t-point counts in distance-regular graphs

Arnold Neumaier

Faculty of Mathematics

University of Vienna, Austria

joint work with

Safet PenjicUniversity of Primorska, Slovenia

These slides are available at

http://www.mat.univie.ac.at/~neum/slides/Bled2019.pdf

1

We consider distance regular graphs Γ of diameter d = dΓ with

vΓ vertices and intersection array

i(Γ) = {b0, . . . , bd−1; c1, . . . , cd}.

We write ni for the number of points at distance i from a point,

n := n1 = b0 for the valency, and

ai := n− bi − ci, λ := a1.

Many well-known inequalities for the intersection array can be

derived in a uniform way using t-point counts [∆], normalized

counts of the number of ordered subsets isomorphic to a template

∆ with certain specified distances between the t vertices of ∆.

In this talk I’ll show that by considering t-point counts with t ≤ 6,

the diameter bounds by Ivanov & Ivanov may be derived.

2

A t-point type is an undirected graph ∆ with nodes 1, . . . , t whose

edges are labelled with integers ∈ {0, 1, . . . , dΓ}; it is called

complete if any two nodes are joined by a labelled edge. ∆µν

denotes the label of the edge µν. In drawings, missing labels are

taken as having the value 1.

A configuration is a finite ordered list z̄ = z1 . . . zt of points of Γ.

A configuration z̄ is of type ∆ if

d(zµ, zν) = ∆µν whenever µ ∼ ν in ∆.

The (rational) number [∆] is the number of configurations of

type ∆ divided by vΓ. Clearly,

[∆] ≥ 0 for all types ∆. (1)

3

[•

•

•.........................................................................

.............

.............

.............

.............................................................. i

jk

]= nkp

kij = nip

ijk = njp

jki. (2)

Elimination of distance 0: e.g.,[• •

•

•..................................................

.........................................................

...........................................................................................................................................................................................

...........................................................................................................i l

j m

0 k]

= δilδjm

[•

•

•.........................................................................

.............

.............

.............

.............................................................. i

jk

]= δilδjmnkp

kij . (3)

Elimination of nodes of valency 2: e.g.,[• •

•

•..................................................

.........................................................

...........................................................................................................................................................................................i l

j m

k

]= pklm

[•

•

•.........................................................................

.............

.............

.............

.............................................................. i

jk

]= nkp

kijp

klm. (4)

Sum over a distance: e.g.,∑h

[• •

•

•..................................................

.........................................................

...........................................................................................................................................................................................

...........................................................................................................i l

j m

h k]

=[• •

•

•..................................................

.........................................................

...........................................................................................................................................................................................i l

j m

k

]. (5)

A t-point count is a number [∆], where ∆ is a complete t-point

type. Isomorphic types have the same count. As in the special case

(5), every [∆] can be written as a sum of t-point counts.

4

Proposition 1. All 4-point counts containing two disjoint edges

can be expressed in terms of the rational numbers

ei :=1

nici

[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i

i

i−1i−1

]≥ 0 : (6)

Ai :=[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i

i

ii

]= ni

(a2i − ci(bi−1 − ei)− bi(ci+1 − ei+1)

),

Bi :=[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i−1

i

i−1i

]= nici

(ai−1−(ci−ci−1−ei)

), Ci :=

[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i−1

i+1

ii

]= nicibi,

Di :=[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i−1

i

ii

]= nici(bi−1 − bi − ei), Ei :=

[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i

i

i−1i−1

]= niciei.

Fi :=[ • •

• •................................................................................................................................................................................................................................................................................................................................................................................

i−1

i

i−1i−1

]= nici(ci − ci−1 − ei),

5

Proof. By summing over a distance (see (5) and (4)) we find

ni+1ci+1ci = nicibi = Ci,

niciai = Bi +Di,

niciai−1 = Fi +Bi,

nicibi−1 = Di + Ei + Ci,

nic2i = Ci−1 + Ei + Fi,

nia2i = Di +Ai + Fi+1.

