Weakly CirclePreserving Maps in Inversive Geometry
Abstract.
Let be the standard sphere embedded in . A mapping , not assumed continuous or even measurable, nor injective, is called weakly circlepreserving if the image of any circle under is contained in some circle in the range space . The main result of this paper shows that any weakly circlepreserving map satisfying a very mild condition on its range must be a Möbius transformation.
1. Introduction
The object of this paper is to give a characterizations of Möbius transformations acting on , under very weak conditions on such a map , which do not assume invertibility or even continuity of the map.
The standard sphere , viewed in is the real algebraic set
The set of Möbius transformations are the set of invertible maps generated by inversions. Such maps send circles to circles and spheres to spheres. The study of geometric properties invariant under such transformations is called inversive geometry.
One can also identify with under sterographic projection. In the space , an inversion (or a reflection) in an sphere is the function defined by . is well defined on , and at these two points, we define and . A reflection in a hyperplane is a usual reflection in and fixes the point in . We define a Möbius transformation on to be a finite composition of reflections in spheres or hyperplanes. The group of all Möbius transformations is called the Generalized Möbius Group , following Beardon [2, Chapter 3]. Note that in dimension , identifying with the Riemann sphere , Möbius transformations include all the linear fractional transformations , which are orientationpreserving maps and also the conjugate ones , which are orientation reversing maps.
1.1. Main Result
We will study mappings satisfying the following very weak version of the circlepreserving property.
Definition 1.1.
A map of the sphere to itself is called weakly circlepreserving if for every circle , lies in some circle.
Definition 1.2.
A map of the sphere to itself is called weakly spherepreserving if for every sphere , lies in some sphere.
In these two definitions we do not assume that is injective or even continuous. There are many such maps, including some that are not Möbius transformations. For example, any map whose image is finite and consists of points or less is automatically weakly spherepreserving and any map on sphere with image consisting of points or less is weakly circlepreserving. Nevertheless such maps are quite restricted when further assumptions are imposed on them.
The key restrictions we consider are the following “general position” conditions on the image of the map.
Definition 1.3.
A subset of is said to lie in circular general position if for any circle , the complement of contains at least two points of .
Definition 1.4.
A subset of is said to lie in spherical general position if for any sphere , the complement of contains at least two points of .
It is obvious from the definition that must contain at least points, and there do exist many point sets in spherical general positon. If lies in spherical general position, then every set containing lies in spherical general position. However, if lies in spherical general position, it is not clear whether it always contains an subset of lies in spherical general position, and this result does always not hold in dimension .
Our main result is as follows.
Theorem 1.5.
For , be a weakly circlepreserving map. If is in spherical general position and there is a sphere with in circular general position, then is a Möbius transformation.
1.2. Previous Results
Rigidity theorems of Möbius transformation have been investigated extensively. It is clear that Möbius transformations take generalized spheres to generalized spheres and a converse was known to Möbius, under the assumption that the map is continuous, see Blair [4, Theorem 5.6]. A map is called conformal if the map preserves angles. One can define it formally using conformal manifolds, see (Kobayashi [14], Blair [4]). Any Möbius transformation is a conformal diffeomorphism on . A result of Liouville[17] in 1850 asserts (in modern form) a local converse: when , any smooth conformal diffeomorphism of a simply connected open domain of into is the restriction of a Möbius transformation. In particular, for , the conformal group on is precisely the generalized Möbius group.
In 1937, Carathéodory [5] proved that a local version of the circlepreserving condition is enough to force a map to be (part of) a Möbius transformation. Given a domain , and a 11 map with , such that fpr any circle in that is contractible in , is a circle, then lies inside a plane and is a restriction of a Möbius transformation in .
More recently in 2001, Beardon and Minda [3] proved that is a Möbius transformation if and only if locally maps each sphere onto sphere. In 2005 Li and Wang [15] showed that is a Möbius transformation if and only if is circle preserving and is not a circle. All the above characterizations assume that is circle (or sphere) preserving, i.e., maps circles (or spheres) onto circles (or spheres).
The weakly circle preserving assumption is much less restrictive than those assumed on maps above. This assumption was introduced in 1979 by the first author with Webb [12]. Unfortunately, that paper used the term “circlepreserving map” to mean “ weakly circlepreserving map” as above. The paper [12] established a local result, under the weakly circle preserving map hypothesis. A special case is stated below in Theorem 2.1.
In this paper we prove only a global result, assuming the map is defined on all of . However we expect that the main result extends to a local version, where one assumes only that is defined on a simply connected open set inside .
1.3. Notation
Given , we will denote to be the smallest dimension sphere in that contains . If is some finite set, e.g., , we will simply write to mean . Similarly, if , we will simply use to mean .
To distinguish the domain and range space, we will use lower case roman letter (e.g., ) to denote a point in the domain space and lower case roman letter with an apostrophe symbol (e.g., ) to denote a point in the range space.
We will also use to denote sphere in and if is a map on , we will use the notation , i.e., the smallest dimension sphere containing .
Throughout this paper, we will identify with , and a sphere with either Euclidean sphere in or a kdimensional affine space in together with the point . We will also often use to mean the subspace in .
