Characterization of intersecting families of maximum size in PSL(2,q)
aa r X i v : . [ m a t h . C O ] M a r CHARACTERIZATION OF INTERSECTING FAMILIES OF MAXIMUMSIZE IN
P SL (2 , q ) LING LONG ∗ , RAFAEL PLAZA, PETER SIN † , QING XIANG ‡ Abstract.
We consider the action of the 2-dimensional projective special linear group
P SL (2 , q ) on the projective line P G (1 , q ) over the finite field F q , where q is an odd primepower. A subset S of P SL (2 , q ) is said to be an intersecting family if for any g , g ∈ S ,there exists an element x ∈ P G (1 , q ) such that x g = x g . It is known that the maximumsize of an intersecting family in P SL (2 , q ) is q ( q − /
2. We prove that all intersectingfamilies of maximum size are cosets of point stabilizers for all odd prime powers q > Introduction
Let n, k be positive integers such that k ≤ n and let [ n ] = { , , . . . , n } . A family of k -subsets of [ n ] is said to be intersecting if the intersection of any two k -subsets in thefamily is non-empty. The Erd˝os-Ko-Rado (EKR) theorem is a classical result in extremalset theory. It states that when k < n/ k -subsets has size at most (cid:0) n − k − (cid:1) ; equality holds if and only if the family consists of all k -subsets of [ n ] containing afixed element of [ n ] (cf. [9]). In this paper, we focus on EKR type problems for permutationgroups. In particular, for any odd prime power q , we consider the natural right action of P SL (2 , q ), the 2-dimensional projective special linear group over the finite field F q , on theset of points of P G (1 , q ), the projective line over F q .Let X be a finite set and G a finite group acting on X . A subset S of G is said to be an intersecting family if for any g , g ∈ S there exists an element x ∈ X such that x g = x g , i.e., g g − stablizes some x ∈ X . In the context of EKR-type theorems, the following problemsabout intersecting families in G are of interest:I (Upper Bound) What is the maximum size of an intersecting family?II (Characterization) What is the structure of intersecting families of maximum size?Extensive research has been done to solve the above problems for different groups. In1977, Deza and Frankl [10] solved Problem I for the symmetric group S n acting on [ n ]. Theyproved that any intersecting family of S n has size at most ( n − S n is an intersecting family of size precisely( n − n − Key words and phrases.
Character table, Erd˝os-Ko-Rado theorem, Hypergeometric function over finitefield, Intersecting family, Legendre sum, Soto-Andrade sum. ∗ Research partially supported by NSF grants DMS-1303292 and DMS-1602047. † Research partially supported by a grant from the Simons Foundation ( ‡ Research partially supported by an NSF grant DMS-1600850. n [19], Meagher and Spiga studied Problem I and II for the group P GL (2 , q ) acting onthe set of points of the projective line P G (1 , q ). These authors proved that the maximumsize of an intersecting family in P GL (2 , q ) is q ( q − P GL (2 , q ) is a cosetof a point stabilizer. In [20], they went one step further to solve Problem I and II for thegroup P GL (3 , q ) acting on the set of points of the projective plane P G (2 , q ).In this paper we study Problem II for the group P SL (2 , q ) acting on P G (1 , q ), where q isan odd prime power. Here we only consider the q odd case since if q is a power of two, we have P SL (2 , q ) = P GL (2 , q ), and both Problem I and II were solved in [19]. It is known, fromthe combined results of [2, 19], that the maximum size of an intersecting family in P SL (2 , q )is q ( q − /
2. (In fact, in a recent paper [21], it is proved that if G ≤ S n is a 2-transitivegroup, then the maximum size of an intersecting family in G is | G | /n . That is, the maximumsize of an intersecting family is the cardinality of a point stabilizer.) However, it is only aconjecture that all intersecting families of maximum size are cosets of point stabilizers when q >
3. (See the second part of Conjecture 1 in [19].) In this paper, we prove that the secondpart of Conjecture 1 in [19] is true for all odd prime powers q > Theorem 1.
Let S be an intersecting family in P SL (2 , q ) of maximum size, where q > isan odd prime power. Then S is a coset of a point stabilizer. Note that when q = 3, we have P SL (2 , q ) ∼ = A , and the action of P SL (2 , q ) on theprojective line P G (1 , q ) is equivalent to the (natural) action of A on { , , , } ; in this case,it was pointed out in [16] that the set S = { (1) , (123) , (234) } (we are using cycle notationfor permutations), is an intersecting family of maximum size in A , but S is not a coset ofany point stablizer. To prove Theorem 1 we apply a general method for solving ProblemII for some 2-transitive groups. This technique was described by Ahmadi and Meagher in[2] and they called it “The Module Method”. This method reduces the characterization ofintersecting families of maximum size to the computation of the C -rank of a matrix whichwe define below. Definition 2.
