A construction for infinite families of semisymmetric graphs revealing their full automorphism group

Abstract

We give a general construction leading to different non-isomorphic families $\varGamma_{n,q}(\mathcal{K})$ of connected q-regular semisymmetric graphs of order 2q ⁿ⁺¹ embedded in $\operatorname{PG}(n+1,q)$, for a prime power q=p ^h, using the linear representation of a particular point set $\mathcal{K}$ of size q contained in a hyperplane of $\operatorname{PG}(n+1,q)$. We show that, when $\mathcal{K}$ is a normal rational curve with one point removed, the graphs $\varGamma_{n,q}(\mathcal{K})$ are isomorphic to the graphs constructed for q=p ^h in Lazebnik and Viglione (J. Graph Theory 41, 249–258, 2002) and to the graphs constructed for q prime in Du et al. (Eur. J. Comb. 24, 897–902, 2003). These graphs were known to be semisymmetric but their full automorphism group was up to now unknown. For q≥n+3 or q=p=n+2, n≥2, we obtain their full automorphism group from our construction by showing that, for an arc $\mathcal{K}$, every automorphism of $\varGamma_{n,q}(\mathcal{K})$ is induced by a collineation of the ambient space $\operatorname{PG}(n+1,q)$. We also give some other examples of semisymmetric graphs $\varGamma _{n,q}(\mathcal{K})$ for which not every automorphism is induced by a collineation of their ambient space.

1 Introduction

In the following, all graphs are assumed to be finite and simple, i.e. they are undirected graphs which contain no loops or multiple edges.

Definition

We say that a graph is vertex-transitive if its automorphism group acts transitively on the vertices. Similarly, a graph is edge-transitive if its automorphism group acts transitively on the edges. A graph is semisymmetric if it is regular and edge-transitive but not vertex-transitive (see [10]).

One can easily prove that a semisymmetric graph must be bipartite with equal partition sizes. Moreover, the automorphism group must be transitive on both partition sets. General constructions of semisymmetric graphs are quite rare.

We construct several infinite families $\varGamma_{n,q}(\mathcal{K})$ of semisymmetric graphs using affine points and some selected lines of a projective space $\operatorname{PG}(n+1,q)$. In Sect. 5 we show that the infinite series of semisymmetric graphs given in [19] is exactly one of the families that we construct in this paper; the graphs in [9] are shown to be part of the same series. Using our construction, in many cases, the structure of the full automorphism group of the graphs $\varGamma_{n,q}(\mathcal{K})$ can be clarified (at least for q≥n+3 and q=p=n+2). This structure was not given in [9, 19] where only part of the automorphism group is constructed, enough to show edge-transitivity.

2 Construction and properties of the graph $\varGamma _{n,q}(\mathcal{K} )$

Let $\operatorname{PG}(n,q)$ denote the n-dimensional projective space over the finite field $\mathbb{F}_{q}$, q=p ^h. Throughout this paper we assume n≥2.

We wish to emphasise the distinction we will make between a subspace and a subgeometry. A subspace of $\operatorname {PG}(n,q)$ is a projective space $\operatorname{PG}(m,q)$ contained in $\operatorname {PG}(n,q)$, m≤n, over the same finite field $\mathbb{F}_{q}$. An n-dimensional subgeometry of $\operatorname{PG} (n,q)$ is a projective space $\operatorname{PG}(n,q_{0})$ contained in $\operatorname{PG}(n,q)$ over the finite field $\mathbb{F}_{q_{0}}$ where $\mathbb{F}_{q_{0}} \subseteq \mathbb{F}_{q}$.

Definition

Let $\mathcal{K}$ be a point set in $H_{\infty}=\operatorname{PG}(n,q)$ and embed H _∞ in $\operatorname{PG}(n+1,q)$. The linear representation $T_{n}^{*}(\mathcal{K})$ of $\mathcal{K}$ is a point-line incidence structure with natural incidence, point set $\mathcal{P}$ and line set ${\mathcal{L}}$ as follows:

$\mathcal{P}$::

affine points of $\operatorname {PG}(n+1,q)$ (i.e. the points of $\operatorname{PG}(n+1,q)\setminus H_{\infty}$),

${\mathcal{L}}$::

lines of $\operatorname{PG}(n+1,q)$ through a point of $\mathcal{K}$, but not lying in H _∞.

For more information on linear representations of geometries, we refer to [8].

Definition

We denote the point-line incidence graph of $T_{n}^{*}(\mathcal{K} )$ by $\varGamma_{n,q}(\mathcal{K})$, i.e. the bipartite graph with classes $\mathcal{P}$ and ${\mathcal{L}}$ and adjacency corresponding to the natural incidence of the geometry.

Whenever we consider the incidence graph $\varGamma_{n,q}(\mathcal{K})$ of some linear representation $T_{n}^{*}(\mathcal{K})$ of $\mathcal{K}$, we still regard the set of vertices as a set of points and lines in $\operatorname{PG}(n+1,q)$. In this way we can use the inherited properties of this space and borrow expressions such as the span of points, a subspace, incidence, and others.

We define the closure of a set of points in $\operatorname{PG}(n,q)$ as follows:

Definition

If a point set S contains a frame of $\operatorname{PG}(n,q)$, then its closure $\overline{S}$ consists of the points of the smallest n-dimensional subgeometry of $\operatorname{PG}(n,q)$ containing all points of S.

The closure $\overline{S}$ of a point set S can be constructed recursively as follows:

• (i) determine the set $\mathcal{A}$ of all subspaces of $\operatorname{PG}(n,q)$ spanned by an arbitrary number of points of S;

• (ii) determine the set $\overline{S}$ of points P for which there exist two subspaces in $\mathcal{A}$ that intersect only at P, if $\overline{S} \neq S$ replace S by $\overline{S}$ and go to (i), otherwise stop.

This construction corresponds to the definition of a closure of a set of points in a projective plane in [16, Chap. XI]. Here the authors show that if S is contained in a projective plane and contains a quadrangle, the points of $\overline{S}$ form the smallest subplane containing all points of S.

Result 2.1

[7, Corollary 4.3]

The graph $\varGamma_{n,q}(\mathcal{K})$ is connected if and only if the span $\langle \mathcal{K}\rangle$ has dimension n.

Remark

Suppose the set $\mathcal{K}$ spans a t-dimensional subspace $\operatorname{PG}(t,q)$ of $H_{\infty}=\operatorname{PG}(n,q)$, t<n. One can check that in this case the graph $\varGamma_{n,q}(\mathcal{K})$ is a non-connected graph with q ^n−t connected components, where each component is isomorphic to the graph $\varGamma _{t,q}(\mathcal{K})$. This explains why we will only consider graphs $\varGamma _{n,q}(\mathcal{K})$ with set $\mathcal{K}$ such that $\langle \mathcal{K}\rangle= H_{\infty}$.

Throughout this paper, we use the following theorems of [4].

Result 2.2

[4]

Let $|\mathcal{K}|\neq q$ or let $\mathcal{K}$ be a set of q points of H _∞ such that every point of $H_{\infty}\backslash\mathcal{K}$ lies on at least one tangent line to $\mathcal{K}$. Suppose α is an isomorphism between $\varGamma _{n,q}(\mathcal{K})$ and $\varGamma_{n,q}(\mathcal{K}')$, for some set $\mathcal{K}'$ in H _∞, then α stabilises $\mathcal{P}$.

Result 2.3

[4]

Let q>2. Let $\mathcal{K}$ and $\mathcal{K}'$ be sets of q points such that $\overline {\mathcal{K}}$ is equal to H _∞ and such that every point of H _∞ lies on at least one tangent line to $\mathcal{K}$. Consider an isomorphism α between $\varGamma_{n,q}(\mathcal{K})$ and $\varGamma _{n,q}(\mathcal{K}')$. Then α is induced by an element of the stabiliser $\mathrm{P}\Gamma \mathrm{L} (n+2,q)_{H_{\infty}}$ of H _∞ mapping $\mathcal{K}$ onto $\mathcal{K}'$.

Result 2.4

[4]

Let q>2 and let $\mathcal{K}$ be a set of q points such that $\overline{\mathcal{K}}$ is equal to H _∞ and such that every point of H _∞ lies on at least one tangent line to $\mathcal{K}$. Then $\operatorname {Aut}(\varGamma_{n,q}(\mathcal{K})) \cong \mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$.

From Result 2.2 we now easily deduce the following corollary.

Corollary 2.5

If $\mathcal{K}$ is a set of q points of H _∞ such that every point of H _∞ lies on at least one tangent line to $\mathcal{K}$, then $\varGamma _{n,q}(\mathcal{K})$ is not vertex-transitive.

Recall that if a group G has a normal subgroup N and the quotient G/N is isomorphic to some group H, we say that G is an extension of N by H. This is written as G=N.H.

An extension G=N.H which is a semidirect product is also called a split extension. This means that one can find a subgroup $\overline{H}\cong H$ in G such that $G = N\overline{H}$ and $N \cap\overline{H} = \{e_{G} \}$ and is denoted by G=N⋊H.

A perspectivity is an element of PΓL(n+2,q) which fixes a hyperplane of $\operatorname{PG}(n+1,q)$ pointwise, this hyperplane is called the axis. Every perspectivity also has a centre, i.e. a point such that every line through it is stabilised. If this centre belongs to the axis, such a perspectivity is called an elation.

The subgroup of PΓL(n+2,q) consisting of all perspectivities with axis H _∞ is written as $\operatorname{Persp}(H_{\infty})$. A perspectivity ϕ is uniquely determined by its axis and centre and one ordered pair (P,ϕ(P)) for a point P different from the centre and not on the axis. Hence, one can easily count $\mathopen{|}\operatorname{Persp} (H_{\infty})|= q^{n+1}(q-1)$.

Result 2.6

[4]

Let $\mathcal{K}$ be a point set spanning $H_{\infty}=\operatorname {PG}(n,q)$. If the setwise stabilisers $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ and $\operatorname{PGL}(n+1,q)_{\mathcal{K}}$, respectively, fix a point of H _∞, then $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}\cong\operatorname {Persp}(H_{\infty})\rtimes\mathrm{P}\Gamma\mathrm{L} (n+1,q)_{\mathcal{K}}$ and $\operatorname{PGL}(n+2,q)_{\mathcal{K}}\cong\operatorname {Persp}(H_{\infty})\rtimes\operatorname{PGL}(n+1,q)_{\mathcal{K}}$, respectively.

Result 2.7

[4]

Let $\mathcal{K}$ be a point set spanning $H_{\infty}=\operatorname {PG}(n,q)$, q=p ^h. If the setwise stabiliser $\operatorname{PGL}(n+1,q)_{\mathcal{K}}$ fixes a point of H _∞, and $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}\cong \operatorname{PGL}(n+1,q)_{\mathcal{K}}\rtimes\operatorname {Aut}(\mathbb{F}_{q_{0}})$, for $q_{0}=p^{h_{0}}$, h ₀|h or $\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}}\cong\operatorname{PGL} (n+1,q)_{\mathcal{K}}$, then $\mathrm{P}\Gamma\mathrm {L}(n+2,q)_{\mathcal{K}}\cong\operatorname{Persp}(H_{\infty})\rtimes \mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$.

The following theorem is easy to prove. We will use it to show the edge-transitivity of the constructed graphs.

