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1URAL MATHEMATICAL JOURNAL, Vol. 8, No. 2, 2022, pp. 127-132
DOI: 10.15826/umj.2022.2.010
ON DISTANCE-REGULAR GRAPHS OF DIAMETER 3 WITH EIGENVALUE 6 = 1
Alexander A. Makhnev^, Ivan N. Belousovt_t, Konstantin S. Efimov^tt
Krasovskii Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, 16 S. Kovalevskaya Str., Ekaterinburg, 620108, Russian Federation
Ural Federal University, 19 Mira str., Ekaterinburg, 620002, Russian Federation
tmakhnev@imm.uran.ru tti_belousov@mail.ru tttkonstantin.s.efimov@gmail.com
Abstract: For a distance-regular graph r of diameter 3, the graph r can be strongly regular for i = 2 or 3. J. Kulen and co-authors found the parameters of a strongly regular graph T2 given the intersection array of the graph r (independently, the parameters were found by A.A. Makhnev and D.V. Paduchikh). In this case, r has an eigenvalue «2 — C3. In this paper, we study graphs r with strongly regular graph r and eigenvalue 0 = 1. In particular, we prove that, for a Q-polynomial graph from a series of graphs with intersection arrays {2c3 + «1 + 1, 2c3, C3 + «1 — C2; 1, C2, C3}, the equality C3 = 4(t2 + t)/(4t + 4 — c2) holds. Moreover, for t < 100000, there is a unique feasible intersection array {9, 6, 3; 1, 2, 3} corresponding to the Hamming (or Doob) graph H(3, 4). In addition, we found parametrizations of intersection arrays of graphs with 02 = 1 and 03 = «2 — C3.
Keywords: Strongly regular graph, Distance-regular graph, Intersection array.
1. Introduction
We consider undirected graphs without loops and multiple edges.
Let r be a connected graph. The distance d(a, b) between two vertices a, b of r is the length of a shortest path between a and b in r. For a vertex a of r, denote by ri(a) the induced subgraph on the set of all vertices at distance i from a in r. Let r be a graph with diameter d and let a and b be vertices of r at distance i (0 < i < d). Then the number of vertices that are at distance j from a and h from b is denoted by pjh(a, b) (0 < i, j, h < d) and is called an intersection number of r. Note that pjh(a, b) = ir(a) nrh(b)|. Consider the numbers ci(a, b) = pi,11(a, b), ai(a, b) = pi1(a, b), and bi(a, b) = pi+1,1(a, b). If the intersection numbers do not depend on the choice of a and b but only on i, then these numbers are denoted simply by (0 < i, j, h < d). In this case, r of diameter d is called a distance-regular graph with intersection array (b0, b1,..., bd— 1; c1,..., cd).
If a and b are vertices of the graph r, then we denote by d(a, b) the distance between a and b. Given a vertex a in a graph r, we denote by ri(a) the subgraph induced by r on the set of all vertices at the distance i from a. The subgraph r1(a) is called the neighbourhood of the vertex a and is denoted by [a], if the graph r is fixed.
Let r be a graph of diameter d and i € {1,2,3,..., d}. The graph ri have the same set of vertices, and vertices u and w are adjacent in ri if dr(u, w) = i. For a subset of vertices Y from r, we denote by ri(Y) the subgraph with the set of vertices Y in which PI vertices u and w are adjacent if dr(u, w) = i.
An incidence system with a set of points P and a set of lines L is called an a-partial geometry of order (s, t) if each line contains exactly s + 1 points, each point lies 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 (the notation is pGa(s,t)).
A 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 see that a point graph of a partial geometry pGa(s, t) is strongly regular with parameters v = (s + 1)(1+si/a), k = s(t + 1), A = (s — 1) + (a — 1)t, and ^ = a(t + 1). A strongly regular graph having the above parameters for some positive integers a, s, and t is called a pseudogeometric graph for pGa(s,t).
