Binomial partial Steiner triple systems with complete graphs: structural problems
Krzysztof Petelczyc, Małgorzata Prażmowska, Krzysztof Prażmowski
aa r X i v : . [ m a t h . C O ] A ug Binomial partial Steiner triple systems with completegraphs: structural problems
Krzysztof Petelczyc, Małgorzata Prażmowska, Krzysztof PrażmowskiJune 7, 2018
Introduction
In the paper we study the structure of hyperplanes of so called binomial partialSteiner triple systems (
BSTS ’s, in short) i.e. of configurations with (cid:0) n (cid:1) points and (cid:0) n (cid:1) lines, each line of the size 3. Consequently, a BSTS has n − M is a maximalproper subspace of M . A more specialized characterization of a (“geometrical”)hyperplane comes from projective geometry: a hyperplane of a (partial linear =semilinear) space M is a proper subspace of M which crosses every line of M . Notethat these two characterizations are not equivalent in general. In the context of in-cidence geometry the second characterization is primarily used (cf. [2] or [20]), andalso in our paper in investigations on some classes of partial Steiner triple systems(in short: PSTS ’s) we shall follow this approach. For a
PSTS M there is a naturalstructure of a projective space with all the lines of size 3 definable on the family of allhyperplanes of M (the so called Veldkamp space of M ). On other side our previousinvestigations on PSTS ’s and graphs contained in them lead us to characterizationsof systems which freely contain complete graphs (one can say, informally and notreally exactly: systems freely generated by a complete graph); these all fall into theclass of so called binomial configurations i.e. (cid:16)(cid:0) ν + κ − ν (cid:1) ν (cid:0) ν + κ − κ (cid:1) κ (cid:17) -configurationswith integers ν, κ ≥
2. A characterization of
PSTS ’s which freely contain at leastgiven number m of complete subgraphs appeared available, and for particular valuesof m a complete classification of the resulting configurations was proved (see [10]).It turned out so, that the structure of complete subgraphs of M says much about M , but fairly it does not determine M .Now, quite surprisingly, we have observed that the complement of such a freecomplete subgraph of a PSTS M is a hyperplane of M . So, our previous classifica-tion is equivalent to characterizations and classifications of binomial PSTS ’s basedon the structure of their binomial hyperplanes. But a
PSTS , if contains a binomialhyperplane, usually contains also other (non-binomial) hyperplanes. So, the struc-ture of all the hyperplanes of a
PSTS M says much more about the structure of M .In the paper we have determined the structure of hyperplanes of PSTS ’s of someimportant classes, in particular of so called generalized Desargues configurations (cf.1 yperplanes of binomial
PSTS ’s PSTS ’s were prepared by Krzysztof Petelczyc. We have alsoshown a general method to characterize all the hyperplanes in an arbitrary
BSTS with at least one ((maximal) free complete subgraph (Theorems 3.5, 3.9).As it was said: the hyperplanes of a
PSTS yield a projective space P . In essence, P = P G ( n,
2) for some integer n , so only n = dim( P ) is an important parameter,but non-isomorphic PSTS ’s may have the same number 2 n +1 − PSTS M does notgive a complete information on the geometry of M .However, if the points of the P G ( n, BSTS , are labelled bythe type of geometry that respective hyperplanes carry, the number of nonisomor-phic realizations of such labelled spaces drastically decreases. It is pretty well seenin the case of 10 -configurations, but one can observe it for all BSTS with arbitraryrank of points.
BSTS A partial Steiner triple system (a PSTS ) is a partial linear space M = h S, Li withthe constant point rank and all the lines of the size 3. A binomial partial Steinertriple system (a BSTS ) is a configuration of the type (cid:16)(cid:0) n (cid:1) n − (cid:0) n (cid:1) (cid:17) for an integer n ≥
4; for short, we write B n for a configuration with these parameters.The symbols ℘ ( X ) and ℘ k ( X ) stand for the subsets and the k -subsets of a set X , resp. A PSTS M freely contains the complete graph K X , X ⊂ S iff for any disjoint 2-subsets { a , a } and { b , b } of X we have a , a ∩ b , b = ∅ ( a denotes the line of M which contains a ) and no 3-subset of X is on a line of M .Let us recall after [13] some basic properties of BSTS ’s.
Proposition . Let n ≥ be an integer. A smallest PSTS that freely containsthe complete graph K n is a B n +1 -configuration. Consequently, it is a BSTS . Proposition . Let M = h S, Li be a minimal PSTS which freely contains acomplete graph K X = h X, ℘ ( X ) i and | X | = n . Then the complement of K X , i.e.the structure M \ X := h S \ X, L \ { e : e ∈ ℘ ( X ) }i (1) is a B n -configuration and a subspace of M .Conversely, let M contain as a subspace a B n -configuration N = h Z, Gi , Then S \ Z yields in M a complete K n -graph freely contained in M , whose complementis N . Proposition . Any two distinct complete K n -graphs freely contained in a B n +1 -configuration share exactly one vertex. yperplanes of binomial PSTS ’s Proposition . Let h X i , ℘ ( X i ) i , i = 1 , , be three distinct K n graphs freelycontained in a B n +1 -configuration M . Let c k ∈ X i ∩ X j for all { k, i, j } = { , , } .Then { c , c , c } is a line of M . Let Z , Z be two subsets of a set S . We write (cf. [16]) Z ⋔ Z := ( Z ∩ Z ) ∪ (cid:0) ( S \ Z ) ∩ ( S \ Z ) (cid:1) (2)= S \ ( Z ÷ Z ) , (3)where ÷ denotes the operation of symmetric difference. Note that identifying asubset Y of S with its characteristic function χ Y , and, consequently, identifying S with the constant function we can compute simply S \ Y = + Y . Afterthat we have Y ⋔ Y = + ( Y + Y ). Simple computations in the Z -algebra ofcharacteristic functions of subsets of S yield immediately the following equationsvalid for arbitrary subsets Y , Y , Y of S : Y ⋔ Y = S, (4) Y ⋔ S = Y , (5) Y ⋔ Y = Y ⋔ Y , (6) Y ⋔ ( Y ⋔ Y ) = Y , (7)( Y ⋔ Y ) ∩ Y = Y ∩ Y , (8)( Y ⋔ Y ) ⋔ ( Y ⋔ Y ) = Y ⋔ Y , (9)Formally, the operation ⋔ depends on the superset S which contains the argumentsof ⋔ . In what follows we shall frequently use this operation without fixing S explic-itly: the role of S will be seen from the context.A hyperplane of a PSTS M is an arbitrary proper subspace of M which crossesevery line of M . Proposition . If H , H are distinct hyperplanes of M then H ⋔ H is ahyperplane of M as well. Proof.
