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4URAL MATHEMATICAL JOURNAL, Vol. 6, No. 2, 2020, pp. 63-67
DOI: 10.15826/umj.2020.2.006
DISTANCE-REGULAR GRAPH WITH INTERSECTION ARRAY {27, 20, 7;1,4, 21} DOES NOT EXIST1
Konstantin S. Efimov
Ural State University of Economics, 62 March 8th Str., Ekaterinburg, 620144, Russia
Ural Federal University, 19 Mira Str., Ekaterinburg, 620002, Russia
konstantin.s.efimov@gmail.com
Alexander A. Makhnev
Krasovskii Institute of Mathematics and Mechanics,
Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620108, Russia
Ural Federal University, 19 Mira Str., Ekaterinburg, 620002, Russia
makhnev@imm.uran.ru
Abstract: In the class of distance-regular graphs of diameter 3 there are 5 intersection arrays of graphs with at most 28 vertices and noninteger eigenvalue. These arrays are {18, 14, 5; 1, 2, 14}, {18, 15, 9; 1, 1,10}, {21, 16, 10; 1, 2,12}, {24, 21, 3; 1, 3, 18}, and {27, 20, 7; 1, 4, 21}. Automorphisms of graphs with intersection arrays {18, 15, 9; 1, 1, 10} and {24, 21, 3; 1, 3,18} were found earlier by A.A. Makhnev and D.V. Paduchikh. In this paper, it is proved that a graph with the intersection array {27, 20, 7; 1, 4, 21} does not exist.
Keywords: Distance-regular graph, Graph r with strongly regular graph T3, Automorphism.
1. Introduction
We consider undirected graphs without loops and multiple edges. For given vertex a of a graph r, we denote by r(a) the subgraph of r induced by the set of all vertices at distance i from a. The subgraph [a] = r1(a) is called the neighbourhood of the vertex a.
Let r be a graph of diameter d and i € {1,2,3,... , d}. The graph r have the same set of vertices, and vertices u and w are adjacent in r if dr(u, w) = i.
If vertices u and w are at distance i in r, then denote by b(u, w) (by c(u, w)) the number of vertices in the intersection of Ti+1(u) (ri-1 (u)) with [w]. A graph r of diameter d is called a distance-regular graph with intersection array {b0, b1,..., bd-1; c1,..., cd} if the values b(u, w) and Cj(u, w) are independent of the choice of the vertices u and w at distance i in r for any i = 0,..., d.
uv
is independent of u and v for all
For such graph and for 0 < i, j, h < d, the number pj =
3 L j
vertices u, v € r with d(u, v) = h. The constants ph are called the intersection numbers of r [1].
The incidence system with set of points P and set of lines L is called a-partial geometry of order (s, t) (and is denoted by pGa(s, t)) if every line contains exactly s + 1 points, every point lies
1This work was supported by RFBR and NSFC (project No. 20-51-53013).
exactly on t + 1 lines, any two points lie on at most one line, and, for any antiflag (a, l) € (P, L), there is exactly a lines passing through a and intersecting l. If a = t + 1, then the geometry is called a dual 2-scheme; and if a = t, then the geometry is called a net.
The point graph of a geometry of points and lines is a graph whose vertices are points of the geometry, and two different vertices are adjacent if they lie on a common line. It is easy to understand that the point graph of a partial geometry pGa(s, t) is strongly regular with parameters
v = (s + 1)(1 + st/a), k = s(t + 1), A = (s - 1) + (a - 1)t, ^ = a(t + 1).
A strongly regular graph having these parameters for some positive integers a, s,t is called a pseudogeometric graph for pGa(s,t).
