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Spaces with congruence
Discrete Mathematics 308 (2008) 207 – 213 www.elsevier.com/locate/disc
Spaces with congruence Alexander Kreuzer, Kay Sörensen Technische Universität München, Zentrum Mathematik, Boltzmannstr. 3, 85747 München, Germany Received 11 October 2004; received in revised form 30 November 2005; accepted 27 November 2006 Available online 2 June 2007
Abstract In this paper we consider an exchange space with congruence (P , L, ≡) not assuming any further geometrical properties. If the dimension of the space is greater than 2, we show that for any line G of a plane E and any point x ∈ G there is a unique perpendicular line through x in E and that any line reflection is a motion. © 2007 Elsevier B.V. All rights reserved. Keywords: Congruence; Motions; Line reflections
1. Introduction It is well known that every Euclidean plane (E, L, , ≡) [2,9] is isomorphic to an affine plane AG(2, K) over an ordered and pythagorean commutative field K (cf. [2] (24.5)). Here (E, L) denotes an affine plane, (E, L, ) an ordered plane, and ≡ denotes the congruence relation on E × E. We may consider E as a quadratic separable field extension of K with the corresponding involutory field automorphism ¯ : E → E and we have (a, b) ≡ (c, d) if and only if (a − b)(a − b) = (c − d)(c − d) for any points a, b, c, d ∈ E. There is a corresponding theorem for hyperbolic planes (cf. [2]). For both proofs one first considers the group of motions, in particular the line reflections. For the definition of a motion and a line reflection the order relation is not necessary. We need only the linear structure of (E, L) and the congruence relation ≡, but for the proof that the line reflection is a motion it seems that additional assumptions on the geometry are necessary. For example Sörensen assumes in [8] that for given lines G1 , G2 , there exist distinct lines H1 , H2 through a common point z which intersect G1 , G2 . Here we give a proof for exchange planes with congruence not using this property. In a plane with congruence (E, L, ≡) two perpendicular lines may have an empty intersection. We show that if in a plane any two perpendicular lines have an empty intersection, then the relation “perpendicular” together with the “identity relation” is transitive (cf. Theorem 2.13). If there are two perpendicular lines with a non-empty intersection, one can show that for any line G and any point x ∈ G we have a perpendicular line through x. But it is open if there is a unique perpendicular line to G through x. This is true for finite or affine planes, but not known in general. In this paper we consider a space with congruence (P , L, ≡) not assuming any further geometrical properties. (For special spaces with congruence cf. [5,6].) We only assume that the space satisfies the exchange property, which for example can be easily shown for ordered spaces. In Section 3 we assume that there are two points with a midpoint. We use the property that if two points have a midpoint, E-mail addresses: [email protected] (A. Kreuzer), [email protected] (K. Sörensen). 0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.11.034
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then they have a midline in every plane. If the dimension of the space is greater than 2, we show that for any line G of a plane E and any point x ∈ G there is a unique perpendicular line through x in E and that any line reflection is a motion. It follows that for any two points b, z there exists a unique point b ∈ z, b\{b} with (z, b) ≡ (z, b ) and also point reflections are motions. For the case that there exist no points with a midpoint, we conjecture that all examples in which the line reflections are motions have dimension 2. 2. Spaces with congruence Let (P , L) denote a linear space or incidence space with the point set P, the line set L and at least three points on every line, i.e., • for any two points there is exactly one line containing them and • for any line L ∈ L we have |L|3. A subspace is a subset U ⊂ P such that for all distinct points x, y ∈ U the unique line passing through x and y, denoted by x, y, is contained in U. Let U denote the set of all subspaces. For every subset X ⊂ P we define the following closure operation U. (1) : P(P ) → U; X → X by X := U ∈U X⊂U
For U ∈ U we call dim U := inf{|X| − 1 : X ⊂ U and X = U } the dimension of U. A subspace of dimension two is a plane. For a set {a, b, c, . . .} we write a, b, c, . . . instead of {a, b, c, . . .}. We introduce the concept of a space (P , L, ≡) with congruence (cf. [8]). We assume that (P , L) is a linear space which satisfies the following exchange condition. (EC) Let S ⊂ P and let x, y ∈ P with x ∈ S ∪ {y}\S. Then y ∈ S ∪ {x} Let ≡ be a congruence relation on P × P , i.e., ≡ is an equivalence relation with (a, b) ≡ (b, a), (a, a) ≡ (b, b) and (a, a) ≡ (b, c) implies b = c. We use the notation (x1 , x2 , x3 ) ≡ (y1 , y2 , y3 ) if and only if (xi , xj ) ≡ (yi , yj ) for i, j ∈ {1, 2, 3}. (P , L, ≡) is a space with congruence if the axioms (W1), (W2) and (W3) are satisfied. (W1) Let a, b, c ∈ P be distinct and collinear, and let a , b ∈ P with (a, b) ≡ (a , b ). Then there exists exactly one c ∈ a , b with (a, b, c) ≡ (a , b , c ). (W2) Let a, b, x ∈ P be non-collinear and let a , b , x ∈ P with (a, b, x) ≡ (a , b , x ). For any c ∈ a, b and c ∈ a , b with (a, b, c) ≡ (a , b , c ) it holds (x, c) ≡ (x , c ). (W3) For a, b, x ∈ P non-collinear there exists exactly one x ∈ a, b, x\{x} with (a, b, x) ≡ (a, b, x ). We call a bijective mapping : P → P a motion, if (x, y) ≡ ((x), (y)) for all x, y ∈ P . Lemma 2.1. (i) If a, b, c are collinear points and a , b , c ∈ P with (a, b, c) ≡ (a , b , c ), then a , b , c are collinear. (ii) Any motion is a collineation. Proof. (i) By (W1) the point c ∈ a , b exists with (a , b , c ) ≡ (a, b, c). If c ∈ / a , b , then by (W2) it would follow (c , c ) ≡ (c, c), hence c = c ∈ a , b . (ii) By (i), and −1 map collinear points onto collinear points. For a subspace U and points a, b ∈ U we define MU (a, b) := {x ∈ U : (a, x) ≡ (b, x)}. We call MU (a, b) a midpoint, a midline, or a midplane of a, b, respectively, if it is a point, a line, or a plane, respectively. Lemma 2.2. (i) Two distinct points a, b have at most one point m ∈ a, b with (a, m) ≡ (b, m). (ii) MU (a, b) is a subspace of U.
