The pseudo-affine transformations in R2

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. ARTICLES .

March 2010 Vol. 53 No. 3: 1–8 doi: 10.1007/s11425-010-0032-8

The pseudo-affine transformations in R2 Dedicated to Professor Yang Lo on the Occasion of his 70th Birthday

LI BaoKui1,∗ , WANG XianTao2 & WANG YueFei3 1Department

of Mathematics, Beijing Institute of Technology, Beijing 100081, China; of Mathematics, Hunan Normal University, Changsha 410081, China; 3Institute of Mathematics, AMSS and KLG Key Laboratory of Mathematics, Chinese Academy of Sciences, Beijing 100190, China Email: henan [email protected], [email protected], [email protected] 2Department

Received September 24, 2009; accepted December 18, 2009

Abstract In this paper, we introduce the pseudo-affine transformations, a new class of maps, and obtain their characterization in terms of g-triangle-reflections. Keywords

affine transformation, pseudo-affine transformation, g-triangle-reflection

MSC(2000):

30C35, 51F99

Citation: Li B K, Wang X T, Wang Y F. The pseudo-affine transformations in R2 . Sci China Math, 2010, 53, doi: 10.1007/s11425-010-0032-8

1

Introduction

Suppose that D is a subset of R2 . We say that ℓ is a line in D if there is a straight line s ⊂ R2 such that ℓ = s ∩ D, and a map f : D 7→ D′ ⊂ R2 is line-to-line if the image of each line in D is contained in a line of D′ . The line-to-line maps in Rn have been investigated for a long time and many results have been in literature. Among them, the following are from [2] and [14] due to Artin and Jeffers, respectively. Theorem A ([2]). Suppose that f : Rn 7→ Rn (n > 1) is a bijection and preserves lines, and suppose that the images of any two parallel lines under f are still parallel lines. Then f is an affine transformation. Here, f is said to preserve lines if the image of each line is still a line. Theorem B ([14, Theorem 4.5]). Suppose that f : Rn 7→ Rn (n > 1) is a bijection and preserves lines. Then f is an affine transformation. In [7], Chubarev and Pinelis showed that the condition “f being injective” in Theorems A and B can be removed, and the condition “f preserving lines” can be replaced by the one “f being line-to-line”. Precisely, we have the following. Theorem C ([7]). transformation.

Suppose that f : Rn 7→ Rn (n > 1) is surjective and line-to-line. Then f is an affine

In [15], the authors proved ∗ Corresponding

author

c Science China Press and Springer-Verlag Berlin Heidelberg 2010 ⃝

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Theorem D ([15, Theorem 3]). Suppose that f : Rn 7→ Rn (n > 1) preserves lines. Then f is an affine transformation if and only if it is non-degenerate. Here, a line-preserving map f : Rn 7→ Rn (n > 1) is non-degenerate if the image f (Rn ) is not contained in a line. See [1, 3–6, 8–13] and [6–22] for other related discussions. We know from Theorem C that if f : R2 7→ R2 is line-to-line and f 2 = id, where id denotes the identical map, then f is an affine transformation. In fact, it can be conjugated to the rotation f (x, y) = (x′ , y ′ ) for any (x, y) ∈ R2 , where x′ = −x and y ′ = −y, or the reflection f (x, y) = (x′ , y ′ ) for any (x, y) ∈ R2 , where x′ = −x and y ′ = y. A natural question is: For a given subset D ⊂ R2 , if f : D 7→ D is line-to-line and satisfies f 2 = id, is f the restriction to D of an affine transformation of R2 ? Here and in the following, id means the restriction to D of the identical map of R2 . We shall construct an example to show that the answer is negative in Section 2. To deal with the problem, we introduce the following concept of pseudo-affine transformation. Definition 1.1. Let D be a subset of R2 with non-empty interior. Suppose that a map f : D 7→ D is line-to-line and satisfies f 2 = id. If f is not the restriction to D of an affine transformation of R2 , then we say that f is a ‘pseudo-affine transformation’. The aim of this paper is to characterize the pseudo-affine transformations. Our main result is the following. Theorem 1.1. Suppose that D is a convex subdomain of R2 . Then a map f is a pseudo-affine transformation if and only if there exists a g-triangle-reflection ϕ such that f = ϕ|D . Here the g-triangle-reflection ϕ is a map, which, roughly speaking, is affinely conjugated to a trianglereflection of the form: ( ) x y η : (x, y) 7→ − , . 1 + Kx 1 + Kx We shall give the precise definition in Section 3.