In the class of distance-regular graphs r of diameter 3, there are 5 hypothetical graphs with at most 28 vertices and non-integer eigenvalues. They have intersection arrays {18,14, 5; 1,2,14}, {18,15, 9; 1,1,10}, {21,16,10; 1, 2,12}, {24, 21, 3; 1, 3.18}, and {27, 20, 7; 1, 4, 21}. Earlier, automorphisms of graphs with intersection arrays {18,15,9; 1,1,10} and {24,21,3; 1,3,18} were found by A.A. Makhnev and D.V. Paduchikh [4], [5].
In this paper, we study the properties of a hypothetical distance-regular graph with intersection array {27,20, 7; 1,4,21} and prove the following theorem.
Theorem 1. A distance-regular graph with intersection array {27,20, 7; 1,4,21} does not exist.
2. Preliminary results
In the proof of Theorem 1, we use triple intersection numbers [2].
Let r be a distance-regular graph of diameter d. If u^u2, and u3 are vertices of r and ri,r2,
, , f u1u2u3 and r3 are non-negative integers not greater than d, then <
I rir2r3
such that
d(w, m) = ri
is the set of vertices w € r
u1u2u3 ( u1u2u^
_ r1r2r3 1 r1r2r^
The numbers
u1u2u3
_ rir2r3
we will write [r1r2r3] instead of
are called triple intersection numbers. For a fixed triple of vertices u1, u2, u3, u1u2u3
Unfortunately, there are no general formulas for the
r1r2r3 J
numbers [r1r2r3]. However, a method for calculating some numbers [r1r2r3] was suggested in [2].
Assume that u, v, and w are vertices of the graph r, W = d(u, v), U = d(v, w), and V = d(u, w). Since there is exactly one vertex x = u such that d(x,u) = 0, the number [0jh] is either 0 or 1. Hence, [0jh] = 5jw$h.v. Similarly, [i0h] = 5iW$hu and [ij0] = 5iUSjv.
Another set of equations can be obtained by fixing the distance between two vertices from {u, v,w} and counting the number of vertices located at all possible distances from the third:
= pUh - [0jh], £[ilh] = pVh - [i0h], £[ijl] = pWW - [j0]
(2.1)
1=1
1=1
1=1
At the same time, some triplets disappear. For |i - j| > W or i + j < W, we have pW = 0; therefore, [ijh] = 0 for all h € {0,..., d}. Let
d
Sijh(u, v, w) = QriQsjQth
r,s,t=0
uvw rst
d
d
d
If Krein's parameter qh = 0, then Sijh(u, v, w) = 0.
H]
We fix vertices u, v, and w of a distance-regular graph r of diameter 3 and put uvw
{ijh} =
ijh
[ijh] =
uvw ijh
[ijh]' =
uwv ihj
[ijh]* =
vuw jih
[ijhr =
wvu hji .
In the cases d(u,v) = d(u,w) = d(v,w) = 2 or d(u,v) = d(u,w) = d(v,w) = 3, the calculation of the numbers
[ijh]' =
uwv ihj
[ijh]* =
vuw jih
[ijhP =
wvu hji .
(symmetrizing an array of triple intersection numbers) can give new relations for the prove of the nonexistence of the graph.
3. Proof of Theorem 1
In this section, we prove Theorem 1.
Let r be a distance-regular graph with intersection array {27,20, 7;1,4,21}. Then r has
1 + 27 + 135 + 45 = 208 vertices, the spectrum 271, (2 + v/13)45, -l117, (5 - 2VT3)45, and the dual matrix Q of eigenvalues
\
1 45 117 45
1 10 T v/13 + 5 -13/3 3 3
1 2 "3 -13/3 2 ,— 5
1 -7 13 -7
/
By [3, Lemma 3], the complement of ra is a pseudo-geometric graph for pG2i(27, 5).
Lemma 1. The intersection numbers of the graph r are :
(1) pli = 6, p2i = 20, pi2 = 35, p\2 = 80, pla = 10;
(2) p2i = 4, p22 = 16, Pia = 7, p22 = 90, p2a = 28, p2a = 10;
(3) p32 = 21, pfa = 6, p32 = 84, p2a = 30, pl3 = 8. Proof. The lemma is proved by direct calculations.
