is
true in any n-dimensional hyperbolic space for 21En<cv ;
it is true in any n-dimensional elliptic or spherical
space if the preserved distance is small enough; and it is
true in any Minkowski plane whose unit circle does not have
a flat spot of length greater than 1. In hyperbolic and
spherical and elliptic spaces, a further generalization
will be shown: if T:M->M. such that there are two
distances a and b such that x,yéEM and d(x,y) = a
implies d(Tx,Ty) = b, then a = b and T is an isometry
(here again in the Spherical and elliptic spaces, a and
b must be small enough). Note that in Euclidean spaces
(and Minkowski spaces) any such map is the product of a
similarity and a distance preserving map.

As noted by Beckman and Quarles [1], this theorem is
not true for E1, the Euclidean line, or Eafl Hilbert
space. In E1 the map T defined by Tx = x+l if x is
an integer point and Tx = x otherwise is a counter-

example. In E“, a counterexample is found as follows:

1

2

let {yi} be a countable everywhere dense set of points.
Define R:E“¥—afyi} so that d(x,Rx) < %3 Define
S:{yi}**+{ai} so that Syi = a1 where ai is the point
in E” with coordinates (ail,a12,...) such that aij =
[ijA/E’ where Jij is the Kronecker delta. Then T = SR
is a map of E” into itself which preserves the distance
1. For if d(x,y) = 1, then Rx # Ry and hence Tx # Ty,
But T is not an isometry.

The theorem is also not true in particular Minkowski
spaces, such as any max norm space (see Section 2). It is
also untrue in spherical spaces for particular distances.
For example, in Sn, if T maps every point onto itself
except the north and south poles and maps these two points

onto each other, then it is clear that T is not an

isometry, but T does preserve the two distances w and

I.
2.

§l. PRELIMINARIES

Generally we will be concerned throughout this thesis
with an injective mapping, T, from one metric space
(M1,d1) to another, (M2,d2). Such a mapping is said to be
distance transforming and to transform distance a into b
if whenever dl(x,y) = a, then d2(Tx,Ty) = b. If a = b,
then T is called distance preserving and we say that T
preserves a. If T preserves ai for each term of the
null sequence {ai} then we will frequently say that T
preserves arbitrarily small distances. Analogously, if the
{ai} approach infinity, T is said to preserve arbitrar-

ily large distances.

Definition. A metric segment is the isometric image

of a real segment.

Definition. A metric Space is called segmentally con-
nected if every two points of the space are the endpoints

of some segment.

Definition. If in a metric space, d(x,y)+d(y,z) =
d(x,z), then y is said to be between x and z. This

relation is denoted by the symbol [xyz].

Definition. A metric space (M,d) has the extension
property if for each two points x,y in M and each real
number r such that d(x,y)< r:§diameter of M, there is a

point z in M such that [xyz] and d(x,z) = r.

4

 

Definition. If in a metric space [xyz], [yzw] and
d(x,z)+d(z,w) is less than the diameter of the space,
imply [xzw], then the betweenness relation is said to be

externaly transitive in this space.

Definition. A metric space (M,d) has the local
isosceles prOperty if there is a p(M):> 0 such that for
any positive r less than p, and any x,y in M with
d(x, y) _<_ 2r, there is a point 2 such that d(x,z) =

d(Z,Y) = r.

Lemma 1.1. ;£_ (M1,d1) §gg_ (M2,d2) g£g_ggg_metric
spgces ghggg, (M1,d1) ig_seggentaly connected ggg_hgg_£hg_
lgggl_isosceles proggrty ggg_ T:M.1——-*M2 transforme a
ig£g_ b, ghgg_d1(x,y)§ ka implies Eggg. d2(Tx,Ty)£ kb,

for k any positive integer greater than 1.

Proof. Let zo,zl,...,zj be points on a segment with

endeints x and y such that x = 20 and d1(zi’zi+1)

a for osiéj-l, and a<d1(zj,y)s 2a. Since (M1,d1)
has the local isoscles property, there is a point zj+1

s dl(z = a. If we denote y

z3+1) j+1’ Y) 3..
= (
then d2(TX,TY) 62(T209sz+2)- 3362(Tzi’Tzi+l)

to

such that d1(zj,
by zj+2’
= (j+2)b. But j+l< k-l, so j+2$k and d2(Tx,Ty)$ kb.0

The following is a generalization of a lemma of Bishop

[3].

5

emma 1.2. ;£_ (M1,d1) and (M2,d2) are two metric
spaces such that (M1,d1) 'ig segmentaly connected, has the

local isosceles ppgperty and the extension property and the

 

 

 

 

diameter g§_ M1 i§_not less than that 9;, M2, and pg
T:M.1--—-=>M2 preserves 3 null segpence pg distances, then T

preserves all distances less than any preserved distance.

Proof. Let a be any preserved distance and let c

 

be less than a. Let x,y be in M with d1(x,y) = c.

1
Let 2 be in M such that [xyz] and d1(x,z) = a.

1
Such a point 2 exists by virtue of the fact that (M1,dl)
has the extension prOperty.
Let u = d2(Tx,Ty) and v = d2(Tx,Tz). We wish to
show that c = u. Let r be a preserved distance under T

such that r< min (c,a-c) and r< p(M Then there exist

1).
integers k and m z 2 such that (m-l)r< c 5mr and
(k-l)r<a-c_<_ kr. By Lemma 1.1, we have ugmr and v.<_kr.
Therefore u-c < r and v-a+c< r. Now, by the triangle
inequality, we have a5;u+v and this combined with the

previous line gives: c-u_<_c+v-a< r. Therefore Ic-ul < r.

Since r can be arbitrarily small it follows that c = u.u

Lemma 1.3. ;£_ (M1,d1), (M2,d2) and T are pp ;p_
Lemma 1.2 and if. ip_addition, the betweenness relation ip_

(M2,d2) i§.externaly transitive, then T ig_gp.isometry.

Proof. Let c = sup{r: r is preserved under T}. If

c =¢x> then for any distance a, there is a preserved

6

distance larger than a, so by Lemma 1.2, a is preserved.
Hence T is an isometry.

If (M1,d1) is bounded and c is its diameter, then
all distances less than c are preserved by Lemma 1.2. If
there is a pair of points x and y in M1 such that
c = d1(x,y), then for any 8 such that c > E > 0, there is
a point 2 such that d1(x,z) = E and d1(z,y) = c-E.
Both 8 and c-E are preserved.

From the triangle inequality we have that

d2(Tx,TY) 2 d2(Ty,Tz) - d2(Tz,Tx) = c-ze.
Now since 9 can be taken arbitrarily small and the
diameter of M2 is no greater than c it follows that
d2(Tx,Ty) = c and T is an isometry.

Now suppose (M1,d1) is bounded but c is less than
its diameter. Let r be a distance in (M1,d1) such that
2rW<c, Note that r and Zr are preserved by T because
f such that

of Lemma 1.2. Consider four points f ,f

3’ 4
and c <1

1’f2

f2 and f3 are both between f1 and f4

d1(f1,f4)< c+r, dl(f3,f4) = r and d1(f2,f4) = 2r.

Since all distances less than c are preserved, sz

is between Tf

is between Tf and Tf . Similarly Tf3 2

1 3

4. Now the diameter of M2 is not less than that

so d2(Tf1,Tf3)+d2(Tf3,Tf4) is not greater than

and Tf

of M1,

the diameter of M2 since it is equal to d1(f1,f4)

d1(f1,f3)+dl(f3,f4). Since the betweenness relation in M2

is transitive, we have:

7
d2(Tf1,Tf4) = d2(Tf1,Tf3) + d2(Tf3,Tf4)
d1(f1,f3) + dl(f3,f4)

= dl(f1,f4).

Therefore the distance d1(f1,f4), which is greater than

c, is preserved under T contradicting the definition of

c. Hence c equals the diameter of M1 and T is an

isometry. U

§2. MINKOWSKI PLANES

 

l and M2 will denote two

Minkowski planes such that the unit circle in M2 does not

have a flat spot of length > 1. Also. T will denote a

Throughout this section. M

mapping of M1 into M2 which preserves the distance 1.
The objective of this section is to show that T must be an
isometry. In this section the distance between points x

and y will be denoted by Hx-yH if they are in M1 and

tar MX-yul if they are in M and we will use the vector

20

notation in M1 and M2. The method used here is similar

to that used by Bishop [3].

Lemma 2.1. If. x,y 6 M2 and “bead" = 1 then there

 

exists a_unique pair pf points a and b such that

 

 

IUX-alll = lily-am = lUy-bfll = Hx-bHI = 1- For these a and

 

b, [Ha-b!” > 1 and x -a =b -y.

ngpf, Let Ck and Cy be the unit circles in M
with centers x and y. Then Ox n Cy # ¢ so let
a E Ck 0 cy. and set b==x4-y-a. Then b E ck 0 cy and
the required norms are = 1.

To show uniqueness. suppose ck n cy contains at least
3 points. Since none of them can be on the line xy, two
of them must lie on the same side of line xy: call these
two points c and d. Then points y-—x+—c and y-x4-d
are on 6y. so the convexity of Cy requires that c.d,
y-x+-c, and y-x4-d be collinear. Then they are on a
segment of length > 1. Hence no such c and d exists.

