/^creative ^commor ARS MATHEMATICA CONTEMPORANEA ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 17 (2019) 447-454 https://doi.org/10.26493/1855-3974.1917.ea5 (Also available at http://amc-journal.eu) The complement of a subspace in a classical polar space Krzysztof Petelczyc , Mariusz Zynel Institute of Mathematics, University of Bialystok, Ciolkowskiego 1 M, 15-245 Bialystok, Poland Received 22 January 2019, accepted 6 July 2019, published online 7 November 2019 Abstract In a polar space, embeddable into a projective space, we fix a subspace, that is contained in some hyperplane. The complement of that subspace resembles a slit space or a semiaffine space. We prove that under some assumptions the ambient polar space can be recovered in this complement. Keywords: Polar space, projective space, semiaffine space, slit space, complement. Math. Subj. Class.: 51A15, 51A45 1 Introduction Cohen and Shult coined the term affine polar space in [4] as a polar space with some hyperplane removed. They prove that from such an affine reduct the ambient polar space can be recovered. In [9] we prove something similar for the complement of a subset in a projective space. Looking at the results of these two papers it is seen that an interesting case has been set aside: the complement of a subspace in a polar space. We are trying to fill this this gap here, although under several specific assumptions: we consider classical polar spaces, i.e. embeddable into projective spaces (cf. [2]), and our subspace is contained in a hyperplane. A projective space with some subspace removed is called a slit space (cf. [5, 6, 8]) so, our complement can be seen as a generalized slit space. Singular subspaces in a polar space are projective spaces, in an affine polar space they are affine spaces (cf. [4]), while in our complement they are semiaffine or projective spaces. Adopting the terminology of [7], where the class of semiaffine spaces includes affine spaces, projective spaces and everything in between, we could say that singular subspaces of our complement are simply E-mail addresses: kryzpet@math.uwb.edu.pl (Krzysztof Petelczyc), mariusz@math.uwb.edu.pl (Mariusz Zynel) ©® This work is licensed under https://creativecommons.org/licenses/by/4.0/ 447 Ars Math. Contemp. 17(2019)447-454 semiaffine spaces. This let us call our complement a semiaffine polar space. Anyway, it is clear that the complement we examine is affine in spirit. A natural parallelism is there and the subspace we remove can be viewed as the horizon. As this paper is closely related to [4] and [9], it borrows some concepts, notations and reasonings from these two works. There are however new difficulties in this case. In contrast to [4], the horizon is not a hyperplane and thus, it induces a partial parallelism (cf. [8]). There are lines disjoint with the horizon in the ambient space and those lines, called non-affine, cannot be parallel to any line in the complement. If we had applied the definition of parallelism from [4] as it is, we would end up with non-affine lines in its equivalence classes. Therefore we use the Veblen condition to express parallelism in terms of incidence in the complement. This method unfortunately is viable only if we have at least 4 points per line in the polar space. Roughly speaking, the points of the horizon are identified with equivalence classes of parallelism, or, in other words, with directions of lines. On the horizon of an affine polar space a deep point emerges as the point which could be reached by no line of the complement. If the removed subspace is not a hyperplane then there is no deep point but a new problem arises. Some lines on the horizon are recoverable in a standard way, as directions of planes. For the others there are no planes in the complement that would reach them. An analogy to a deep point is clear, so we call them deep lines. To overcome the problem we introduce the following relation: a line K is anti-euclidean to a line L iff there is no line intersecting K that is parallel to L. Based on this relation is a ternary collinearity of points on deep lines. We do not know whether every subspace of a polar space is contained in a hyperplane. Any subspace can be extended to a maximal one, but does it have to be a hyperplane? If that is the case our assumptions could be weakened significantly. 