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Proof. (i) For m, m ∈ a, b we have (a, m, m ) ≡ (b, m, m ), and (W1) implies m = m . (ii) Assume a = b and let x, y ∈ MU (a, b) be distinct points, hence (a, x, y) ≡ (b, x, y). By (W1) the points a, b, x, y are not collinear. For w ∈ x, y by (W2) it follows (a, w) ≡ (b, w), i.e., w ∈ MU (a, b). Using that every plane satisfies the exchange condition, we can show the following important lemma for a space with congruence: Lemma 2.3. Let a, b, x ∈ P be non-collinear points and let x ∈ a, b, x\{x} with (a, b, x) ≡ (a, b, x ). Then for any c ∈ a, b, x it holds (x, c) ≡ (x , c) if and only if c ∈ a, b. / ME (x, x ) it follows ME (x, x ) = E. By Lemma 2.2(ii), Proof. For E := a, b, x we have a, b ∈ ME (x, x ). Since x ∈ ME (x, x ) is a subspace of E, and since E is an exchange plane and a, b ∈ ME (x, x ), it follows ME (x, x ) = a, b. For a line L ∈ L, x ∈ P \L and a, b ∈ L with a = b there exists by (W3) the unique point x ∈ L ∪ {x}\{x} with (a, b, x) ≡ (a, b, x ). By Lemma 2.3, x is independent of the choice of a, b ∈ L, hence we may denote x = L(x). We call the following mapping line reflection ∼ x if x ∈ L, : P → P ; x → L L(x) if x ∈ / L. ∼
Clearly L is an involutory bijection with z = L(z) if and only if z ∈ L. We remark that by the notations above we ∼
can write L(x) or L(x) as well. We prefer L(x). Lemma 2.4. Let L be a line of a plane E and let p := L(p) for p ∈ E. (i) If (p, q) ≡ (p , q ) for p, q ∈ E, then (p, q, x) ≡ (p , q , x ) for every x ∈ p, q. (ii) Let a, b, c, d ∈ E be points with (p, q) ≡ (p , q ) for all p, q ∈ {a, b, c, d}. Then (x, y) ≡ (x , y ) for x ∈ a, b and y ∈ c, d. Proof. (i) By (W1) there exists a point x1 ∈ p , q with (p, q, x) ≡ (p , q , x1 ). If x = x1 , then by Lemma 2.3 x ∈ L. For any point u ∈ L , (p, q, u) ≡ (p , q , u), hence by (W2) (u, x) ≡ (u, x1 ) and therefore x1 = x . (ii) By (i) we have (a, b, x) ≡ (a , b , x ) and (c, d, y) ≡ (c , d , y ). If c ∈ a, b, then (c, x) ≡ (c , x ) by (i). For c∈ / a, b, the assumptions give (a, b, c) ≡ (a , b , c ) and (W2) implies (c, x) ≡ (c , x ) . Also (d, x) ≡ (d , x ). If x ∈ c, d, then by (i), (x, y) ≡ (x , y ). If x ∈ / c, d with (c, d, x) ≡ (c , d , x ) it follows by (W2) (x, y) ≡ (x , y ). For the proof that the restriction of a line reflection to a plane is a motion, we have to consider the closure of three points. We define for a subset X ⊂ P x, y, [X]1 := X and [X]n+1 := [[X]n ] for n ∈ N. [X] := x,y∈X
Clearly [X]n ⊂ X and since
n∈N [X]n
is a linear space, X =
n∈N [X]n .
Theorem 2.5. Let L = u, v be a line and let w, z be points and w = L(w), z = L(z). If (z, w) ≡ (z , w ) and ∼
U := u, v, w, z, then L |U is a motion. Proof. With u = u and v = v , by definition (u, v, w, z) ≡ (u , v , w , z ). We prove the theorem by induction. For x, y ∈ [u, v, w, z]1 = {u, v, w, z} clearly (x, y) ≡ (x , y ). Assume this property for [u, v, w, z]n and let x, y ∈ [u, v, w, z]n+1 . Then a, b, c, d ∈ [u, v, w, z]n exist with x ∈ a, b and y ∈ c, d. By assumption a, b, c, d satisfy the conditions of Lemma 2.4(ii) and we have (x, y) ≡ (x , y ). For X = {u, v, w, z} we have U = X = n∈N [X]n and it follows (x, y) ≡ (x , y ) for any x, y ∈ U . ∼
Theorem 2.6. Let L be a line of a plane E. Then L |E is an involutory motion with z = L(z) if and only if z ∈ L.
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Proof. Clearly L is involutory and z = L(z) only for z ∈ L. Let u, v, w be points with L = u, v and E = u, v, w. We set z = w and Theorem 2.5 implies the assumption. Lemma 2.7. Let L be a line of a plane E and let |E be a motion of E. ∼
(i) Let a, b ∈ L be distinct points with (a) = a, (b) = b. Then |E = id|E or |E = L |E . ∼ E = L −1 |E . (ii) (L)| Proof. (i) For any z ∈ L, (a, b, z) ≡ (a, b, (z)), hence (z) = z by (W1) and Lemma 2.1. For any p ∈ E with p = (p), we have (p) = L(p) by (W3). If there exists a point p ∈ E with p = (p) = L(p), then for any q ∈ E with (q) = q, it follows (p, q) ≡ (L(p), q) and q ∈ L by Lemma 2.3. (ii) is a consequence of (i). We define for lines A, B ∈ L: ∼
A ⊥ B ⇐⇒ A(B) = B
and
A = B.
Lemma 2.8. (i) If A ⊥ B, then A, B are coplanar.
∼ ∼
(ii) For lines A, B of a plane E we have A ⊥ B if and only if A B |E is an involutory motion. (iii) If A ⊥ B, then B ⊥ A. (iv) Let A ⊥ B and let A∪B be a motion, then (A) ⊥ (B). ∼ ∼ ∼∼ Proof. (ii) follows by Lemma 2.7(ii) since A B A |E = A(B)|E . (iii) is a consequence of (ii), and (iv) follows by (ii) and Lemma 2.7(ii).