Let L be a line of M . Set H = H ⋔ H . Write L = { q , q , q } . It is seenthat, up to a numbering of variables, one of the following must occur:(i) q ∈ H , H , q , q / ∈ H ∪ H : then L ⊂ H .(ii) q ∈ H , H , q , q ∈ H \ H .(iii) q , q , q ∈ H , H : clearly, L ⊂ H .(iv) q ∈ H \ H , q ∈ H \ H , q / ∈ H , H : then q ∈ H .In each case L crosses H , and if L has two points in H then L is contained in H .Let H ( M ) be the set of hyperplanes of M . Note, in addition, that the structure V ( M ) := h H ( M ) , (cid:8) { H , H , H ⋔ H } : H , H ∈ H ( M ) , H = H (cid:9) i (10)is a projective space P G ( n, n = − , , the Veldkamp space of M (cf. [2], [22],(or [20])). yperplanes of binomial PSTS ’s for each PSTS M , | H ( M ) | = 2 n +1 − for an integer n ≥ − X and ∅ 6 = A ′ , A ′′ ⊂ X we write H ( A ′ | A ′′ ) := ℘ ( A ′ ) ∪ ℘ ( A ′′ ) . (11)The following set-theoretical formula H ( A | X \ A ) ⋔ H ( B | X \ B ) = H (cid:16) ( A ∩ B ) ∪ (cid:0) ( X \ A ) ∩ ( X \ B ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:0) A ∩ ( X \ B ) (cid:1) ∪ (cid:0) B ∩ ( X \ A ) (cid:1)(cid:17) = H ( A ÷ B | X \ ( A ÷ B )) (12)is valid for any distinct ∅ 6 = A, B ( X .Let us fix Z ( X , X – finite. The following is just a simple though importantobservation. Remark . The set (cid:8) H ( { i }| X \ { i } ) : i ∈ Z (cid:9) generates via ⋔ the subalgebra D ( Z ) = (cid:8) H ( A | X \ A ) : ∅ 6 = A ⊂ Z (cid:9) of the ⋔ -algebra D ( X ) = (cid:8) H ( A | X \ A ) : ∅ 6 = A ( X (cid:9) . Clearly, D ( Z ) determines as in (10) a Fano projective space P G ( n, D ( X ). Both algebras andboth projective spaces up to an isomorphism depend entirely on the cardinalities | Z | and | X | . Next, we continue investigations of Subsection 1.1, but now we concentrate uponthe ‘complementary configurations’ contained in a
BSTS ; in view of 1.2, this is anequivalent approach.Let us begin with a few words on basic properties of such a complementaryconfiguration.
Proposition . Let X be a K n -graph freely contained in a B n +1 -configuration M = h S, Li and let Y = S \ X be the corresponding B n -subconfiguration, comple-mentary to K X . Then Y is a hyperplane of M . Proof.
Let L ∈ L , if x ∈ L ∩ Y or x, y ∈ L ∩ Y are given, then L ∩ Y = ∅ :by assumptions. Suppose there are two x, y ∈ L , x, y / ∈ Y . So, x, y ∈ X , byassumptions on X we have x, y \ X ∈ L ∩ Y .Roughly speaking, establishing the structure which maximal complete graphsyield in a B n +1 - BSTS consists in establishing the structure which maximal binomialsubspaces yield in the configuration, which can be equivalently reformulated as yperplanes of binomial
PSTS ’s hyperplanes arrangements in binomial partial Steiner triple systems. The question if each hyperplane is acomplete-graph-complement has, generally a negative solution: indeed (cf. 4.1), if | A | , | X \ A | > H ( A | X \ A ) defined by (11) is not a binomial configuration,though it happens to be even a hyperplane. A first counterexample is given in 4.1.A more general argument follows by 1.5 and the following observation. Proposition . Let Y , Y be complements of distinct maximal complete graphs X , X freely contained in a BSTS M = h S, Li , | S | > . Then Y ⋔ Y is ahyperplane of M which is not the complement of any complete subgraph of M . Proof.
