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ЧЕБЫШЕВСКИЙ СБОРНИК
Том 20. Выпуск 2.
УДК 517 DOI 10.22405/2226-8383-2019-20-2-198-206
Нет строго регулярного графа локально Хивуд
Aleksandar Jurisic, Janos Vidali
Jurisic Aleksandar — Faculty of Computer Science and Informatics, University of Ljubljana, and Institute of Mathematics, Physics and Mechanics (Ljubljana, Slovenia). e-mail: [email protected]
Vidali J. — Faculty of Mathematics and Physics, University of Ljubljana, and Institute of Mathematics, Physics and Mechanics (Ljubljana, Slovenia). e-mail: janos. [email protected]
Аннотация
Мы исследуем, когда сильно регулярный граф локально Хивуд. Мы фокусируемся на предполагаемом сильно регулярном графе с параметрами (v,k,X,^) = (85,14, 3, 2), который является единственным кандидатом на такой график. Предполагая, что граф является локально Хивудом, мы анализируем его структуру, в конце концов приходя к противоречию, которое позволяет нам заключить, что никакой сильно регулярный граф не является локально Хивудом.
Ключевые слова: сильно регулярные графы, локальные графы, граф Хивуд.
Библиография: 6 названий.
Для цитирования:
A. Jurisic, J. Vidali. Нет строго регулярного графа локально Хивуд // Чебышевский сборник, 2019, т. 20, вып. 2, с. 198-206.
CHEBYSHEVSKII SBORNIK Vol. 20. No. 2.
UDC 517 DOI 10.22405/2226-8383-2019-20-2-198-206
No strongly regular graph is locally Heawood
Aleksandar Jurisic, Janos Vidali
Jurisic Aleksandar — Faculty of Computer Science and Informatics, University of Ljubljana, and Institute of Mathematics, Physics and Mechanics (Ljubljana, Slovenia). e-mail: [email protected]
Vidali J. — Faculty of Mathematics and Physics, University of Ljubljana, and Institute of Mathematics, Physics and Mechanics (Ljubljana, Slovenia). e-mail: janos. [email protected]
Abstract
We investigate when a strongly regular graph is locally Heawood. We focus on a putative strongly regular graph with parameters (v, k, A, ^) = (85,14,3, 2), which is the only candidate for such a graph. Assuming that the graph is locally Heawood, we analyze its structure, finally-arriving to a contradiction, which allows us to conclude that no strongly regular graph is locally Heawood.
Keywords: strongly regular graphs, local graphs, Heawood graph.
Bibliography: 6 titles.
For citation:
A. Jurisic, J. Vidali, 2019, "No strongly regular graph is locally Heawood", Chebyshevskii sbornik, vol. 20, no. 2, pp. 198-206.
1. Introduction
Let r be a fc-regular graph with v vertices such that each pair of adjacent vertices has A common neighbours, and each pair of nonadjacent vertices has ^ common neighbours. Such a graph is said to be strongly regular with parameters (n, k,X, or SRG(n, fc, A, y) for short. Note that for a given parameter set, there may be one strongly regular graph, or there may be more or none at all. Looking at the tables of feasible parameters of strongly regular graphs by Andries E. Brouwer [2], one finds only 10 open cases on at most 100 vertices, see Table 1.
n к Л M a т mT graph # of edges
01 65 32 15 16 3.531 -3.531 32 32 2-graph\*? 1040
02 69 20 7 5 5 -3 23 45 690
03 85 14 3 2 4 -3 34 50 595
04 85 30 11 10 5 -4 34 50 S(2, 6,51)? 1275
05 85 42 20 21 4.110 -5.110 42 42 2-graph\*? 1785
Об 88 27 6 9 3 -6 55 32 1188
07 96 35 10 14 3 -7 63 32 Pg(5, 6, 2) 2030
08 99 14 1 2 3 -4 55 44 693
09 gg 42 21 15 9 -3 21 77 2079
010 100 33 8 12 3 -7 66 33 1650
Таблица 1: Feasible parameters of strongly regular graphs on at most 100 vertices, with their parameters, spectrum, some graph information and the number of edges.