Let X be a finite set and G a finite group acting on X . An element g ∈ G is said to be a derangement if its action on X is fixed-point-free. The derangement matrix of G acting on X is the (0 , M , whose rows are indexed by the derangements of G , whose columns are indexed by the ordered pairs of distinct elements in X , and for anyderangement g ∈ G and ( a, b ) ∈ X × X with a = b , the ( g, ( a, b ))-entry of M is defined by M ( g, ( a, b )) = (cid:26) , if a g = b, , otherwise.The Module Method states that, under certain conditions, if the rank of the derangementmatrix M of G acting on X is equal to ( | X | − | X | − G . This technique has been appliedto show that the cosets of point stabilizers are the only intersecting families of maximumsize for the symmetric group [13], the alternating group [3], P GL (2 , q ) [19], and many othergroups [2].Thus, in order to prove Theorem 1 by applying the Module Method, it is enough to showthat the rank of the derangement matrix M of P SL (2 , q ) acting on P G (1 , q ) is equal to q ( q − heorem 3. Let M be the derangement matrix of P SL (2 , q ) acting on P G (1 , q ) , where q > is an odd prime power. Then the C -rank of M is q ( q − . Exactly the same statement for
P GL (2 , q ) is proved in [19, Prop. 9], so we must first ex-amine why the proof does not immediately carry over to P SL (2 , q ). In [19] the matrix M ⊤ M represents a certain P GL (2 , q )-module endomorphism of a permutation module. The maincalculation is to show, for each irreducible constituent character of this module, that theimage of M ⊤ M is not annihilated by the corresponding central idempotent. Consequently,the image also contains the character as a constituent, and the rank result follows due to thefact that the module in question is almost multiplicity-free, in the sense that, with one ex-ception, each irreducible constituent character occurs with multiplicity one. If one attemptsto follow the same procedure for P SL (2 , q ) one runs immediately into the problem that the P SL (2 , q )-constituents of the permutation module have high multiplicity. Fortunately, thisobstacle can be sidestepped by observing that although we are working in P SL (2 , q ), oursets and permutation modules admit the action of P GL (2 , q ), and for the larger group thepermutation module has the property of being almost multiplicity-free. A more serious diffi-culty arises when one attempts to show that the central idempotents have nonzero images inthe permutation module. As for P GL (2 , q ), the problem boils down to showing that certainsums of character values are not zero. For P GL (2 , q ), these sums could be estimated byelementary arguments. However, the sums for P SL (2 , q ) appear to be much harder to dealwith, and our proof proceeds by reformulating the sums as character sums over finite fieldsand applying some deep results on hypergeometric functions over finite fields. The finite fieldcharacter sums which appear are Legendre and Soto-Andrade sums (see Section 2.4). Thisis not a surprise; it is well known that these sums appear in connection with the complexrepresentation theory of P GL (2 , q ) [14]. To prove that these character sums are not equalto zero the following facts will be crucial:(1) The Legendre and Soto-Andrade sums (see Definitions 7 and 8) on F q form an or-thogonal basis in the inner product space ℓ ( F q , m ) [14], where m is the measureassigning mass q + 1 to the points ± P GL (2 , q ), Legendre and Soto-Andrade sums, and hypergeo-metric functions over finite fields. In Section 3, we show that the rank of the derangementmatrix M is equal to the dimension of the image of a P GL (2 , q )-module homomorphism.We use this fact to reduce the problem of computing the rank of M to that of showingsome explicit character sums over P GL (2 , q ) are not equal to zero. In Section 4, we findsome formulas to express those character sums over P GL (2 , q ) in terms of Legendre andSoto-Andrade sums. In Section 5, we prove Theorem 3. In Section 6, we conclude with someremarks and open problems. 2. Background
We start by recalling standard facts about the groups
P GL (2 , q ) and P SL (2 , q ) and theircomplex characters, introducing our notation in the process. We shall assume that the reader s familiar with the general terminology and basic results from the representation theory offinite groups over the complex field, as can be found in many textbooks, and we shall use[26] for specific references when necessary.2.1. The groups
P GL (2 , q ) and P SL (2 , q ) . Let F q be the finite field of size q and F q its unique quadratic extension. We denote by F ∗ q and F ∗ q the multiplicative groups of F q and F q , respectively. Let GL (2 , q ) be the group of all invertible 2 × F q and SL (2 , q ) the subgroup of all invertible 2 × Z ( GL (2 , q )) of GL (2 , q ) consists of all non-zero scalar matrices and we define P GL (2 , q ) = GL (2 , q ) /Z ( GL (2 , q )) and P SL (2 , q ) = SL (2 , q ) / ( SL (2 , q ) ∩ Z ( GL (2 , q ))). If q is odd then P SL (2 , q ) is a subgroup of P GL (2 , q ) of index 2, while if q is even then P GL (2 , q ) = P SL (2 , q ).We denote by P G (1 , q ) the set of 1-dimensional subspaces of the space F q of row vectorsof length 2. Thus, P G (1 , q ) is a projective line over F q and its elements are called projectivepoints. An easy computation shows that P G (1 , q ) has cardinality q + 1. From the abovedefinitions, it is clear that the GL (2 , q )-action on F q by right multiplication induces a naturalright action of the groups P GL (2 , q ) and P SL (2 , q ) on P G (1 , q ). The action of the subgroup P SL (2 , q ) is 2-transitive, that is, given any two ordered pairs of distinct points there is agroup element sending the first pair to the second. The action of P GL (2 , q ) is sharply -transitive , that is, given any two ordered triples of distinct points there is a unique groupelement sending the first triple to the second.2.2. The character table of
P GL (2 , q ) . We assume in this section and throughout thispaper that q is an odd prime power. We briefly describe the character table of P GL (2 , q ).We refer the reader to [23] for a complete study of the complex irreducible characters of P GL (2 , q ). We start by describing its conjugacy classes. By abuse of notation we willdenote the elements of P GL (2 , q ) by 2 × F q .First note that, the elements of P GL (2 , q ) can be collected into four sets: The set consistingof the identity element only; the set consisting of the non-scalar matrices with only oneeigenvalue in F q ; the set consisting of matrices with two distinct eigenvalues in F q ; andthe set of matrices with no eigenvalues in F q . Recall that the elements of P GL (2 , q ) areprojective linear transformations so if { x , x } are eigenvalues of some g ∈ P GL (2 , q ) then { ax , ax } are also eigenvalues of g for any a ∈ F ∗ q . Hence, the eigenvalues of elements in P GL (2 , q ) are defined up to multiplication by elements of F ∗ q .The identity of P GL (2 , q ), denoted by I , defines a conjugacy class of size 1. Every non-identity element of P GL (2 , q ) having only one eigenvalue in F ∗ q is conjugate to u = (cid:18) (cid:19) . The conjugacy class of u contains q − F q are conjugate to d x = (cid:18) x
00 1 (cid:19) for some x ∈ F ∗ q \ { } . Moreover, d x and d y are conjugate if and only if x = y or x = y − .The size of the conjugacy class containing d x is q ( q + 1) for x ∈ F ∗ q \ {± } and q ( q + 1) / able 1. Character table of
P GL (2 , q ) I u d x d − v r v i λ λ − δ ( x ) δ ( − δ ( r ) δ ( i ) ψ q − − ψ − q δ ( x ) δ ( − − δ ( r ) − δ ( i ) η β q − − − β ( r ) − β ( r q ) − β ( i ) ν γ q + 1 1 γ ( x ) + γ ( x − ) 2 γ ( −
1) 0 0for x = −
1. Finally, the elements of
P GL (2 , q ) with no eigenvalues in F ∗ q are conjugate to v r = (cid:18) − r q r + r q (cid:19) for some r ∈ F ∗ q \ F ∗ q . The matrices v r have eigenvalues { r, r q } . Hence, v r and v r lie in thesame conjugacy class if and only if r F ∗ q = r F ∗ q or r F ∗ q = r − F ∗ q . The size of the conjugacyclass containing v r is q ( q −
1) if r ∈ F ∗ q \ ( F ∗ q ∪ i F ∗ q ) and q ( q − / r ∈ i F ∗ q , where i is anelement of F ∗ q \ F ∗ q such that i ∈ F ∗ q .The complex irreducible characters of P GL (2 , q ) are described in Table 1. They also comein four families. First the characters λ and λ − correspond to representations of degree 1.Here λ is the principal character and the values of λ − depend on a function δ which isdefined as follows: δ ( x ) = 1 if d x ∈ P SL (2 , q ) and δ ( x ) = − δ ( r ) = 1if v r ∈ P SL (2 , q ) and δ ( r ) = − ψ and ψ − correspond to representations of degree q . The char-acter ψ is the standard character which is an irreducible character of P GL (2 , q ). Thus, forevery g ∈ P GL (2 , q ), the value of ψ ( g ) is equal to the number of projective points fixed by g in P G (1 , q ) minus 1. The values of ψ − depend on the function δ defined above.The third family is known as the cuspidal characters of P GL (2 , q ). They correspond torepresentations of degree q − F q .In fact, the label β in Table 1 runs through all homomorphism β : F ∗ q / F ∗ q → C ∗ of ordergreater than 2 up to inversion. Note that every β corresponds to a unique multiplicativecharacter of F q which is trivial on F ∗ q .Finally, the fourth family of irreducible characters is known as the principal series of P GL (2 , q ). These characters correspond to representations of degree q + 1 and their valuesdepend on multiplicative characters of F q . In fact, the label γ in Table 1 runs through allthe homomorphism γ : F ∗ q → C ∗ of order greater than 2 up to inversion.Throughout this paper we denote by Γ and B a fixed selection of characters γ and β , asdefined above, up to inversion of size ( q − / q − /
2, respectively. Therefore, theprincipal series and cuspidal irreducible characters of
P GL (2 , q ) are given by { ν γ } γ ∈ Γ and { η β } β ∈ B , respectively. .3. Hypergeometric functions over finite fields.