Lemma 2.9. (i) For distinct points b, b of a plane E, there exists at most one line L in E with L(b) = b . (ii) Let a, b, b be non-collinear points with (a, b) ≡ (a, b ). Then there exists exactly one line L ⊂ a, b, b through a with b = L(b). (iii) For distinct lines L, G of a plane E, L(b) = G(b) implies b = L ∩ G. Proof. (i) follows by Lemma 2.3. (ii) For G := b, b , a := G(a) and L := a, a we have L ⊥ G and since (a, b, a ) ≡ (a, b , a ) also L(b) = b . (iii) Since G = L, (i) implies b = L(b) = G(b), hence b = L ∩ G. By [8]: Theorem 2.10. Let A, B, C be lines of a plane E through a point a. Then there exists a line D ⊂ E through a with ∼ ∼ ∼
∼
A B C |E = D |E (cf. [8] (1.15)). Theorem 2.11. For a plane E and distinct points a, m, c ∈ E with (a, m) ≡ (c, m), there exists a unique line X ⊂ E with m ∈ X and X(a) = c, i.e., X = ME (a, c) (cf. [8] (1.7.2), (1.8), (4.2)). We call X the midline of a, c in E. Lemma 2.12. The following propositions are equivalent (cf. [8] (4.3)): (a) (b) (c) (d)
There are three collinear points a, b, b with (a, b) ≡ (a, b ). For any points x, y there is a point y ∈ x, y\{y} with (x, y) ≡ (x, y ). There are two perpendicular lines A, B with A ∩ B = ∅. For a line L of any plane E and x ∈ L, there exists a line G ⊂ E through x with L ⊥ G.
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It is an open problem for a plane with congruence, if for x ∈ L there exists only one perpendicular line G ⊥ L through x (but for spaces see Corollary 3.5(i)). Theorem 2.13. Assume that for any two perpendicular lines A, B we have A ∩ B = ∅. Then for distinct lines G, H, L of a plane E with L ⊥ G, H we have G ⊥ H . ∼
∼
Proof. Let a ∈ L. Since G |E , H |E , are motions by Theorem 2.6, (G(a), H (a)) ≡ (GH G(a), G(a)). Because L ⊥ G, H , clearly a, H (a), GH G(a) ∈ L, and by assumption and Lemma 2.12, there exits only one point x ∈ L with (x, G(a)) ≡ (H (a), G(a)), i.e., GH G(a) = H (a). ∼ ∼ ∼ ∼ ∼ Now by Lemma 2.7(ii), G H G |E = G(H )|E , and since a ∈ / H , G(H ) = H is by Lemma 2.9(i) the unique midline of a and H (a) in E. Hence G ⊥ H . 3. Three-dimensional spaces with congruence In this section let (P , L, ≡) be a linear space with congruence and dim P 3 satisfying the exchange condition. Let A ∈ L be any line. For any two points x, y ∈ P there is a three-dimensional subspace U of P with A ∪ {x, y} ⊂ U . ∼
∼
To show that A is a motion it suffices to show that the restriction of A to U is a motion. Therefore in the following we consider a three-dimensional subspace U of P and a line A ⊂ U . Remarks. We recall that if there are two points in P with a midpoint, then by Lemma 2.12 for any distinct a, m ∈ A there exists a point c ∈ A with (a, m) ≡ (c, m). Further we remark that for affine spaces (P , L) with dim P 3 one can show that perpendicular lines have a nonempty intersection. With Lemma 2.12 it follows that for affine spaces (P , L) with congruence and with dim P 3 on every line there exist points a, c with a midpoint. By [8, (2.3)] finite planes with congruence are affine planes. Further finite affine spaces with congruence have dimension 2. Therefore finite spaces with congruence are affine planes. Lemma 3.1. Let a, c ∈ U be distinct points, then dim MU (a, c) 2. If a, c have a midpoint m ∈ a, c, then MU (a, c) is a plane of U. Proof. Since a, c ∈ / MU (a, c), dim MU (a, c) 2. If m is the midpoint of a, c, then by Theorem 2.11 in every plane E of U containing a, c there exists a line L ⊂ MU (a, c), hence dim MU (a, c) = 2. Lemma 3.2. Let A, B ⊂ U be perpendicular lines, let a ∈ A\B and b ∈ B\A. If the point m = A ∩ B exists, then X := MU (a, B(a)) ∩ MU (b, A(b)) is a line with m ∈ X. Proof. The point m is the midpoint of a, B(a) and b, A(b). By Lemma 3.1, MU (b, A(b)) is a plane with A ⊂ MU (b, A(b)) and by Theorem 2.11 there is a line X ⊂ MU (b, A(b)) with m ∈ X and X(a) = B(a), hence X = MU (a, B(a)) ∩ MU (b, A(b)). Lemma 3.3. Let A, B ⊂ U be lines with A ⊥ B and for a ∈ A\B and b ∈ B\A let X := MU (a, B(a))∩MU (b, A(b)) be a line. Then: (i) For distinct points u, v ∈ X the set {a, b, u, v} is a basis of U. (ii) It holds A(b) = X(b) and B(a) = X(a). ∼
(iii) X |U is a motion. ∼ (iv) A |U is a motion if and only if A(x) = B(x) for any x ∈ X\(A ∪ B). Proof. (i) Since b ∈ / MU (b, A(b)) we have b ∈ / X and B, X ⊂ MU (a, B(a)) implies that MU (a, B(a))=B ∪ X=b, u, v is a plane. Now a ∈ / MU (a, B(a)) shows that a, b, u, v are independent and form in the exchange space U a basis of U. (ii) Because X=MU (a, B(a))∩MU (b, A(b)), MU (a, B(a))=B ∪ X and MU (b, A(b))=A ∪ X we have A(b)=X(b) and B(a) = X(a).
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(iii) Since A, B are motions of A ∪ B, it follows (a, b) ≡ (a, A(b)) ≡ (B(a), A(b)) ≡ (X(a), X(b)). Now for the ∼
basis {u, v, a, b} of U we have (a, b) ≡ (X(a), X(b)) and Theorem 2.5 implies that X |U is a motion.