2 Generalities A point-line structure M = (S, L}, where the elements of S are called points, the elements of L are called lines, and where L c 2S, is said to be a partial linear space, or a point-line space, if two distinct lines share at most one point and every line is of size (cardinality) at least 2 (cf. [3]). A line of size 3 or more will be called thick. If all lines in M are thick then M is thick. M is said to be nondegenerate if no point is collinear with all others, and it is called singular if any two of its points are collinear. It is called Veblenian iff for any two distinct lines Li, L2 through a point p and any two distinct lines Ki, K2 not through the point p whenever each of L1, L2 intersects both of K1, K2, then K1 intersects K2. A subspace of M is a subset X C S that contains every line, which meets X in at least two points. A proper subspace of M that shares a point with every line is said to be a hyperplane. If M satisfies exchange axiom, then a plane of M is a singular subspace of dimension 2. A partial linear space satisfying one-or-all axiom, that is for every line L and a point a G L, a is collinear with one or all points on L, will be called a polar space. The rank of a polar space is the maximal number n for which there is a chain of singular subspaces 0 = X1 c X2 c ■ ■ ■ C Xn (n = -1 if this chain is reduced to the empty set). For a G S by a^ we denote the set of all points collinear with a, and for X C S we put X^ = Q{a^ : a G X}, radX = X n X K. Petelczyc and M. Zynel: The complement of a subspace in a classical polar space 449 As an immediate consequence of one-or-all axiom we get (cf. [4]): Fact 2.1. For any point a G S the set is a hyperplane of p. Following [10], a subset X of S is called • spiky when every point a G X is collinear with some point b G X, • flappy when for every line L C X there is a point a G X such that L C aL. 2.1 Complement Let M = (S, L} be a thick partial linear space and let W be a proper subspace of M. By the complement of W in M we mean the structure Dm(W) := (Sw, £w}, where Sw := S \W and Lw := {k n Sw : kG LA k £ W}. The subspace W will be called the horizon of (W). Note that the complement (W) is a partial linear space. Following a standard convention we call the points and lines of the complement DM(W) proper, and points and lines of W are said to be improper. Given a subspace X of DM(W) its closure X is a subspace of M with X C X. We say that two lines K, L G Lw are parallel, and we write K ||W L iff K n L nW = 0. (2.1) This is always an equivalence relation. Its domain is Lw only in case W is a hyperplane, or in other words, a line L G Lw with L = L cannot be parallel to any line. A line L G Lw with the property that L ||w L will be called an affine line. The set of all affine lines, the domain of ||w, will be denoted by L*. For affine line L we write for the point of L in W, i.e. the point at infinity. A point a G W is said to be a deep point if there is no line L G Lw such that a = LTO. A plane of DM(W) containing an affine line is said to be a semiaffine plane. By we denote the set of points at infinity of a semiaffine plane n, i.e. n~ = {M~ : M G L* and M C n}. Note that n~ is a line iff n is an affine plane. A line L C W is said to be a deep line if there is no plane in Dp(W) with L = nTO. 3 Complement in a polar space Let p = (S, L} be a thick, nondegenerate polar space of rank at least 3. In the remainder of the paper we deal with