The direct problem in the theory of distance-regular graphs is, given an intersection array, to find the parameters of a symmetric structure corresponding to a graph with this intersection array. The inverse problem is finding the intersection array of a distance-regular graph given the parameters of the corresponding symmetric structure.
If, for a distance-regular graph r of diameter 3, the graph r3 is strongly regular, then, by [1, Lemma 3], the graph r3 is pseudogeometric for pGC3(k,6i/c2). Conversely, for the graph r3, which is pseudogeometric for pGa(1, t), the graph r has an intersection array {l, tc2, l—a+1; 1, c2, a}, where l > tc2 > l — a + 1 and c2 < a.
Let r be a non-bipartite distance-regular graph of diameter 3. By [2, Lemma 3.1], the graph r2 is strongly regular if and only if r has the eigenvalue d = a2 — c3.
The inverse problem was solved by A.A. Makhnev and D.V. Paduchickh. Let r be a distance-regular graph of diameter 3, for which r2 is a strongly regular graph with parameters (v, k, A,^) and eigenvalues k, r, and —s. Then for x = b2 + c2 < rs and ^x = rs(r + 1)(s — 1) the parameters of the intersection array of the graph r are expressed in terms of k,^, r, —s, and x ([3, Theorem 2]).
We continue the study of distance-regular graphs r of diameter 3 with strongly regular graph r2 and eigenvalue 02 = 1.
The following result is obtained in [2, Lemma 4.5].
Proposition 1. Let r be a non-bipartite distance-regular graph of diameter 3 with eigenvalue #2 = a2 — c3 = 1. The following statements hold:
(1) the eigenvalues and 03 are integer, + 03 = a1;
(2) 03(02 + 2) = —(0i + 1)(03 + 1);
(3) r has the intersection array {2c3 + a1 + 1,2c3, c3 + a1 — c2; 1, c2, c3}.
By Proposition 1, the graph r with 02 = a2 — c3 = 1 and n = a2 + 4(c2 + 2)c3 + 4a1 + 4 has non-principal eigenvalues 1 and a\/2 ± \/ñ, where the multiplicity of 1 is equal to
(2a1 — C2 + 4c3 + 2)(a1 + 2c3 + 1)03/(0203 + 2a1 + 2c3). This implies that n is a square and the multiplicity of a\/2 ± \/ñ is equal to
4(2a1 — 02 + 403 + 2)(a1 — 02 + 03)(a1 + 203 + 1)(a1 + 203)/((2a? — a2o2 + 2a? 03
+8a,ic2c3 - 4c¡c3 + 8c2c¡ + \fn,(2a\ - a,ic2 + 2a\c¿ + 2c2c3 + 2c2) +8a1 — 4a102 + 24a103 — 80203 + 16^ + 8a1 — 402 + 803)02).
Theorem 1. Let r be a Q-polynomial distance-regular graph of diameter 3 with strongly regular graph r2. If r has an eigenvalue d = a2 — e3 = 1, then e3 = 4(t2 + t)/(4t + 4 — e2) and r has the intersection array {(g2 + 4g2 + 4t + 4)(t + 1)/(4t + 4 — g2 ), 8(t + 1)t/(4t + 4 — g2 ), (g2 +1 + 2)g2/(4t + 4 — 02); 1,02, 4(t2 + t)/(4t + 4 — C2)}.
For t < 100000, there is only one feasible intersection array {9,6,3; 1,2,3} (t = c2 = 2) corresponding to the Hamming graph H(3,4) or the Doob graph with the same parameters.
We found parametrizations of distance-regular graphs of diameter 3 with eigenvalues = 1 =
$3 = a2 - C3.
Theorem 2. Let r be a distance-regular graph of diameter 3 with strongly regular graph r2. If r has the eigenvalue = 1 = a2 — c3, then r has the intersection array {(2n + r)t + 1,2(n — 1)t, r(t — 1); 1, n + r + 1,2nt} or {(2n + r)t + n + r + 1, (n — 1)(2t +1), r(2t — 1); 1, n + 2r +1, n(2t +1)}.
