We will improve the upper bound ofc2 in (4.1.10). We need the following lemma.
Lemma 4.2.1. Let Γ denote a distance-regular graph which has classical parameters (D, b, α, β), where D ≥ 3. Assume the intersection numbers a1 = 0 and a2 = 0. Let Ω be a weak-geodetically closed subgraph of diameter 2 in Γ. Let r > s denote the nontrivial eigenvalues of the strongly regular graph Ω. Then the following (i), (ii) hold:
(i) The multiplicity of r is
f = (bα − 1)(bα − 1 − α)(bα − 1 + α)
(α − 1)(α + 1) . (4.2.1)
(ii) The multiplicity of s is
g = −b(bα − 1)(bα − 2)
(α − 1)(α + 1) . (4.2.2)
Proof. Let v=|Ω| and k be the valency of Ω. Note that c2(Ω) = (1 +b)(1 + α) by (2.4.1),k(Ω) = (1 + b)(1 − αb) by (4.1.4), and v = (1 + b)(bα − 2)(bα − 1 − α)/(1 + α) by (4.1.8). Now (4.2.1) and (4.2.2) follow from (2.1.3) and (2.1.4).
Corollary 4.2.2. Let Γ denote a distance-regular graph which has classical parameters (D, b, α, β), where D ≥ 3. Assume Γ has intersection numbers a1 = 0 anda2 = 0. Then
b(b + 1)2(b + 2)
c2 (4.2.3)
and are integers since f, g, b and ρ are integers.
Proposition 4.2.3. Let Γ denote a distance-regular graph which has classical parame-ters (D, b, α, β), where D ≥ 3. Assume Γ has intersection numbers a1 = 0 and a2 = 0.
Then c2 ≤ b(b + 1).
Proof. Recall c2 ≤ b2+b + 2 by (4.1.10). First, suppose
c2 =b2+b + 2. (4.2.5)
Then the integral condition (4.2.3) becomes b2+ 3b + −4b
b2+b + 2. (4.2.6)
Since 0 < −4b < b2 +b + 2 for b ≤ −5, we have −4 ≤ b ≤ −2. For b = −4 or −3, expression (4.2.6) is not an integer. The remaining case b=−2 implies α = −5 by (4.1.6),v = 28 by (4.1.8) and g = 6 by (4.2.2). This contradicts to v ≤ 12g(g + 3) [22,
The results come from Corollary 4.2.2 and Proposition 4.2.3.
Chapter 5
3-bounded Property
Let Γ denote a distance-regular graph which has classical parameters (D, b, α, β) and D ≥ 3. Assume the intersection numbers a1 = 0 and a2 = 0. Note that Γ contains no parallelograms of any length by Theorem 3.2.1. We have known that Γ is 2-bounded.
We shall prove that Γ is 3-bounded in this chapter.
5.1 Weak-geodetically Closed with respect to a Ver-tex
First we give a definition.
Definition 5.1.1. For any vertex x ∈ X and any subset C ⊆ X, define
[x, C] := {v ∈ X | there exists z ∈ C, such that ∂(x, v) + ∂(v, z) = ∂(x, z)}.
Throughout this section, fix two vertices x, y ∈ X with ∂(x, y) = 3. Set C := {z ∈ Γ3(x) | B(x, y) = B(x, z)}
and
Δ = [x, C]. (5.1.1)
We shall prove Δ is a regular weak-geodetically closed subgraph of diameter 3. Note that the diameter of Δ is at least 3. If D = 3 then C = Γ3(x) and Δ = Γ is clearly a regular weak-geodetically closed graph. Thereafter we assume D ≥ 4. By referring to Theorem 2.2.4, we shall prove Δ is weak-geodetically closed with respect tox, and the subgraph induced on Δ is regular with valencya3+c3.
Lemma 5.1.2. For adjacent vertices z, z ∈ Γi(x), where i ≤ D, we have B(x, z) = B(x, z).
Proof. By symmetry, it suffices to show B(x, z) ⊆ B(x, z). Suppose contradictory there exists w ∈ B(x, z) \ B(x, z). Then ∂(w, z) = i + 1. Note that ∂(w, z) ≤
∂(w, x) + ∂(x, z) = 1 +i and ∂(w, z)≥ ∂(w, z) − ∂(z, z) =i. This implies ∂(w, z) =i and wxzz forms a parallelogram of length i + 1, a contradiction.
It is known that Γ is 2-bound by Theorem 2.2.7. For two vertices z, s in Γ with
∂(z, s) = 2, let Ω(z, s) denote the regular weak-geodetically closed subgraph containing z, s of diameter 2.