A (generalized) hypergeometric func-tion with parameters a i , b j is defined by n +1 F n (cid:20) a a · · · a n +1 b · · · b n ; x (cid:21) = X k ≥ ( a ) k · · · ( a n +1 ) k ( b ) k · · · ( b n ) k x k k ! , where a = 1 and for k ≥
1, ( a ) k = a ( a + 1) · · · ( a + k −
1) is called the Pochhammer symbol.Hypergeometric functions over finite fields were introduced independently by John Greene[12] and Nicholas Katz [15]. Note that the two definitions differ only in a normalizing factorfor cases related to our discussion.In this section and throughout this paper we denote by ǫ and φ the trivial and quadraticmultiplicative characters of F q , respectively. Also throughout this paper we adopt the con-vention of extending multiplicative characters by declaring them to be zero at 0 ∈ F q . Forany multiplicative character γ , we use γ to denote its complex conjugation. A Gauss sumof γ is defined by g ( γ ) := P x ∈ F q γ ( x ) θ ( x ) where θ is any nontrivial additive character of F q .Let γ , γ , γ be multiplicative characters of F q and x ∈ F q . Greene defines the followingfinite field analogue of a hypergeometric sum(1) F (cid:20) γ γ γ ; x ; q (cid:21) := ǫ ( x ) γ γ ( − q X y ∈ F q γ ( y )( γ γ − )(1 − y ) γ − (1 − xy ) . Since the seminal work of Greene and Katz a lot of work has been done on special functionsover finite fields, in particular generalized hypergeometric functions. In this section, we recallsome definitions and results that we will use later in this paper.Following Greene [12], we introduce other n +1 F n functions inductively as follows. Formultiplicative characters A , A , . . . , A n and B , . . . , B n of F q and x ∈ F q , define n +1 F n (cid:20) A A · · · A n B · · · B n ; x ; q (cid:21) := A n B n ( − q X y ∈ F q n F n − (cid:20) A A · · · A n − B · · · B n − ; xy ; q (cid:21) A n ( y ) A n B n (1 − y ) . See § Lemma 4.
For any non-trivial multiplicative character γ of F q , q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) = X z ∈ F q φ ( z ) F (cid:20) φ φǫ ; z ; q (cid:21) F (cid:20) γ γ − ǫ ; z ; q (cid:21) , where φ ( · ) denotes the quadratic character of F q . roof. The lemma follows from the recursive definition of n +1 F n . First, q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) = φ ( − X x ∈ F ∗ q φ ( x ) φ (1 − x ) F (cid:20) γ γ − φǫ ǫ ; x ; q (cid:21) = 1 q X x ∈ F ∗ q X y ∈ F ∗ q φ ( x ) φ (1 − x ) φ ( y ) φ (1 − y ) F (cid:20) γ γ − ǫ ; xy ; q (cid:21) . Now replacing xy by z , q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) = 1 q X x ∈ F ∗ q X z ∈ F ∗ q φ (1 − x ) φ (1 − z/x ) φ ( z ) F (cid:20) γ γ − ǫ ; z ; q (cid:21) . Letting w = 1 /x and using (1) we get, q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) = X z ∈ F ∗ q q X w ∈ F ∗ q φ (1 − w ) φ (1 − zw ) φ ( z ) F (cid:20) γ γ − ǫ ; z ; q (cid:21) = X z ∈ F ∗ q q X w ∈ F ∗ q φ ( − φ ( w ) φ (1 − w ) φ (1 − zw ) φ ( z ) F (cid:20) γ γ − ǫ ; z ; q (cid:21) = X z ∈ F q φ ( z ) F (cid:20) φ φǫ ; z ; q (cid:21) F (cid:20) γ γ − ǫ ; z ; q (cid:21) . (cid:3) Like their classical counterparts hypergeometric functions over finite fields satisfy manytransformation formulas [11, 12]. In particular, the next one will be useful for our purpose.
Lemma 5. (Greene, [12] ) For x ∈ F q with x = 0 we have, F (cid:20) φ φǫ ; x ; q (cid:21) = φ ( x ) F (cid:20) φ φǫ ; 1 x ; q (cid:21) . Proposition 6.
Let n = 2 , , or , F q be any finite field of size q that is congruent to n , and γ be any order n multiplicative character of F q . Then (cid:12)(cid:12)(cid:12)(cid:12) q · F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) + φ ( − γ ( − q (cid:12)(cid:12)(cid:12)(cid:12) ≤ q / . Proof.
