Unit circle preserving mappings

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UNIT-CIRCLE-PRESERVING MAPPINGS

SOON-MO JUNG and BYUNGBAE KIM

Received 6 April 2004

We prove that if a one-to-one mappingf:Rn_→_Rn_(n_≥_{2) preserves the unit circles, then}

fis a linear isometry up to translation.

2000 Mathematics Subject Classification: 51K05.

1. Introduction. LetXandY be normed spaces. A mappingf:X→Y is called an isometry iff satisfies the equality

_{f (x)}₋_{f (y)}₌_x₋_y _(1.1)

for allx,y ∈X. A distance r >0 is said to be preserved (conserved) by a mapping

f:X→Y if

_{f (x)}₋_{f (y)}₌_r _∀_x,y_∈_Xwithx−y =r . (1.2)

Iff is an isometry, then every distancer >0 is conserved byf, and vice versa. We can now raise a question whether each mapping that preserves certain distances is an isometry. Indeed, Aleksandrov [1] had raised a question whether a mappingf:X→X preserving a distancer >0 is an isometry, which is now known to us as the Aleksandrov problem. Without loss of generality, we may assumer=1 whenXis a normed space (see [16]).

Beckman and Quarles [2] solved the Aleksandrov problem for finite-dimensional real Euclidean spacesX=Rn_{(see also [}₃_,₄_,₅_,₆_,₇_,₈_,₁₁_,₁₂_,₁₃_,₁₄_,₁₅_,₁₇_,₁₈_,₁₉_,₂₀_]).

Theorem1.1(Beckman and Quarles). If a mappingf:Rn→Rn(2≤n <∞)preserves a distancer >0, thenfis a linear isometry up to translation.

Recently, Zaks [25] proved the rational analogues of the Beckman-Quarles theorem. Indeed, he assumes thatn=4k(k+1)for somek≥1 orn=2m2₋_{1 for some}_m_≥_3,

and he proves that if a mappingf:Qn_→_Qn_{preserves the unit distance, then}_f_{is an}

isometry (see also [21,22,23,24]).

It seems interesting to investigate whether the “distancer >0” in the Beckman-Quarles theorem can be replaced by some properties characterized by “geometrical figures” without loss of its validity.

In [9], the first author proved that if a one-to-one mappingf:Rn_→_Rn_(n_≥₂₎_maps

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y1

yn

a

b C1

C2

T

cosϕ0

S1

Figure_2.1

the same type with side lengthb >0, then there exists a linear isometryI:Rn_→_Rn_up

to translation such that

f (x)=_abI(x). (1.3)

Furthermore, the first author proved that if a one-to-one mappingf:R2_→_R2_maps

every unit circle onto a unit circle, thenfis a linear isometry up to translation (see [10]). In this connection, we will extend the result of [10] to then-dimensional cases; more precisely, we prove in this paper that if a one-to-one mappingf:Rn_→_Rn_(n_≥₂₎_maps

every unit circle onto a unit circle, thenfis a linear isometry up to translation.

2. Preliminaries. We start with any two distinct pointsaand b inRn _{with the}

dis-tance between the two less than 2. Let their disdis-tance be

2c=2 sinϕ0 with 0< ϕ0<π

2,0< c <1. (2.1)

Given such two distinct points whose distance is less than 2, we can choose a coordinate

(y1,...,yn)forRnsuch that

a=0,...,0,sinϕ0, b=0,...,0,−sinϕ0. (2.2)

Let the(n−2)-dimensional unit sphere contained in the space orthogonal to theyn -direction be

Y=y1,...,yn−1,0|y12+···+yn2−1=1

. (2.3)

If we call the center of any unit circle passing through the two points (aand b) oand

the origin of the coordinate o, then the vectoroo→is perpendicular to theyn-axis and

its length must be cosϕ0and therefore

→

oo∈˜Y=cosϕ0Y, seeFigure 2.1. It means that

any unit circle passing through the pointsaand b has its center in ˜Y=cosϕ0Y. Let

T be the set of union of all the unit circles passing through the pointsaand b. More precisely, if we define the following set:

T=cosϕ+cosϕ0

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then it is clear that this is the set of union of all the unit circles which are centered at cosϕ0yfor each fixedy∈Y and which pass throughaand b whenϕ=π∓ϕ0(see

Figure 2.1).