Related to the open case 08, John H. Conway asked if there is a graph on 99 vertices in which every edge belongs to a unique triangle and every nonedge to a unique quadrilateral (see [5], cf. [1])
Out of all these 10 cases, 03 has the smallest number of edges. The graph induced bv the neighbours of a vertex is a cubic graph (i.e., all vertices have degree 3) on 14 vertices. It is known that there are precisely 509 connected cubic graphs on 14 vertices [6]. If we add the restriction that each pair of nonadjacent vertices has at most one common neighbour, then we are left with only 36 potential candidates. Among them, the Heawood graph has the largest girth, namely 6. One can describe it as the cycle C14 (labeled with elements of Z14), where we add the following chords: x ~ x + 5 for all even x £ Z14, адd x ~ x — 5 to all оdd x £ Z14. Three more possibilities are shown in Figure l(b, c, d). There are also three more possibilities in the case the local graph is disconnected. In each case, one component is K4, while the other component is either (a) the Petersen graph, (b) the cycle C9 (labeled with the elements of Zg), with 3 triangles (x ~ x + 2 for x = 0,3,6) and the vertex 9 adjacent to 1, 4 and 7, (c) the 3-prism with vertical edges subdivided and the new vertices adjacent to one more vertex.
We will prove that no strongly regular graph is locally Heawood.
(a) (b) (c) (d)
Рис. 1: (a) The Fano plane, (b, c, d) Some trivalent graphs on 14 points.
2. Preliminaries
It is easy to see that a strongly regular graph is either a disjoint sum of complete graphs (if Л = к — 1), or a connected graph with diameter 2. We may generalize the latter case.
Let Г be a connected graph of diameter d and assume that there exist constants ai,bi,ci (0 ^ i ^ d) such that for each pair of vertices u, v at distance i, there are precisely ci neighbours of v at distance i — 1 from u, ai neighbours of v at distance i from u, and bi neighbours of v at distance г + 1 from u. Such a graph is said to be distance-regular with intersection array {b0, b1,..., bd-1; c1,c2,..., cd}. Clearly, such a graph is fc-regular for к = b0, and к = ai + bi + ci for all г (0 ^ i ^ d). ^rthermore, we see that we always have a0 = c0 = bd = 0 and c1 = 1. Equivalently, we could say that a graph is distance-regular if all of its distance, partitions (the partitions of vertices with respect to the distance from a fixed vertex) are equitable (the number of neighbours a vertex и has in a part С only depends on С and the part containing u) with the same parameters (which are precisely the numbers ai, fy, ci).
Given an intersection array of a distance-regular graph, it is possible to compute the intersection numbers p^ (0 ^ h,i,j ^ d) counting the number of vertices at distances г and j from any pair of vertices at distance h [3]. Just like in the case of strongly regular graphs, there might be zero, one or multiple distance-regular graphs with a given intersection array. It is easy to see that a graph SRG(n, k, X, p) of diameter 2 is distance-regular with intersection array {к, к — X — 1; 1, p,} - conversely, every strongly regular graph with Л = к — 1 arises in this way. For more on strongly regular and distance-regular graphs, see Brouwer, Cohen & Neumaier [3].
Let Г be a graph with n vertices, and let и be a vertex of Г. We define ri(«) as the set of vertices at distance i from и in Г (also called the i-th subconstituent). We abbreviate Г^и) as Г(и). For vertices u, v of Г, we also define Г(и, v) = Г(и) ПГ(^). By abuse of notation, we will also denote bv Г(и) the local graph at u, i.e., the graph induced bv the set Г(и). If there is a graph A such that Г(и) is isomorphic to A for every vertex и of Г, then the graph Г is said to be locally A. If и and v are vertices of Г at distance 2, then the graph induced by Г(и, v) is called a y-gmph of Г.
Let A £ {0,1}raxra be the adjacency matrix of Г An eigenvalue of the graph Г is a number в such that there exists a vector ж for which Ax = вх holds. The subspace of all such vectors is called the eigenspace of the eigenvalue 0; its dimension is the multiplicity of в. The multiset of all
ГГ in the form d1mi 02m2 ... 0fmi, where the eigenvalues 9i (1 ^ i ^ £) are given in decreasing order, and the numbers mi represent their multiplicities.
For sets P of points and L of lines and ш incidence relation 1С P x L, we define an incidence structure as the triple (P,L, 1). A pair (p,£) (p £ P, 1 £ L) is called a flag Hp X I and an antiflag otherwise. A projective geometry PG(d, q) is ад incidence structure (P, L, £) in which the points are 1-dimensional subspaces of Fd (i.e., the d-dimensional vector space over the finite field of size q), the lines are 2-dimensional subspaces of F^ and the incidence relation is inclusion (i.e., for p £ P and 1 £ L, p £ I holds whenever p ^ Such ад incidence relations has precisely q2 + q + 1 points and q2 + q + 1 lines.
