Cameron-Liebler line classes with parameter x=[Formula presented]

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Advances

in

Mathematics

www.elsevier.com/locate/aim

Cameron-Liebler

line

classes

with

parameter

x

=

Tao Fenga,1_{, Koji Momihara}b,2_{, Morgan Rodgers}c_,

Qing Xiangd,∗,3_{, Hanlin Zou}e

a_{SchoolofMathematicalSciences,ZhejiangUniversity,38ZhedaRoad,Hangzhou}

310027,Zhejiang,China

b

FacultyofEducation,DivisionofNaturalScience,FacultyofAdvancedScience andTechnology,KumamotoUniversity,2-40-1Kurokami,Kumamoto860-8555, Japan

c_{DepartmentofMathematics,IstinyeUniversity,Istanbul,Turkey}

d_{DepartmentofMathematicsandSUSTechInternationalCenterforMathematics,}

SouthernUniversityofScienceandTechnology,Shenzhen518055,China

e_{DepartmentofMathematicalSciences,UniversityofDelaware,Newark, DE}

19716,USA

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received10February2020 Receivedinrevisedform18April 2021

Accepted21April2021 Availableonlinexxxx

CommunicatedbytheManaging Editors

Keywords:

Cameron-Lieblerlineclass Gausssum

Kleinquadric

Cameron-Liebler line classes were introduced in [5], and motivatedby aquestion about orbitsof collineationgroups of PG(3,q). These line classes have appeared in different contextsunderdisguised namessuchas Booleandegreeone functions,regularcodesofcoveringradiusone,andtightsets. In this paper we construct an infinite family of Cameron-Liebler line classes in PG(3,q) with new parameter x = (q +1)2_{/3 forallprimepowersq congruentto2modulo3.The} examplesobtainedwhenq isanoddpoweroftworepresentthe

* Correspondingauthor.

E-mailaddresses:tfeng@zju.edu.cn(T. Feng),momihara@educ.kumamoto-u.ac.jp(K. Momihara), morgan.rodgers@istinye.edu.tr(M. Rodgers),xiangq@sustech.edu.cn(Q. Xiang),hanlin@udel.edu (H. Zou).

1 _{ResearchpartiallysupportedbytheNationalNaturalScienceFoundationofChinagrant11771392.} 2 _{ResearchpartiallysupportedbyJSPSunderGrant-in-AidforYoungScientists(B)17K14236,Scientific} Research(B)15H03636,andScientificResearch(C)20K03719.

3

ResearchpartiallysupportedbytheNationalNaturalScienceFoundationofChinagrant12071206. https://doi.org/10.1016/j.aim.2021.107780

Spread Tightset

firstinfinitefamilyofCameron-LieblerlineclassesinPG(3,q), q even.

©2021ElsevierInc.Allrightsreserved.

1. Introduction

Let q be aprimepower and letPG(3,q) bethe 3-dimensional projective spaceover

thefinitefieldFqoforderq.Aspread inPG(3,q) isasetofitslineswhichpartitionsits

points. LetL beasetoflines ofPG(3,q) with|L|= x(q2_{+ q + 1),x apositive integer.}

We say that L is aCameron-Liebler line class with parameter x if |L∩ S| = x for all

spreadsS ofPG(3,q).Forexample,letL be eithertheset ofalllinespassingthrougha

fixedpointP ofPG(3,q) orthesetofalllinesinaplaneπ ofPG(3,q);thenweseethat

L isaCameron-Lieblerlineclasswithparameter1.Furthermoretheunionofthesetwo

sets oflines forP /∈ π forms aCameron-Lieblerline classwith parameterx= 2.These

examples of Cameron-Liebler line classeswith parameter x= 1 or 2 are called trivial.

Also,thecomplementofaCameron-Lieblerlineclasswith parameterx inthesetof all

linesofPG(3,q) isaCameron-Lieblerlineclasswithparameter(q2+ 1)− x.Sowithout

loss ofgenerality wemayassumethatx≤ (q2_{+ 1)/2 whendiscussingCameron-Liebler}

line classesofparameterx inPG(3,q).

Cameron-LieblerlineclasseswerefirstintroducedbyCameronandLiebler[5] intheir

study of collineation groups of PG(3,q) having the same number of orbits on points

and linesofPG(3,q).Penttila[26,27] coinedtheterm“Cameron-Lieblerlineclass”and

studiedtheseobjectsinsomedepth.BruenandDrudge[3] constructedthefirstinfinite

family ofCameron-Liebler lineclasseswith parameterx= (q2_{+ 1)/2 foralloddprime}

powers q. After much study of Cameron-Liebler line classes in PG(3,q), the notionof

Cameron-Liebler line classes hasbeen generalizedto Cameron-Liebler k-classes [29] in

PG(2k + 1,q),andto Cameron-Lieblersets ofgeneratorsinfinite classicalpolarspaces

[11]. In fact, Cameron-Liebler sets can be introduced for any distance-regular graph;

thiswasdonepreviouslyundervariousnames:Booleandegreeonefunctions,completely

regularcodes ofstrength0andcoveringradius1,and tightsets.Wereferthereaderto

[19] formoredetailsontheseconnections.InthispaperwewillfocusonCameron-Liebler

line classesinPG(3,q).

ThecentralproblemconcerningCameron-LieblerlineclassesinPG(3,q) is:forwhich

valuesoftheparameterx,1≤ x≤ (q2+1)/2,dothereexistCameron-Lieblerlineclasses

withparameterx?Onthenonexistenceside,thestate-of-the-artresultsarethosein[24]

and[21].Inparticular,itisshownin[24] thattherearenoCameron-Liebler lineclasses

withparameterx inPG(3,q) if3≤ x≤ q3

q/2− 2q/3.Intermsofconstructiveresults,

infinite families of Cameron-Liebler line classes with parameter x = (q2_{+ 1)/2 and}

x = (q2_{− 1)/2 havebeen constructed in [3,7–10,18,22]. Even though there have been}

parametersoftheknowninfinitefamiliesarerestrictedtoeither(q2+ 1)/2 or(q2− 1)/2.

Inparticular, no infinitefamiliesof nontrivial Cameron-Liebler line classes inPG(3,q)

are known when q is a power of 2 (note that there are a few examples of

Cameron-LieblerlineclassesknowninPG(3,4),PG(3,8),PG(3,32),andPG(3,128);see[20,28]).

In this paper, we construct Cameron-Liebler line classes in PG(3,q) with parameter

x= (q + 1)2_{/3 for allq congruent to 2modulo3. Inparticular,the firstinfinitefamily}

ofCameron-Lieblerline classesinPG(3,q),q anoddpowerof2,isconstructedhere.

We give an overview of our construction here. The initial step is to prescribe an

automorphism groupfor the Cameron-Liebler line classes thatwe intend to construct;

oncethis isdone, theCameron-Liebler lineclasseswe wanttoconstruct willbe unions

oforbitsoflinesundertheactionoftheprescribedautomorphismgroup.Forthechoices

ofautomorphismgroups,wefollowtheideain[28];thatis,wewillchooseacyclicgroup

oforderq2+ q + 1 astheprescribedautomorphismgroup.ExamplesofCameron-Liebler

classeswithparameter(q + 1)2_{/3 havebeenfoundinthisway byusingacomputerfor}

allq < 150 withq ≡ 2mod 3 (see [28]). Thedifficultylies inhow to come upachoice

of orbits for generalq which will always give aCameron-Liebler line class in PG(3,q)

withparameter(q + 1)2_{/3.Theexamplesin[28] providedvitalcluesforageneralchoice;}

alsothecomputationsofadditivecharactersums(neededtoprovethattheunionofthe

chosen orbitsis aCameron-Liebler line class)gave us hints for makingcorrect choices

of orbits. In Section 3, we come up with an explicit choice of orbits that will give a

Cameron-Liebler line classes with parameter x = (q + 1)2_{/3 for all q congruent to 2}

modulo3.