Set Y = Y ⋔ Y ; it suffices to note that the complement X = S \ Y of Y is not a complete graph. It is seen that X = ( X ∩ Y ) ∪ ( X ∩ Y ).If x, y ∈ X i , x = y then, by definition, there is z ∈ Y i such that { x, y, x } is a line of M ; we write z = { x, y } ∞ i . Take any distinct x, y ∈ S .If x, y ∈ X , Y then x, y are joinable; let z = x ⊕ y be the third element of x, y .Then z ∈ x, y ⊂ Y and z = { x, y } ∞ ∈ Y .Analogously, if x, y ∈ X , Y then the third point z = x ⊕ y of x, y lies on Y ∩ Y .Take x ∈ X , Y , y ∈ X , Y and suppose there is a line through x, y ; again we take z = x ⊕ y . By the above, z / ∈ X ∩ Y , X ∩ Y , Y ∩ Y . So, only the case z ∈ X ∩ X remains to be examined.By 1.3, X ∩ X = { c } for a point c . So, finally, we take x ∈ X ∩ Y collinear with c , and y ∈ X ∩ Y , y / ∈ x, c and then x, y ∈ X are not collinear in M . -configurations To give intuitions how the hyperplanes in well known configurations look like weenclose this Section. The following can be proved by a direct inspection of all the10 configurations. Some of these facts follow from more general theory developedin next sections, but we give them right at the beginning to give intuitions how thetheory looks like. It is known that there are exactly ten 10 -configurations (see e.g.[12], [1], [11], [13]); names of the configurations in question are used mainly after[8]). Proposition . (schemes of Veldkamp spaces of the configurations enumeratedbelow are presented in figures 1-6) (i) Desargues configuration has fifteen hyperplanes (cf. [21]). (ii)
The Kantor G -configuration (Fig. 1) and the nightcap configuration(Fig. 6) have seven hyperplanes each. (iii) The fez configuration (Fig. 4) and the headdress configuration (Fig. 5)contain three hyperplanes each. (iv)
The basinet configuration (Fig. 2) and the overseashat configuration (Fig.3) contain exactly one hyperplane each. (v)
Every of the remaining three configurations does not contain any hy-perplane. yperplanes of binomial PSTS ’s G -Configuration V (3)Figure 2: The Veldkamp space of thebasinet configuration Figure 3: The Veldkamp space of theoverseas-cap configurationFigure 4: The Veldkamp space of the fez configurationFigure 5: The Veldkamp space of the headdress configurationAs a consequence we can formulate Remark . There are binomial configurations (even quite small: 10 -configura-tions) with exactly one hyperplane. This one may be the complement of a complete yperplanes of binomial PSTS ’s -configuration with exactly one Veblen subconfiguration) or not (a 10 -configuration whose unique hyperplane consists of a point and a line).Figure 6: The Veldkamp space of the nightcap configuration PSTS ’swith a maximal complete graph inside
Let us begin with the formal construction which makes more precise the statementsof 1.2. Let | X | = n and 0 / ∈ X , W = X ∪ { } . Every BSTS M with (cid:0) n +12 (cid:1) verticesand K X freely contained in it can be presented in the form K X + µ V , defined below:let V be a B n -configuration and µ be a bijection of ℘ ( X ) onto the point setof V . The point set of K X + µ V is the union of the set of vertices of K X and thepoint set of V . The set of lines of K X + µ V is the union of the set of lines of V andthe family (cid:8) { x, y, µ ( { x, y } ) } : x, y ∈ X, x = y (cid:9) . Up to an isomorphism, K X + µ V can be in a natural way defined on the set ℘ ( W ) as its point set: we identify each x ∈ X with the set { , x } , and identify each point µ ( { x, y } ) of V with the set { x, y } ,suitably transforming the line set of V and putting, formally, µ ( { x, y } ) = { x, y } .Frequently, we write ( x, y ) ∞ = x, y ∞ = µ ( { x, y } ) for distinct x, y ∈ X .In the first step we shall characterize hyperplanes in a configuration M = K X + µ V . So, let H be a hyperplane of M . Let V be the point set of V ; then H = H ∩ V is a hyperplane of V or H = V . We begin with several technical lemmas. For x, y ∈ X we write x ∼ y when x = y and ( x, y ) ∞ ∈ H or x = y ∈ X . Lemma . The relation ∼ is an equivalence relation. Proof.
It is evident that ∼ is symmetric and reflexive. So it remains to prove thetransitivity of ∼ . Let x, y, z ∈ X be pairwise distinct. Assume that ( x, y ) ∞ , ( x, z ) ∞ ∈ H and suppose that ( y, z ) ∞ / ∈ H . Then H ∩ y, z ⊂ { y, z } , as H crosses every line yperplanes of binomial PSTS ’s M . Assume y ∈ H ; from y, ( x, y ) ∞ ⊂ H we infer x ∈ H and then z ∈ H follows.Finally, y, z ⊂ H , so ( y, z ) ∞ ∈ H . Lemma . Let x ∈ X . If there is z ∈ [ x ] ∼ ∩ H then [ x ] ∼ ⊂ H . Proof.
From assumptions, [ x ] ∼ = [ z ] ∼ . Let y ∼ z be arbitrary. Then ( y, z ) ∞ ∈ H and z ∈ H yield y ∈ H .Write X = X (cid:30) ∼ . From 3.2 we know thatfor every a ∈ X , either a ⊂ H or a ∩ H = ∅ . Lemma . For every a , b ∈ X if a , b ⊂ H then a = b . Proof.
Let x, y ∈ X such that a = [ x ] ∼ and b = [ y ] ∼ . Let x = y . From assump-tions, x, y ∈ H and then x, y ⊂ H gives ( x, y ) ∞ ∈ H i.e. x ∼ y , as required. Lemma . For every distinct a , b ∈ X , a ⊂ H or b ⊂ H . Proof.
Let x, y ∈ X such that a = [ x ] ∼ and b = [ y ] ∼ . From assumptions, x y i.e. ( x, y ) ∞ / ∈ H . But x, y crosses H , so x ∈ H or y ∈ H . From 3.2 we get theclaim.Now, we are in a position to prove the first (main) characterization. Theorem . Let H be a hyperplane of M = K X + µ V defined on the set ℘ ( W ) ,as introduced at the beginning of the section. Then there is a subset A of W suchthat H = H ( A | W \ A ) . Proof.