8

 

 

 

 

 

x+a-b a y
x Y
Figure 2-1 Figure 2-2
Finally, IHa-bm > 1 since if Wa-bHI < 1, then

x4-a-b would be inside 6x, which cannot be since ex
is convex and a 6 OX, y 6 Ox; and if IHa-bm = 1, then
x4—a-b E Ox and hence the segment from y to x4—a-b

is a flat spot of length 2. 0

Lemma 2.2. If x,y 6 M and Hx-yH 1 then there

 

1

exists a and b 6 M1 such that Hx-aH

Hb-aH and b-a = x-y. Also, a and b depend continu-

Hy-aH = Hx-bH =

 

 

ously 93’ x if y ig held fixed.

Egggf. There are two points a, b on the unit circle
with center x such that b-a = x-y. Then linearity
gives that the needed norms = 1. Note that if a is
restricted so that triangle xya has positive orientation,
then a and b are unique by the same argument used in

Lemma 2.1. This implies that a and b are continuous

10

b,,,/’*——-~—~\~,,e

x /y
Figure 2-3

functions of x if y is held fixed. 0

 

 

 

Lemma 2.3. 1;: a,b,x,y 6 M1, b 79 y, llx-yl = :y-all =
Hb-aH = Hx-bH = Hx-aH = 1, then there exists §,S,§ 6 M1
such that

IIy-L-Zu = Illa-'13“ = H’s—ml = His—2;“ = lib-Bu = ”SE-y” = 1.

II
|-‘
‘

Proof. By Lemma 2.2, for any § 6 M1 with Hx-yH

there is 3. b so that Ha-yH = Ha-bH = H3-§n = u§-Su

II
I-‘
O

0‘)

-.
—— —..

 

Figure 2-4

But 3 is a continuous function of E and b = b when

rd

§ = x and b = b = 2y-b when § = 2y-x, then since

ll

Hb-bH = 0 and Hb-b“ 2Hb-yH 2 2, there is some choice

of E so that Hb-bH 1” a

Lemma 2.4. If T has arbitrarily large and arbitrarily

 

small preserved distances, then T ie ea isometry.

 

Proof. This is immediate from Lemma 1.2. 0

Lemma 22;. If a,b,c,d G M1 such that Ha-bH = Ha-cH =

Hd-bH = Hb-CH = Ha-dH = 1 then Tc # Td.

 

Proof. By Lemma 2.3 we have a,b,d 6 M1 so that

\J N N rV

HB-cll = H’b—dll = HIE—’5” = IIE—cll = Ila-bll = lld—a’ll = 1.

Now since T preserves the distance 1, we have

1 = mTa-TbHI = mTa-TCHI = WTb-TCHI = mTa-Tdm ==|HTb-Tdm
= IIITE-TB’III = IIIT'S—Tclll = HITS-Tell! = HITS-Tall! = HITS-T3“!
= lHTd-TEHI.

d b Tb

 

(spun
\Jr
0
fiii

0)
a
m

Figure 2-5

Then by Lemma 2.1, either Td = or Td = Tc (where

N

x
x = Ta+—Tb-Tc). Also either Td = x or Td = To (where

12

32’ = T§+Tb-Tc).

But ud-En = 1 =¢’|HTd-Tam = 1 while Lemma 2.1==a
|”x-TCH| > 1 and mx-TCHI > 1. Also MTc-TCHl = 0. So
the only possibility for Td and Ta is Td = x and
T3 = fi.

Hence Tc # Td. o

 

 

 

Lemma 23g. .3: xl,x2,...,xn 6 M1 such that xl - x2 -
xk - xk+l for k = l,2,...,n-1 and Hxl-xZH = 1 then
MTxl-TXZHI = l and Txl - Tx2 = Txk - Txk+1 for

k: l,2,...,n-l.

2 6 M1 such that

bl - b2 = x1 - x2 and Hxl-bln = ”bl—b,” = “bl-X2“ =

Proof. By Lemma 2.2, there is b1,b

Hbz-XZH = 1. For k 2 3, define bk = bzi-(k-2)(b2-bl).

 

 

 

Figure 2-6
Then
ka-xk+lH = ka-ka = ku-xk+1H
= ”bk’bk+1“
”xk+1'bk+1“ 1

and

= Hb

k

for 1.2,:-

Then by Lemma 2

3 ‘ sz

and by induction on

Tx = sz - Tb

k = l,2,'--,n-l. 0
Mar “242 °

preserved by T.

 

Proof. This is

”X

_ I =
n+1 Xl‘

”n(TXZ—Tx1

Lemma_21§.

= ubk’bk+1H = HXk+1'bk+l*

k+1'xk+2~

.l,

1 O

)H

For any integer

13

H = uxk+l-xk+2H

H = 1

-,n-2.

Tb - Tb -

2 1 _

therefore

k: - TX

TXk+l

For every integer n 2 l,

 

immediate from Lemma 2.6 since

”n(XZ-X1)H = nHXZ—Xl H = nHsz-Txlu

—Tx

= HTXn+1

In.

1

n 2 l. is preserved

 

 

p1 T.

Proof. Let c

unit circles with centers

and b==y+-(n-l)(Y—c).

UnX—nYH
Hb-CH = n-

and Tc-—Tb

on-yH = 1

Therefore Lemma 2.6 implies

be a point of intersection of the two

x and y. Let a==x+-(n-l)(x—c)

Then Hc-yH = Hc—xu = 1, Ha-b“ =

, Ua—CU = an—nc“ = on—CH n,

Tc-—Ta n(Tc—Tx)

n(Tc-Ty).

14

 

ya
x
c
Y
‘ b
Figure 2—7
Therefore "‘Tx-Tyul = %-“‘Ta—Tb"| = %- since Ha—bH = 1.

This gives us the main result of this section:

Theorem 1. If_ T: M1 -—9 M2 1 e_mapping between two

 

Minkowski planes M and M where the unit circle ip_ M

 

 

 

 

 

l 2 2
has pg flat spot pf length > 1 and T preserves the
distance 1, then T i§_pp_isomet£y.

Proof. This is immediate from Lemmas 2.4, 2.7, and
2.8. 0

This theorem is not true if M1 = M2 = 2m as the fol-
lowing example shows: let M1 = M2 = {(x,y) | x,y real}
and Wlth max norm, “(xl,yl)-(x2,y2)H = max(|x1-x2|,|y1-y2I).

Let T be defined by T(x,y) = ([x]+{x}2,[y]+{y}2) where
[z] is the largest integer s z and {z} = z - [2]. Then T
preserves all integer distances, but is not an isometry.

A similar example can be found in any Minkowski or
Banach space which has a flat spot of length 2. This

includes all £1 and 1m spaces.

15

It is not known what the situation is in Minkowski
planes with a flat spot of length > 1 and < 2 or in
Minkowski spaces of dimension 2 3 without flat spots of

length 2.

§3. HYPERBOLIC SPACE

 

Let Hn be n-dimensional hyperbolic space and let

T: H“ + H“

be a mapping so that there are two distances a
and b such that d(x,y) = a =%:d(Tx,Ty) = b. The aim of
this section is to show that T must then be an isometry.
The method used here is similar to that used by Bishop [3]

in En for n z 3, especially the idea behind Lemma 3.6.

In this section, the following notation will be used:

hn(c) = the distance between the two remaining points of
two c-equilateral n-simplices having n points

in common;

example: n = 2

 

AC - h2(c)
Figure 3-1
Hn(c) = cosh(hn(c));
cn(c) = distance between the centroid of a c-equilateral
n—simplex and any of the vertices;
Cn(c) = cosh(cn(c));
d (C) = J'h (C) - c (c)-
n if n. n '
Dn(C) = cosh(dn(c));
xn(c) = the angle between two adjacent (n-l)-simplices

which are faces of a c-equilateral n-simplex.

16

o — o = l o =
Note. c0(c) — 0, c1(c) 7c, so C0(c) 1, and

 

_ l + cosh c

 

When there is no ambiguity in the value of the argument

of the functions h , H , c , C , d , D , and X ; they
n n n n n n

n
will be written without the argument.

= 2n(cosh c)2
l + (n-l)cosh c

 

Lemma 3.1. Hn(c)

__.—..._. ’—_-

— 1 for n 2 1.

Proof. Let A = cosh G. Let fl,el,e2,...,en and

f2,e1,e2,...,en be two c-equilateral n-simplices with

f1 # f2. Let 91 be the midpoint of the segment flf2 and

let g2 be the centroid of the simplex f1,e1,e2,...,en.

example: n = 2

 

 

 

e2

Figure 3-2

Then gl is the centroid of the c-equilateral (n-l)-

simplex el,e2,...,en; and d(e1,gl) = Cn-l; d(el,gz)

_. -1. ___.
d(fl,g2) — on, d(f1,gl)- 71%“ d(g1,gz) dn' and angle

f is "/2.

19181

Hence

18

Agzgle1 is a right triangle==e

 

cosh Cn = cosh cn_1 cosh dn =e:Dn = Cn/Cn-l’
Similarly
Af g e is a right triangle==¢
l l l 2
coshc=cosh£h coshc ==>H = 2A -1.
2 r1 n-l n C2
n-l
Therefore
(1 = —l-h -c
n 2 11 11

cosh-i-hn cosh cn - sinh%-hn sinh cn;

/H +1 /H -1
==% C /C = D = n C - n \/C2-l ;
n n-l n 2 11 2 n

1i
0
O
U)
D“
Q.
:3
ll

 

 

 

 

 

 

 

 

 

H -l 2 H +1 1 V2'VH +1 2
==> n (C -1) ‘ n + ———— --—————9—— C ;
2 n 2 C2 Cn-l n
n-l
2 H -1 1 v2 x/rTTi
: C = n "1 _ + n
n 2 2 C
n-1
n-1
A2 _ 1
2 2 2
_ Cn-l _ A Cn-l
‘ 1 A " 2 ‘
‘1 "' 5-2-— + 2 ET- 2A "‘ Cn-l 1
n-1 n-1
Since 00 = O, CO = 1, by induction on n it follows
that:
C2 = nA+l for n > 0
n n+1 ’ ’

Then

19

 

 

 

2 2
= 2A — :1: = .._._2__n_A_.—__ -- >
Hn 2 l 9’ Hn 1+(n-1)A 1 for n _ l. u
n-l
Hn(c)-l . .

