Brouwer, A.E. (Andries)

Neumaier, A.

English

Brouwer, Andries

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Strongly regular graphs where mu equals two and lambda is large

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stichting mathematisch centrum AFDELING ZUIVERE WISKUNDE (DEPARTMENT OF PURE MATHEMATICS) A.E. BROUWER & A. NEUMAIER zw 151/81 STRONGLY REGULAR GP~PHS WHERE MU EQUALS TWO AND LAMBDA IS LARGE ~ MC MAART kruislaan 413 1098 SJ amsterdam Plvlnted a.:t :the Ma.:thema:Uc.al Cent!Le, 413 Ktr.u.l6laa.n, Am6:teJul.am. The Ma.:themati.c.al Cent!Le , 6ounded :the 11-:th 06 Febll.u.a/l.y 1946, -l6 a. non-plto 6U w:tltuti.o n a.i.mlng a.:t :the pJtomo:Uo n o 6 puJLe ma.:themaUC-6 a.nd w a.ppllc.a:Uon6. 1:t ,l6 .6pon601ted by :the Ne:theJcla.n.c:l6 GoveJtnment :th/tough :the Ne:theJcla.nd.6 01r.ga.rilzaUo n 6 OIL :the Adva.nc.ement o 6 PUite Re6 ea.Jte.h ( Z. W. 0. ) • Strongly regular graphs where mu equals two and lambda is large by . *) A.E. Brouwer & A. Neumaier ABSTRACT Certain strongly regular graphs withµ= 2 are killed by observing that the first constituents are partial linear spaces. KEY WORDS & PHRASES: strongly regular graphs *) Insitut fur Angewandte Mathematik Universitat Freiburg, D-7800 Freiburg, BRD 1 0. INTRODUCTION Let G be a strongly regular graph with parameters v,k,).,µ whereµ= 2. Let x be a fixed vertex, and H = r(x) the graph induced on the neigh-bours of x. Then H has k vertices, is regular of valency A and two nonadjacent vertices have at most one common neighbour. Consequently each edge of His contained in a unique maximal clique, and His the pointgraph of a partial linear space (H,L) which has the property that if y EH, LE Landy i L then y is on at most one line intersecting L. [i.e., we have a partial quad-rangle where lines do not necessarily have constant size.] Looking at the structure of Hone can kill graphs that just escape the · 210 CLAW bound: for example, ifµ= 2 ands= -4 then f = 24). + ).+G so that ). E {0,1,4,8,9,15,24,29,36,64,99,204}. The CLAW bound kills the graphs with ). ~ 24. Here we shall show that also A E {8,9,15} is impossible, thus leaving the parameter sets (v,k,A) = (56,10,0), (99,14,1), (300,26,4). The first one corresponds to the GEWIRTZ graph, the other two are unknown. 1. PARTIAL LINEAR SPACES OF GIRTH 5 In view of the application to strongly regular graphs we shall use k for the number of points and A for the valency (of the pointgraph) of a partial linear space. THEOREM. A connected partial linear space with girth at least 5 and more than one line (lines possibly of varying size) in which every point has A neighbours, contains k ~ ~).().+3) points. PROOF. Let L be a line of size l. Denote by T the set of points at distance at least two from L. Then ITI = k-l().+2--l), and l:;;; A since a line of size ).+1 would be a component. Let x. be the number of points in T having exactly 1 i neighbours at distance one from L. We have (i) Ix. = 1 i k - l().+2-l), 2 .(ii) I ix. ~ .l(A+l-.l) (.l-1), i l. (iii) I i ::;; (}) (A+l-.l) 2 • (2) x. i l. Hence O::;;}: (i-(A+2-.l)) 2x. . l. l. = 2}:C~)x. - (2A+3-2.l)}:ix. + (A+2-'.l) 2 }:x·.-. l. . l. . l. l. l. l. · 2 2 ::;; .l(.l-1) (A+l-.l) - .l(.l-1) (A+l-.l) (2A+3-2.l) + (A+2-.l) (k-.l(A+2-.l)), whence k(A+2-.l) 2 ~ f(A+2-£.) ((A+2) (A+l-£.)+1) which can be written as k ~ ½A (A+3) + (A+2) (A-l) (2£...;.3..;.A) 2(A+2-.l) It follows that if there is a line of size£. with£.~ ½(A+3) then k ~ ~A(A+3), (with strict inequality unless £. = A or £. = ~_(A+3)). If£. is relatively small then we can imporve on estimate (ii). Let m be the size of the longest line intersecting L. We have (ii) I I ix. l. i ~ l(A+l-l) (A+l-m). Hence, evaluating O::;; }:Ci-£.) (i-l-l)x. we find . l. l. k ~ (A+l) 2 - £.A - 2£.(m..;..ll (A+l..;.£.) £.