From 3.3 and 3.4 we get that either ∼ has exactly two equivalence classes a ⊂ H, b ⊂ X \ H or X = { X } . In the second case, V ⊂ H , and if there were x ∈ H ∩ X then H is the point set of M . So, H = V = ℘ ( X ) = H ( { }| W \{ } ). Letus pass to the first case. Note that H is the union of three sets: µ ( ℘ ( a )) = ℘ ( a ), µ ( ℘ ( b )) = ℘ ( b ), and a which, under identification introduced before, correspondsto (cid:8) { , x } : x ∈ a (cid:9) . So, finally, H can be written in the form ℘ ( b ) ∪ ℘ ( a ∪ { } ) = H ( b | W \ b ).Next, we are going to determine which “bipartite” sets H ( A | X \ A ) are hyper-planes of suitable BSTS ’s. To this aim one should know more precisely what is thenumber of complete graphs inside a given configuration.Let us recall the following constructionLet I = { , . . . , m } be arbitrary, let n > m be an integer, and let X be a setwith n − m + 1 elements. Let us fix an arbitrary B n − m +1 -configuration B = h Z, Gi .Assume that we have two maps µ, ξ defined: µ : I −→ Z℘ ( X ) and ξ : I × I −→ S X ,such that ξ i,i = id, ξ i,j = ξ − j,i , and µ i is a bijection for all i, j ∈ I . Set S = Z ∪ ( X × I ) ∪ ℘ ( I ) (to avoid silly errors we assume that the given three sets arepairwise disjoint). On S we define the following family L of blocks L = G (13) ∪ the lines of G ( I ) (14) ∪ {{{ i, j } , ( x, i ) , ( ξ i,j ( x ) , j ) } : { i, j } ∈ ℘ ( I ) , x ∈ X } (15) ∪ {{ ( a, i ) , ( b, i ) , µ i ( { a, b } ) } : { a, b } ∈ ℘ ( X ) , i ∈ I } . (16) yperplanes of binomial PSTS ’s m ⊲⊳ µξ B = h S, Li . It needs only a straightforward (though quite tedious)verification to prove that M := m ⊲⊳ µξ B is a B n +1 -configuration .For each i ∈ I we set Z i = X × { i } , S i = { e ∈ ℘ ( I ) : i ∈ e } , and X i = Z i ∪ S i .Then M freely contains m K n -graphs; these are X , . . . , X m . It is seen that thepoint { i, j } is the “perspective center" of two subgraphs Z i , Z j of M . So, we callthe arising configuration a system of perspectives of ( n − m + 1) -simplices . Define µ i : ℘ ( Z i ) −→ Z by the formula µ i ( { ( x, i ) , ( y, i ) } ) = µ ( i )( { x, y } ); the configuration B is the common ‘axis’ of the configurations h Z i , ℘ ( Z i ) i + µ i B contained in M . Letus denote W = X ∪ I . Without loss of generality we can assume that Z = ℘ ( X )and each ( x, i ) ∈ X × I can be identified with the set { x, i } . After this identification ℘ ( W ) becomes the point set of M .The following is crucial: Theorem ([13]) . Let M be a B n +1 -configuration. M freely contains (at least) m K n -graphs iff M ∼ = m ⊲⊳ µξ B for a B n − m +1 -configuration B and a pair ( µ, ξ ) ofsuitable maps. Combining the results of [13] and [10] it is not too hard to prove the followingcriterion
Proposition . Let M = m ⊲⊳ µξ B for a B n − m +1 -configuration B and a pair ( µ, ξ ) of suitable maps. The following conditions are equivalent (i) M freely contains at least m + 1 K n graphs. (ii) M contains at least m + 1 B n -subconfigurations. (iii) There is x ∈ X such that ξ ( i, j )( x ) = x for each pair i, j ∈ I . From now on we assume that B has ℘ ( X ) as its point set and M = m ⊲⊳ µξ B , defined on the point set ℘ ( W ),freely contains exactly m K n -subgraphs;this means that H ( i ) = H ( { i }| W \ { i } ) with i ∈ I are the unique hyperplanes of M of the size (cid:0) n (cid:1) .From the above, 1.5, and 1 we infer immediately Proposition . (i) Every set H ( J | W \ A ) with J ⊂ I is a hyperplane of M . In particular, H ( I | X ) is a hyperplane of M . (ii) There is no a ∈ X such that H ( { a }| W \ { a } ) is a hyperplane of M . With the help of (12) we get H ( A ∪ J (cid:12)(cid:12) W \ ( A ∪ J )) = H ( A | W \ A ) ⋔ H ( J | W \ J )for every J ⊂ I , A ⊂ X . So, from 3.8 we getif J ⊂ I , A ⊂ X then H ( A ∪ J | W \ ( A ∪ J )) is a hyperplane of M iff H ( A | W \ A ) is a hyperplane of M . Theorem . Let A ⊂ X . Then H ( A | W \ A ) is a hyperplane of M iff thefollowing conditions are satisfied: (i) ℘ ( A ) is a hyperplane of B . yperplanes of binomial PSTS ’s A is invariant under every ξ ( i, j ) , i, j ∈ I . (iii) A is invariant under every µ i , i ∈ I , which means the following: if µ i ( x, y ) = { u, v } then { x, y } ⊂ A or { x, y } ⊂ X \ A iff { u, v } ⊂ A or { u, v } ⊂ X \ A . Note . In some applications there is no way to present, in a natural way, theunderlying B n − m +1 -configuration B as a structure defined on the family of 2-subsetsof a ( n − m + 1) -element set; natural from the point of view of the geometry of B .Then one can take one of µ i ’s as basic and replace B as its coimage under µ i definedon ℘ ( X ) . Under this stipulation the condition (iii) of 3.9 is read as follows:(iii’) if µ i ( x, y ) = µ j ( u, v ) for some i, j ∈ I then { x, y } ⊂ A or { x, y } ⊂ X \ A iff { u, v } ⊂ A or { u, v } ⊂ X \ A . Proof.