Lemma 3.2. For all c, cosh<: _ 1 > 2, ‘ip_particular,

hn(c) > c.
2

Proof Let A = cosh c Then H = ——3§§——— - l==$

--———° ° n l+(n-l)A
Hn-l nA+l

 

.A-l =‘2 TEZITKII > 2'

So Hn > 2A-l > A, (since A > 1) hence

hn(c) > C. 0

Lemma 3.3. I£_ d(x,y) = a =%’ d(Tx,Ty) = b, then

d(x,y) = hn(a)==$' d(Tx,Ty) = hn(b).

Proof. Let fl,e1,e2,...,en and f2,el,e2,...,en be

two a-equilateral n-simplices with fl # f2. Then
hn(a) = d(f1,f2).
Now hn(a) > a implies that the (n-l)-sphere with

center fl and radius hn(a) has point f2 so that

d(f2,f2) = a. Then there exists points e1....,en such

that £1,31,32,...,En and 22.31.32,....En are a-

equilateral n-simplices.
Then Tfl, Tel, Te2,...,Ten and Tf2,Te1,Te2,...,Ten
are b—equilateral n-simplices, so d(Tfl,Tf2) = 0 or

hn(b) depending on whether or not Tf1 = sz. Likewise

Tfl,Tel,Tez,...,TEn and T?2,T31,TE2,...,TED are b-

equilateral n-simplices and d(Tf sz) = 0 or hn(b)

1!

depending on whether or not Tf = T? . But d(f2,f

1 2 ) = a

2

20

so d(Tf2,Tf2) = b. So it cannot happen that Tf1 = sz =

sz since then d(Tf2,Tf2) would be 0; also Tf = sz #

1
sz cannot be since then d(Tf2,Tf2) would be d(Tf1,Tfé)=

hn(b) > b; and similarly Tf1 = sz # sz cannot be, so

Tfl g Tf and Tfl e sz. Hence d(Tfl,Tf2) = hn(b). 0

2

Corollary 3.4. f T has e_preserved distance, then T

 

has arbitrarily large preserved distances.

 

Proof. Let a be a preserved distance of T and de-

0

fine ak k-l)

preserved distance of T for all k 2 l, and by Lemma 3.2;

= hn(a for k > 1. By Lemma 3.3, ak is a

0 -1)2k. Hence cosh ak + a as k + w.

Hence ak 4 w as k + w. 0

cosh ak - l > (cosh a

cosh c
(n-l) cosh c + l ‘

 

Lemma 3.5. cos xn(c)

Proof. ln(c) is the angle opposite the base of a

triangle whose sides are éhn-lw)’ %-hn_l(c), and c.

The law of cosines yields:

 

cosh c = cosh2%-hn_l(c) —sinh2%hn_l(c) cos ln(c).
Let A = cosh c, Hn-l = cosh hn_1(c); then
Therefore
H +1-2A
cos Kn = 3-1 -¥l_’ = TH:{%A:1" o

21

example: n = 3
in (c) =d(e e)
2 2 l' 3
= d(e2,e3)

c = d(el,e2)

 

X3 = Lele3e2.

Figure 3-3

Lemma 3L6. If d(x,y) = a =e> d(Tx,Ty) = b, and

Zn . . N
x b) 15 223.22.1232EEEI then there are a and B such

n __
that d(x,y) = a =e d(Tx,Ty) = B’ and cosh 8—1 <

0.6(coshb-l) .

Proof. Let e1....,en_l,fo,f1 and el""'en-l’fl’f2

be a-equilateral n-simplices so that f0 # f and by

2;

 

Figure 3-4

22

induction let f be another such simplex

el""'en-l’fk-l' k
50 that fk_2 F fk for k 2 2. Since d(x,y) = are»

d(Tx,Ty) = b; Te1,...,Ten_1,Tfk_1,Tf is a b-equilateral

k
n-simplex for all k 2 1. Then the angle between the

(n-l)-Simplices Tel,...,Ten_1,Tf and Tel,...,Ten_1,Tfk

0

is kxn(b). Now, if Xj¥b7 is not an integer, there exists
n

< 1
k so that 0< |2n-kxn(b)| _ 711nm).

Now, d(f0,f2) = hn(a), therefore d(Tfo,Tf2) = hn(b)
by Lemma 3.3. Likewise, for all i, 0 s i 5 k-2 we have
d(Tfi,Tfi+2)
Hence d(x,y) = d(f0,fk) =e> d(Tx,Ty) = d(Tfo,Tfk) for

= hum.

all x, y in Hn since any such x and y can be made

the f0 and fk points of some such e1,...,en_l,fo,...,f€

Let a = d(fo,fk) and B = d(Tfo,Tfk). Let B be the
(smallest) angle between the (n-l)-simplices

Te ,...,Te ,Tf and Tel"'°'Ten-1’Tfk' Then

1 0
0 < B = [211 -kxn(b)| .<_ é—an) and (3 is the angle oppo-

n-l

site the base of a triangle whose sides are %hn_1(b) ,

‘b‘ r Tfo

U?

  

 

 

'ih
2 n_1(b) 0 Tf

Figure 3-5

23

%hn_l (b) , and '15.

The law of cosines yields:

~ _ 21 _ . 21
cosh b - cosh 2hn—l(b) Slnh -2—hn_l(b) cosB .

Let B denote cosh b. Then:

~ _ . 2 1 _
cosh b 1 — (Slnh -2-hn_l(b)) (1 cos (3 )

N < ' 2 l .. l
cosh b - 1 _ (Slnh 2hn-1(b)) (1 cos-z-ln (10))

 

 

 

Hn_l(b)-l 1 + cosln(b)
2 1" 2

2
2(n-l)cosh b
= 1—4 Th-2)c03h B'- 2 ( ._V/1.+-cos n(b))
’2

 

 

 

 

 

 

 

2
2
2(n-l)B _ 2 l + B
= l-tTn-2)B 1 _ l4-(n-1)B
2 2

 

 

 

_ (B—l)((n-l)B4-1) l _‘l; «f nBi-l
_ (n-2)B+1 V2 (n-l)B+-l °
Now if n > 2:

cosh b - 1

IA

(B-l) :17};- (l -\/-;:) < (B-l)2(l -/-§)

< 0.6(3-1).

If n 2, then:

24

(13-1) (B+l)(l - / % fig?)
(B-l) B + 1 — / %\/ZBZ+BB+1).

cosh b - 1

IA

 

Define p(B) = B + l - / % V2B24-3B4-l. Then

= 1 _ 1 4B+3 .

92
dB
2V5 7232+3b+1

Now: 1632 + 243 + 8 < 1632 + 24B + 9; so

8(ZBZ+BB+1) = (4B+3)27

from which it follows that

 

1 < 1 4B+l ,
2V2 VZB§+3B+1
hence %§-< 0 for all B > 0;

B>l=>p(B)<p(l)=2-/-;:V€=2-V3< 0.6;

therefore cosh b - l < 0.6(B-l). 0

Corollary 3.7. If d(x,y) = a =#:d(Tx,Ty) = b, then

N

either there i§_some a and 3 such that d(x,y) = a =e~

 

 

 

d(Tx,Ty) = b, b s b and ip pp integer, 2; there

N

)‘n (b)

are arbitrarily small 3 and 3 such that d(x,y) = a =¢>

 

 

d (TXITY) = 15.

Proof. If we assume that there is no 3 and 3 such

25

 

that d(x,y) = a *9 d(Tx,Ty) = b, b .<_ b and 21L is an
xn(b)

integer, then we can define a sequence of ak's and bk's

by letting a0 = a and b0 = b and then define ak and

bk inductively for k 2 l by setting ak = ak-l and bk =

bk-l where ak-l and bk-l are the results obtained from

Lemma 3.6 using a and b in place of a and b.

k-l k-l

Then cosh b - 1 2 0.6(cosh b

k - l) for k 2 1, so

k-l

cosh b - 1 < (0.6)k(cosh b

k —l)l

0

hence

cosh bk - 1 + 0 as k + on; therefore bk + 0 as k + m.

For any k > 0, d(x,y) = a -#> d(Tx,Ty) = b Let x

k k'
and y be two points in Hn such that d(x,y) = a. Let ik

a ; then by Lemma

be the smallest integer greater than 5—

k
1.1, b = d(Tx,Ty) s ikbk.' and so
i 2-5L, which + w as k + w since b + 0 as k + w.
k bk k
But then
5L + w as k + m,
ak
therefore

ak also + O as k + w. a

26

Lemma 3.8. If d(x,y) = a =%> d(Tx,Ty) = b > 0 and

. 2n 2n .
either or are integers, then a = b.
lnla) Xn(B)

2n . 2n

Proof. Let k = 1:737, i = XETBT° Then either k
or i (or both) is an integer. Let x,yo,yl,... 6 Hn such
that d(x,yj) = a for all j and d(yj’yj+l) = a and
yj # yj+2 for all j.

