On the equivalence of linear sets

Abstract

Let L be a linear set of pseudoregulus type in a line $\ell$ in $\varSigma ^*={\mathrm {PG}}(t-1,q^t)$, $t=5$ or $t>6$. We provide examples of q-order canonical subgeometries $\varSigma _1,\, \varSigma _2 \subset \varSigma ^*$ such that there is a $(t-3)$-subspace $\varGamma \subset \varSigma ^*\setminus (\varSigma _1 \cup \varSigma _2 \cup \ell )$ with the property that for $i=1,2$, L is the projection of $\varSigma _i$ from center $\varGamma$ and there exists no collineation $\phi$ of $\varSigma ^*$ such that $\varGamma ^{\phi }=\varGamma$ and $\varSigma _1^{\phi }=\varSigma _2$. Condition (ii) given in Theorem 3 in Lavrauw and Van de Voorde (Des Codes Cryptogr 56:89–104, 2010) states the existence of a collineation between the projecting configurations (each of them consisting of a center and a subgeometry), which give rise by means of projections to two linear sets. It follows from our examples that this condition is not necessary for the equivalence of two linear sets as stated there. We characterize the linear sets for which the condition above is actually necessary.

1 Introduction

If V is a vector space over the finite field ${\mathbb {F}}_{q^t}$, then ${\mathrm {PG}}_{q^t}(V)$ denotes the projective space whose points are the one-dimensional ${\mathbb {F}}_{q^t}$-subspaces of V. If V has dimension n over ${\mathbb {F}}_{q^t}$, then ${\mathrm {PG}}_{q^t}(V)={\mathrm {PG}}(n-1,q^t)$. For a point set $T\subset {\mathrm {PG}}(n-1,q^t)$ we denote by $\langle T \rangle$ the projective subspace of ${\mathrm {PG}}(n-1,q^t)$ spanned by the points in T. For $m \mid t$ and a set of elements $S\subset V$ we denote by $\langle S \rangle _{q^m}$ the ${\mathbb {F}}_{q^m}$-vector subspace of V spanned by the vectors in S. For the rest of the paper we assume that $q=p^e$ is a power of the prime p.

Let $R={\mathbb {F}}_{q^t}^r$. The field reduction map [8] ${\mathcal {F}}_{r,t,q}$ is a map from the points of ${\mathrm {PG}}_{q^t}(R)={\mathrm {PG}}(r-1,q^t)$ to the $(t-1)$-subspaces of ${\mathrm {PG}}_q(R)={\mathrm {PG}}(rt-1,q)$. Let $P={\mathrm {PG}}_{q^t}(T)$ be a point of ${\mathrm {PG}}(r-1,q^t)$, where T is a one-dimensional ${\mathbb {F}}_{q^t}$-subspace of R. Then ${\mathcal {F}}_{r,t,q}(P):={\mathrm {PG}}_q(T)$. As $\dim _{{\mathbb {F}}_q}(T)=t$, it follows that ${\mathcal {F}}_{r,t,q}(P)$ is a $(t-1)$-dimensional subspace of ${\mathrm {PG}}(rt-1,q)$. Denote the point set of ${\mathrm {PG}}(r-1,q^t)$ by ${\mathcal {P}}$. Then ${\mathcal {F}}_{r,t,q}({\mathcal {P}})$ is a Desarguesian $(t-1)$-spread of ${\mathrm {PG}}(rt-1,q)$ [13]. Let S be a subspace of ${\mathrm {PG}}(rt-1,q)={\mathrm {PG}}_q(R)$, then

$$\begin{aligned} {\mathcal {B}}(S):=\left\{ P \in {\mathrm {PG}}\left( r-1,q^t\right) :{\mathcal {F}}_{r,t,q}(P)\cap S \ne \emptyset \right\} . \end{aligned}$$

When S is an ${\mathbb {F}}_q$-vector space of R, then we define ${\mathcal {B}}(S)$ analogously. A point set $L \subseteq {\mathrm {PG}}(r-1,q^t)$ is said to be ${\mathbb {F}}_q$ -linear (or just linear) of rank n if $L={\mathcal {B}}(S)$ for some $(n-1)$-subspace $S\subseteq {\mathrm {PG}}(rt-1,q)$. If L has size $(q^n-1)/(q-1)$ (which is the maximum size for a linear set of rank n), then L is a scattered linear set. For further information on linear sets see [8, 17].

Definition 1

Let ${\mathrm {PG}}_{q^t}(V)={\mathrm {PG}}(n-1,q^t)$ and let W be an ${\mathbb {F}}_q$-vector subspace of V. If $\dim _{\,{\mathbb {F}}_q}(W)=\dim _{\,{\mathbb {F}}_{q^t}}(V)=n$ and $\langle W \rangle _{q^t}=V$, then $\varSigma '=\{ \langle w \rangle _{q^t} :w\in W^*\}$ is a canonical subgeometry of ${\mathrm {PG}}_{q^t}(V)$.

Let $\varSigma '\cong {\mathrm {PG}}(n-1,q)$ be a canonical subgeometry of $\varSigma ^*={\mathrm {PG}}(n-1,q^t)$. Let $\varGamma \subset \varSigma ^* \setminus \varSigma '$ be an $(n-1-r)$-subspace and let $\varLambda \subset \varSigma ^* \setminus \varGamma$ be an $(r-1)$-subspace of $\varSigma ^*$. The projection of $\varSigma '$ from center $\varGamma$ to axis $\varLambda$ is the point set

$$\begin{aligned} L=p_{\,\varGamma ,\,\varLambda }\left( \varSigma '\right) :=\left\{ \langle \varGamma , P \rangle \cap \varLambda :P\in \varSigma '\right\} . \end{aligned}$$

(1)

In [15] Lunardon and Polverino characterized linear sets as projections of canonical subgeometries. They proved the following.

Theorem 1

([15, Theorems 1 and 2]) Let $\varSigma ^*$, $\varSigma '$, $\varLambda$, $\varGamma$ and $L=p_{\,\varGamma ,\,\varLambda }(\varSigma ')$ be defined as above. Then L is an ${\mathbb {F}}_q$-linear set of rank n and $\langle L \rangle =\varLambda$. Conversely, if L is an ${\mathbb {F}}_q$-linear set of rank n of $\varLambda ={\mathrm {PG}}(r-1,q^t)\subset \varSigma ^*$ and $\langle L \rangle =\varLambda$, then there is an $(n-1-r)$-subspace $\varGamma$ disjoint from $\varLambda$ and a canonical subgeometry $\varSigma '\cong {\mathrm {PG}}(n-1,q)$ disjoint from $\varGamma$ such that $L=p_{\,\varGamma ,\,\varLambda }(\varSigma ')$.

Note that when $r=n$ in Theorem 1, then $L=p_{\,\varGamma ,\,\varLambda }(\varSigma ')=\varSigma '$, hence L is a canonical subgeometry. We rephrase here a theorem by Lavrauw and Van de Voorde.

Theorem 2

([6, Theorem 3]) For $i=1,2$, let $\varSigma _i \cong {\mathrm {PG}}(n-1,q)$ be two canonical subgeometries of $\varSigma ^*={\mathrm {PG}}(n-1,q^t)$ and let $\varGamma _i$ be two $(n-1-r)$-subspaces of $\varSigma ^*$, such that $\varGamma _i \cap \varSigma _i = \emptyset$. Let $\varLambda _i \subset \varSigma ^*\setminus \varGamma _i$ be two $(r-1)$-subspaces and let $L_i=p_{\,\varGamma _i,\,\varLambda _i}(\varSigma _i)$. Suppose that $L_i$ is not an ${\mathbb {F}}_q$-linear set of rank $k<n$. Then the following statements are equivalent.

• (i) There exists a collineation $\alpha :\varLambda _1 \rightarrow \varLambda _2$, such that ${L_1}^{\alpha }=L_2$,

• (ii) there exists a collineation $\beta$ of $\varSigma ^*$ such that ${\varSigma _1}^{\beta }=\varSigma _2$ and ${\varGamma _1}^{\beta }=\varGamma _2$,

• (iii) for all canonical subgeometries $\varSigma \cong {\mathrm {PG}}(n-1,q)$ in $\varSigma ^*$, there exist $\delta ,\phi ,\psi$ collineations of $\varSigma ^*$, such that¹ $\varSigma ^{\delta }=\varSigma$, ${\varGamma _1}^{\phi \delta }={\varGamma _2}^{\psi }$, ${\varSigma _1}^{\phi }=\varSigma$ and ${\varSigma _2}^{\psi }=\varSigma$.

We are going to prove that the implication (i) $\Rightarrow$ (ii) does not hold. In Sects. 3 and 4, we will show this in the case when $L_1=L_2$ is a linear set of pseudoregulus type in ${\mathrm {PG}}(1,q^n)$, for $n=5$ or $n>6$. In Sect. 5 a characterization will be given of the linear sets for which (i) $\Rightarrow$ (ii) holds.

In this paper, the results which lead to counterexamples are presented in a general setting. In the remainder of this introduction some hints are given to a better understanding of the key facts.

1.1 A minimal counterexample

The idea of a counterexample arose from the investigation of the linear sets of pseudoregulus type [3, 14]. A minimal counterexample to Theorem 2 (i) $\Rightarrow$ (ii) can be obtained as follows. In ${\mathrm {PG}}(4,q^5)$, with coordinates $X_1,X_2,\ldots ,X_5$, let $\varGamma$ be the plane of equations $X_1=X_2=0$, and let $\ell$ be the line $X_3=X_4=X_5=0$. The sets

$$\begin{aligned} \varSigma _1=\left\{ \left\langle \left( \lambda ,\lambda ^q,\lambda ^{q^2},\lambda ^{q^3},\lambda ^{q^4}\right) \right\rangle _{q^5}:\lambda \in {\mathbb {F}}_{q^5}^*\right\} \end{aligned}$$

and

$$\begin{aligned} \varSigma _2=\left\{ \left\langle \left( \lambda ,\lambda ^{q^2},\lambda ^q,\lambda ^{q^3},\lambda ^{q^4}\right) \right\rangle _{q^5}:\lambda \in {\mathbb {F}}_{q^5}^*\right\} \end{aligned}$$

are canonical subgeometries. Defining ${\mathrm {PG}}(4,q)=\{\langle x\rangle _q :x\in {\mathbb {F}}_{q^5}^*\}$, the map $\varphi _1:{\mathrm {PG}}(4,q)\rightarrow \varSigma _1$ defined by $\langle x\rangle _q^{\varphi _1}=\langle x,x^q,x^{q^2},x^{q^3},x^{q^4}\rangle _{q^5}$ is a collineation, as well as the map $\varphi _2:{\mathrm {PG}}(4,q)\rightarrow \varSigma _2$ similarly defined. By [3, 14], the projections $p_{\,\varGamma ,\,\ell }(\varSigma _1)$ and $p_{\,\varGamma ,\,\ell }(\varSigma _2)$ coincide, and are a linear set of pseudoregulus type, say ${\mathbb {L}}=\{\langle (\lambda ,\lambda ^q,0,0,0)\rangle _{q^5}:\lambda \in {\mathbb {F}}_{q^5}^*\}$.

Now assume that as stated in Theorem 2 (ii) a collineation $\beta$ of ${\mathrm {PG}}(4,q^5)$ exists such that $\varGamma ^\beta =\varGamma$, and $\varSigma _1^\beta =\varSigma _2$. It is shown in Theorem 5 that a projectivity exists having the same two properties; call it $\phi$. By the arguments in Lemma 1 in Sect. 4, the collineation $\varphi _2\phi ^{-1}\varphi _1^{-1}$ of ${\mathrm {PG}}(4,q)$ is the composition of a projectivity and either the map $\langle x\rangle _q\mapsto \langle x^{\theta _1}\rangle _q$ (cfr. (6)), or the map

$$\begin{aligned} \left\langle x\right\rangle _q\mapsto \left\langle x^{-\theta _1}\right\rangle _q=\left\langle x^{\theta _4-\theta _1}\right\rangle _q=\left\langle x^{q^2\theta _2}\right\rangle _q, \end{aligned}$$

which in turn is the composition of a projectivity and $\langle x\rangle _q\mapsto \langle x^{\theta _2}\rangle _q$. However, by Corollary 2 in Sect. 3, no map of type $\langle x\rangle _q\mapsto \langle x^{\theta _s}\rangle _q$ with $0<s<t=5$ is a collineation, and this is a contradiction.

1.2 Construction of counterexamples using condition (A) in Section 5

In Sect. 5 counterexamples are found with essentially distinct arguments. Theorem 7 states that a counterexample to Theorem 2 (i) $\Rightarrow$ (ii) can always be constructed in case the following condition does not hold: (A) For any two $(n-1)$-subspaces $U,U' \subset {\mathrm {PG}}(rt-1,q)$, such that ${\mathcal {B}}(U)=L={\mathcal {B}}(U')$, a collineation $\gamma \in {\mathrm {P}}\Gamma {\mathrm {L}}(rt,q)$ exists, such that $U^{\gamma }=U'$, $\gamma$ preserves the Desarguesian spread ${\mathcal {F}}_{r,t,q}({\mathcal {P}})$, and the induced map on ${\mathrm {PG}}(r-1,q^t)$ is a collineation.

If $L={\mathcal {B}}({\mathcal {Q}}_{t-1,q})$, where ${\mathcal {Q}}_{t-1,q}\subset {\mathrm {PG}}(2t-1,q)$ is the nonsingular hypersurface of degree t studied in [5], then L is a linear set of pseudoregulus type in ${\mathrm {PG}}(1,q^t)$.

In [5] all linear subspaces contained in ${\mathcal {Q}}_{t-1,q}$ are described for $q\ge t$. For $t\ge 4$ there exist two $(t-1)$-subspaces, say U and $U'$, that are contained in ${\mathcal {Q}}_{t-1,q}$, and such that for any element $\gamma$ of the stabilizer of ${\mathcal {Q}}_{t-1,q}$, preserving the standard Desarguesian spread, it holds $U^\gamma \ne U'$. For $t\ge 5$, $t\ne 6$, it is possible to choose U and $U'$ with the additional property ${\mathcal {B}}(U)=L={\mathcal {B}}(U')$. This leads to the counterexample of the previous subsection and its generalisations.

1.3 On the proof of [6, Theorem 3]

In the opinion of the authors of this paper, the proof of [6, Theorem 3] contains a wrong argument. The map $\delta$ defined at p. 93, line 21, is declared to be a projectivity, but is dealt with as a linear map e.g. in (1), as well as $p_2$ is. Such a $\delta$ is assumed to satisfy both conditions (i) $\delta$ maps an ${\mathbb {F}}_q$-vector space associated with $\varSigma _1^\chi$ onto an ${\mathbb {F}}_q$-vector space associated with $\varSigma _2$, and (ii) $p_2(a_i)=p_2(a_i^\delta )$, $i=0,\ldots ,m$. However, in the case of the minimal example in Sect. 1.1, no such $\delta$ exists.

2 Linear sets of pseudoregulus type in a projective line

A family of scattered ${\mathbb {F}}_q$-linear sets of rank tm of ${\mathrm {PG}}(2m-1,q^t)$, called of pseudoregulus type, have been introduced in [16] for $m=2$ and $t=3$, further generalized in [7] for $m\ge 2$ and $t=3$ and finally in [14] for $m\ge 1$ and $t\ge 2$ (for $t=2$ they are the same as Baer subgeometries, see [14, Remark 3.4]). We will only consider linear sets of pseudoregulus type in ${\mathrm {PG}}(1,q^t)$. It has been proved in [14, Sect. 4] and in [3, Remark 2.2] that all linear sets of pseudoregulus type in ${\mathrm {PG}}(1,q^t)$ are ${\mathrm {PGL}}(2,q^t)$-equivalent and hence we can define them as follows.

Definition 2

([3, 14]) A point set L of ${\mathrm {PG}}_{q^t}({\mathbb {F}}_{q^t}^2)={\mathrm {PG}}(1,q^t)$, $t\ge 2$, is called a linear set of pseudoregulus type if L is projectively equivalent to

$$\begin{aligned} {\mathbb {L}}:=\left\{ \langle \left( \lambda ,\lambda ^q\right) \rangle _{q^t} :\lambda \in {\mathbb {F}}_{q^t}^*\right\} . \end{aligned}$$

(2)

Definition 3

Let $t\ge 2$ be an integer and let $\pi$ be a permutation of $N_{t-1}:=\{0,1,2,\ldots ,t-1\}$. We define $\varSigma _{\pi }$ as the following set of points in ${\mathrm {PG}}_{q^t}({\mathbb {F}}_{q^t}^t)={\mathrm {PG}}(t-1,q^t)$.

$$\begin{aligned} \varSigma _{\pi }:=\left\{ \left\langle \left( \alpha ^{q^{\pi (0)}}, \alpha ^{q^{\pi (1)}},\alpha ^{q^{\pi (2)}},\ldots ,\alpha ^{q^{\pi (t-1)}}\right) \right\rangle _{q^t} :\alpha \in {\mathbb {F}}_{q^t}^*\right\} . \end{aligned}$$

(3)

Proposition 1

Consider ${\mathbb {F}}_{q^t}$ as a vector space over ${\mathbb {F}}_q$ and let $\varSigma ={\mathrm {PG}}_q({\mathbb {F}}_{q^t})\cong {\mathrm {PG}}(t-1,q)$. Let $\pi$ be a permutation of $N_{t-1}$. Then the following statements hold.

• 1. $S_{\pi }:=\{(\alpha ^{q^{\pi (0)}},\alpha ^{q^{\pi (1)}},\alpha ^{q^{\pi (2)}},\ldots ,\alpha ^{q^{\pi (t-1)}}) :\alpha \in {\mathbb {F}}_{q^t}\}$ is an ${\mathbb {F}}_q$-subspace of ${\mathbb {F}}_{q^t}^t$.

• 2. $\phi _{\pi }' :{\mathbb {F}}_{q^t} \rightarrow S_{\pi } :\alpha \mapsto (\alpha ^{q^{\pi (0)}},\alpha ^{q^{\pi (1)}},\alpha ^{q^{\pi (2)}},\ldots ,\alpha ^{q^{\pi (t-1)}})$ is a vector space isomorphism.

• 3. $\langle \varSigma _{\pi } \rangle ={\mathrm {PG}}(t-1,q^t)$.

• 4. The map

  $$\begin{aligned} \phi _{\pi } :\varSigma \rightarrow \varSigma _{\pi } :\langle \alpha \rangle _q \mapsto \left\langle \left( \alpha ^{q^{\pi (0)}},\alpha ^{q^{\pi (1)}}, \alpha ^{q^{\pi (2)}},\ldots ,\alpha ^{q^{\pi (t-1)}}\right) \right\rangle _{q^t} \end{aligned}$$

  (4)

  is a collineation.

Proof

1. and 2. are trivial, and 1., 2., 3. together imply 4., so it is enough to show 3. Let $\alpha _1,\alpha _2,\ldots ,\alpha _t\in {\mathbb {F}}_{q^t}$ and denote by $M_{\pi }$ the $t \times t$ matrix over ${\mathbb {F}}_{q^t}$ whose jth row is $\phi _{\pi }'(\alpha _j)$ for $j=1,2,\ldots ,t$. According to [12, Lemma 3.51], $\det (M_{id})\ne 0$ if and only if $\alpha _1,\alpha _2,\ldots ,\alpha _t$ are linearly independent over ${\mathbb {F}}_q$. As $M_{\pi }$ can be obtained by permuting the columns of $M_{id}$, it follows that $\det (M_{\pi })\ne 0$ if and only if $\det (M_{id})\ne 0$. As $\dim _{{\mathbb {F}}_q}({\mathbb {F}}_{q^t})=t$, it follows that $\dim _{{\mathbb {F}}_{q^t}}(S_{\pi })=t$ and hence 3. follows. $\square$

Let $\varSigma ^*={\mathrm {PG}}(t-1,q^t)$, and denote the coordinates in $\varSigma ^*$ by $X_1$, $X_2$, $\ldots ,X_t$. Let $\varLambda \cong {\mathrm {PG}}(1,q^t)$ be the line $X_3=\ldots =X_{t}=0$ and let $\varGamma \cong {\mathrm {PG}}(t-3,q^t)$ be the $(t-3)$-subspace $X_1=X_2=0$ in $\varSigma ^*$. Let $\pi$ be a permutation of $N_{t-1}$, it is easy to see that if $\gcd (\pi (1)-\pi (0),t)=1$, then $p_{\, \varGamma ,\, \varLambda }(\varSigma _{\pi })={\mathbb {L}}$, see for example [14, Remark 4.2]. If $\phi$ is a collineation of $\varSigma ^*$ such that $\varSigma _{id}^{\phi }=\varSigma _{\pi }$, then

$$\begin{aligned} \widetilde{\phi } :=\phi _{\pi }\phi ^{-1}\phi _{id}^{-1} \end{aligned}$$

(5)

[see (4)] is a collineation of $\varSigma$. To study the action of $\widetilde{\phi }$ we need some results regarding the $\varTheta _s$ map defined in the next section.

3 The $\varTheta _s$ map

As before, let $\varSigma$ denote the ${\mathrm {PG}}(t-1,q)$ associated to ${\mathbb {F}}_{q^t}$.

Definition 4

For an integer $s\ge -1$ let $\theta _s$ denote the number of points of ${\mathrm {PG}}(s,q)$, that is

$$\begin{aligned} \theta _s=\frac{q^{s+1}-1}{q-1}. \end{aligned}$$

(6)

The map $\varTheta _s :\varSigma \rightarrow \varSigma$ is defined as follows. For each $x\in {\mathbb {F}}_{q^t}^*$ let

$$\begin{aligned} \varTheta _s(\langle x \rangle _q)=\langle x^{\theta _s} \rangle _q. \end{aligned}$$

Remark 1

If $s,\,d \ge -1$ are two integers, then $\varTheta _s=\varTheta _d$ if and only if $s\equiv d \pmod t$. As we do not use this result, the proof is left to the reader.

In Sect. 4 we show the following. If $\phi$ is a collineation such that $\varSigma _{id}^{\phi }=\varSigma _{\pi }$ and $\varGamma ^{\phi }=\varGamma$, then $\widetilde{\phi }=\varTheta _s \varOmega$ [see (5)] for some collineation $\varOmega$ and integer $0\le s < t$. Note that $\varTheta _0$ is the identity. In Corollary 2, we prove that $\varTheta _s$, with $0< s <t$, is not a collineation of ${\mathrm {PG}}(t-1,q)$. It will follow that in most cases $\widetilde{\phi }$ is not a collineation and hence such $\phi$ cannot exist.

To determine whether $\varTheta _s$ is a collineation or not we need some results regarding difference sets. A $(v,k,\lambda )$ -difference set is a k-subset D of a group G of order v such that the list of non-zero differences (if the group is written additively) contains each non-zero group element exactly $\lambda$ times. If the group is cyclic etc., then we say that the difference set is cyclic etc. For our purposes here, it is enough to consider cyclic difference sets, so from now on we assume that G is cyclic. Let D be a $(v,k,\lambda )$-difference set of G and let j be an integer. By $D+j$ and jD we mean $\{d + j :d \in D \}$ modulo v and $\{ jd :d\in D\}$ modulo v. Note that $D+j$ is also a $(v,k,\lambda )$-difference set of G. If $\gcd (j,v)\ne 1$, then let $t:=v/\gcd (j,v)$ and let $d_1-d_2=t$, where $d_1,\,d_2\in D$. As $v \mid tj$, it follows that $jd_1 \equiv jd_2 \pmod v$ and hence $|jD|<|D|$. On the other hand, if $\gcd (j,v)=1$, then jD is also a $(v,k,\lambda )$-difference set of G. The integer m is a (numerical) multiplier of D if $mD=D+g$ for some integer g. The trace of $\beta \in {\mathbb {F}}_{q^t}$ is ${{\mathrm{Tr}}}(\beta )={{\mathrm{Tr}}}_{{\mathbb {F}}_{q^t}/{\mathbb {F}}_q}(\beta )=\sum _{i=0}^{t-1}\beta ^{q^i}$.

Theorem 3

([18, Theorem 2.1.1]) Let $\alpha$ be a generator of the multiplicative group of ${\mathbb {F}}_{q^t}$, $t\ge 3$. The set of integers $D:=\{i :0 \le i < \theta _{t-1}, {{\mathrm{Tr}}}(\alpha ^i)=0 \}$ modulo $\theta _{t-1}$ forms a (cyclic) difference set in ${\mathbb {Z}}_{\theta _{t-1}}$ (written additively), with parameters $\left( \theta _{t-1},\theta _{t-2},\theta _{t-3}\right)$. The set of points of ${\mathrm {PG}}_q({\mathbb {F}}_{q^t})\cong {\mathrm {PG}}(t-1,q)$ which corresponds to the set D, i. e. the point set ${\mathcal {H}}_D:=\{\langle \alpha ^i \rangle _q :i \in D\}$, is a hyperplane of ${\mathrm {PG}}(t-1,q)$. Moreover, the hyperplanes of ${\mathrm {PG}}(t-1,q)$ are the sets ${\mathcal {H}}_{D+j}:=\{\langle \alpha ^i \rangle _q :i \in D+j\}$, where $0\le j < \theta _{t-1}$.

The difference sets constructed in Theorem 3 are called classical Singer difference sets. The next theorem describes the multipliers of these difference sets.

Theorem 4

([18, Corollary 1.3.4 and Proposition 3.1.1]) Let D be a classical Singer difference set with parameters as in Theorem 3. Then the multipliers of D are the powers of p modulo $\theta _{t-1}$.

For a subset ${\mathcal {X}}\subseteq \varSigma$ and an integer m, let ${\mathcal {X}}^m= \{\langle x^m \rangle _q :\langle x \rangle _q \in {\mathcal {X}}\}$.

Corollary 1

Let $t\ge 3$ and m be two integers. The $\varSigma \rightarrow \varSigma$ map, $\langle x \rangle _q \mapsto \langle x^m \rangle _q$ is a collineation if and only if $m\equiv p^h \pmod {\theta _{t-1}}$, for some $h\in {\mathbb {N}}$.

Proof

The map $\langle x \rangle _q \mapsto \langle x^m \rangle _q$ is a collineation if and only if it maps hyperplanes into hyperplanes. Let $\alpha$, D and ${\mathcal {H}}_{D+j}$ for $0\le j <\theta _{t-1}$ be defined as in Theorem 3. For any hyperplane ${\mathcal {H}}_{D+j}$ of $\varSigma$ the point set ${\mathcal {H}}_{D+j}^m=\{ \langle \alpha ^{mi} \rangle _q :i\in D+j \}$ is a hyperplane if and only if $\{mi :i\in D+j \}=mD+mj$ is a translate of D, i.e when $mD+mj=D+k$ for some k and hence $mD=D+g$ (with $g=k-mj$), which is equivalent to saying that m is a multiplier of D. The assertion follows from Theorem 4. $\square$

Corollary 2

Let $t\ge 3$. For $0<s<t$ the map $\varTheta _s$ is not a collineation.

Proof

According to Corollary 1, it is enough to show that for any $0<s<t$ there is no integer h such that $\theta _s \equiv p^h \pmod {\theta _{t-1}}$. Note that $q^t=(q-1)\theta _{t-1}+1$, thus $p^0=1\equiv q^t \pmod {\theta _{t-1}}$. It is therefore enough to show the assertion when $p^h < q^t$. Suppose, contrary to our claim, that $\theta _s \equiv p^h \pmod {\theta _{t-1}}$ for some $p^h<q^t$ and $0<s<t$. Since $p^h \not \equiv 0 \pmod {\theta _{t-1}}$, we have $s\le t-2$ and

$$\begin{aligned} \theta _s q^t/p^h \equiv q^t\equiv 1 \pmod {\theta _{t-1}}. \end{aligned}$$

(7)

It is easy to see that $p^h\ne 1$. If $1<p^h<\theta _{t-1}$, then $p^h=\theta _s$, which cannot be as p does not divide $\theta _s$. So we may assume $\theta _{t-1} < p^h < q^t$. In this case $p^h < q^t(p^h-\theta _{t-1})$, thus $q^t \theta _{t-1}<p^h(q^t-1)$ and hence $q^t/p^h < (q^t-1)/\theta _{t-1}=q-1$. On the other hand we have $1 < q^t/p^h$ and hence

$$\begin{aligned} 1< \theta _s q^t/p^h < \theta _{t-2} (q-1) = q^{t-1}-1 < \theta _{t-1}, \end{aligned}$$

contrary to (7). $\square$

Proposition 2

For $0 \le s<t$, the map $\langle x \rangle _q \mapsto \langle x^{-\theta _s} \rangle _q$ is the composition of $\varTheta _{t-s-2}$ and a projectivity of $\varSigma$.

Proof

For each $x\in {\mathbb {F}}_{q^t}^*$ we have $x^{\theta _{t-1}}\in {\mathbb {F}}_{q}^*$, thus

$$\begin{aligned} \left\langle x^{-\theta _s}\right\rangle _q=\left\langle x^{\theta _{t-1}-\theta _s}\right\rangle _q=\varTheta _{t-s-2}\left\langle x^{q^{s+1}} \right\rangle _q. \end{aligned}$$

As $\langle x \rangle _q \mapsto \langle x^{q^h} \rangle _q$ is a projectivity for each h, the assertion follows. $\square$

4 Linear sets obtained via non-equivalent projections

Definition 5

For an integer h let $\varPhi _h$ denote the collineation of $\varSigma ^*={\mathrm {PG}}(t-1,q^t)$, defined as

$$\begin{aligned} \left\langle \left( \alpha _0,\alpha _1,\ldots , \alpha _{t-1}\right) \right\rangle _{q^t}^{\varPhi _h}= \left\langle \left( \alpha _0^{p^h},\alpha _1^{p^h},\ldots ,\alpha _{t-1}^{p^h}\right) \right\rangle _{q^t}\!. \end{aligned}$$

Proposition 3

For each integer h and permutation $\pi$ of $N_{t-1}$, $\varPhi _h$ fixes $\varSigma _{\pi }$.

Proof

For each $\alpha \in {\mathbb {F}}_{q^t}^*$ we have

$$\begin{aligned} \left\langle \left( \alpha ^{q^{\pi (0)}},\alpha ^{q^{\pi (1)}},\ldots ,\alpha ^{q^{\pi (t-1)}}\right) \right\rangle _{q^t}^{\varPhi _h}=\left\langle \left( \beta ^{q^{\pi (0)}},\beta ^{q^{\pi (1)}},\ldots ,\beta ^{q^{\pi (t-1)}}\right) \right\rangle _{q^t}\!, \end{aligned}$$

where $\beta =\alpha ^{p^h}$. $\square$

Proposition 4

Let $\pi$ be a permutation of $N_{t-1}$ and let $\omega$ be the permutation defined such that $\omega (i)\equiv \pi (i)-\pi (0) \pmod t$ for each $i\in N_{t-1}$. Then $\varSigma _{\pi } = \varSigma _{\omega }$.

Proof

Let $h=-\pi (0)e$, where $q=p^e$, p prime. It is easy to see that $\varPhi _h(\varSigma _{\pi })=\varSigma _{\omega }$. As $\varPhi _h$ fixes $\varSigma _{\pi }$ the assertion follows. $\square$

Lemma 1

Let $\varGamma \cong {\mathrm {PG}}(t-3,q^t)$ be the $(t-3)$-space $X_1=X_2=0$ in $\varSigma ^*={\mathrm {PG}}(t-1,q^t)$, $t\ge 3$. If $\phi$ is a projectivity of $\varSigma ^*$ and $\pi$ is a permutation of $N_{t-1}$ such that $\varSigma _{id}^{\phi }=\varSigma _{\pi }$ and $\varGamma ^{\phi }=\varGamma$, then $\pi (1)-\pi (0)\equiv \pm 1 \pmod t$, or $\gcd (\pi (1)-\pi (0),t)>1$.

Proof

Suppose $\gcd (\pi (1)-\pi (0),t)=1$. Let $M=(m_{ij})_{1\le i,j\le t}\in {\mathrm {GL}}(t,q^t)$ be a matrix associated with $\phi$. As $\phi$ fixes $\varGamma$, we have $m_{ij}=0$ when $i=1,2$ and $3 \le j \le t$. Denote the $2\times 2$ matrix $(m_{ij})_{1\le i,j\le 2}$ by A. As M is non-singular, the same holds for A. According to Proposition 4 we may assume $\pi (0)=0$. Let $\mu =\pi (1)$, whence $\gcd (\mu ,t)=1$. As $\varSigma _{id}^{\phi }=\varSigma _{\pi }$, for each $\alpha \in {\mathbb {F}}_{q^t}^*$ there exist $\delta _{\alpha }, t_{\alpha } \in {\mathbb {F}}_{q^t}^*$ such that

$$\begin{aligned} \delta _{\alpha }t_{\alpha }= & {} m_{11}\alpha + m_{12}\alpha ^q, \end{aligned}$$

(8)

$$\begin{aligned} \delta _{\alpha }t_{\alpha }^{q^{\mu }}= & {} m_{21}\alpha + m_{22}\alpha ^q. \end{aligned}$$

(9)

Let N denote the norm function from ${\mathbb {F}}_{q^t}$ to ${\mathbb {F}}_q$, that is $N(x)=x^{\theta _{t-1}}$. As $N(\delta _{\alpha }t_{\alpha }^{q^{\mu }})=N(\delta _{\alpha }t_{\alpha })N(t_{\alpha }^{q^{\mu }-1})=N(\delta _{\alpha }t_{\alpha })$, it follows that

$$\begin{aligned} N\left( m_{11}\alpha + m_{12}\alpha ^q\right)= & {} N\left( m_{21}\alpha + m_{22}\alpha ^q\right) \!,\\ N\left( m_{11} + m_{12}\alpha ^{q-1}\right)= & {} N\left( m_{21} + m_{22}\alpha ^{q-1}\right) \!, \end{aligned}$$

and hence for each $(q-1)$th power z of ${\mathbb {F}}_{q^t}^*$ we have

$$\begin{aligned} P(z):=N(m_{11} + m_{12}z)-N(m_{21} + m_{22}z)=0, \end{aligned}$$

where P(z) is a polynomial of degree at most $\theta _{t-1}$.

We claim $m_{11}m_{21}=0$ and $m_{12}m_{22}=0$. First suppose $m_{11}m_{21}\ne 0$. Then the coefficient of $z^q$ in P(z) is

$$\begin{aligned} m_{12}^q N(m_{11})/m_{11}^q-m_{22}^q N(m_{21})/m_{21}^q. \end{aligned}$$

(10)

The coefficient of $z^{q+1}$ is

$$\begin{aligned} m_{12}^{q+1}N(m_{11})/m_{11}^{q+1}-m_{22}^{q+1}N(m_{21})/m_{21}^{q+1}. \end{aligned}$$

(11)

As P(z) vanishes for each z with $z^{\theta _{t-1}}=1$, we have $P(z)=a(z^{\theta _{t-1}}-1)$ for some $a\in {\mathbb {F}}_{q^t}$. It follows that (10) and (11) are zero, thus if $m_{12}m_{22}\ne 0$, then

$$\begin{aligned} m_{12}/m_{11}=m_{22}/m_{21}, \end{aligned}$$

and hence $\det (A)=0$, a contradiction. On the other hand if one of $\{m_{12},m_{22}\}$ is zero, then both of them are zero, thus we obtain again $\det (A)=0$. Now suppose $m_{12}m_{22}\ne 0$. Then the coefficient of $z^{\theta _{t-1}-q}$ in P(z) is

$$\begin{aligned} m_{11}^q N(m_{12})/m_{12}^q-m_{21}^q N(m_{22})/m_{22}^q. \end{aligned}$$

(12)

As before, it follows that (12) is zero. Together with $m_{11}m_{21}=0$, it means that $m_{11}$ and $m_{21}$ are both zero, and hence $\det (A)=0$, a contradiction.

First we consider the case when A is diagonal, i.e. $m_{12}=m_{21}=0$. We may assume $m_{11}=1$. Then (8) and (9) yield

$$\begin{aligned} t_{\alpha }^{q^{\mu }-1}=m_{22}\alpha ^{q-1}\!. \end{aligned}$$

The last equation implies that a $\rho \in {\mathbb {F}}_{q^n}^*$ exists such that ${\rho }^{q-1}=m_{22}$. Then for any $x\in {\mathbb {F}}_{q^t}^*$ the following holds:

$$\begin{aligned} \left\langle \frac{x^{\theta _{\mu -1}}}{\rho }\right\rangle _q^{\phi _{id}\phi }= & {} \left\langle \left( \frac{x^{\theta _{\mu -1}}}{\rho },\frac{x^{q\theta _{\mu -1}}}{\rho ^q},\ldots , \frac{x^{q^{t-1}\theta _{\mu -1}}}{\rho ^{q^{t-1}}}\right) \right\rangle ^{\phi }_{q^t}\\= & {} \left\langle \left( \frac{x^{\theta _{\mu -1}}}{\rho },\frac{x^{q\theta _{\mu -1}}}{\rho },* \right) \right\rangle _{q^t}=\left\langle \left( 1,x^{q^\mu -1},*\right) \right\rangle _{q^t}\in \varSigma _\pi ;\\ \langle x\rangle _q^{\phi _{\pi }}= & {} \left\langle \left( x,x^{q^\mu },*\right) \right\rangle _{q^t}=\left\langle \left( 1,x^{q^\mu -1},*\right) \right\rangle _{q^t}\in \varSigma _\pi \!. \end{aligned}$$

Since $\mu$ and t are coprime, $\varSigma _\pi$ contains exactly one element of the form $\langle (1,x^{q^\mu -1},*)\rangle _{q^t}$. It follows that for any $x\in {\mathbb {F}}_{q^t}^*$

$$\begin{aligned} \left\langle x^{\theta _{\mu -1}}/\rho \right\rangle _q^{\phi _{id}\phi }= \langle x\rangle _q^{\phi _{\pi }}\!, \end{aligned}$$

and hence $\phi _{\pi }\phi ^{-1}\phi _{id}^{-1} :\varSigma \rightarrow \varSigma$ which maps $\langle x \rangle _q$ to $\langle x^{\theta _{\mu -1}}/\rho \rangle _q$ is a collineation. Since the map $\langle y \rangle _q \mapsto \langle y/\rho \rangle _q$ is a collineation of $\varSigma$, Corollary 2 yields that $\phi _{\pi }\phi ^{-1}\phi _{id}^{-1}$ is a collineation only if $\mu =1$.

Now consider the case when $m_{11}=m_{22}=0$. We may assume $m_{12}=1$. Then (8) and (9) yield

$$\begin{aligned} t_{\alpha }^{q^{\mu }-1}=m_{21}\alpha ^{1-q}\!. \end{aligned}$$

This implies that a $\rho \in {\mathbb {F}}_{q^t}^*$ exists such that $\rho ^{q-1}=m_{21}$. As before, it follows that

$$\begin{aligned} \left\langle x^{-\theta _{\mu -1}}\rho \right\rangle _q^{\phi _{id}\phi }=\langle x\rangle _q^{\phi _{\pi }}\!, \end{aligned}$$

for each $x\in {\mathbb {F}}_{q^t}^*$ and hence $\phi _{\pi }\phi ^{-1}\phi _{id}^{-1} :\varSigma \rightarrow \varSigma$ which maps $\langle x \rangle _q$ to $\langle x^{-\theta _{\mu -1}}\rho \rangle _q$ is a collineation. Proposition 2 and Corollary 2 yield that this happens only if $\mu =t-1$. $\square$

Theorem 5

Let $\varGamma \cong {\mathrm {PG}}(t-3,q^t)$ be the $(t-3)$-space $X_1=X_2=0$ in $\varSigma ^*={\mathrm {PG}}(t-1,q^t)$, $t\ge 3$. If $\phi$ is a collineation of $\varSigma ^*$ such that $\varSigma _{id}^{\phi }=\varSigma _{\pi }$ and $\varGamma ^{\phi }=\varGamma$, then $\pi (1)-\pi (0)\equiv \pm 1 \pmod t$, or $\gcd (\pi (1)-\pi (0),t)>1$.

Proof

According to the Fundamental Theorem of Projective Geometry we may assume $\phi =\varPhi _h\varOmega$, where $\varOmega$ is a projectivity and $\varPhi _h$ is as in Definition 5. As $\varPhi _h$ fixes $\varGamma$ and $\varSigma _{id}$, it follows that $\varOmega$ satisfies the conditions of Lemma 1 and hence the assertion follows. $\square$

Corollary 3

If $t=5$ or $t> 6$, then any linear set of pseudoregulus type in ${\mathrm {PG}}(1,q^t)$ can be obtained as the projection of two different subgeometries, $\varSigma _1\cong \varSigma _2\cong {\mathrm {PG}}(t-1,q)$, from a center $\varGamma \cong {\mathrm {PG}}(t-3,q^t)$ to an axis $\varLambda \cong {\mathrm {PG}}(1,q^t)$ in the ambient space $\varSigma ^*={\mathrm {PG}}(t-1,q^t)$, such that there exists no collineation $\phi$ of $\varSigma ^*$ with $\varGamma ^{\phi }=\varGamma$ and $\varSigma _1^{\phi }=\varSigma _2$.

Proof

Let $\varLambda$ be the line $X_3=\cdots =X_{t}=0$ and let $\varGamma$ be the $(t-3)$-subspace $X_1=X_2=0$ in $\varSigma ^*$. As $t=5$ or $t>6$, we have $\varphi (t)\ge 3$, where $\varphi$ is the Euler function. Thus we may choose a permutation $\pi$ of $N_{t-1}$ such that $\gcd (\pi (1)-\pi (0),t)=1$ and $\pi (1)-\pi (0)\notin \{-1,1\}$. Let $\varSigma _1=\varSigma _{id}$ and $\varSigma _2=\varSigma _{\pi }$. Then $p_{\,\varGamma ,\,\varLambda }(\varSigma _1)=p_{\,\varGamma ,\,\varLambda }(\varSigma _2)\cong {\mathbb {L}}$ and the assertion follows from Theorem 5. $\square$

Note that ${\mathbb {L}}$ is not a linear set of rank $k<t$, thus (i) $\Rightarrow$ (ii) in Theorem 2 cannot hold without further conditions.

5 Equivalence of linear sets

We say that the pair (L, n), where L is a linear set of rank n in ${\mathrm {PG}}(r-1,q^t)$, satisfies condition (A), if the following holds.

• (A) For any two $(n-1)$-subspaces $U,U' \subset {\mathrm {PG}}(rt-1,q)$, such that ${\mathcal {B}}(U)=L={\mathcal {B}}(U')$, a collineation $\gamma \in {\mathrm {P}}\Gamma {\mathrm {L}}(rt,q)$ exists, such that $U^{\gamma }=U'$, $\gamma$ preserves the Desarguesian spread ${\mathcal {F}}_{r,t,q}({\mathcal {P}})$, and the induced map on ${\mathrm {PG}}(r-1,q^t)$ is a collineation (which will again be denoted by $\gamma$).

In the next theorem we follow the proof of [2, Theorem 2.4] by Bonoli and Polverino.

Theorem 6

For $i=1,2$, let $\varSigma _i \cong {\mathrm {PG}}(n-1,q)$ be two canonical subgeometries of $\varSigma ^*={\mathrm {PG}}(n-1,q^t)$ and let $\varGamma _i$ be two $(n-1-r)$-subspaces of $\varSigma ^*$, such that $\varGamma _i \cap \varSigma _i=\emptyset$. Let $\varLambda _i \subset \varSigma ^*\setminus \varGamma _i$ be two $(r-1)$-subspaces and let $L_i=p_{\,\varGamma _i,\,\varLambda _i}(\varSigma _i)$. Suppose that $(L_2,n)$ satisfies condition (A). Then (i) $\Rightarrow$ (ii), where (i) and (ii) are the conditions stated in Theorem 2.

Proof

The collineation $\alpha$ in Theorem 2 can be extended to an element $\widehat{\alpha }\in {\mathrm {P}}\Gamma {\mathrm {L}}(n,q^t)$ such that $\varGamma _1^{{\widehat{\alpha }}}=\varGamma _2$. Then $L_2=L_1^{{\widehat{\alpha }}}=p_{\,\varGamma _2,\,\varLambda _2}(\varSigma _1^{{\widehat{\alpha }}})$. Hence it is sufficient to prove the existence of a collineation $\phi \in {\mathrm {P}}\Gamma {\mathrm {L}}(n,q^t)$ such that $\varGamma _2^{\phi }=\varGamma _2$ and $\varSigma _1^{{\widehat{\alpha }}\phi }=\varSigma _2$. If such $\phi$ exists, then $\beta :={\widehat{\alpha }}\phi$.

So it is enough to prove the assertion when $\varGamma _1=\varGamma _2=:\varGamma$, $\varLambda _1=\varLambda _2=:\varLambda$ and $L_1=L_2=:L$. Let $\varSigma ^*={\mathrm {PG}}_{q^t}({\mathbb {F}}_{q^t}^n)$ and for $j=1,2$, let $\varSigma _j={\mathcal {B}}(V_j)$, where $V_j$ is an n-dimensional ${\mathbb {F}}_q$-subspace of ${\mathbb {F}}_{q^t}^n$. As $\varSigma _j$ is a canonical subgeometry, we have $\langle V_j \rangle _{q^t}={\mathbb {F}}_{q^t}^n$, for $j=1,2$. Let $V_1=\langle v_1,v_2,\ldots v_n \rangle _q$. Also let $\varGamma ={\mathrm {PG}}_{q^t}(Z)$ and $\varLambda ={\mathrm {PG}}_{q^t}(R)$. Note that $V_j':=(V_j+Z)\cap R$ is an n-dimensional ${\mathbb {F}}_q$-vector space in R. As $p_{\,\varGamma ,\,\varLambda }(\varSigma _1)=p_{\,\varGamma ,\,\varLambda }(\varSigma _2)=L$, it follows that ${\mathcal {B}}(V_1')={\mathcal {B}}(V_2')$. As (L, n) satisfies condition (A), it follows that there exists a non-singular ${\mathbb {F}}_{q^t}$-semilinear map $\gamma ' :R \rightarrow R$ such that $V_1'^{\gamma '}=V_2'$. The map $\gamma '$ can be extended to a non-singular semilinear map ${\widehat{\gamma }}:{\mathbb {F}}_{q^t}^n \rightarrow {\mathbb {F}}_{q^t}^n$ such that $Z^{{\widehat{\gamma }}}=Z$. Then for each vector $v\in {\mathbb {F}}_{q^t}^n$ and $\mu \in {\mathbb {F}}_{q^t}$ we have $(\mu v)^{{\widehat{\gamma }}}=\mu ^{p^k}v^{{\widehat{\gamma }}}$ for some integer k. Since all of $R+Z={\mathbb {F}}_{q^t}^n$ and $V_j+Z$ ($j=1,2$) are direct sums, for any $v\in V_j\setminus \{0\}$ and $z\in Z$, the intersection $(\langle v\rangle _q+Z)\cap R$ is a one-dimensional ${\mathbb {F}}_q$-subspace, say $\langle v'\rangle _q$, with $v'\in V_j'$; this implies $hv'=v+z'$ for some $z'\in Z$ and $h\in \mathbb F_q$, whence $v+z=hv'+(z-z')\in V_j'+Z$. This implies $V_j+Z\subseteq V'_j+Z$ and therefore $V_j+Z=V'_j+Z$. Then we have

$$\begin{aligned} V_2+Z=V_2'+Z=V_1'^{\hat{\gamma }}+Z^{\widehat{\gamma }}=\left( V_1'+Z\right) ^{\widehat{\gamma }}=\left( V_1+Z\right) ^{{\widehat{\gamma }}}, \end{aligned}$$

and hence for each $i=1,2,\ldots ,n$, $v_i^{{\widehat{\gamma }}}=v_i'+z_i$ for some $v_i'\in V_2$ and $z_i\in Z$. We claim that $v_1',v_2',\ldots ,v_n'$ is an ${\mathbb {F}}_q$-basis of $V_2$. To see this, suppose $\sum _{i=1}^n \lambda _i v_i'=0$ with $\lambda _i \in {\mathbb {F}}_q$ for $i=1,2,\ldots ,n$. Then $\sum _{i=1}^n \lambda _i v_i^{{\widehat{\gamma }}}=\sum _{i=1}^n \lambda _i z_i$. As on the left-hand side $v_1^{{\widehat{\gamma }}},v_2^{{\widehat{\gamma }}},\ldots , v_n^{{\widehat{\gamma }}}$ are ${\mathbb {F}}_q$-independent, it follows that either $\lambda _i=0$ for $i=1,2,\ldots ,n$, or there is a non-zero vector $z\in V_1^{{\widehat{\gamma }}}\cap Z$. The map ${\widehat{\gamma }}$ fixes Z and hence in the latter case $z^{{\widehat{\gamma }}^{-1}}\in V_1\cap Z$, a contradiction because of $V_1 \cap Z =\{0\}$. As $\varSigma _2$ is a canonical subgeometry, it follows that $v_1',v_2',\ldots ,v_n'$ are linearly independent over ${\mathbb {F}}_{q^t}$. Let f be the ${\mathbb {F}}_{q^t}$-semilinear map of ${\mathbb {F}}_{q^t}^n$ such that $v_i^f=v_i'$ for each $i=1,2,\ldots ,n$ and $(\mu v)^f=\mu ^{p^k}v^f$ for each $\mu \in {\mathbb {F}}_{q^t}$ and $v\in {\mathbb {F}}_{q^t}^n$. If $P=\langle z \rangle _{q^t}\in \varGamma$, then we have $z=\sum _{i=1}^n a_iv_i$ for some $a_i\in {\mathbb {F}}_{q^t}$. Then $z^f=\sum _{i=1}^n a_i^{p^k}v_i'=\sum _{i=1}^n a_i^{p^k}(v_i^{{\widehat{\gamma }}}-z_i)=z^{{\widehat{\gamma }}}-\sum _{i=1}^n a_i^{p^k}z_i\in Z$, and hence the collineation induced by f fixes $\varGamma$ and maps $\varSigma _1$ to $\varSigma _2$. $\square$

Remark 2

In Theorem 6, it is not assumed, contrary to Theorem 2, that the linear sets are not of rank $<$ n.

Remark 3

By [2, Proposition 2.3], condition (A) holds for any ${\mathbb {F}}_q$-linear blocking set of exponent e (where $p^e=q$) in ${\mathrm {PG}}(2,q^t)$.

Remark 4

Up to the knowledge of the authors, no linear set is known which is not of pseudoregulus type in ${\mathrm {PG}}(1,q^t)$ and does not satisfy condition (A).

Theorem 7

If L is a linear set of rank n in $\varLambda ={\mathrm {PG}}(r-1,q^t)$, $\langle L\rangle ={\mathrm {PG}}(r-1,q^t)$ and (L, n) does not satisfy condition (A), then in ${\mathrm {PG}}(n-1,q^t)\supset \varLambda$ there are a subspace $\varGamma =\varGamma _1=\varGamma _2$ disjoint from $\varLambda$, and two q-order canonical subgeometries $\varSigma _1,\varSigma _2\subset {\mathrm {PG}}(n-1,q^t)\setminus \varGamma$ such that $L=p_{\,\varGamma ,\,\varLambda }(\varSigma _1)=p_{\,\varGamma ,\,\varLambda }(\varSigma _2)$, and such that condition (ii) in Theorem 2 does not hold.

Proof

Let $U_1$ and $U_2$ be two $(n-1)$-subspaces of ${\mathrm {PG}}_q(R)$, where R is the r-dimensional ${\mathbb {F}}_{q^t}$-subspace of ${\mathbb {F}}_{q^t}^n$ such that $\varLambda ={\mathrm {PG}}_{q^t}(R)$. Assume that ${\mathcal {B}}(U_1)=L={\mathcal {B}}(U_2)$. Let $W_i=\langle w_1^{(i)},w_2^{(i)},\ldots ,w_n^{(i)}\rangle _q$, $i=1,2$, be the n-dimensional vector ${\mathbb {F}}_q$-subspaces of R whose associated projective subspaces in ${\mathrm {PG}}(rt-1,q)$ are $U_1$ and $U_2$, respectively. From $\langle L\rangle ={\mathrm {PG}}(r-1,q^t)$ we may assume that $w_1^{(1)},w_2^{(1)},\ldots ,w_r^{(1)}$ are ${\mathbb {F}}_{q^t}$-linearly independent, and also that $w_1^{(2)},w_2^{(2)},\ldots ,w_r^{(2)}$ are ${\mathbb {F}}_{q^t}$-linearly independent. Let $\varGamma$ be an $(n-r-1)$-subspace in ${\mathrm {PG}}(n-1,q^t)$ disjoint from $\varLambda$, associated with the vector subspace $Z=\langle z_1,z_2,\ldots , z_{n-r}\rangle _{q^t}$. For $i=1,2$, let $\varSigma _i$ be the ${\mathbb {F}}_q$-linear set defined by the following ${\mathbb {F}}_q$-subspace of ${\mathbb {F}}_{q^t}^n$:

$$\begin{aligned}V_i:= \left\langle w_1^{(i)},w_2^{(i)},\ldots ,w_r^{(i)}, w_{r+1}^{(i)}+z_1,w_{r+2}^{(i)}+z_2,\ldots ,w_n^{(i)}+z_{n-r}\right\rangle _q\!\!. \end{aligned}$$

Since those n vectors of ${\mathbb {F}}_{q^t}^n$ are also ${\mathbb {F}}_{q^t}$-linearly independent, $\varSigma _i$ is a canonical q-order subgeometry. Furthermore, $w_1^{(i)}$, $w_2^{(i)}$, $\ldots$, $w_n^{(i)}$, $z_1$, $z_2$, $\ldots$, $z_{n-r}$ are ${\mathbb {F}}_q$-linearly independent, and this implies $\varSigma _i\cap \varGamma =\emptyset$. We summarize here some properties of our construction, which we will use later. For $i=1,2$ we have the following.

• 1. For each $w\in W_i$, there exists a unique $z_w\in Z$ such that $w+z_w\in V_i$,

• 2. if w and $w'$ are two ${\mathbb {F}}_q$-independent vectors in $W_i$, then $w+z_w$ and $w'+z_{w'}$ are ${\mathbb {F}}_{q^t}$-independent vectors in $V_i$.

If $w=\sum _{j=1}^n \alpha _j w_j^{(i)}$ with $a_j\in {\mathbb {F}}_q$, $j=1,2,\ldots ,n$, then let $z_w=\sum _{j=r+1}^n \alpha _j z_{j-r}$. The unicity of $z_w$ follows from $\varSigma _i \cap \varGamma =\emptyset$. To see 2., note that $\varPsi _i :U_i \rightarrow \varSigma _i :\langle w \rangle _q \mapsto \langle w+z_w \rangle _{q^t}$ is a bijection, as $|U_i|=|\varSigma _i|=\theta _{n-1}$.

Now assume that condition (ii) in Theorem 2 holds. Let $\beta ':{\mathbb {F}}_{q^t}^n\rightarrow {\mathbb {F}}_{q^t}^n$ be the semilinear map associated with the collineation $\beta$; so, an integer $\nu$ exists such that for any $a\in {\mathbb {F}}_{q^t}$ and $v\in {\mathbb {F}}_{q^t}^n$ there holds $(av)^{\beta '}=a^{p^{\nu }} v^{\beta '}$. Denote by $\pi$ the canonical projection from $Z\oplus R$ to R. Fix an element $w\in W_1$. Then 1. and the condition $\varSigma _1^{\beta }=\varSigma _2$ imply the existence of $z_w\in Z$ and $\lambda \in {\mathbb {F}}_{q^t}^*$ depending on $z_w$ such that $\xi _{\lambda }:=\lambda (w+z_w)^{\beta '}\in V_2$. Let $T_{\lambda }=\{v \in W_1 :\lambda v^{\beta ' \pi }\in W_2\}$. Note that $T_{\lambda }$ is an ${\mathbb {F}}_q$-vector subspace of R. As $\lambda (w+z_w)^{\beta '}\in V_2$, we have $\lambda (w+z_w)^{\beta '\pi }=\lambda w^{\beta ' \pi }\in V_2^{\pi }=W_2$, thus $w\in T_{\lambda }$. We are going to show that $W_1=T_{\lambda }$. Suppose to the contrary that there exists $w'\in W_1 \setminus T_{\lambda }$. As $T_{\lambda }$ is an ${\mathbb {F}}_q$-vector space, w and $w'$ are ${\mathbb {F}}_q$-independent. The same argument as above yields the existence of $z_{w'}\in Z$ and $\mu \in {\mathbb {F}}_{q^t}^*$ such that $\xi _{\mu }:=\mu (w'+z_{w'})^{\beta '}\in V_2$ and hence $\mu w'^{\beta ' \pi }\in W_2$. Then 2. yields that $\langle w+z_w \rangle _{q^t}$ and $\langle w'+z_{w'} \rangle _{q^t}$ are distinct points of $\varSigma _1$, and hence $\langle \xi _{\lambda } \rangle _{q^t}$ and $\langle \xi _{\mu } \rangle _{q^t}$ are distinct points of $\varSigma _2$. First we prove

$$\begin{aligned} \langle \xi _{\lambda }, \xi _{\mu } \rangle _{q^t} \cap V_2 = \langle \xi _{\lambda }, \xi _{\mu } \rangle _{q}. \end{aligned}$$

(13)

Denote by X de left hand side of (13). The inclusion $\supseteq$ trivially holds. Suppose to the contrary $\dim _{{\mathbb {F}}_q}(X)\ge 3$, then there exist three ${\mathbb {F}}_q$-independent vectors, $v_1,v_2,v_3\in X$. We can extend the set $\{v_1,v_2,v_3\}$ to a basis of $V_2$. As $\dim _{{\mathbb {F}}_{q^t}}(\langle v_1,v_2,v_3\rangle _{q^t})=2$, it follows that $\langle V_2 \rangle _{q^t}\ne {\mathbb {F}}_{q^t}^n$, a contradiction since $\varSigma _2$ is a canonical subgeometry.

Let $P_1=\langle w+z_w \rangle _{q^t}$ and $P_2=\langle w'+z_{w'} \rangle _{q^t}$. Denote by $\ell$ the q-order subline $\langle P_1,P_2 \rangle \cap \varSigma _1$. Then $\langle (w+z_w)^{\beta '}+ (w'+z_{w'})^{\beta '} \rangle _{q^t}=\langle \xi _{\lambda } + \lambda \mu ^{-1}\xi _{\mu } \rangle _{q^t}\in \ell ^{\beta } \cap \varSigma _2$. It follows that there exists some $\delta \in {\mathbb {F}}_{q^t}^*$ such that $\delta (\xi _{\lambda }+\lambda \mu ^{-1}\xi _{\mu })\in V_2$. It follows from (13) that $\delta \in {\mathbb {F}}_q^*$ and $\delta \lambda \mu ^{-1}\in {\mathbb {F}}_q^*$. Hence $\lambda \mu ^{-1} \in {\mathbb {F}}_q^*$. This is a contradiction, as in this case $(\lambda \mu ^{-1})\mu w'^{\beta ' \pi }= \lambda w'^{\beta ' \pi } \in W_2$, contradicting the choice of $w'$. Hence $T_\lambda =W_1$, that is, $\lambda w^{\beta '\pi }\in W_2$ for any $w\in W_1$.

Now define the map $\varphi ' :R \rightarrow R :v \mapsto \lambda v^{\beta ' \pi }$. As $\beta '$ is semilinear, $\pi$ is linear and $v \mapsto \lambda v$ is also linear, it follows that $\varphi '$ is ${\mathbb {F}}_{q^t}$-semilinear. Also, $\varphi '$ is non-singular. Let $\varphi$ be the associated collineation of $\varLambda$. Then

$$\begin{aligned} U_1^{\varphi }=\left\{ \left\langle \lambda w^{\beta ' \pi }\right\rangle _q :w\in W_1^*\right\} ={\mathrm {PG}}_q\left( W_2\right) = U_2, \end{aligned}$$

because of $T_{\lambda }=W_1$.

We have constructed an example for which, if condition (ii) in Theorem 2 is satisfied, then also condition (A) must be satisfied: this concludes the proof. $\square$

Scattered linear sets of pseudoregulus type in a line are, under the field reduction map, embeddings of minimum dimension of ${\mathrm {PG}}(t-1,q)\times {\mathrm {PG}}(t-1,q)$, which have been studied in [5]. These embeddings are projections of Segre varieties. A similar representation of the clubs as projections of Segre varieties can be found in [11] (see also [9]). By the arguments in the proof of [5, Theorem 16], $({\mathbb {L}},t)$ does not fulfill condition (A), for $t=5$ or $t>6$, where ${\mathbb {L}}$ is the scattered linear set of pseudoregulus type in ${\mathrm {PG}}(1,q^t)$. For in this case it is possible to choose a subspace in the family $\mathcal{S}_1$ and a subspace in $\mathcal{S}_i$, $i\in \{2,3,\ldots ,t-2\}$, $\gcd (i,t)=1$. By Theorem 7 this yields the counterexamples as shown in the previous sections. Many results in [6] are on linear sets in ${\mathrm {PG}}(1,q^3)$; hence, the following result can be of interest:

Proposition 5

Condition (A) is fulfilled by linear sets of rank three in ${\mathrm {PG}}(1,q^3)$, $q>2$, for $n=3$.

Proof

A linear set L of rank three in ${\mathrm {PG}}(1,q^3)$ is either a point, or a club, or a scattered linear set.

If L is a point, then from

$$\begin{aligned} {\mathcal {B}}(U)={\mathcal {B}}\left( U'\right) ,\quad \dim (U)=\dim \left( U'\right) =2, \end{aligned}$$

(14)

one obtains $U=U'={\mathcal {F}}_{2,3,q}(L)$. This implies (A).

Assume now that L is a club. Denote by H its head, which is uniquely defined [4, 6]. Assume that the subspaces U and $U'$ satisfy (14). Take a point $P=\langle (v_0,v_1) \rangle _{q^3}\in L\setminus H$ and let $A={\mathcal {F}}_{2,3,q}(P)\cap U$ and $X={\mathcal {F}}_{2,3,q}(P)\cap U'$. Then $A=\langle \alpha (v_0,v_1) \rangle _q$ and $X=\langle \beta (v_0,v_1) \rangle _q$ for some $\alpha ,\beta \in {\mathbb {F}}_{q^3}^*$, and the map $\gamma :\langle (x_0,x_1) \rangle _q \mapsto \langle \beta \alpha ^{-1}(x_0,x_1) \rangle _q$ is a projectivity of ${\mathrm {PG}}(5,q)$ belonging to the elementwise stabilizer of the Desarguesian spread, and mapping A into X. Then $X\in U^\gamma \cap U'\setminus {\mathcal {F}}_{2,3,q}(H)$. The element of ${\mathrm {P}}\Gamma {\mathrm {L}}(2,q^3)$ associated with $\gamma$ is the identity map. Denote by $\ell _1$ and $\ell _2$ the lines $U^{\gamma }\cap {\mathcal {F}}_{2,3,q}(H)$ and $U'\cap {\mathcal {F}}_{2,3,q}(H)$, respectively. If $\ell _1=\ell _2$, then $U^{\gamma }=\langle \ell _1, X \rangle =U'$. Otherwise let $\ell '$ be a line in $U'$ through X such that $\ell _1 \cap \ell _2 \notin \ell '$. The club L does not contain irregular sublines [6, Corollary 13], [4, Proposition 5]. Applying this result to the q-order subline ${\mathcal {B}}(\ell ')$, we have that a line $\ell$ in $U^\gamma$ exists such that ${\mathcal {B}}(\ell )={\mathcal {B}}(\ell ')$. The point X is on $\ell$. The lines $\ell$, $\ell '$ of ${\mathrm {PG}}(5,q)$ are transversal lines to the regulus ${\mathcal {F}}_{2,3,q}({\mathcal {B}}(\ell ))$, having a common point, whence $\ell =\ell '$. This implies $U^\gamma =\langle \ell , \ell _1 \cap \ell _2 \rangle =U'$ and (A).

Up to projectivities there is a unique scattered linear set in ${\mathrm {PG}}(1,q^3)$. To see this, fix a q-order canonical subgeometry $\varSigma$ of $\varSigma ^*={\mathrm {PG}}(2,q^3)$ and denote by ${\mathcal {I}}$ the set of points of $\varSigma ^*$ not contained in a line of $\varSigma$. Denote by G the stabilizer of $\varSigma$ in the group of projectivities of $\varSigma ^*$. Each scattered linear set of rank three of ${\mathrm {PG}}(1,q^3)$ can be obtained as a projection of $\varSigma$ from a point in ${\mathcal {I}}$. In Theorem 2 the implication (ii) $\Rightarrow$ (i) holds and hence the number of projectively non-equivalent scattered linear sets of rank three in ${\mathrm {PG}}(1,q^3)$ is at most the number of orbits of G on the points of ${\mathcal {I}}$. It follows from [2, Proposition 3.1] that G is transitive on ${\mathcal {I}}$ and hence there is a unique scattered linear set of rank three in ${\mathrm {PG}}(1,q^3)$. An alternate proof can be obtained from [1], where the uniqueness of an exterior splash is proved, taking into account that the exterior splashes dealt with in that paper are precisely the scattered linear sets of rank three [10].

Assume that L is a scattered linear set of rank three, hence a linear set of pseudoregulus type, and that (14) holds. Then U and $U'$ are planes contained in the hypersurface ${\mathcal {Q}}_{2,q}$ defined in [5, Eqs. (7), (8)]. By [5, Corollary 10], $U=S_{h,k}$, $U'=S_{h',k'}$ for $h,h'\in \{1,2\}$, $N(k)=N(k')=1$. A collineation $\gamma$ satisfying (A) can be obtained in the form $\varphi _{0,0}(a,b)$ if $h=h'$, and $\psi _{0,0}(a,b)$ if $h\ne h'$, for suitable $a,b\in {\mathbb {F}}_{q^3}$. $\square$