The intersection of T and they1-ynplane consists of two circles, say C1(wheny1=1,

i.e.,y=(1,0,...,0)) and C2(wheny1= −1, i.e.,y=(−1,0,...,0), seeFigure 2.1). In the

following contexts, we will consider the casesy1=1 and−1 in connection with T as the

circles C1and C2, respectively. Call S1the(n−1)-dimensional unit sphere containing

the circle C1. If we let the center of C1be O and the center of S1be ˜O, then it is obvious

that O=_O.˜

(To see this, choose any point A∈C1and its antipodal point B in C1. Then, by the

definition of the antipodal points that they lie exactly the opposite with respect to the center of the circle C1whose center is at O, and because they are of the same length 1,

we have the following condition that

→

OA= −OB→, AB→=AO→ +OB→=2OB→. (2.5)

On the other hand, we have, since the two points A and B lie also on the unit sphere S1

with its center at ˜O,

2=AB→=A˜→O+OB˜→≤A˜→O+OB˜→=1+1=2. (2.6)

Therefore, by the Cauchy-Schwarz inequality, →

A˜O is a positive multiple of → ˜

OB, which

means →

A˜O=_{OB because their lengths are both 1. So,}˜→

→

AB=A˜→O+_OB˜→₌₂_OB˜→_, _(2.7)

and therefore ˜O=O.)

Now, we first show that S1and T intersect only at C1. To make computation simpler

we use a new coordinatexforRn_{, where}

x=y−cosϕ0,0,...,0

. (2.8)

In this coordinate (seeFigure 2.2), S1becomes the unit sphereScentered at the origin,

S1=S=

x1,...,xn

|x2

1+···+xn2=1, (2.9)

T=x=cosϕ+cosϕ0y+(0,...,0,sinϕ)

−cosϕ0,0,...,0|y∈Y,0≤ϕ <2π.

(2.10)

With the help of this coordinate we show the following lemma.

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y1

yn

a

b C1

C2

T

cosϕ0

S1

-y−(cosϕ0,0,...,0)

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q

x1

xn

a

b C1

C2

T

S1=S

Figure_2.2

Proof. If any element in T has distance 1 from the origin of thex-coordinate, then we have

1=cosϕ+cosϕ0

y1−cosϕ0

2

+cosϕ+cosϕ0

2

y2 2

+···+cosϕ+cosϕ0

2

y2

n−1+sin2ϕ

=cosϕ+cosϕ0

2

−2 cosϕ0

cosϕ+cosϕ0

y1+cos2ϕ0+sin2ϕ

=1+2 cos2ϕ01−y1+2 cosϕ0cosϕ1−y1.

(2.11)

Therefore, we have

0=2 cosϕ0

1−y1

cosϕ+cosϕ0

. (2.12)

Withy1=1, T in (2.10) represents the unit circle C1in thex1-xnplane. If

cosϕ= −cosϕ0, i.e.,ϕ=π∓ϕ0, (2.13)

then it follows from (2.10) that

T=x=−cosϕ0,0,...,0,±sinϕ0

= {a,b} (2.14)

which also belong to C1.

Now, consider, as inFigure 2.3, the origin e and ˜e=(−2,0,...,0)in thex-coordinate and the unit circle C1passing through e and ˜e in thex1-xnplane. Choose a point d∈C1,

d∈ {e,˜e}. We parameterize all the unit circles passing through the points e and d. We assume thexn-coordinate of d is negative.

By triangle inequality, the distance between e and d is less than 2, say 2 sinϕ0, with

0< ϕ0< π/2. Choose a new coordinateyforRnand consider two points

e=0,...,0,sinϕ0, d=0,...,0,−sinϕ0, (2.15)

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q

x1

xn

e ˜e

d C1

C3

Figure_2.3

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q

y1

yn

e

d ˜e

C₃

C₁

-M

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q

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_x₁

xn

e ˜

e

d C1

C3

Figure_2.4

To get a parameterization of the unit circles passing through e and d, we consider the mappingMdefined by

x=My=          

cosϕ0 0 ··· 0 sinϕ0

0 1 ··· 0 0 ..