T
_, Y2 .1' 7/ 0 "Y&\
/2?/2i1\\\ //.' \,'Ty1
// \ , —
\ / h
/ Y1 /
. /
T i
. “‘9’ K Tx TY i
X :1]
\ fyoayk \ ' YO
\ ,.‘K /
\ “ a ,/
\‘\L_,’, 7 Yk-l \\-_2..-:—. ‘ TYi-i
Figure 3-6

We have two cases:

Case 1: i is an integer, k is not. Then Ty0 = Tyi and

Case 2: k is an integer. Then y0 = y and hence Ty0 =

kl
Tyk. Thus, 1 is an integer and i s k since i is the
minimum integer such that Tyo = Tyi. If i = k, then
a = b, since ln(c) is a monotonic function of c. If

i < k, then y0 # yi, but TyO = Tyi.

So in either case we have either a = b or for any two

27

pOints yo and yi such that d(yo,yi) = d(yo,yi); Tyo =
T91, hence d(Tyo,T§i) = 0. But then by Lemma 1.1; for any
w,z e Hn, d(Tw,Tz) = 0, hence the image of T is one

point and hence b = 0. Therefore b > 0 =e>a = b. 0

Lemma 3.Q. If d(x,y) = a =e> d(Tx,Ty) = b, then the

image under T of a circle of radius -]2'-hn_1(a) i_s_ con-

 

(b).

 

I O I I 1
tained pp some Circle pf radius -—hn_

2 1
2 n 2
Proof. Let H C H be some plane, and let 6 C H be
a circle in that plane with center x and radius 2hn-1 (a) .

There are points e1,e2,...,e such that d(ei,ej) = a

n-l
for i % j, and point x is the centroid of the (n-2)-
simplex e1,e2,...,en_l, and plane H2 is normal to the
(n-2)~plane formed by el""'en-l’ Then C is the locus

of points which can be appended to e]_,...,en_._1 to form an

example: n = 3

 

 

 

 

Figure 3-7

a-equilateral (n—l)-simplex. But the image of an
a-equilateral (n-l)-simplex is b-equilateral (n-l)-

simplex, so the image of C under T is contained in the

28

circle whose center is the centroid of the (n—2)-simplex
Te1,Te2,...,Ten_1, whose radius is b, and whose plane is

normal to the (n-2)-plane formed by Te ,Te ,...,Ten

1 2 -1' 0

Lemma 3.10. ‘1: there are two seguences pf distances

 

 

 

 

 

{ai}i=0' and {bi}i=0 such that for all i, d(x,y) = ai

==» d(Tx,Ty) = bi and ai + 0 and bi + 0 pg i + w,
sinh % hn_l (ai)

then is constant over i.

 

. ll -——
811111 -2- hn—l (bi)

Proof. T is continuous since ai + O and bi + 0 as
i + a. Let x 6 Hn and let C be some circle with center
x and radius %hn_1(ai). Fix y E C. Consider the func-
tion given by f(w) = Lwa'Ty for w E C, where x' is
the center of circle TC. Then T continuous =%» f is
continuous. But for any y1,y2 6 C such that d(y1,y2)
ai, we have d(Ty1,Ty2) = bi’ so that f(yl) - f(yz) is

. = - ~ ~

always 2 LTylx Ty2 t(f(y1) f(y2)) as yl and y2 vary
over. C with L§1x§2 = Lylxyz. But continuity of f==s
f(y1)-—f(y2) = f(yl)-f(y2) for all such yl, y2. Hence if
y1,y2,y3,... is a sequence of points on C such that

d(yk,yk+1) = a1 and Lykxyk+1 = Lylxyz, then Tyl, Ty2,...

_. ' =
are on TC such that d(Tyk'Tyk+1) - hi and LTka TYk+1

For any i 2 0, let y1,y2,...,ym. he points on C
i
such that for all k < mi, d(yk'yk+l) = ai, Lykxyk+l =
Lylxy2 and d(yl,yk) > ai, and d(y1,ym ) s ai; then
i
d(Ty1,Tym ) s 2b1 by Lemma 1.1.
i

Y

29

Figure 3-8

Figure 3-9

TY

TY3

Ty2

3O

 

 

Let
c = circumference of C = lim.m.a.
1+” i i
= 2n sinh-l-h (a)
:2 n-l O ’
and let
3 = circumference of TC = lim mibi
1+»
= 2n sinhl'h (b )
'2 nrl 0 '
. l
Slnhfh _l(ao) ai
So that 1 = lim 5_ ; likewise,
Slnhi-hn l(b0) 1+» 1
sinh%—hn_l(ai) a.
1 = lim 5; by using a1 and bi instead of

sinh-—h (b.) i+~ i
:2 nrl i . 1
511111 2hn-l (ai)

 

and b above. Therefore is constant

a
O O . 1
811111 7 hn-l (bi)

over i. 0

Lemma 3.11. If there are two sequences 92 distances

 

 

such that for all i, d(x,y) = a.

{a 1

i}i=1’ EEQ- {bi}i=l

=e> d(Tx,Ty) = hi and ai + 0 and bi + 0 pg 1 + ”I

 

then there lg pp integer N such that for i > N, ai = bi'

 

 

Proof. For any i 2 l, we have by Lemma 3.3 that
d(x,y) = hn(ai) =a: d(Tx,Ty) = hn(bi), so if we let
a0 = hn(ai)' bo = hn(bi)' the“ {ai}i=0’ {bi}i=0

satisfies the hypothesis of Lemma 3.10 so we have

. l . l
Slnh2hn-l(ai) _ Sinh-z-hn_1(hn(ai)) Therefore

 

 

. l — . 1
Sinh-Z—hn_l (bi) Slnh 2hn-l (hn(bi))

31

 

 

. l . l

sinh- h _ (a.) Sinh-h _ (b.)

. 12n2____11 = . lgnll forall i.
81.th hn-l(hn(ai)) Sinh-z-hn_1(hn(bi))

Let A = cosh a, and define p by

2

 

 

 

 

. l
p(A) = ( sinh-i-hn_l(a) = cosh hn-Jlja) -l .
sinh‘lz‘hn_1(hn(a)) cosh hn_1(hn(a)) -l

2(n-1)A2

By Lemma 3.1, cosh hn_1(a) = 1—: (n—2)A - l,
2

2 (rm-1) (3(a))
cosh hn_1(hn(a)) = 1 + (n-2)Hn(a) - 1, and
2nA2
Hn(a) = 1 + (n-1)A - 1, so it is easily seen that p(A)

is a rational fraction in A. There are only a finite
number of roots of %§ which are ) 1, so there is some A
such that p(A) is monotone in the interval (1,3). For i
sufficiently large, cosh a1 and cosh bi are in (1,3),
hence there is an N such that for i > N, p(cosh ai) =

p(cosh bi) only if cosh ai = cosh bi' Therefore ai =bi
for all i > N. D

Corollary 3, 2. _§, d(x,y) = a2==:’d(Tx,Ty) = b, then

there i§_p distance Sfiéa such that d(x,y) a 3' ===9

 

d (Tx, Ty) = 3’.

Proof. This is immediate from Corollary 3.7 and Lemmas
3.8 and 3.11. D

For the remainder of this section we assume that a is

a preserved distance, that is, that

32

d(x,y) = a =e’ d(Tx,Ty) = a.

V
U1
0
H

Corollary 3.13. .2: T preserves a and n _

 

n = 3, then T ‘ig ep_isometry.

Proof. If n 2 5, by Lemma 3.5, cos xn(a)

 

1+-(:31?c:sh a . Now cosh a > 1, so 0 < %-< cos Xn(a) <
fiéf S %.. Then
%F' > Xn(a) > cos.1 % > %g .
Hence
4 < 2" < 5 for all a so 2" is non-inte er.

Therefore by Corollary 3.7 and Lemma 3.8, T has arbitrar-
ily small preserved distances. Then by Corollary 3.4 and

Lemma 1.2, T is an isoemtry.

 

- _ cosh a
If n — 3, then cos 13(a) - l4-2cosh a . So
1 1
'3' < COS X3 (a) < 72' .
So that
%; > cos.1 %- > 13(a) > cos-1 % = %g .

2n 2n . .
Therefore 5 < 13737 < 6, and 13737 is non integer.

Again T has arbitrarily small preserved distances and

is an isometry. 0

Lemma 3.14. If n = 4 and T preserves a, then T

 

£2.22 isometpy.

33

 

 

Proof. For any 5: cos 14(5) = COShEi . Then
1-+3cosh a
% < cos 14(5) < %'
Hence
-1 1 . -l 1
cos 4 > 14(a) > cos 3.
Zn . . A 2n
SO that is an integer, x (a) = —— , and
. 4 5
MM!)

cos 14(5) .e 0.30902,
cosh a x 4.2361

5 e 2.1226.

Now if a < a, then by Corollary 3.7 and Lemma 3.11, T
has arbitrarily small preserved distances. Also if a 2 a,
then either T has arbitrarily small preserved distances or
T preserves 5.

But if 5 is preserved by T, then by Lemma 3.3,
h4(3) is preserved by T, and by using the construction of
Lemma 3.6 with h4(5) replacing a and b in the lemma,
we get a preserved distance 3 which by calculation is
about 0.241226. So 3 is a preserved distance smaller than
5, so by Corollary 3.7 and Lemma 3.11, T has arbitrarily
small preserved distances.