+1 It follows that if the longest line in our partial linear space has length at most ~(A+l) (putting m ::;; £. ::;; ~(A+l)) then k ~ ~(A+l) (A+2) = ~A(A+3)+1. In case the longest line has length ~(A+2), we have to estimate somewhat 3 more carefully. If there is a line L of length l ~ ½A such that each line intersecting L has length at most ½A then k ~ ½~2 + 2A + 1. This shows that for smaller k there are many lines of size _½A + 1 ~ in fact too many. Write IMI = l+sM for each line M. Considering the lines M distinct from L passing through a point x EL we see that xis at distance two from l SM(A-sM) points in T. But l SM= A+ 1 - l, so where n X Let = A(A+l-l) - _(A+l--l) 2 + · l SMSN M#N ~ Cl-1) (A+l-l) + (n -1) (2 (A+l--l)-n ) X · X = l 1 is the number of lines intersecting Lin the point x. n M= 1 for j points of L, so that n ~ 2 for the remaining l-j X X points of L. Then (ii)'' l ix. ~ l(l-1) (A+l-l) + 2(l-j) (A-l). l. In particular, for l = ½A+l we find, evaluating O ~ L(i-l) (i-l+l)x., that l. On the other hand, if for some j E :N each line of size l = ½A+l intersects at least j+l others of this size then considering j(j+l) lines of size l intersecting such lines intersecting a given line L we find ITI = k - l 2 ~ ½j(j+l) (l-1) = \Aj(j+l). This shows that if the linear space contains lines of size l = ½A+l then (*) k ~ max min {½A2 + 3A - 3 - j ¥-, _\Aj(j+l) + \A2 +A+ 1}. O~j~ Hence, for A= 2,4,6,8,10 we find k ~ 5,15,27,46,67 respectively, and r ,.7 in general putting j = vA we find k > ½A_(A+3) for A > 6. This proves the theorem. D REMARK. Equality holds in the theorem iff A= 2 and k = 5. (This partial linear space exists - it is a pentagon.) For: if there is a line of size 4 ½A+ 1 then k > ½A(A+3) unless A= 2 or A= 6. But if A= 6 and k = 27 one sees that each line of size 4 intersects exactly three others and that each point in T has three or four neighbours at distance one of L (where l = 4) - hence lines of size 4 do not intersect in T and !Tl 2 18, k 2 34, contra-diction. Hence there are no lines of size ½A+1, but each line of size at most ½(A+1) intersects a longer line, so there are lines of size ½(\+3) or A. In the former case (l = ½(A+3)) we may suppose A> 3. We see that j = l, i.e., each line of size ½(A+3) intersects only lines of size ½(A+1). Let there to be a lines of size ½(A+3). Then there are ½(A-a) (A+3) points not in one of these lines, and \a(A-1) (A+3) incidences of such points with ½(A+1)-lines. But each point is in at most two ½(A+1)-lines, so (A-a) (A+3) 2 \a(A-1) (A+3) and a$ 3. If a 2 1 then we find a line L with l = ½(A+1) intersecting only one line of size ½(A+3), i.e., with j = 1. From 0 $ 2(i-l-1) 2x. one finds l $ 1, contradiction. 1. Hence there are no lines of size ½(A+3) and all lines have size 2 or A. If some point is only in lines of size 2 then it has A neighbours and A(A-1) points at distance 2 so that k 2 A2+1. On the other hand, if A> 2 and each point is in a line of size A, then let L be a line of size A. Each of its A neighbours is in a line of length A, and these lines cannot inter-sect, so !Tl 2 A(A-1) and k 2 A(A+1). This proves our claim. REMARK. We do not know the right order of magnitude of the lower bound. The theorem gives something of order ½A 2 - on the other hand, the Moore graphs 2 of diameter two are examples with k = A + 1. For small A we have: A= 2, k = 5, the pentagon. A= 3, k = 10, the Petersen graph. A = 4, k = 15: There is a unique partial linear space on 15 points with 10 lines of size 3 and girth 5. Its point-graph is distance regular with parameters i(4,2,1;1,1,4). It is obtained from the generalised quadrangle GQ(2,2) by deleting a parallel class of lines. It is the line graph of the Petersen graph. 5 Now these three examples are regular: all lines have the same size. But for regular partial linear spaces these same methods yield stronger bounds: we have k ;::-: >..(>..-l+2) + 1 if all lines have size l. If A :S: l(l-1) this can be strengthened to 3 k;::,: .t2 (>..-2l+3) + l(l-l) A (In case A -- l(l-1), equality would mean that we have a strongly regular graph withµ= 1 and discriminant (l-1)/s, hence equality occurs only for l = 2. In general equality means that we have a strongly regular graph with µ = 1 in the first case and distance regular graph of diameter at most three in the second case - the linegraph of a system satisfying one bound with equality, satisfies the other with equality. This yields very strong con-ditions on the parameters, and only finitely many examples are known.) An infinite family of regular examples if provided by the incidence graphs of finite projective planes: they have k = 2(>.. 