∼
∼
(iv) If A(x) = B(x), then A = MU (x, B(x)) ∩ MU (b, X(b)) and A |U is a motion by (ii). On the other hand, if A |U is a motion, then (A(b), A(x)) ≡ (b, x) ≡ (X(b), x) = (A(b), x) ≡ (b, A(x)). Hence b, A(b) ∈ MU (x, A(x)) and this implies B = b, A(b) ⊂ MU (x, A(x)), so A(x) = B(x). Theorem 3.4. Let (P , L, ≡) be a space with congruence and dim P 3. If there are two points in P with a midpoint, then any line reflection is a motion. ∼
Proof. Let A ∈ L, x, y ∈ P , and U := A ∪ {x, y}. As mentioned before, it suffices to show that A |U is a motion. By Lemma 2.12 there are distinct points a, m, c ∈ A with (a, m) ≡ (c, m). Let B ⊂ U be a midline of a, c, i.e. B ⊥ A and A ∩ B = m (cf. Theorem 2.11). For b ∈ B\A by Lemma 3.2 the line X := MU (a, B(a))∩MU (b, A(b)) exists with m ∈ X. Now let x ∈ X\{m}. Since ∼
X ⊥ A, B, by Lemma 3.2 the line B = MU (a, X(a)) ∩ MU (x, A(x)) exists with m ∈ B . By Lemma 3.3(iii), X |U and ∼
B |U are motions. Hence for b ∈ B \{m}, by Lemma 3.3(iv), A(b ) = X(b ), i.e., A = MU (b , X(b )) ∩ MU (x, B (x)) ∼
and by Lemma 3.3(iii), A |U is a motion.
Corollary 3.5. Let A, B ∈ L be lines in the three-dimensional subspace U with A ⊥ B and m = A ∩ B. Then: (i) B is the unique line through m in A ∪ B with A ⊥ B. (ii) For a ∈ A\{m} there exists at most one point b ∈ A\{a} with (a, m) ≡ (b, m). (iii) There exists only one line X ⊂ U with X ⊥ A, B and it holds X ⊥ C for any line C ⊂ A ∪ B with m ∈ C. Proof. (i) Let a ∈ A\B, b ∈ B\A and X := MU (a, B(a)) ∩ MU (b, A(b)) (cf. 3.2). For x ∈ X\(A ∪ B), by Lemma 3.3 and Theorem 3.4, A(x) = B(x) and A ∪ B = MU (x, A(x)). Let B ⊂ MU (x, A(x)) through m with B ⊥ A, i.e., ∼
B (x) = A(x) and B ⊥ A, X. Since B is a motion, B (a) = X(a) = B(a) and B = B by Theorem 2.11. (ii) Assume b , b ∈ A\{a} with (b, m) ≡ (b , m) ≡ (a, m) and let L, L ⊂ A ∪ B be the lines through m with L(a) = b, L (a) = b (cf. Theorem 2.11). Now (i) shows L = L and b = b . (iii) Let Y ⊥ A, B. Then Y ⊂ / A ∪ B (else Y has two perpendicular lines through m in A ∪ B), hence m ∈ A ∪ Y ∩B ∪ Y =Y and for a ∈ A\B, b ∈ B\A by (ii) it follows A(b)=Y (b) and B(a)=Y (a), i.e., Y =MU (a, B(a))∩ MU (b, A(b)) = X is uniquely determined. Clearly A ∪ B = MU (x, A(x)) for any point x ∈ X\m. Therfore for any ∼
line C ⊂ A ∪ B through m we have C (X) = X, hence C ⊥ X.
Lemma 3.6. Let z, a, b ∈ U be distinct points, let a ∈ z, a\{a} and b ∈ z, b\{b} with (z, a) ≡ (z, a ) and (z, b) ≡ (z, b ), then (a, b) ≡ (a , b ). Proof. Let H := a, z and E be a plane with a, b, z ∈ E. By Lemma 2.12 there is a line G ⊂ E with z ∈ G and ∼
G ⊥ H . For p ∈ G\{z}, by Lemma 3.2 the line X = MU (a, a ) ∩ MU (p, H (p)) exists, and by Lemma 3.3(iii), X is a motion. Since a, z, b, z ⊂ E, by Corollary 3.5(iii) it follows X ⊥ b, z, a, z, and by Corollary 3.5(ii) X(a) = a and X(b) = b . Hence we have (a, b) ≡ (X(a), X(b)) = (a , b ). In the following we assume, that there are two perpendicular lines with a non-empty intersection. Then by Lemma 2.12 and Corollary 3.5 for any two points a, b there is a unique point b ∈ a, b\{b} with (a, b) ≡ (a, b ). Now we consider point reflections. For distinct points a, x ∈ P we denote with a(x) the unique point a(x) ∈ a, x\{x} with (a, x) ≡ (a, a(x)). We call the following mapping point reflection: x if x = a, a : P → P; x → a(x) if x = a. Now Lemma 3.6 implies:
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Theorem 3.7. Every point reflection a is an involutory motion with x = a (x) if and only if x = a. 4. Addition of points In this section we assume that any two distinct points a, a have a midpoint m ∈ a, a with (a, m) ≡ (a , m). For a fixed point 0 ∈ P we denote for any point a ∈ P \{0} the unique midpoint of 0 and a by a/2. We denote 0 = 0/2. Then for the point reflection corresponding to a/2 we have a/2(0) =a
and
a/2(a) = 0.
We define for points a, b the addition + on the point set P by (cf. [1,4]). ◦ a + b := a/2 0(b) = a/2 0(b). Theorem 4.1 (Karzel [1], Konrad [3], Kreuzer [4]). Let (P , L, ≡) be a space with congruence of dim P 3 such that any two distinct points have a midpoint. Then fixing a point 0 ∈ P , an addition + can be defined on P such that (P , +) is a Bruck loop (for a definition cf. [7]) with the neutral element 0. The point −a := 0(a) is the inverse of a ∈ P . (P , +) is associative if and only if for three points a, b, c ∈ P the product a b c is a point reflection, too. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
H. Karzel, Recent developments on absolute geometries and algebraization by K-loops, Discrete Math. 208/209 (1999) 387–409. H. Karzel, K. Sörensen, D. Windelberg, Einführung in die Geometrie, UTB Vandenhoeck, Göttingen, 1973. A. Konrad, Nichteuklidische Geometrie und K-loops, Ph.D. Thesis, Technische Universität München, 1995. A. Kreuzer, Reflection loops and linear spaces with hyperbolic incidence structure, Comment Math. Univ. Carolinae 45 (2004) 303–320. H.-J. Kroll, K. Sörensen, Pseudo-Euklidische Ebenen und Euklidische Räume, J. Geom. 8 (1976) 95–115. H.-J. Kroll, K. Sörensen, Hyperbolische Räume, J. Geom. 64 (1998) 141–149. H.O. Pflugfelder, Quasigroups and Loops: Introduction, Heldermann, Berlin, 1990. K. Sörensen, Ebenen mit Kongruenz, J. Geom. 22 (1984) 15–30. K. Sörensen, Eine Bemerkung über absolute Ebenen, J. Geom. 64 (1999) 160–166.