Therefore T is an isometry by Lemma 1.3. 0

Lemma 3.15. If n = 2 and a i§_preserved py T, and

 

ie pp integer, then T i§,ep.isomet£y.

 

2n
1 a)
n

 

Proof. Let k = .

. ~ _ 1.
Case 1. k is even. Let k- 3k. Let ei, fij he

points for i 2 l, l S j s k-l so that for all i, and

S k-l, e e f

1+1 11' ei+1ei+2

are equilateral triangles.

for 2 S f and

3 i ik-l’
ei+lfij-lfij

 

 

Figure 3-10

Then h2(a) is preserved by T=%>T is a motion on this
set of points. Also e1,e2,... are collinear since k is
even. Hence there exists a distance a (= a) such that
for any collinear set of points {ei} such that
d(ei,ei+1) = a, then T is a motion on the set {ei}; so

that m3 is a preserved distance for all positive integers

m.

IV
|-‘

Case 2: k is odd. Let k: %(k+l). For i

i S j s k+1, let ei, fij be points so that e f f and

[\J

filfijfij+l for 2 E j S k, and ei+lfijfij+l for

k S j S k are a-equilateral triangles and ei+1 # fil'
Then h2(a) is preserved by T =e>T is a motion on

this set of points. Also k is odd w---—--=>el,e2,... are

35

 

 

 

f f
f14 f24
_____ _____ez _____ e f28
e1 f11 fie -f31
f
f 21 f
example: 15 f 25 f2?
f 17 f 6
k = 7 16 2

Figure 3-11

collinear. Hence there exists a distance a (= a-th2(a))
such that for any collinear set of points {ei} with

N

d(ei,ei+l) = a, then T is a motion of the set {ei}.

So whether k is even or odd, there is a preserved dis-
tance a such that m3 is preserved by T for all posi-
tive integers m.

Consider five points e,f f such that

1' 2'91'92

d(erfl) = d(erfz) = d(gl'gz) = g and d(elgl) = d(ergz) =

mg and d(f1,gl) - d(f2,gz) = (m-l)'5 for an integer m22.
Then T is a motion on this point set, hence 3 =

d(f1,f2) is a preserved distance. Let B be the angle

Lglegz. The law of cosines for triangle eflf2 yields:

e 2rv . 2 ~
cosh a = cosh a - 31111"! a c038,

and applied to the triangle eglg2 it yields:

~ 2 rv . 2 .e
cosh a = cosh ma - Slnh ma cos B.

36

 

 

Figure 3-12

 

So
cosh 3 - l = sinh2 m3 (1 — cos B)
and
cosh 3 - l = sinh2 S (l - cos 8).
Therefore
.. sinhz'a’ (cosh '5 - l)
cosh a = 1 + .

. 2 ~
Slnh ma
Now as m + w,

sinhza' (cosh '5 - 1)

1+ +1,

 

O 2 N
Sinh ma
hence a + 0 as m + w, and T is an isometry by Lemma

1.3. D

gprollary 3:16. .1: n = 2 and a is e preserved

 

distance pf T, then T ig pp isometry.

Proof. By Corollaries 3.7 and 3.12, either T has

37

arbitrarily small preserved distances (making T an isom-

etry by Lemma 1.3) or there is a preserved distance 3 such

 

that. 21’ is an integer, again making T an isometry by
12(a)
Lemma 3.15. 0
Therefore:

Theo em 2. f n 2 2, Hn ig n-dimensional hyperbolic

_" —

 

h

 

space, T: H 4 H", and d(x,y) = a =€> d(Tx,Ty) = b # 0,

 

then a = b and T must pe pp isometry.

§4. SPHERICAL AND ELLIPTIC SPACES

 

11

Let T: M + M where M is either 8 the n-

dimensional sphere in En+1 (with the shortest arc metric),
or E“, the n-dimensional elliptic space. Let T have
the property that there are two distances a and b such
that d(x,y) = a =s> d(Tx,Ty) = b. Our aim in this section

is to show that T must be an isometry if is small

a
enough. Let a be restricted so that a S % if M is a
spherical space and a < % if M is an elliptic space,
then cos a 2 0 and a-equilateral n-simplices have unique

centroids.

In this section, the following notation will be used:

hn(c) = the distance between the two remaining points of
two c-equilateral n—simplices having n points
in common;

Hn(c) = cos(hn(c));

cn(c) = distance between the centroid of a c-equilateral
n-simplex and any of the vertices;

Cn(c) = cos(cn(c));

d (c) = in (c) - c (c)-

n 2 r) n ’

Dn(c) = cos(dn(c));

xn(c) = angle between two adjacent (n—l)-simp1ices which

are faces of a c-equilateral n-simplex.

38

39

.. . — .1; e =
Note: c0(c) — 0, cl(c) — 2c, so C0(c) l, and

_ 1 + cos c
C1(C) ‘x/F 2 '

When there is no ambiguity in the value of the argument

 

 

of the functions hn' Hn, c , Cn' d , Dn’ and 1n, they

n n

will be written without argument.
2
2n(cos c) 1

= _ >
Lgmflg=ééi' Hn(c) 1-t(n-l)cos c £95 n ‘ 1'

 

_.- .__ ,.._,__

Proof. Let A = cos c and let f1,f2,gl,g2,el,...,en

be as in Lemma 3.1.

_ _ 2 2 _
Then Cn/Cn-l - DD and Hn — 2A /Cn-1 1, the same as

in Lemma 3.1.

 

 

 

 

 

 

Also
d == 111 - c
I) 2 n n
_ 1 . l .
==> cos d — cos-—h cos c + Sln-h Sin c ;
n 2 :1 r1 2 r: n
C H +1 1-H
-~—> n = D = / n C + n\/1-C2
C n 2 n 2 n
n-l
H-l n+1 V'z—VH +1
n 2 _ n 1 n 2
==e -———— (C -1) - + - C .
2 n 2 2 C n
C n-1
n-1
From this point on, the proof is the same as in Lemma
3.1. D

2
Lemma 4.2. ;£ 8n2A3 + (-n2+10n-1)A + (-2n+2)A - 1 2 0

h—“——

 

then a circle of radius hn has two points pp distance c

from each other.

 

 

4O

 

 

Figure 4-1
Proof. What we need is a sufficient condition that
there exists an angle 8 such that

2 .
cos a = cos hn + Sin2 hn cos 8.

That condition is that -l 5 cos B S l where cos 8 =
2 .

 

 

cos a - cos h
n
sin2 hn
M:
Then cos B = ———§ so that H < l and A < 1==e
n
l-H
cos B < 1. Now
A-Hn
cos 8 = -———
1-32
n
A ___2_n£_-12
= l+(n-1)A
2nA2 2

1 - 1+(n-ISA - 1

2

-4n A4+(5n2—6n+l)A3+(—n +8n-3)A2+(-2n+3)A-l
‘ 2 4 *2

-4n A +(4n -4n)A3:+4nA2

2

 

41

(l-A)(4n2A3+(-n2+6n-1)A2+(-2n+2)A—1)
(1-A)(4n2A3+4nA2)

All we need now is -1 5 cos 8.

But
0 s 8n2A3 + (-n2+10n-1)A2 + (-2n+2)A - 1
==e -4n2A3-4nA2 s 4n2A3-+(-h2+6n-1)A2-+(-2n+2)A—1
==> -l 5 cos B. 0

Remark: For any positive integer n > 1, the cubic (in

2 3 2

A): p(A) = 8n A +(-n +10n-1)A2+(-2n+2)A-1 has exactly one

positive root rn. For all n 2 2, “r > l and r < 31 .

n 8 n O
In fact: r2 m .2553, rn is monotone decreasing as n
increases, and lim r =-l.
n+~ n 8
Thus, p(A) 2 0 <=> A 2 rn. Therefore if n 2 2, a
sufficient condition that p(A) > 0 is A 2 r2 3 .2553,

that is, c S arccos r2 m 1.313.

For the rest of this section we assume that

a 5 arccos rn (s arccos %-m 1.445).

Lemma 522° lf' d(x,y) = a ==> d(Tx,Ty) = b and

a < arccos rn, then d(x,y) = hn(a) =2» d(Tx,Ty) = hn(b).

Proof. The proof is the same as that of Lemma 3.3,

using Lemma 4.2 in place of Lemma 3.2. D

COS C

Lemma-4;4. cos Xn(C) = 1-+(n—l)cos c

-.~.—¢ ..———. 2...-

 

 

42

Proof. Xn(c) is the angle opposite the base of a tri-

angle whose sides are %hn-l' 3hn-1' and c (see Figure
3.3 at Lemma 3.5).

The law of cosines yields:

 

 

cos c = cosz-l-h + sinzlh cos I
I! n-l 2 :1 r1

H +1 H -l

===e cos c = n 1 - n-l cos kn
2 2

H +l-2cos c

===>cos )‘n = n1 = cosc . D
Hn-l-l l-+(n-1)cos c

Lemma 4._5__. _I_f_ d(x,y) = a =9 d(Tx,Ty) = b and

a < arccos rn and T311153- i_s_ not _a_n_ integer, then there are
n

'V

a and 13 such that d(x,y) = 3 => d(Tx,Ty) = b and

 

(1 - cos b) < 0.6(1- cos b).