The paper is organized as follows. In Section 2, we review background material on

Cameron-Liebler line classes, x-tight sets in Q+(5,q), and character sums over finite

fields.InSection3,weintroducetwomultisetsD1andD2ofF_q∗3 whichwillbecrucialfor choosingorbits.InSection4,wegivetheproofsthatourchoiceoforbitswillindeedgive

Cameron-Lieblerlineclasses;sincetheprescribedgroupisacyclicone,thecomputations

of additivecharactersums necessarilyinvolve Gauss sums. InSection5, we determine

thestabilizersofourCameron-LieblerlineclassesinPSL(4,q).IntheAppendix,wegive

thedetailedcomputationsofexponentialsumsneededintheproofofourmaintheorem.

2. Preliminaries

2.1. Cameron-Lieblerlineclassesin PG(3,q) and tight setsinQ+_(5,q)

ToinvestigateCameron-LieblerlineclassesinPG(3,q),itisoftenusefultotranslate

theirdefinitiontothesettingofQ+_{(5,q) usingtheKleincorrespondence(here Q}+_(5,q)

isthe5-dimensionalhyperbolicorthogonalspace,alsoknownastheKleinquadric).Let

x be apositive integer.Asubset M ofthepoints ofQ+_{(5,q) iscalled an x-tightset if}

foreverypointP ∈ Q+_(5,q),|P⊥_{∩ M|= x(q + 1)+ q}2 _{orx(q + 1) accordingasP isin}

M ornot,where ⊥ isthepolaritydetermined byQ+_{(5,q). Thegeometriesof PG(3,q)}

which maps the lines of PG(3,q) bijectively to the points of Q+(5,q), cf. [23,25]. Let

L be aset of lines of PG(3,q) with |L| = x(q2+ q + 1), x a positive integer, and let

M be the image of L under the Klein correspondence. Then it is known that L is a

Cameron-Liebler line class withparameterx in PG(3,q) if andonlyif M isan x-tight

set of Q+_{(5,q).Moreover, ifL isaCameron-Liebler line classwithparameter x,by[1,}

Theorem 12] (see,also[24,Theorem2.1(b)])itholdsthat|P⊥∩ M|= x(q + 1)+ q2for

any point P ∈ M and |P⊥∩ M| = x(q + 1) forany point P /∈ M (here P can be in

theexteriorofQ+_{(5,q));consequentlyM isaprojectivetwo-intersectionsetinPG(5,q)}

withintersectionsizesh1= x(q + 1)+ q2andh2= x(q + 1) (cf.[4]).Wesummarizethese

knownfactsasfollows.

Result2.1.LetL beasetofx(q2_{+ q + 1) linesinPG(3,q) with1≤ x≤ (q}2_{+ 1)/2,and}

let M betheimage of L undertheKleincorrespondence. ThenL is aCameron-Liebler line classwith parameterx if andonly if M isan x-tightset in Q+_{(5,q);moreover,in}

thecase whenL is aCameron-Lieblerlineclass,we have

|P⊥_{∩ M| =}



x(q + 1) + q2_{, if P ∈ M,}

x(q + 1), otherwise.

Let L be a Cameron-Liebler line class with parameter x in PG(3,q) and let M ⊂

Q+_{(5,q) be the image of L under the Klein correspondence. By Result 2.1, M is a} projectivetwo-intersectionsetinPG(5,q).LetF_q∗= Fq\ {0} bethemultiplicativegroup

of Fq. Define D = {λv : λ ∈ Fq∗,v ∈ M}, which is asubset of (Fq6,+). Let ψ be a

non-principaladditivecharacterof(F6

q,+).Then ψ isprincipalonauniquehyperplane

P⊥ forsomepointP ∈ PG(5,q).Wehave

ψ(D) =  v∈M  λ∈F∗ q ψ(λv) =  v∈M (q1P⊥(v ) − 1) =−|M| + q|P⊥∩ M| =  −x + q3_{, if P ∈ M,} −x, otherwise,

where 1_P⊥(v ) is the characteristic function taking value 1 if v ∈ P⊥, and 0

oth-erwise. Conversely, for each point P ∈ PG(5,q), there is a non-principal character ψ

thatisprincipalonthehyperplaneP⊥,andthesize ofP⊥∩ M canbe computedfrom

ψ(D).ThereforethecharactervaluesofD reflectthesizesofintersectionofM withthe

hyperplanes ofPG(5,q).Tosummarize,wehavethefollowingresult.

Result2.2.LetL beasetofx(q2+ q + 1) linesinPG(3,q) with1≤ x≤ (q2+ 1)/2,and

let M betheimage of L undertheKlein correspondence.Define D ={λv : λ ∈ F_q∗,v ∈ M} ⊂ (F_q6, +).

ThenL is aCameron-Lieblerlineclasswith parameterx if andonly if|D|= (q3− 1)x

andforany P ∈ PG(5,q),

ψ(D) =



−x + q3_{, if P ∈ M,}

−x, otherwise,

where ψ is any non-principal character of (F6

q,+) that is principal on the hyperplane

P⊥.

2.2. Gausssums

We collect some auxiliary results on Gauss sums as a preparation for computing

additive charactervalues of asubset of vectors of a vector space over Fq. We assume

thatthe readeris familiar with thebasic theoryof characters of finite fields as can be

foundinChapter 5of[17]

Letq = pn withp aprimeandn≥ 1,andletζp= exp(2π √

−1

p ).Furthermore,letψFq

bethecanonical additive characterofFqdefinedbyψFq(x)= ζ Trq/p(x)

p , ∀x∈ Fq,where

Trq/pistheabsolutetracefromFqtoFp.Foranymultiplicativecharacterχ ofFq,define

theGausssum by

Gq(χ) =



x∈F∗ q

ψFq(x)χ(x).

ThefollowingaresomebasicpropertiesofGausssums:

(i) Gq(χ)Gq(χ) = q ifχ is non-principal;

(ii) Gq(χ−1)= χ(−1)Gq(χ);

(iii) Gq(χ)=−1 ifχ isprincipal.

Gausssumsareinstrumentalinthetransitionfromtheadditivetothemultiplicative

structure(or theotherway around)ofafinitefield.Thiscanbe seenmoreprecisely in

thenextlemma.

Lemma 2.3. By orthogonality of characters, thecanonical additive character ψFq of Fq

canbe expressedas alinearcombination ofthemultiplicative characters: ψFq(x) = 1 q− 1  χ∈ F∗ q Gq(χ−1)χ(x), ∀x ∈ Fq∗, (2.1)

whereFq∗ isthecharactergroupof Fq∗.Ontheotherhand,eachnontrivial multiplicative

χ(x) = 1 qGq(χ)  a∈F∗ q χ−1(−a)ψFq(ax), ∀x ∈ F ∗ q.

Lemma 2.4.Let C0 be asubgroup of Fq∗ of indexN , and letχ be a characterof Fq∗ of

order N . Thenforanyx∈ Fq∗ wehave

1 N N−1_ j=0 Gq(χ−j)χj(x) =  a∈C0 ψFq(xa).

Proof. Letθ beacharacterofF_q∗ oforderq− 1,andletχ= θ(q−1)/N_{.By(2.1),wehave}  a∈C0 ψFq(xa) = 1 q− 1 q−1  i=0 Gq(θ−i)θi(x)  a∈C0 θi(a). (2.2) LetC0={1,wN,w2N,. . . ,w( q−1

N −1)N},wherew isaprimitiveelementofF_q.Wheni≡ 0

(mod q−1_N ),θi_{isprincipalonC}

0,sotheinnersum_a∈C0θi(a)= (q− 1)/N;wheni≡ 0 (mod q_N−1),wehaveθi_(wN_{)= 1, andso theinnersum}

a∈C0θ

i_(a)= θ(q−1)i(w)−1

θi_(wN₎₋₁ = 0. Therefore _a∈C0ψFq(xa) equals

1

N

N−1

j=0 Gq(χ−j)χj(x) asdesired. 

Thefollowing resultonthecharactervaluesofaSingerdifferenceset willbe usedin

theproof ofourmain theorem.

Lemma 2.5 ([14, Theorem 2.1]).Let L be a complete set of coset representatives of F_q∗

in F_q∗3.Let

S = {x ∈ L | Trq3_/q(x) = 0}.

If χ is anonprincipal characterof F_q∗3 whose restrictionon F_q∗ isprincipal,then

χ(S) = Gq3(χ)/q.