2

Triangle-reflections

In this section, we construct a class of pseudo-affine transformations which are called triangle-reflections. We shall show that such maps are indeed not affine transformations in certain situations. For the sake of convenience, in the following, we always use A, B, . . . to denote the points in R2 , A′ , ′ B , . . . their images under the map we consider, LAB the line determined by A and B, AB the segment with the endpoints A and B. Let △ABC denote the closure of the triangle domain in R2 bounded by the segments AB, BC and AC. Definition 2.1. We say that δ : △ABC 7→ R2 is a ‘triangle-reflection’ of △ABC along AD for some D ∈ (BC)◦ if it is line-to-line and satisfies the conditions: δ(P ) = P for any P ∈ AD,

δ(B) = C = B ′ and δ(C) = B = C ′ .

Proposition 2.1. If δ is a triangle-reflection of △ABC along AD, then δ(△ABC ) = △ABC and δ : △ABC 7→ △ABC is homeomorphic with δ 2 = id. Proposition 2.2. For a triangle △ABC and D ∈ (BC)◦ , if the triangle-reflection of △ABC along AD exists, then it must be unique. The proofs of Propositions 2.1 and 2.2 easily follow from the following determination of the image for any point in △ABC . For any P ∈ △ABC , we come to determine the image P ′ of P under δ. If P ∈ AD, then P ′ = P . If P ∈ △ABC \(AD ∪ BC), we choose two different points E, F ∈ AD such that LBE ∩ LCF = {P }. Then P ′ is the intersection point of LCE and LBF . Obviously, P ′ ∈ △ABC and it is unique.

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If P ∈ BC\AD, then we choose E ∈ AD and F ∈ (△ABC )◦ \AD such that LEF ∩ LBC = {P }. Then P ′ is the intersection point of LEF ′ and LBC , where F ′ is the image of F under δ. Also P ′ ∈ △ABC and it is unique. Lemma 2.1. Suppose that δ is a triangle-reflection of △ABC along AD for some D ∈ (BC)◦ , and that g : R2 7→ R2 is an affine transformation. Let g(A) = A1 , g(B) = B1 , g(C) = C1 and g(D) = D1 . Then the conjugation δ1 = g · δ · g −1 : △A1 B1 C1 7→ △A1 B1 C1 is a triangle-reflection of △A1 B1 C1 along A1 D1 . Lemma 2.2. For a triangle △ABC with D ∈ (BC)◦ , there exist an affine transformation g and a triangle △A1 B1 C1 with D1 ∈ (B1 C1 )◦ such that (1) g(A) = A1 , g(B) = B1 , g(C) = C1 and g(D) = D1 ; (2) g(△ABC ) = △A1 B1 C1 ; (3) A1 D1 is orthogonal to B1 C1 ; (4) |BD| = |B1 D1 | and |DC| = |D1 C1 |. Proof. Under the conjugation of some euclidean isometry of R2 , we may assume that D is the origin, LBC the x-axis with B(−|BD|, 0), C(|CD|, 0). Also we assume that A(a, b) with b > 0. Let g be the affine transformation as follows: g : △ABC 7→ △A1 B1 C1 (x, y) 7→ (x′ , y ′ ), where x′ = x − ab y and y ′ = y. Obviously, g(B) = B1 (−|BD|, 0), g(C) = C1 (|CD|, 0), g(D) = D1 (0, 0) and g(A) = A1 (0, b). Hence g and the triangle △A1 B1 C1 are the desired. Example 2.1. Let A(0, a), B(b, 0), C(−c, 0), D(0, 0), where a, b, c ∈ R+ . Define γ : △ABC 7→ △ABC y x , y ′ = 1+Kx and K = c−b as: For any P = (x, y) ∈ △ABC , P ′ = γ(P ) = (x′ , y ′ ), where x′ = − 1+Kx bc . Then Proposition 2.3. γ is a triangle-reflection of △ABC along AD and γ is an affine transformation if and only if K = 0, i.e., |BD| = |CD|. Here and in the following, “a map in a domain Ω is an affine transformation” means “it is the restriction to Ω of an affine transformation of R2 ”. Proof. The proof for the second part is obvious. For the proof of the first part, it suffices to prove that γ is line-to-line. In order to prove this, let P1 (x1 , y1 ), P2 (x2 , y2 ), P3 (x3 , y3 ) ∈ △ABC be distinct and collinear. Then there exists some constant λ ∈ / {0, −1} such that x3 =

x1 + λx2 y1 + λy2 and y3 = . 1+λ 1+λ

If we use P1′ (x′1 , y1′ ), P2′ (x′2 , y2′ ) and P3′ (x′3 , y3′ ) to denote the images of P1 , P2 and P3 under γ, respectively, then elementary computations show that x′3 = − and y3′ = Let λ′ =

