Imprimitivity on Distance Regular Graphs
Gabriel Coutinho
2012
Abstract
We characterize the DRGs which are imprimitive. Based on Chapter 4 of BCN [1] and on Godsil
and Hensel [2].
1
Primitivity
Given a distance regular graph G with distance matrices A1 ,...,Ad , we define the distance graphs Gi
to be the graph whose adjacency matrix is Ai .
1.1
Examples
Example 1. We start with the cube G:
G1
G2
G3
G2
G3
Example 2. Consider the icosahedral G:
G1
1
Example 3. The following graph is a generalized hexagon (2,1). It’s the graph obtained from a
particular geometry of points and lines such that every point is in 2 lines and every line has 3 points.
Parameters (4,2,2;1,1,2)
G1
G2
G3
Example 4. The following graph is the Coxeter graph (3,2,2,1; 1,1,1,2) - (its definition is not trivial)
G1
G2
G3
Example 5. Finally, the dodecahedral:
2
G4
G2
1.2
G3
G4
G5
Discussion
The graph G is said to be primitive if Gi is connected for all i, and imprimitive otherwise. Among
the examples above, 3 and 4 are primitive.
Note that if G is bipartite then G2 is disconnected. This was observed in the Example 1. G is said
to be antipodal if Gd is the disjoint union of cliques (examples 1, 2 and 5). As a trivial observation,
observe that distance-regularity (parameter pddd ) implies that the cliques of Gd must be of the same
size.
Theorem 1 (see BCN [1], Chapter 4). If G is an imprimitive DRG of valency at least 3, then G is
either bipartite or antipodal (or both).
Note that a distance regular graph of valency 2 is a cycle. Every cycle is distance regular - even
cycles are both bipartite and antipodal, while odd cycles of non-prime order offer the only examples
of imprimitive DRGs that are neither bipartite nor antipodal.
Proof. By a (i, j, k) triangle, we mean a set of vertices u, v, w such that d(u, v) = i, d(v, w) = j and
d(u, w) = k.
Suppose G is imprimitive and suppose G2 is not connected. If a1 = 0, then the neighbourhood
of each vertex of G is formed by vertices at distance 2, hence it lies in the same component of G2 .
Hence G has at least 2 classes - it must be bipartite. If a1 > 0, suppose that d > 2 and let (u, v, w, x)
be a shortest u − x path. Let z be a common neighbour of u and v. If z is not adjacent to w, then
(u, z, w) form a (1, 2, 2) triangle. If it is, then (v, z, x) form a (1, 2, 2) triangle. In any case, G must
contain a (1, 2, 2) triangle, so p122 6= 0, implying that any two adjacent vertices of G are in the same
component of G2 . This contradicts the fact that G2 is not connected because G is connected. Hence
d = 2, G is strongly regular, and the only imprimitive strongly regular graphs are either complete
multipartite (antipodal) or disconnected.
Now let us suppose that G2 is connected. Let Gr be disconnected for some d > r > 2. Using
an argument similar to what we did above for the case a1 > 0, we can see that ar must be 0 (note
that d > 3 now). Likewise, because we are supposing that G2 is also connected, we see that p2rr = 0.
This implies that br−1 ≤ 1 and also cr+1 ≤ 1. Now br ≤ br−1 ≤ 1 and also cr ≤ cr+1 ≤ 1. Hence
b0 = ar + br + cr ≤ 2.
Finally, if all Gr are connected except for Gd . Again, if there are vertices u, v and w in G such
that d(u, v) = d(u, w) = d but d(v, w) = r < d, then prdd > 0 and every pair of vertices at distance r
in G will be in some (d, d, r) triangle, therefore in the same component of Gd . This contradicts the
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fact that Gr is connected but Gd is not. Thus any two vertices in the same component of Gd are
adjacent and so G is antipodal.
1.3
New graphs - halved graphs and folded graphs
If a DRG G is bipartite, then G2 has two components and the graphs induced on these components
are called the halved graphs of G (they need not to be isomorphic). They will have diameter bd/2c.
In the example given, the halved graphs are the complete graphs K4 . Consider the following example:
Example 6. The Hadamard graph H(4) is constructed as follows. Consider a Hadamard matrix of
+
+
−
−
order 4. Define a graph on 16 vertices {ri± , c±
j }. Define edges ri ∼ cj and ri ∼ cj if hij = 1. Define
+
−
−
+
edges ri ∼ cj and ri ∼ cj if hij = −1.
G2
G3
G4
Note that in this case, the halved graphs are strongly regular of parameters (8, 6, 4, 6). In general,
halved graphs are distance regular and non-bipartite.
If a DRG G is antipodal, then we can construct the folded graph G as the graph whose vertices
are the equivalence classes of G, two of them adjacent if and only if there is an edge in G between
the classes. Observe the hypercube and the folded hypercube:
Example 7.
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In general the folded graph G is not antipodal, unless G has diameter 4 and is bipartite, in which
case G is complete bipartite.
With the two operations above, it is possible in general to obtain primitive distance regular graphs
from imprimitive ones after folding and halving at most once (except for a particular family).
2
Antipodal DRGs of diameter 3
In this section, we will study the imprimitive DRGs of diameter 3. Because bipartite DRGs of
diameter 3 can be easily described in terms of square designs, we will focus on the antipodal case.
Theorem 2 (see Godsil, Hensel [2]). Suppose G is antipodal with diameter d > 2 and intersection
array (b0 , ..., bd−1 , c1 , ..., cd ).
(a) If there is an edge between two given fibres of G, then each vertex in one fibre has a unique
neighbour in the other.
(b) If the distance between two fibres in G is i ≤ d2 , then each vertex in the first is at distance i
from a vertex in the second fibre, and is at distance d − i from every other vertex there.
(c) Let Q be the graph which has the fibres of G as its vertices, two fibres adjacent if and only if
there is one edge between them in G. This is a distance regular graph with intersection array
(b0 , ..., bm−1 , 1, c2 , ..., γcm ), where γ is the size of a fibre if d = 2m and γ = 1 if d = 2m + 1.
(d) Every eigenvalue of Q is also an eigenvalue of G with the same multiplicity.
Proof. Clearly a vertex cannot have two neighbours in a given fibre, otherwise d = 2. Given two
fibres with an edge between them, say uv, consider a vertex w at distance d from u. Observe that
d(w, v) = d − 1, and this is true for all vertices in the same fibre as u. Because of the edge uv and
because of distance regularity, pd−1
1 d ≥ 1. By symmetry of this coefficients, it follows that there must
be a vertex at distance 1 from w and distance d from v, hence w has a (unique) neighbour on the
fibre of v.
The key thing now is that if a vertex u is at distance ` from a vertex v in a fibre V , then there
is a vertex w ∈ V at distance d − ` from u. In fact all the coefficients bj are non-zero, so there is a
path of length d that starts at v, passes at u and ends at w.
Now suppose the minimum distance between two fibres U and V is i, and let u ∈ U and v ∈ V be
at distance i. Clearly the distance between u and any other vertex in V is at least d − i and cannot
be larger, otherwise there would be a vertex in V at distance smaller than i from u. By symmetry
(applying the paragraph above), (b) holds.
Observe that (a) implies that two adjacent fibres in Q are connected in G by a matching. Consider
arbitrary vertices U and V that are at distance i ≤ m in Q. Let u and v be the pair of vertices at
distance i in G belonging to U and V respectively. Observe that there is no vertex at distance m + 1
of U , hence bm = 0. If i < m, observe that each neighbour of v at distance i + 1 from u contributes
to a neighbour of V at distance i + 1 from U because no shortest path in G uses two vertices of the
same fibre. On the other hand, consider a neighbour W of V at distance i + 1 from U . There is
exactly one vertex in W which is a neighbour of v, and this will be at distance i + 1 from u.
Therefore the number of vertices in Q which are neighbours of V and are at distance i + 1 from
U is bi , and this depended only on d(U, V ) = i.
The proof for the coefficients ci with i < d2 is similar - every vertex at distance i − 1 from u and
neighbour from v corresponds to a vertex in Q at distance i − 1 from U and neighbour of V and
vice-versa.
For i = m = d2 , suppose U and V are vertices of Q at distance d/2. Then there will be γ · cm
vertices at distance m − 1 from u and neighbour of some vertex in V . Each of these vertices must be
in a different fibre, because u can be at distance m − 1 to just one vertex in a fibre. Likewise, each
vertex W neighbour of V and at distance m − 1 from U must contain one vertex at distance m − 1
from u.
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Note finally that the adjacency matrix of Q is a multiple of the adjacency matrix of the quotient
graph of G regarding the equitable partition π into fibres. Every eigenvalue of such graph is an
eigenvalue of G, hence (d) follows.
Let G be a graph and suppose that there exists a partition π of the vertices of G into cells such
that:
• Each cell is an independent set.
• Between any two cells either are there no edges or there is a matching.
For example, the line graph of the Petersen Graph:
Define G/π to be the graph whose vertices are cells of π, two vertices adjacent if and only if there
is a matching between the corresponding cells in G. The graph G is said to be covering graph of
G/π. Observe that antipodal distance regular graphs are a particular cases of covering graphs (those
that are distance regular).
If G/π is connected, each cell has the same number of vertices, which is called the index of the
covering. If the index is r, G is called an r-fold covering of G/π.
2.1
Covers of the complete graph
Let G be an antipodal distance regular graph of diameter 3. As we saw, G is a cover of a distance
regular graph of diameter 1, which must be the complete graph. Suppose G is an r-fold cover of Kn .
This implies that G has valency n − 1. We will now determine the intersection array of G. Let u be
a fixed vertex and define Gi (u) as the vertices at distance i from u.
• c3 = n − 1, as every neighbour of a vertex at distance 3 from u must be at distance 2 from u.
• b2 = 1, for the distance between two vertices at distance 3 from u is 3, never 2.
• There are r − 1 vertices at distance 3 from u. Each has n − 1 neighbours in G2 (u), no two share
the same neighbour. Because every vertex in G2 (u) must have a neighbour in the partition of
u, and none of such can be u, we have that:
|G2 (u)| = (r − 1)(n − 1)
• Each vertex in G1 (u) has b1 neighbours in G2 (u), and each vertex in G2 (u) has c2 neighbours
in G1 (u). As |G1 (u)| = n − 1, we have that:
(n − 1)b1 = (n − 1)(r − 1)c2 ⇒ b1 = (r − 1)c2
• The intersection array of G is therefore:
{b0 , b1 , b2 ; c1 , c2 , c3 } = {n − 1, (r − 1)c2 , 1; 1, c2 , n − 1}
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We’ve seen above that once G is said to be an antipodal distance regular graph of diameter 3, its
intersection array is determined by three parameters. The “converse” is also true:
Proposition 1 (see Godsil, Hensel [2]). If G is an r-fold covering of the complete graph Kn and c2
is a positive integer such that two non-adjacent vertices of G from different fibres have c2 common
neighbours, then G is an antipodal distance regular graph of diameter 3 with parameters (n, r, c2 ).
Proof. Start observing that the graph must be regular of valency b0 = n − 1.
Suppose u and v are in the same fibre. Because fibres are independent and edges between two
fibres form a matching, u and v are at least at distance 3. If w is a neighbour of v, then it is in a
different fibre and is not adjacent to u, therefore they have c2 common neighbours. Hence u and v
are in the same fibre in and only if they are at distance 3.
The coefficient c3 = n − 1 is therefore well defined for any pair of vertices.
Moreover, c2 as it was defined on the statement now only applies to vertices at distance 2, hence
it has the standard meaning for distance regular graphs. Clearly c1 = 1.
If u and w are at distance 2, observe that w has a unique neighbour on the fibre of u (because G
is a cover and is a cover of a complete graph). Hence b2 = 1 is well defined. It remains to prove that
b1 is well defined.
Let u and v be neighbours. Then v is at distance 2 from any vertex in the fibre of u, so it shares c2
neighbours with each of them. Moreover, any neighbour of v at distance 2 from u must be adjacent
to one (and only one) vertex in the fibre of u. So there are precisely (r − 1)c2 neighbours of v at
distance 2 from u, and b1 is determined.
References
[1] A. E. Brouwer, A. M. Cohen, and A. Neumaier. Distance-Regular Graphs. Springer-Verlag,
Berlin, 1989.
[2] C. D. Godsil and A. D. Hensel. Distance regular covers of the complete graph. J. Combin. Theory
Ser. B, 56(2):205 – 238, 1992.
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