2.3. Cubicpolynomials overFq

Let q = pn _{be aprimepower, where p= 3 isaprime. Letf (X)= X}3_{+ cX + d be} acubic polynomialoverFq,and letγ1,γ2,γ3 beits rootsinsomeextensionfieldof Fq.

Thediscriminant off isdefinedby

Δ(f ) := (γ1− γ2)2(γ2− γ3)2(γ3− γ1)2,

whichequals−4c3_−27d2_{forallq.Hencef hasnorepeatedrootsifandonlyifΔ(f )= 0.}

In particular,when q iseven,we haveΔ(f )= d2.We shallneedthefollowing theorem

Theorem2.6. [13,30] Letp= 3 beaprimeandq = pn.Supposethatf (X)= X3+cX +d

isapolynomialover Fq withdiscriminantΔ(f )= 0.

(i) If q is odd, f has exactly one root in Fq if Δ(f ) isa nonsquare in Fq, and 0 or3

roots inFq otherwise.

(ii) If q is even, f has exactly one root in Fq if Trq/2(c3d−2) = Trq/2(1), and 0 or 3

roots inFq otherwise.

Our construction of newCameron-Liebler line classes is based on theimage sets of

certaincubicpolynomialsasshowninthenextsection.Thisideawaspreviouslyusedin

[12] forconstructingnewdifferencesetswith Singerparameters.

3. Cameron-Lieblerlineclasseswith parameterx= (q + 1)2_/3

3.1. The setE

Throughout the rest of the paper, we always assume that q is a prime power such

thatq≡ 2 (mod 3).Wedefine

T0={x ∈ Fq∗3 : Trq3_/q(x) = 0},

L0={x ∈ T0: Nq3_/q(x) = 1},

where Trq3_/q and N_q3_/q are the relative trace and norm from F_q3 to F_q, respectively. Then |T0|= q2− 1,|L0| = q + 1 andL0· Fq∗ = T0. (Here, forany two subsets A,B of

F_q∗3, define A· B := {ab : a∈ A, b ∈ B}.) Since gcd(q− 1,q2+ q + 1) = 1, we have

C0· Fq∗= Fq∗3,whereC0 isthesubgroupofFq∗3 oforder q2+ q + 1.

Lemma3.1. If q≡ 2 (mod 3) withq odd, then−3 is anonsquareinFq.

Proof. Writeq = pn _{withp anoddprime.Thenp≡ 2 (mod 3) andn isodd.Itsuffices}

toshowthat−3 isanonsquareinFp.Bythequadraticreciprocitywehave

 −3 p  =  −1 p  ·  3 p  = (−1)(p−1)/2· (−1)(p−1)/2·(3−1)/2· p 3 =−1.

Here,(_p·) istheLegendresymbol.Theproofiscomplete. 

Lemma3.2. If z isanelement ofF_q∗3 suchthat Trq3_/q(z)= 0, thenTr_q3_/q(z1+q)= 0. Proof. We havez /∈ Fq, sinceotherwise 3z = Trq3_/q(z) = 0.If Tr_q3_/q(z1+q) = 0, then

theminimalpolynomialofz overFqisX3−c,wherec= Nq3_/q(z).Sinceq≡ 2 (mod 3),

wehavegcd(q− 1,3)= 1, soX3= c hasexactlyonerootinFq:acontradictionto the

Sinceq≡ 2 (mod 3),wehavegcd(q−1,3)= 1 andsothemapy → y3isapermutation

of Fq.Wewrite y → y1/3 fortheinverseofthemap y → y3.Wedefine twomultisetsas

follows: D1= [xNq3_/q(λ + xq− xq 2 )1/3: x∈ L0, λ∈ Fq], D2= [β−1xNq3_/q(λ + xq− xq 2 )−1/3: x∈ L0, λ∈ Fq], where β =−3−1 ∈ Fq.Setγ := β−3 =−27.

Lemma 3.3.Letx∈ L0 andsetz := xq− xq

2

.Foreach α∈ F_q∗,set

cα:=|{λ ∈ Fq: Nq3_/q(λ + z) = α}| + |{λ ∈ F_q: γN_q3_/q(λ + z)−1= α}|.

Then cα= 1 or4.

Proof. Write a := Trq3_/q(z1+q), b := N_q3_/q(z), and set u := −1

3a. It is clear that Trq3_/q(z) = 0, so a = 0 by Lemma 3.2. The minimal polynomial of x over F_q is

g(X):= X3_{+ eX− 1 with e= Tr}

q3_/q(x1+q). Thediscriminantof thepolynomialg(X) isz2+2q+2q2 = b2bydefinition,anditequals−4e3− 27.Sob2=−4e3+ γ.Ontheother

hand,wehave a = Trq3_/q((xq− xq 2 )1+q) = Trq3_/q(xq+1− x2) = Trq3_/q(xq+1+ x(xq+ xq 2 )) = 3e.

Wethushaveb2_{− 4u}3_{= γ,whichisanonsquareinF}

q whenq isoddbyLemma3.1.

Foranyelementα∈ F_q∗,letf1(X):= X3+aX +b−α,f2(X):= X3+aX +b−β−3α−1.

Uponexpansionwededucethat

cα=|{λ ∈ Fq: f1(λ) = 0}| + |{λ ∈ Fq: f2(λ) = 0}|.

Thediscriminants ofthetwocubicpolynomials f1 andf2 are

Δ1=−4a3− 27(b − α)2= γ((α− b)2− 4u3),

Δ2=−4a3− 27(b − β−3α−1)2= γ((γα−1− b)2− 4u3),

respectively. We claim that either none or both of Δ1,Δ2 are zero, and if the latter

occurs thenα2_{− 2bα + γ = 0.Wecomputethat}

Δ1Δ2=γ2((α− b)2− 4u3)((γα−1− b)2− 4u3) =γ2(α2− 2αb + γ)(γ2α−2− 2γα−1b + γ)

=γ3(α + γα−1)2+ 4bγ3(b− α − γα−1) =γ3(α + γα−1− 2b)2.

IfΔ1= 0,thenΔ1Δ2= 0 and soα− b= b− γα−1,whichinturnimpliesthatΔ2= 0.

Theconverseisalso true.Theclaimisnow established.

WefirstconsiderthecasewhereΔ1= Δ2= 0,sothatα2− 2bα + γ = 0.Inthiscase,

f1(X) has arepeated root η. If η is not inFq, then ηq wouldalso be arepeated root

of f1(X), contradicting the factthat deg(f1)= 3. Henceall the roots off1(x) = 0 lie

inFq. Thethree roots of f1 cannotbe allequal: iff1(X) = (X− η)3, then η = 0 by

comparingthecoefficientsofX2_{,andsoa= 0:acontradiction.Toconclude,f}

1(X) has

twodistinctrootsinFq.Thesameistrueforf2 bythesamereasoning.Wethusdeduce

thatcα= 4 inthis case.

WenextconsiderthecaseΔ1Δ2= 0.Inthecasewhereq isodd,Δ1Δ2isanonsquare

in Fq by Lemma 3.1. That is, exactlyone of Δ1 and Δ2 is anonsquare of Fq. By (i)

of Theorem 2.6,we conclude thatoneoff1 and f2 hasexactlyonezero inFq and the

otherhas 0 or3 zeros inFq. Itfollows thatcα= 1 or4 as desired.In thecasewhere q

iseven, wehaveγ = β = 1 and z = x. Itfollows thata= Trq3_/q(x1+q)= Tr_q3_/q(x−q 2

)

and b = Nq3_/q(x) = 1 by the fact x ∈ L₀. The two cubic polynomials take the form

X3_{+ aX + 1+ α,X}3_{+ aX + 1+ α}−1_{,respectively.Sincex∈ L}

0,wehaveTrq3_/q(x)= 0 andx1+q+q2= 1.Wecomputethat

Trq/2  a3 1 + α2  + Trq/2  a3 1 + α−2  = Trq/2(a3) =Trq/2 (x−1+ x−q+ x−q2)3= Trq3_/2(x−3+ xq−1+ xq 2₋₁ ) =Trq3_/2(xq 2_+q+1−3 + 1) = Trq3_/2(xq−1· xq 2₋₁ ) + 1 =Trq3_/2(xq−1x−1(x + xq)) + 1 = Tr_q3_/2(x2(q−1)+ xq−1) + 1 =1.