Proof. Constructing the points e1....,en_1,fo,...,fk

as in the proof of Lemma 3.6, we get 3 = d(fo,fk), b =

d(Tfo,Tfk) and the law of cosines yields:

~ _ 21 . 21
cos b — cos 2hn-l(b) + Sin 2hn-l(b) cos B

_ _ ~ _ . 21 _
-2 1 cos b — Sln 2hn_1(b) (1 COS B)

1 - Hn_l(b)

s 2 (1 - cos%—ln(b))

 

=—e l - cos b

\

(l-cosb)(1+(n-l)cosb) (1__1__/ l+ncosb )
l + (n-2)cos b V? l + (n+l)cos b .

43

Then just as in Lemma 3.6, we get:

if n > 2: (1-cos’f5) < (l—cosb)2(1-‘/-—;—) < 0.6(1-cosb);
if n = 2: cos b > 0 es>p(cos b) < p(O) = l —V/3 < 0.6;
therefore 1 - cos b < 0.6(1 - cos b). 0

Corollary 4.6. If. d(x,y) = a =e> d(Tx,Ty) = b and

a < arccos rn, then either there i5 some 3 229 3' such
Zn
In (’5)
pp there are arbitrarily small 3 222. 3 such that

 

 

that d(x,y) = 3 =e> d(Tx,Ty) = b and

 

ie pp integer

 

rJ

d(x,y) = ’5’ =3 d(Tx,Ty) = b.

Proof. The proof is the same as that of Corollary 3.7

with the 'cos' function replacing the 'cosh' function. 0

Lemma 4.7. If d(x,y) = a =e> d(Tx,Ty) = b and either

 

 

2n 2n .
or are integers, then_ a = b.
hula) —— ln(b)

Proof. The proof is the same as that of Lemma 3.8. 0

Lemma 4.8. If d(x,y) = a =e> d(Tx,Ty) = b, then the

 

image under T of a circle of radius %hn_1(a) _i_§_ con-

 

 

tained ip some circle g radius %hn_1(b).
Proof. The proof is the same as that of Lemma 3.9. 0

Lemma 4.9. If there are two sequences g£_distances

 

 

 

{ai}i=0' and {bi}i=0 such that for all i, d(x,y) = a

ea; d(Tx,Ty) = b1 and ai + 0 and bi + 0 pg 1 e w:

44

. 1
Sin -2-hn_l(ai)
sin%—h

then

 

i5 constant over i.

 

(bi)

n-l
Proof. The proof is the same as that of Lemma 3.10 with

the function 'sin' replacing the function 'sinh.‘ 0

Lemma 4.10. If there are two sequences pf distances

 

 

 

 

{ai}i=l' and {bi}i=1 such that fer all i, d(x,y) = ai
=e> d(Tx,Ty) = b1 and a1 + 0 and bi e 0 pg 1 e m,
then there pg 3p integer N such that for i > N, ai = bi'

 

 

Proof. The proof is the same as that of Lemma 3.11,
with 'sin' and 'cos' replacing 'sinh' and 'cosh' and noting

that g% has only finitely many roots < l. 0

Corollary 4.11. If d(x,y) = a =e» d(Tx,Ty) = b, and

 

o . N
a < arccos rn, then there pp 3 distance a s a such that

 

d(x,y) = '5 =9 d(Tx,Ty) = 'a’.

Proof. This is immediate from Lemmas 4.6, 4.7 and

4.10. D

For the remainder of this section we assume that a is
a preserved distance, that is, that d(x,y) = a =e> d(Tx,Ty)

=3.

Corollary 4.12. If_ n 2 4, a lg e preserved distance

 

under T, and a < arccos rn, then T is an isometry.

cos a .
l + (n-1)cos a’

 

Proof. cos a < l and cos ln =

me 0<COSX<iS-]-'
n n 4

2n 2w
7" T>ln>s
n
211. .
Hence if. is not an integer.

n
So by Corollary 4.6 and Lemmas 4.7 and 4.10, T has

arbitrarily small preserved distances; hence T is an

isometry by Lemma 1.3. 0

Corollary 4.13. pp n = 3, a is p preserved distance

 

under T, and a < %-, then .T i a isometry.

 

 

 

cos 21
Egg.’ If a< 115" then COS a > COSH§= 5 2" _
1 - 2cos 1r
1 \/§ 1 1 _ cosa 2n_
-4- T. Then §>§>COSA3—l+§cosa>cos_§._
n
cos 3

. Therefore

 

1 + 2cos %

Also arccos r3 3 1.3396 > %' so a < arccos r3.

By Corollary 4.6 and Lemmas 4.7 and 4.10, T has
arbitrarily small preserved distances; hence T is an

isometry by Lemma 1.3. 0

Corollary 4.14. if n = 2, a i p preserved distance

cos %%
under T, and a < arccos 2 g 1.108, then T

l - cos 1%

 

 

 

pp pp isometry.

 

 

46

 

Proof. If a < arccos 2" , then
1 ’ COS :-
2w Zn . .
7f > 12 > 7; and a < arccos r2 m 1.313; so T is again

an isometry by Corollary 4.6 and Lemmas 4.7, 4.10, and

1.3. 0
Then we get:

gheorem 3..£§ an Sn—+ Sn, and there are two distances

-_._._____.—-

 

a and b such that d(x,y) = a =e> d(Tx,Ty) = b and

 

IV
.5

either n and a < arccos rn,

n
or n = 3 and a < E ,

T1
COS 2—

5
pp n = 2 and a < arccos 2" ,
l - cos 7?

 

then a

II
G
In:
:3
Q;
*3
P
U)

.i_ pp isometry.

 

Since In > g , for all n 2 2, the following theorem

holds for elliptic spaces:

 

Theorem.4, If’ an pp_ n-dimensional elliptic space,

n 2 2, T: 8n-e-En, and a and b are two distances such

 

that d(x,y) = a fie» d(Tx,Ty) = b, and

either n = 3 and a <

U1|:l
‘

or n # 3 and a <

00':
‘

then T pp pp isometry.

 

The situation for large values of a is not so clear.

47

An example was given in the introduction to show that the

. fl .
theorem is not true for a = 5 or n in the case of

spherical spaces. However, these are the only distances

for which I have counterexamples in Sn. However, this

counterexample does not carry over to an. In elliptic

spaces, the situation for large distances is unknown.

At present, the only result I have concerning large

n

distances in S is the following:

Theorem 5. I_f_ T: Sn—VSnleZ. and a=1r(§1-E-L:T) is 2

preserved distance under T, then T pp pp_isometry for

 

 

positive integer k 2 5.

 

Proof. For any x in Sn, let 3 denote the point

antipodal to x. Let x 6 Sn, and let C be a great

0

circle passing through x Let x0,31,x2,33,...,x2k,3

0‘ 2k+1'
3o,x1,32,x3,...,32k,x2k+1 be points around C such that

x0 = x2k+1 and the distance between adjacent points in the

. . fl .
order listed above is 7k:l . Then for each 1, xi and

31 are antipodal and d(xi,x. = a for 0 s i 5 2k.

1+1)

Then: d(Tx.,Tx. = a for 0 s i 5 2k. Also

1 1+1)

0 S d(Txi,Tx.

fl .
1+1) < 2 m 210']. Since

for 0 s i

M

d(Txi,Txi+1) E d(Txi,Txi+1) + d(Txi+l'Txi+2)
S d(Txi’Txi+l) + d(Txi+1,Txi+2)
5 —1L— + —3L— .

2k+1 2k+1

 

‘.
.‘C

he

 

Figure 4-2

 

Now d(x2k,x0) = d(XZk'x2k+l) = a, so d(x2k,x0) = 2k+l ,

 

_ _ fl ._ an .
hence d(x0,x2k) — n ikil — 2k:1 . But if for any even
. . T!
i, 0 S i < 2k-l, d(Txi,Txi+2) < 2 2k:l , then
an
d(Tx0,Tx2k) S . X d(Txi,Txi+2) < 2k+l ,
i even
OSiSZk-Z
. . . . , _ 2n _

which is a contradiction. Hence d(Txi,Txi+2) — 2k+1 -

 

d(xi,xi+2) for all even i, 0 S i S 2k-2.

Hence 2331 is also a preserved distance under T. So

 

by the previous theorems, if k 2 5, then a = 2k+1 is

small enough so T must be an isometry. a

§5. DIRECTIONAL RESTRICTIONS

 

Greenwell and Johnson [7] considered the question of
restricting the hypothesis so that if d(x,y) = l and vector
xy is in one of certain directions, then d(Tx,Ty) = 1. Let
£3 be any subset of S, where S is the set of all unit
vectors in En. Greenwell and Johnson [7] showed the follow—

ing two results:

_m

Thegrem. I; the cardinalitypp .0 .pp less than that pf

 

 

S, then there exists T: En e En such that T pp not pp

 

 

isometry, but for any x,y 6 En such that xy 6.0,

 

 

HTx-TyH = 1.

Theorgp. 1; the cardinality pp S-uD pp less than that

-—_._.—-———.

 

 

pp 3 then if T: En + En such that xy €4U==’”Tx-Ty” = 1,

 

 

then T is pp isometry.

 

This section gives a little more information on how small
#5 can be and still force T to be an isometry. For the
remainder of this section we consider E2 and assume that
D is a subset of 3 so that the interior of D is non-void
(in the topology of 3 relative to E2). Let L be an
equilateral unit lattice, for example the points n; + my

1 and 3

. -+ ->
where n and m are any integers and ”x” = My“

and § are at an angle of fi/3.