Spaces with congruence Alexander Kreuzer, Kay Sörensen Technische Universität München, Zentrum Mathematik, Boltzmannstr. 3, 85747 München, Germany Received 11 October 2004; received in revised form 30 November 2005; accepted 27 November 2006 Available online 2 June 2007
Abstract In this paper we consider an exchange space with congruence (P , L, ≡) not assuming any further geometrical properties. If the dimension of the space is greater than 2, we show that for any line G of a plane E and any point x ∈ G there is a unique perpendicular line through x in E and that any line reflection is a motion. © 2007 Elsevier B.V. All rights reserved. Keywords: Congruence; Motions; Line reflections
1. Introduction It is well known that every Euclidean plane (E, L, , ≡) [2,9] is isomorphic to an affine plane AG(2, K) over an ordered and pythagorean commutative field K (cf. [2] (24.5)). Here (E, L) denotes an affine plane, (E, L, ) an ordered plane, and ≡ denotes the congruence relation on E × E. We may consider E as a quadratic separable field extension of K with the corresponding involutory field automorphism ¯ : E → E and we have (a, b) ≡ (c, d) if and only if (a − b)(a − b) = (c − d)(c − d) for any points a, b, c, d ∈ E. There is a corresponding theorem for hyperbolic planes (cf. [2]). For both proofs one first considers the group of motions, in particular the line reflections. For the definition of a motion and a line reflection the order relation is not necessary. We need only the linear structure of (E, L) and the congruence relation ≡, but for the proof that the line reflection is a motion it seems that additional assumptions on the geometry are necessary. For example Sörensen assumes in [8] that for given lines G1 , G2 , there exist distinct lines H1 , H2 through a common point z which intersect G1 , G2 . Here we give a proof for exchange planes with congruence not using this property. In a plane with congruence (E, L, ≡) two perpendicular lines may have an empty intersection. We show that if in a plane any two perpendicular lines have an empty intersection, then the relation “perpendicular” together with the “identity relation” is transitive (cf. Theorem 2.13). If there are two perpendicular lines with a non-empty intersection, one can show that for any line G and any point x ∈ G we have a perpendicular line through x. But it is open if there is a unique perpendicular line to G through x. This is true for finite or affine planes, but not known in general. In this paper we consider a space with congruence (P , L, ≡) not assuming any further geometrical properties. (For special spaces with congruence cf. [5,6].) We only assume that the space satisfies the exchange property, which for example can be easily shown for ordered spaces. In Section 3 we assume that there are two points with a midpoint. We use the property that if two points have a midpoint, E-mail addresses: [email protected] (A. Kreuzer), [email protected] (K. Sörensen). 0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2006.11.034
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then they have a midline in every plane. If the dimension of the space is greater than 2, we show that for any line G of a plane E and any point x ∈ G there is a unique perpendicular line through x in E and that any line reflection is a motion. It follows that for any two points b, z there exists a unique point b ∈ z, b\{b} with (z, b) ≡ (z, b ) and also point reflections are motions. For the case that there exist no points with a midpoint, we conjecture that all examples in which the line reflections are motions have dimension 2. 2. Spaces with congruence Let (P , L) denote a linear space or incidence space with the point set P, the line set L and at least three points on every line, i.e., • for any two points there is exactly one line containing them and • for any line L ∈ L we have |L|3. A subspace is a subset U ⊂ P such that for all distinct points x, y ∈ U the unique line passing through x and y, denoted by x, y, is contained in U. Let U denote the set of all subspaces. For every subset X ⊂ P we define the following closure operation U. (1) : P(P ) → U; X → X by X := U ∈U X⊂U
For U ∈ U we call dim U := inf{|X| − 1 : X ⊂ U and X = U } the dimension of U. A subspace of dimension two is a plane. For a set {a, b, c, . . .} we write a, b, c, . . . instead of {a, b, c, . . .}. We introduce the concept of a space (P , L, ≡) with congruence (cf. [8]). We assume that (P , L) is a linear space which satisfies the following exchange condition. (EC) Let S ⊂ P and let x, y ∈ P with x ∈ S ∪ {y}\S. Then y ∈ S ∪ {x} Let ≡ be a congruence relation on P × P , i.e., ≡ is an equivalence relation with (a, b) ≡ (b, a), (a, a) ≡ (b, b) and (a, a) ≡ (b, c) implies b = c. We use the notation (x1 , x2 , x3 ) ≡ (y1 , y2 , y3 ) if and only if (xi , xj ) ≡ (yi , yj ) for i, j ∈ {1, 2, 3}. (P , L, ≡) is a space with congruence if the axioms (W1), (W2) and (W3) are satisfied. (W1) Let a, b, c ∈ P be distinct and collinear, and let a , b ∈ P with (a, b) ≡ (a , b ). Then there exists exactly one c ∈ a , b with (a, b, c) ≡ (a , b , c ). (W2) Let a, b, x ∈ P be non-collinear and let a , b , x ∈ P with (a, b, x) ≡ (a , b , x ). For any c ∈ a, b and c ∈ a , b with (a, b, c) ≡ (a , b , c ) it holds (x, c) ≡ (x , c ). (W3) For a, b, x ∈ P non-collinear there exists exactly one x ∈ a, b, x\{x} with (a, b, x) ≡ (a, b, x ). We call a bijective mapping : P → P a motion, if (x, y) ≡ ((x), (y)) for all x, y ∈ P . Lemma 2.1. (i) If a, b, c are collinear points and a , b , c ∈ P with (a, b, c) ≡ (a , b , c ), then a , b , c are collinear. (ii) Any motion is a collineation. Proof. (i) By (W1) the point c ∈ a , b exists with (a , b , c ) ≡ (a, b, c). If c ∈ / a , b , then by (W2) it would follow (c , c ) ≡ (c, c), hence c = c ∈ a , b . (ii) By (i), and −1 map collinear points onto collinear points. For a subspace U and points a, b ∈ U we define MU (a, b) := {x ∈ U : (a, x) ≡ (b, x)}. We call MU (a, b) a midpoint, a midline, or a midplane of a, b, respectively, if it is a point, a line, or a plane, respectively. Lemma 2.2. (i) Two distinct points a, b have at most one point m ∈ a, b with (a, m) ≡ (b, m). (ii) MU (a, b) is a subspace of U.