Asinthecasewhereq isodd,wededucethatcα= 1 or4 byusing(ii)ofTheorem2.6. 

Proposition3.4. Thereisasubset E⊆ T0 of size (q+1)

2

3 such that

D1+ D2= 3E + T0

inthegroupring Z[F_q∗3].

Remark3.5. Herewesayafewwordsaboutournotationused intheaboveproposition

and its proof below.A multiset M on F_q∗3 is apair (F_q∗3,ν), where ν : F_q∗3 → N ∪ {0}

is a function; that is, for any x ∈ F_q∗3, ν(x) is the multiplicity of x appearing in the

M := _x∈F∗

q3ν(x)x ∈ Z[F

∗

q3], i.e., thecoefficient of x of the groupring element M is

themultiplicityofx appearinginthemultisetM .Forexample,theabovepropositionis

stating thateachelement inthemultisetunion ofD1 andD2 (whichisidentified with

thegroupringelementD1+ D2)hasmultiplicity 4 or1 accordingasx isinE or not.

Proof. For x∈ L0,setz := xq− xq 2

,anddefineamultiset

Wx:= [Nq3_/q(λ + z) : λ∈ F_q]∪ [γN_q3_/q(λ + z)−1: λ∈ F_q]. (3.1)

For any element α ∈ F_q∗, its multiplicity in Wx equals cα, where cα is as defined in

Lemma 3.3. We have cα ∈ {1,4} by the same lemma. Therefore, there is a subset

Nx of Fq∗ such that Wx = Fq∗+ 3Nx as elements in the group ring Z[Fq∗]. Applying

the principal character of Fq∗ to both sides of this group ring equation, we find that

|Nx| =

2q−|Fq∗|

3 =

q+1

3 . Since gcd(q − 1,3) = 1, we see that y → y1/3 is a bijection

from F_q∗ to itself. Define Lx ={y1/3 | y ∈ Nx}, ∀x ∈ L0. Then Lx is a subset of Fq∗,

|Lx|=|Nx|=q+1₃ ,and

Wx= Fq∗+ 3L(3)x ,

where L(3)x =_z∈Lxz3.

It isroutineto check thatD1+ D2 =



x∈L0xW (1/3)

x . SetE :=∪x∈L0xLx,whichis

asubsetofT0ofsize(q + 1)2/3.Thentheclaiminthepropositionfollowsfromthefact

Wx= Fq∗+ 3L

(3)

x ∈ Z[F_q∗3] forx∈ L0,andL0· Fq∗= T0.Theproof isnow complete. 

3.2. Theset M

LetV = Fq3× F_q3,whichisviewedasa6-dimensionalvectorspaceoverF_q.Definea

map Q: V → Fqby

Q((x, y)) = Trq3_/q(xy), ∀(x, y) ∈ V.

It is easy to check that Q is a non-degenerate hyperbolic quadratic form on V . The

quadric definedby Q willbe our modelforQ+_{(5,q) whosepoints canbe expressed as}

{(x,y) : Q(x,y)= 0}.Thepolarform f : V × V → Fq ofQ isgivenby

f ((x, y), (a, b)) = Trq3_/q(bx + ay).

ForapointP =(x0,y0) ,itspolarhyperplaneP⊥ isgivenby

P⊥={(x, y) : Trq3_/q(xy₀+ x₀y) = 0}. Letw beaprimitiveelementofFq3.LetC₀bethesubgroupofF∗

q3oforderq2+ q + 1.

(x, y) → (μx, μ−1_{y) .}

Then, C0 is embedded as a subgroup i(C0) of PGO+(6,q), and it acts semi-regularly

on thepoints of Q+(5,q). So the points of Q+(5,q) are partitionedinto orbitsof this

action;eachorbithaslengthq2_{+ q + 1,andthenumberof orbitsisq}2_{+ 1.Wedenote}

theorbitcontaining thepoint (a,b) byO(a,b). Then, theorbitsare O(0,1) andO(1,z),

z∈ Fq3 withTr_q3_/q(z)= 0.Thefollowingis ourmain theorem.

Theorem 3.6.Let q ≡ 2 (mod 3) be a primepower. LetM =_z∈EO(1,z), where E is

definedinProposition3.4.Then,thelinesetL inPG(3,q) correspondingtoM underthe KleincorrespondenceformsaCameron-Lieblerlineclasswithparameterx= (q + 1)2_/3.

Remark3.7.When q = 2,theabovetheorem produces aCameron-Liebler line classin

PG(3,q) ofparameter3,whichisthecomplementofatrivialCameron-Lieblerlineclass

ofparameter2.SinceatrivialCameron-Lieblerlineclassofparameter2isstabilizedby

PSL(3,2),the Cameron-Liebler line class inPG(3,2) of parameter 3 produced by the

above theorem is also stabilized by PSL(3,2). In whatfollows, we will always require

q > 2 (and q ≡ 2 (mod 3)) since we said in Section 1 that we will concentrate on

Cameron-Lieblerline classesinPG(3,q) ofparameterx with1≤ x≤ (q2_{+ 1)/2.}

LetM bedefinedasinTheorem3.6.Clearly|M|= x(q2+ q + 1) withx= (q + 1)2/3.

SetD ={λv : λ∈ F_q∗,v ∈ M}.Then

|D| = (q − 1)|M| = (q − 1)|E|(q2_{+ q + 1) = (q}3_{− 1)}(q + 1)2

3 .

To prove Theorem 3.6, we need to show that D has the correct character values as

specified inResult 2.2. Each additivecharacter of (V,+) is of the form ψa,b for some

(a,b)∈ V ,where

ψa,b((x, y)) = ψFq(f ((a, b), (x, y))) = ψFq3(bx + ay), ∀(x, y) ∈ V

Here ψFq and ψFq3 arethecanonical additivecharacters ofFq and Fq3,respectively.It isclearthatψa,b istrivialonthehyperplane(a,b) ⊥.ByResult2.2, inorder toprove

Theorem3.6itsufficestoprovethefollowingclaim:foranynonzeroelement(a,b) ofV ,

wehave ψa,b(D) =  −(q+1)2 3 + q3, if (a, b)∈ D, −(q+1)2 3 , otherwise. (3.2)

4. Proofofthe maintheorem

LetM bedefinedasinTheorem3.6,andletD bethecorrespondingsubsetofnonzero

vectors inV = Fq3× F_q3. Takethesamenotationas introduced inSubsection 3.2,and

set N := q2_{+ q + 1. To simplify notation, we write G(χ) for the Gauss sum G}

q3(χ),

where χ isamultiplicativecharacterofF_q∗3. Weneedto evaluatethecharactersum

ψa,b(D) =



z∈E

ψa,b(Fq∗O(1,z))

for nonzero(a,b)∈ V ,where Fq∗O(1,z) ={(yμ,yμ−1z): y∈ Fq∗,μ∈ C0}.Forz∈ E,we have ψa,b(Fq∗O(1,z)) =  y∈F∗ q  μ∈C0 ψF_q3(byμ + ayμ−1z). (4.1)

The inner suminthe right handside of(4.1) is an incompleteKloosterman sum.It is

ingeneral verydifficultto evaluateincomplete Kloostermansumsexactly.Here weare

dealingwithcertainsumsψa,b(D) of incompleteKloostermansums;andforthese sums

we canevaluatethemexactly.

Lemma 4.1.If ab= 0 but(a,b)= (0,0), thenψa,b(D)=−(q+1)

2 3 . Proof. Recall thatF_q∗3 = C0· Fq∗.Whena= 0 andb= 0,

ψ0,b(Fq∗O(1,z)) =  θ∈F∗ q  μ∈C0 ψF_q3(bθμ) =  x∈F∗ q3 ψF_q3(bx) =−1.

The computations inthe case where a = 0 and b = 0 are similar. This completes the

proof. 

From now on, we assume that ab = 0. Let χ be a generator of the multiplicative

charactergroupofF_q∗3,andset

χ1:= χq−1, χ2:= χN.