We use notation of the definition of a system of perspectives of simplicespresented in this Section. The symbol a ⊕ b means the third point a, b \ { a, b } onthe line through a, b (if the line exists). Note that ℘ ( W \ A ) = ℘ ( I ) ∪ ℘ ( X \ A ) ∪ ( X \ A ) ⊠ I , where ( X \ A ) ⊠ I := (cid:8) { b, i } : i ∈ I, b ∈ X \ A (cid:9) . Therefore, H = H ( A (cid:12)(cid:12) W \ A ) = ℘ ( A ) ∪ ℘ ( I ) ∪ ℘ ( X \ A ) ∪ ( X \ A ) ⊠ I ,where A ⊂ X . In view of 3.8(i) without loss of generality we can assume that A = X .Since ℘ ( I ) ⊂ H ,if a line L of M has two points common with ℘ ( I ) then L ⊂ H, and each line of M of the form (14) crosses H. (17)Assume that H is a hyperplane of M . Since ℘ ( X ) is the point set of B , and B is a subspace of M right from definition, H ( A | X \ A ) = ℘ ( A ) ∪ ℘ ( X \ A ) is ahyperplane of B or ℘ ( X ) = ℘ ( A ). The latter means A = X , which contradictsassumptions. So, (i) follows.On the other hand, converting the above reasoning we easily prove that (i)impliesif a line L of M has two points common with ℘ ( A ) ∪ ℘ ( X \ A ) then L ⊂ H, and each line of M of the form (13) crosses H. (18)Next, let us pass to the lines of M of the form (15). Suppose that a / ∈ A , i , i ∈ I , i = i . Then { a, i } , { i , i } ∈ H , so { a, i }⊕{ i , i } = { ξ ( i , i )( a ) , i } ∈ H . Consequently, ξ ( i , i )( a ) / ∈ A . This justifies condition (ii).Considering all the points expressible in the form { a, i } , { i , i } , { a ′ , i } ∈ H ,i.e. with a, a ′ / ∈ A , i , i ∈ I , i = i and afterwards considering their ‘product’ { a, i } ⊕ { i , i } and { a, i } ⊕ { a ′ , i } we see that, conversely, (iii) impliesif a line L of M has two points common with H, two in ( X \ A ) ⊠ I or one in ( X \ A ) ⊠ I and the second in ℘ ( I ) then L ⊂ H, and each line of M of the form (15) crosses H. (19)Finally, we pass to the lines of M of the form (16). Let p = { a , a } ∈ ℘ ( X ),and b ∈ X , i ∈ I , q = { i, b } . Suppose p, q are collinear in M ; this means µ i ( b, b ′ ) = yperplanes of binomial PSTS ’s { a , a } for a point b ′ ∈ X and r := p ⊕ q = { i, b ′ } . Set d = { b, b ′ } . Assume that p ∈ H i.e. p ⊂ A or p ⊂ X \ A . Then q ∈ H iff r ∈ H i.e. q ∈ ( X \ A ) ⊠ I iff r ∈ ( X \ A ) ⊠ I . Finally: b ∈ A iff b ′ ∈ A , so d ⊂ A or d ⊂ ( X \ A ) follows.Next, let q ∈ H . Then p ∈ H iff r ∈ H yields that the implication ( b, b ′ ∈ A or b, b ′ / ∈ A = ⇒ p ⊂ A or p ⊂ X \ A ) holds. So, finally, we have proved (iii).Converting the reasonings above we see that (iii) impliesif a line L of M has two points common with H, two in ( X \ A ) ⊠ I or one in ( X \ A ) ⊠ I and the second in ℘ ( A ) ∪ ℘ ( X \ A ) then L ⊂ H, and each line of M of the form (16) crosses H. (20)Gathering together the conditions (17), (18), (19), and (20) we obtain that theconjunction (i) & (ii) & (iii) implies that H is a hyperplane of M .There do exist PSTS ’s which satisfy the assumptions (i)-(iii); as examples knownin the literature we can quote quasi Grassmannians, comp. Subsect. 4.2 and, inparticular, 4.4. Another class of examples is shown in 3.10.
Example . Let I = { , . . . , m } , X = { a, a, b, b ′ } and B = G ( X ) be theVeblen configuration. Set ξ ( i, j ) = ξ ( j, i )( a, a ′ , b, b ′ ) = ( a ′ , a, b ′ , b ), µ i ( x, y ) = { x, y } for all i, j ∈ I , x, y ∈ X . Let us put M := m ⊲⊳ µξ B . Then M is a system ofperspectives of m tetrahedrons. It freely contains exactly m graphs K m +3 , so itcontains exactly m hyperplanes of the form H ( { x }| X \ { x } ) = H ( x ) with x ∈ I ∪ X .However, it contains the hyperplane H ( { a, a ′ }|{ b, b ′ } ∪ I ) which is not ⋔ -generatedfrom the H ( x )’s.We close this section with a characterization of geometries on hyperplanes H ( A ∪ J | B ∪ E ) of M , where { A, B } is a decomposition of X and { J, E } is a decompositionof I . So, let us assume that (i)-(iii) of 3.9 hold. Proposition . Let k = | J | , m = | I | . (i) H ( J | W \ J ) is the union of the generalized Desargues configuration G ( J ) and the system k ⊲⊳ µ ↾ Jξ ↾ J × J B of perspectives of k simplices K X . (ii) H ( A | W \ A ) is the union of the restriction B ↾ ℘ ( A ) =: B ′ and the system m ⊲⊳ µ ′ ξ ′ B ′ , where ξ ′ ( i, j ) = ξ ( i, j ) ↾ A and µ ′ ( i ) = µ ( i ) ↾ ℘ ( A ) for all i, j ∈ I , ofperspectives of m simplices K A . BSTS ’s of someknown classes
Recall: a B n -configuration M freely contains n graphs K n − (the maximal possibleamount) iff M is isomorphic to the generalized Desargues configuration G ( X ) = h ℘ ( X ) , { ℘ ( Z ) : Z ∈ ℘ ( X ) }i for a X with | X | = n (cf. [13], [10]). The class ofgeneralized Desargues configurations appears in many applications, even in physics:[3], [4]. yperplanes of binomial PSTS ’s Theorem . Let H ⊂ ℘ ( X ) . Write H = G ( X ) . The following conditions areequivalent (i) H is a hyperplane of H . (ii) There is a proper non void subset Z of X such that H = H ( Z | X \ Z ) .Consequently, V ( G ( n )) = P G ( n − , (comp. [21]). Proof.