Theorem 2.8

If the stabiliser $\mathrm{P}\Gamma \mathrm{L}(n+1,q)_{\mathcal{K}}$ of $\mathcal{K}$ in the full collineation group of H _∞ acts transitively on the points of $\mathcal{K}$, then $\varGamma_{n,q}(\mathcal{K})$ is an edge-transitive graph.

Proof

Consider two edges (R _i ,L _i ), i=1,2, where $R_{i}\in\mathcal{P}$, $L_{i} \in{\mathcal{L}}$, R _i ∈L _i , we will construct a mapping from one edge to the other. Let P _i be L _i ∩H _∞. Since $\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}}$ acts transitively on $\mathcal{K}$, we may take an element β of $\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}}$ such that β(P ₁)=P ₂. This element extends to an element β′ of $(\mathrm {P}\Gamma\mathrm{L} (n+2,q)_{H_{\infty}})_{\mathcal{K}}$ mapping P ₁ onto P ₂.

If β′(R ₁)=R ₂, then β′(L ₁)=L ₂, hence the statement follows. If β′(R ₁)≠R ₂, then let S be the point at infinity of the line β′(R ₁)R ₂. There is a (unique) elation γ with centre S and axis H _∞ mapping β′(R ₁) to R ₂. This elation maps β′(L ₁) onto L ₂. Since γ∘β′ is an element of $(\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{H_{\infty}})_{\mathcal{K}}$ mapping (R ₁,L ₁) onto (R ₂,L ₂), the statement follows. □

The main goal of this paper is the construction of infinite families of semisymmetric graphs. The results of [4] introduced in this section will enable us in some cases to explicitly describe the automorphism group of the constructed graphs. Note that, since a semisymmetric graph is regular, any graph $\varGamma_{n,q}(\mathcal{K})$ that is semisymmetric, necessarily has $|\mathcal{K}|=q$. For this reason, we will investigate point sets of size q in $\operatorname{PG}(n,q)$. Considering Theorem 2.8 we need point sets $\mathcal{K}$ of H _∞ such that $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ acts transitively on the points of $\mathcal{K}$.

In Sect. 3, considering Result 2.4, we will look for point sets $\mathcal{K}$ such that the closure $\overline {\mathcal{K}}$ is equal to H _∞. In Sect. 4 we will look for point sets $\mathcal{K}$ spanning H _∞ such that the closure $\overline {\mathcal{K} }$ is equal to a subgeometry of H _∞.

We give a brief overview of all constructions to come. We use the abbreviation NRC for a normal rational curve, for its definition see Sect. 3. When the size of the automorphism group is given, all automorphisms are geometric, i.e. induced by a collineation of the ambient space. If the size is larger than a given bound, this means there exist automorphisms that are not geometric.

$\mathcal{K}$	Condition	$\mathopen{|}\operatorname{Aut}(\varGamma _{n,q}(\mathcal{K}))\mathclose{|}$	Reference
Basis	q=n+1	hq n+1(q−1)q!	Section 3.1
Frame	q=n+2	hq n+1(q−1)n q!	Section 3.1, [19]
⊂NRC	q≥n+3	hq n+2(q−1)2	Section 3.2, [9, 19] (q=p)
⊂non-classical arc	q>4 even	hq 5(q−1)2	Section 3.3
⊂Glynn-arc	q=9	9682	Section 3.4
⊂Q −(3,q)	q>4 square	>2hq 5(q−1)2	Section 4.1
⊂Tits-ovoid	q=22(2e+1)	${>} hq^{5}(q-1)(\sqrt{q}-1)$	Section 4.2
⊂Q +(3,q)	q>4 square	${> }2hq^{5}(q-1)(\sqrt{q}-1)^{2}$	Section 4.3
⊂cone $V{\mathcal{O}}$	q=q 0 h	${>} hq^{2n+1}(q-1)^{2} |\mathrm{P}\Gamma\mathrm{L} (n,q_{0})_{\mathcal{O}}|$	Section 4.4

3 Families of semisymmetric graphs arising from arcs

We are in search of point sets $\mathcal{K}$ such that the closure $\overline {\mathcal{K} }$ is equal to H _∞ and such that $\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}}$ acts transitively on the points of $\mathcal{K}$. An arc of size q turns out to be an excellent choice.

Definition

A k-arc in $\operatorname{PG}(n,q)$ is a set of k points, k≥n+1, such that no n+1 points lie on a hyperplane.

If $\mathcal{A}$ is a k-arc in $\operatorname{PG}(n,q)$, then k≥n+1, hence, we will only consider the case where q≥n+1. If q=n+1, then it is easy to see that an arc of size q in $\operatorname{PG}(n,q)$ is a basis, if q=n+2, then every arc of size q is a frame. Hence, when q=n+1 or q=n+2, all arcs of size q in $\operatorname{PG}(n,q)$ are PΓL-equivalent. Because of the isomorphism of the graph $\varGamma_{n,q}(\mathcal{K})$ with other graphs (see Sect. 5), we will explicitly investigate these cases, but the more interesting examples occur when q≥n+3.

It is conjectured that an arc in $\operatorname{PG}(n,q)$, 3≤n≤q−3, has at most q+1 points (this is the well-known MDS-conjecture, in view of its coding-theoretical description). An example of an arc of size q+1 is given by the normal rational curve.

Definition

[15, Sect. 27.5]

A normal rational curve in $\operatorname{PG}(n,q)$, 2≤n≤q, is a (q+1)-arc $\operatorname{PGL}$-equivalent to the (q+1)-arc $\{(0,\ldots,0,1)\}\cup\{ (1,t,t^{2},t^{3},\ldots,t^{n}) \mid t\in \mathbb{F} _{q}\}$

Remark

There are results showing that, if n is sufficiently large w.r.t. q, an arc of size q in $\operatorname{PG}(n,q)$ can be extended to an arc of size q+1. Moreover, other results show that for many values of q and n, all (q+1)-arcs in $\operatorname{PG}(n,q)$ are normal rational curves. The combination of these results leads to the understanding why there are not many known examples of q-arcs in $\operatorname {PG}(n,q)$ that are not contained in a normal rational curve. For an overview, we refer to [14].

We will construct different families of graphs, arising from non-PΓL-equivalent arcs of size q. Since these arcs satisfy the conditions of Result 2.3, we see that the obtained graphs are non-isomorphic.

In view of Result 2.4, our first goal is to show that the closure of a set of q points of an arc in $\operatorname{PG}(n,q)$, q≥n+3 or q=p=n+2 prime, is H _∞. When n=2, this follows immediately. In the following lemmas, we deal with the case n≥3.

Lemma 3.1

Let $\mathcal{K}$ be an arc of size q in $\operatorname{PG} (n,q)$, n≥3. Let P ₁ and P ₂ be any two points of $\mathcal{K}$;

• if q=n+2, there is at least one additional point in $\overline{\mathcal{K}}$ (the closure of $\mathcal{K}$) on the line P ₁ P ₂,

• if q≥n+3, there are at least q/2 additional points in $\overline{\mathcal{K}}$ on the line P ₁ P ₂.

Proof

Note that a k-space π, k≤n−2, with k+1 points of $\mathcal{K}$, different from P ₁ and P ₂, does not intersect P ₁ P ₂, since otherwise 〈π,P ₁ P ₂〉 would be a (k+1)-space containing k+3 points of $\mathcal{K}$, contradicting the arc condition.

Let P ₃,…,P _n+2 be n points of $\mathcal{K}$, different from P ₁ and P ₂. The space 〈P ₃,…,P _n+2〉 is a hyperplane of H _∞, hence, it meets the line P ₁ P ₂ in a point Q. This point Q is contained in $\overline{\mathcal{K}}$ but not contained in $\mathcal{K}$ since $\mathcal{K}$ is an arc. If q=n+2, there is exactly one set {P ₃,…,P _n+2} of n points of $\mathcal{K}$, different from P ₁ and P ₂, yielding an extra point in $\overline{\mathcal {K}}$ on P ₁ P ₂.

If n+3≤q≤2n+2, then let {P ₃,…,P _n+3} be a set of n+1 points of $\mathcal{K}$, different from P ₁ and P ₂. Any subset with n points of {P ₃,…,P _n+3} defines a hyperplane intersecting P ₁ P ₂ in a point Q≠P ₁,P ₂ contained in $\overline{\mathcal{K}}$. These points Q are all different since any two considered hyperplanes intersect in a (n−2)-space with n−1 points of $\mathcal{K}$, and hence this space does not intersect P ₁ P ₂. There are n+1 such subsets, so the line P ₁ P ₂ contains q/2≤n+1≤q−2 additional points in $\overline{\mathcal{K}}$ different from P ₁ and P ₂.

If q≥2n+2, then let P ₃,…,P _n+1 be n−1 points of $\mathcal{\mathcal{K}}$, different from P ₁ and P ₂. Clearly, 〈P ₃,…,P _n+1〉 is disjoint from P ₁ P ₂. There are q−n−1 points of $\mathcal{\mathcal{K}}$ different from all P _i , i=1,…,n+1. For every such point R, the hyperplane 〈P ₃,…,P _n+1,R〉 intersects P ₁ P ₂ in a point of $\overline{\mathcal{K}}$ different from P ₁ and P ₂. Again, all these points are different since two such hyperplanes intersect in 〈P ₃,…,P _n+1〉. The line P ₁ P ₂ contains q−n−1≥q/2 points of $\overline{\mathcal{K}}$ different from P ₁ and P ₂. □

Lemma 3.2

Let $\mathcal{K}$ be an arc of size q in $\operatorname{PG} (n,q)$, n≥2. Let q≥n+3 or q=p=n+2 and let μ _∞ be a plane containing 3 points of $\mathcal{K}$. Then every point of μ _∞ is contained in  $\overline{\mathcal{K}}$.

Proof

Let P ₁,P ₂,P ₃ be 3 points of $\mathcal{K}$ and let μ _∞ be the plane $\langle P_{1},P_{2},P_{3}\rangle= \operatorname{PG}(2,q)$. Consider q≥n+3. By Lemma 3.1, we know that there exist at least q/2 points in $\overline{\mathcal{K}}$ on each of the lines P ₂ P ₃, P ₁ P ₃ and P ₁ P ₂, different from P ₁, P ₂ and P ₃. Consider the set S containing all these points and the points P ₁, P ₂ and P ₃. Its closure $\overline{S}$ forms a subplane π of μ _∞ consisting of only points of $\overline {\mathcal{K}}$. Since a proper subplane of $\operatorname{PG}(2,q)$ contains at most $\sqrt{q}+1 < q/2+2$ points of the line P ₁ P ₂, we see that π must be μ _∞.

If q=n+2 is prime, by Lemma 3.1, we find an extra point $Q_{i} \in\overline{\mathcal{K}}$, i=2,3, on the line P ₁ P _i . The closure of {P ₁,P ₂,P ₃,Q ₂,Q ₃} forms a subplane with all points in $\overline{\mathcal{K}}$. By the fact that q is prime, this subplane equals $\mu_{\infty}= \operatorname{PG}(2,q)$. □

Lemma 3.3

Let L be a line such that every point is in $\overline{\mathcal{K}}$, let π _∞ be a plane of H _∞ through L, containing at least two points R ₁ and R ₂ of $\overline{\mathcal{K}}$ outside L. Then every point in the plane π _∞ is in $\overline{\mathcal{K}}$.

Proof

The closure of the set of points of $\overline{\mathcal {K}}$ on the line L, together with the points R ₁ and R ₂ is clearly the plane π _∞ itself. □

Lemma 3.4

For n≥2, let q≥n+3 or q=p=n+2 and let $\mathcal{K}$ be an arc of size q in $\operatorname{PG}(n,q)$, then $\overline{\mathcal{K}}=\operatorname{PG}(n,q)$.

Proof

For n=2, this easily follows. Let P ₁,…,P _q be the points of $\mathcal{K}$. By Lemma 3.2, we know that every point of 〈P ₁,P ₂,P ₃〉 is in $\overline{\mathcal{K}}$. Suppose, by induction, that every point in 〈P ₁,…,P _k 〉, k≤n is in $\overline {\mathcal{K}}$. The point P _k+1 is not contained in 〈P ₁,…,P _k 〉. There exists an additional point Q in $\overline{\mathcal{K}}$ on the line P ₁ P _k+1 by Lemma 3.1. Let S be a point of 〈P ₁,…,P _k+1〉, not on the line P ₁ P _k+1, and let R be the intersection of the line SP _k+1 with 〈P ₁,…,P _k 〉. Since every point on the line RP ₁ is in $\overline{\mathcal{K}}$, and 〈RP ₁,P _k+1〉 contains the points Q and P _k+1 of $\overline{\mathcal{K}}$, Lemma 3.3 implies that the point S is in $\overline{\mathcal {K}}$, as are the points of P ₁ P _k+1. This shows that every point in 〈P ₁,…,P _k+1〉 is in $\overline{\mathcal{K}}$. The lemma follows by induction and the fact that H _∞=〈P ₁,…,P _n+1〉. □

Theorem 3.5

Let n=2 and q odd or n≥3, q≥n+3 or q=p=n+2, and let $\mathcal{K}$ be an arc in $\operatorname{PG}(n,q)$, then $\operatorname{Aut}(\varGamma_{n,q}(\mathcal{K}))\cong\mathrm {P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$.

Proof

It is clear that if n=2, q odd or n≥3, then every point of H _∞ lies on a tangent line to the arc. By Lemma 3.4, $\overline{\mathcal{K}}$ equals $\operatorname{PG}(n,q)$. The theorem follows from Result 2.4. □

3.1 $\mathcal{K}$ is a q-arc in $\operatorname{PG}(n,q)$ with q=n+1 or q=n+2

As noted before, a q-arc in $\operatorname{PG}(n,q)$ with q=n+1 is a basis, a q-arc in $\operatorname{PG}(n,q)$ with q=n+2 is a frame. In these cases, the linear representation of a q-arc gives rise to a semisymmetric graph, however, the description of the automorphism group is different from the case q≥n+3. In the following proof, we cannot use the same techniques as in [4] to show that $\mathrm{P}\Gamma\mathrm {L}(n+2,q)_{\mathcal{K}}$ splits over $\operatorname{Persp}(H_{\infty})$.

We introduce some definitions.

Definition

A permutation matrix is a square binary matrix that has exactly one entry 1 in each row and each column and 0’s elsewhere. A monomial matrix or generalised permutation matrix has exactly one non-zero entry in each row and each column. The monomial matrices form a group.

Let $\operatorname{PMon}(q)$ denote the quotient group of the monomial matrices over $\mathbb{F}_{q}$ by the scalar matrices. Let S _k denote the symmetric group of degree k, meaning the group of all permutations of {1,2,…,k}.

Theorem 3.6

If $\mathcal{K}$ is a q-arc in $\operatorname{PG}(n,q)$, n≥2, q=n+1 or q=n+2, with (n,q)≠(2,4) then $\varGamma _{n,q}(\mathcal{K})$ is a semisymmetric graph. The group $\mathrm{P}\Gamma\mathrm {L}(n+2,q)_{\mathcal{K}}$ is a subgroup of $\operatorname{Aut}(\varGamma_{n,q}(\mathcal{K}))$ and is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$, where $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ is isomorphic to

• (i) $\mathrm{S}_{q} \rtimes\operatorname{Aut}(\mathbb{F}_{q})$ if q=n+2, having size hq ⁿ⁺¹(q−1)q!;

• (ii) $\operatorname{PMon}(q)\rtimes\operatorname {Aut}(\mathbb{F}_{q})$ if q=n+1, having size hq ⁿ⁺¹(q−1)ⁿ q!.

Moreover, if q=n+2 and q is prime, then $\operatorname{Aut}(\varGamma _{n,q}(\mathcal{K}))$ is isomorphic to $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$.

Proof

(i) If q=n+2, then $\mathcal{K}$ is $\operatorname{PGL}$-equivalent to the frame $\mathcal{K}'$ of $\operatorname{PG}(n,q)$ with points P ₁,…,P _n+2, where P _i has coordinates v _i , and v ₁=(1,0,…,0),v ₂= (0,1,0,…,0), …,v _n+1=(0,…,0,1),v _n+2=(−1,−1,…,−1). Let B _k =(b _ij ) _k , 1≤k≤n+1, be the matrix with b _ii =1, i≠k, 1≤i≤n+1, b _ik =−1, 1≤i≤n+1, and b _ij =0 for all other i,j. The considered action of B _k on the points of $\operatorname{PG}(n,q)$ is by left-multiplication on the column vector of their coordinates. Let G _per denote the subgroup of permutation matrices of $\operatorname{GL}(n+1,q)$, and consider the subgroup G of $\operatorname{GL}(n+1,q)$, generated by the elements of G _per and the matrices B _k , 1≤k≤n+1.

For every matrix B=(b _ij ), 1≤i,j≤n+1, in G, we can define a matrix A=(a _ij ), 0≤i,j≤n+1, as the (n+2)×(n+2) matrix with a ₀₀=1, a _i0=a _0j =0 for i,j≥1 and a _ij =b _ij for 1≤i,j≤n+1. Let $\widetilde{G}$ be the group obtained by extending all matrices of G in this way. It is clear that the elements of G are exactly the permutations of the elements of {v ₁,…,v _n+2} and hence that $\widetilde{G}$ is isomorphic to $\operatorname{PGL}(n+1,q)_{\mathcal{K}}\cong\mathrm{S}_{q}$.

It follows that the only element of $\widetilde{G}$ fixing $\mathcal{K}$ pointwise corresponds to the identity matrix, which implies that any element of $\operatorname{Persp}(H_{\infty})$ contained in $\widetilde {G}$ is trivial. Hence, $\operatorname{PGL}(n+2,q)_{\mathcal{K}}$ is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes \operatorname{PGL} (n+1,q)_{\mathcal{K}}$. Clearly, $\operatorname{PGL}(n+1,q)_{\mathcal{K}}$ acts transitively on the points of $\mathcal{K}$, hence by Theorem 2.8 the graph $\varGamma _{n,q}(\mathcal{K})$ is edge-transitive.

(ii) Now suppose q=n+1. The group $\operatorname{PSL}(n+1,q)$ is a subgroup of $\operatorname{PGL} (n+1,q)$ and a quotient of $\operatorname{SL}(n+1,q)$. When q=n+1, all three groups have the same order and thus are all isomorphic. Hence, $\operatorname{PGL}(n+1,q)$ can be embedded in $\operatorname {PGL}(n+2,q)_{H_{\infty}}$ by taking all matrices B=(b _ij ), 1≤i,j≤n+1, of $\operatorname {SL}(n+1,q)$ and, as before, defining A=(a _ij ), 0≤i,j≤n+1, with a ₀₀=1, a _i0=a _0j =0 for i,j≥1 and a _ij =b _ij for 1≤i,j≤n+1. An element of $\operatorname{Persp} (H_{\infty})$ corresponds to a matrix of the form D=(d _ij ), 0≤i,j≤n+1, with d _0j =λ _j , 0≤j≤n+1, d _ii =μ, 1≤i≤n+1, for some $\lambda_{j},\mu\in\mathbb{F}_{q}$, and d _ij =0 otherwise. This implies that the group $\widetilde{G}$ of matrices A defined in this way meets $\operatorname{Persp}(H_{\infty})$ trivially. Hence, $\operatorname{PGL}(n+2,q)_{\mathcal{K}}$ is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes\operatorname {PGL}(n+1,q)_{\mathcal{K}}$.

Since q=n+1, the curve $\mathcal{K}$ is $\operatorname {PGL}$-equivalent to the set $\mathcal{K}'$ of points P ₁,…,P _n+1 in $\operatorname{PG}(n,q)$, where P _i has coordinates v _i , and v ₁=(1,0,…,0), v ₂=(0,1,0,…,0), …,v _n+1=(0,…,0,1). Using this, it is clear that $\operatorname{PGL} (n+1,q)_{\mathcal{K}}$ is isomorphic to the quotient group of monomial matrices by scalar matrices and that $\operatorname{PGL}(n+1,q)_{\mathcal{K}}$ acts transitively on $\mathcal{K}$. Hence, $\varGamma_{n,q}(\mathcal{K})$ is an edge-transitive graph.

In both cases, it is clear that $\mathcal{K}'$ is stabilised by the Frobenius automorphism, hence, using Result 2.6, it also follows that $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}\cong\operatorname {Persp}(H_{\infty})\rtimes(\operatorname{PGL}(n+1,q)_{\mathcal{K}} \rtimes\operatorname{Aut}(\mathbb{F}_{q}))$. The observation on the sizes follows from |S _q |=q! and $\mathopen{|}\operatorname{PMon}|=|\mathrm {S}_{q}|.|(\mathbb{F} _{q}^{*})^{n}|/(q-1)=q!(q-1)^{n-1}$.

Since through every point of H _∞ there is a tangent line to $\mathcal{K}$, Corollary 2.5 shows that $\varGamma_{n,q}(\mathcal{K})$ is not vertex-transitive. Since $\mathcal{K}$ spans H _∞ and $|\mathcal{K}|=q$, we get that $\varGamma_{n,q}(\mathcal{K})$ is semisymmetric.

The last part of the statement follows from Theorem 3.5. □

Remark

For n=2,q=3, and $\mathcal{K}$ a basis is $\operatorname{PG}(2,3)$, we have shown, by using the computer program GAP [12], that all automorphisms are induced by a collineation of $\operatorname {PG}(3,3)$ so we have that the automorphism group of $\varGamma_{2,3}(\mathcal{K})$ is isomorphic to $\mathrm{P}\Gamma\mathrm{L}(4,3)_{\mathcal{K}}$. For n=3,q=4, however, again using the computer, we find that $[\operatorname{Aut}(\varGamma_{n,q}(\mathcal{K})):\mathrm {P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}]=8$. This implies that there exist automorphisms of the graph $\varGamma _{3,4}(\mathcal{K})$ that are not collineations of $\operatorname{PG}(4,4)$. For n=4,q=5, this index is already 7776. This might indicate that the general description of the full automorphism group of $\varGamma_{n,q}(\mathcal{K})$, with n+1=q and $\mathcal{K}$ a basis, is a hard problem.

3.2 $\mathcal{K}$ is contained in a normal rational curve and q≥n+3

We will use the following theorem by Segre.

Result 3.7

[22]

If q≥n+2 and S is a set of n+3 points in $\operatorname{PG}(n,q)$, no n+1 of which lie in a hyperplane, then there is a unique normal rational curve in $\operatorname{PG}(n,q)$ containing the points of S.

Corollary 3.8

If $\mathcal{K}$ is a set of q points of a normal rational curve $\mathcal{N}$ in $\operatorname{PG}(n,q)$, q≥n+3, then $\mathcal{N}$ is the unique normal rational curve through the points of $\mathcal{K}$.

The following theorem is well known, a proof can be found in e.g. [15, Theorem 27.5.3].

Result 3.9

If q≥n+2 and $\mathcal{N}$ is a normal rational curve in $\operatorname{PG}(n,q)$, then the stabiliser of $\mathcal{N}$ in PΓL(n+1,q) is isomorphic to PΓL(2,q) (in its faithful action on q+1 points).

These results enable us to give a construction for the following infinite two-parameter family of semisymmetric graphs.

The subgroup of PΓL(2,q) fixing one point in its natural action is isomorphic to the affine semilinear group AΓL(1,q) in dimension 1.

Theorem 3.10

If $\mathcal{K}$ is a set of q points, contained in a normal rational curve of $\operatorname{PG}(n,q)$, q=p ^h, n≥3, q≥n+3, or n=2, q odd, then $\varGamma_{n,q}(\mathcal{K})$ is a semisymmetric graph.

Moreover, $\operatorname{Aut}(\varGamma_{n,q}(\mathcal{K}))$ is isomorphic to $\operatorname{Persp} (H_{\infty})\rtimes\mathrm{A}\Gamma \mathrm{L}(1,q)$ and has size hq ⁿ⁺²(q−1)².

Proof

Since $|\mathcal{K}|=q$, the graph $\varGamma_{n,q}(\mathcal {K})$ is q-regular. The set $\mathcal{K}$ is an arc spanning the space $\operatorname{PG}(n,q)$. It is clear that if n≥3, or if q is odd, every point of $\operatorname{PG}(n,q)$ lies on at least one tangent line to $\mathcal{K}$. Hence, by Result 2.1, Corollary 2.5 and Theorem 3.5, $\varGamma_{n,q}(\mathcal{K})$ is a connected non-vertex-transitive graph for which $\operatorname{Aut}(\varGamma_{n,q}(\mathcal{K}))\cong\mathrm {P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$. By Corollary 3.8, $\mathcal{K}$ extends by a point P to a unique normal rational curve $\mathcal{N}$. Since P must be fixed by the stabiliser of $\mathcal{K}$ and PΓL(2,q) _P ≅AΓL(1,q), we get $\mathrm {P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}\cong\operatorname{Persp} (H_{\infty})\rtimes\mathrm{A}\Gamma \mathrm{L}(1,q)$, by Result 2.6. The size of this group follows when considering that $\mathopen{|}\operatorname {Persp}(H_{\infty})|=q^{n+1}(q-1)$ and |AΓL(1,q)|=hq(q−1). By Theorem 2.8 the graph $\varGamma_{n,q}(\mathcal{K})$ is edge-transitive and thus semisymmetric. □

3.3 $\mathcal{K}$ is contained in a non-classical arc in $\operatorname{PG}(3,q)$, q even

The (q+1)-arcs in $\operatorname{PG}(3,q)$, q even, have been classified, each of them has the same stabiliser group as the normal rational curve.

Result 3.11

[6]

In $\operatorname{PG}(3, q)$, q=2^h, h>2, every (q+1)-arc is $\operatorname{PGL}$-equivalent to some $\mathcal{C}(\sigma)=\{(1,x,x^{\sigma },x^{\sigma+1}) \mid x \in\mathbb{F}_{q} \} \cup\{(0, 0, 0, 1)\}$ where σ is a generator of $\operatorname{Aut}(\mathbb{F}_{q})$.

Result 3.12

[20]

In $\operatorname{PG}(3, q)$, q=2^h, h>2, the stabiliser of $\mathcal{C}(\sigma)$ in PΓL(4,q) is isomorphic to PΓL(2,q) (in its faithful action on q+1 points).

The case q=4 is already discussed in Sect. 3.1.

Result 3.13

[5]

For any k-arc of $\operatorname{PG}(3,q)$, q=2^h, h>1, we have k≤q+1.

Result 3.14

[3]

Let $\mathcal{K}$ be any k-arc in $\operatorname{PG}(3, q)$, q=2^h. If k>(q+4)/2, then $\mathcal{K}$ is contained in a unique complete arc.

Corollary 3.15

Consider a (q+1)-arc $\mathcal{C}(\sigma)$ of $\operatorname {PG}(3,q)$, q=2^h, h>2. If $\mathcal{K}$ is a set of q points contained in $\mathcal{C}(\sigma )$, then there is a unique (q+1)-arc through the points of $\mathcal{K}$, namely $\mathcal{C}(\sigma)$.

Proof

Using Result 3.14, since q>(q+4)/2, q>4, we find a unique complete arc through $\mathcal{K}$. This arc has size at most q+1 by Result 3.13 and thus is equal to $\mathcal{C}(\sigma)$. □

Theorem 3.16

If $\mathcal{K}$ is a set of q points contained in any (q+1)-arc of $\operatorname{PG}(3,q)$, q≥8 even, then $\varGamma _{3,q}(\mathcal{K})$ is a semisymmetric graph.

Moreover, $\operatorname{Aut}(\varGamma_{3,q}(\mathcal{K}))$ is isomorphic to $\operatorname{Persp} (H_{\infty})\rtimes\mathrm{A}\Gamma \mathrm{L}(1,q)$ and has size hq ⁵(q−1)².

Proof

The proof goes in exactly the same way as the proof of Theorem 3.10, by making use of Corollary 3.15 and Results 3.11 and 3.12. □

3.4 $\mathcal{K}$ is contained in the Glynn arc in $\operatorname{PG}(4,9)$

In [11] Glynn constructs an example of an arc of size 10 in $\operatorname{PG}(4,9)$, which is not a normal rational curve. We call this 10-arc the Glynn arc (of size 10). He also shows that an arc in $\operatorname{PG} (4,9)$ of size 10 is a normal rational curve or a Glynn arc.

Result 3.17

[11]

The stabiliser in PΓL(5,9) of the Glynn arc of size 10 in $\operatorname{PG}(4,9)$ is isomorphic to $\operatorname{PGL}(2,9)$.

Result 3.18

[2]

A k-arc in $\operatorname{PG}(n, q)$, n≥3, q odd and $k \geq\frac{2}{3}(q - 1) + n$ is contained in a unique complete arc of $\operatorname{PG}(n,q)$.

Corollary 3.19

If $\mathcal{K}$ is a set of 9 points contained in a Glynn 10-arc $\mathcal{C}$ of $\operatorname{PG}(4,9)$, then $\mathcal{K}$ is contained in a unique 10-arc, namely $\mathcal{C}$.

Theorem 3.20

If $\mathcal{K}$ is a 9-arc contained in a Glynn 10-arc of $\operatorname{PG}(4,9)$, then $\varGamma_{4,9}(\mathcal{K})$ is a semisymmetric graph.

Moreover, $\operatorname{Aut}(\varGamma_{4,9}(\mathcal{K}))$ is isomorphic to $\operatorname{Persp} (H_{\infty})\rtimes\operatorname{AGL}(1,9)$ and has size 9⁶8².

Proof

Since $|\mathcal{K}|=9$, $\varGamma_{4,9}(\mathcal{K})$ is a 9-regular graph. The set $\mathcal{K}$ is an arc spanning the space $\operatorname{PG}(4,9)$. It is clear that every point of $\operatorname{PG}(4,9)$ lies on at least one tangent line to $\mathcal{K}$. Hence, by Result 2.1, Corollary 2.5 and Theorem 3.5, $\varGamma_{4,9}(\mathcal{K})$ is a connected non-vertex-transitive graph for which $\operatorname{Aut}(\varGamma _{4,9}(\mathcal{K}))\cong \mathrm{P}\Gamma\mathrm{L}(6,9)_{\mathcal{K}}$. By Corollary 3.19, $\mathcal{K}$ extends by a point P to a unique Glynn 10-arc $\mathcal{C}$. By Result 2.6 we have $\mathrm{P}\Gamma\mathrm {L}(6,9)_{\mathcal{K}}\cong\operatorname{Persp}(H_{\infty}) \rtimes\mathrm{P}\Gamma\mathrm{L}(5,9)_{\mathcal{K}}$. Since $\operatorname{PGL}(2,9)_{P}\cong\operatorname{AGL}(1,9)$, we find $\mathrm{P}\Gamma\mathrm{L}(6,9)_{\mathcal{K}}\cong \operatorname{Persp}(H_{\infty}) \rtimes\operatorname{AGL}(1,9)$. As before, the size easily follows. By Theorem 2.8 the graph $\varGamma_{4,9}(\mathcal{K})$ is edge-transitive and thus semisymmetric. □

3.5 Using the dual arc construction

Let $\mathcal{K}= \{P_{1}, \ldots, P_{k}\}$ be a k-arc in $\operatorname{PG}(n, q)$, k≥n+4. Consider the respective coordinates (a _0j ,…,a _nj ) of P _j , 1≤j≤k, then the (n+1)×k-matrix A=(a _ij ) determines a vector space (an MDS code) V ₁=V(n+1,q), which is a subspace of V(k,q). The space V ₁ has a unique orthogonal complement V ₂=V(k−n−1,q) in V(k,q). Then V ₂ is also an MDS code [21, p. 319]. A k-arc $\hat{\mathcal{K}}=\{Q_{l}, \ldots, Q_{k}\}$ of $\operatorname{PG}(k - n - 2, q)$ with respective coordinates (b _0j ,…,b _k−n−2,j ) of Q _j , 1≤j≤k, such that the (k−n−1)×k-matrix B=(b _ij ) generates V ₂, is called a dual k-arc $\hat{\mathcal{K}}$ of the k-arc $\mathcal{K}$ [27].

It should be noted that duality for arcs is a 1−1-correspondence between equivalence classes of arcs, rather than a correspondence between arcs: with another ordering of $\mathcal{K}$ and choosing other coordinates for the points of $\mathcal{K}$, we obtain the same set of dual k-arcs.

Result 3.21

[26, Theorem 2.1]

A k-arc $\mathcal{K}$ in $\operatorname{PG}(n, q)$, k≥n+4, and a dual k-arc $\hat {\mathcal{K}}$ of $\mathcal{K}$ in $\operatorname{PG}(k - n - 2, q)$ have isomorphic collineation groups and isomorphic projective groups.

The duality transformation maps normal rational curves to normal rational curves and non-classical arcs to non-classical arcs. This implies that the arcs in Sects. 3.3 and 3.4 give rise to a different family of semisymmetric graphs. This follows from the following theorem.

Theorem 3.22

Let $\mathcal{K}$ be a q-arc in $H_{\infty}=\operatorname{PG} (n,q)$, q≥n+4, and let $\hat{\mathcal{K}}$ be a dual arc of $\mathcal{K}$ in $\hat {H}_{\infty}=\operatorname{PG}(q-n-2,q)$. Suppose that one of the groups $\mathrm{P}\Gamma\mathrm{L} (n+1,q)_{\mathcal{K}}$ or $\mathrm{P}\Gamma\mathrm{L}(q-n-1,q)_{\hat {\mathcal{K}}}$ fixes a point outside $\mathcal{K}$, $\hat{\mathcal{K}}$, respectively, and acts transitively on the points of $\mathcal{K}$, $\hat{\mathcal{K}}$, respectively, then $\varGamma_{n,q}(\mathcal {K})$ and $\varGamma _{q-n-2,q}(\hat{\mathcal{K}})$ are semisymmetric, $\operatorname {Aut}(\varGamma_{n,q}({\mathcal{K} }))\cong \operatorname{Persp}(H_{\infty})\rtimes\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}}$ and $\operatorname{Aut}(\varGamma _{q-n-2,q}(\hat{\mathcal{K}}))\cong\operatorname{Persp}(\hat {H}_{\infty})\rtimes\mathrm{P}\Gamma\mathrm{L} (n+1,q)_{\mathcal{K}}$.

Proof

In the same way as before, using Result 2.1, Corollary 2.5 and Theorem 3.5, we see that $\varGamma _{n,q}(\mathcal{K})$ and $\varGamma_{q-n-2,q}(\hat{\mathcal{K}})$ are connected non-vertex-transitive graphs for which $\operatorname{Aut}(\varGamma _{n,q}(\mathcal{K}))\cong \mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$ and $\operatorname {Aut}(\varGamma_{q-n-2,q}(\hat{\mathcal{K}}))\cong \mathrm{P}\Gamma\mathrm{L} (q-n,q)_{\hat{\mathcal{K}}}$.

Suppose w.l.o.g. that $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ fixes a point outside $\mathcal{K}$, then by Result 2.6, $\mathrm{P}\Gamma\mathrm {L}(n+2,q)_{\mathcal{K}}\cong\operatorname{Persp}(H_{\infty})\rtimes\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$. The embedding of $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ in $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$ used to show this result was constructed by adding a 1 at the lower right corner of every matrix B corresponding to an element (B,θ) of $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$, for some $\theta\in \operatorname{Aut} (\mathbb{F}_{q})$ to obtain a matrix B′ corresponding to an element (B′,θ) of $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$. This subgroup meets $\operatorname{Persp}(H_{\infty})$ trivially, which implies that in the group of matrices defining elements of $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$, no proper scalar multiple of the identity matrix occurs. Now, from the isomorphism of Result 3.21, it follows that the group $\operatorname{PGL}(q-n-1,q)_{\hat{\mathcal{K}}}$, which is isomorphic to $\operatorname{PGL} (n+1,q)_{\mathcal{K}}$, also contains no proper scalar multiple of the identity matrix. Hence, by embedding $\mathrm{P}\Gamma\mathrm{L}(q-n-1,q)_{\hat {\mathcal{K}}}$ in PΓL(q−n,q) in the same way (by adding a 1 at the lower right corner), we see that it meets $\operatorname{Persp}(\hat{H}_{\infty})$ trivially. This implies that $\mathrm{P}\Gamma\mathrm{L}(q-n,q)_{\hat{\mathcal {K}}}\cong\operatorname{Persp}(\hat {H}_{\infty})\rtimes\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$.

We know that $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ and $\mathrm{P}\Gamma\mathrm{L}(q-n-1,q)_{\hat{\mathcal{K}}}$ are permutation isomorphic, hence, if one of them acts transitively on the points of $\mathcal{K}$ or $\hat{\mathcal{K}}$, so does the other. By Theorem 2.8, the graphs $\varGamma_{n,q}(\mathcal{K})$ and $\varGamma _{q-n-2,q}(\hat{\mathcal{K}})$ are edge-transitive and hence semisymmetric. □

If we restrict ourselves in the previous theorem to elements of the projective groups, using Result 2.7 we get the following corollary.

Corollary 3.23

Let $\mathcal{K}$ be a q-arc in $H_{\infty}=\operatorname{PG}(n,q)$, q≥n+4, and let $\hat{\mathcal{K}}$ be a dual arc of $\mathcal{K}$ in $\hat{H}_{\infty}=\operatorname{PG} (q-n-2,q)$. Suppose that one of the groups $\operatorname {PGL}(n+1,q)_{\mathcal{K}}$ or $\operatorname{PGL} (q-n-1,q)_{\hat{\mathcal{K}}}$ fixes a point outside $\mathcal{K}$, $\hat{\mathcal{K}}$, respectively, and acts transitively on the points of $\mathcal{K}$, $\hat{\mathcal{K}}$, respectively. Suppose $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}\cong\operatorname{PGL}(n+1,q)_{\mathcal{K}} \rtimes \operatorname{Aut}(\mathbb{F}_{q_{0}})$ or $\mathrm{P}\Gamma\mathrm {L}(q-n-1,q)_{\hat{\mathcal{K}}}\cong\operatorname{PGL} (q-n-1,q)_{\hat{\mathcal{K}}} \rtimes\operatorname{Aut}(\mathbb {F}_{q_{0}})$, respectively, for $q_{0}=p^{h_{0}}$, h ₀|h or $\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}}\cong\operatorname{PGL}(n+1,q)_{\mathcal{K}}$, $\mathrm{P}\Gamma\mathrm{L}(q-n-1,q)_{\hat{\mathcal{K}}}\cong \operatorname{PGL}(q-n-1,q)_{\hat{\mathcal{K}}}$, respectively. Then $\varGamma_{n,q}(\mathcal{K})$ and $\varGamma _{q-n-2,q}(\hat{\mathcal{K}})$ are semisymmetric, $\operatorname{Aut}(\varGamma_{n,q}({\mathcal {K}}))\cong\operatorname{Persp}(H_{\infty})\rtimes\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ and $\operatorname{Aut}(\varGamma_{q-n-2,q}(\hat{\mathcal{K} }))\cong \operatorname{Persp}(\hat{H}_{\infty})\rtimes\mathrm{P}\Gamma \mathrm{L}(n+1,q)_{\mathcal{K}}$.

Consider the Glynn 10-arc contained in $\operatorname{PG}(4,9)$ and take any point P of this 10-arc; if we project the arc from P onto a $\operatorname{PG}(3, 9)$ skew to P, then we obtain a complete 9-arc of $\operatorname {PG}(3, 9)$. In [11] the author also shows that all complete 9-arcs in $\operatorname{PG} (3, 9)$ can be obtained in this way, i.e. all complete 9-arc of $\operatorname{PG} (3, 9)$ are PΓL-equivalent. It follows from [25] that the complete 9-arc in $\operatorname{PG}(3,9)$ is the dual of a 9-arc that is contained in the Glynn arc in $\operatorname{PG}(4,9)$. If we apply Theorem 3.22 to the Glynn 10-arc, we obtain the following corollary. The size of the automorphism group follows as before.

Corollary 3.24

If $\mathcal{K}$ is a complete 9-arc of $\operatorname{PG}(3,9)$, then $\varGamma _{3,9}(\mathcal{K})$ is a semisymmetric graph. Moreover, $\operatorname{Aut}(\varGamma_{3,9}(\mathcal{K}))$ is isomorphic to $\operatorname{Persp} (H_{\infty})\rtimes\operatorname{AGL}(1,9)$ and has size 9⁵8².

We can also apply Theorem 3.22 to the arcs of Sect. 3.3.

Corollary 3.25

Let $\mathcal{K}$ be an arc of size q contained in any (q+1)-arc of $\operatorname{PG}(q-4,q)$, q=2^h>8, then $\varGamma _{q-4,q}(\mathcal{K})$ is a semisymmetric graph.

Moreover, $\operatorname{Aut}(\varGamma_{q-4,q}(\mathcal{K}))$ is isomorphic to $\operatorname{Persp} (H_{\infty})\rtimes\mathrm{A}\Gamma \mathrm{L}(1,q)$ and has size hq ^q−2(q−1)².

4 Families of semisymmetric graphs arising from other sets

By Result 2.4, if $\mathcal{K}$ is a set of points such that its closure $\overline{\mathcal{K}}$ is the whole space H _∞, then every automorphism of the graph $\varGamma_{n,q}(\mathcal{K})$ is induced by a collineation of its ambient space $\operatorname{PG}(n+1,q)$. However, we do not need this property for the construction of semisymmetric graphs. From the results and theorems of Sect. 2, the following theorem clearly follows.

Theorem 4.1

Let $\mathcal{K}$ be a point set of $H_{\infty}=\operatorname {PG}(n,q)$ of size q spanning H _∞ such that every point of $H_{\infty}\backslash\mathcal{K}$ lies on at least one tangent line to $\mathcal{K}$, and such that $\mathrm {P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ acts transitively on the points of $\mathcal{K}$. Then the graph $\varGamma _{n,q}(\mathcal{K})$ is a connected semisymmetric graph.

The subgroup of the automorphism group of the graph $\varGamma _{n,q}(\mathcal{K})$ for which the elements are induced by collineations of the space $\operatorname{PG} (n+1,q)$ will be called the geometric automorphism group of $\varGamma_{n,q}(\mathcal{K})$.

We now give some examples of semisymmetric graphs for which $\overline {\mathcal{K}}$ is a subgeometry of H _∞. In the first three examples $\overline{\mathcal{K}}$ is a Baer subgeometry, obviously this only works if we look at a projective space over a field of square order. We will also construct their geometric automorphism group.

4.1 $\mathcal{K}$ is contained in an elliptic quadric

Let π be a Baer subgeometry $\operatorname{PG}(3,\sqrt{q})$ embedded in $H_{\infty}=\operatorname{PG}(3,q)$, q a square. Let $\mathcal{K}$ denote the set of points of an elliptic quadric $Q^{-}(3,\sqrt{q})$ in π with one point removed. This set $\mathcal{K}$ has q points and clearly every point not in $\mathcal{K}$ lies on at least one tangent line to $\mathcal{K}$.

We introduce the definition of a cap and some results.

Definition

A k-cap in $\operatorname{PG}(n,q)$ is a set of k points such that no 3 points lie on a line.

Result 4.2

[1]

A q-cap in $\operatorname{PG}(3,\sqrt{q})$, q an odd square, is uniquely extendable to an elliptic quadric $Q^{-}(3,\sqrt{q})$.

Result 4.3

[23, Chap. IV]

In $\operatorname{PG}(3,\sqrt{q})$, q>4 an even square, a k-cap with $q-\sqrt [4]{q}/2 +1 < k < q+1$ lies on a unique complete (q+1)-cap.

Result 4.4

[13, Sect. 15.3]

The stabiliser in $\mathrm{P}\Gamma\mathrm{L}(4,\sqrt{q})$ of an elliptic quadric in $\operatorname{PG} (3,\sqrt{q})$ is $\mathrm{P}\Gamma\mathrm{O}^{-}(4,\sqrt{q})$, which is isomorphic to PΓL(2,q) (in its faithful action on q+1 points).

Theorem 4.5

The graph $\varGamma_{3,q}(\mathcal{K})$, q>4 square, is semisymmetric. Moreover, the geometric automorphism group is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes(\mathrm{A}\Gamma\mathrm {L}(1,q)\rtimes2)$ and has size 2hq ⁵(q−1)².

Proof

Since $\mathcal{K}$ consists of q points spanning $\operatorname {PG}(3,q)$, the graph $\varGamma _{3,q}(\mathcal{K})$ is q-regular and it is connected by Result 2.1. The graph $\varGamma_{3,q}(\mathcal{K})$ is not vertex-transitive by Corollary 2.5. The geometric automorphism group of $\varGamma _{3,q}(\mathcal{K})$ is $\mathrm{P}\Gamma\mathrm{L}(5,q)_{\mathcal{K}}$. By Results 4.2 (q odd) and 4.3 (q even), the cap $\mathcal{K}$ extends uniquely to an elliptic quadric in $\operatorname{PG}(3,\sqrt{q})$ by a point P. This point is obviously fixed by the stabiliser of $\mathcal{K}$ and hence, by Result 2.6, we find $\mathrm{P}\Gamma\mathrm{L}(5,q)_{\mathcal{K}}\cong \operatorname{Persp}(H_{\infty}) \rtimes \mathrm{P}\Gamma\mathrm{L}(4,q)_{\mathcal{K}}$. The group stabilising $\mathcal{K}$ stabilises the subgeometry $\overline{\mathcal{K}}$, hence $\mathrm{P}\Gamma\mathrm{L}(4,q)_{\mathcal{K}}\cong\mathrm {P}\Gamma\mathrm{L}(4,\sqrt{q})_{\mathcal{K}}\rtimes(\operatorname{Aut} (\mathbb{F} _{q})/\operatorname{Aut}(\mathbb{F}_{\sqrt{q}}))\cong\mathrm {P}\Gamma\mathrm{L}(4,\sqrt{q})_{\mathcal{K}}\rtimes2$. The stabiliser of $\mathcal{K}$ stabilises the elliptic quadric and fixes its point P, hence we find $\mathrm{P}\Gamma\mathrm{L}(4,\sqrt{q})_{\mathcal{K}}\cong\mathrm{P}\Gamma\mathrm{O} ^{-}(4,\sqrt {q})_{P} \cong\mathrm{P}\Gamma\mathrm{L}(2,q)_{P} \cong\mathrm {A}\Gamma\mathrm{L}(1,q)$. Since AΓL(1,q) acts transitively on the points of $\mathcal{K}$, the graph is semisymmetric. The size of this group follows from $\mathopen{|}\operatorname{Persp}(H_{\infty})|=q^{4}(q-1)$ and |AΓL(1,q)|=hq(q−1). □

4.2 $\mathcal{K}$ is contained in a Tits-ovoid

Let π be a Baer subgeometry $\operatorname{PG}(3,\sqrt{q})$ embedded in $H_{\infty}=\operatorname{PG}(3,q)$, q=2^2(2e+1), e>0. Let $\mathcal{K}$ denote the set of points of a Tits-ovoid in π with one point removed. This set $\mathcal {K}$ has q points and forms a cap in $\operatorname{PG}(3,q)$.

The canonical form of a Tits-ovoid in $\operatorname{PG}(3,\sqrt {q})$, $\sqrt {q}=2^{{2e+1}}$ is

$$\bigl\{\bigl(1,s,t,st+s^{\sigma+2}+t^{\sigma}\bigr) \bigm| s,t \in \mathbb{F}_{\sqrt {q}} \bigr\} \cup \bigl\{(0,0,0,1)\bigr\}, $$

where $\sigma: \mathbb{F}_{\sqrt{q}} \rightarrow\mathbb{F}_{\sqrt {q}} : x \mapsto x^{2^{e+1}}$. Let the set $\mathcal{K}$ correspond to the points of this ovoid minus the point (0,0,0,1), then $\mathcal{K}$ is clearly stabilised by $\operatorname{Aut}(\mathbb{F}_{q})$.

Result 4.6

[28]

The stabiliser of $\mathcal{K}$ in $\operatorname{PGL}(4,\sqrt{q})$ is the 2-transitive Suzuki simple group $\operatorname{Sz}(\sqrt{q})$.

Following the notation of [17, Chap. 11], the point stabiliser of $\operatorname{Sz}(\sqrt{q})$ will be denoted by $\mathfrak{FH}$. Since $\operatorname{Sz}(\sqrt{q})$ is 2-transitive, the group $\mathfrak {FH}$ is transitive.

Theorem 4.7

The graph $\varGamma_{3,q}(\mathcal{K})$, q=2^2(2e+1), e>0, is semisymmetric. Moreover, the geometric automorphism group is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes(\mathfrak{FH}\rtimes \operatorname{Aut}(\mathbb{F}_{q}))$ and has size $hq^{5}(q-1)(\sqrt{q}-1)$.

Proof

The proof works in almost the same way as for the elliptic quadric. The size of the group follows when considering that $\mathopen{|}\operatorname {Persp}(H_{\infty})|=q^{4}(q-1)$ and $|\mathfrak{FH}|=q(\sqrt{q}-1)$. □

4.3 $\mathcal{K}$ is contained in a hyperbolic quadric Q ⁺(3,q)

Let π be a Baer subgeometry $\operatorname{PG}(3,\sqrt{q})$ embedded in $H_{\infty}=\operatorname{PG}(3,q)$, q>4 square. Let $\mathcal{K}$ denote the set of points of a hyperbolic quadric $Q^{+}(3,\sqrt{q})$ in π with two lines of different reguli removed. This set $\mathcal{K}$ has q points.

Result 4.8

[13, Sect. 15.3]

The stabiliser in $\mathrm{P}\Gamma\mathrm{L}(4,\sqrt{q})$ of a hyperbolic quadric in $\operatorname{PG} (3,\sqrt{q})$ is $\mathrm{P}\Gamma\mathrm{O}^{+}(4,\sqrt{q})$, which is isomorphic to $((\operatorname{PGL}(2,\sqrt{q}) \times\operatorname{PGL}(2,\sqrt {q}))\rtimes2)\rtimes\operatorname{Aut} (\mathbb{F} _{\sqrt{q}})$ for $\sqrt{q}>2$.

Corollary 4.9

For $\sqrt{q}>2$, the stabiliser in $\mathrm{P}\Gamma\mathrm {L}(4,\sqrt{q})$ of a hyperbolic quadric in $\operatorname{PG}(3,\sqrt{q})$ fixing two lines of different reguli is isomorphic to $((\operatorname{AGL}(1,\sqrt{q}) \times \operatorname{AGL}(1,\sqrt {q}))\rtimes2)\rtimes\operatorname{Aut}(\mathbb{F}_{\sqrt{q}})$.

Theorem 4.10

The graph $\varGamma_{3,q}(\mathcal{K})$, q=p ^h>4 square, is semisymmetric. Moreover, the geometric automorphism group is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes((\operatorname {AGL}(2,\sqrt{q})\times\operatorname{AGL}(2,\sqrt {q}))\rtimes2)\rtimes\operatorname{Aut}(\mathbb{F}_{q})$ and has size $2hq^{5}(q-1)(\sqrt{q}-1)^{2}$.

Proof

Since $\mathcal{K}$ consists of q points spanning $\operatorname {PG}(3,q)$, $\varGamma _{3,q}(\mathcal{K} )$ is q-regular and is connected by Result 2.1. Clearly, every point of $\operatorname{PG}(3,q)$ not in $\mathcal{K}$ lies on at least one tangent to $\mathcal{K}$, hence $\varGamma_{3,q}(\mathcal{K})$ is not vertex-transitive by Corollary 2.5. The geometric automorphism group is $\mathrm{P}\Gamma \mathrm{L}(5,q)_{\mathcal{K}}$. Clearly, $\mathcal{K}$ extends uniquely to a hyperbolic quadric in $\operatorname{PG} (3,\sqrt {q})$ by adding the missing line of each regulus. Since the intersection point of these lines will be fixed by the stabiliser of $\mathcal{K}$, we find by Result 2.6 that $\mathrm{P}\Gamma\mathrm {L}(5,q)_{\mathcal{K}}\cong\operatorname{Persp} (H_{\infty}) \rtimes\mathrm{P}\Gamma\mathrm{L}(4,q)_{\mathcal{K}}$. Since the group stabilising the hyperbolic quadric also stabilises the subgeometry $\overline{\mathcal {K}}=\operatorname{PG} (3,\sqrt{q})$ and the canonical form of $Q^{+}(3,\sqrt{q})$ is fixed by $\operatorname{Aut}(\mathbb{F}_{q})$, we find $\mathrm{P}\Gamma \mathrm{L}(4,q)_{\mathcal{K}}\cong\operatorname{PGL}(4,\sqrt {q})_{\mathcal{K}} \rtimes \operatorname{Aut}(\mathbb{F}_{q}) \cong(( \operatorname{AGL}(1,\sqrt {q})\times\operatorname{AGL}(1,\sqrt {q}))\rtimes2 )\rtimes\operatorname{Aut}(\mathbb{F}_{q})$, by Result 2.7. Since $( \operatorname{AGL} (1,\sqrt {q})\times\operatorname{AGL}(1,\sqrt{q}))\rtimes2$ acts transitively on the points of $\mathcal{K}$, the graph is semisymmetric. □

4.4 $\mathcal{K}$ is contained in a cone

Let Π be a subgeometry $\operatorname{PG}(n,q_{0})$ embedded in $H_{\infty}=\operatorname{PG} (n,q)$, $q=q_{0}^{h}$. Let π be a hyperplane of Π. Consider a set ${\mathcal{O}}$ of $q_{0}^{h-1}$ points of π. Let V be a point of Π∖π and let $V{\mathcal{O}}$ denote the set of points of the cone in Π with vertex V and base ${\mathcal{O}}$. This set minus its vertex V has q points.

For a vertex v in a graph Γ and a positive integer i we write Γ _i (v) for the set of vertices at distance i from v.

Lemma 4.11

Let $\mathcal{K}$ be the cone $V{\mathcal{O}}$ of Π minus its vertex V, such that every point of $\pi\backslash{\mathcal{O}}$ lies on at least one tangent line to ${\mathcal{O}}$, then $\forall P \in\mathcal{P}, \forall L \in {\mathcal{L}}: \varGamma _{n,q}(\mathcal{K} )_{4}(P) \ncong\varGamma_{n,q}(\mathcal{K})_{4}(L)$.

Proof

Let $\varGamma=\varGamma_{n,q}(\mathcal{K})$. We will prove that, for every line $L \in{\mathcal{L}}$, the set of vertices Γ ₄(L) contains more than q−1 vertices that have all their neighbours in Γ ₃(L), while for every point $P \in\mathcal{P}$, there are exactly q−1 vertices in the set Γ ₄(P) that have all their neighbours in Γ ₃(P).

To prove the first claim, consider a line $L \in{\mathcal{L}}$ with $L \cap H_{\infty}= P_{1} \in\mathcal{K}$. Choose an affine point Q on L and a point $P_{2} \in\mathcal{K}$ different from P ₁. Take a point R on QP ₂, not equal to Q or P ₂, then clearly the line RP ₁∈Γ ₄(L). We will show that RP ₁ has all its neighbours in Γ ₃(L). Consider a neighbour S of RP ₁, i.e S∈RP ₁∖{P ₁}. The line SP ₂ meets L in a point T. Since T∈Γ ₁(L) and TP ₂∈Γ ₂(L), it follows that S∈Γ ₃(L). Clearly, any line $M \in{\mathcal{L}}$ through P ₁, such that 〈M,L〉∩H _∞ contains at least two points in $\mathcal{K}$, belongs to Γ ₄(L) and has all its neighbours in Γ ₃(L). Since the points of $\mathcal{K}$ do not lie on one line, there are more than q−1 such lines M.

Consider now a point $P \in\mathcal{P}$ and a point T∈Γ ₄(P). Look at the following minimal path of length 4 from T to P: the point T, a line Q ₁ P ₁∈Γ ₃(P) containing T for some $P_{1} \in\mathcal{K}$, an affine point Q ₁∈Γ ₂(P), the line PP ₂∈Γ ₁(P) containing Q ₁, for some $P_{2} \in\mathcal{K}$ different from P ₁, and finally the point P. Consider the point R=PT∩H _∞, then it follows from our construction that R lies on the line P ₁ P ₂. Since PR∉Γ ₁(P), we have R not in $\mathcal{K}$. First, suppose there is a tangent line of $\mathcal{K}$ through R, say RP ₃, with $P_{3}\in\mathcal{K}$. The line TP ₃ is a neighbour of T. If TP ₃ belongs to Γ ₃(P), then there exists a line PT′ through a point $P_{4}\in\mathcal{K}$, with T′ on TP ₃, which implies that RP ₃ contains the point $P_{4}\in\mathcal{K}$, a contradiction. Hence in this case there are neighbours of T that do not belong to Γ ₃(P). Now suppose there is no tangent line of $\mathcal{K}$ through R, then by construction, R is the vertex V of the cone. A line through V either contains 0 or q ₀ points of $\mathcal{K}$, so in this case, any neighbour of T belongs to Γ ₃(P). There are exactly q−1 points on the line VP different from P and V. □

Corollary 4.12

The graph $\varGamma_{n,q}(\mathcal{K})$ is not vertex-transitive.

Proof

Since any graph automorphism preserves distance and hence neighbourhoods, no automorphism of $\varGamma_{n,q}(\mathcal{K})$ can map a vertex in $\mathcal{P}$ to a vertex in ${\mathcal{L}}$. □

Denote the subgroup of PΓL(n+1,q) consisting of the perspectivities with centre V by $\operatorname{Persp}(V)$.

Lemma 4.13

Consider $\mathcal{K}$, the point set of the cone $V{\mathcal{O}}$ in $\operatorname{PG}(n,q_{0})$, minus its vertex V, where ${\mathcal{O}}$ spans π. If $\mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}$ and $\operatorname{PGL}(n,q_{0})_{\mathcal{O}}$, respectively, fix a point of π, then $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal {K}} \cong\operatorname{Persp}(V)\rtimes \mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}\rtimes (\operatorname{Aut}(\mathbb{F}_{q}) / \operatorname{Aut}(\mathbb {F}_{q_{0}}))$ and $\operatorname{PGL} (n+1,q)_{\mathcal{K}} \cong\operatorname{Persp}(V)\rtimes \operatorname{PGL}(n,q_{0})_{\mathcal{O}}$, respectively.

Proof

First, it should be noted that the $\mathbb{F}_{q_{0}}$-span of ${\mathcal{O}}$ is π, the $\mathbb{F}_{q_{0}}$-span of $\mathcal{K}$ is Π, so $\mathrm {P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ and $\operatorname{PGL}(n+1,q)_{\mathcal{K}}$ stabilise the subgeometry Π of H _∞. This implies that $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}\cong (\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\varPi})_{\mathcal{K}}$, and $\operatorname{PGL}(n+1,q)_{\mathcal{K}}\cong(\operatorname {PGL}(n+1,q)_{\varPi})_{\mathcal{K}}$, respectively. Since PΓL(n+1,q) _Π is clearly isomorphic to $\operatorname{PGL}(n+1,q_{0})\rtimes (\operatorname{Aut}(\mathbb{F}_{q}) / \operatorname{Aut}(\mathbb {F}_{q_{0}}))$, we have that $\mathrm{P}\Gamma\mathrm {L}(n+1,q)_{\mathcal{K}} \cong \mathrm{P}\Gamma\mathrm{L}(n+1,q_{0})_{\mathcal{K}}\rtimes (\operatorname{Aut}(\mathbb{F}_{q}) / \operatorname{Aut}(\mathbb {F}_{q_{0}}))$. Also, since $\operatorname{PGL}(n+1,q)_{\varPi}$ is isomorphic to $\operatorname {PGL}(n+1,q_{0})$, we have that $\operatorname{PGL}(n+1,q)_{\mathcal{K}}\cong\operatorname {PGL}(n+1,q_{0})_{\mathcal{K}}$.

Let ϕ be an element of $\mathrm{P}\Gamma\mathrm {L}(n+1,q_{0})_{\mathcal{K}}$, then ϕ preserves the lines through V. Define the action of ϕ on π to be the mapping taking L∩π to ϕ(L)∩π.

The kernel of this action of $\mathrm{P}\Gamma\mathrm {L}(n+1,q_{0})_{\mathcal{K}}$ on π is clearly isomorphic to $\operatorname{Persp}(V)$, as it consists of all collineations fixing the lines through V. The image of the action is isomorphic to $\mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}$, showing that $\mathrm{P}\Gamma\mathrm{L}(n+1,q_{0})_{\mathcal{K}}$ is an extension of $\operatorname{Persp}(V)$ by $\mathrm{P}\Gamma\mathrm {L}(n,q_{0})_{\mathcal{O}}$. To show that this extension splits, we embed $\mathrm{P}\Gamma\mathrm {L}(n,q_{0})_{\mathcal{O}}$ in $\mathrm{P}\Gamma\mathrm{L}(n+1,q_{0})_{\mathcal{K}}$ in such a way that it intersects trivially with $\operatorname{Persp}(V)$. By assumption, $\mathrm{P}\Gamma \mathrm{L}(n,q_{0})_{\mathcal{O}}$ fixes a point P∈π. W.l.o.g. let π be the hyperplane with equation X ₀=0 and let V be the point (1,0,…,0). Suppose that P has coordinates (0,c ₁,c ₂,…,c _n ), where the first non-zero coordinate equals one. This implies that for each $\beta\in\mathrm {P}\Gamma\mathrm{L} (n,q_{0})_{\mathcal{O}}$, there exists a unique n×n matrix B=(b _ij ), 1≤i,j≤n, and $\theta\in\operatorname{Aut}(\mathbb {F}_{q_{0}})$ corresponding to β, such that (c ₁,c ₂,…,c _n )^θ.B=(c ₁,c ₂,…,c _n ). Moreover, the obtained matrices B form a subgroup of ΓL(n,q ₀). Let A _β =(a _ij ), 0≤i,j≤n, be the (n+1)×(n+1) matrix with a ₀₀=1, a _i0=a _0j =0 for i,j≥1 and a _ij =b _ij for 1≤i,j≤n. It is clear that the semi-linear map (A _β ,θ) defines an element of $\mathrm{P}\Gamma\mathrm{L} (n+1,q_{0})_{\mathcal{K}}$, corresponding to a collineation α acting in the same way as β on H _∞. If θ is not the identity , then α is not a perspectivity. If , then α fixes every point on the line through P and V, thus fixes at least two affine points and hence is not a perspectivity. This implies that the elements α form a subgroup of $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ isomorphic to $\mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}$ and intersecting $\operatorname{Persp}(V)$ trivially. This implies that $\mathrm{P}\Gamma\mathrm{L} (n+1,q_{0})_{\mathcal{K}}\cong\operatorname{Persp}(V)\rtimes\mathrm {P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}$, and we have seen before that $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}\cong\mathrm{P}\Gamma\mathrm{L}(n+1,q_{0})_{\mathcal{K}} \rtimes (\operatorname{Aut}(\mathbb{F}_{q}) / \operatorname{Aut}(\mathbb {F}_{q_{0}}))$. Since $\operatorname{Persp}(V)$ intersects trivially with the standard embedding of $\operatorname{Aut}(\mathbb{F}_{q}) / \operatorname{Aut}(\mathbb{F}_{q_{0}})$, the claim follows.

The claim for $\operatorname{PGL}(n+1,q)_{\mathcal{K}}$ can be proved in the same way. □

The following corollary follows easily when we take into account that $\operatorname{Persp}(V)$ acts transitively on the points of each line through V.

Corollary 4.14

If $\mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}$ acts transitively on ${\mathcal{O}}$, then $\mathrm{P}\Gamma\mathrm{L} (n+1,q)_{\mathcal{K}}$ acts transitively on $\mathcal{K}$.

Theorem 4.15

Suppose that ${\mathcal{O}}$ spans π, that every point of $\pi\backslash{\mathcal{O}}$ lies on a tangent line to ${\mathcal {O}}$ and that $\mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}$ acts transitively on ${\mathcal{O}}$. Then the graph $\varGamma_{n,q}(\mathcal{K})$ is semisymmetric. Moreover, the geometric automorphism group is isomorphic to $\operatorname{Persp}(H_{\infty})\rtimes\operatorname{Persp} (V)\rtimes\mathrm{P}\Gamma\mathrm{L}(n,q_{0})_{\mathcal{O}}\rtimes (\operatorname{Aut}(\mathbb{F}_{q}) / \operatorname{Aut}( \mathbb{F}_{q_{0}}))$.

Proof

Since $\mathcal{K}$ consists of q points spanning $\operatorname {PG}(n,q)$, $\varGamma _{n,q}(\mathcal{K} )$ is q-regular and is connected by Result 2.1. The graph $\varGamma_{n,q}(\mathcal{K})$ is not vertex-transitive by Lemma 4.11. Clearly, $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ stabilises the point V, so we find by Result 2.6 that $\operatorname{PGL}(n+2,q)_{\mathcal{K}}\cong \operatorname{Persp}(H_{\infty}) \rtimes \operatorname{PGL}(n+1,q)_{\mathcal{K}}$. The expression for the geometric automorphism group follows from Lemma 4.13. Since $\mathrm{P}\Gamma\mathrm{L}(n+1,q)_{\mathcal{K}}$ acts transitively on the points of $\mathcal{K}$, by Theorem 2.8, the graph is edge-transitive, and hence semisymmetric. □

5 Isomorphisms of $\varGamma_{n,q}(\mathcal{K})$ with other graphs

In this section, we will show that the graphs constructed by Du, Wang and Zhang [9], and the graphs of Lazebnik and Viglione [19] belong to the family $\varGamma_{n,q}(\mathcal{K})$, where $\mathcal{K}$ is a q-arc contained in a normal rational curve (see Sect. 3.2).

5.1 The graph of Du, Wang and Zhang

If q=p prime, then the point of $\operatorname{PG}(n,p)$ with coordinates (0,…,0,1) and the orbit of the point P with coordinates (1,0,…,0) under the element $\phi\in\operatorname {PGL}(n+1,p)$ of order p, defined by the matrix A _ϕ , form a normal rational curve $\mathcal{N}$ in $\operatorname {PG}(n,p)$ (see [24]):

$$A_\phi= \left [ \begin{matrix}1&1&0&0&\cdots&0&0\\ 0&1&1&0&\cdots&0&0\\ 0&0&1&1&\cdots&0&0\\ &&&\cdots&&&\\ 0&0&0&0&\cdots&1&1\\ 0&0&0&0&\cdots&0&1 \end{matrix} \right ]. $$

When we use the orbit of P for the point set $\mathcal{K}$ at infinity, we obtain a reformulation of the construction of the semisymmetric graphs found by Du, Wang and Zhang in [9]. This shows that our construction of the graph $\varGamma_{n,p}(\mathcal{K})$, with $\mathcal {K}$ a set of p points, contained in a normal rational curve, contains their family (and extends their construction to the case where q is not a prime). Moreover, the edge-transitive group of automorphisms described by the authors is not the full automorphism group of the graph: they only consider automorphisms induced by the group 〈ϕ〉 of order p acting on the points of $\mathcal{K}$, together with $\operatorname {Persp}(H_{\infty})$.

5.2 The graph of Lazebnik and Viglione

In [19], the authors define the graph Λ _n,q as follows. Let $\mathcal{P}_{n}$ and $\mathcal{L}_{n}$ be two (n+1)-dimensional vector spaces over $\mathbb{F}_{q}$, q=p ^h. The vertex set of Λ _n,q is $\mathcal{P}_{n}\cup\mathcal{L}_{n}$, and we declare a point (p)=(p ₁,p ₂,…,p _n+1) adjacent to a line [l]=[l ₁,l ₂,…,l _n+1] if and only if the following n relations on their coordinates hold.

$$\begin{aligned} l_2+p_2 =& p_1l_1, \\ l_3+p_3 =& p_1l_2, \\ \vdots& \\ l_{n+1}+p_{n+1} =&p_1l_{n}. \end{aligned}$$

In the following theorem, we will show that the graph Λ _n,q is isomorphic to the graph $\varGamma_{n,q}(\mathcal{K})$, where $\mathcal{K}$ is contained in a normal rational curve; hence, $\varGamma_{n,q}(\mathcal{K})$ provides an embedding of the Lazebnik–Viglione graph in $\operatorname {PG}(n+1,q)$. It should be noted that in [19], the authors provide some automorphisms, acting on the graph Λ _n,q , to show that this graph is semisymmetric. From the isomorphism with $\varGamma_{n,q}(\mathcal{K})$ it follows that $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$ is also the full automorphism group of the Lazebnik–Viglione graph when q≥n+3 or q=p=n+2.

Theorem 5.1

$\varLambda_{n,q}\cong\varGamma_{n,q}(\mathcal{K})$, where $\mathcal{K}$ is a q-arc contained in a normal rational curve.

Proof

The graph Λ _n,q is isomorphic to the graph $\varLambda_{n,q}'$ obtained by reversing the role of the points and the lines in the definition of Λ _n,q . So, $\varLambda_{n,q}'$ is the bipartite graph with parts $\mathcal {P}_{n}$ and $\mathcal{L} _{n}$, where $(p_{1},\ldots,p_{n+1}) \in\mathcal{P}_{n}$ is incident with $(l_{1},\ldots,l_{n+1}) \in\mathcal{L}_{n}$ if and only if p _i+1+l _i+1=l ₁ p _i for all 1≤i≤n. Let ℓ=(l ₁,…,l _n+1) be a vertex of $\varLambda_{n,q}'$, then the points, incident with ℓ form a line of $\operatorname {AG}(n+1,q)$: suppose (p ₁,…,p _n+1) and $(p_{1}',\ldots,p_{n+1}')$ are vertices, adjacent with ℓ, then so is the vertex $(p_{1}+\lambda(p_{1}'-p_{1}),\ldots,p_{n+1}+\lambda(p_{n+1}'-p_{n+1}))$, for any $\lambda\in\mathbb{F}_{q}$.

Now let (p ₁,…,p _n+1) and $(p_{1}',\ldots,p_{n+1}')$ be vertices of $\varLambda_{n,q}'$ and embed these points of $\operatorname{AG}(n+1,q)$ in $\operatorname{PG}(n+1,q)$, by identifying (p ₁,…,p _n+1) with (1,p ₁,…,p _n+1). The line L determined by these points meets the hyperplane at infinity with equation X ₀=0 of $\operatorname{AG}(n+1,q)$ in the point $P_{\infty}=(0,p_{1}-p_{1}',\ldots ,p_{n+1}-p_{n+1}')$. Now the affine point set of L is a vertex of $\varLambda_{n,q}'$ if and only if there is an element $(l_{1},\ldots,l_{n+1})\in\mathcal {L}_{n}$ such that for all 1≤i≤n:

$$\begin{aligned} p_{i+1}+l_{i+1} =&l_1p_i, \\ p_{i+1}'+l_{i+1} =&l_1p_i'. \end{aligned}$$

This implies that there exists some $l_{1}\in \mathbb{F}_{q}$ such that $p_{i+1}-p_{i+1}'=l_{1}(p_{i}-p_{i}')$ for all 1≤i≤n. Hence, the point P _∞ has coordinates $(0,1,l_{1},l_{1}^{2},\ldots,l_{1}^{n})$, which implies that all the vertices (l ₁,…,l _n+1) of $\varLambda_{n,q}'$ define a line in $\operatorname{PG}(n+1,q)$ through a point of the standard normal rational curve $\mathcal{K}$, minus the point (0,…,0,1). This is exactly the description of the graph $\varGamma_{n,q}(\mathcal{K})$. □

Corollary 5.2

The automorphism group $\operatorname{Aut}(\varLambda _{n,q})$ of the graph Λ _n,q is isomorphic to the edge-transitive group $\mathrm{P}\Gamma\mathrm{L}(n+2,q)_{\mathcal{K}}$. Moreover:

• If q≥n+3, q=p ^h, p prime, n≥3 or n=2 and q odd, then $\operatorname{Aut}(\varLambda_{n,q})$ has size hq ⁿ⁺²(q−1)²;

• If q=p=n+2, then $\operatorname{Aut}(\varLambda_{n,q})$ has size q ⁿ⁺¹(q−1)q!.

5.3 The graph of Wenger and cycles in $\varGamma_{n,q}(\mathcal{K})$

We use the symbol C ^k for a cycle of length k. The infinite family of graphs H _n (q) introduced in [18] and [29] are clearly isomorphic to the graphs Λ _n−1,q of Sect. 5.2, and thus isomorphic to the graphs $\varGamma _{n-1,q}(\mathcal{K} )$, where $\mathcal{K}$ is a q-arc contained in a normal rational curve. Wenger [29] proved that the graphs H ₂(p), H ₃(p), H ₅(p) do not contain a C ⁴, C ⁶, C ¹⁰, respectively, for any prime p. In [18] the authors notice that, for a prime power q (implicitly assuming n≥5), the graph H _n (q) contains no C ¹⁰ and prove it has girth 8 for n≥3.

We now prove a similar theorem for the graph $\varGamma_{n,q}(\mathcal {K})$ using its geometric properties.

Theorem 5.3

Let $\mathcal{K}$ be any arc in $\operatorname{PG}(n,q)$, q≥n+1, then the graph $\varGamma_{n,q}(\mathcal{K})$ does not contain a C ⁴, C ⁶ and has girth 8. For n=2 and $|\mathcal{K}|\geq4$ or n=3, q>4 and $|\mathcal{K}|\geq5$, the graph $\varGamma_{n,q}(\mathcal {K})$ contains cycles of length 10. If n≥4, the graph $\varGamma_{n,q}(\mathcal{K})$ is C ¹⁰-free.

Proof

Since $\varGamma_{n,q}(\mathcal{K})$ is bipartite, every cycle has even length. Note that a cycle C ^2k of $\varGamma_{n,q}(\mathcal {K})$ contains k points of $\mathcal{P}$ and k lines of ${\mathcal{L}}$. Since there is at most one line of ${\mathcal{L}}$ through any two affine points, the graph does not contain a C ⁴. Suppose $\varGamma_{n,q}(\mathcal{K})$ contains a C ⁶, R ₁∼R ₁ R ₂∼R ₂∼R ₂ R ₃∼R ₃∼R ₃ R ₁, $R_{i} \in\mathcal {P}$, $R_{i} R_{j} \in{\mathcal{L}}$. Clearly, the affine points R ₁,R ₂,R ₃ are not collinear. The plane 〈R ₁,R ₂,R ₃〉 intersects H _∞ in a line. The lines R ₁ R ₂, R ₂ R ₃ and R ₃ R ₁ define three different points of $\mathcal{K}$, all lying on this line, a contradiction since $\mathcal{K}$ is an arc.

Consider two points $P_{1}, P_{2} \in\mathcal{K}$ and a plane π through P ₁ P ₂ not contained in H _∞. For i=1,2 consider distinct lines L _i through P ₁ and distinct lines M _i through P ₂, different from P ₁ P ₂. Define the intersection points R _ij =L _i ∩M _j . The path R ₁₁∼L ₁∼R ₁₂∼M ₂∼R ₂₂∼L ₂∼R ₂₁∼M ₁ is a cycle C ⁸. Since $\varGamma_{n,q}(\mathcal{K})$ does not contain a C ⁴ or C ⁶, it has girth 8.

Let $\mathcal{K}$ be an arc in $\operatorname{PG}(2,q)$ and let P ₁,P ₂,P ₃,P ₄ be four points of $\mathcal{K}$. Let R ₁ be an affine point. Let π be a plane through P ₃ P ₄, not through R ₁. Let R ₂ be π∩R ₁ P ₂ and R ₅ be π∩R ₁ P ₁. For q>2, we can choose an affine point R ₃ on R ₂ P ₃, different from R ₂, but not lying on the line P ₄ R ₅. Let R ₄ be the point R ₃ P ₄∩R ₅ P ₃. Then R ₁∼R ₁ R ₂∼⋯∼R ₅∼R ₅ R ₁, $R_{i} \in\mathcal{P}$, $R_{i} R_{j} \in{\mathcal{L} }$, is a cycle of length 10.

Suppose n=3 and $|\mathcal{K}|\geq5$. For q>4, one can consider five points P ₁,…,P ₅ in $\operatorname{PG}(4,q)$ disjoint from H _∞ forming a basis of $\operatorname{PG}(4,q)$. The five points P ₁ P ₂∩H _∞,…,P ₄ P ₅∩H _∞,P ₅ P ₁∩H _∞ form a frame of H _∞. Any five points of an arc in $\operatorname{PG}(3,q)$ form a frame and all frames are $\operatorname{PGL}$-equivalent. Hence, we can assume w.l.o.g. that these five points of H _∞ belong to $\mathcal{K}$. It is clear that P ₁∼P ₁ P ₂∼⋯∼P ₅∼P ₅ P ₁, $P_{i} \in\mathcal{P}$, $P_{i} P_{j} \in{\mathcal{L}}$, is a cycle of length 10.

Now let n≥4, let $\mathcal{K}$ be an arc and assume $\varGamma _{n,q}(\mathcal{K})$ contains a C ¹⁰, R ₁∼R ₁ R ₂∼⋯∼R ₅∼R ₅ R ₁, $R_{i} \in\mathcal{P}$, $R_{i} R_{j} \in{\mathcal{L}}$. Note that two lines at distance 2 intersect H _∞ in different points of $\mathcal{K}$; hence the five lines intersect H _∞ in at least three different points of $\mathcal{K}$. The space π=〈R ₁,R ₂,R ₃,R ₄,R ₅〉 has dimension at most 4 and at least 3, so intersects H _∞ in at most a 3-space, containing at most 4 points of $\mathcal{K}$. Hence there are at least two lines of our set intersecting in a point of $\mathcal{K}$, these lines are not at distance two of each other, so without loss of generality, assume these are the lines R ₁ R ₂ and R ₃ R ₄. It follows that π is a 3-space, intersecting H _∞ in a plane containing 3 points of $\mathcal{K}$. However, the points R ₁,R ₂,R ₃ and R ₄ lie in a plane containing two points P ₁ and P ₂ of $\mathcal {K}$. The point R ₅ does not lie in this plane, so the lines R ₄ R ₅ and R ₅ R ₁ intersect H _∞ in two new points P ₃ and P ₄. The points P ₁,P ₂,P ₃ and P ₄ lie in a plane of H _∞, a contradiction since $\mathcal{K}$ is an arc. □