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Proof. (i) For m, m ∈ a, b we have (a, m, m ) ≡ (b, m, m ), and (W1) implies m = m . (ii) Assume a = b and let x, y ∈ MU (a, b) be distinct points, hence (a, x, y) ≡ (b, x, y). By (W1) the points a, b, x, y are not collinear. For w ∈ x, y by (W2) it follows (a, w) ≡ (b, w), i.e., w ∈ MU (a, b). Using that every plane satisfies the exchange condition, we can show the following important lemma for a space with congruence: Lemma 2.3. Let a, b, x ∈ P be non-collinear points and let x ∈ a, b, x\{x} with (a, b, x) ≡ (a, b, x ). Then for any c ∈ a, b, x it holds (x, c) ≡ (x , c) if and only if c ∈ a, b. / ME (x, x ) it follows ME (x, x ) = E. By Lemma 2.2(ii), Proof. For E := a, b, x we have a, b ∈ ME (x, x ). Since x ∈ ME (x, x ) is a subspace of E, and since E is an exchange plane and a, b ∈ ME (x, x ), it follows ME (x, x ) = a, b. For a line L ∈ L, x ∈ P \L and a, b ∈ L with a = b there exists by (W3) the unique point x ∈ L ∪ {x}\{x} with (a, b, x) ≡ (a, b, x ). By Lemma 2.3, x is independent of the choice of a, b ∈ L, hence we may denote x = L(x). We call the following mapping line reflection ∼ x if x ∈ L, : P → P ; x → L L(x) if x ∈ / L. ∼
Clearly L is an involutory bijection with z = L(z) if and only if z ∈ L. We remark that by the notations above we ∼
can write L(x) or L(x) as well. We prefer L(x). Lemma 2.4. Let L be a line of a plane E and let p := L(p) for p ∈ E. (i) If (p, q) ≡ (p , q ) for p, q ∈ E, then (p, q, x) ≡ (p , q , x ) for every x ∈ p, q. (ii) Let a, b, c, d ∈ E be points with (p, q) ≡ (p , q ) for all p, q ∈ {a, b, c, d}. Then (x, y) ≡ (x , y ) for x ∈ a, b and y ∈ c, d. Proof. (i) By (W1) there exists a point x1 ∈ p , q with (p, q, x) ≡ (p , q , x1 ). If x = x1 , then by Lemma 2.3 x ∈ L. For any point u ∈ L , (p, q, u) ≡ (p , q , u), hence by (W2) (u, x) ≡ (u, x1 ) and therefore x1 = x . (ii) By (i) we have (a, b, x) ≡ (a , b , x ) and (c, d, y) ≡ (c , d , y ). If c ∈ a, b, then (c, x) ≡ (c , x ) by (i). For c∈ / a, b, the assumptions give (a, b, c) ≡ (a , b , c ) and (W2) implies (c, x) ≡ (c , x ) . Also (d, x) ≡ (d , x ). If x ∈ c, d, then by (i), (x, y) ≡ (x , y ). If x ∈ / c, d with (c, d, x) ≡ (c , d , x ) it follows by (W2) (x, y) ≡ (x , y ). For the proof that the restriction of a line reflection to a plane is a motion, we have to consider the closure of three points. We define for a subset X ⊂ P x, y, [X]1 := X and [X]n+1 := [[X]n ] for n ∈ N. [X] := x,y∈X
Clearly [X]n ⊂ X and since
n∈N [X]n
is a linear space, X =
n∈N [X]n .
Theorem 2.5. Let L = u, v be a line and let w, z be points and w = L(w), z = L(z). If (z, w) ≡ (z , w ) and ∼
U := u, v, w, z, then L |U is a motion. Proof. With u = u and v = v , by definition (u, v, w, z) ≡ (u , v , w , z ). We prove the theorem by induction. For x, y ∈ [u, v, w, z]1 = {u, v, w, z} clearly (x, y) ≡ (x , y ). Assume this property for [u, v, w, z]n and let x, y ∈ [u, v, w, z]n+1 . Then a, b, c, d ∈ [u, v, w, z]n exist with x ∈ a, b and y ∈ c, d. By assumption a, b, c, d satisfy the conditions of Lemma 2.4(ii) and we have (x, y) ≡ (x , y ). For X = {u, v, w, z} we have U = X = n∈N [X]n and it follows (x, y) ≡ (x , y ) for any x, y ∈ U . ∼
Theorem 2.6. Let L be a line of a plane E. Then L |E is an involutory motion with z = L(z) if and only if z ∈ L.
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Proof. Clearly L is involutory and z = L(z) only for z ∈ L. Let u, v, w be points with L = u, v and E = u, v, w. We set z = w and Theorem 2.5 implies the assumption. Lemma 2.7. Let L be a line of a plane E and let |E be a motion of E. ∼
(i) Let a, b ∈ L be distinct points with (a) = a, (b) = b. Then |E = id|E or |E = L |E . ∼ E = L −1 |E . (ii) (L)| Proof. (i) For any z ∈ L, (a, b, z) ≡ (a, b, (z)), hence (z) = z by (W1) and Lemma 2.1. For any p ∈ E with p = (p), we have (p) = L(p) by (W3). If there exists a point p ∈ E with p = (p) = L(p), then for any q ∈ E with (q) = q, it follows (p, q) ≡ (L(p), q) and q ∈ L by Lemma 2.3. (ii) is a consequence of (i). We define for lines A, B ∈ L: ∼
A ⊥ B ⇐⇒ A(B) = B
and
A = B.
Lemma 2.8. (i) If A ⊥ B, then A, B are coplanar.