Let H be as required in (ii) and let L = ℘ ( A ) for a A ∈ ℘ ( X ) be a lineof H . If A ⊂ Z or A ⊂ X \ Z then L ⊂ H . Assume that A Z, X \ Z . Then thereare i, j ∈ A , i ∈ / ∈ Z , j / ∈ ( X \ Z ) So: i ∈ X \ Z , j ∈ Z . Write A = { i, j, l } . If l ∈ Z then { j, l } ∈ L ∩ H , if l ∈ X \ Z then { i, l } ∈ L ∩ H . Finally, we note that if L ∩ ℘ ( Z ) = ∅ then there is no point in L ∩ ℘ ( X \ Z ). And similarly conversely.This proves that H is a subspace of G , so, finally, (i) is valid.The implication (i) = ⇒ (ii) is immediate after 3.5.Finally, there are n − = 2 n − − X ; this determinesdim( V ( G ( X ))).Recall after [17] that each set S ( i ) = { a ∈ ℘ ( X ) : i ∈ a } (21)is a complete graph freely contained in G ( X ). In consequence of 1 and 4.1, thehyperplanes H ( i ) = H ( { i }| X \ { i } ) with i ∈ X (‘binomial hyperplanes’ of G ( X ))generate via the operation ⋔ all the hyperplanes of G ( X ).Note that each of the two components A = ℘ ( A ) and A ′ = ℘ ( A ′ ) with A ′ = X \ A of H ( A | A ′ ) is a binomial configuration. These two components are complementary (unconnected) in the following sense:if a ∈ A and a ′ ∈ A ′ then a, a ′ are uncollinear in G ( X ); A ′ consists of the pointsthat are uncollinear with every point in A , and conversely. First, we recall after [19] a construction of quasi Grassmannians.Let us fix two sets: Y such that | Y | = 2( k −
1) for an integer k and X suchthat X ∩ Y = ∅ , X = { , } or X = { , , } . We put X = X ∪ Y ; Then n := | X | = 2 k or n = 2 k + 1, resp. The points of the quasi Grassmannian R n are the elements of ℘ ( X ). The lines of R n are of two sorts: the lines of G ( X )which miss { , } =: p remain unchanged. The class of lines of G ( X ) through p (i.e. the sets ℘ ( Z ) with 1 , ∈ Z ∈ ℘ ( X )) is removed; instead, we add thefollowing sets (cid:8) { , } , { , j + 2 } , { , j + 1 } (cid:9) , (cid:8) { , } , { , j + 1 } , { , j + 2 } (cid:9) (weadopt a numbering of Y so that Y = { , , , , . . . , k } ). It is seen that R n is a B n -configuration. Proposition ([19]) . The maximal complete K n − -graphs contained in R n are exactly all the sets S ( i ) (as defined by (21) ) with i ∈ X . Write D ( i ) := ℘ ( X ) \ S ( i ) = H ( { i }| X \ { i } ). Then the following is immediate Fact . When n = 2 k + 1 , then D (0) ∼ = R k . For every n , D (1) and D (2) yieldin R n (binomial) subconfigurations isomorphic to G ( n − . yperplanes of binomial PSTS ’s Theorem . Set T = { , , . . . , k } , q t = { t, t − } for t ∈ T . Then Y = S t ∈ T q t . The family H ( R n ) consists of the sets H (cid:16) A ∪ S t ∈ J q t (cid:12)(cid:12)(cid:12) ( X \ A ) ∪ S s ∈ T \ J q s (cid:17) (22) with arbitrary A ⊆ X , J ⊆ T such that ( A, J ) = ( ∅ , ∅ ) , ( X , T ) . Consequently, V ( R k ) = P G ( k − , and V ( R k +1 ) = P G ( k, . Proof.
An elementary computation shows that each set of the form (22) is ahyperplane of R n . Let us represent R n as a suitable system of perspectives. Define,first, for i ∈ X and distinct x, y ∈ Y : µ i ( { x, y } ) = { x, y } . Next, we set ξ (1 , j +1 , j + 2) = (2 j + 2 , j + 1) for every j = 1 , ..., k − ξ (1 ,
0) = ξ (2 , x ) = x for x ∈ Y , if 0 ∈ X . We have obtained two maps µ : X −→ ℘ ( Y ) ℘ ( Y ) and ξ : X × X −→ S Y . It is seen that R n ∼ = m ⊲⊳ µξ G ( X ). A ξ -invariant subset of Y is the union of several sets of the form q t , t ∈ T . It is seen that such a union is µ -invariant. From 3.9 we infer that each hyperplane of R n has form (22).To complete the proof it suffices to note that there are 2 | X | decompositions of X , 2 k − decompositions of T , and | X | + k − − | X | + k − − W which yield a hyperplane. Substituting | X | = 2 and | X | = 3 we get | H ( R k ) | =2 k − | H ( R k +1 ) | = 2 k +1 −
1, which closes the proof.
Recall another fact: a B n -configuration freely contains n − K n − if it isa (simple) multi-veblen configuration . The multi-veblen configurations can be alsodefined by means of a direct construction. Let us recall, briefly, after [15] thisconstruction.Let X be an n -set disjoint with a two-element set p and P be a graph defined on X . The points of the configuration M ( X, p, P ) are the following: p , s i with s ∈ p , i ∈ X , and c i,j with { i, j } ∈ ℘ ( X ). The lines are: the sets of the form { p, a i , b i } , thesets { a i , a j , c i,j } , { b i , b j , c i,j } for { i, j } ∈ P , p = { a, b } , and { a i , b j , c i,j } , { b i , a j , c i,j } for { i, j } / ∈ P , p = { a, b } , finally: the sets c u , c v , c w , where { u, v, w } is a lineof G ( X ). It is seen that after the identification s i ↔ { s, i } for s ∈ p, i ∈ X and c i,j ↔ { i, j } for i, j ∈ X the structure M ( X, p, P ) can be defined on the set ℘ ( X ∪ p ) and then the configuration is easily seen to be a B n +2 -configuration. Onecan observe that each of the sets S ( i ) = { s i , c i,j : j ∈ X \ { i } , s ∈ p } ↔ { q ∈ ℘ ( X ∪ p ) : i ∈ q } is a complete K n +1 -graph freely contained in M ( X, p, P ). Moreover, the comple-ment H ( i ) = { p, s j , c j,l : s ∈ p, j, l ∈ X \ { i } ↔ { q ∈ ℘ ( X ∪ p ) : i / ∈ q } of S ( i ) is a multi-veblen configuration M ( X \ { i } , p, P ↾ ( X \ { i } )). So, we can writesimply H ( i ) = H ( { i }| X \ { i } ). It is known that a mutliveblen B n +2 -configurationis either a generalized Desargues configuration or it has exactly n maximal freelycontained complete graphs. yperplanes of binomial PSTS ’s Proposition . Assume that M ( X, p, P ) =: M is not a generalized Desarguesconfiguration. Then every binomial hyperplane of M has the form H ( { i }| X ∪ { p } \{ i } ) with i ∈ X . Each hyperplane of M has form H ( A | ( X \ A ) ∪ p ) for ∅ 6 = A ⊂ X . Proof.