□
We fix vertices u, v, and w of the graph r and put
{ijh} = ^ ^jh^ [ijh] =
uvw _ ijh _
Let A = r2(u) and A = A2. Then A is a regular graph of degree 90 on 135 vertices.
Lemma 2. Let d(u,v) = d(u,w) = 2 and d(v,w) = 1. Then the triple intersection numbers
are:
(1) [111] = r4, [112] = [121] = -r4 + 4, [122] = -ri + ra + r4 + 5; [123] = [132] = ri - ra + 7,
[133] = -ri + ra ;
(2) [211] = -r2-r4+6, [212] = [221] = r2+r4+9, [222] = ri-r2-r4+53, [223] = [232] = -ri+28, [233] = n;
(3) [311] = r2, [312] = [321] = -r2 + 7, [322] = r2 - r3 + 21, [323] = [332] = r3, [333] = -r3 + 10, where r1, r3 € {0,1,... , 10}, r2 € {0,1,... , 6}, and r4 € {0,1,... , 4}.
Proof. Let [111] = r4. Then [113] = 0 and [111] + [112] = C2 = 4; thus, [112] = -r4 + 4. Similarly, [121] = -r4 + 4.
Let [311] = r2. Then [313] = 0 and [311] + [312] = p23 = 7; thus, [312] = -r2 + 7.
Using formulas (2.1), we obtain all the equalities. □
By Lemma 2, we have 43 < [222] = r1 - r2 - r4 + 53 < 63. Since {v, w} U A(v) U A(w) contains 182 - [222] vertices, we have 182 - [222] < 135; hence, 47 < [222] < 63 and -ri + r2 + r4 < 6.
Lemma 3. Let d(u, v) = d(u, w) = 2 and d(v,w) = 3. Then the triple intersection numbers
are :
(1) [113] = r5 + re + r7 + rs - rg - 26, [121] = -r5 - re - rz - rs + r10 + 30,
[122] = r5 + re + rz + rs - rg - r10 - 14, [123] = rg, [131] = r5 + re + rz + rs - r10 - 26, [132] = no, [133] = -r5 - re - n - rs + 33;
(2) [212] = r5 + rz + rs - rg - 9, [213] = -r5 - rz - rs + rg + 25, [221] = r5 + re + rs - no - 9,
[222] = -r5 - re - rz - 2rs + rg+no+97, [223] = n + rs - rg + 2, [231] = -r5 - re - rs + no + 25, [232] = re + rs - no + 2, [233] = n;
(3) [312] = re, [313] = -re + 7, [321] = rz, [322] = rs, [323] = -rz - rs + 28, [331] = -rz + 7, [332] = -re - rs + 28, [333] = re + rz + rs - 25,
where r5 € {0,1,... , 8}, re, rz € {1, 2,... , 7}, rs € {11,12,... , 27}, and rg, r1o € {0,1,... , 7}. Proof. Using (2.1), we arrive at relations (1)-(3) of the Lemma 3. □
By Lemma 3, we have 47 < [222] = -r5 - re - rz - 2rs + rg + r1o + 97 < 90. Consider the appropriate symmetrization. Let d(u, v) = d(u, w) = 2 and d(v, w) = 3. Then the following equalities are true: [123] = rg = [132]' = r'io, [233] = r5 = r'5, re = [312] = [321]' = r'7, [322] = rs = rs. Further, rz + rs - rg + 2 = [223] = [232]' = r'6 + r'8 - r'io + 2.