λ(1+Kx′1 ) 1+Kx′2 .

x3 (1 + Kx′2 )x′1 + λ(1 + Kx′1 )x′2 = , 1 + Kx3 (1 + Kx′2 ) + λ(1 + Kx′1 )

(1 + Kx′2 )y1′ + λ(1 + Kx′1 )y2′ y3 = . 1 + Kx3 (1 + Kx′2 ) + λ(1 + Kx′1 )

Then

y1′ + λ′ y2′ x′1 + λ′ x′2 ′ and y = , 3 1 + λ′ 1 + λ′ which implies that P1′ , P2′ and P3′ are collinear. By Lemma 2.4, Example 2.1 and Proposition 2.3, we have x′3 =

Theorem 2.1. For a triangle △ABC and any D ∈ (BC)◦ , there exists a triangle-reflection δ of △ABC along AD, and δ is an affine transformation if and only if |BD| = |CD|.

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g-triangle-reflections

In this section, we first introduce the definition of g-triangle-reflections and then discuss their elementary properties. Let K be a positive constant, La = {(x, y) ∈ R2 : x = 0}, { } 1 Lb = (x, y) ∈ R2 : x = − , K Le = {(x, y) ∈ R2 : y = 0}, and Ω = R2 \Lb . Definition 3.1.

We define the map η : Ω 7→ Ω as follows. For any P = (x, y) ∈ Ω, P ′ = η(P ) = (x′ , y ′ ),

y x where x′ = − 1+Kx and y ′ = 1+Kx . If we use Ω1 to denote the connected component of Ω containing La and Ω2 the other one, then the following is obvious. 2 Proposition 3.1. η is line-to-line, η 2 = id and the fixed point set fix(η) of η is La ∪ {P0 (− K , 0)}. Moreover, Ω1 and Ω2 are two completely invariant components of η.

Definition 3.2. We say that two lines ℓ1 , ℓ2 in Ω are ‘parallel’ if there are two parallel straight lines s1 and s2 in R2 such that ℓ1 = s1 ∩ Ω and ℓ2 = s2 ∩ Ω. Then Proposition 3.2. parallel to La .

For any line ℓ in Ω, if it is parallel to La , then the image η(ℓ) of ℓ under η is still

Proposition 3.3 For any E ∈ Ω\(La ∪ P0 ), the line LEE ′ contains the point P0 . Furthermore, each line in Ω containing P0 is invariant under η. Proof. If the line LEP0 is parallel to La , then by Proposition 3.2, the image line LE ′ P0 is parallel to La , from which one can obtain P0 ∈ LEE ′ . If LEP0 has an intersection point with La , which is denoted by Q. Then E ′ ∈ LP0 Q , which implies that P0 ∈ LEE ′ . Proposition 3.4. Suppose that ℓ denotes the line in Ω parallel to La and passing through P0 . Then for any P ∈ ℓ\{P0 }, P and its image P ′ are symmetric with respect to P0 . Proposition 3.5. Suppose that two lines s1 , s2 ⊂ R2 have an intersection point in Lb . Then the image line η(ℓ1 ) is parallel to η(ℓ2 ), where ℓi = si ∩ Ω (i = 1, 2). Proposition 3.5 is a corollary of the following result. Proposition 3.6. Suppose that s is a straight line in R2 with the slope k. Then the line in R2 containing 1 k the image η(s∩Ω) goes through the point (− K , K ) which is the intersection point of Lb and the line passing through P0 with the slope k. Proof. For any point P (x, y) ∈ s, x and y satisfy the relation y = kx + d, where d is a constant. 1 k , K + d) in Ω, which implies that the straight line s′ in Then s contains the points P1 (0, d) and P2 ( K 1 k 2 R determined by η(s ∩ Ω) contains the points P1′ (0, d) and P2′ (− 2K , 2K + d2 ). Hence the equation of (u, v) ∈ s′ is v = (dK − k)u + d, from which the proof easily follows. Definition 3.3. We call the map η : Ω 7→ Ω defined as above an ‘s-triangle-reflection’. Also we call the line La the ‘axis’, the only isolated fixed point P0 the ‘base’ point, the exceptional line Lb the ‘boundary’, and the line Le orthogonal to the axis and through the base point P0 the ‘equator’. The ‘main component’ Ω1 is the connected component containing La of Ω and the ‘slave component’ Ω2 the other one. ′ For any P (x, y) ∈ Ω\La , if P ′ (x′ , y ′ ) denotes its image, then − x+x xx′ = K. Then by Proposition 3.2, the following equalities are obvious (see Figure 3.1).