This proposition is a corollary of Theorem 2 of [18]. The background is about charac-ter sums in the perspective of hypergeometric motives [5, 15, 25] and we will only point outhow to obtain our claim. Under the assumption on n , the choice of γ is unique up to complexconjugation and γ ( −
1) is independent of the choice of γ . For each n , let α = { n , n − n , , } and β = { , , , } and ω be any order ( q −
1) multiplicative character of F q . Thus either γ or γ − is ω ( q − /n . The normalized Katz version of hypergeometric sum is defined as (seeDefinition 1.1 of [5])(2) H q ( α , β ; λ ) := 11 − q q − X k =0 Y α ∈ α g ( ω k +( q − α ) g ( ω ( q − α ) Y β ∈ β g ( ω − k − ( q − β ) g ( ω − ( q − β ) ω k (cid:0) ( − m λ (cid:1) . e take λ = 1 here. Then the conversion between Greene and the normalized Katz versionsof hypergeometric finite sums says(3) − q · F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) = H q ( α , β ; 1) , independent of the choice of γ . Then Theorem 2 in [18] implies that there are two imaginaryquadratic algebraic integers A ,q and A ,q (depending on both n and q ) both of complexabsolute values q / such that H q ( α , β ; 1) = φ ( − γ ( − q + A ,q + A ,q . Thus (cid:12)(cid:12)(cid:12)(cid:12) q · F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) + φ ( − γ ( − q (cid:12)(cid:12)(cid:12)(cid:12) = | A ,q + A ,q | ≤ q / . The proof is now complete. (cid:3)
The vector space ℓ ( F q , m ) . Let m : F q → C be m ( x ) = 1 + qD ( x ) + qD − ( x )where D a ( x ) is 1 if x = a and 0 otherwise. We denote by ℓ ( F q , m ) the vector space ofcomplex-valued functions on F q equipped with the Hermitian form h f , f i := X x ∈ F q f ( x ) f ( x ) m ( x ) . Note that the following character sums are elements of ℓ ( F q , m ). Definition 7.
For any multiplicative character γ of F q , the Legendre sum with respect to γ is defined as P γ ( a ) := 1 q X x ∈ F ∗ q γ ( x ) φ ( x − ax + 1) , for all a ∈ F q . Definition 8.
For any multiplicative character β of F q , the Soto-Andrade sum with respectto β is defined as R β ( a ) := 1 q ( q − X r ∈ F ∗ q β ( r ) φ (( r + r q ) − a + 1) r q ) , for all a ∈ F q . The Legendre and Soto-Andrade sums have appeared several times in the literature inconnection with the irreducible representations of
P GL (2 , q ) [14]. In fact, we will encounterthem in Section 4 in our study of some character sums over P GL (2 , q ). In this section, werecall some properties of these sums that will be useful for us in the coming sections.The next lemma shows that the Legendre and Soto-Andrade sums form an orthogonalbasis of ℓ ( F q , m ). Lemma 9. (Kable, [14] ) The set L := (cid:26) P ǫ − q − q , P φ , P γ , R β : γ ∈ Γ , β ∈ B (cid:27) is an orthogonal basis for the space ℓ ( F q , m ) , where Γ and B were defined in the end ofSection 2.2 with | Γ | = q − and | B | = q − . The square norm of the elements of this basis are s follows: (cid:13)(cid:13)(cid:13)(cid:13) P ǫ − q − q (cid:13)(cid:13)(cid:13)(cid:13) = q − q , k P φ k = q − q , k P γ k = q − q , k R β k = q + 1 q . If we normalize the basis given by Lemma 9 then we can easily obtain an orthonormalbasis of ℓ ( F q , m ). We denote the elements of this orthonormal basis by { P ′ ǫ , P ′ φ , P ′ γ , R ′ β : γ ∈ Γ , β ∈ B } .The next lemmas list some elementary properties of the Legendre and Soto-Andrade sumsthat we will need later. Lemma 10 implies that the Legendre sum with respect to the trivialcharacter is easy to evaluate. This is not true for Legendre sums with respect to charactersof higher orders. On the other hand, Lemma 11 shows that the Legendre and Soto-Andradesums are easy to evaluate at ± Lemma 10.
The values of the Legendre sum with respect to ǫ are, P ǫ ( a ) = (cid:26) q − q , if a = ± , − q , if a = ± . Lemma 11.
Let γ and β be characters from the sets Γ and B , respectively. Then P γ (1) = − /q and R β (1) = 1 /q . Moreover, P γ ( −
1) = − γ ( − q , R β ( −
1) = − β ( i ) q where i ∈ F ∗ q \ F q such that i ∈ F ∗ q . Lemma 12.
The values of the Legendre and Soto-Andrade sums are real numbers. Moreover,for every γ ∈ Γ , β ∈ B and a ∈ F q we have P γ − ( a ) = P γ ( a ) and R β − ( a ) = R β ( a ) . The following result establishes a relation between Legendre sums and hypergeometricsums over finite fields. This fact will be crucial later in this paper.
Lemma 13. (Kable, [14] ) If γ is a nontrivial character of F q and a ∈ F q \ {± } then P γ ( a ) = F (cid:20) γ γ − ǫ ; 1 − a q (cid:21) . A P GL (2 , q ) -module homomorphism In this section we show that the rank of the derangement matrix M of P SL (2 , q ) is equalto the dimension of the image of a certain P GL (2 , q )-module homomorphism. Actually, wewill show that N = M ⊤ M is a matrix representation of a P GL (2 , q )-module homomorphism.We will use this fact to compute the rank of M . .1. The matrix N . We identify the points of the projective line
P G (1 , q ) with elementsof the set F q ∪ {∞} , by letting a ∈ F q denote the point spanned by (1 , a ) ∈ F q and denotingby ∞ the point spanned by (0 , P GL (2 , q ) on P G (1 , q ). Let a ∈ F q ∪ {∞} and g ∈ P GL (2 , q ). We use a g to denote the element in P G (1 , q ) obtained by applying g to a . The action of P GL (2 , q ) on P G (1 , q ) is faithful.Hence, we can associate with each element of P GL (2 , q ) a permutation of the q + 1 elementsof P G (1 , q ). Moreover, recall that an element g ∈ P GL (2 , q ) is said to be a derangement ifits associated permutation is fixed-point-free. Definition 14.
Let Ω be the set of ordered pairs of distinct projective points in
P G (1 , q ).The matrix N is a q ( q + 1) by q ( q + 1) matrix whose rows and columns are both indexed bythe elements of Ω; for any ( a, b ) , ( c, d ) ∈ Ω we define N ( a,b ) , ( c,d ) := the number of derangements of P SL (2 , q ) sending a to b and c to d. Note that the above definition of N agrees with our former definition, N = M ⊤ M . Hence,basic linear algebra implies that rank C ( M ) = rank C ( N ). The next lemma gives informationabout the entries of N . Lemma 15.
