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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
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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.
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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)
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[•
•
•.........................................................................
.............
.............
.............
.............................................................. 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.
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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),
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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. �
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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. �
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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)
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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.
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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). �
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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,
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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)
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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.
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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.
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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. �
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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. �
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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)
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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.
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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.
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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. �
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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. �
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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. �
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Thank you for your attention!
These slides are available at
http://www.mat.univie.ac.at/~neum/slides/Bled2019.pdf
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