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La Y

6

A A   A D A   D′   A  P A  E A  ′  A XXX P0 B  O  A r A X  X @
′XXXX B  X  @
E Q XXXX  X @


-X

Figure 3.1

Proposition 3.7.

(1) Suppose that D ∈ Ω1 , P ∈ La ∩ LP0 D and D′ ∈ P P0 . Then 1 |P D||P D′ | |P P0 | = . 2 |P D| − |P D′ |

(2) Suppose that E ∈ Ω2 , Q ∈ La ∩ LP0 E and E ′ ∈ QP0 . Then 1 |QE||QE ′ | |P0 E||P0 E ′ | |QP0 | = = . 2 |QE| + |QE ′ | |P0 E| − |P0 E ′ | By the definition, the following two results are obvious. Lemma 3.1. Any two of the axis La , the boundary Lb and the base point P0 determines the third and the corresponding s-triangle-reflection. Lemma 3.2 Suppose that η : Ω 7→ Ω is an s-triangle-reflection with the axis La , the boundary Lb and the base point P0 . If g : R2 7→ R2 is an affine transformation, then η1 = g · η · g −1 : g(Ω) 7→ g(Ω) satisfies the following. (1) η1 is line-to-line and η12 = id; (2) fix(η1 ) = g(La ) ∪ g(P0 ); (3) The boundary of η1 is g(Lb ); (4) The equalities in Proposition 3.7 still hold under the conjugation of g. From Lemma 3.2, we generalize the concept “s-triangle-reflection” as follows. Definition 3.4. A map ϕ is called a ‘g-triangle-reflection’ if there are an affine transformation g : R2 7→ R2 and an s-triangle-reflection η such that ϕ = g ◦ η ◦ g −1 . Also we call g(La ) the ‘axis’ of ϕ, g(P0 ) the ‘base point’ and g(Lb ) the ‘boundary’. Obviously, Proposition 3.8. Let ϕ and ψ be two g-triangle-reflections. Then there exists an affine transformation g : R2 7→ R2 such that ψ = g · ϕ · g −1 . Lemma 2.2 and Theorem 2.1 imply that Proposition 3.9. For any triangle-reflection γ : △ 7→ △ which is not an affine transformation, there exists a unique g-triangle-reflection ϕ such that γ = ϕ|△ . The following easily follows from Proposition 2.3 and Lemma 3.2. Corollary 3.1.

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Each g-triangle-reflection is a pseudo-affine transformation.

The proof of Theorem 1.1

Obviously, the sufficiency of Theorem 1.1 follows from Corollary 3.1. The proof of the necessity follows from the following series of lemmas.

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Suppose that D is a convex domain in R2 , R2 \D ̸= ∅, and that f : D 7→ D is a pseudo-affine reflection. Lemma 4.1.

f maps each line in D onto a line.

Definition 4.1.

We say that a line ℓ in D is ‘complete’ if it itself is a straight line in R2 .

Obviously, if D contains some complete line, then for any point P ∈ D, there exists one and only one complete line passing through P , which can be denoted by LP . This implies that all the complete lines in D are parallel to each other, which shows that D is either a strip domain or a half plane. Hence for any line ℓ in D, ℓ either is parallel to LP or intersects with LP in D. If ℓ is not a complete line in D, then there exist two lines L1 and L2 with an intersection point in D such that (L1 ∪ L2 ) ∩ l ∩ D = ∅. From this the following easily follows. Lemma 4.2.

A line ℓ in D is complete if and only if its image ℓ′ under f is complete.