. ... . .. ... ... 0 0 ··· 1 0 −sinϕ0 0 ··· 0 cosϕ0

         

y+cosϕ0,0,...,0t−(1,0,...,0)t.

(2.16)

This transformationMis an isometry (since it is a composition of a rotation and trans-lations) and sends

y=0,...,0,±sinϕ0

= {e,d} (2.17)

to

x=(0,...,0), x=cos−2ϕ0−1,0,...,0,sin−2ϕ0= {e,d} (2.18)

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Therefore, by comparingFigure 2.4withFigure 2.1and considering (2.4), all the unit circles passing through e and d can be parameterized as

x=My|y=cosϕ+cosϕ0

y₊₍₀_,...,₀_,_sin_{ϕ), y}_∈_Y_,₀_≤_{ϕ <}₂_π_. _(2.19)

With the help of this parameterization, we are ready to show the following lemma.

Lemma2.2. Ford∈C₁,d∈ {e,˜e}, any unit circle inRn, which passes throughdande, has some point whosex1-coordinate is positive, except the circleC1.

Proof. _{Without loss of generality, we can assume the}xn_{-coordinate of d is negative.} Note that withϕ=π∓ϕ0in (2.19),y=(0,...,0,±sinϕ0)are the points eor din the

y-coordinate and furtherϕ=π∓ϕ0means that

x=(0,...,0)=e, x=cos−2ϕ0

−1,0,...,0,sin−2ϕ0

=d (2.20)

in thex-coordinate, regardless ofy_∈_{Y. Any unit circle passing through e and d is} given asx=Mywithygiven as in (2.19), that is,

          x1 x2 .. .

xn−1

xn           =          

cosϕ0 0 ··· 0 sinϕ0

0 1 ··· 0 0 ..

. ... . .. ... ... 0 0 ··· 1 0 −sinϕ0 0 ··· 0 cosϕ0

                   

cosϕ+cosϕ0

y

1+cosϕ0

cosϕ+cosϕ0

y 2 .. .

cosϕ+cosϕ0

y

n−1

sinϕ           −           1 0 .. . 0 0           . (2.21)

The first coordinate is

x1=cosϕ0cosϕ+cosϕ0y1+cos2ϕ0+sinϕ0sinϕ−1. (2.22)

We show that fory

1≠−1 (y1= −1 means the circle C1in they-coordinate and the

circle C1in thex-coordinate, seeFigure 2.4), there is always someϕnearπ−ϕ0(i.e.,

near the point e) such that the abovex1becomes positive.

Let

θ=π−ϕ0

−ϕ=π−ϕ+ϕ0

, (2.23)

and so

ϕ=π−θ+ϕ0

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Then, the above is

x1= −cosϕ0cos

θ+ϕ0

y

1+cos2ϕ0

1+y

1

+sinϕ0sin

θ+ϕ0

−1 = −cosϕ0

cosθcosϕ0−sinθsinϕ0

y

1+sinϕ0

sinθcosϕ0+cosθsinϕ0

−1+cos2ϕ0

1+y

1

=sinθsinϕ0cosϕ0

1+y

1

+cosθsin2ϕ0−cosθcos2ϕ0y1

−1+cos2_ϕ 01+y1

=sinθsinϕ0cosϕ0

1+y

1

+cosθ−cosθcos2_ϕ 0

1+y

1

−1−cos2ϕ0

1+y

1

=sinθsinϕ0cosϕ01+y1

−1−cos2ϕ01+y1

(1−cosθ).