14
14
'10
12
(a)
(b)
Рис. 2: Distance partitions of (a) SRG(85,14,3,2), spectrum 141 434 -350, bi = 10 a,2 = 12, k2 = 70 = 57, P22 = 60, ..., and (b) the Heawood graph, which is the (3,6)-cage and is bipartite with spectrum 31 л/2 — л/2 — 31.
i
3
2
0
0
3. Locally Heawood
Let F = (P, L, T) be the Fano plane PG(2, 2) and H its incidence graph, which is known as the Heawood graph (see Figure 1(a)). It is the unique distance-regular graph with intersection array [bo, b\,b2; 1, c2, C3} = {3, 2, 2; 1,1, 3}. We try to construct a strongly regular graph with parameters (n,k,\,p) = (85,14,3,2) denoted by r that is locally Heawood. Figure 2 shows the distance partitions of r and H.
Let ^ be a vertex of r. As r is locally Heawood, we identify the vertices of the local graph r(rc>) with points and lines of F, with a point and line adjacent when they are incident in F. Now consider the second subconstituent graph r2(rc>). It has 70 vertices that are divided corresponding to their ^-graphs into
(i) 42 = 2 ■ 7 ■ 3 vertices that correspond to signed flags of F: by A = 3, there are 2 vertices per
each flag, which we denote by (p, l)+ and (p, l)-, where p G l, and
(ii) 28 = 7 ■ 4 vertices that correspond to the antiflags of F: by p = 2, there is one such vertex for each antiflag, and we denote it by (p,£), where p G l-
Therefore, VГ = {то} U (P U L) U (A U В), where
A :={(p,£) | p G P,l G L,pG 1} В :={(p,£)S p G P,l G L,p G 1,5 G {+, —}}
(the antiflags), and (the signed flags)
and we have established all adjacencies when both vertices do not belong to r2(to):
V p G P
V 1 G L V p G P,l G L
то ~ p то ~ I p ~ 1
& p G 1
V p G P, (q,m) G A : p ~ (q,m) & p = q
V l G L, (q,m) G A : l ~ (q,m) & l = m
V p G P, (q, m)s G B : p ~ (q, m)s & p = q
V l G L, (q,m)s G B : l ~ (q,m)s & l = m
However, we can also say something about adjacencies in r2(to).
Lemma 1. Let e = (p,k) and f = (q,l) be flags of F, and g = (r,m) and h = (s,n) be antiflags of F. Then the following statements (i)-(iv) hold:
(i) signed flags corresponding to the same, point or line (or both) are. not adjacent, i.e.,
(p = q V k = l) ^ e5 / f£ for alI S,£ G {+, -},
(ii) antiflags corresponding to the same point or line are not adjacent, i.e.,
(r = s V m = n) ^ g / h, (Hi) a signed flag is adjacent to two of the four antiflags corresponding to the same point or line,
(iv) any antiflag is adjacent to precisely one of the signed flags f + and f- corresponding to the same point or line, i.e., (q = r V I = m) ^ (f + ~ g ■ f- / g),
(v) two distinct antiflags corresponding to the same point or line have precisely one common neighbour in ^(ro), which is a signed flag corresponding to the common point or line.
(vi) the signed flags f + and f- have no common neighbours in ^(ro),
In particular, the second subconstituent graph r2(TO) has diameter 3 and only the opposite signed flags are at distance 3 in it.
PROOF. Let us consider the local graph r(p) (resp. r(l)). Then
{to} U (r(p) n r(TO)) U (r(P) n B) U (r(P) n A)
is its distance partition as a Heawood graph H, where the last two sets consist of signed flags and antiflags corresponding to the point p, respectively. A similar distance partition can be obtained for a line I. As H is bipartite, (i) and (ii) follow immediately. The parameter b2 = 2 implies (iii), and the parameters c2 = 1 and c2 = 3 of H then imply (iv). Considering (ii), (v) also follows. Finally, the signed flags f ^d f- are nonadjacent by (i) and their common neighbours are q,£ £ r(ro), so (vi) follows bv p = 2. Let f = (p, 1) be an antiflag or a flag of F, i.e., f£ £ r2(to), where e £ {0, +, —}. Suppose that there exists an antiflag or signed flag gs £ r2(ro) (where S £ {0, +, —}) that is not adjacent to f£ and has no common neighbours with it in r2(rc>). Then r(/£,gs) = {p,£} C r(^) and r(p,£) = {to, f£, gs}. Since \r(p,£) = 3, we conclude f is a flag and g = f, 5 = —e. Therefore, each antiflag is at distance at most 2 from every other vertex in r2(rc>), so the diameter of the latter graph is 3 by (vi). □
Lemma 2. Let g = (q, m) be an antiflag of F.