For any line ℓ in D, ℓ divides D into two parts. We use U1 , U2 to denote the two connected components of D\ℓ, respectively. Choose P1 ∈ U1 . Also the image ℓ′ of ℓ under f divides D into two connected parts. We use U1′ to denote the part containing P1′ and U2′ the other one. Then Lemma 4.3.

f (Ui ) = Ui′ (i = 1, 2).

Proof. We prove this lemma by contradiction. Suppose that there exist P and Q ∈ U1 such that ′ ′ P ∈ U1 and Q′ ∈ U2′ . We divide the discussions into two cases. Case I. ℓ is not a complete line. Then there are two lines ℓ1 and ℓ2 in D with P ∈ ℓ1 and Q ∈ ℓ2 such that ℓ1 ∩ ℓ2 ̸= ∅, (ℓ1 ∪ ℓ2 ) ∩ ℓ = ∅. Then ℓ′1 ⊂ U1′ and ℓ′2 ⊂ U2′ . This is the desired contradiction since ℓ′1 ∩ ℓ′2 ̸= ∅. Case II.

ℓ is a complete line.

Let ℓ1 be a line which is not complete, ℓ1 ∩ ℓ = {K}, ℓ1 ∩ LP = {P1 } and ℓ1 ∩ LQ = {Q1 }. Choose a line ℓ2 such that it is not complete, ℓ2 ∩ ℓ = {K} and P1 , Q1 lie in the same connected component of D\ℓ2 . Then P1′ and Q′1 must lie in different components of D\ℓ′2 . Since ℓ′2 is not complete, we see from Case I that it is impossible. Lemma 4.4. For {E, F } ⊂ D, if the points E, E ′ , F and F ′ are non-collinear, then there exists a unique convex (closed) quadrilateral domain with vertexes E, E ′ , F and F ′ , which is denoted by P. Proof. The assumptions imply that any three of E, E ′ , F and F ′ are not collinear. Hence it suffices to prove that any one of E, E ′ , F and F ′ , say E ′ , is not contained in the closed triangle ∆EF F ′ with the vertices E, F and F ′ . Suppose for the contrary that E ′ ⊂ ∆EF F ′ . Then E ′ , F ′ are in the same component of D\LEF , but their images E, F under f are in the different components of D\LE ′ F ′ . By Lemma 4.3, this is a contradiction. Corollary 4.1. For {E, F } ⊂ D, if the points E, E ′ , F and F ′ are non-collinear, then the convex (closed) quadrilateral domain P with vertexes E, E ′ , F and F ′ is invariant under f . As shown in Figure 4.1, we only need to consider two cases for the sides of the quadrilateral domain P determined by E, E ′ , F and F ′ . They are EF, F E ′ , E ′ F ′ , F ′ E or EE ′ , E ′ F, F F ′ , F ′ E. For the former case, we use PEF E ′ F ′ to replace P; for the latter, we replace P by PEE ′ F F ′ . First, we assume that PEF E ′ F ′ ⊂ D. ′ HH E r H  HH  HHF ′  F O r r r @ H @ H  @ @ @ @ @r E@

HH F r H HH   HHE ′ ′  F O r r r @ H @ H  @ @ @ @ @r E@

Figure 4.1

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Lemma 4.5. The intersection point O of LEE ′ and LF F ′ is the only fixed point of f in PEF E ′ F ′ , and for any P ∈ PEF E ′ F ′ , the line LOP is invariant under f . Proof.

Obviously, O is a fixed point of f in PEF E ′ F ′ . Lemma 4.3 shows that it is the only one.

For the proof of the second part, suppose for the contrary that there exists some P ∈ P\{O} such that P ′ ̸∈ LP O . Then the intersection point of LP P ′ and LEE ′ or the one of LP P ′ and LF F ′ is a fixed point in PEF E ′ F ′ which is different from O. This is the desired contradiction. Lemma 4.6. Suppose that D contains a parallelogram PABA′ B ′ with the diagonals AA′ and BB ′ . Then f is a rotation with the rotation angle π. Proof. Denote the intersection point of AA′ and BB ′ by M . Lemma 4.5 shows that M is the only fixed point of f in PABA′ B ′ . Obviously, it suffices to prove that for any P ∈ D, P and its image P ′ under f are symmetric with respective to M . We divide the discussions into three cases. Case I. P ∈ ∂PABA′ B ′ , the boundary of PABA′ B ′ . Without loss of generality, we may assume that P ∈ AB. Then P ′ is the intersection point of the lines LP M and LA′ B ′ , which completes the proof. Case II.