2. Two dimensional case
The proof will use some results from the twodimensional case which were obtained in previous papers, [12] and [11]. They state that if is weakly circlepreserving, and satisfies some conditions on the image of the map , then will automatically become continuous and bijective, in fact, will be a Möbius transformation.
In 1979 the first author with Webb [12, Theorem 1] proved a ”sixpoint theorem” for locally defined maps. The following theorem is the special case of that theorem. Notice that we use the term Möbius transformation for inversive transformation.
Theorem 2.1.
(“Sixpoint theorem”) Let be a weakly circlepreserving map from into with which satisfies the following conditions.

Every circle in the codomain does not contain at least two points in the image , i.e. is in circular general position in .

The image contains at least six distinct points.
Then is a 2sphere and is a Möbius transformation.
A result that we proved in [11] allows us to strengthen the result above, as follows.
Theorem 2.2.
(“Fivepoint theorem”) Let be a weakly circlepreserving map from into with which satisfies the following conditions.

Every circle in the codomain does not contain at least two points in the image , i.e. is in circular general position in .

The image contains five or more distinct points.
Then is a 2sphere and is a Möbius transformation.
Remark 2.3.
Proof.
This was proved in [11, Theorem 4.4]. ∎
3. Proof of Main Theorem 1.5.
Before proving this theorem, we need to prove several lemmas. The first two lemmas are geometric facts about intersecting spheres, which will be used a lot in proving the other lemmas.
Lemma 3.1.
If and are sphere and msphere in , assume contains at least two points, then dimension of the sphere can be to .
Proof.
To see this, we simply use a Möbius transformation to map two intersection points to and , then and are mapped to two vector subspace in union the point . We apply basic dimension theorem in linear algebra and get the result. ∎
Lemma 3.2.
For , , let be a sphere in and be two points in . Assume that , then any circle through intersects in at most one point.
Proof.
Let , and suppose for contradiction that there is a circle through that intersect at at least two points. Let be two of the intersection points. Notice that , so . So we have . This is a contradiction to . ∎
The following lemma shows that if is a Möbius transformation on a small dimensional sphere, will be Möbius transformation in a larger dimensional sphere provided some condition on the image of this larger dimensional sphere. This lemma gives us a tool to build up a chain of spheres with increasing dimension and is a Möbius transformation on each of them.
Lemma 3.3.
Let be a weakly circle preserving map, and are sphere and sphere in respectively. Suppose that is a Möbius transformation and , then is a Möbius transformation.
Proof.
We will first show is injective on . Let , and fix two points .
Claim 1.
There is a 2sphere through , and intersecting in a circle.
Proof.
(of Claim 1)
Since has codimension 1 in , divides into two components.
If and are on the opposite side of , then any 2sphere through will intersect in a circle.
Otherwise, choose on the opposite side of and , then any 2sphere through will intersect in a circle. ∎
Notice that the image of contains and a whole circle, so is in circular general position. By the fivepoint theorem (Theorem 2.2), is a Möbius transformation on . Also notice that must intersect both components in divided by , so we let on the opposite side of , and be its image. Since is a Möbius transformation on , we have and . Now given any two points , we form a 2sphere through , then this 2sphere must intersect in a circle as it contains and . Therefore, the image of this 2sphere consists of and a whole circle, which means the image is in circular general position. Therefore, is a Möbius transformation on this 2sphere by fivepoint theorem (Theorem 2.2), in particular, . This shows that is injective on .
We will then prove that lies in some sphere. Suppose not, let be two points in such that . Let be a 2sphere through and intersecting in a circle , then is in circular general position, so is a Möbius transformation on . But contains and , so , then . But this is a contradiction to the assumption . Therefore lies in some sphere.
We will now prove this lemma by induction.
The base case : Under a Möbius transformation, we may assume that , and . Given any 2sphere which is 2dimensional plane through origin with in other than , then is a circle , so the image contains a whole circle . Moreover, since is injective on , also consists of 2 distinct points off , so is in circular general position. By the fivepoint theorem (Theorem 2.2), is a Möbius transformation on . Now given arbitrary 2sphere in , it must intersect some plane through origin in a circle. By the same argument, is a Möbius transformation on . By the corollary 6.3 in Beardon and Minda [3] (namely, let , if restrict to any sphere is a Möbius transformation, then is a Möbius transformation), we conclude that is a Möbius transformation on .
The induction step: Assume that the lemma holds for , we prove the case for . Again, under a Möbius transformation, we may assume that , and . Given any sphere which is (k+1)dimensional plane through origin with in other than , is a sphere . Notice is a Möbius transformation as is a Möbius transformation on . Moreover, since is injective on , also consists of 2 distinct points off . By induction hypothesis, is a Möbius transformation on . Now given arbitrary sphere in , it must intersect some hyperplanes in through origin in a sphere. By the same argument, is a Möbius transformation on . By the corollary 6.3 in Beardon and Minda [3], we conclude that is a Möbius transformation on .
By induction, we conclude that the lemma holds for every . ∎
The following lemma shows that if the hypothesis of the Lemma 3.3 fails for every possible spheres, there will be some strict restriction on all higher dimensional spheres.