(i) The graph induced on r(g) n A is an induced subgraph of a hexagon.
(ii) Let f = (p,£) be a flag of F such that f5 is adjacent to g for some 5 £ {+, —}. Then there is a signed flag (r, n)£ £ B with q = r or m = n for some e £ {+, —} that is adjacent to f5 and 9-
(iii) Let, f = (p,£) be an antiflag of r such th at g ~ f. The n p £ m or q £ £.
(iv) Let e = (s,k) an d f = (p, I) be flags of F such th at e£, f5 an d g are mutually adjacent for some S,e £ {+, —}■ If s = q, then p £ m. If k = m, then q £ £.
(v) Let e = (s, k) and e! = (s', k') be distinct flags of F, and Iet f = (p, I) and f' = (p', £r) be flags or antiflags of F such th at g, e£, f& an d g, e'£ , f'5 induce triangles for some 5, 5' £ {0, +, —} and e, e' £ {+, —}■ If s = s' = q and I = m, or k = k' = m and p = q, then p = p' and I = I'.
PROOF. Consider the local graph r(g) = H. ft contains the nonadjacent vertices ^d m, whose common neighbours to and g are both out side r(g). Therefo re, ^d m are at distance 3 in r(g). Let be the set of vertices at d istances i and j from q md mm r(g). Since the Heawood graph has p\2 = = 3, we have |^2| = |^2| = |^2| = |^2| =3, and these four sets cover all the vertices of r(g) except ^d m. Hy c2 = 1, there are matchings between D)> and D2, D^ md D"2, and between D2 md D^. b2 = 2 that each vertex from or D^ has two neighbours in D2, or
D"2, respectively.
Bv Lemma l(iv), the antiflag g is adjacent to six signed flags (q, ni)£i £ D}^ md (ri, m)c,i £ D2 for i = 1, 2, 3 with ni = nj md ri = rj if i = j. by Lemma l(ii), the vertices in U D2
are all flags, it follows that r(g) n A C ^ u D". The graph indueed on U D" is bipartite on 6 2
Let f be a flag of F such that f5 ~ g for some 5 G {+, —}. Then we have fs G D2 U^2 U D2 U-D". so f5 must be adjacent to a vertex in D^ U D\ and (ii) follows.
Let f = (p, l) be a flag or antiflag such that f5 is in D2 (resp. D3) for some 5 G [0, +, —}. Then there is a flag e = (s, k) such th at s = q and e£ G (resp. k = m and e£ G D\) and e£ is adjacent to fs for some e G {+, —}. Each signed flag in D"2 (resp. D2) is nonadjacent to f6, and their p = 2 common neighbours are g and another flag or antiflag of r(g). Therefore, p = ri (resp. l = ni) for i = 1, 2, 3, i.e., p G ^ (resp. q G l), so (iii) and (iv) follow.
Now let f' = (p',l') be another flag or antiflag such that f"5 is in D3 (resp. D") for some S' G {0, +,-}. Then f ^d f'5 are at distance 2 in r(g), so their p = 2 common neighbours are g and another flag or antiflag of r(g). Therefore, p = and l = l', which proves (v). □
Lemma 3. Let f = (p, l) be a flag of F and 5 G {+, -}.
(i) The graph induced on r(f5) n B is an induced subgraph of an octagon.
(ii) Let g = (q,m), g' = (q',m') and h = (r, n) be three distinct antiflags adjacent to f5 with either p = q = q' and m = n, or l = m = m' and q = r. Then h ~ g'.
(Hi) Let g = (q,m) and g' = (q',m') be two distinct antiflags adjacent to fs, and e = (s,k) and e! = (s', k') be two flags with f5, g ~ e£ and f5, g' ~ e'£ for some e,e' G {+, —}. If either p = q = q', m = k and m' = k', or l = m = m', q = s and q' = s', then s = s' and k = k!.