It suffices to present M in the form n ⊲⊳ µξ G ( p ). Indeed, we observe,first, that G ( p ) is a trivial structure with a single point and no line. Next, weput µ i ( a i , b i ) = p for all i ∈ I , { a, b } = p . Finally, ξ ( i, j )( a, b ) = ( a, b ) when { i, j } ∈ P and ξ ( i, j )( a, b ) = ( b, a ) otherwise. Comparing definitions we see that M ( X, p, P ) ∼ = n ⊲⊳ µξ G ( p ).To complete the proof we make use of 3.5 and 3.9: a hyperplane of M has form H ( J | ( I \ J ) ∪ p ) ⋔ H ( A | ( p \ A ) ∪ I ) for a subset J of I and an ( µ, ξ )-invariant subset A of p . From 3.7 we get that a non void proper subset of p is a one-element set, andsuch a subset of p is invariant only when M is a generalized Desargues configuration,which closes our proof.Let us apply 4.5 to the particular case P = N X (the empty graph defined on X ); it isknown that M (( X, p, P ) is the structure V ∗| X | (3) dual to the combinatorial Verone-sian V ( X ) (see [14] and Section 4.4). So, V ∗ ( n ) has all its binomial hyperplanesof the same geometrical type: the dual Veronesian V ∗ ( n − Next, let us pay attention to the class of combinatorial Veronese spaces defined in[14]. Write X := { a, b, c } for pairwise distinct a, b, c . Generally, if f = a i b j c m is amultiset with the elements in X we put | f | = i + j + m .The combinatorial Veronese space V k (3) = V k ( { a, b, c } ) is the configurationwhose points are the multisets a i b j c m , i + j + m = k : the elements of y k ( { a, b, c } ),and whose lines have form eX i , i + | e | = k . It is a (cid:16)(cid:0) k +22 (cid:1) k (cid:0) k +23 (cid:1) (cid:17) -configuration.It is known that V ( X ) ∼ = G (4) so the hyperplanes of V ( X ) are, generally,known.Let M = V k ( X ). It is known (cf. [5], [13]) that K k +1 -graphs freely containedin M have form y k ( A ), where A ∈ ℘ ( X ), the complement of such a graph is theset z y k − ( X ), where { z } = X \ A , so it yields a (binomial) subspace of V k ( X )isomorphic to V k − ( X ). Remark . Note that the set H = { a c, b a, c b } yields a hyperplane in everyVeblen subconfiguration contained in V (3), but H is not a hyperplane of V (3):a 3-element anticlique of a 10 -configuration ’suffices’ for at most 3 × ⋔ the hyperplanes, starting form the binomial hyperplanesof a V k (3). • There are three hyperplanes H ( u ) = u y k − ( X ) and three their complements H ( u ) = y k ( { x, y } ), X = { u, x, y } . • Let us compute: H ( x ) ⋔ H ( y ) = { u k } ∪ xy y k − ( X ) =: H ( u ), where X = { x, y, u } .The complement of H ( u ) has the form H ( u ) = x y k − ( { u, x } ) ∪ y y k − ( { u, y } ). yperplanes of binomial PSTS ’s • Let us compute: H ( a ) ⋔ H ( b ) = H ( c ). • The properties of ⋔ yield H ( x ) ⋔ H ( y ) = H ( u ) for x = y and X = { x, y, u } . • Let us compute again: H ( x ) ⋔ H ( x ) = X k ∪ abc y k − ( X ).We have got seven hyperplanes of V k ( X ). Theorem . The above are all the hyperplanes of V k ( X ) . So, V ( V k (3)) = P G (2 , . Proof.