Dp(W), where W is a proper subspace contained in some hyperplane of p. Let us emphasize, that we do not mean one particular hyperplane and it is not fixed in our reasonings in any way. If there was a unique hyperplane H containing W we would be able to recover the ambient space applying Proposition 2.7 from [4], which says that every automorphism of the complement Dp(H) can be uniquely extended to an automorphism of p. It is not however doable as there could be many hyperplanes containing W and none of them can be distinguished in terms of the complement. In polar spaces deep points appear only on hyperplanes and there could be at most one deep point on a hyperplane. 450 Ars Math. Contemp. 17(2019)447-454 Lemma 3.1. (i) If W is a hyperplane, then there is at most one deep point on W and it is in rad W. (ii) If W is not a hyperplane, then there are no deep points on W, that is W is spiky. Proof. (i): By Corollary 1.3 (ii) in [4]. (ii): Assume that a is a deep point in W. Then aL C W, and by Fact 2.1 we get that W contains a hyperplane. A contradiction, as a hyperplane in P is a maximal proper subspace (cf. [4, Lemma 1.1]). □ Lemma 3.2. Let P be an embeddable polar space and K,L e Lw be two distinct lines such that K || w L. The subspace W can be extended to a hyperplane of P not containing K and L. Proof. If W is a hyperplane of P then W itself is the required hyperplane. Assume that W is not a hyperplane. Let H be a hyperplane containing W, N be a projective space embracing p, and f be an embedding of P into N. Consider the projective subspace G spanned by f (H). By Proposition 5.2 from [4] G is a hyperplane of N. If f (K), f (L) £ G then the hyperplane H = f-1(G n f (S)) is the required one. Let H be the family of all hyperplanes in G containing f (W). _Now, assume that f (K) C G and f (L) £ G. Take aK e f (K) \ f (W) and aL e f (L) \ f (W) and choose a hyperplane G0 e H with aK e G0. Note that aK, aL meets G in aK. Take b e aK ,aL distinct from aK and aL. Assume that there is a line through b that intersects f (L) \ f (W) in some point c and meets G0 in a point d. Note that d e f (K) as otherwise we would have aK e G0. Lines aL, d and f (K) are on a plane spanned by the triangle aL, b, c. Therefore the line aL, d intersects f (K) in some point e distinct from d. Then d,e C G, and thus aL e G, a contradiction. Hence, G' = (G0,b) is a hyperplane of N such that f (L) £ G'. We have also aK, b n G0 = 0 since aK, b £ G. Thus f (K) £ G'. Finally, H' := f-1(G' n f (S)) is the hyperplane we are looking for. The case with f (K) £ G and f (L) C G goes the same way. Now, let f (K) C G and f (L) C G. As in the previous case we take aK e f (K) \ f (W), aL e f (L) \ f (W), but this time choose a hyperplane G0 e H with aK, aL e G0. Let b e G. Note that 0~b n G0 = 0 for i = K, L. So, if we set G' = (G0, b) then f (K) £ G' and f (L) £ G'. Moreover, G' is a hyperplane of N. Again, H' := f-1(G' n f (S)) is the required hyperplane. □ Lemma 3.3. Let K,L e Lw be two distinct lines such that K ||w L. There is a sequence ni,..., nn of planes in Dp(W) such that K= e n for i = 1,... ,n and K C ni, L C nn, and n, nj+1 share a line for j = 1,... ,n — 1. Proof. By Lemma 3.2 we can extend W to a hyperplane H of P such that K, L £ H. Take the point a = KBy (2.1) we have a = LTO. Now, take in P the bundle of all the lines together with all the planes through a. This structure is actually the quotient space a±/a, and it is, up to an isomorphism, a polar space (cf. [1, Lemma 2.1]), that we denote by P'. The set H', consisting of all the lines through a contained in H, is a hyperplane in P' induced by H. Then Dp/ (H') is an affine polar space, that in itself is connected (cf. [4]). So there is in Dp/ (H') a sequence of intersecting lines joining K and L as points of Dp/ (H'). However, lines of Dp/ (H') are planes of Dp(H). As W C H these planes are also planes of Dp (W). □ K. Petelczyc and M. Zynel: The complement of a subspace in a classical polar space 451 3.1 Parallelism Let K2 € Lw . Then K ||* K2 iff Ki n K2 = 0 and there are two distinct lines Li, L2 € Lw crossing both of Ki; K2, such that (3.1) Li n L2 = 0 and Li n L2 n K = 0 for i = 1, 2. In case there are are exactly 3 points per line in our polar space p, no two lines Ki, K2 on an affine plane in Dp(W) such that Ki || w K2 satisfy the right hand side of (3.1), as the required lines Li, L2 had to be of size 4. This is why from now on we assume that there are at least 4 points on every line of p. Let || be the transitive closure of ||*. It is clearly seen that || C L* x L*. Lemma 3.4. The relation || is reflexive on L*. Proof. Given a line Ki € Lw , considering that the rank of p is at least 3, take a plane n containing Ki in a maximal singular subspace through Ki. There are lines K2, Li, L2 on n such that Ki n K2 = 0 (that is K° = K2°°), Li = L2, Li n L2 = 0, and K, n Lj = 0 for i, j = 1, 2. Thus Ki ||* K2 by (3.1). This means that Ki || K2 and K2 || Ki, which by transitivity implies that Ki || Ki. □ Proposition 3.5. Let W be a subspace of p. The relation ||w defined in (2.1) and the relation || coincide on the set of lines of Dp(W). Proof. Let Ki, K2 € Lw. If Ki = K2, then Ki ||w K2 and Ki || K2. So, assume that Ki = K2. _ _ Consider the case where Ki ||w K2. By (2.1) it means that K n L n W = 0, and consequently K° = K°° = a for some a € W. This implies that Ki n K2 = 0. Assume that Ki and K2 are coplanar, and n is the plane of Dp(W) containing both of Ki, K2. The plane n is, up to an isomorphism, a projective plane, so it is Veblenian. Thus, by (3.1), Ki || * K2 .If Ki and K2 are not coplanar, then by Lemma 3.3 there is a sequence of planes ni,..., nn such that Ki C ni, K2 C nn, a € n, for i = 1,..., n, and nj, nj+i share a line for j = 1,..., n - 1. Let nj n nj+i = Mj. Note that a € Mi,..., Mn-i and Mj, Mj+i are coplanar. Therefore Mj ||* Mj+i. Moreover, Ki ||* Mi and M„_i ||* K2 by the same reasons. So finally we get Ki || K2. Now, assume that Ki ||* K2. Then Ki, K2 are disjoint and coplanar. Thus Ki, K2 meet in the closure of some plane, this means that they meet in W. By (2.1) it gives Ki ||w K2. If Ki || K2 then there is a sequence of proper lines Li,..., Ln such that Ki ||* Li ||* ••• ||* Ln ||* K2. So, from the previous reasoning we get Ki Li • • • || w Ln || w K2. As the relation ||w is transitive we have Ki || w K2. □ As an immediate consequence of Proposition 3.5 we get Corollary 3.6. Affine lines can be distinguished in the set Lw as those parallel to themselves. 452 Ars Math. Contemp. 17(2019)447-454 3.2 Recovering If W is a hyperplane it follows by [4, Proposition 2.7] that: Proposition 3.7. Let p be a thick nondegenerate polar space of rank at least 2 and let H be its hyperplane. The polar space p can be recovered in the complement Dp(H). So, from now on we additionally assume that W is not a hyperplane. By Proposition 3.5 the relation ||w can be expressed purely in terms of Dp(W). Recall that our parallelism is partial: it is defined only on affine lines. However it is not a problem in view of Corollary 3.6. From Lemma 3.1(ii) there is a bijection between the sets W = {LTO : L e C*} and {[L]|| : L e C*}. Thus we can recover W pointwise in a standard way: points of the horizon W are identified with equivalence classes of parallelism i.e. directions of affine lines of the complement Dp(W). Let us introduce a relation ~CC* x C* defined by the following condition: K1 - K2 iff for all M e C* if M n K = 0 then M ft K2. (3.2) In the sense of Euclid's Fifth Postulate it could be read as anti-euclidean parallelism. A lot more useful for us is its derivative = C C*/| x C*/| defined as follows: [Ki]y = [K2]| iff for all M e [Ki]| ,N e [K2]|: M - N and N - M. (3.3) Lemma 3.8. Let M, N be two nonparallel affine lines. The following conditions are