Proof. Let f = (p,l) be a flag of F, and consider the local graph r(/5) = H. It contains the adjacent vertices p and l, two antiflags (p,m), (p,m') adjacent to p, and two antiflags (q,l), (q',l) adjacent to l. Let Dzj be the set of vertices at distances i and j from p and l in r(/5). Then Dl, = {(p,m), (p,m')}, D" = {(q,l), (q',l)} and |^f| = 1^31 = 4 by p23 = 4- These sets cover all vertices of T(/5). By b2 = 2 and c2 = 1, each vertex in D2 or D" has two neighbours in D" or D2 and one neighbour in D^ or D2, respectively.
Since the vertices in D2 U are all antiflags, it follows that r(/5) n B C U D\. The graph induced on D2 U D" is bipartite on 8 vertices of valency 2, and since r(f5) is square-free by c2 = 1, it follows that it must be an octagon. This proves (i).
Let g., g' be the two antifl ags in D^ (resp. an d h an antiflag in f5 corresponding to the same line (resp. point) as g. Since g and h are nonadjacent by Lemma 1 and their p = 2 common neighbours are fs and the common corresponding line (resp. point), it follows that they must be at distance 3 in T(/5). The vertices at distance 3 from g are in D\ U D2 (resp. D)> U D"). As r = p and n = l by Lemma 1 (iii), we have h G U D2. ^teefore, h must ^e adjacent to g' and (ii) follows
Let g., g' be the two antifl ags in (resp. ^^^^ an d e£ and e'£ ^e two signed flags in D2 (resp. D2) adjacent to g and g', respectively, with each adjacent pair corresponding to the same point (resp. line). As r(f5) is bipartite, e£ and e'£ must be at even distance in the octagon induced on D\ U D\. But as r(/5) has girth 6 and C" = 3, two vertices at distance 4 in this octagon must have a common neighbour in D)> U D2. Therefore, e£ and e'£ ^re at distance 2 in the octagon. They are nonadjacent, with their p = 2 common neighbours being f5 and an antiflag or signed flag in
the octagon. It follows that e£ and e'£ cannot correspond to the same point or line, so (iii) holds. □
Lemma 4. The following hold.
(i) Let, e = (p, l) and f = (q, m) be flags and g = (r, n) be an antiflag sueh that e£ ~ g ~ f5 for some S,e G {+, —}■ If p = r or l = n, and q = r or m = n, the n r(e£) n r(fs) n r(g) n A = 0.
(ii) Each antiflag g G A has 3 neighbours in A and 9 neighbours in B.
(iii) Each signed flag f5 £ B has 6 neighbours in A and 6 neighbours in B, of which no two correspond to the same point or line.
(iv) Let g = (r,n) an d h = (s,k) be two adjacent antiflags. Then the re are flags e = (p,£) with p = r or I = n and f = (q, m) with q = s or m = k such th at g and h are adjacent to both e£ and f5 for some 5,e £ {+, —}.
Proof. Let e = (p,£) md f = (q, m) be flags and g = (r, n) be an antiflag such that e£ ~ g ~ f5 for some 5,e £ {+, —}. If p = q = r (resp. I = m = n), the common neighbours of e£ and f5 are p (resp. £) and g - in particular, no antiflag adjacent to g is among them. If p = r = q and m = n = I, then e£ and fs are at odd distance in r(g), so they do not share a common neighbour with g. This proves (i).
Consider two distinct signed flags (p,£)s and (p, m)£. Bv Lemma l(i), they are nonadjacent and their p = 2 common neighbours are p and an antiflag corresponding to p (or I if I = m and S = —e).
Therefore, a signed flag can be adjacent to at most one signed flag corresponding to a chosen point (and by a similar argument, a chosen line), i.e., a signed flag can have at most 6 neighbours in B and therefore at least 6 neighbours in A Thus, there are at least 42 ■ 6 = 252 edges between A and B.
Now consider the local graph r(/5). By Lemma 3(i), the at most 6 adjacent signed flags of f5 are on an octagon, and by Lemma l(i), they are not adjacent to q and m, which are also in the local graph. The remaining vertices must then be antiflags. Consider only the four of them that are adjacent to q or m. There are at 1 east 2 antiflags in r(f) adjacent to one of them. By Lemma l(iv) and picking all possibilities for f6, each antiflag occurs six times in this position (for signed flags corresponding to the same point or line), and in each local graph r(/5) it has distinct adjacent antiflags by (i). Therefore, there are at least 42 ■ 2 = 84 ordered pairs of adjacent antiflags, so the average valency on the graph induced on A is at least 84/28 = 3. Thus, there are at most 28 ■ (12 — 3) = 252 edges between A and B.