Lemma 3.4.
Let be a weakly circlepreserving map and be a fixed sphere in such that is a Möbius transformation. Assume that for any sphere containing , , then the following holds:
For any sphere () containing such that (recall here is the smallest dimension sphere containing ), we have
(1) 
and
(2) 
Proof.
Notice that from the hypothesis , it is necessary that . Hence, to prove (1), it is sufficient to prove . The proof of this lemma is by induction from . However, the base cases for are treated differently.
Case :
(1): This is immediate from the hypothesis of the lemma.
(2): Let and fix . Suppose for contradiction that there is with .
Claim 1.
There is a circle through and intersect at two points and .
Proof.
(of Claim 1)
Since has codimension 1 in , divides into two components.
If are in opposite side of , then any circle containing intersects at two points. Let and be two different circles through , then (as three points determines a circle). Since is injective on , so . Hence, for at least one of .
Otherwise, choose two different point and in the opposite side and form . Exact same argument shows that at least one of the two circles will satisfy the desired property. ∎
Now let be the intersection point of the circle with , and be its image respectively. Since is weakly circle preserving and determines a circle (as no two of them are equal), we have lies on the circle which is a contradiction to .
Case :
(1): Given containing , and suppose for contradiction that . Choose and fix in its preimage, i.e. , and let for . Notice that as , so by the (2) of , we have . Now given , and be its image under .
Claim 2.
There is a circle though and intersect for all .
Proof.
(of Claim 2)
Under a Möbius transformation, we may assume that and and . Then we have is the union of the point and a (k+1)dimensional plane in containing . If we consider the 2dimensional plane spanned by and where is the usual basis in , then is a line passing through . It is clear that we can choose a circle in through that not tangent to the three lines (). This circle will intersect at another point other then , which is not in , so we proved the claim (see Figure 1). ∎
Since , we have . Since is weakly circle preserving, , in particular, . This is true for any , so which is a contradiction.
(2): Let . Fix in its preimage and let . Again, , and .
Suppose for contradiction that there exists such that , let . Notice that it is necessary that .
Claim 3.
.
Proof.
(of Claim 3)
Again, under a Möbius transformation, we may assume that and .
Suppose for contradiction, then there exists on the opposite side of in such that . Let be the circle through , and .
Now replace by in claim 2, and exact same argument shows that which is a contradiction.
Case :
Induction hypothesis: Assume the lemma holds for .
(1): Given containing and . Suppose for contradiction that .
Let such that and determine a sphere, then and any subset of determine a sphere. Let , and choose a point in that not in the sphere determined by and , say this point is . Fix in its preimage, and let and .
We let , then it is clear that , and , so by induction hypothesis, we have . This implies that and .
Similarly, we have and .
Notice that , so contains at least two points. By lemma 3.1, we know . This means that there exists . Notice that . Therefore, contains at least points, which is a contradiction.
(2): Let , and fix in its preimage, and let . Exact same argument in the proof of (1) of case , we have and .
Suppose for contradiction that there exists such that . Let . Notice that and . So by the induction hypothesis for (2) of , we have . Hence, .
Notice that , so contains at least two points. By lemma 3.1, we know . This means that there exists . Notice that , say . Then is a (k+2)sphere containing with (as it contains and ), so by induction hypothesis, we have which is a contradiction to . ∎
Proof.
(of the theorem 1.5)
We will prove the theorem by building up a chain , where is sphere in , such that restrict to is a Möbius transformation.
We will build this chain by induction. Base case : By assumption, there is a two sphere with in circular general position, so by fivepoint theorem (Theorem 2.2), we conclude that is a Möbius transformation on .
The induction step: Assume that we have build the chain (), with being a Möbius transformation on each sphere, we will build sphere .
Case 1: There is an sphere containing such that consists of at least two points. By the lemma 3.3, we know that is a Möbius transformation on , so we let .
Case 2: There is no sphere containing such that consists of at least two points. In other words, for any sphere containing , we have . Then the hypothesis for lemma 3.4 is satisfied, so we conclude that in particular, (notice ). But on the other hand, as is in general position. So this is a contradiction, which means case 2 cannot happen.
Therefore, the theorem follows. ∎
4. Weakly SpherePreserving Maps
We first show that weakly spherepreserving maps are automatically weakly circlepreserving maps.
Lemma 4.1.
Suppose is weakly spherepreserving, and assume that is not contained in some sphere, i.e., . Then maps spheres into spheres for all dimensions with . In particular, is weakly circlepreserving.
Proof.
We will prove by downwards induction on that maps spheres into spheres. The base case is true by the weakly spherepreserving hypothesis.
For the induction step, assume that maps spheres into spheres, for a fixed . Given a sphere , choose a point , let , then by induction hypothesis, (recall here means the smallest dimension sphere containing and and ). Since is not contained in some sphere, there is an image point . We fix , and denote . Notice that , so that
But as , so is a sphere of dimension less than or equal to . This proves that lies in some sphere, and completes the induction step. ∎
Remark 4.2.
Notice that if