The following examples of graphs with eigenvalues = 1 = = a2 — c3 are known:
(1) {21,10,3; 1,6,15}, half 7-cube with spectrum 211, 97,121, —335, v = 1 + 21 + 35 + 7 = 64, and r2 is a graph with parameters (64,35,18,20);
(2) {111, 88, 9; 1,12, 99} with spectrum 1111, 21148,1444, —9407, v = 1 + 111 + 814 + 74 = 1000, and r2 is a strongly regular graph with parameters (1000,814,663,660).
For graphs from Theorem 2 for n < 350, t < 1000, we have only feasible intersection arrays {21,10, 3; 1, 6,15}, {111, 88, 9;1,12, 99}, {561, 448, 54; 1,12, 504}, and {561, 448, 75; 1, 21, 480}.
Let r be a Q-polynomial distance-regular graph of diameter 3 with eigenvalue 02 = a2 — c3 = 1. By Proposition 1, the graph r has integer eigenvalues.
Lemma 1. a1 = (c2 + 2)c3/t — t — 2 for some positive integer t.
Proof. We have
2. Proof of Theorem 1
(ai + 4(c2 + 2)c3 + 4ai + 4) = u2, where u is a positive integer. Solving the Diophantine equation
u2 — (ai + 2)2 = 4(c2 + 2)c3,
2
we get
u = (C2 + 2)c3/t +1, ai = (C2 + 2)c3/t — t — 2
for some positive integer t.
□
Lemma 2. The inequality c3 > t holds.
P r o o f. We have
k = (C2C3 + 2c3t — t2 + 2c3 — t)/t,
hence
(C2C3 + 2c31 — t2 + 2c3 — t) > 0.
Further,
k3 = 2(C2C3 + 2C3t — t2 + 2C3 — t)(c2 + t + 2)(C3 — t)/(c212),
hence c3 > t.
□
Lemma 3. The graph r is not Q-polynomial with respect to E2.
Proof. Suppose that r is a Q-polynomial graph with respect to E2. Then, by [4], the equality
-2(c2es + 2c3t - t2 + 2c3 - 2t)(c2 + 2t + 2)(2cs - i)(cs + 1)/((c2C3 + 2cs - 2t)(t + 2)t) = -(C2C3 + 2cst - t2 + 2cs - 2t)(c2 + 2t + 2)(2cs - t)(cs + 1)/((c2C3 + 2cs - 2t)(t + 2)t)
holds and either c3 = (t2 + 2t)/(c2 + 2t + 2), or c3 = t/2, or c3 = -1.
In any case, we have a contradiction. □
Lemma 4. If r is not Q -polynomial with respect to Ei; then c3 = 4(t2 + t)/(4t + 4 - c2).
Proof. Let r be a Q-polynomial graph with respect to Ei. Then, by [4], the following equality holds:
-(c2c2 - c2c3t - C2C3t2 + 4c2c2 - 4C2C3t + 2C312 - 2t3 + 4c3 - 4c3t)(c2C3 + 2c3t
-t2 + 2c3 - 2t)(c2 + 2t + 2)(2c3 - t)/((c2C3 + t2 + 2c3)(c2C3 + 2c3 - 2t)c212) = -(c4c3 + 4c2c3t - 5c3c3t2 + 4c2c3t2 + c3c3t3 - 10c2c3t3 + 4c2c314 - 4c2c3t4 +4c2C3t5 + 8c2c3 - 6c3c2t + 24c2c3t - 42c2c3t2 + 16c2c3t2 + 16c2c3t3 - 40c2c3t3
+24c2C3t4 + 24c2c3 - 36c2c|t + 48c2c3t + 12c2c3t2 - 108c2c2t2 + 16c312 +68c2c3t3 - 40c3t3 - 8c2t4 + 32c314 - 8t5 + 32c2c3 - 72c2c2t + 32c31 + 48c2c3t2 -88c3t2 - 8c2t3 + 80c313 - 24t4 + 16c3 - 48c31 + 48c3t2 - 16t3)(c2c3 + 2c31 - t2 + 2c3 -2t)(2c3 - t)/((c2c3 + 2c3t - 2t2 + 2c3 - 2t)(c2C3 +12 + 2c3)(C2C3 + 2c3 - 2t)c2t2).