Let a, b, c, d ∈ F q ∪ {∞} . Then,(1) N ( a,b ) , ( a,b ) = ( q − , ∀ ( a, b ) ∈ Ω .(2) N ( a,b ) , ( c,d ) = 0 , if a = c, b = d or a = c, b = d .(3) N ( a,b ) , ( b,a ) = (cid:26) , if q ≡ mod , ( q − / , if q ≡ mod , ∀ ( a, b ) ∈ Ω .(4) (a) N (0 , ∞ ) , (1 , = (cid:26) ( q − / , if q ≡ mod , ( q − / , if q ≡ mod . (b) N (0 , ∞ ) , (1 ,d ) = q − − φ (1 − d )2 − X x ∈ F ∗ q φ (( x + x − ) − d ) , ∀ d = 0 , , ∞ . Moreover, the value of N ( a,b ) , ( c,d ) for any ( a, b ) , ( c, d ) ∈ Ω is given by one of the above expres-sions.Proof. Let g be an arbitrary element in P GL (2 , q ). Note that for every h ∈ P SL (2 , q )sending a to b and c to d , the element g − hg ∈ P SL (2 , q ) sends a g to b g and c g to d g . Hencethe entries of N satisfy the following property(4) N ( a,b ) , ( c,d ) = N ( a g ,b g ) , ( c g ,d g ) , because P SL (2 , q ) is a normal subgroup of P GL (2 , q ) and the set of derangements in P SL (2 , q ) is closed under conjugation. To prove Lemma 15 we proceed case by case. • Case 1 .Recall that N ( a,b ) , ( a,b ) is the number of derangements in P SL (2 , q ) sending a to b .From Equation (4) and the 2-transitivity of P GL (2 , q ) we conclude that N ( a,b ) , ( a,b ) = N ( c,d ) , ( c,d ) for any ( a, b ) , ( c, d ) ∈ Ω. The total number of derangements in
P SL (2 , q )is q ( q − / q ( q − X b ∈ P G (1 ,q ) b = a N ( a,b ) , ( a,b ) , for any fixed a ∈ P G (1 , q ) , hich implies that N ( a,b ) , ( a,b ) = ( q − / a, b ) ∈ Ω. • Case 2 .Every element of
P SL (2 , q ) is related to a permutation of projective points in P G (1 , q ). This implies N ( a,b )( a,d ) = 0 and N ( a,b )( c,b ) = 0 whenever b = d and a = c . • Case 3 .Using the 2-transitivity of
P GL (2 , q ) and Equation (4) we can assume withoutloss of generality that a = 0 and b = ∞ . The elements g λ ∈ P SL (2 , q ) sending 0 to ∞ and ∞ to 0 are of the form g λ := (cid:18) λ − λ − (cid:19) , λ ∈ F ∗ q . This representation of elements in
P SL (2 , q ) is redundant because g λ and g − λ represent the same element of P SL (2 , q ). Let ξ be an element in F ∗ q such that h ξ i = F ∗ q . Hence, the set { g λ : λ = ξ i , i = 1 , . . . , ( q − / } corresponds preciselyto the ( q − / P SL (2 , q ) sending 0 to ∞ and ∞ to 0.Recall that g λ is a derangement if and only if its eigenvalues are not in F q . Thus, g λ is a derangement if and only if its characteristic polynomial, p λ ( t ) := det (cid:12)(cid:12)(cid:12)(cid:12) − t λ − λ − − t (cid:12)(cid:12)(cid:12)(cid:12) = t + 1 , is irreducible over F q .If q ≡ − F q . Thus, p λ ( t ) is reducible for every λ ∈ F ∗ q . Hence N ( a,b ) , ( b,a ) = N (0 , ∞ ) , ( ∞ , = 0 in this case. On the other hand, if q ≡ − F q . This implies that p λ ( t ) is irreducible for every λ ∈ F ∗ q . Therefore, N ( a,b ) , ( b,a ) = N (0 , ∞ ) , ( ∞ , = ( q − / • Case 4 .Every element of
P SL (2 , q ) sending 0 to ∞ and 1 to d is of the form g λ := (cid:18) − λλ − λ − d + λ (cid:19) , λ ∈ F ∗ q . Again note that g λ and g − λ represent the same element of P SL (2 , q ). The matrix g λ is a derangement if and only if its characteristic polynomial, p λ ( t ) := det (cid:12)(cid:12)(cid:12)(cid:12) − t − λλ − λ − d + λ − t (cid:12)(cid:12)(cid:12)(cid:12) = t − ( λ − d + λ ) t + 1 , is irreducible over F q . To compute N (0 , ∞ )(1 ,d ) it is enough to count the number ofvalues of λ such that p λ ( t ) is reducible.If p λ ( t ) is reducible then there exist x and y in F ∗ q such that p λ ( t ) = t − ( λ − d + λ ) t + 1 = ( t − x )( t − y ) = t − ( x + y ) t + xy. Hence, xy = 1 and x + y = λ − d + λ . Assume without loss of generality that y = x − .If there exist values of λ such that g λ has eigenvalues { x, x − } , then they have tosatisfy the following quadratic equation(5) λ − ( x + x − ) λ + d = 0 . Case 4 (a) :If we assume d = 0 then λ = 0 is a solution of Equation (5), however, thatsolution is not admissible by the definition of g λ . Hence, we just consider thesolution λ = x + x − for every x ∈ F ∗ q . Moreover, note that x and x − generatethe same value of λ . In fact, we can relate to each set { x, x − } a unique valueof λ .Let q ≡ k ∈ F ∗ q be an element of order 4. Note that the set { k, k − } does not generate any admissible value of λ . Thus, the number of valuesof λ such that p λ ( t ) is reducible is ( q − /
2. Therefore, N (0 , ∞ ) , (1 , = 12 (cid:18) q − − q − (cid:19) = q − . On the other hand, if q ≡ F ∗ q does not have an element oforder 4. This implies that every set { x, x − } ⊂ F ∗ q generates an admissible valueof λ . Thus, the number of values for λ such that p λ ( t ) is reducible is ( q + 1) / N (0 , ∞ ) , (1 , = ( q − / – Case 4 (b) :The number of solutions of Equation (5) in F q is given by 1+ φ (( x + x − ) − d ).In this case, x and x − leads to the same value of λ . Thus, the number of valuesof λ ∈ F ∗ q such that p λ ( t ) is reducible is2(1 + φ (1 − d )) + 12 X x ∈ F ∗ q x =1 , − (1 + φ (( x + x − ) − d )) . Therefore, for d = 0 , , ∞ , N (0 , ∞ ) , (1 ,d ) = 12 ( q − − φ (1 − d )) + 12 X x ∈ F ∗ q x =1 , − (1 + φ (( x + x − ) − d )) = q − − φ (1 − d )2 − X x ∈ F ∗ q x =1 , − φ (( x + x − ) − d )which gives the desired formula for N (0 , ∞ ) , (1 ,d ) . (cid:3) Corollary 16.