Lemma 5.1.3. Suppose stuzw is a pentagon in Γ, where s, u ∈ Γ3(x) and z ∈ Γ2(x).
Pick v ∈ B(x, u). Then ∂(v, s) = 2.
Proof. Suppose contradictory ∂(v, s) = 2. Note ∂(z, s) = 1, since a1 = 0. Note that z, w, s, t, u ∈ Ω(z, s). Then s ∈ Ω(z, s) ∩ Γ2(v) and u ∈ Ω(z, s) ∩ Γ4(v) = ∅. Hence
∂(v, z) = ∂(v, s) + ∂(s, z) = 2 + 2 = 4 by Theorem 2.2.8. A contradiction occurs since
∂(v, x) = 1 and ∂(x, z) = 2.
Lemma 5.1.4. Suppose stuzw is a pentagon in Γ, where s, u ∈ Γ3(x) and z ∈ Γ2(x).
Then B(x, s) = B(x, u).
Proof. Since |B(x, s)| = |B(x, u)| = b3, it suffices to show B(x, u) ⊆ B(x, s).
By Lemma 5.1.3,
B(x, u) ⊆ Γ3(s) ∪ Γ4(s).
Suppose
|B(x, u) ∩ Γ3(s)| = m,
|B(x, u) ∩ Γ4(s)| = n.
Then
m + n = b3. (5.1.2)
By Theorem 2.3.1, by Theorem 2.2.8, which contradicts to∂(x, s) = 3. Taking the inner product of s with both side of (5.1.3) and evaluating the result using (2.3.13), we have
mθ∗3+nθ4∗− b3θ∗3 =b3θ1∗− θ∗4
θ0∗− θ∗3(θ3∗− θ∗2). (5.1.4) Solve (5.1.2) and (5.1.4) to obtain
n = b3(θ2∗− θ∗3)
Lemma 5.1.6. The subgraph Δ is weak-geodetically closed with respect to x.
Proof. Clearly C(z, x) ⊆ Δ for any z ∈ Δ. It suffices to show A(z, x) ⊆ Δ for any
The result comes immediately from Lemma 5.1.5.
5.2 3-bounded Property
Theorem 5.2.1. Let Γ denote a distance-regular graph which has classical parameters (D, b, α, β) and D ≥ 3. Assume the intersection numbers a1 = 0 and a2 = 0. Then Γ is 3-bounded.
Proof. By Theorem 2.2.4 and Lemma 5.1.6, it suffices to show that Δ defined in (5.1.1) is regular with valencya3+c3. Clearly from the construction and Lemma 5.1.6,|Γ1(z)∩
Δ| = a3+c3for anyz ∈ C. First we show |Γ1(x)∩Δ| = a3+c3. Note thaty ∈ Δ∩Γ3(x) by construction of Δ. For any z ∈ C(x, y) ∪ A(x, y),
∂(x, z) + ∂(z, y) ≤ ∂(x, y) + 1.
This impliesz ∈ Δ by Definition 2.2.3 and Lemma 5.1.6. Hence C(x, y) ∪ A(x, y) ⊆ Δ.
Suppose B(x, y) ∩ Δ = ∅. Choose t ∈ B(x, y) ∩ Δ. Then there exists y ∈ Γ3(x) ∩ Δ such that t ∈ C(x, y). Note thatB(x, y) = B(x, y). This leads to a contradiction to t ∈ C(x, y). Hence B(x, y) ∩ Δ = ∅ and Γ1(x) ∩ Δ = C(x, y) ∪ A(x, y). Then we have
|Γ1(x) ∩ Δ| = a3+c3.
Since each vertex in Δ appears in a sequence of vertices x = x0, x1, x2, x3 in Δ, where ∂(x, xj) =j and ∂(xj−1, xj) = 1 for 1≤ j ≤ 3, it suffices to show
|Γ1(xi)∩ Δ| = a3+c3 (5.2.1)
for 1≤ i ≤ 2. For each integer 0 ≤ i ≤ 2, we show
|Γ1(xi)\ Δ| ≤ |Γ1(xi+1)\ Δ|
by counting the number of pairs (s, z) for s ∈ Γ1(xi)\Δ, z ∈ Γ1(xi+1)\Δ and ∂(s, z) = 2 in two ways. For a fixed z ∈ Γ1(xi+1)\ Δ, we have ∂(x, z) = i + 2 by Lemma 5.1.6, so ∂(xi, z) = 2 and s ∈ A(xi, z). Hence the number of such pairs (s, z) is at most
|Γ1(xi+1)\ Δ|a2.