Hence,
C3 € {4(t2 + t)/(4t + 4 - c2), (2t3 + (t2 + 2t)c2 + 4t2 + 4t)/(c2 + 2c2(t + 2) + 2t2 + 4t + 4),
(t2 + 2t)/(c2 + 2t + 2), 1/2t}.
The latter three cases contradict Lemma 2. □
Theorem 1 is proved. □
3. Proof of Theorem 2
Let r be a non-bipartite distance-regular graph of diameter 3 with eigenvalues
61 = ai - 1, 6*2 = 1, 63 = a2 - C3.
By [2, Lemma 3.1(v)], we have bi = (a2 - c3 + 1)c3/(a2 - c3). This implies the following statement.
Lemma 5. One of the following equalities holds :
(1) c3 = (c3 - a2)m, where m is a positive integer not exceeding 1;
(2) k = 62 + C2 + C3 + 1;
(3) k = 62 + C2 + C3 - 1.
In the second case, we have a2 — c3 = 1. In the third case, we have a2 — c3 = —1, a contradiction with [2, Lemma 3.1(b)]. Hence,
c3 = (c3 — a2)m, a2m = c3(m — 1), a2 = (m — 1)n,
61 = mn — m for some positive integer n greater than 1.
The non-principal eigenvalues a1 — 1 and 1 are roots of the quadratic equation
x2 — (62 + c2 + m — n — 1)x + c2m — (m — 1)n — 62 — c2 = 0.
Hence,
a1 = k — a2 + m — n — 1
and
a1 — 1 = c2m — (m — 1)n — k + a2.
Hence
k = a1 + 1 + mn — m, k + a1 — 1 = c2m, 2a1 = m(c2 — n + 1). If m = 2t, then c2 = n + r + 1, a1 = t(r + 2), 61 = 2t(n — 1), and r has the intersection array
{t(2n + r) + 1, 2t(n — 1), rt — r; 1, n + r + 1, 2nt}
and the non-principal eigenvalues rt + 2t — 1,1, and —n of multiplicities
(2nt + rt + n + 1)(2nt + rt + 1)(2n + r) (t — 1) (n — 1)/((rt + n + 2t — 1)(rt + 2t — 2)(n + r + 1)n), (2nt + rt + n + 1)(2nt + rt + 1)(nt — t + 1)(n — 1)r/((rt + 2t — 2)(n + r + 1)(n + 1)n), 2(2nt + rt + 1)(nt — t + 1)(2n + r)t/((rt + n + 2t — 1)(n + 1)n),
respectively.
If m = 2t + 1, then
C2 = n + 2r + 1, a1 = (2t + 1)r, 61 = (2t + 1)(n — 1),
and r has the intersection array
{(2t + 1)(n + r — 1) + 1, (2t + 1)(n — 1), 2rt — r; 1, n + 2r + 1, 2nt + n}
and the non-principal eigenvalues r(2t + 1) + 2t, 1, and —n of multiplicities
(2nt + 2rt + 2n + r + 1)(2nt + 2rt + n + r + 1)(n + r)(n — 1)(2t — 1)/((2rt + n + r + 2t)
x(2rt + r + 2t — 1)(n + 2r + 1)n), (2nt + 2rt + 2n + r + 1)(2nt + 2rt + n + r + 1)(2nt + n — 2t + 1)(n — 1)r/((2rt + r + 2t — 1)
x (n + 2r + 1) (n + 1)n), (2nt + 2rt + n + r + 1)(2nt + n — 2t + 1)(n + r)(2t + 1)/((2rt + n + r + 2t)(n + 1)n),
respectively.
Theorem 2 is proved. □
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