The ordersof χ1 andχ2 areN andq− 1,respectively.For asubset Y ofFq∗3 (possibly

amultiset) andamultiplicativecharacterχi, wewrite χi(Y ):=_x∈Y χi(x). Itis well

knownthat χk(C0) =  N, if k≡ 0 (mod N), 0, otherwise, (4.2) and

χk(F_q∗) = 

q− 1, if k≡ 0 (mod q − 1),

0, otherwise. (4.3)

Weintroducetwo auxiliaryexponential sums:

S1= 1 q3_{− 1} N−1_ =0 G(χ−₁ )2χ₁(ab)χ₁(E) and S2= 1 q3− 1 q−2  i=1 N−1_ =0

G(χi2χ−1 )G(χ−i2 χ−1 )χ1(ab)χi2(ab−1)χi2χ1(E).

Lemma4.2. Itholdsthat ψa,b(D)= S1+ S2.

Proof. ByLemma2.3, wehave

ψa,b(Fq∗O(1,z)) =  θ∈F∗ q  μ∈C0 ψF_q3(bθμ + aθμ−1z) = 1 (q3_{− 1)}2 q3₋₂  i=0 q3₋₂  j=0  θ∈F∗ q  μ∈C0 G(χ−i)χi(bθμ)G(χ−j)χj(aθμ−1z) = 1 (q3− 1)2 q3₋₂  i=0 q3₋₂  j=0  θ∈F_q∗ G(χ−i)G(χ−j)χi(bθ)χj(aθz)  μ∈C0 χi−j(μ) (4.2) = N (q3− 1)2 q_3−2 j=0 q−2  h=0  θ∈F∗ q G(χ−j−Nh)G(χ−j)χj+N h(bθ)χj(aθz) = N (q3_{− 1)}2 q_3−2 j=0 q−2  h=0 G(χ−j−Nh)G(χ−j)χj+N h(b)χj(az)χ2j+N h(F_q∗). By(4.3),χ2j+N h_(F∗

q)= q− 1 ifandonlyif2j + N h≡ 0 (mod q − 1), i.e., h≡

2j(q−2) 3

(mod q− 1).Wethushave

ψa,b(Fq∗O(1,z)) = 1 q3− 1 q3₋₂  j=0 G(χ−j−2j(q−2)N3 )G(χ−j)χj+ 2j(q−2)N 3 (b)χj(az).

SinceN = q2_{+ q + 1 andgcd(N,q− 1)= 1,anyintegerj ∈ [0,q}3_{− 2] canbe uniquely}

writtenasj = N i+ (q− 1) forsomeintegers0≤ i≤ q − 2 and 0≤ ≤ q2_{+ q by the}

j + 2j(q− 2)N

3 ≡ −Ni + (q − 1) (mod q

3_{− 1)}

bythefact (q−2)₃ N ≡ −1 (mod q − 1).Byrewritingj asN i+ (q− 1) wededucethat

ψa,b(Fq∗O(1,z)) = 1 q3_{− 1} q−2  i=0 N_−1 =0

G(χi₂χ−₁ )G(χ−i₂ χ−₁ )χ−i₂ χ₁(b)χi₂χ₁(az)

= 1 q3_{− 1} N−1_ =0 G(χ−₁ )2χ₁(b)χ₁(az) + 1 q3_{− 1} q−2  i=1 N_−1 =0

G(χi₂χ−₁ )G(χ−i₂ χ−₁ )χ−i₂ χ₁(b)χi₂χ₁(az).

Takingsummationoverz∈ E,wegetthedesiredequalityψa,b(D)= S1+ S2. 

We now explicitly evaluate S1 and S2. Write ab = wN s0+(q−1)t0 for some s0 ∈

{0,1,. . . ,q− 2} andt0∈ {0,1,. . . ,q2+ q}.

Lemma 4.3.It holdsthat

S1= ⎧ ⎪ ⎨ ⎪ ⎩ (q + 1)(q3_{− q}2_{+ 1)} 3(q− 1) , if Trq3/q(w (q−1)t0_{) = 0,} −(q + 1)2 3 , otherwise.

Proof. From theproof of Proposition3.4, we seethat E =_x∈L

0xLx,where each Lx

is asubsetofF_q∗ ofsize q+1₃ .ByLemma2.5,forany 1≤ ≤ N − 1 we have

χ₁(E) =  x∈L0  y∈Lx χ₁(xy) = q + 1 3  x∈L0 χ₁(x) = (q + 1) 3q G(χ  1).

Togetherwith thefactG(χ−₁ )G(χ

1)= q3 for1≤ ≤ N − 1,wehave S1= |E| q3_{− 1}+ 1 q3_{− 1} N−1_ =1 G(χ−₁ )2χ₁(ab)χ₁(E) =(q + 1) 2 3(q3− 1) + q + 1 3q(q3− 1) N−1_ =1 G(χ−₁ )G(χ−₁ )G(χ1)χ1(ab) =(q + 1) 2 3(q3_{− 1)} + (q + 1)q2 3(q3_{− 1)} N_−1 =1 G(χ−₁ )χ₁(ab).

S1= (q + 1)2 3(q3− 1)+ (q + 1)q2 3(q− 1)  ψF_q3(abFq∗) + 1 N  . Wecompute ψF_q3(abFq∗) = ψFq(w N s0_Tr q3_/q(w(q−1)t0)F_q∗) =  q− 1, if Trq3_/q(w(q−1)t0) = 0, −1, otherwise. (4.4)

Hencewe finallyobtain

S1= _(q+1)(q3_−q2 +1) 3(q−1) , if Trq3_/q(w(q−1)t0) = 0, −(q+1)2 3 , otherwise.

Thiscompletestheproofofthelemma. 

Nextwe evaluateS2.BythedefinitionofE, wehave

χi₂χ₁(E) = 1 3  χi₂χ₁(D1) + χi2χ1(D2)− χi2χ1(T0)  . ThereforeS2= Σ1+ Σ2+ Σ3,where Σ1= 1 3(q3− 1) q−2  i=1 N−1_ =0

G(χi₂χ−₁ )G(χ−i₂ χ−₁ )χ₁(ab)χi₂(ab−1)χi₂χ₁(D1),

Σ2= 1 3(q3_{− 1)} q−2  i=1 N_−1 =0

G(χi₂χ−₁ )G(χ−i₂ χ−₁ )χ₁(ab)χi₂(ab−1)χi₂χ₁(D2),

Σ3= −1 3(q3− 1) q−2  i=1 N−1_ =0

G(χi2χ−1 )G(χ−i2 χ−1 )χ1(ab)χi2(ab−1)χi2χ1(T0).

Write ab = wN s0+(q−1)t0_{, ab}−1 _{= w}N u0+(q−1)v0_{, and set z}

0 = wN u0+(q−1)t0 and

z1= w−Nu0+(q−1)t0.Thenextpropositiongivesthevaluesof Σ1,Σ2 andΣ3.

Proposition4.4. Foreach z∈ T0,define

μz:=|{(y, λ) ∈ C0× Fq: y− (zq− zq 2 ) + λ = 0}|, μ_z:=|{(y, λ) ∈ C0× Fq: y− β(zq− zq 2 ) + λ = 0}|,

whereβ =−3−1∈ Fq.Thenitholdsthat

(Σ1, Σ2, Σ3) =  (q₃3μz0− q4 3(q−1), q 3 3μz1− q4 3(q−1), 0), if w(q−1)t0 ∈ T0, (0, 0, 0), otherwise.

The proof ofProposition4.4 involves very complicatedcomputations of exponential

sums.Tostreamlinethepresentationofthepaper,wedelaytheprooftotheAppendix.

Proof of Theorem3.6. AsmentionedinSubsection 3.2,it sufficesto establish(3.2) for

nonzeroelement (a,b) of V ,i.e.,

ψa,b(D) =  −(q+1)2 3 + q 3_{, if (a, b)∈ D,} −(q+1)2 3 , otherwise. (4.5)

If either a = 0 or b = 0, then (a,b) ∈ D and/ the claim has been established in

Lemma4.1.

Wenexttreatthecasewhenab= 0. Writeab= wN s0+(q−1)t0_,ab−1_{= w}N u0+(q−1)v0_, and set x0 = w(q−1)t0, θ0 = wN u0. It follows that a2 = wN (u0+s0)+(q−1)(t0+v0). In the

rest oftheproof,weusethenotationintroducedinthestatementofProposition4.4.In

particular, z0= x0θ0 andz1= x0θ−10 .

We claim that (a,b) ∈ D if and onlyif z1 ∈ E.We write a= xμ for some x∈ Fq∗

and μ ∈ C0. From a2 = wN (u0+s0)+(q−1)(t0+v0) we deduce that x2 = wN (u0+s0). It

is straightforward to check that ab = x2_z

1, so b = (xμ)−1x2z1 = xμ−1z1. That is,

(a,b)= (xμ,xμ−1z1).Theclaimnowfollowsfrom thedefinitionofD.

ByLemma4.3 andProposition4.4,wehave

ψa,b(D) = _(q+1)(q3_−q2₊₁₎ 3(q−1) + q3 3(μz0+ μz1)− 2q4 3(q−1), if x0∈ T0, −(q+1)2 3 , otherwise.

Hence, we need to compute μz0 + μz1 under the assumption that x0 ∈ T0. Set z˜0 :=

xq₀− xq₀2.Wenowhave μz0+ μ  z1=|{(y, λ) ∈ C0× Fq: y = (z q 0− z q2 0 )− λ}| +|{(y, λ) ∈ C0× Fq: y = β(z1q− z q2 1 )− λ}| =|{λ ∈ Fq: Nq3_/q(λ + θ₀z˜₀) = 1}| +|{λ ∈ Fq: Nq3_/q(λ + θ−1₀ β ˜z₀) = 1}| =|{λ ∈ Fq: Nq3_/q(λ + ˜z₀) = θ₀−3}| +|{λ ∈ Fq: β−3Nq3_/q(λ + ˜z₀)−1= θ−3₀ }|,

whichisthemultiplicityofθ−3₀ inthemultisetWx0.ByProposition3.4,thismultiplicity

is equal to 1 or 4, and correspondingly ψa,b(D) = −(q+1)

2 3 or q3− (q+1)2 3 . Moreover μz0+ μz1= 4 ifand onlyifθ −1

0 ∈ Lx0,i.e.,z1= x0θ−10 ∈ E.Tosumup,wehaveshown that

ψa,b(D) =  q3₋(q+1)2 3 , if z1∈ E, −(q+1)2 3 , otherwise.

Alsowehaveshownthat(a,b)∈ D ifandonlyifz1∈ E.Theproofisnowcomplete. 

5. ThestabilizerofM inPΩ(V )

Letq beaprimepowersuchthatq > 2 andq≡ 2 (mod 3).RecallthatV = Fq3×F_q3,

and Q : V → Fq defined by Q((x,y)) = Trq3_/q(xy), ∀(x,y) ∈ V , is a nonsingular

hyperbolic quadratic form on V . Let Ω(V ) be the derived subgroup of the isometry

groupof the quadratic space (V,Q), and let PΩ(V ) bethe quotient groupmodulo its

center.Inthis section,wedeterminethestabilizerofM in PΩ(V ).

SetW :={x∈ Fq3: Tr_q3_/q(x)= 0},U₁:= F_q3× {0},and U₂:={0}× F_q3.LetE be thesubsetofF_q∗3 as inProposition3.4. Foreachx∈ L0,there existsasubsetLx ofFq∗

ofsize q+1₃ suchthatE =∪x∈L0xLxbytheproofofProposition3.4.

Let betheset ofsquaresofFq∗ (sointhecasewhenq iseven,= Fq∗).Asbefore,

forany two subsetsA, B ofF_q∗3, define A· B := {ab : a∈ A,b ∈ B}.In particular, if

A={a},wewrite aB forA· B.

Lemma5.1. AsFq-vector spaces,Fq3= W ⊕ F_q.Furthermore F_q∗· E = W \ {0}.

Proof. Sinceq≡ 2 (mod 3),weseethat3 |q,consequentlyW andFqintersecttrivially;

andthefirstclaimfollows.Thesecondis clearfrom theabovedescriptionofE. 

Foreachu∈ L0,wedefineBu:={yq

2

uq_−yq_uq2

: y∈ L0\{u}}.Itisroutinetocheck thattheelements ofBu lieinFq∗ bythefactTrq3_/q(y)= 0 fory∈ L₀.

Lemma5.2. Foru∈ L0,wehave|Bu|= 2₃(q + 1)− 1 andBu· Lu= Fq∗.

Proof. WehaveshownthatW andFqintersecttrivially,sou∈ F/ q andW =u,uq Fq.

Moreover, uq−1 ∈ F/ q∗ by the fact thatgcd(q− 1,q2+ q + 1) = 1. We thus haveL0 =



uq_+λu

Nq3 /q(uq+λu)1/3 : λ∈ Fq 

∪ {u}. It follows thatBu=



uq+1_−u2q2

Nq3 /q(uq+λu)1/3 : λ∈ Fq 

, and

itssize equals thatof {Nq3_/q(uq−1+ λ): λ∈ F_q}.Writex:= λ+1

3Trq3_/q(uq−1)∈ F_q. Then

Nq3_/q(uq−1+ λ) = x3+ bx + c

for some b,c ∈ Fq. The polynomial X3 + bX + c ∈ Fq[X] has no roots in Fq, so it

is irreducible over Fq. It follows that bc = 0. The polynomial X3 + bX is a Dickson

polynomial of degree 3, and its value set over Fq has size _{2·gcd(3,q−1)}q−1 +_{2·gcd(3,q+1)}q+1 =

2

3(q + 1)− 1 by Theorem10andTheorem 10’of[6].Foranya∈ Fq∗,wehave

|{ay−1_{: y∈ L}

This shows that a ∈ Bu · Lu. Hence Bu · Lu = Fq∗. The proof of the lemma is now

complete. 

Thegeneratorsof(V,Q) fallintotwoequivalenceclasses;twogeneratorsU andUare

equivalentifandonlyifU∩ Uhasdimension1,cf.[15,Theorem1.39].ThegroupΩ(V )

stabilizes eachequivalence class, cf. [16,p.30]. Thetwo subspacesU1 and U2 are both

generatorsofthequadraticspace(V,Q),andtheyareindifferent equivalenceclasses.

Lemma 5.3.The only generatorsof(V,Q) that are disjointfromM are U1 andU2.

Proof. It is clear thatU1 and U2 are disjoint from M. Suppose that U isa generator

other thanU1,U2.WewillshowthatU intersectsM nontrivially.

We first consider the case where U andU2 are equivalent. In this case, U∩ U2 is a

projective point P . By applying the action of some element ini(C0) ≤ PGO+(6,q) if

necessary, we mayassume without loss of generality thatP = (0,1) . It is clear that

P⊥= W×Fq3.SinceF_q3 = W⊕F_q,weidentifyW×W withP⊥/P naturally.Inthisway,

W× W becomesaquadraticspaceQ+_{(3,q) whoseinheritedquadraticform isthesame}

as the restrictionofQ toW × W .We haveM∩ P⊥ ={(y,y−1x) : x∈ E,y ∈ L0},

and the corresponding set inW × W is MP :={(y,τy(x)) : x∈ E,y ∈ L0},where

τy(x):= y−1x−1₃Trq3_/q(y−1x).Notethatτ_y : F_q3 → F_q3isF_q-linear.Itisstraightforward

to checkthat

ker(τy) = Fq· y, Im(τy)≤ W, τy(W )≤ {z ∈ W : Trq3_/q(yz) = 0} = yq− yq 2

Fq.

We thus have τy(W ) = yq− yq

2

Fq by comparing dimensions. Let U

 _{be the totally}

singular lineofW × W corresponding toU .To showthatU intersectsM nontrivially,

itsufficestoshowthatUintersectsMP nontrivially.Thereare2(q + 1) totallysingular

linesofQ+_{(3,q),theseare}

y=(y,0),(0,yq−yq

2

) withy∈ L0, a={(x,axq−axq

2 ) :

x∈ W } with a∈ Fqand ∞={0}× W . Thelastline ∞ correspondsto thegenerator

U2, soU= _∞.

(1) IfU = y for somey ∈ L0,thenthe point(y,τy(x)) withx∈ E is inU ifthere

existsλ∈ F_q∗ suchthatτy(λx)= yq− yq

2

.ByLemma5.1,wehaveF_q∗· E = W \ {0}.

The existence ofsuch x∈ E and λ∈ F_q∗ now follows from the factthat τy(W ) =

yq_{− y}q2

Fq.

(2) If U = _a forsomefixed a∈ Fq, then U∩ MP = ∅ ifthere exists y ∈ L0,u∈ L0 and c∈ Lu such thatτy(uc)= ayq− ayq

2

.The lefthandside equals zq− zq2 with

z = −1₃(y−1u)q_c+1

3(y−1u)q 2

c, so ay − z ∈ Fq. By taking the relative trace, we

see that it equals 0. By the fact that Nq3_/q(y) = 1 for y ∈ L₀, we deduce that

−3a= (yq2

uq_{− y}q_uq2

)c. When a= 0,we cansimplytakey = u ∈ L0 andc ∈ Lu

arbitrarily.Whena= 0,wetakeu tobeanyelement ofL0andtheexistenceofthe

Inboth cases,we haveshown thatU intersects MP nontrivially. This establishes the

claiminthecaseU isinthesameequivalenceclassas U2.

We next consider the case where U and U1 are equivalent. Observe thatO(1,xa) =

O(xa−1,1) for x ∈ C0 and a ∈ Fq∗, so M = ∪x∈EO_(x,1), where E = ∪x∈L0xLx with

L_x={a−1: a∈ Lx}.Theargumentisexactlythesameasinthepreviouscase. 

Let K be the stabilizer of U1 and U2 in Ω(V ), i.e., K = {α ∈ Ω(V ) : α(U1) =

U1and α(U2)= U2}.By[16,Lemma4.1.9],K consistsof

κ(h, h∗) : V → V, (x, y) → (h(x), h∗(y)),

where both h and h∗ are bijective Fq-linear transformations of Fq3 such that det(h),

det(h∗) ∈  and Q((x,y)) = Q((h(x),h∗(y))) for all x,y ∈ Fq3. Here, det(h) is the

determinantofh withrespect toany Fq-basisofFq3. ForeachbijectiveF_q-linear

trans-formationh of Fq3 with det(h)∈ , there is aunique h∗ such thatκ(h,h∗)∈ K, and

viceversa.

Wenowdescribesomespecialelements ofK.Fora∈ F_q∗3,define

ha: Fq3 → F_q3, x → ax,

andsetκa:= κ(ha,ha−1).Anelementz∈ C0 (whichweidentifywiththecorresponding

elementini(C0)) actsonV inexactlythesameway asκz.

Lemma5.4. Fora∈ F_q∗3,κa isinK if andonly ifa is asquarein Fq∗3.

Proof. The linear transformation κa clearly has determinant1 and stabilizes the

gen-erators U1 and U2, so it suffices to show thatdet(ha)∈  if and only ifa is asquare

in F_q∗3. For a ∈ C0, we have hq 2 +q+1 a = idF_q3, so det(ha)q 2 +q+1 _{= 1. It follows that}

det(ha)= 1 fromthefactgcd(q2+ q + 1,q− 1)= 1.Fora∈ Fq∗,wehavedet(ha)= a3,

whichis asquare if and onlyif a is.The claimthen follows readilyfrom the factthat

F_q∗3= C0· Fq∗. 

Wedefineι: K→ PGL(3,q) suchthatι(g) isthequotientimageofg|U1 inPGL(3,q),

where g|U1 is therestriction of g to U1. Since gcd(3,q− 1)= 1, we have PGL(3,q) =

PSL(3,q).Thehomomorphismι issurjectivebytheabovedescriptionofK.

Lemma5.5. Wehaveker(ι)= κ_,where κ_={κa: a∈ }.

Proof. If κ = κ(h,h∗) ∈ ker(ι), then h = ha for some a ∈ Fq∗ and correspondingly

κ= κa. Theclaimisnow aneasyconsequenceofLemma5.4. 

Letσ betheFq-lineartransformationofV suchthatσ((x,y))= (xq,yq).Ithasorder

Lemma 5.6.We haveσ∈ K,andσ(M)=M.

Proof. ThefirstclaimfollowsbythesameargumentasintheproofofLemma5.4.The

second claim is equivalent toσ(E)= E, or equivalently,Lxq ={aq: a∈ L_x} for each

x∈ L0.This isclearfromthedefinitionofLx intheproofofProposition3.4. 

LetG bethestabilizerofM inΩ(V ).Letα∈ G.FromU1∩ M=∅ andU2∩ M=∅,

weobtainα(U1)∩ M=∅ andα(U2)∩ M=∅.ByLemma5.3andthefactthatU1and

U2areindifferentequivalenceclasses,wededucethatα(U1)= U1andα(U2)= U2,and

soα∈ K.WehaveshownthatG≤ K.Moreover,G containsthesubgroupH generated

byσ and i(C0).

Lemma 5.7.Letq > 2 andq≡ 2 (mod 3).Thenthegroupι(G) has order3(q2+ q + 1).

Proof. The groupι(H) has order3(q2+ q + 1) andisamaximalsubgroupofPSL(3,q)

by[2,Table8.3].Henceeitherι(G)= PSL(3,q) orι(G)= ι(H).

Supposethatι(G)= PSL(3,q). Fixanelement u∈ L0,and takeλ to beaprimitive

elementofF_q∗.Letg = κ(h,h∗) betheelementofK suchthath∗(1)= 1,h∗(u)= λu and

h∗(uq_{)= λ}−1_uq_{.Bylettingx= 1,u,u}q_{,respectively,inQ((1,x))= Q((h(1),h}∗_(x))),we

obtain Trq3_/q(h(1))= 3, Tr_q3_/q(h(1)u)= 0,and Tr_q3_/q(h(1)uq)= 0, respectively.Since

{1,u,uq} isanFq-basisofFq3,theabovethreeequationsuniquelydetermineh(1).Since

1∈ Fqsatisfiesthesethreeequations,wededucethath(1)= 1.Byourassumptionthere

exists a ∈  such thatκag ∈ G, i.e., κag stabilizes M. The image of {(1,x) : x ∈

E}⊆ M underκag is{(1,a−2h∗(x)) : x∈ E},so wehaveE = a−2h∗(E).Comparing

both sides,wededucethat

uLu= a−2λuLu, (5.1)

and

uqLuq = a−2λ−1uqL_uq. (5.2)

Multiplyingalltheelements intheset oneachside of (5.1) (respectively,(5.2)),weget (a−2λ)(q+1)/3= 1, and(a−2λ−1)(q+1)/3= 1, respectively.It followsthatλ2(q+1)/3= 1. Ifq > 5,then 2(q+1)₃ < q− 1,andtheequalityλ2(q+1)/3_{= 1 contradictstheassumption} thatλ isprimitive.Ifq = 5,wehavea4_{= 1 andthisleadstoλ}2_{= 1,againcontradicting}

theassumption thatλ isprimitive.Theproofisnow complete. 

Lemma 5.8.If q is odd,then |Lx∩ |= q+1₆ foreach x∈ L0.

Proof. Fromthe proofofProposition3.4, weknow thatWx= Fq∗+ 3L

(3)

x inthegroup

ring Z[Fq∗], where Wx is the sameas in(3.1). Let ρ be thequadratic character of Fq∗,

Sinceγ =−27 isanonsquareinFq∗,wededucethatρ(Wx)= 0.Itfollowsthatρ(Lx)= 0,

i.e.,Lx hasthesamenumberofsquaresasnonsquares.Thiscompletestheproof. 

Theorem5.9. Letq beaprime powersuchthat q > 2 and q≡ 2 (mod 3). The groupG

has order3(q2_{+ q + 1)s,wheres= 1 ors= gcd(2,}q−1

2 ) accordingasq is evenor odd.

Proof. ByLemma5.7,G lies inthe groupH × κ_,where κ_ is as inLemma5.5. We

haveshownthatH ≤ G,soG= H×(G∩κ_).Itnowsufficestodeterminethestabilizer

ofM inκ_.

Supposethatκastabilizes M,wherea isasquareofFq∗.Theconditionκa(M)=M

isequivalentto a2E = E,i.e., a2Lx= Lxfor eachx∈ L0.Takingtheproduct overthe

set oneachside, wededucethata2(q+1)/3_{= 1. Ifq iseven, then theorderof a divides}

gcd(2(q + 1)/3,q− 1)= 1,implying a= 1.If q≡ 3 (mod 4), then gcd(q−1₂ , 2(q+1)₃ )=

1 and we also get a = 1. If q ≡ 1 (mod 4), then from a2_L

x = Lx we deduce that

a2_(L

x∩ )= Lx∩ .ByLemma5.8wehave|Lx∩ |= q+1₆ .Bythesameargumentwe

geta(q+1)/3= 1.Inthiscase,we havegcd(q+1₃ ,q−1₂ )= 2,so a2= 1,i.e.,a=±1. Since

−1 isin,wesee thatindeedκ₋₁ is inG.Thiscompletestheproof. 

Corollary 5.10. Let q > 2 and q ≡ 2 (mod 3), and let L be the Cameron-Liebler line

class in PG(3,q) corresponding to M.Then the stabilizer of L in PSL(4,q) has order

3(q2+ q + 1)s,where s= 1 ors= gcd(2,q−1₂ ) according asq is evenorodd.

Proof. ByTheorem 5.9, thestabilizer of M inPΩ(V ) has order 3(q2_{+ q + 1)s,where}

s= 1 ors= gcd(2,q−1₂ ) accordingasq isevenorodd.BytheisomorphismPΩ+_(6,q)∼₌

PSL(4,q),we see thatthestabilizer ofthecorresponding Cameron-Liebler line classin

PSL(4,q) hasorder3(q2+ q + 1)s,wheres= 1 ors= gcd(2,q−1₂ ) accordingasq iseven

orodd. 

6. Concludingremarks

In this paper, we have constructed Cameron-Liebler line classes in PG(3,q) with

parameter x = (q + 1)2_{/3 for all primepowers q congruent to 2 modulo 3. This is a}

contribution to the study of the central problem about Cameron-Liebler line classes

in PG(3,q). Besides the trivial examples with x = 1,2, all known infinite families of

Cameron-Liebler line classespriorto our workhaveparametersx= (q2_{− 1)/2 orx=}

(q2_{+ 1)/2,uptocomplement.}

Most notably, we have constructed the first infinite family of nontrivial

Cameron-LieblerlineclassesinPG(3,q) withq even.Incontrast,thefirstnontrivialinfinitefamily

of Cameron-Liebler line classes in PG(3,q) for odd q was constructed by Bruen and

Drudge [3] twenty years ago. The major obstaclein the characteristic two case seems

to be that such line classes, if they exist, tend not to be highly symmetric. In our

sizes. This factmakesit difficultto giveaneat geometricdescription oftheobjects we

haveconstructed.Ourproofisveryalgebraic,dueto thenatureofourconstruction.

In Section5, we havedetermined the stabilizers ofour Cameron-Liebler line classes

inPSL(4,q).It willbeof particularinteresttofindinfinitefamiliesofCameron-Liebler

line classeswhosestabilizersinPSL(4,q) donotgrowasq increases.Arelatedquestion

is whethertheCameron-Liebler line class L arising from Theorem 3.6is aline classof

asymmetric tactical decompositionofPG(3,q). Wehavechecked byusing acomputer

thattheanswerisnoinbothcaseswhenq = 5 and8,andwebelievethattheansweris

noingeneral.

Appendix A. ProofofProposition 4.4

Inthisappendix,wewillproveProposition4.4.RecallthatN = q2+ q + 1.Westart

with anobservationonGauss sums.Let S beanysubset ofF_q∗3,and setTS :={(s,t):

0≤ i≤ N − 1,0≤ t≤ q − 2,ws(q−1)+tN _{∈ S}.BythedefinitionofGausssums,forany}

integersi and and ,δ∈ {1,−1} wehave

G(χi₂χδ₁ )χi₂χ₁(S) =  y∈F∗ q3  z∈S χi₂χδ₁ (y)χi₂χ₁(z)ψF_q3(y) =  y∈F∗ q3  (s,t)∈TS

χi₂(yws(q−1)+tN)χδ₁(ywδs(q−1)+δtN)ψF_q3(y).

(A.1) Since χ2(ws(q−1))= 1 andχ1(wtN)= 1,continuingfrom (A.1),wehave

G(χi₂χδ₁ )χi₂χ₁(S) = 

y∈F∗ q3

 (s,t)∈TS

χi₂(ywδs(q−1)+tN)χδ₁(ywδs(q−1)+tN)ψF_q3(y)

=  z∈F∗ q3  (s,t)∈TS χi₂χδ₁(z)ψF_q3(zw−δs(q−1)−tN). (A.2)

This identitywillbe usedintherest oftheproof.

LetD3= βD2 = [xNq3_/q(λ+ xq− xq 2

)−13 : x∈ L₀,λ∈ F_q].To evaluateΣ₁,weneed

thefollowing observation:By(A.2) wehave

G(χi₂χ−₁ )χi₂χ₁(D1) = 

z∈F∗ q3

χi₂χ−₁ (z)ψF_q3(zD3) (A.3)

LemmaA.1. LetR ={λ+ (hq2− hq): λ∈ Fq,h∈ L0}.Forz∈ Fq∗3,itholds that

ψF_q3(zD3) = 

q2_{+ q, if z∈ F}∗

q,

−1 + ψF_q3(eC0), if z∈ eR for some e ∈ Fq∗.

Proof. WefirstnotethatR isasystemofcosetrepresentativesfor(F_q∗3\F_q∗)/F_q∗.Assume thatthere are λ1,λ2 ∈ Fq, d∈ Fq∗ and h1,h2 ∈ L0 such thatλ1+ (hq

2 1 − h

q

1)= dλ2+

d(hq₂2− hq₂).Thenbytakingtraceofbothsides,wehaveλ1= dλ2.Notethathq 2 1 − h

q

1=

d(hq₂2− hq₂) impliesthathq₁2− dhq₂2 = hq₁− dhq₂,i.e.,h1− dh2∈ Fq.Hence,wehave

0 = Trq3_/q(h₁)− dTr_q3_/q(h₂) = Tr_q3_/q(h₁− dh₂) = 3(h₁− dh₂),

whichimpliesthath1= dh2.Bytakingnormofbothsides,wehaved= 1,λ1= λ2 and

h1= h2.ItisclearthatnoneoftheelementsofR isinFq∗.HenceR isasystemofcoset

representativesof(F_q∗3\ F_q∗)/F_q∗.

Next we evaluate ψF_q3(zD3). Let ηq−1 be a fixed multiplicative character of order

q− 1 ofFq.Then,wehave ψF_q3(zD3) =  λ∈Fq  x∈L0 ψF_q3(zxNq3_/q(λ + xq− xq 2 )−1/3) = 1 q− 1  c∈F_q∗  λ∈Fq  x∈L0 q−2  i=0 ψF_q3(zxc−1)η_q−1i ((λ + xq− xq 2 )N)η_q−1−3i(c) = 1 q− 1  c∈F∗ q  λ∈Fq  x∈L0 q−2  i=1 ψF_q3(zxc−1)η_q−1i ((λ + xq− xq 2 )N)η_q−1−3i(c) (A.4) + 1 q− 1  c∈F∗ q  λ∈Fq  x∈L0 ψF_q3(zxc−1). (A.5)

Denotethesummandsin(A.4) and(A.5) byW1andW2,respectively.Then,ψF_q3(zD3)=

W1+ W2.Here,itiseasytoseethat

W2= 1 q− 1  c∈F_q∗  d∈Fq  x∈C0 ψF_q3(zxc−1+ xd) = 1 q− 1  d∈Fq  x∈F∗ q3 ψF_q3(x(z + d)) = 1 q− 1  q3_{− q, if z∈ F} q, −q, if z /∈ Fq.

WenextevaluateW1.Letρq−1 be theliftofηq−1 toFq∗3,i.e., ρq−1(x)= ηq−1(xN).We notethatforanys∈ F_q∗,ρq−1(s)= ηq−1(sN)= ηq−1(s3).Then