On the other hand, we show this number is exactly |Γ1(xi)\ Δ|a2. Fix an s ∈ Γ1(xi)\ Δ. Observe ∂(x, s) = i + 1 by Lemma 5.1.6. Observe ∂(xi+1, s) = 2 since a1 = 0. Pick any z ∈ A(xi+1, s). We shall prove z ∈ Δ. Suppose contradictory z ∈ Δ in the following arguments and choose any w ∈ C(s, z).
Case 1: i = 0.
Observe ∂(x, z) = 2, ∂(x, s) = 1 and ∂(x, w) = 2. This forces s ∈ Δ by Lemma 5.1.6, a contradiction.
Case 2: i = 1.
Observe ∂(x, z) = 3, otherwise z ∈ Ω(x, x2) and this implies s ∈ Ω(x, x2) ⊆ Δ by Lemma 2.2.5 and Lemma 5.1.6, a contradiction. This also implies s ∈ Δ by Definition 2.2.3 and Lemma 5.1.6, a contradiction.
Case 3: i = 2.
Observe ∂(x, z) = 2 or 3. Suppose ∂(x, z) = 2. Then B(x, x3) =B(x, s) by Lemma 5.1.4 (with x3 = u, x2 = t). Hence s ∈ Δ, a contradiction. So z ∈ Γ3(x). Note
∂(x, w) = 2, 3, otherwise s ∈ Δ by Lemma 5.1.4 and Lemma 5.1.6 respectively. Hence
∂(x, w) = 4. Then by applying Ω = Ω(x2, w) in Theorem 2.2.8 we have ∂(x2, z) = 1, a contradiction to a1 = 0.
From the above counting, we have
|Γ1(xi)\ Δ|a2 ≤ |Γ1(xi+1)\ Δ|a2 (5.2.2)
for 0≤ i ≤ 2. Eliminating a2 from (5.2.2), we find
|Γ1(xi)\ Δ| ≤ |Γ1(xi+1)\ Δ|, (5.2.3)
or equivalently
|Γ1(xi)∩ Δ| ≥ |Γ1(xi+1)∩ Δ| (5.2.4) for 0≤ i ≤ 2. We have known previously |Γ1(x0)∩ Δ| = |Γ1(x3)∩ Δ| = a3+c3. Hence (5.2.1) follows from (5.2.4).
Remark 5.2.2. The 3-bounded property is enough to obtain the main result of this thesis. The 4-bounded property seems to be much harder to prove.
Chapter 6
A Constant Bound of c 2
Let Γ = (X, R) be a distance-regular graph which has classical parameters (D, b, α, β) with D ≥ 3. Assume the intersection numbers a1 = 0 and a2 = 0. We shall show that c2 ≤ 2, and if c2 = 1 then (b, α, β) = (−2, −2, ((−2)D+1− 1)/3).
6.1 Preliminary Lemmas
Let Γ = (X, R) be a distance-regular graph with diameter D ≥ 3 and intersection numbers ai, ci, bi for 0≤ i ≤ D. Assume that Γ is D-bounded. By Theorem 2.2.9, for any x, y ∈ X with ∂(x, y) = t, there exists a unique weak-geodetically closed subgraph Δ(x, y) containing x, y of diameter t, and Δ(x, y) is a distance-regular graph with the intersection numbers
ai(Δ(x, y)) = ai, (6.1.1)
ci(Δ(x, y)) = ci, (6.1.2)
bi(Δ(x, y)) = bi− bt (6.1.3)
for 0 ≤ i ≤ t by Theorem 2.2.4 and (2.1.1). In particular, Δ(x, y) is a clique of size 1 +b0− b1 =a1+ 2 when t = 1.
Lemma 6.1.1. [27, Lemma 4.10] Let Γ denote a distance-regular graph which has clas-sical parameters (D, b, α, β). Let Δ denote a regular weak-geodetically closed subgraph
of Γ. Then Δ is a distance-regular graph which has classical parameters (t, b, α, β),
Proof. By Theorem 2.2.4, Δ is distance-regular with intersection numbers ci(Δ) = ci =
Lemma 6.1.2. Let Γ = (X, R) denote a D-bounded distance-regular graph with D ≥ 3.
Let Λ be a weak-geodetically closed subgraph of Γ with diameters, where 0 ≤ s ≤ D−1.
Suppose x, y ∈ Λ with ∂(x, y) = s. Then the following (i)-(iii) hold.
