Unit-circle-preserving Mappings

We prove that if a one-to-one mapping f : R n → R n (n ≥ 2) preserves the unit circles, then f is a linear isometry up to translation.


Introduction.
Let X and Y be normed spaces.A mapping f : X → Y is called an isometry if f satisfies the equality for all x, y ∈ X.A distance r > 0 is said to be preserved (conserved) by a mapping f : If f is an isometry, then every distance r > 0 is conserved by f , 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 mapping f : X → X preserving a distance r > 0 is an isometry, which is now known to us as the Aleksandrov problem.Without loss of generality, we may assume r = 1 when X is a normed space (see [16]).
Theorem 1.1 (Beckman and Quarles).If a mapping f : R n → R n (2 ≤ n < ∞) preserves a distance r > 0, then f is a linear isometry up to translation.
Recently, Zaks [25] proved the rational analogues of the Beckman-Quarles theorem.Indeed, he assumes that n = 4k(k + 1) for some k ≥ 1 or n = 2m 2 − 1 for some m ≥ 3, and he proves that if a mapping f : Q n → Q n preserves the unit distance, then f is an isometry (see also [21,22,23,24]).
It seems interesting to investigate whether the "distance r > 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 mapping f : R n → R n (n ≥ 2) maps every regular triangle (quadrilateral or hexagon) of side length a > 0 onto a figure of -6 &% '$ q q q &% '$ the same type with side length b > 0, then there exists a linear isometry I : R n → R n up to translation such that Furthermore, the first author proved that if a one-to-one mapping f : R 2 → R 2 maps every unit circle onto a unit circle, then f is a linear isometry up to translation (see [10]).
In this connection, we will extend the result of [10] to the n-dimensional cases; more precisely, we prove in this paper that if a one-to-one mapping f : R n → R n (n ≥ 2) maps every unit circle onto a unit circle, then f is a linear isometry up to translation.

Preliminaries.
We start with any two distinct points a and b in R n with the distance between the two less than 2. Let their distance be Given such two distinct points whose distance is less than 2, we can choose a coordinate (y 1 ,...,y n ) for R n such that a = 0,...,0, sin ϕ 0 , b = 0,...,0, − sin ϕ 0 . (2.2) Let the (n − 2)-dimensional unit sphere contained in the space orthogonal to the y ndirection be If we call the center of any unit circle passing through the two points (a and b) o and the origin of the coordinate o, then the vector → oo is perpendicular to the y n -axis and its length must be cosϕ 0 and therefore It means that any unit circle passing through the points a and b has its center in Ỹ = cos ϕ 0 Y.Let T be the set of union of all the unit circles passing through the points a and b.More precisely, if we define the following set: then it is clear that this is the set of union of all the unit circles which are centered at cos ϕ 0 y for each fixed y ∈ Y and which pass through a and b when ϕ = π ∓ ϕ 0 (see Figure 2.1).
The intersection of T and the y 1 -y n plane consists of two circles, say C 1 (when y 1 = 1, i.e., y = (1, 0,...,0)) and C 2 (when y 1 = −1, i.e., y = (−1, 0,...,0), see Figure 2.1).In the following contexts, we will consider the cases y 1 = 1 and −1 in connection with T as the circles C 1 and C 2 , respectively.Call S 1 the (n − 1)-dimensional unit sphere containing the circle C 1 .If we let the center of C 1 be O and the center of S 1 be Õ, then it is obvious that O = Õ.
(To see this, choose any point A ∈ C 1 and its antipodal point B in C 1 .Then, by the definition of the antipodal points that they lie exactly the opposite with respect to the center of the circle C 1 whose center is at O, and because they are of the same length 1, we have the following condition that (2.5) On the other hand, we have, since the two points A and B lie also on the unit sphere S 1 with its center at Õ, and therefore Õ = O.) Now, we first show that S 1 and T intersect only at C 1 .To make computation simpler we use a new coordinate x for R n , where x = y − cos ϕ 0 , 0,...,0 . (2.8) In this coordinate (see Figure 2.2), S 1 becomes the unit sphere S centered at the origin, T = x = cos ϕ + cos ϕ 0 y + (0,...,0, sin ϕ) With the help of this coordinate we show the following lemma.
Proof.If any element in T has distance 1 from the origin of the x-coordinate, then we have (2.11) Therefore, we have 0 = 2 cos ϕ 0 1 − y 1 cos ϕ + cos ϕ 0 . (2.12) With y 1 = 1, T in (2.10) represents the unit circle C 1 in the x 1 -x n plane.If then it follows from (2.10) that which also belong to C 1 .Now, consider, as in Figure 2.3, the origin e and ẽ = (−2, 0,...,0) in the x-coordinate and the unit circle C 1 passing through e and ẽ in the x 1 -x n plane.Choose a point d ∈ C 1 , d ∈ {e,ẽ}.We parameterize all the unit circles passing through the points e and d.We assume the x n -coordinate of d is negative.
By triangle inequality, the distance between e and d is less than 2, say 2 sin ϕ 0 , with 0 < ϕ 0 < π/2.Choose a new coordinate y for R n and consider two points e = 0,...,0, sin ϕ 0 , d = 0,...,0, − sin ϕ 0 , ( (see Figure 2.4). -6 To get a parameterization of the unit circles passing through e and d, we consider the mapping M defined by This transformation M is an isometry (since it is a composition of a rotation and translations) and sends y = 0,...,0, ± sin ϕ 0 = {e , d } (2.17With the help of this parameterization, we are ready to show the following lemma.Proof.Without loss of generality, we can assume the x n -coordinate of d is negative.Note that with ϕ = π ∓ ϕ 0 in (2.19), y = (0,...,0, ± sin ϕ 0 ) are the points e or d in the y-coordinate and further ϕ = π ∓ ϕ 0 means that x = (0,...,0) = e, x= cos − 2ϕ 0 − 1, 0,...,0, sin in the x-coordinate, regardless of y ∈ Y. Any unit circle passing through e and d is given as x = My with y given as in (2.19), that is, The first coordinate is We show that for y 1 ≠ −1 (y 1 = −1 means the circle C 1 in the y-coordinate and the circle C 1 in the x-coordinate, see Figure 2.4), there is always some ϕ near π − ϕ 0 (i.e., near the point e) such that the above x 1 becomes positive.Let and so Then, the above is (2.25) θ = 0 (ϕ = π − ϕ 0 ) means the intersection point e and the above x 1 becomes 0 as it should.Assume Then, x 1 is positive if and only if (2.28) (recall y 1 ≠ −1 and 0 < ϕ 0 < π/2).In other words, the x 1 -coordinate is positive if and Therefore, for y 1 ≠ −1 (i.e., except the circle C 1 ), the x 1 -coordinate is positive for small enough θ > 0.