Let d ∈ F q , d = 0 , . The number of derangements of P SL (2 , q ) sending to ∞ and to d can be expressed in terms of the Legendre sum with respect to φ . Specifically, (6) N (0 , ∞ ) , (1 ,d ) = q − − φ (1 − d )2 − q P φ (2 d − . roof. To prove this corollary, we compute X x ∈ F ∗ q φ (( x + x − ) − d ) = X x ∈ F ∗ q φ ( x ) φ (( x + x − ) − d )= X x ∈ F ∗ q φ ( x − d − x + 1) . Next we replace x by y . If y ∈ F ∗ q is not a square, then 1 + φ ( y ) = 0; on the other hand, if y ∈ F ∗ q is a square, then x = y has 1 + φ ( y ) = 2 solutions. It follows that X x ∈ F ∗ q φ (( x + x − ) − d ) = X y ∈ F ∗ q (1 + φ ( y )) φ ( y − d − y + 1)= X y ∈ F ∗ q φ ( y − d − y + 1) + X y ∈ F ∗ q φ ( y ) φ ( y − d − y + 1)= − X y ∈ F q φ ( y − d − y + 1) + qP φ (2 d − . Applying Theorem 5.48 from [17] it follows that, X y ∈ F q φ ( y − d − y + 1) = − . Thus, the above computations imply that(7) X x ∈ F ∗ q φ (( x + x − ) − d ) = − qP φ (2 d − . Now, Corollary 16 follows from part 4(b) of Lemma 15 and Equation (7). (cid:3)
A permutation
P GL (2 , q ) -module. In this section we define a
P GL (2 , q )-module V and a P GL (2 , q )-module homomorphism T N from V to V . We use the subscript N toemphasize that N is the matrix associated with T N with respect to a certain basis of V .Recall that we denote by Ω the set of ordered pairs of distinct projective points in P G (1 , q ).Let V be the C -vector space spanned by the vectors { e ω } ω ∈ Ω . The dimension of V is q ( q + 1).We define a right action of P GL (2 , q ) on the basis { e ω } of V . Specifically, if ω = ( a, b )then e ω · g = e ω g = e ( a g ,b g ) for any g ∈ P GL (2 , q ). Thus, V is a right permutation P GL (2 , q )-module. The next lemmashows that V has a very simple decomposition into irreducible modules; apart from V λ − and V ψ each irreducible module of P GL (2 , q ) appears exactly once.Let ( χ, ψ ) denote the inner product of the characters χ and ψ of P GL (2 , q ) (see [26,Section 2.3]). Lemma 17.
Let V χ denote an irreducible module of P GL (2 , q ) with character χ . Then thedecomposition of V into irreducible constituents is given by, V ∼ = V λ ⊕ V ψ ⊕ V ψ − ⊕ M β ∈ B V η β ⊕ M γ ∈ Γ V ν γ . roof. Let π be the character afforded by the P GL (2 , q )-module V . By definition we have π ( g ) := |{ ω ∈ Ω : ω g = ω }| hence the character π has an easy description given by the following table1 u d x v r π q ( q + 1) 0 2 0 .Now let V χ be an irreducible representation of P GL (2 , q ) and χ its irreducible character.It is known ([26, Chapter 2, Theorem 4]) that the multiplicity of V χ in V is equal to thecharacter inner product ( π, χ ). Thus, the lemma follows by direct calculation using thecharacter table of P GL (2 , q ). (cid:3) For a, b ∈ P G (1 , q ) with a = b , consider the following vectors in V, l a,b := X p ∈ P G (1 ,q ) p = a,b ( e ( a,p ) − e ( b,p ) ) + e ( a,b ) − e ( b,a ) , (8) r a,b := X p ∈ P G (1 ,q ) p = a,b ( e ( p,a ) − e ( p,b ) ) + e ( b,a ) − e ( a,b ) . (9)We use these vectors to define the following vector subspaces of V , V := span C { l a,b : a, b ∈ P G (1 , q ) , a = b } and V := span C { r a,b : a, b ∈ P G (1 , q ) , a = b } . In fact, the next lemma shows that V and V are P GL (2 , q )-submodules of V . Lemma 18.
The vector subspaces V and V satisfy the following properties:(1) dim C ( V ) = dim C ( V ) = q ,(2) V ∩ V = { } ,(3) V and V are P GL (2 , q ) -submodules of V ,(4) V ∼ = V as P GL (2 , q ) -modules.Proof. Note that the vectors defined in Equations (8) and (9) satisfy the following relations, l a,b − l a,c = l c,b and r a,b − r a,c = r c,b for all a, b, c ∈ P G (1 , q ) with a = b = c . Hence, fixing a ∈ P G (1 , q ) we see that { l a,b : b ∈ P G (1 , q ) , b = a } and { r a,b : b ∈ P G (1 , q ) , b = a } are basis for V and V , respectively.To prove the conclusion in part (2) we proceed by contradiction. Assume there exists v ∈ V ∩ V with v = 0. Hence we can write(10) v = X p ∈ P G (1 ,q ) p = a α p l a,p = X p ∈ P G (1 ,q ) p = a β p r a,p where not all α p and β p are equal to zero.For a fixed b ∈ P G (1 , q ), the vector l a,b is the only one in the set { l a,p } p ∈ P G (1 ,q ) ,p = a thatcontains e ( b,a ) . On the other hand, every vector of the form r a,p with p = a contains e ( b,a ) .Therefore, using Equation (10) we get α b = X p ∈ P G (1 ,q ) p = a β p , hich implies that the values of the coefficients α p in Equation (10) are all the same. Analo-gously, we can show that the values β p in Equation (10) are the same. Thus, we can rewriteEquation (10) as follows, α X p ∈ P G (1 ,q ) p = a l a,p = β X p ∈ P G (1 ,q ) p = a r a,p where α = P p = a β p and β = P p = a α p . This implies that α = qβ = q α , a contradiction,because q is not equal to one.To prove part (3) it is enough to note that l a,b · g = l a g ,b g and r a,b · g = r a g ,b g for all a, b ∈ P G (1 , q ) with a = b . For part (4) consider the function θ from V to V defined by θ ( l a,b ) = r a,b for all a, b ∈ P G (1 , q ) with a = b ; we extend the definition of θ to all elementsof V linearly. Now, from the definition of θ we see that clearly θ ( l ( a,b ) · g ) = θ ( l ( a,b ) ) · g for all g ∈ P GL (2 , q ) and ( a, b ) ∈ Ω. Therefore, θ is a P GL (2 , q )-module isomorphism. Thiscompletes the proof of part (4). (cid:3) Lemma 19.