(i) For any w ∈ X, let M(w) = {m − {w} | m ⊆ X is a clique of size a1 + 2 containing w}. Then M(w) is a partition of Γ1(w) with |M(w)| = b0
a1+ 1. (ii) If z ∈ B(y, x), then Δ(x, z) ⊇ Λ and Δ(x, z) has diameter s + 1.
(iii) If Δ is a weak-geodetically closed subgraph of Γ with diameter s + 1 and contains Λ, then Δ = Δ(x, z) for some z ∈ B(y, x).
Proof. Note that Λ = Δ(x, y) by Theorem 2.2.9.
(i) The 1-bounded property implies each edge is contained in a clique of sizea1+ 2.
Since there are b0 edges in Γ containing a fixed vertex w, we have (i).
(ii) Note that Δ(x, z) ∩ Λ is a weak-geodetically closed subgraph of Γ and y ∈ Δ(x, z) ∩ Λ since y ∈ C(z, x). This implies the diameter of Δ(x, z) ∩ Λ is s and we have Δ(x, z) ∩ Λ = Λ by Theorem 2.2.9. Hence Δ(x, z) ⊇ Λ. The diameter of Δ(x, z) is s + 1 since ∂(x, z) = s + 1.
(iii) Suppose that Δ is a weak-geodetically closed subgraph of Γ with diameters+1 and contains Λ. Note thatx, y ∈ Δ. Choose z ∈ Δ and z ∈ B(y, x). Then Δ = Δ(x, z) by (ii).
Lemma 6.1.3. Let Γ denote a D-bounded distance-regular graph with D ≥ 3. Let Λ, Λ be two weak-geodetically closed subgraphs of Γ with diameter s, s + 3 respectively and Λ ⊆ Λ, where 0 ≤ s ≤ D − 3. Let P and B be the sets of weak-geodetically closed subgraphs of Λ which contain Λ, with diameter s + 1 and s + 2 respectively. Let I = {(p, B) | p ∈ P, B ∈ B, and p ⊆ B}. Then (P, B, I) is a 2-(v, κ, 1) design, where
Simplifying (6.1.4) by (6.1.3) we have
|P| = bs(Λ)
bs(Ω) = bs− bs+3 bs− bs+1. Fix Δ∈ B. Using the same technique as above, there are
bs− bs+2
bs− bs+1
distinct elements ofP incident with Δ. Note that the number is independent of choice of Δ.
Fix any distinct Ω, Ω∈ P. Pick z ∈ B(y, x) ∩ Ω. Then Ω = Δ(x, z) by Theorem 6.1.2. Pick w ∈ Ω1(x) − Ω. Note that w ∈ B(x, z). Then Δ(w, z) ∈ B containing Ω and Ω. Suppose that Δ ∈ B is another block incident with Ω and Ω. Observe
that Ω, Ω⊆ Δ(w, z) ∩ Δ ⊆ Δ(w, z). This implies that the diameter of Δ(w, z) ∩ Δ
6.2 An Application of 3-bounded Property
Let Γ = (X, R) be a distance-regular graph which has classical parameters (D, b, α, β) with D ≥ 3. Suppose the intersection numbers a1 = 0 and a2 = 0. Then α < 0 and b < −1 by Lemma 3.1.2. Now we are ready to prove the main theorem of this chapter.
Theorem 6.2.1. Let Γ denote a distance-regular graph which has classical parameters (D, b, α, β) and D ≥ 3. Assume the intersection numbers a1 = 0 and a2 = 0. Then c2 ≤ 2.
Proof. It was shown in Theorem 5.2.1 that Γ is 3-bounded. Fix a vertex x ∈ X and a weak-geodetically closed subgraph Δ containing x of diameter 3. By (6.1.1)-(6.1.3), and Lemma 6.1.1 we find a1(Δ) = 0 and Δ has classical parameters (3, b, α, β) where
by applyinga1(Δ) = 0 to (2.4.4). LetP denote the set of all maximal cliques containing x in Δ, and B be the set of all weak-geodetically closed subgraphs of diameter 2 containing x in Δ. Let I = {(p, B) | p ∈ P, B ∈ B, and p ⊆ B}. Then (P,B, I) is a
by (2.4.2) and (6.2.1). Applying (6.2.2), (6.2.3), and Corollary 2.5.3 to the design, we have
(1 +b)(1 − αb − αb2− α)
(1− αb − α) ≥ (1 + b)(1 − αb). (6.2.4) Note that
(1− αb − α) = b1(Δ)− b2(Δ)
b < 0 (6.2.5)
sinceb1(Δ)−b2(Δ) > 0 by Theorem 2.2.10 and b < −1. By (6.2.4), (6.2.5), and b < −1 we have
(1− αb − αb2− α) ≥ (1 − αb)(1 − αb − α). (6.2.6) Simplifying (6.2.6) we have
αb(αb + α + b − 1) ≤ 0. (6.2.7)
Observe that αb > 0 since α < 0 and b < −1. Then
αb + α + b − 1 ≤ 0. (6.2.8)
Note that αb + α + b − 1 = c2− 2 by (2.4.1) and hence c2 ≤ 2.