(2.25)

θ=0 (ϕ=π−ϕ0) means the intersection point e and the abovex1becomes 0 as it

should. Assume

θ≠0 −π−ϕ0< θ <0,0< θ≤π−ϕ0

. (2.26)

Then,x1is positive if and only if

sinθsinϕ0cosϕ01+y1

>1−cos2_ϕ 01+y1

(1−cosθ), (2.27)

that is,

sinθ

1−cosθ >

1−cos2_ϕ 0

1+y

1

sinϕ0cosϕ01+y1

(2.28)

(recally

1≠−1 and 0< ϕ0< π/2). In other words, thex1-coordinate is positive if and

only if

cotθ 2>

1−cos2_ϕ 01+y1

sinϕ0cosϕ0

1+y

1

. (2.29)

Therefore, fory1≠−1 (i.e., except the circle C1), thex1-coordinate is positive for

small enoughθ >0.

3. Main theorem. In the previous section, we introduced all preliminary lemmas for the main result of this paper. Now, we prove our main theorem.

Theorem3.1. If a one-to-one mappingf :Rn→Rn maps every unit circle onto a unit circle, thenf is a linear isometry up to translation.

Proof. We showfpreserves the distance 2. Suppose the distance betweena=f (A) and b=f (B)is less than 2, while the distance between A and B is 2—seeFigure 3.1. Then, we show it leads to a contradiction.

Let the distance betweenaand b be 2c (0< c <1). Choose any unit circle C passing through A and B and letf (C)=C1. Choose a coordinate foraand b as inFigure 3.1

such that C1lies in thex1-xnplane and

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A

˜ E B

E C

-f

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x1

xn

a

b

e ˜e

d C1

C3

T

Figure_3.1

Let

e=(0,...,0), ˜e=(−2,0,...,0). (3.2)

Letf (E)=e and ˜E the antipodal point (in C) of E and letf (˜_E₎₌_{d. Let the union of all} the unit circles passing throughaand b be T and the (n−1)-dimensional unit sphere passing through A and B be S and the (n−1)-dimensional unit sphere passing through e and ˜e be S1.

Then, it is clear that any point P on S (P∈ {A,B}) lies in some unit circle determined by the three points A, B, and P. To see this, if we call O the common center of C and S, and let

→

OP,OA→=sinϕ0

−π₂ < ϕ0<π

2

, (3.3)

then the unit circle determined by these three points is parameterized as

→

OV(ϕ)=cosϕ

→

OP−sinϕ0

→ OA cosϕ0

+sinϕOA→ (−π < ϕ≤π). (3.4)

Note that

→

OP−sinϕ0

→ OA cosϕ0

,OA→

(3.5)

are orthonormal to each other and

→ OVϕ0=

→

OP, OV→ _π

2

=OA→, →

OV

−π₂

= −OA→ =OB→.

(3.6)

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It is also obvious that thex1-coordinate of any point in T is nonpositive. (Note that

the center of any unit circle passing throughaand b has coordinate

1−c2_y₋₁₊₁₋_c2_,₀_,...,₀ _{for some}_y_∈_Y_, _(3.7)

(see (2.4)) and the distance between this center and anyx=(x1,...,xn)is

x1+1+

1−c2₁₋_y 1

2

+··· (3.8)

and because

1−c2₁₋_y 1

≥0, (3.9)

positivex1makes the distance larger than 1, which means that ifx1>0, we havex∈T.)

Now, if d=˜e, then the image of any unit circle passing through E and ˜E lies in both T and S1. However, byLemma 2.1, T∩S1=C1and this fact contradicts the injectivity

off.

On the other hand, if d≠˜e, the image of any unit circle, except the circle C, passing through E and ˜E is a unit circle passing through e and d. This unit circle is not C1since

f is one-to-one, and byLemma 2.2it cannot stay completely in T, a contradiction. Consequently,f preserves the distance 2. According to the well-known theorem of Beckman and Quarles,f is a linear isometry up to translation.

Acknowledgment. This work was supported by Korea Research Foundation Grant KRF-2003-015-C00023.

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Soon-Mo Jung: Mathematics Section, College of Science and Technology, Hong-Ik University, 339-701 Chochiwon, Korea

E-mail address:[email protected]

Byungbae Kim: Mathematics Section, College of Science and Technology, Hong-Ik University, 339-701 Chochiwon, Korea

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