The submodules V and V are isomorphic to V ψ .Proof. This result follows directly from Lemmas 17 and 18. If we consider the decompositionof V into irreducible constituents, we note that each irreducible representation appears onlyonce, except for V ψ . Therefore, because V is isomorphic to V , we must have V ψ ∼ = V ∼ = V . (cid:3) We now define a linear transformation T N from V to V . We first define T N on the basis { e ω } ω ∈ Ω of V by T N ( e ( a,b ) ) := X ω ∈ Ω N ω, ( a,b ) e ω for any ( a, b ) ∈ Ω, and then extend the definition of T N to all elements of V linearly. Itfollows from the definition of T N that N is the matrix associated with T N with respect tothe basis { e ω } ω ∈ Ω of V . Therefore, the dimension of the image of T N is equal to the rank ofthe derangement matrix M of P SL (2 , q ) acting on P G (1 , q ). Lemma 20.
The linear transformation T N defined above is a P GL (2 , q ) -module homomor-phism from V to V .Proof. To prove the lemma we have to show that the linear transformation T N respects theaction of P GL (2 , q ) on V ; that is, for each g ∈ P GL (2 , q ) and each ( a, b ) ∈ Ω,(11) T N ( e ( a,b ) · g ) = T N ( e ( a,b ) ) · g. First, consider the left hand side of Equation (11). From the definition of T N it followsthat T N ( e ( a,b ) · g ) = T N ( e ( a g ,b g ) ) = X ω ∈ Ω N ω, ( a g ,b g ) e ω . Now, note that the right hand side of Equation (11) can be written as T N ( e ( a,b ) ) · g = X ω ∈ Ω N ω, ( a,b ) e ω g = X ω g − ∈ Ω N ω g − , ( a,b ) e ω . urthermore, recall that N ( a,b ) , ( c,d ) = N ( a g ,b g ) , ( c g ,d g ) for all g ∈ P GL (2 , q ). Therefore, X ω g − ∈ Ω N ω g − , ( a,b ) e ω = X ω g − ∈ Ω N ω, ( a g ,b g ) e ω = X ω ∈ Ω N ω, ( a g ,b g ) e ω which implies that Equation (11) holds. This completes the proof of the lemma. (cid:3) The image of T N . Recall that the rank of the derangement matrix M of P SL (2 , q )acting on P G (1 , q ) is equal to the dimension of the image of T N . Since T N is a P GL (2 , q )-module homomorphism (Lemma 20) we can use some tools from representation theory tocompute the dimension of the image of T N . We start by observing that the submodules V and V are in the kernel of T N . Lemma 21.
The subspaces V and V lie in the kernel of T N .Proof. First, recall that the derangement matrix M is a q ( q − / q + 1) q matrixwhose rows are indexed by the derangements of P SL (2 , q ) and whose columns are indexedby elements of Ω. For any derangement g ∈ P SL (2 , q ) and ( a, b ) ∈ Ω we have M ( g, ( a, b )):= (cid:26) , if a g = b, , otherwise.Furthermore, also by definition, we have N = M ⊤ M . Thus, the lemma follows from thefollowing observation M l a,b = 0 and
M r a,b = 0 for all a, b ∈ P G (1 , q ) , with a = b, and the fact that for a fixed a ∈ P G (1 , q ) the sets { l a,b : b ∈ P G (1 , q ) , b = a } and { r a,b : b ∈ P G (1 , q ) , b = a } are basis of V and V , respectively. (cid:3) From Lemma 19 and 21, we conclude that the restriction of T N to 2 V ψ is the zero map.It follows that the dimension of the image of T N is at most q ( q − T N onto the other irreducible constituents of V . To do that we apply Schur’slemma.Let χ be the irreducible character corresponding to an irreducible representation of P GL (2 , q )appearing as a constituent of V . Schur’s lemma implies that, T N ( V χ ) ∼ = V χ or T N ( V χ ) = { } . Thus, either the dimension of the restriction of T N to V χ is zero or is equal to the dimensionof V χ . Hence, to study the image of V χ under T N for any χ ∈ { λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } weproceed in the following way:(1) Consider the vector e (0 , ∞ ) ∈ V .(2) Project e (0 , ∞ ) onto V χ using the following scalar multiple of a central primitive idem-potent E χ := X g ∈ P GL (2 ,q ) χ ( g − ) g. Therefore, the projection of e (0 , ∞ ) onto V χ is equal to E χ ( e (0 , ∞ ) ) = X g ∈ P GL (2 ,q ) χ ( g − ) e (0 g , ∞ g ) = X ( a,b ) ∈ Ω " X g = a, ∞ g = b χ ( g − ) e ( a,b ) . here g in the inner sum runs over all elements in P GL (2 , q ) sending 0 to a and ∞ to b .(3) To prove that T N ( V χ ) ∼ = V χ it is enough to show that the (0 , ∞ ) coordinate of T N ( E χ ( e (0 , ∞ ) )) is not equal to zero. This is equivalent to showing that the followingcharacter sum is not equal to zero:(12) T N,χ := T N ( E χ ( e (0 , ∞ ) )) (0 , ∞ ) = X ( a,b ) ∈ Ω " X g = a, ∞ g = b χ ( g − ) N (0 , ∞ ) , ( a,b ) , where g in the inner sum runs over all elements in P GL (2 , q ) sending 0 to a and ∞ to b .Therefore, we get the following lower bound on the rank of the derangement matrix M ,(13) X χ dim( V χ ) ≤ rank( M ) , where χ in the sum on the left hand side of (13) runs through { λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } such that T N,χ = 0. In particular, if T N,χ is not zero for all χ ∈ { λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } then the rank of the derangement matrix M is equal to q ( q − T N,χ with χ ∈ { λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } are not equal to zero. This will be our objective in the nexttwo sections.4. The character sums X g = ∞ , ∞ g =0 χ ( g − ) and X g = ∞ , g = d χ ( g − )The sums T N,χ are character sums over
P GL (2 , q ). In general, it is not easy to gettight bounds on the values of characters sums over non-abelian groups. Fortunately, theclose relationship between the irreducible characters of P GL (2 , q ) and the multiplicativecharacters of F q and F q allows us to conclude in Section 5 that the expressions T N,χ arenot equal to zero. In this section, we show that we can express the sums T N,χ in terms ofcharacters sums over finite fields for every χ ∈ { λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } .First, we consider T N,χ when χ = λ . In this case, we know that λ ( g ) = 1 for any g ∈ P GL (2 , q ). Moreover, there are precisely q − P GL (2 , q ) sending 0 to a and ∞ to b for any a, b ∈ P G (1 , q ). Therefore, we can compute (12) explicitly for χ = λ : T N,λ = ( q − X ( a,b ) ∈ Ω N (0 , ∞ )( a,b ) = ( q − q + 1) ( q − , where we have used Lemma 15 to obtain the last equality. Thus, from the analysis given inSection 3.3 we conclude that T N ( V λ ) ∼ = V λ .The other irreducible characters of P GL (2 , q ) are not so easy to handle. The next lemmagives an expression for T N,χ with χ ∈ { ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } which will be helpful to writeEquation (12) in terms of character sums over finite fields. Lemma 22.