Lemma 4. Let d(u, v) = d(u, w) = d(v,w) = 2. Then the triple intersection numbers are:
(1) [111] = rg + rio - rii - 24, [112] = [121] = ri5, [113] = [131] = rn, [122] = -rio - ri5 + 16,
[123] = [132] = rio, [133] =7 - rn - rio;
(2) [211] = ri5, [212] = [221] = -rio - ri5 + 16, [213] = rio, [222] = 2rg + 2rio - 11,
[223] = [232] =28 - rg - rio, [231] = rio, [233] = rg;
(3) [311] = rii, [312] = [321] = rio, [313] = [331] =7 - rn - rio, [322] = -rio - ri5 + 16, [323] = [332] = rg, [333] = rii + rio + 3,
where r11 + 24 < rg + rio < 28, r11 + rio < 7, rio + r15 < 16, and r12 < 22.
P r o o f. Using formulas (2.1), we get the equalities:
[111] = -rii - ri2 + 4, [112] = ri5, [113] = rn, [121] = rio + ri2 + ri5 + rie - 28, [122] = -rio - ri5 + 16, [123] = -ri2 - rie + 28, [131] = -rio + rn - ri2 - rie + 28, [132] = rio, [133] = -rii + ri2 + rie - 21;
[211] = ri2 + ri3 + ri4 + ri5 - 28, [212] = -ri3 - ri5 + 16, [213] = -ri2 - ri4 + 28, [221] = -rg - rio - ri2 - ri3 - ri5 + 44, [222] = rg + rio + ri3 + ri4 + 45, [223] = ri2, [231] = rg + rio - ri4, [232] = -rg - rio + 28, [233] = rM;
[311] = rii -ri2 -ri3 -ri4 + 28, [312] = ri3, [313] = -rn + ri2 + ri4 -21, [321] = rg + ri3 - rie, [322] = -rg - ri3 + 28, [323] = rie, [331] = -rg - rn + ri2 + ri4 + rie - 21, [332] = rg, [333] = rii - ri2 - ri4 - rie + 31.
Now consider symmetrization. The following equalities are true:
[112] = ri5 = ri5, [113] = rii = rii, [223] = ri2 = ri2, [233] = rM = r'M, [323] = rie = r^, [332] = rg = r9*, rio = [132] = [312]* = r^.
Further, rg+rio+ri3+rM+45 = [222] = [222]* = r*+r*o+r*3+r*4+45 = rg+ri3+rio+r*4+45; therefore, [233] = r14 = r1 4 = [323] = r1e.
We have [111] = -r 1 1 - r1 2 + 4; hence r1 1 + r1 2 = r' 1 + r'2 = r^i + rJ2 Similarly, [122] = rio-r 15+16; therefore, rio+ri5 = r'io+r'15, [123] = -r 12-rie+28, and r 12+rie = r'2+r'e.
Finally, [133] = -r + r 2 + r e - 21 = -r ' + r '2 + r 'e - 21; thus, r = r ' , r 2 = r '2, and r 1 e = r'e. Hence r 11 = [113] = [131] = -rio + r 11 - r 1 2 - rie + 28 and r 1 o + r 12 + rie = 28. Further, r 12 = [223] = [232] = -rg - r 1 o + 28, r 12 + rg + r 1 o = 28, and rg = rie.
The equalities [113] = [131] = r 11, r 1 1 = r * 1, and [311] = r 11 + rio - r 1 3 imply that r 1 o = r 13. Hence, we obtain the equalities from the conclusion of the lemma. □
By Lemma 4, we have r11 +24 < rg+rio < 28; hence 45 < [222] = 2rg+2rio-11 < 56-11 = 45. Thus, A is an edge-regular graph with parameters (135,90,45).
In view of Lemmas 2 and 3, the following inequalities hold for the number of edges e between A(w) and A - ({w} U A(w)):
2068 = 47 ■ 16 + 47 ■ 28 < e = 63 ■ 16 + 90 ■ 28 < 3528.
Contrariwise, we have e = 90 ■ 89 - Ei[222]i; therefore, 2068 < e = 90 ■ 89 - Ei[222]i < 3528, 4482 < Ei[222]i < 5942, and 49.8 < Ei[222]i/90 < 66.03.
The resulting contradiction completes the proof of Theorem 1.
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