2-1.+1), for A= q+1, q a primepower. Concerning irregular examples there are precisely two others with A = 3 , k = lO name 1 y and For A= 2, k = 5 and A= 4, k = 15 there are no irregular examples. There are no examples with A = 5, k = 21. 6 2. STRONGLY REGULAR GRAPHS WITHµ= 2. As aln~ady remarked in the introduction, if G is any graph such that two nonadjacent vertices have at most two common neighbours then the collection of neighbours of a fixed vertex x 0 carries a partial linear space of girth at least 5. Now if G is moreover regular and edge-regular then if not k ~ ~A(A+3) then by the theorem f(x0 ) is a disjoint union of lines of si:ze A + 1. It follows that (A+1) lk and that G itself is a partial linear space. Thus: COROLLARY • .?l. strongly regular graph withµ= 2 and k < ~A(A+2) is a partial quadrangle; in particular it satisfies the divisibility condition (A+l) lk. This corollary rules out infinitely many feasible sets of parameters of strongly regular graphs e.g. (v,k,A) = (736,42,8), (875,46,9) and (1961,70,15), the three cases mentioned in the introduction. Each of these is the first in an infinite series, e.g. (v,k,A) = (t2 (6t+11) (3t+2), 6t2+10t-2,5t-2) is admissible fort= 1,2,4 or 6 (mod 7) but excluded by the corollary fort~ 2. Lookin,g at the table we found one parameter set that just escaped the bound, namely (v,k, A) = (1944,67, 10). But also this one is easily killed. Returning to the proof of the theorem in the special case k = 67, A= 10 we see that there are no lines of size 7,8 or 9 and that each line of length 6 intersects at least 4 and hence exactly 4 other lines of size 6. If there is a line of size 6 then there are at least 1 + 4 + 12 = 17 such lines, together containing at least (2 + 4.~).17 = 68 points, a contradiction. Hence there are only lines of size 2 or 10 and k ~ A2 + 1 = 101, again a contradiction. Clearly our result can also be applied to other distance regular graphs, but we have no examples at present. strongly 1·egular graphs with ! n=4 k=2 lb=0 mu=2 ! n=16 k = ~, lb=0 mu=2 k = 1 0 lb=5 mu.=E, n=~,6 k=10 lb=0 mu=2 k =c 4~, lb=36 rnu<:E; n=B~, K = 1 4 lb=3 ~) mu=c_ k=?0 1 b= ~, 7 rnu=60 n=99 k = 14 lb=1 - .-l mu-,::. k=34 lb=71 mu=72 t n=243 k=22 lb=1 n,u=2 k=220 lb=199 mu=200 n=300 k=25 16=4 mu=2 k=273 lb=24B mu=2~,2 n=352 k=25 lb=0 mu=2 k =32~, 16=300 ITIU'"'3(l(l n=455 k=3S 16=10 mu=2 n=530 k=37 lb=4 mu=2 k=S92 lb=~,~,6 mu=~i50 :I= / ,:. lj k=37 16=0 -, mu=,::. k=555 16=630 mu=530 n=7::5 k=42 lb=8 . -, mu.= c: . n=87~, k=45 16=9 - -) mu-c n=1175 k=S0 lb=4 mu=2 k=1125 16=1075 mu=1080 n=1275 k=S0 lb=0 .-, mu=.:::. k=122~, 16=1175 mu=1175 n=1525 k=58 16=3 mu=2 k=1565 lb=1509 mu=1512 n=1944 k=57 lb=10 mu=2 n=1951 k=70 16=15 mu=2 \= n <= 2000 and mu=2 out not with the parameters of a net (OAJ complete bipartite graph r=1 s=-3 f=10 g=S CLEBSCH Hub C12-{2.2} vL-5{1} T~O r=2 s=-2 f=S g=l0 Kag{4) vL-'.:;{2J TWO r=2 s=-4 f=35 g=20 GEWIRTZ Hub S Ouasi-stmmetric design r=3 s=-3 f=20 g=35 r=4 s=-3 f=34 g=S0 r=2 s=-5 f=S0 g=34 r=3 :::;=-4 f=54 g=44 <::;pg(3,?,1,2)7 r=3_ s=-4 f=44 g=S4 r=4 :::;=-::, f 00 132 q=ll0 BERLEKAMF'-·'.:,EIDEL-vanLINT- _::;pg(3, 11, I ,2) r=L1 ·:::;=-~, f=11U g=13:2 r=6 s=-4 f=117 g=182 r=3 :::;=-7 f=lE!~: g=11l r=4 s=-5 f=208 g=143 r=S s~-5 f=143 q=208 r=11 s=-3 f=35 ;=360 CLAW r=7 s=-5 f=259 g=370 r=4 s=-8 f=370 g=259 r=S s~-7 f=407 g=296 r=5 s=-6 f=295 g=407 r=10 s=-4 f=20/ g=528 AEB r=11 s=-4 f=230 q=E,44 AEB r=B :=:.=--E, f='.::,00 g=E;7~, '.:::,pg(E,,!IJ,1,2)"i r=S s=-9 f=675 g=S00 r=6 s=-8 f=725 g=SS0 r=7 s=-7 f=550 g=725 r=8 s=-7 f=754 g=870 r=6 s=-9 f=870 g=/54 r=13 s=-5 f=536 q=1407 AN & AEB r=17 s=-q f=370 g=l590 AEB -..J 

Strongly regular graphs where mu equals two and lambda is large

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