49

 

Figure 5—1

~

Let D denote the interior of ,0.

Lemma 5.1. Let p lattice point 6 L pp chosen pp orighh

 

then the set pf unit vectors

8

V = {-—— I B E L and H8" is irrational}

IIBII

 

 

pp dense pp 3.
Proof. Consider 3 and § so that

-> -> ,
L = {nx-tmy I n, m are any integers}.

 

\/n2+nm+m2.

Then ”n; + my”

2 2

Then if n 1 (mod 4), we get that n +nm+m a

m

 

3 (mod 4) hence VbZ-tnmntmz is irrational. It is clear
that any point in the plane is within distance 4 of a

point n§-tm§ with n s s 1 (mod 4).

<43

Let any unit vector and any a > 0 be given.
Choose a point n§th§ such that n s m s 1 (mod 4) and

.’
-V

 

< 8’

 

||n§+m§- (4 + g) ‘6” < 4. Then

 

 

 

 

51

 

 

since:
anz-tnmrtmz - (4 + g) < 4,
so
0 < \/n2+nm+m2-§8— < 8,
and
g < $3+hm+m2.
Therefore
->+ ->
nx my _ 3

 

 

 

2

 

 

 

fl/nz +nm+m

 

 

 

 

 

 

 

= 1 ”hi? + m3; - \/r12+nm+w;2 3”
\/n2+nm + m2
s g-“n'i + m3; - x/n2+nm+mj2 {I'll
_<_ EénniE-tmiF- ”+243”+-§—||(\/n2+nm+m2-4-§)VH
s §4+§—(\/n2+nm+m2-4-§)
6
s g-(Vbz-rnmmtmz - g)
8
E g 8. D

pemma 5.2. l: T restricted pp L pp the identity, and

 

 

a e t) then x E L and x-y = a implies Tx-Ty = a.

Proof. Let the point x 6 L be used as the origin, then
we must show that Ta = a. Assume the contrary, that Ta;£a.

Now a (,0' implies that "Ta" = 1. Let 2 be the foot

of the perpendicular to line xa passing through point Ta.

52

Let b 6 L be chosen so that B = angle axb is small
enough so that any vector within an angle of 28 of a
will be in .D' and so that Ta is on the opposite side of

-> ->
xa as b, and Nb” .is irrational and ”b" > 2. This can

be done by Lemma 5.1 since .5 is open.

 

 

 

Figure 5-2

Let c be the point on line xb, # x so that H3-3H==l.

N

-> -) -) 4 .p -) .
Let d = b/Hb”. Then c-a and d are in .D’ by our
I -. 4 -+ -’ I I o
chOIce of b. Let p = c-d, y = the intersection of line
xc and line zTa (if Ta = 2, this is just the perpen-

dicular to xa at z).

 

Then Hp" = (V2 - 2cos 28 - 1) and u?“ = Hszec B. In

particular, Ta # a==e z # a=ee y is on the same side of

p as x is for 8 small enough (since lim n?” = ”3”
9+0
and lim “E“ = ”3” = 1 and u?" < 1 unless E = -a).
3+0

Then p is between y and mb for any m 2 1. Let n
and m be positive integers such that %'< n§-§n and
_, -> _, 1 -> -> -> —> -> 1
"CH +n > mllbll > llCll HI -5 - Then ||C+nd|| > mllbll > ||C+nd|| ~11.- -
Let g be the point on segment xc such that

4+
Hmb-q" = n-tl. Then q is between y and p since:

SO

and

ele

the

on

im;

th

of

it

«X

%X

S3

”sum” < nan-Ln” = “sum-k
m m
.5
< mllbll < "Sum = IIEII+n+L
SO
+ -> ‘* —>
llyll = uyu+n+1-n-1 < mubn-n-l = 1an
and
-> ‘* 4 ->
uqn = mubn-n-l < npu+n+1-n-1 = up”.

Now ”mg-3N < H3+n3-3H < n-tl and mb-3 is a sum of
elements of .5, so "mg-T3" s n+1 hence Ta must lie on
the same side of the perpendicular to xc at q as does
mb. But the choice of b, n and m require that Ta be
on the other side of this line. Hence Ta ¢ a is

impossible. 0

Lemma 5.3. pp T restricted pp L pp the identipy,

 

 

then T pp the identipypp all pp E2.

 

gpppp. By Lemma 5.2, T is the identity on all points
of the form 3+3 where 3 E L and 3 6J5. Then replac-
ing the set L with the set a-tL = {3+3 I 3 6 L}, Lemma
5.2 gives us that T is the identity on all points
3+3+3' where 3 E L and 3,3' 6 .5. By induction we get
that T is the identity on all points of the form

3 + 22:1 3. where 3 6 L and 31 (,5.

i
It is easily seen that for any point b in the plane,

-+ -+ -> —) —) ~ . .
the set {b+a1+a2 I a1,a2 6,0} contains a disc. If the

radius of this disc is 6 and k is any integer larger

54

~

than 1/0 . then the set {b+2i=l ai I ai 6 ,6} Will con-

tain a point y 6 L . Then there are 31 in B so that

~

+ -> k
b = y + 21:1 (-é'i) . But from the definition of b it is

clear that 3 6 f7 implies —3 6 D , hence T is the iden-

tity on b. Therefore T is the identity on all of E2. D

Corollpg 2.3. _I_f_ T restricted pp any lattice _jp_ p
motion then T is a motion on all of E2 .

 

Proof. If T restricted to a lattice is the motion M,
then M- T is the identity on that lattice and so by Lemma

5.3. m" T is the identity on all of 32. Hence T=M. a

W. _I_p' ,D’contains pp Ojen set and there are

seven points a,b,c,d,e,f,g such that a—b 6 40’. a-g E 95,

b-g65,a—f€5.a-e€fl,e-fé fi.c-d€D, then

 

T _i___p_n isometpy.

Proof. The same argument as in Lemma 2.5 shows that

 

Ta # Tc, hence if T restricted to a,b,g is the motion
M, then “beg“ = “3+3“ = 1 requires that To = Me.
Repeating this argument on the equilateral rhombi contained
in the lattice which contains points a,b,c,g gives that T
restricted to this lattice is a motion, and hence by Corol-
lary 5.4, T is a motion and hence an isometry. a

It seems likely that a similar theorem can be found in
En for n 2 3‘, but the proof is likely to be more diffi—
cult since for n 2 3, equilateral n-simplices do not

tesselate as simply as triangles in E2.

§6. COUNTEREXAMPLES AND OPEN QUESTIONS

 

In the introduction, counterexamples were given to Show
that in the spaces El and E", a transformation with a
preserved distance need not be an isometry. But what hap-
pens if the mapping is required to be continuous?

In E1, the transformation x + [x] + {x}2 (where [x]
is the integer part of x and {x} = x - [x]) is continuous
and preserves the distance 1 but is not an isometry. How-
ever, in E“, it is not so clear what happens; it seems
that the additional restriction of continuity should force
the map to be an isometry but no proof is known.

Also, in the introduction an example was given in Sn
that preserved the distances %: and u but was not an
isometry. However, this transformation was not continuous.
It is easy to construct a continuous map Of Sn into itself
which preserves just the distance n, since preserving the
distance fl just means preserving the relation of two
points being antipodal and this can be done with a continu-
ous map. However, it is still not known whether or not the
distance % can be preserved by a continuous map which is
not an isometry. Also in spherical and elliptic spaces, it
is not known whether or not the additional requirement Of
continuity placed on a map which preserves large distances
forces that map to be an isometry.

The combination of continuity and distance preserving

seems to be powerful enough to make the following conjecture

seem likely:

55

56

CONJECTURE. If T: M—VM where M is a convex,
finitely compact metric space with unique segments and any
segment in M has a prolongation, and M has topological
dimension > 1 and T is continuous and preserves some

distance, then T is an isometry.

The following shows an example of a map T: M-4'M
where M is a convex, finitely compact metric space of
topological dimension 2 and T is continuous and preserves
all distances less than or equal to 2. M also has the
prOperty that segments haveprolongations, but they need not
have unique prolongations, nor are segments unique. Let

M c E3 be defined by:

M = {(X.y.0) IX.y real}U[(X.y.1) IX.y. real}

U U I(k,y,z) Iy real, O<z<l}
k=1

with the metric defined by the length of the shortest path
between two points which is contained in M. Let T be the
map T(x,y,z) = (x+1,y,z). Then in Figure 6—1,

«3.3) a! d(T3, TS) but it is clear that T is continuous
any distance 5 2 is preserved by T but T is not an

isometry.

 

 

 

 

NU?
\
\

 

1)

Figure 6—1

 

diffil

of sp

tinui
with
prese

ture

spher
whic?

isome

Cle V

ix

lsOme

57

It does seem that the above conjecture would be very
difficult to show when M is not one of the special types
of spaces considered so far.

In spherical and elliptic spaces, it seems as if con-
tinuity combined with distance preserving yields an isometry
with the exception of the case when the distance n is
preserved in a spherical space, hence the following conjec-

ture seems reasonable:

CONJECTURE. If T: M + M where M is either a

 

spherical or elliptic space and T is a continuous map
which preserves some distance not = n, then T is an

isometry.

Once again, no proof is known.
The following conjecture would be an interesting

generalization of the results of Sections 3 and 4:

CONJECTURE. If M is a locally Euclidean manifold of

 

finite dimension 2 2, then there is a distance a such
that for any h < a, and any map T: M + M, T preserves

b implies that T is an isometry.