Let χ be any irreducible character of P GL (2 , q ) from the set { ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } .Let h be the unique element of P GL (2 , q ) sending to , to ∞ , and ∞ to . If q ≡ mod hen T N,χ = ( q − − q − X g = ∞ , ∞ g =0 χ ( g − ) + ( q − X b ∈ F ∗ q b =1 X g = ∞ , g = b h χ ( g − ) N (0 , ∞ ) , (1 ,b ) , and if q ≡ mod then T N,χ = ( q − X g = ∞ , ∞ g =0 χ ( g − ) + ( q − X b ∈ F ∗ q b =1 X g = ∞ , g = b h χ ( g − ) N (0 , ∞ ) , (1 ,b ) . Proof.
We start by presenting some results on character sums over
P GL (2 , q ) that we willneed.We denote by P GL (2 , q ) , ∞ the subgroup of P GL (2 , q ) fixing 0 and ∞ . Analogously, P GL (2 , q ) denotes the subgroup of P GL (2 , q ) fixing 0. Applying the Frobenius ReciprocityTheorem [26, Chapter 7, Theorem 13], we have:(Res( χ ) , P GL (2 ,q ) , ∞ = ( χ, π ) P GL (2 ,q ) and (Res( χ ) , P GL (2 ,q ) = ( χ, λ + ψ ) P GL (2 ,q ) where π is the permutation character defined in the proof of Lemma 17 and 1 is the trivialcharacter of the groups P GL (2 , q ) , ∞ and P GL (2 , q ) , respectively. Using these equalitiesand the decomposition of π in terms of irreducible characters (which was given in Lemma17), we evaluate the following character sums: X g =0 , ∞ g = ∞ χ ( g − ) = ( q − χ ) , P GL (2 ,q ) , ∞ = ( q − χ, π ) P GL (2 ,q ) = q − , and X g =0 χ ( g − ) = q ( q − χ ) , P GL (2 ,q ) = q ( q − χ, λ + ψ ) P GL (2 ,q ) = 0 . Note that χ ( kgk − ) = χ ( g ) for any k ∈ P GL (2 , q ) since χ is a character, hence a classfunction. This fact implies many relations between character sums over P GL (2 , q ). Inparticular,(14) X a g = b χ ( g − ) = X ( a k ) g =( b k ) g χ ( g − ) , and(15) X a g = b,c g = d χ ( g − ) = X ( a k ) g = b k , ( c k ) g = d k χ ( g − ) . We claim that P g = ∞ χ ( g − ) = 0. To prove this claim, recall that χ is a non-trivial characterof P GL (2 , q ). Therefore,0 = X g ∈ P GL (2 ,q ) χ ( g − ) = X g =0 χ ( g − ) + X a ∈ P G (1 ,q ) a =0 X g = a χ ( g − ) . ince P g =0 χ ( g − ) = 0, we conclude that0 = X a ∈ P G (1 ,q ) a =0 X g = a χ ( g − ) = q X g = ∞ χ ( g − ) , where Equation (14) is used to obtain the last equality.Moreover, it follows from the above equations and the 2-transitivity of the action of P GL (2 , q ) on P G (1 , q ) that X ∞ g = ∞ χ ( g − ) = 0 and X ∞ g =0 χ ( g − ) = 0 . Now, we are ready to prove Lemma 22. From Equation (12) and Lemma 15 we get, T N,χ = ( q − X g =0 , ∞ g = ∞ χ ( g − ) + " X g = ∞ , ∞ g =0 χ ( g − ) N (0 , ∞ ) , ( ∞ , + X b ∈ F ∗ q " X g = ∞ , ∞ g = b χ ( g − ) N (0 , ∞ ) , ( ∞ ,b ) + X a ∈ F ∗ q " X g = a, ∞ g =0 χ ( g − ) N (0 , ∞ ) , ( a, + X a,b ∈ F ∗ q a = b " X g = a, ∞ g = b χ ( g − ) N (0 , ∞ ) , ( a,b ) . First, assume that q ≡ N (0 , ∞ ) , ( ∞ ,b ) = N (0 , ∞ ) , ( a, = ( q − / a, b ∈ F ∗ q , and N (0 , ∞ ) , ( ∞ , = 0 . Hence, using the above analysis we can write, X b ∈ F ∗ q " X g = ∞ , ∞ g = b χ ( g − ) N (0 , ∞ ) , ( ∞ ,b ) = q − X b ∈ F ∗ q " X g = ∞ , ∞ g = b χ ( g − ) = q − " X g = ∞ χ ( g − ) − X g = ∞ , ∞ g =0 χ ( g − ) = − ( q − X g = ∞ , ∞ g =0 χ ( g − ) , and using the same ideas we get X a ∈ F ∗ q " X g = a, ∞ g =0 χ ( g − ) N (0 , ∞ ) , ( a, = − ( q − X g = ∞ , ∞ g =0 χ ( g − ) . Let a, b ∈ F ∗ q with a = b . Using the 3-transitivity of the action of P GL (2 , q ) on P G (1 , q )and (4) we conclude that N (0 , ∞ ) , ( a,b ) = N (0 , ∞ )(1 ,b k ) where k ∈ P GL (2 , q ) is the unique element ending 0 to 0, ∞ to ∞ and a to 1. Moreover, applying Equation (15) we obtain X g = a, ∞ g = b χ ( g − ) = X g =1 , ∞ g = b k χ ( g − ) . Putting all these facts together we conclude that X a,b ∈ F ∗ q a = b " X g = a, ∞ g = b χ ( g − ) N (0 , ∞ ) , ( a,b ) = ( q − X b ∈ F ∗ q b =1 " X g =1 , ∞ g = b χ ( g − ) N (0 , ∞ ) , (1 ,b ) = ( q − X b ∈ F ∗ q b =1 X g = ∞ , g = b h χ ( g − ) N (0 , ∞ ) , (1 ,b ) . Thus, Lemma 22 is proved for the case where q ≡ q ