Also available at http://amc.imfm.si
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 3 (2010) 49–58
On vertex-stabilizers of
bipartite dual polar graphs
Štefko Miklavič
Faculty of Mathematics, Natural Sciences and Information Technologies
University of Primorska, SI-6000 Koper, Slovenia
Received 18 September 2009, accepted 15 December 2009, published online 3 February 2010
Abstract
Let X,Y denote vertices of a bipartite dual polar graph, and let GX and GY denote
the stabilizers of X and Y in the full automorphism group of this graph. In this paper, a
description of the orbits of GX ∩GY in the cases when the distance between X and Y is 1
or 2, is given.
Keywords: Dual polar graphs, automorphism group, quadratic form, isotropic subspace.
Math. Subj. Class.: 05E18, 05E30
1 Preliminaries and introductory remarks
Let q denote a prime power, letGF (q) denote a finite field with q elements, and let d denote
a positive integer. Let V = GF (q)2d denote the vector space overGF (q) of dimension 2d,
consisting of column vectors with entries in GF (q). We define a map Q : V → GF (q) as
follows. For u = (u1, u2, ..., u2d)t ∈ V we let
Q(u) =
d∑
i=1
u2i−1u2i. (1.1)
The form Q is a quadratic form on V , that is, Q(λu) = λ2Q(u) (λ ∈ GF (q), u ∈ V ),
and
f(u, v) = Q(u+ v)−Q(u)−Q(v) (u, v ∈ V ) (1.2)
is a symmetric bilinear form on V . The form Q is usually called hyperbolic quadric. Note
that for vectors u = (u1, u2, ..., u2d)t and v = (v1, v2, ..., v2d)t of V we have
f(u, v) =
d∑
i=1
(u2i−1v2i + u2iv2i−1). (1.3)
E-mail address: stefko.miklavic@upr.si (Štefko Miklavič)
Copyright c© 2010 DMFA Slovenije
50 Ars Math. Contemp. 3 (2010) 49–58
A vector v ∈ V is called isotropic, if Q(v) = 0. A subspace U of V is called isotropic,
if Q(u) = 0 for every u ∈ U , and it is called maximal isotropic, if it is maximal (with
respect to inclusion) in the set of all isotropic subspaces of V . It turns out that the dimension
of every maximal isotropic subspace is d (see, for example, [1, Theorem 3.10] or [10,
Lemma 3]). Observe that if u, v ∈ V belong to the same isotropic subspace of V , than
Q(λu+ µv) = 0 for every λ, µ ∈ GF (q). Furthermore,
f(u, v) = Q(u+ v)−Q(u)−Q(v) = 0. (1.4)
Conversely, if u and v are isotropic with f(u, v) = 0, then 〈u, v〉 is an isotropic subspace
of V . Indeed, for λ, µ ∈ GF (q) we have
Q(λu+ µv) = λ2Q(u) + µ2Q(v) + λµf(u, v) = 0. (1.5)
We now define the dual polar graph Dd(q) on V . The vertex-set V (Dd(q)) of Dd(q) is the
set of all maximal isotropic subspaces of V . Vertices X,Y ∈ V (Dd(q)) are adjacent in
Dd(q) if and only if the dimension ofX ∩Y is d−1. Let ∂ denote the path-length distance
function on Dd(q). It is easy to see that ∂(X,Y ) = i if and only if dim(X ∩ Y ) = d − i
(X,Y ∈ V (Dd(q))). The following facts about Dd(q) can be found, for example, in [2,
Section 9.4]. The graph Dd(q) is bipartite with diameter d and with
∏d−1
i=0 (q
d−i−1 + 1)
vertices. For convenience let
bi = qi
qd−i − 1
q − 1
, ci =
qi − 1
q − 1
and ki =
b0b1 · · · bi−1
c1c2 · · · ci
(1.6)
for 0 ≤ i ≤ d. The graph Dd(q) is regular with valency b0 = k1. For X ∈ V (Dd(q)) and
an integer 0 ≤ i ≤ d we set Si(X) = {Z ∈ V (Dd(q)) | ∂(X,Z) = i}.
Let GL(V ) denote the general linear group of V . Then σ ∈ GL(V ) is called isometry
of V , if Q(σ(v)) = Q(v) for every v ∈ V . It follows from (1.2) that if σ is an isometry of
V , then f(u, v) = f(σ(u), σ(v)) for u, v ∈ V . The group of all isometries of V is called
the orthogonal group for Q, and is denoted by O+2d(q). Note that every σ ∈ O
+
2d(q) acts on
V (Dd(q)) as an automorphism of Dd(q). The full automorphism group G of Dd(q) acts
distance-transitively on V (Dd(q)), that is, for X,Y, Z,W ∈ V (Dd(q)) with ∂(X,Y ) =
∂(Z,W ) there exists σ ∈ G such that σ(X) = Z and σ(Y ) = W (see, for example, [2,
Table 6.1]). Recall that every distance-transitive graph is also distance-regular in the sense
of [2, Section 4.1].
Pick X,Y ∈ V (Dd(q)) and let GX and GY denote the stabilizers of X and Y in
G, respectively. Since G acts distance-transitively on V (Dd(q)), the orbits of GX are
precisely the sets Si(X) (0 ≤ i ≤ d). In this paper we examine the orbits of GX ∩ GY .
These orbits play an important role in the theory of Terwilliger algebras of Dd(q). This
role is especially important in the case when ∂(X,Y ) ∈ {1, 2}, see [6]. For the definition
and more background on Terwilliger algebras of distance-regular graphs see [3, 4, 7, 8, 9].
In this paper we give a description of the orbits of GX ∩GY when ∂(X,Y ) ∈ {1, 2}.
To do this, we consider the following situation for the rest of this paper.
Notation 1.1. Let q denote a prime power, let GF (q) denote a finite field with q elements,
and let d denote a positive integer. Let V = GF (q)2d denote the vector space over GF (q)
of dimension 2d, consisting of column vectors with entries in GF (q). Let Q and f be
as defined in (1.1) and (1.2). Let Dd(q) denotes the bipartite dual polar graph over V ,
Š. Miklavič: On vertex-stabilizers of bipartite dual polar graphs 51
and let bi, ci and ki be as in (1.6). Fix X,Y ∈ V (Dd(q)). For 0 ≤ i, j ≤ d let Dij =
Dij(X,Y ) = Si(X) ∩ Sj(Y ). Let GX and GY denote the stabilizers of X and Y in the
full automorphism group G of Dd(q).
Our paper is organised as follows. In Section 2 we state some results about maximal
isotropic subspaces that we need later. In Section 3 (Section 4, respectively) we describe
the orbits of GX ∩ GY in the case when ∂(X,Y ) = 1 (∂(X,Y ) = 2, respectively). In
what follows we use the same symbols (capital letters) for the vertices of Dd(q) and for the
maximal isotropic subspaces of V ; this should cause no confusion.
2 Maximal isotropic subspaces
In this section we state some results about maximal isotropic subspaces of V that we need
later. The first one is known as Witt’s lemma (see, for example, [1, Theorem 3.9]).
Lemma 2.1. With reference to Notation 1.1, let U and W be subspaces of V , and let
σU : U → W be a bijective linear map satisfying Q(σU (u)) = Q(u) for every u ∈ U .
Then there is an isometry of V which extends σU .
Lemma 2.2. With reference to Notation 1.1, let U and W be maximal isotropic subspaces
of V with dim(U ∩W ) = d − i for some 1 ≤ i ≤ d. Pick linearly independent vectors
u1, . . . , ui ∈ U \W and linearly independent vectors w1, . . . , wi ∈ W \ U . Let F be the
i× i matrix with (j, `)-entry equal to f(uj , w`). Then the determinant of F is nonzero.
Proof. First note that F is a nonzero matrix. Namely, if f(uj , w`) = 0 for every 1 ≤
j, ` ≤ i, then a subspace generated by U and W is isotropic subspace of dimension d + i,
a contradiction. Suppose now that det(F ) = 0. Then the columns of F are linearly
dependent vectors of GF (q)i, that is, there exist scalars λj (1 ≤ j ≤ i) which are not all
equal to zero, such that for each 1 ≤ ` ≤ i we have
0 = λ1f(u`, w1)+λ2f(u`, w2)+ · · ·+λif(u`, wi) = f(u`, λ1w1 +λ2w2 + · · ·+λiwi).
Note that w = λ1w1 + λ2w2 + · · · + λiwi is nonzero, since w1, w2, . . . , wi are lin-
early independent. Multiplying the above equation with an arbitrary scalar µ` gives us
µ`f(u`, w) = 0. Adding the obtained equations we get
i∑
`=1
µ`f(u`, w) = f(µ1u1 + µ2u2 + · · ·+ µiui, w) = 0.
This implies that f(u,w) = 0 for every u ∈ U . By (1.5), the subspace generated by U and
w is isotropic with dimension d+ 1, a contradiction. Therefore, det(F ) 6= 0.
Lemma 2.3. With reference to Notation 1.1, let U,U1,W and W1 be maximal isotropic
subspaces of V with dim(U ∩W ) = dim(U1 ∩W1) = d − i for some 1 ≤ i ≤ d. Let
u1, u2, . . . , ud be a basis of U such that ui+1, . . . , ud is a basis of U∩W . Letw1, . . . , wi ∈
W be such that w1, . . . , wi, ui+1, . . . , ud is a basis of W . Let v1, v2, . . . , vd be a basis
of U1 such that vi+1, . . . , vd is a basis of U1 ∩ W1. Let z1, . . . , zi ∈ W1 be such that
z1, . . . , zi, vi+1, . . . , vd is a basis of W1. Then there exists an isometry σ of V , such that
σ(uj) = vj (1 ≤ j ≤ d) and σ(wj) ∈ 〈z1, . . . , zi〉 (1 ≤ j ≤ i).
52 Ars Math. Contemp. 3 (2010) 49–58
Proof. We first define a bijective linear map σ from a subspace generated by U and W to
a subspace generated by U1 and W1, such that σ(uj) = vj (1 ≤ j ≤ d) and σ(wj) ∈
〈z1, . . . , zi〉 (1 ≤ j ≤ i). We will then show that σ extends to an isometry of V . We
now define σ(wj) (1 ≤ j ≤ i). Let F denote an i × i matrix with (j, `)-entry equal to
f(vj , z`). For 1 ≤ ` ≤ i consider the following system of linear equations in variables
α`1, α
`
2, . . . , α
`
i :
F (α`1, α
`
2, . . . , α
`
i)
t = (f(u1, w`), f(u2, w`), . . . , f(ui, w`))t. (2.1)
Note that this system has a unique solution since F is nonsingular by Lemma 2.2. For
convenience, we denote the solutions of this system also by α`1, α
`
2, . . . , α
`
i . For 1 ≤ ` ≤ i
we let
σ(w`) = α`1z1 + α
`
2z2 + · · ·+ α`izi. (2.2)
We extend σ to a linear map from 〈U,W 〉 to 〈U1,W1〉 in a natural way:
σ(λ1u1 + · · ·+ λdud + µ1w1 + · · ·+ µiwi) =
λ1σ(u1) + · · ·+ λdσ(ud) + µ1σ(w1) + · · ·+ µiσ(wi)
for λ1, . . . , λd, µ1, . . . , µi ∈ GF (q).
We now show that σ is a bijection. To do this, it is enough to show that σ(w`) (1 ≤ ` ≤
i) are linearly independent. Let A be an i× i matrix with (j, `)-entry equal to α`j . Observe
that σ(w`) (1 ≤ ` ≤ i) are linearly independent if and only if A is nonsingular. Let F1
denote an i × i matrix with (j, `)-entry equal to f(uj , w`). The matrix F1 is nonsingular
by Lemma 2.2. Furthermore, it follows from (2.1) that F · A = F1, implying that A is
nonsingular.
We now show that σ preserves Q. Pick arbitrary v ∈ 〈U,W 〉:
v =
d∑
j=1
αjuj +
i∑
j=1
βjwj .
By (1.2) and (1.4),
Q(v) =
i∑
r=1
i∑
s=1
αrβsf(ur, ws).
Let us now compute Q(σ(v)). By (1.2) and (1.4) we first get
Q(σ(v)) =
i∑
r=1
i∑
s=1
αrβsf(σ(ur), σ(ws)).
By (2.2) and since σ(ur) = vr we further find
f(σ(ur), σ(ws)) = f(vr, αs1z1 + · · ·+ αsi zi) = αs1f(vr, z1) + · · ·+ αsi f(vr, zi).
Finally, by (2.1), the above expression is equal to f(ur, ws). Therefore, Q(v) = Q(σ(v)).
By Lemma 2.1 there exists an isometry σ of V which extends σ. This completes the
proof.
Š. Miklavič: On vertex-stabilizers of bipartite dual polar graphs 53
Lemma 2.4. With reference to Notation 1.1, let U be a (d − 1)-dimensional isotropic
subspace of V . Then U is contained in exactly two maximal isotropic subspaces of V .
Proof. By [2, p. 274], the number of isotropic k-dimensional subspaces of V containing a
given isotropic (k − 1)-dimensional subspace of V is (qd−k+1 − 1)(qd−k + 1)/(q − 1).
The result follows.
3 The case ∂(X, Y ) = 1
With reference to Notation 1.1, in this section we describe the orbits of GX ∩ GY when
∂(X,Y ) = 1. We first determine the size of the Dij (0 ≤ i, j ≤ d).
Lemma 3.1. With reference to Notation 1.1 assume that ∂(X,Y ) = 1. Then the following
(i), (ii) hold.
(i) |Dii−1| = |D
i−1
i | = ciki/b0 (1 ≤ i ≤ d).
(ii) Dij = ∅ if |i− j| 6= 1 (0 ≤ i, j ≤ d).
Proof. (i) This follows from [5, Lemma 4.1(i)].
(ii) By the triangle inequality we find Dij = ∅ if |i − j| ≥ 2. Since Dd(q) is bipartite, we
also have Dii = ∅.
Lemma 3.2. With reference to Notation 1.1 assume that ∂(X,Y ) = 1. Pick u ∈ X \ Y
and v ∈ Y \X . Then f(u, v) 6= 0. In particular, u and v are not contained in a common
isotropic subspace.
Proof. Suppose on the contrary that f(u, v) = 0. Pick λ, µ ∈ GF (q) and w ∈ X ∩ Y .
Consider λu+ w + µv ∈ 〈X,Y 〉. By (1.2) and (1.4) we have
Q(λu+w+ µv) = Q(λu+w) +Q(µv) + f(λu+w, µv) = λµf(u, v) + µf(w, v) = 0.
This shows that 〈X,Y 〉 is an isotropic subspace of dimension d+ 1, a contradiction.
Theorem 3.3. With reference to Notation 1.1 assume that ∂(X,Y ) = 1. Then the following
(i), (ii) hold for 1 ≤ i ≤ d.
(i) For every Z,Z ′ ∈ Di−1i there exists σ ∈ GX ∩GY which maps Z to Z ′.
(ii) For every Z,Z ′ ∈ Dii−1 there exists σ ∈ GX ∩GY which maps Z to Z ′.
Proof. (i) If i = 1 then the result is clear. Assume now that i ≥ 2. Since dim(X ∩ Z) =
d− i+ 1 and dim(Y ∩ Z) = d− i, it follows from Lemma 3.2 that X ∩ Y ∩ Z = Y ∩ Z
with dim(X ∩ Y ∩ Z) = d− i, and X ∩ Z = 〈X ∩ Y ∩ Z, u〉 for some u ∈ X \ Y . Pick
w ∈ Y \ X . Let v1, . . . , vd−1 be a basis of X ∩ Y , such that vi, . . . , vd−1 is a basis of
X ∩ Y ∩ Z. Let z1, . . . , zi−1 ∈ Z be such that u, vi, . . . , vd−1, z1, . . . , zi−1 is a basis of
Z. Note that u, v1, . . . , vd−1 is a basis of X and that w, v1, . . . , vd−1 is a basis of Y .
Similarly as above, let u′ ∈ X \ Y be such that X ∩ Z ′ = 〈X ∩ Y ∩ Z ′, u′〉. Let
v′1, . . . , v
′
d−1 be a basis of X ∩ Y , such that v′i, . . . , v′d−1 is a basis for X ∩ Y ∩ Z ′. Let
z′1, . . . , z
′
i−1 ∈ Z ′ be such that u′, v′i, . . . , v′d−1, z′1, . . . , z′i−1 is a basis for Z ′. Observe that
u′, v′1, . . . , v
′
d−1 is a basis for X and that w, v
′
1, . . . , v
′
d−1 is a basis for Y .
Applying Lemma 2.3 (with U = X = 〈u, v1, . . . , vd−1〉, W = Z = 〈u, vi, . . . ,
vd−1, z1, . . . , zi−1〉, U1 = X = 〈u′, v′1, . . . , v′d−1〉 and W1 = Z ′ = 〈u′, v′i, . . . , v′d−1,
54 Ars Math. Contemp. 3 (2010) 49–58
z′1, . . . , z
′
i−1〉) we find that there exists an isometry σ such that σ(u) = u′, σ(vj) = v′j (1 ≤
j ≤ d − 1), and σ(zj) ∈ 〈z′1, . . . , z′i−1〉 (1 ≤ j ≤ i − 1). Clearly, σ preserves X (and
thus also X ∩ Y ), and maps Z to Z ′. To finish the proof we have to show that σ preserves
Y . Observe that X ∩ Y is a (d− 1)-dimensional isotropic subspace of V . By Lemma 2.4,
the only two maximal isotropic subspaces containing X ∩ Y are X and Y . Since X and
X ∩ Y are both preserved by σ, also Y is preserved by σ.
(ii) Similar as (i) above.
Proposition 3.4. With reference to Notation 1.1 assume that ∂(X,Y ) = 1. Then the
following (i), (ii) hold.
(i) The set Di−1i (1 ≤ i ≤ d) is an orbit of GX ∩GY .
(ii) The set Dii−1 (1 ≤ i ≤ d) is an orbit of GX ∩GY .
Proof. It is clear that two vertices from different sets from (i) and (ii) above could not be
in the same orbit of GX ∩GY . The result now follows from Theorem 3.3.
4 The case ∂(X, Y ) = 2
With reference to Notation 1.1, in this section we describe the orbits of GX ∩ GY when
∂(X,Y ) = 2. We first determine the size of the sets Dij (0 ≤ i, j ≤ d). The proposition
below follows from [5, Lemma 4.1(ii)–(iv)].
Proposition 4.1. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. Then the
following (i)–(iv) hold.
(i) |Dii−2| = |D
i−2
i | = kici−1ci/(b0b1) (2 ≤ i ≤ d);
(ii) |D00| = 0 and |Dii| = ki(ci(bi−1 − 1) + bi(ci+1 − 1))/(b0b1) (1 ≤ i ≤ d− 1);
(iii) |Ddd| = kd(bd−1 − 1)/b1;
(iv) |Dij | = 0 if |i− j| 6∈ {0, 2} (0 ≤ i, j ≤ d).
Lemma 4.2. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. Then the following
(i), (ii) hold.
(i) Let u1, u2 ∈ X \ Y be linearly independent, and let w ∈ Y \X . Then u1, u2 and w
are not contained in a common isotropic subspace of V .
(ii) Let w1, w2 ∈ Y \X be linearly independent, and let u ∈ X \ Y . Then w1, w2 and
u are not contained in a common isotropic subspace of V .
Proof. (i) Suppose on contrary that u1, u2 and w are contained in a common isotropic
subspace. Pick λ1, λ2, µ ∈ GF (q) and v ∈ X ∩ Y . Consider λ1u1 + λ2u2 + v + µw ∈
〈X,w〉. By (1.2) and (1.4) we have
Q(λ1u1 +λ2v2 +v+µw) = Q(λ1u1 +λ2u2 +v)+Q(µw)+f(λ1u1 +λ2u2 +v, µw) =
λ1µf(u1, w) + λ2µf(u2, w) + µf(v, w) = 0.
Therefore, 〈X,w〉 is an isotropic subspace of dimension d+ 1, a contradiction.
(ii) Similar as (i) above.
Š. Miklavič: On vertex-stabilizers of bipartite dual polar graphs 55
Theorem 4.3. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. Then the following
(i), (ii) hold for 2 ≤ i ≤ d.
(i) For every Z,Z ′ ∈ Di−2i there exists σ ∈ GX ∩GY which maps Z to Z ′.
(ii) For every Z,Z ′ ∈ Dii−2 there exists σ ∈ GX ∩GY which maps Z to Z ′.
Proof. (i) Note that the result is clear if i = 2. Namely, for i = 2 we have Z = Z ′ = X .
Assume now i ≥ 3. By Lemma 4.2, there exists a basis v1, . . . vd−2 of X ∩ Y , vectors
u1, u2 ∈ X , vectors w1, w2 ∈ Y , and vectors z1, . . . zi−2 ∈ Z, such that vi−1, . . . , vd−2
is a basis of X ∩ Y ∩ Z, u1, u2, v1, . . . vd−2 is a basis of X , w1, w2, v1, . . . vd−2 is a basis
of Y , and u1, u2, vi−1, . . . vd−2, z1, . . . , zi−2 is a basis of Z. Without loss of generality we
can assume that f(u1, w1) = 0 (otherwise we replace w1 by w1 + λw2 for an appropriate
λ ∈ GF (q)). This implies that 〈X ∩ Y, u1, w1〉 is maximal isotropic subspace.
Similarly, there exists a basis v′1, . . . v
′
d−2 of X ∩ Y , vectors u′1, u′2 ∈ X and vectors
z′1, . . . z
′
i−2 ∈ Z ′, such that v′i−1, . . . , v′d−2 is a basis of X ∩ Y ∩Z ′, u′1, u′2, v′1, . . . v′d−2 is
a basis of X , w1, w2, v′1, . . . v
′
d−2 is a basis of Y , and u
′
1, u
′
2, v
′
i−1, . . . v
′
d−2, z
′
1, . . . , z
′
i−2 is
a basis of Z ′. Without loss of generality we can assume that f(u′1, w1) = 0 (otherwise we
replace u′1 by u
′
1 + λu
′
2 for an appropriate λ ∈ GF (q)). This implies that 〈X ∩ Y, u′1, w1〉
is maximal isotropic subspace.
Applying Lemma 2.3 (with U = 〈u1, w1, v1, . . . , vd−2〉, U1 = 〈u′1, w1, v′1, . . . , v′d−2〉,
W = Z and W1 = Z ′) we find that there exists an isometry σ of V , such that σ(u1) = u′1,
σ(w1) = w1, σ(vj) = v′j for 1 ≤ j ≤ d − 2, and σ(u2), σ(zj) ∈ 〈u′2, z′1, . . . , z′i−2〉 (1 ≤
j ≤ i−2). Clearly, σ maps Z to Z ′. It remains to show that σ preservesX and Y . Consider
the subspace W = 〈X ∩ Y, u1〉. Note that W is a (d− 1)-dimensional isotropic subspace
of V . By Lemma 2.4, the only two maximal isotropic subspaces containing W are X
and 〈W,w1〉. Isometry σ maps W to W ′ = 〈X ∩ Y, u′1〉. By Lemma 2.4, the only two
maximal isotropic subspaces containing W ′ are X and 〈W ′, w1〉. Since σ maps 〈W,w1〉
to 〈W ′, w1〉, it must map X to X . Similarly we show that σ maps Y to Y . It follows that
σ ∈ GX ∩GY , completing the proof of (i).
(ii) Similarly as (i) above.
Let us now consider the sets Dii (1 ≤ i ≤ d). Pick Z ∈ Dii . By Lemma 4.2, two
essentially different situations can occur: either dim(X ∩ Y ∩ Z) = d − i (and therefore
X ∩ Z = Y ∩ Z = X ∩ Y ∩ Z), or dim(X ∩ Y ∩ Z) = d − i − 1 (and therefore
X ∩ Z 6= Y ∩ Z).
Definition 4.4. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. LetZ ∈ Dii (1 ≤
i ≤ d). We say Z is of positive (negative, respectively) type, whenever dim(X ∩Y ∩Z) =
d− i (dim(X ∩ Y ∩ Z) = d− i− 1, respectively).
Observe that all vertices of D11 are of negative type, and that all vertices of D
d
d are of
positive type. Moreover, every Dii (2 ≤ i ≤ d− 1) is a disjoint union of the set of vertices
of Dii of positive type, and the set of vertices of D
i
i of negative type.
Remark 4.5. In [6], the definition of the vertices of positive (negative, respectively) type
is different from Definition 4.4 above. Namely, Z ∈ Dii is defined to be of positive type,
whenever all vertices in D11 are at distance i−1 from Z. On the other hand, Z is defined to
be of negative type, if there exists a vertex in D11 which is at distance i− 1 from Z, and all
other vertices inD11 are at distance i+1 from Z. However, these definitions are equivalent.
56 Ars Math. Contemp. 3 (2010) 49–58
If dim(X ∩ Y ∩ Z) = d − i, then Z is at distance at most i from every vertex in D11 . By
the triangle inequality and since Dd(q) is bipartite, Z is at distance i− 1 from every vertex
of D11 . On the other hand, if dim(X ∩ Y ∩ Z) = d − i − 1, then pick u ∈ (X ∩ Z) \ Y
and v ∈ (Y ∩Z) \X . Then W = 〈X ∩ Y, u, v〉 is a vertex of Dd(q), which belongs to D11
and is at distance i− 1 from Z. Furthermore, all other vertices in D11 are at distance i+ 1
from Z.
Lemma 4.6. ([6, Theorem 5.3(iv),(v) and Proposition 6.3]) With reference to Notation 1.1
assume that ∂(X,Y ) = 2. Then the following (i), (ii) hold for 2 ≤ i ≤ d− 1.
(i) |{z ∈ Dii | z is of positive type}| = ki(q − 1)cici−1/(b0b1);
(ii) |{z ∈ Dii | z is of negative type}| = kibicic2/(b0b1).
Theorem 4.7. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. Let Z,Z ′ ∈
Dii (1 ≤ i ≤ d−1) and assume Z,Z ′ are of negative type. Then there exists σ ∈ GX ∩GY
which maps Z to Z ′.
Proof. Let v1, . . . , vd−2 be a basis of X ∩ Y such that vi, . . . , vd−2 is a basis of X ∩
Y ∩ Z. Let u1 ∈ X and w1 ∈ Y be such that u1, vi, . . . , vd−2 is a basis of X ∩ Z and
such that w1, vi, . . . , vd−2 is a basis of Y ∩ Z. Let u2 ∈ X and w2 ∈ Y be such that
u1, u2, v1, . . . , vd−2 is a basis of X and such that w1, w2, v1, . . . , vd−2 is a basis of Y .
Finally, let z1, . . . , zi−1 ∈ Z be such that u1, w1, z1, . . . , zi−1, vi, . . . , vd−2 is a basis of Z.
Similarly, let v′1, . . . , v
′
d−2 be a basis of X ∩ Y such that v′i, . . . , v′d−2 is a basis of
X ∩ Y ∩ Z ′. Let u′1 ∈ X and w′1 ∈ Y be such that u′1, v′i, . . . , v′d−2 is a basis of X ∩ Z ′
and such that w′1, v
′
i, . . . , v
′
d−2 is a basis of Y ∩ Z ′. Let u′2 ∈ X and w′2 ∈ Y be such
that u′1, u
′
2, v
′
1, . . . , v
′
d−2 is a basis of X and such that w
′
1, w
′
2, v
′
1, . . . , v
′
d−2 is a basis of Y .
Finally, let z′1, . . . , z
′
i−1 ∈ Z ′ be such that u′1, w′1, z′1, . . . , z′i−1, v′i, . . . , v′d−2 is a basis of
Z ′.
Applying Lemma 2.3 (with U = 〈u1, w1, v1, . . . , vd−2〉, U1 = 〈u′1, w′1, v′1, . . . , v′d−2〉,
W = Z and W1 = Z ′) we find that there exists an isometry σ such that σ(u1) = u′1,
σ(w1) = w′1, σ(vj) = v
′
j (1 ≤ j ≤ d− 2), and σ(zj) ∈ 〈z′1, . . . , z′i−1〉 for 1 ≤ j ≤ i− 1.
Clearly, σ maps Z to Z ′. It remains to show that σ preserves X and Y . Note that W =
〈X ∩Y, u1〉 is a (d−1)-dimensional isotropic subspace of V . By Lemma 2.4, the only two
maximal isotropic subspaces containing W are X and 〈W,w1〉. Note that σ maps W to
W ′ = 〈X ∩ Y, u′1〉, which is a (d− 1)-dimensional isotropic subspace of V . The only two
maximal isotropic subspaces containing W ′ are X and 〈W ′, w′1〉. Since σ maps 〈W,w1〉
to 〈W ′, w′1〉, it must map X to X . Similarly we show that σ maps Y to Y . Therefore
σ ∈ GX ∩GY and the proof is completed.
Theorem 4.8. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. Let Z,Z ′ ∈
Dii (2 ≤ i ≤ d) and assume Z,Z ′ are of positive type. Then there exist σ ∈ GX ∩ GY
which maps Z to Z ′.
Proof. Let v1, . . . , vd−2 be a basis of X ∩ Y such that vi−1, . . . , vd−2 is a basis of X ∩
Y ∩ Z. Let u1, u2 ∈ X and w1, w2 ∈ Y be such that u1, u2, v1, . . . , vd−2 is a basis of X
and w1, w2, v1, . . . , vd−2 is a basis of Y . Without loss of generality we can assume that
f(u1, w1) = 0 (otherwise we replacew1 byw1+λw2 for an appropriate λ ∈ GF (q)). Note
that 〈X∩Y, u1, w1〉 ∈ D11 . Since Z is of positive type we have dim(〈X∩Y, u1, w1〉∩Z) =
d− i+1. Therefore, there exist α, β ∈ GF (q) and v ∈ X ∩Y such that 〈X ∩Y, u1, w1〉∩
Š. Miklavič: On vertex-stabilizers of bipartite dual polar graphs 57
Z = 〈αu1 + βw1 + v, vi−1, . . . , vd−2〉. Since dim(X ∩ Z) = dim(Y ∩ Z) = d − i, we
have α 6= 0 and β 6= 0. Without loss of generality we can therefore assume that 〈X ∩
Y, u1, w1〉 ∩Z = 〈u1 +w1, vi−1, . . . , vd−2〉 (otherwise we replace u1 by αu1 + v and w1
by βw1). Finally, let z1, . . . , zi−1 ∈ Z be such that z1, . . . , zi−1, u1 + w1, vi−1, . . . , vd−2
is a basis of Z.
Similarly, Let v′1, . . . , v
′
d−2 be a basis of X ∩ Y such that v′i−1, . . . , v′d−2 is a basis
of X ∩ Y ∩ Z ′. Let u′1, u′2 ∈ X and w′1, w′2 ∈ Y be such that u′1, u′2, v′1, . . . , v′d−2 is a
basis of X and w′1, w
′
2, v
′
1, . . . , v
′
d−2 is a basis of Y . Without loss of generality we can
assume that f(u′1, w
′
1) = 0 and that 〈X ∩ Y, u′1, w′1〉 ∩ Z ′ = 〈u′1 + w′1, v′i−1, . . . , v′d−2〉.
Let z′1, . . . , z
′
i−1 ∈ Z ′ be such that z′1, . . . , z′i−1, u′1 + w′1, v′i−1, . . . , v′d−2 is a basis of Z ′.
Applying Lemma 2.3 (with U = 〈u1, u1 +w1, v1, . . . , vd−2〉, W = Z, U1 = 〈u′1, u′1 +
w′1, v
′
1, . . . , v
′
d−2〉 and W1 = Z ′) we find that there exists an isometry σ of V such that
σ(u1) = u′1, σ(u1 + w1) = u
′
1 + w
′
1 (and therefore also σ(w1) = w
′
1), σ(vj) = v
′
j (1 ≤
j ≤ d − 2), and σ(zj) ∈ 〈z′1, . . . , z′i−1〉 for 1 ≤ j ≤ i − 1. Clearly, σ maps Z to Z ′. It
remains to show σ preserves X and Y .
Note thatW = 〈X∩Y, u1〉 is a (d−1)-dimensional isotropic subspace of V . By Lemma
2.4, the only two maximal isotropic subspaces containing W are X and 〈W,w1〉. Note that
σ maps W to W ′ = 〈X ∩ Y, u′1〉, which is a (d− 1)-dimensional isotropic subspace of V .
The only two maximal isotropic subspaces containing W ′ are X and 〈W ′, w′1〉. Since σ
maps 〈W,w1〉 to 〈W ′, w′1〉, it must map X to X . Similarly we show that σ maps Y to Y .
Therefore σ ∈ GX ∩GY and the proof is complete.
Proposition 4.9. With reference to Notation 1.1 assume that ∂(X,Y ) = 2. Then the
following (i)–(iii) hold.
(i) Each of D11 , D
d
d is an orbit of GX ∩GY .
(ii) For 2 ≤ i ≤ d the sets Dii−2 and D
i−2
i are orbits of GX ∩GY .
(iii) For 2 ≤ i ≤ d − 1 the set of vertices in Dii that are of positive type (resp. negative
type) is an orbit of GX ∩GY .
Proof. Observe that two vertices of Dd(q), which are contained in distinct sets listed in (i),
(ii) and (iii) above, cannot belong to the same orbit of GX ∩ GY . The result now follows
from Theorems 4.3, 4.7 and 4.8.
5 Acknowledgement
The author would like to thank to Klavdija Kutnar and Pablo Spiga for many valuable
suggestions.
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