We have thus established that there are precisely 252 edges between ^d B, i.e., on average, an antiflag g £ A has 3 neighbours in A and 9 neighbours in while each signed flag f5 £ B has precisely 6 neighbours in each of ^d B - this proves (iii). Consider the quotient matrix M of the partition of the vertices of r into {to} r(ro), A md B
0 14 0 0
1 3 4 6
0 2 3 9
0 2 6 6
142 42 —32
spectrum of r tightly. The aforementioned partition is therefore equitable (see [4]). Hence, (ii) holds.
Let g = (r, n) be an antiflag. The above argument shows that each edge between g and another antiflag lies in the local graph r(e£) to precisely one choice of e = (p,£) with p = r or I = n, and e £ {+, —}. Let h = (s, k) be an antiflag that is adjacent to g and one of the aforementioned signed flags e£. By applying the same argument to h, we see that there is precisely one flag f = (q, m)
with q = s or m = k such th at f5 is adjacent to both g and h for som e 5 £ {+, —}. This proves □
We are now ready to prove the main result.
Theorem 1. No strongly regular graph is locally Heawood.
Proof. We have k = 14 Mid A = 3, therefore p | k(k — 1 — A) = 140 and hence p £ {2, 4, 5, 7,10}, i.e., n £ {85, 50, 43, 35, 29}. By examining the tables by Brouwer [2], we see that only the first possibility is feasible. We therefore continue with the labels introduced above.
Consider an antiflag g = (p, l) and a point q = p such that q G l (resp. a line m = l such that p G m)- They are not adjacent, and no point or line or to is among their common neighbours, which must then be two antiflags or signed flags corresponding to the point q (resp. line m). Since these are not adjacent to p or l and they are distinct for the 3 possible choices of q (resp. m), it follows that the 6 neighbours of g not equal or adjacent to p or l are, by Lemma 4(ii, iii), three antiflags and three signed flags, each one of which corresponds to one choice of q and one choice of m. Therefore, any antiflag (q, m) adjacent to g must have p G m and q G l, and there are three such neighbours (actually all three antiflags satisfying the condition), so the graph induced on the
{3, 2, 2, 1; 1, 1, 1, 2}
Since the Coxeter graph is triangle-free, the three common neighbours of two adjacent antiflags g = (p,l) and h = (q, m) are all signed flags: by Lemma 4(iv), one corresponding to p or l and one corresponding to g or m, and by Lemma 4(i), another signed flag f 5 for f = (r,n), r = p, q, n = l, m and some 5 G {+, —}. In the focal graph r(fs), the antifl ags g and h are not adjacent to r or n, so they are adjacent to an antiflag g' = (r,l') and an antiflag h' = (q',n), thus forming a path g' ~ g ~ h ~ h' or g' ~ h ~ g ~ h'. Since there are 28 ■ 3/2 = 42 edges in the Coxeter graph, it follows that such a path of four antiflags occurs in the local graphs at each of the | B | =42 signed flags.
Consider a signed flag f5 for f = (r, n) and some S G {+, —}, and a line l' such that r G l' (resp. a point q' such that q' G n). They are not adjacent, and by Lemma 4(iii), precisely one of their two common neighbours is a signed flag corresponding to l' (resp. qr), so the second common neighbour is an antiflag also corresponding to l' (resp. h' corresponding to qr). For a fixed choice of fLemma 1(iii) implies that the antiflag g' is (r, ll) (resp. h' = (q',n)) for precisely two of the four possible choices of l' (resp. qr). Therefore, the two antiflags g = (p,l) and h = (q, m) adjacent to f5 with r = p,q, n = l,m have r G l, m and p,q G n. ^v the above argument, g and h are adjacent, so p G m and q G l also holds. As p, q, l, m ^^^ct all ^^^^^s and fines of the Fano plane F,
we are forced to conclude that r and n ^re not on the plane F, contradiction. Thus, the statement □
Acknowledgements
Aleksandar Jurisic and Janos Vidali are supported by the Slovenian Research Agency research program Pl-0285. Janos Vidali is also supported by Slovenian Research Agency project Jl-8130.
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Получено 27.06.2019 г. Принято в печать 12.07.2019 г.