◦ P ∈ PABA ′ B ′ , the interior of PABA′ B ′ .

We may assume that P is in the interior of the triangle△AA′ B . Let ℓ be the line in D passing through P and parallel to LBA′ . Let C be the intersection point of LA′ P with AB. So C ′ is the intersection point of LCM with A′ B ′ . Then P lies in the boundary of the parallelogram PACA′ C ′ whose diagonals are AA′ and CC ′ with AA′ ∩ CC ′ = {M }. From Case I, the result follows. Case III.

P ∈ D\PABA′ B ′ .

For each i ∈ {1, 2}, let ℓi be the line which passes through P and intersects two different sides of the parallelogram ABA′ B ′ , and we use Pi1 and Pi2 to denote the intersection points. Then P ′ is the ′ ′ P′ ′ P ′ . Obviously, P and P intersection point of the lines LP11 and LP21 are symmetric with respective 12 22 to M . Similar discussions as in the proof of Lemma 4.6 also show Lemma 4.7. Suppose that D contains a quadrilateral domain P0 with diagonals CC ′ and DD′ . Then f : D 7→ D is uniquely determined by the vertices C, D, C ′ , D′ . Now we are ready to prove the following result. Lemma 4.8.

There is a g-triangle-reflection ϕ such that f = ϕ|D .

Proof. By Lemma 4.6, PEF E ′ F ′ is not a parallelogram domain in D since f is a pseudo-affine transformation. We may assume that |EO| > |E ′ O| and |F O| > |F ′ O|. Then there is a unique A ∈ EF ′ such that |AO| = |A′ O|. Choose B ∈ ∂PEF E ′ F ′ such that BO is orthogonal to LAA′ . By Lemma 4.6, we may assume that |BO| > |B ′ O|. Choose P ∈ LBB ′ such that B ′ ∈ P O and |P O| =

2|BO||B ′ O| . |BO| − |B ′ O|

Denote the straight line in R2 , passing through P and parallel to AA′ , by La . Let ϕ be the g-triangle-reflection with the base point O and the axis La . The definition of ϕ and Propositions 3.4 and 3.7 imply that ϕ(A) = A′ and ϕ(B) = B ′ . Denote PABA′ B ′ by P1 . By Lemma 4.7, the restriction f |P1 is uniquely determined by A, B, A′ , B ′ , and f |P1 (P1 ) = P1 . Hence f |P1 = ϕ|P1 , which together with the assumption “f being line-to-line in D” yields that f = ϕ|D . The proof is complete. Second, we assume that PEE ′ F F ′ ⊂ D. Lemma 4.9.

There exist P ∈ EE ′ and Q ∈ F F ′ such that {P, Q} ⊂ fix(f ).

Proof. Obviously, the intersection point EF and E ′ F ′ is a fixed point of f . For A ∈ EE ′ , if A′ = A, we take P as A. If A′ = ̸ A, then the intersection point B of F A and F ′ A′ or F ′ A and F A′ is fixed by f .

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We take P as the intersection point of LBO and EE ′ . Since LP O is invariant under f , we take Q as the intersection point of LP O and F F ′ . Lemma 4.10.

P Q ⊂ fix(f ).

Proof. Suppose, for the contrary, that there is some C ∈ (P Q)◦ such that C ′ = ̸ C. Then E, E ′ , C, ′ and C are non-collinear. Lemma 4.4 shows that it is impossible. It follows from the discussions in Section 2 that there is a triangle-reflection of △P F F ′ along P Q: γ : △P F F ′ 7→ △P F F ′ . Propositions 2.1, 2.2 and their proofs show that f |△P F F ′ = γ. Since f is a pseudo-affine transformation, it follows from Proposition 2.3 that γ is not an affine transformation. Hence Proposition 3.9 shows that there exists a unique g-triangle-reflection ϕ such that γ = ϕ|△P F F ′ . Since ϕ|D is line-to-line with ϕ2 |D = id, the following is implied. Lemma 4.11.

There is a g-triangle-reflection ϕ such that f = ϕ|D .

Lemmas 4.8 and 4.11 show that the necessity of Theorem 1.1 holds.

Acknowledgements

The authors are supported in part by NSF and 973 Project of China, respectively.

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