∼ ∼
(ii) For lines A, B of a plane E we have A ⊥ B if and only if A B |E is an involutory motion. (iii) If A ⊥ B, then B ⊥ A. (iv) Let A ⊥ B and let A∪B be a motion, then (A) ⊥ (B). ∼ ∼ ∼∼ Proof. (ii) follows by Lemma 2.7(ii) since A B A |E = A(B)|E . (iii) is a consequence of (ii), and (iv) follows by (ii) and Lemma 2.7(ii).
Lemma 2.9. (i) For distinct points b, b of a plane E, there exists at most one line L in E with L(b) = b . (ii) Let a, b, b be non-collinear points with (a, b) ≡ (a, b ). Then there exists exactly one line L ⊂ a, b, b through a with b = L(b). (iii) For distinct lines L, G of a plane E, L(b) = G(b) implies b = L ∩ G. Proof. (i) follows by Lemma 2.3. (ii) For G := b, b , a := G(a) and L := a, a we have L ⊥ G and since (a, b, a ) ≡ (a, b , a ) also L(b) = b . (iii) Since G = L, (i) implies b = L(b) = G(b), hence b = L ∩ G. By [8]: Theorem 2.10. Let A, B, C be lines of a plane E through a point a. Then there exists a line D ⊂ E through a with ∼ ∼ ∼
∼
A B C |E = D |E (cf. [8] (1.15)). Theorem 2.11. For a plane E and distinct points a, m, c ∈ E with (a, m) ≡ (c, m), there exists a unique line X ⊂ E with m ∈ X and X(a) = c, i.e., X = ME (a, c) (cf. [8] (1.7.2), (1.8), (4.2)). We call X the midline of a, c in E. Lemma 2.12. The following propositions are equivalent (cf. [8] (4.3)): (a) (b) (c) (d)
There are three collinear points a, b, b with (a, b) ≡ (a, b ). For any points x, y there is a point y ∈ x, y\{y} with (x, y) ≡ (x, y ). There are two perpendicular lines A, B with A ∩ B = ∅. For a line L of any plane E and x ∈ L, there exists a line G ⊂ E through x with L ⊥ G.
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It is an open problem for a plane with congruence, if for x ∈ L there exists only one perpendicular line G ⊥ L through x (but for spaces see Corollary 3.5(i)). Theorem 2.13. Assume that for any two perpendicular lines A, B we have A ∩ B = ∅. Then for distinct lines G, H, L of a plane E with L ⊥ G, H we have G ⊥ H . ∼
∼
Proof. Let a ∈ L. Since G |E , H |E , are motions by Theorem 2.6, (G(a), H (a)) ≡ (GH G(a), G(a)). Because L ⊥ G, H , clearly a, H (a), GH G(a) ∈ L, and by assumption and Lemma 2.12, there exits only one point x ∈ L with (x, G(a)) ≡ (H (a), G(a)), i.e., GH G(a) = H (a). ∼ ∼ ∼ ∼ ∼ Now by Lemma 2.7(ii), G H G |E = G(H )|E , and since a ∈ / H , G(H ) = H is by Lemma 2.9(i) the unique midline of a and H (a) in E. Hence G ⊥ H . 3. Three-dimensional spaces with congruence In this section let (P , L, ≡) be a linear space with congruence and dim P 3 satisfying the exchange condition. Let A ∈ L be any line. For any two points x, y ∈ P there is a three-dimensional subspace U of P with A ∪ {x, y} ⊂ U . ∼
∼
To show that A is a motion it suffices to show that the restriction of A to U is a motion. Therefore in the following we consider a three-dimensional subspace U of P and a line A ⊂ U . Remarks. We recall that if there are two points in P with a midpoint, then by Lemma 2.12 for any distinct a, m ∈ A there exists a point c ∈ A with (a, m) ≡ (c, m). Further we remark that for affine spaces (P , L) with dim P 3 one can show that perpendicular lines have a nonempty intersection. With Lemma 2.12 it follows that for affine spaces (P , L) with congruence and with dim P 3 on every line there exist points a, c with a midpoint. By [8, (2.3)] finite planes with congruence are affine planes. Further finite affine spaces with congruence have dimension 2. Therefore finite spaces with congruence are affine planes. Lemma 3.1. Let a, c ∈ U be distinct points, then dim MU (a, c) 2. If a, c have a midpoint m ∈ a, c, then MU (a, c) is a plane of U. Proof. Since a, c ∈ / MU (a, c), dim MU (a, c) 2. If m is the midpoint of a, c, then by Theorem 2.11 in every plane E of U containing a, c there exists a line L ⊂ MU (a, c), hence dim MU (a, c) = 2. Lemma 3.2. Let A, B ⊂ U be perpendicular lines, let a ∈ A\B and b ∈ B\A. If the point m = A ∩ B exists, then X := MU (a, B(a)) ∩ MU (b, A(b)) is a line with m ∈ X. Proof. The point m is the midpoint of a, B(a) and b, A(b). By Lemma 3.1, MU (b, A(b)) is a plane with A ⊂ MU (b, A(b)) and by Theorem 2.11 there is a line X ⊂ MU (b, A(b)) with m ∈ X and X(a) = B(a), hence X = MU (a, B(a)) ∩ MU (b, A(b)). Lemma 3.3. Let A, B ⊂ U be lines with A ⊥ B and for a ∈ A\B and b ∈ B\A let X := MU (a, B(a))∩MU (b, A(b)) be a line. Then: (i) For distinct points u, v ∈ X the set {a, b, u, v} is a basis of U. (ii) It holds A(b) = X(b) and B(a) = X(a). ∼
(iii) X |U is a motion. ∼ (iv) A |U is a motion if and only if A(x) = B(x) for any x ∈ X\(A ∪ B). Proof. (i) Since b ∈ / MU (b, A(b)) we have b ∈ / X and B, X ⊂ MU (a, B(a)) implies that MU (a, B(a))=B ∪ X=b, u, v is a plane. Now a ∈ / MU (a, B(a)) shows that a, b, u, v are independent and form in the exchange space U a basis of U. (ii) Because X=MU (a, B(a))∩MU (b, A(b)), MU (a, B(a))=B ∪ X and MU (b, A(b))=A ∪ X we have A(b)=X(b) and B(a) = X(a).