A few examples of the form T: M1 + M2 where M1 # M2

are of interest.

If M1 is the Euclidean line and M2 is the unit cir-
cle with are metric and T is the "wrapping" function Tx =
eix, then T preserves any distance 5 u, but is not an

isometry.

isc

be

(I)

58

If T: En + Em preserves some distance, then clearly
n E m (since Em has equilateral n-simplices if and only
if n S m). Beckman and Quarles [1] showed that T is an
isometry if 1 < n = m < w. But if n < m, T might not

be an isometry if m is too large.

Theorem 2. For any integer n 2 1, there exists an

 

 

integer M(n) such that m 2 M(n) implies that there

 

 

exists i_map T: En + Em which preserves the distance 1

 

but is not an isometry.

Proof. Let En be partitioned into a set of regions

{R }i=1

i such that each region Ri has diameter 5 l (with

the distance 1 not being assumed) and so that any closed

n—sphere of radius 1 intersects S k of these regions.

Then an integer M(n) can be found so that the regions R1
+
{s~}§(n’ 1

k k=1
that if x 6 Ri' y E Rj and R1 and Rj are in the same

can be partitioned into M(n)-+1 sets such

SN, then d(x,y) # 1. Then define T: En + Em for
k

m 2 M(n) by mapping each set UR ES Ri into a different
i

f‘.J

vertex of a unit equilateral M(n)-simplex in Em. Then

d(x,y) = 1 =s>x and y are not in the same set UR ES Ri
1 T6
hence d(Tx,Ty) = 1. Then T is a map from En to Em

which preserves the distance 1 but is not an isometry. o

Hadwige[8] gave the following partitioning of the plane
into hexagon shaped regions where each region has diameter

1 and contains only its lower boundary (see Figure 6-1A).

59

Then let these regions be partitioned into 7 sets accord-
ing to the pattern shown in Figure 6-lB. It is then easily
seen that if d(x,y) = 1, then x and y are not assigned
to the same one of the seven sets. Hence, if {a1}:=1
an equilateral unit simplex in E6, then T defined by

is

 

Figure 6-2

Tx = a1 (where x lies in set number i) will preserve
the distance 1. However, T is not an isometry.

It is not known if M(2) = 6 is the smallest dimension
for which such a T exists. In particular, it is not even
known whether or not there is a distance preserving map

2

T: E 4 E3 which is not an isometry.

Also, it is still an open question whether or not there

b0

. . n . . .
is a continuous map T: E + Em for m > n which is dis-

 

tance preserving but not an isometry.

Another possible way to place an added restriction on
the mapping is to require it to preserve two distances. The
example of a mapping of 82 onto 82 given in the intro-
duction shows that in at least one case, preserving two
distances does not imply an isometry. However, the follow-

ing theorem does hold:

Theorem 8. if T: MI —> M2 where M1 2 are

is strictly convex, and T

and M

Banach spaces such that M

 

 

2
preserves the two distances a and ka for some integer

 

 

k 2 2, then T is an isometry.

 

Note that this theorem includes the case M1 = M2 = E ,

which is not true if T preserves only one distance.

grggf. Let x,y,z 6 M1 such that Hx-y” = Hx-z” = ka,
and lly-zll = a. Define w = 1%)! + ESLy and
v = %x + 533-2. Then llx-wll = llx-vll = a and llw-vll =
l-Jé-lly-zll = flea. Then T preserves a and ka =>|lTx-Tw|| =
HTx-Tv" = ”Ty-T2” = ka and HTx—Tw” = ”Tx-Tv” = "Ty-T2" = a.
Since Hy-w" = (k-l)a and Hz-v" = (k-l)a, then HTy-Tw" s
(k-l)a and HTz-Tv" s (k-1)a by Lemma 1.1. But since M2
is strictly convex, "Tx-Tyfl = ka, ”Tx-Twfl = a and
llTy-Twll s (k-1)a =~>Tw = jléTx + bi—lTy. Likewise, Tv =
%Tx + hl'Z—sz. Hence llTw-Tvll = i-llTy-Tzll = i—a. Hence T

1 . . 1
preserves Ea Since for any w, v With llw-vll = Ea,

61

there are x, y, 2 as defined above. Then by induction, T
preserves (-]lE-)la for all integers i 2 O, and hence T is

an isometry by Lemma 1.3. o

The assumption that the two preserved distances have
integer ratio was important in the proof of this theorem.
However, when T preserves two distances with a non-integer

ratio and M1 and M2 are as above, it is not known

whether or not T must be an isometry, except for those
cases covered by the theorems of Section 2 or the work of

Beckman and Quarles [3], or the case when M1 = M2 has a

flat spot on the unit sphere of length 2 (when an example

similar to that given for a“ at the end of Section 2 gives

a counterexample).

CONJECTURE. If M2 is any Banach space without a flat

spot of length 2 on the unit sphere, or any elliptic or

 

hyperbolic space (including spaces of dimension on), and
T: M1 + M2 has two preserved distances, then T is an

isometry.

CONJECTURE. If T: Sn + Sn (where n may be m), and

 

and n, then T is an

NI:

T preserves two distances not

isometry.

A similar idea is to consider maps of the type
T: M1 -> M2 where d(x,y) < 1, x,y e Ml<=¢d('rx,'ry) < 1.
This type of map may come up in some types of psychological

or biological measurement where the quantity under

62

consideration can only be measured indirectly, that is, only
after some transformation has been applied to it; and even
then the only measurement available is to determine whether
or not some threshold is equaled or surpassed.

For example, in an experiment where an individual is
asked to judge whether or not two given color samples are
the same color, the two colors must differ by some thres-
hold amount or they will be seen as the same color. Then
the responses of the subject are the result of transforming
the physical colors of the samples according to the charac-
teristics of the subject's eye. Now if there is a map
T: M1 + M2 where M1 is a metric space representing the
actual physical colors and M2 is a metric space repre-
senting the responses of the subject's eye to colors, such

that x,y E M d(x,y) < l<=>d(Tx,Ty) < l, but T is not

1'

an isometry, then the problem of deducing the nature of M1

from the responses of the subject can become more compli-
cated.

Similar restriction on T are:

(1) d(x,y) > l<==>d(TX.Ty) > 1.
(2) d(x,y) > 1 =>d(TX.Ty) > 1.

and d(x,y) < l =s>d(Tx,Ty) < l,
or even

(3) a < d(x,y) < 9 =9“ < d(Tx,Ty) < 5.

63
The only progress on the question of preserving inequal-

ities is the following two lemmas:

Lemma 6.3. ii T: M1 4 M2 where M1 and M2 satisfy

the hypothesis 2i Lemma 1.1 and d(x,y) < 1, x,y 6 M1 =%’

 

d(Tx,Ty) < l and d(x,y) > 1 =#’d(Tx,Ty) > 1 and T is

onto, then .k < d(x,y) < k-tl, x,y 6 M =g>k S d(Tx,Ty) <

l

 

ki-l for integer k 2 0.

 

Proof. Clearly d(x,y) < k~+l =fi>d(Tx,Ty) < kttl by
an argument similar to Lemma 1.1. If d(Tx,Ty) < k, then
there eXlSt z0,zl,...,zk 6 M1 such that z = x, zk = y

0
i s k-l. Then

IA

and d(Tzi,Tzi+1) < l for all i, 0

d(x,y) S Z§;$ d(zi,z ) S k contrary to the hypothesis.

i+l
Therefore k 5 d(Tx,Ty) < k-tl. D

n

Lemma 6.4. 'ii T: En + E for 2 S n < m and

d(x,y) = l ==>d(Tx,Ty) < l and d(x,y) > 1 ==ed(Tx,Ty) > 1

and T is onto, then T is is isometry.

gissi. If u,v 6 En and d(u,v) = l and T is onto
then there exists x,y 6 En such that u = Tx and v = Ty.
Then d(x,y) must be 1 by the hypothesis. Hence T‘lis
a (possibly multivalued) map from En to En which pre-

serves the distance 1 and is then an isometry (Beckman and

Quarles [1]). Therefore T is an isometry. 0

However, nothing seems to be known about the particular

inequality restriction that seems most relavent to the

64

physics of perception theory: that is, d(x,y) < 1.¢=>

d(Tx,Ty) < l.

BIBLIOGRAPHY

BIBLIOGRAPHY

Beckman, E.S., and D.A. Quarles, On Isometries of
Euclidean Spaces, Proc. Amer. Math. Soc. 4(1953)
810-815. '

Benson, R.V., Euclidean Geometgy and Convexity,
McGraw Hill, New YOrk (1966).

 

Bish0p, R., Characterizing Motions by Unit Distance
Invariance, Math. Mag. 46(1973) 148-151.

Blumenthal, L.M., Theory and Applications of Distance
Geometry, Oxford University Press, London (1953).

 

Busemann, H., The Geometry of Geodesics, Academic
Press, New YOrk (1955).

 

Gans, D., An Introduction to non—Euclidean Geometgy,
Academic Press, New York (1973).

Greenwell, D., and P. Johnson, Functions that Preserve
Unit Distance, Math. Mag. 49(1976) 74-79.

Hadwiger, H., and H. Debrunner, Combinatorial Geometry
in the plane, Holt, Rinehart and Winston, New YOrk
(1964).

 

 

Valentine, F.A., Convex Sets, McGraw Hill, New York
(1964).

 

65

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