Now (6) implies that Ei = niciei, and solving the resulting

triangular linear system of equations gives the above formulas. �

6

Corollary 1. Let x+ = max(x, 0). Then

bi−1 ≥ bi, ci ≥ ci−1, (Biggs) (7)

ci(ai − ai−1)+ + bi(ai − ai+1)+ ≤ a2i , (Brouwer&Lambeck) (8)

max(

0, ci − ci−1 − ai−1, bi−1 − bi −a2i

ci, ci − ci−1 −

a2i−1

bi

)≤ ei, (9)

ei ≤ min(bi−1 − bi, ci − ci−1), (10)

Proof. (7), (11), (12), and (8) follow from

0 ≤ Di + Ei = nici(bi−1 − bi),

0 ≤ Ei + Fi = nici(ci − ci−1),

Di − Fi = nici(ai − ai−1),

(Di − Fi)+ + (Fi+1 −Di+1)+ ≤ Di + Fi+1 +Ai = nia2i .

(9) follows from Di ≥ 0, Fi ≥ 0, and (10) from Ei, Bi, Ai ≥ 0. �

7

Corollary 2.

bi−1 = bi ⇔ Di = Ei = 0, (11)

ci−1 = ci ⇔ Ei = Fi = 0. (12)

These relations show that Di, Ei and Fi are the most restricted and

hence the most interesting counts among those of Proposition 1.

When the lower bound for ei in (9) is positive, we may apply the

following results.

Theorem 1. We have

ei > 0, cs+1 > min(ci − ci−1, bi−1 − bi) ⇒ ci+s > ci, (13)

ei > 0, i > 1 ⇒ c2i−1 > ci. (14)

ei > 0 ⇒ b1 ≥ bi + ci−1, (15)

ei > 0 ⇒ bi−1 − bi + ci − ci−1 ≥ λ+ 2. (16)

8

Proof. If ei > 0 then by (9),

e := min(ci − ci−1, bi−1 − bi) ≥ ei > 0.

Hence we may choose vwxy consistent with

• • • •.......................................................................................................................................................................................................................................................................................................................................................................... ................

.........................................................................

u v w xi−1

y

i−1 i

.................................................................................................................................................................................i+ s

..............................................................

..............................................................

....................................................

• • • •......................................................................................................................................................................................................................................................................................................................................................................................................................... ................

.........................................................................

u v′ w xi−1

y

ii+s−1

.................................................................................................................................................................................i+ s

................................

................................

.........

.

All pisi+1 choices for u yield d(u, y) = i+ s− 1. Now the number of

v′ ∈ Γ(w)∩ Γs(u) with d(v′, y) = i− 1 is ≤ e. If also cs+1 > e, there

is some v′ with d(v′, y) ≥ i. Now consider uvxy to get ci+s > ci.

Thus (13) holds, and (14) follows by taking s = i− 1 > 0.

9

To derive (15) and (16) we use the new counts

.............

.............

.............

.............

........................................................

........................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................•

•

•

•

•

..................................................................................................................

..................................................................................................................................................................................................................................................

i−1

i

i

i−1j

m

k

=:

Hjk if m = 2,

Kjk if m = 1,(17)

which give

Hii+1 = Eibi, Hi−2i−1 = Eici−1,

Hi−1i−1 +Hi−1i +Hii−1 +Hii = Ei(b1 − bi − ci−1), (18)

Ki−1i−1 +Hi−1i−1 +Kii−1 +Hii−1 = Ei(ci − ci−1 − 1), (19)

Kii−1 +Hii−1 +Kii +Hii = Ei(bi−1 − bi − 1), (20)

Ki−1i−1 +Ki−1i +Kii−1 +Kii = Eiλ. (21)

Swapping the two edges gives

Ki−1i = Kii−1. (22)

Now (15) follows from the nonnegativity of (18), and (16) follows

from (19)+(20)≥(21) which holds in view of (22). �

10

To derive the diameter bound by I & I we first look at certain

4-point counts with a single distance 1 only. We define

Dsi :=

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

i+si

s+1 i−1.......................................................................................................................i+s

]=

[• • •

•................................................................................................................................................

....................................

..................................................................................................................................................................................

i+si+ss+1

i−1 .......................................................................................................................i

],

Esi :=

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

i+s−1i

s+1 i−1.......................................................................................................................i+ s

], Gsi :=

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

i+s−1i−1

s+1 i−1.......................................................................................................................i+ s

].

Then

Dsi + Esi =

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

≤i+si

s+1 i−1.......................................................................................................................i+ s

]= ni+sp

i+si−1s+1(bi−1 − bi+s),

Esi +Gsi =

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

i+s−1≥i−1

s+1 i−1.......................................................................................................................i+ s

]= ni+sp

i+si−1s+1(ci+s − ci−1).

D0i = Di, E0

i = Ei,

11

We also define

F si :=

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

i+s−1i

s+1 i−1.......................................................................................................................i+s−1

]=

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

i+s−1i−1

s+1 i.......................................................................................................................i+s−1

]=

[• • •

•......................................................................................................................................................................................................................................................................................................................................................................

s+1i+s−1 i+s−1

i−1.......................................................................................................................i

],

F 0i = G0

i = Fi,

Proposition 2. Let l > s ≥ τ ≥ 1. If

bl = bl−s > bs+1 − cl−1−s, (23)

cl+1−t − cl−t < ct+1 for t = 0, . . . , τ, (24)

then we have

Dtl−t = F tl+1−t = 0 and al ≤ al−1−t for t = 0, . . . , τ.

Proof. (24) for t = 0 implies cl+1 = cl, hence F 0l+1 = Fl+1 = 0, and

(23) implies bl = bl−1, hence D0l = Dl = 0. We now prove by

induction on t ≤ τ that

Dtl−t = F tl+1−t = 0. (25)

12

Suppose (25) holds for t− 1 in place of t. If F tl+1−t > 0 then we can

choose vwxy consistent with the diagram

..................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................................................................................................................................................................................................• • •

•

•

w x y

v

u

j k

l−t

.............................................................................................................................................................................l + 1− t

l−t

t

...........................................................

..................................................................

.............................................................

t+ 1

.

There are ct+1 choices for u ∈ Γ(w) ∩ Γt(v). If k = l − t then

j ≥ l − t (consider uvw); hence (consider uwxy) this can happen at

most cl+1−t − cl−t times. By (24), this implies that we can choose

u such that k ≥ l − t+ 1. Now if j = l − t then vuxy gives

Dt−1l+1−t > 0, if j = l + 1− t then uwxy gives 0 < El+1−t

contradicting bl+1−t > bl−t (note that t ≥ 1), and if j = l + 2− tthen uxyv gives F t−1

l−t > 0. In all cases, the induction assumption is

violated. Thus we must have F tl+1−t = 0.

13

If Dtl−t > 0 then one can choose uvwy consistent with the diagram

............................................................................................................................................................................................................................................................................................................................................................................................... .....................

..........................................

..........................................

...........................................................................• • • •

•y

u v w xl−1−t.............................................................................................................................................................................

l + 1− t

l l t+1 j

and then choose x ∈ Γ(w) ∩ Γl+1−t(u). Among the bl+1−t = be

choices for x there are at most bs+1 − cl−1−s choices with j = t+ 2.

Hence (23) implies at least one choice with j ≤ t+ 1. But then

uwxy gives F tl+1−t > 0, contradiction. Thus we must have

Dtl−t = 0. This shows that (25) holds generally.

14

Now consider

............................................................................................................................................................................................................................................................................................................................................ ....................

........................................

..........................

• • •

•

v w x

u

t+1 l−1−t..................................................................................................................................................

l

lj

The triangle inequality forbids j = l − 2− t, and j = l − t is

forbidden by Dtl−t = 0. Hence any of the al choices for

u ∈ Γl(v) ∩ Γ(x) must be among the al−1−t choices for

u ∈ Γl−1−t(w) ∩ Γ(x), giving al ≤ al−1−t. �

15

Corollary 3. If ci+s+1 = ci, bi+s = bi > bs+1 − ci−1 then

Di = Dsi = F si+1 = 0 and ai ≤ ai−1.

Proof. The assumption implies (23) and (24) for l = i+ s, τ = s;

moreover ai+s = ai. Hence it suffices by Proposition 2 to show that

Di = 0. Suppose that Di > 0. Extend a configuration of type Di

by a vertex u as indicated below.

............................................................................................................................................................................................................................................................................................................................................... ..................

..........................................................

• • • •

•

u v w x

y

s i−1

i i

.............................................................................................................................................................................i+ s

................................................

................................................

Clearly d(u, y) ∈ {i+ s− 1, i+ s}. But the first choice turns uwxy

into a configuration of type Dsi and the second choice contradicts

ci+s = ci. �

16

Theorem 2. Suppose that bs+1 < bi + ci−1.

(i) If bi−1 > bi and cs+1 > min(ci − ci−1, bi−1 − bi) then ci+s+1 > ci

or bi+s > bi.