For the case c2 = 1, we have the following result.
Theorem 6.2.2. Let Γ denote a distance-regular graph which has classical parameters (D, b, α, β) and D ≥ 3. Assume the intersection numbers a1 = 0, a2 = 0 and c2 = 1.
Then (b, α, β) = (−2, −2, ((−2)D+1− 1)/3).
Proof. Substituting a1 = 0 and c2 = 1 into (2.4.4), (2.4.1), and (2.4.3) we have
α = −b
1 +b, (6.2.9)
β = bD+1− 1
b2− 1 . (6.2.10)
Let Ω ⊂ Δ be two weak-geodetically closed subgraphs of Γ with diameters 2 and 3 respectively. Note that Ω is a strongly regular graph with a1(Ω) = 0, c2(Ω) = 1 by (6.1.1) and (6.1.2). Substituting this into (2.1.1) and (2.1.2) we have
|Ω| = 1 + k1(Ω) +k2(Ω) = 1 + (b0(Ω))2. (6.2.11)
Hence we have
b0(Ω) = 2, 3, 7, 57 (6.2.12)
by Lemma 2.1.2. Note that
b0(Ω) =b0 − b2 = 1 +b + b2 (6.2.13)
by (6.1.3), (2.1.13), (6.2.9), and (6.2.10). Solving (6.2.12) with (6.2.13) for integer b < −1 we have b = −2, −3, or −8. By (2.1.2), (6.1.2), and (6.1.3) we have
k3(Δ) = (b0− b3)(b1− b3)(b2− b3)
c1c2c3 . (6.2.14)
Evaluating (6.2.14) using (2.4.1)-(2.4.3), (6.2.9), and (6.2.10) we find k3(Δ) = b3(b2+ 1)(b2+b + 1)(b3+b2+ 2b + 1)
1− b . (6.2.15)
The number k3(Δ) is not an integer when b = −3 or −8. Hence b = −2 and α = −2, β = ((−2)D+1− 1)/3 by (6.2.9) and (6.2.10) respectively.
Example 6.2.3. [9] Hermitian forms graphs Her2(D) are the distance-regular graphs which have classical parameters (D, b, α, β) with b = −2, α = −3, and β = −(−2)D−1, which have a1 = 0, a2 = 0, and c2 = (1 +α)(b + 1) = 2. This is the only known class of examples that satisfies the assumptions of Theorem 6.2.1 with c2 = 2.
Example 6.2.4. [22, p. 237] Gewirtz graph is the distance-regular graph with inter-section numbers a1 = 0, a2 = 8, and c2 = 2, which has classical parameters (D, b, α, β) with D = 2, b = −3, α = −2, and β = −5. It is still open if there exists a class of distance-regular graphs which have classical parameters (D, −3, −2, (−1 − (−3)D)/2) for D ≥ 3.
Example 6.2.5. [3, Table 6.1] Witt graph M23 is the distance-regular graph which has classical parameters (D, b, α, β) with D = 3, b = −2, α = −2, and β = 5, which has a1 = 0, a2 = 2, and c2 = 1. This is the only known example that satisfies the assumptions of Theorem 6.2.1 withc2 = 1.
For summary, we list the parameters in the following table.
name a1 a2 c2 D b α β
Petersen graph 0 2 1 2 −2 −2 −3
Witt graph M23 0 2 1 3 −2 −2 5
?? 0 2 1 D ≥ 4 −2 −2 (−2)D+13 −1
Hermitian forms graph Her2(D) 0 3 2 D −2 −3 −((−2)D+ 1)
Gewirtz graph 0 8 2 2 −3 −2 −5
?? 0 8 2 D ≥ 3 −3 −2 −1−(−3)2 D
We close our thesis with two conjectures.
Conjecture 6.2.6. (With graph M23 does not grow.) There is no distance-regular graph which has classical parameters (D, −2, −2, (−2)D+1− 1
3 ) with D ≥ 4.
Conjecture 6.2.7. (Gewirtz graph does not grow.) There is no distance-regular graph which has classical parameters (D, −3, −2, −1 + (−3)D
2 ) with D ≥ 3.
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