In the first step we present M := V k ( X ) as a system of perspectives ofsimplices. Recall that V k ( X ) is a B k +2 -configuration. In what follows we shallkeep a fixed cyclic order ≺ , say ( a ≺ b ≺ c ≺ a ) of the elements of X . Note that G ( X ) ∼ = V ( X ) is a single 3-element line. Set B = abc V k − ( X ), it is a B k − -subconfiguration of V k ( X ). Moreover, it is the intersection of three complements ofthe three maximal complete subgraphs y k ( { x, y } ), { x, y } ∈ ℘ ( X ) of M . As usually,we write ⊕ for the (partial) binary operation ‘the third point on the line through’.Frequently, writing x, yz, z below we mean any x, y, z such that X = { x, y, z } .Next, let Z = { , . . . , k − } , then | Z | = k −
1. For every z ∈ X we define ν z : Z ∋ s x s y k − s , where x ≺ y, { z, y, z } = X. (23)So, M contains three copies: F z = y k ( { x, y } ) \ ( y k ( { x, z } ) ∪ y k ( { y, z } )) = ν z ( Z )( X = { x, y, z } ) of K Z .Next, for z ∈ Z and distinct i, j ∈ Z we define µ z ( { i, j } ) = µ z ( i, j ) = ν z ( i ) ⊕ ν z ( j ) . (24)It is easy to compute that µ z ( i, j ) = x i y i z k − i ∈ abc V k − ( X ), so we have defined asurjection µ z : ℘ ( Z ) −→ B .Finally, for distinct x, y ∈ X and s ∈ Z we define the map ξ x,y : Z −→ Z by theformula ξ x,y ( s ) = k − s (25)and we set ξ x,x = id. The following holds for { x, y, z } = X and i, j ∈ Z : ν x ( i ) , ν y ( j ) collinear in M ⇐⇒ j = ξ x,y ( i ); ν x ( i ) ⊕ ν y ( k − i ) = z k . So, in fact, for each { x, y, z } = X we have a perspective ξ x,y : F x −→ F y with thecentre z k determined by the formula ξ x,y ( ν x ( i )) = ν y ( ξ x,y ( i )). Then V k ( X ) ∼ = 3 ⊲⊳ µξ B . Suppose that M contains a hyperplane H of the form H ( A, X ∪ ( Z \ A )) with A ⊂ Z . In view of 3.9, A is a ( µ, ξ )-invariant subset of Z . Without loss of generalitywe can assume that 1 ∈ A and then { , k − } ⊂ A . We get µ z (1 , k −
1) = xyz k − = µ y ( k − , ∈ A , because A is µ -invariant (here, we make use of 3.9(iii’), infact). Consequently, k − ∈ A as well.Step by step, we end up with { i, k − i } ⊂ A for every i ∈ Z , so A = Z , which,by 3.5 and 3.9 proves the theorem. yperplanes of binomial PSTS ’s . . . As we see, the number of free subgraphs of a
BSTS M does not determine M . Also,the number of of its hyperplanes and the types of geometry on hyperplanes do notdetermine M . Clearly, V ( M ) says only about | H ( M ) | .Recall that if M is a B n -configuration with a free K n − -subgraph then eachhyperplane of M is either a B n − or the union of two unconnected B k and B k -subconfigurations of M with k + k = n , k , k ≥ k < n we have the list M k of B k -configurations. Let T ( M ) be V ( M ) with itspoints labelled by the types of respective hyperplanes, i.e. by symbols from M k − or unordered pairs of symbols from M k × M n − k . It seems that T ( M ) may uniquelycharacterize M . In all the examples which were examined in the paper a hyperplane of a
BSTS (if exists) is either connected, and then it is a binomial maximal subspace, or itis the union of two unconnected (in a sense: mutually complementary) binomialsubspaces. Is this characterization valid for arbitrary
BSTS . References [1]
D. Betten, U. Schumacher , The ten configurations , Rostock. Math. Kolloq. (1993), 3–10.[2] F. Bukenhout, A. M. Cohen , Diagram Geometry (Chapter 10.2). Springer Verlag;2009.[3]
A. Doliwa , The affine Weil group symmetry of Desargues maps and the non-commutative Hirota-Miwa system , Phys. Lett. A (2011), 1219–1224.[4]
A. Doliwa , Desargues maps and the Hirota-Miwa equation , Proc. R. Soc. A (2010),1177–1200.[5]
I. Golonko, M. Prażmowska , Adjacency in Generalized Projective Veronese Spaces ,Abh. Math. Sem. Univ. Hamb. (2006), 99–114.[6] H. Gropp , Configurations and their realization , Combinatorics (Rome and Montesil-vano, 1994) , Discrete Math. , no. 1–3 (1997), 137–151.[7]
S. Kantor , Die Konfigurationen (3 , , Sitzungsberichte Wiener Akad. (1881),1291–1314.[8] K. T. Lanman , The taxonomy of Various combinatoric and Geometric Configurations ,Proceedingsof the 2001 Butler University Undergraduate Research Conference. HolcombResearch Institute.[9]
K. Petelczyc, M. Prażmowska , Twisted Fano spaces and their classification, linearcompletions of systems of triangle perspectives , Des. Codes Cryptogr. (2010), 241–251.[10] , A complete classification ofthe (15 ) -configurations with at least three K -graphs , Discrete Math. 338 (2015),no. 7, 1243-1251.[11] K. Petelczyc, M. Prażmowska , 10 -configurations and projective realizability ofmultiplied configurations , Des. Codes Cryptogr. , no. 1 (2009), 45–54. yperplanes of binomial PSTS ’s [12] B. Polster , A geometrical picture book , Springer Verlag, New York, 1998.[13]
M. Prażmowska, K. Prażmowski , Binomial configurations which contain completegraphs , arXiv:1404.4064[14]
M. Prażmowska, K. Prażmowski , Combinatorial Veronese structures, their geom-etry, and problems of embeddability , Results Math. (2008), 275–308.[15] M. Prażmowska, K. Prażmowski , Some generalization of Desargues and Veroneseconfigurations , Serdica Math. J. (2006), no 2–3, 185–208.[16] M. Prażmowska, K. Prażmowski , Semi-Pappus configurations; combinatorial gen-eralizations of the Pappus configuration , Des. Codes Cryptogr. 61 (2011), no. 1, 91-103.[17]
M. Prażmowska , Multiple perspectives and generalizations of the Desargues configu-ration , Demonstratio Math. (2006), no. 4, 887–906.[18] M. Prażmowska , On the existence of projective embeddings of some multiveblen con-figurations , Bull. Belg. Math. Soc. Simon-Stevin (2010), 1–15.[19] M. Prażmowska , On some regular multi-veblen configurations, the geometry of com-binatorial quasi Grassmannians , Demonstratio Math. (2009), no.2 387–402.[20] M. Ronan , Embeddings and hyperplanes of discrete geometries , European J. Combin.8 (1987), 179–185.[21]
M. Saniga, F. Holweck, P. Pracna , Cayley-Dickson Algebras and Finite Geome-try , arXiv:1405.6888.[22]
E. Shult , On Veldkamp Lines , Bull. Belg. Math. Soc. Simon Stevin 4 (1997), 299–316.[23]
E. Steinitz , Konfigurationen der projektiven Geometrie , Encyclop. d. math. Wiss. III(Geometrie) 1910, 481–516.Authors’ address:Krzysztof Petelczyc, Małgorzata Prażmowska, Krzysztof PrażmowskiInstitute of Mathematics, University of BiałystokK. Ciołkowskiego 1M15-245 Białystok, Polande-mail: [email protected] , [email protected] ,,