Main theorem.
In the previous section, we introduced all preliminary lemmas for the main result of this paper.Now, we prove our main theorem.Theorem 3.1.If a one-to-one mapping f : R n → R n maps every unit circle onto a unit circle, then f is a linear isometry up to translation.
Proof.We show f preserves the distance 2. Suppose the distance between a = f (A) and b = f (B) is less than 2, while the distance between A and B is 2-see Figure 3.1.Then, we show it leads to a contradiction.
Let the distance between a and b be 2c (0 < c < 1).Choose any unit circle C passing through A and B and let f (C) = C 1 .Choose a coordinate for a and b as in Figure 3.1 such that C 1 lies in the x 1 -x n plane and Let f (E) = e and Ẽ the antipodal point (in C) of E and let f ( Ẽ) = d.Let the union of all the unit circles passing through a and 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 ẽ be S 1 .
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 then the unit circle determined by these three points is parameterized as are orthonormal to each other and Since the image of this unit circle lies in T, it follows that the image of the whole S under f lies in T.
It is also obvious that the x 1 -coordinate of any point in T is nonpositive.(Note that the center of any unit circle passing through a and b has coordinate 1 − c 2 y − 1 + 1 − c 2 , 0,...,0 for some y ∈ Y, ( (see (2.4)) and the distance between this center and any x = (x 1 ,...,x n ) is and because positive x 1 makes the distance larger than 1, which means that if x 1 > 0, we have x ∈ T.) Now, if d = ẽ, then the image of any unit circle passing through E and Ẽ lies in both T and S 1 .However, by Lemma 2.1, T ∩ S 1 = C 1 and this fact contradicts the injectivity of f .
On the other hand, if d ≠ ẽ, the image of any unit circle, except the circle C, passing through E and Ẽ is a unit circle passing through e and d.This unit circle is not C 1 since f is one-to-one, and by Lemma 2.2 it 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.

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Lemma 2 . 2 .
For d ∈ C 1 , d ∈ {e,ẽ}, any unit circle in R n , which passes through d and e, has some point whose x 1 -coordinate is positive, except the circle C 1 .