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(iii) Since A, B are motions of A ∪ B, it follows (a, b) ≡ (a, A(b)) ≡ (B(a), A(b)) ≡ (X(a), X(b)). Now for the ∼
basis {u, v, a, b} of U we have (a, b) ≡ (X(a), X(b)) and Theorem 2.5 implies that X |U is a motion.
∼
∼
(iv) If A(x) = B(x), then A = MU (x, B(x)) ∩ MU (b, X(b)) and A |U is a motion by (ii). On the other hand, if A |U is a motion, then (A(b), A(x)) ≡ (b, x) ≡ (X(b), x) = (A(b), x) ≡ (b, A(x)). Hence b, A(b) ∈ MU (x, A(x)) and this implies B = b, A(b) ⊂ MU (x, A(x)), so A(x) = B(x). Theorem 3.4. Let (P , L, ≡) be a space with congruence and dim P 3. If there are two points in P with a midpoint, then any line reflection is a motion. ∼
Proof. Let A ∈ L, x, y ∈ P , and U := A ∪ {x, y}. As mentioned before, it suffices to show that A |U is a motion. By Lemma 2.12 there are distinct points a, m, c ∈ A with (a, m) ≡ (c, m). Let B ⊂ U be a midline of a, c, i.e. B ⊥ A and A ∩ B = m (cf. Theorem 2.11). For b ∈ B\A by Lemma 3.2 the line X := MU (a, B(a))∩MU (b, A(b)) exists with m ∈ X. Now let x ∈ X\{m}. Since ∼
X ⊥ A, B, by Lemma 3.2 the line B = MU (a, X(a)) ∩ MU (x, A(x)) exists with m ∈ B . By Lemma 3.3(iii), X |U and ∼
B |U are motions. Hence for b ∈ B \{m}, by Lemma 3.3(iv), A(b ) = X(b ), i.e., A = MU (b , X(b )) ∩ MU (x, B (x)) ∼
and by Lemma 3.3(iii), A |U is a motion.
Corollary 3.5. Let A, B ∈ L be lines in the three-dimensional subspace U with A ⊥ B and m = A ∩ B. Then: (i) B is the unique line through m in A ∪ B with A ⊥ B. (ii) For a ∈ A\{m} there exists at most one point b ∈ A\{a} with (a, m) ≡ (b, m). (iii) There exists only one line X ⊂ U with X ⊥ A, B and it holds X ⊥ C for any line C ⊂ A ∪ B with m ∈ C. Proof. (i) Let a ∈ A\B, b ∈ B\A and X := MU (a, B(a)) ∩ MU (b, A(b)) (cf. 3.2). For x ∈ X\(A ∪ B), by Lemma 3.3 and Theorem 3.4, A(x) = B(x) and A ∪ B = MU (x, A(x)). Let B ⊂ MU (x, A(x)) through m with B ⊥ A, i.e., ∼
B (x) = A(x) and B ⊥ A, X. Since B is a motion, B (a) = X(a) = B(a) and B = B by Theorem 2.11. (ii) Assume b , b ∈ A\{a} with (b, m) ≡ (b , m) ≡ (a, m) and let L, L ⊂ A ∪ B be the lines through m with L(a) = b, L (a) = b (cf. Theorem 2.11). Now (i) shows L = L and b = b . (iii) Let Y ⊥ A, B. Then Y ⊂ / A ∪ B (else Y has two perpendicular lines through m in A ∪ B), hence m ∈ A ∪ Y ∩B ∪ Y =Y and for a ∈ A\B, b ∈ B\A by (ii) it follows A(b)=Y (b) and B(a)=Y (a), i.e., Y =MU (a, B(a))∩ MU (b, A(b)) = X is uniquely determined. Clearly A ∪ B = MU (x, A(x)) for any point x ∈ X\m. Therfore for any ∼
line C ⊂ A ∪ B through m we have C (X) = X, hence C ⊥ X.
Lemma 3.6. Let z, a, b ∈ U be distinct points, let a ∈ z, a\{a} and b ∈ z, b\{b} with (z, a) ≡ (z, a ) and (z, b) ≡ (z, b ), then (a, b) ≡ (a , b ). Proof. Let H := a, z and E be a plane with a, b, z ∈ E. By Lemma 2.12 there is a line G ⊂ E with z ∈ G and ∼
G ⊥ H . For p ∈ G\{z}, by Lemma 3.2 the line X = MU (a, a ) ∩ MU (p, H (p)) exists, and by Lemma 3.3(iii), X is a motion. Since a, z, b, z ⊂ E, by Corollary 3.5(iii) it follows X ⊥ b, z, a, z, and by Corollary 3.5(ii) X(a) = a and X(b) = b . Hence we have (a, b) ≡ (X(a), X(b)) = (a , b ). In the following we assume, that there are two perpendicular lines with a non-empty intersection. Then by Lemma 2.12 and Corollary 3.5 for any two points a, b there is a unique point b ∈ a, b\{b} with (a, b) ≡ (a, b ). Now we consider point reflections. For distinct points a, x ∈ P we denote with a(x) the unique point a(x) ∈ a, x\{x} with (a, x) ≡ (a, a(x)). We call the following mapping point reflection: x if x = a, a : P → P; x → a(x) if x = a. Now Lemma 3.6 implies:
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Theorem 3.7. Every point reflection a is an involutory motion with x = a (x) if and only if x = a. 4. Addition of points In this section we assume that any two distinct points a, a have a midpoint m ∈ a, a with (a, m) ≡ (a , m). For a fixed point 0 ∈ P we denote for any point a ∈ P \{0} the unique midpoint of 0 and a by a/2. We denote 0 = 0/2. Then for the point reflection corresponding to a/2 we have a/2(0) =a
and
a/2(a) = 0.
We define for points a, b the addition + on the point set P by (cf. [1,4]). ◦ a + b := a/2 0(b) = a/2 0(b). Theorem 4.1 (Karzel [1], Konrad [3], Kreuzer [4]). Let (P , L, ≡) be a space with congruence of dim P 3 such that any two distinct points have a midpoint. Then fixing a point 0 ∈ P , an addition + can be defined on P such that (P , +) is a Bruck loop (for a definition cf. [7]) with the neutral element 0. The point −a := 0(a) is the inverse of a ∈ P . (P , +) is associative if and only if for three points a, b, c ∈ P the product a b c is a point reflection, too. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
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