equivalent: (i) [M]| = [N]|, (ii) there is a deep line L C W, such that MNe L. Proof. (i) ^ (ii): From one-or-all axiom, Mmust be collinear with at least one point of the line N. Moreover, Mcannot be collinear with a proper point of N, as [M] | = [N] |. Thus Mis collinear with the unique improper point of N, which is N Let L be the line through MNAssume, that n is a semiaffine plane with L = n~. Then, there are some affine lines M1, N1 C n with M~ = Mf° and N~ = Nf°. So, either M1 || N1 or M1 and N1 share a proper point. In view of (3.3), in both cases we get [M]| = [N]|. (ii) ^ (i): Assume that [M]| = [N]|. Due to (3.2) and (3.3) there is a proper point a e M and an affine line K such that a e K || N (or the symmetrical case holds). This means that a and Nare collinear in p. The one-or-all axiom implies, that either there are no other points on M that are collinear with Nor Nis collinear with all points on M. In the first case Nis not collinear with Min the latter (NM} ^ W is the plane containing the line MTO,N. □ One can note, that the relation = defined by (3.3) and the relation = introduced in [4] coincide, though their definitions are expressed differently. Besides, our relation is not transitive, but the reflexive closure of its analogue in [4] is an equivalence relation. This benefit is the result of some hyperplane properties (see Lemma 3.1(i)). Nevertheless, we can overcome this inconvenience and define ternary relation of collinearity on the horizon W. K. Petelczyc and M. Zynel: The complement of a subspace in a classical polar space 453 Lemma 3.9. If Ki, K2, K3 are pairwise nonparallel affine lines such that [Kj]|| = [K(i+i) mod 3]|| for i = 1, 2, 3, then points K°, K°°, K°° are on a line. Proof. Let a = K°, b = K°°, c = K°. By Lemma 3.8 there are improper lines L = a, b, M = b, c, N = cTa. Let H be a hyperplane containing W. If in Dp(H) there is a plane, which closure contains one of the lines L, M or N, then we also have such a plane in Dp(W), that contradicts Lemma 3.8. Thus, L, M, N C H are deep lines in relation to Dp(H). By Lemma 2.3 of [4] this means that each of L, M and N contains a point of rad H. Let d G rad H. Then, by Corollary 1.3 of [4], H = d— {d} = rad H, and d is the unique deep point of H. As we have d G L, M, N, it must be L = M = N. □ Lemma 3.10. Let Ki, K2, K3 be pairwise nonparallel affine lines. Points K°, K°°, K°° are on a line iff one of the following holds: (i) there are affine lines Mi || Ki, M2 || K2, M3 || K3 such that M^ M2, M3form a triangle in Dp(W), (ii) [Ki]|| = [K2]|, [K2]| = [K3]|, and [K3]|| = [Ki]||. Proof. Assume that K°, K°°, K°° are on a line L. If (i) does not hold, then there is no plane n in Dp(W) with L = n°. This means that L is a deep line and by Lemma 3.8 we get (ii). Now, assume that (i) is the case. Take a plane n spanned by the triangle M^ M2, M3. Then K b K2, K3 C n and K°, K2°°, K°° are on a line n°°. If (ii) is fulfilled then K°, K°°, K°° are on a line directly by Lemma 3.9. □ The meaning of Lemma 3.10 is that we are able to recover improper lines regardless of whether W is flappy or not. Let [[K]|h [L]H]_ := {[M: [M= [K, [L] J. Then new lines can be grouped into two sets: C := | [[K]||, [L]|:[K]| = [L]| and K tf l}, C" := {n° : n is a semiaffine plane of Dp(W)}. All our efforts in this paper essentially amount to the following isomorphism P = (sw uc7||, cw u C uc", |). A new point [K] | is incident to a line L G cw iff K || L. It is incident to a line L G C iff there is M g cw such that [[K] |, [M] || ] = = L. Eventually, it is incident to a line L G C" iff K C n and L = n°. " Theorem 3.11. Let P be a nondegenerate, embeddable polar space of rank at least 3, with at least 4 points per line, and W be its subspace, that is contained in a hyperplane. The polar space P can be recovered in the complement Dp(W). References [1] J. 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