(ii) If ai 6= ai−1, bi−1 > ci − ci−1 and

cs+2 > ci − ci−1 + min(bi−1 − bi, ai) then ci+s+1 > ci or bi+s+1 < bi.

Proof. (i) If ei > 0 then Theorem 1 shows that ci+s > ci. Moreover,

ci+s+1 > ci by (7). If ei = 0 the Di > 0 by definition of Di,

contradicting Corollary 3.

(ii) Suppose the conclusion is not valid. Then Corollary 3 implies

ai < ai−1, Di = Dsi = F si+1 = 0. (26)

17

To get a configuration

..............................................................................................................................................................................................................................................................................

........................................................

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................• • • •

• •

u v w x

v′ w′

i−1

i−1i−1 j

k+i−1

i

......................................................................................................................................................................................................................................i+ 1

we can choose uwvxw′v′ for fixed u in

nicibi(ai−1 − ai)(bi−1 − (ci − ci−1)) > 0

ways since w′ ∈ Γ(w) ∩ (Γi−1(v) \ Γi(u)) and

v′ ∈ Γ(v) ∩(

Γi(w′) \ (Γi−1(x) \ Γi−2(w))

); hence the configuration

exists. The triangle inequality for wvv′ and wv′w′ gives j ≤ i and

j ≥ i− 1.

18

Now we define

Oti := Dti + F t+1

i =

[• • •

•................................................................................................................................................

....................................

..................................................................................................................................................................................

i+ti+t≤t+2

i−1 .......................................................................................................................i

]. (27)

If j = i then v′uw′w gives O0i > 0. And if j = i− 1 then v′wx

shows that k = i; then xwv′v′ gives O0i > 0. Thus we always have

O0i > 0, and since pisi+s > 0, a configuration

.......................................................................................................................................................................................................

........................................................

........................................................................................................................ ..................

....................................

....................................

....................................

....................................

...............

• • •

•

•

u v wi−1

≤2 i

x

y

i

s

...............................................................................................................................................................i

.................................................................................................................................................................................................

i+s

exists. Since ci+s = ci we have d(y, w) 6= i+ s− 1, hence

d(y, w) = i+ s, and yuvw gives Osi > 0. Since Dsi = 0 by (26),

hence (27) implies F s+1i > 0.

19

Now we fix uwxy consistent with the configuration

................................................................................................................................................................................................................................................................................................................................................................................................ ..................

..........................................................

• • • •

•

u v w x

y

s+1 i−1

i+si

.......................................................................................................................s+ 2

.............................................................................................................................................................................i+ s

................................

................................

......

,

which exists since F s+1i > 0. Among the cs+2 choices for

v ∈ Γ(w) ∩ Γs+1(u) we can have at most

ci − ci−1 + min(bi−1 − bi, ai) choices with d(v, y) ≤ i and d(v, x) = i

or i+ 1. By assumption on cs+2, there is at least one additional

choice, which therefore must have d(v, y) = i, d(v, x) = i− 1 or

d(v, y) = i+ 1, d(v, x) = i. Now uvxy yields Dsi > 0 in the first case

and F si+1 > 0 in the second case, contradiction. �

20

Corollary 4. (Ivanov & Ivanov)

If s > 0 and d ≥ 2s+ 2 then

bs+1 < bs ⇒ c2s+2 > cs+1 or b2s+1 < bs+1, (28)

bs+1 = bs, cs+1 > cs ⇒ c2s+2 > cs+1 or b2s+2 < bs+1. (29)

Proof. The Taylor–Levingston inequality bs ≥ cs+1 implies the

hypothesis of Theorem 2 for i = s+ 1. Now the first implication

follows from (i), and the second implication from (ii) of the

preceding theorem. �

21

Theorem 3.(Ivanov & Ivanov).

Suppose that cr − br > c1 − b1 for some r > 1. Then the valency n

is bounded by

n ≥ c2ir − b2ir > c1 − b1 + i.

In particular,

d ≤ 22n−2−λr,

Proof. By Corollary 4 and the Biggs inequalities (7),

t > 1, ct − bt > ct−1 − bt−1 ⇒ c2t − b2t > ct − bt.

Hence the first part follows by induction. If d ≥ 22n−2−λr − 1 then

we can choose i = 2n− 2− λ and get a contradiction; this yields

the second part. �

22

Thank you for your attention!

These slides are available at

http://www.mat.univie.ac.at/~neum/slides/Bled2019.pdf

23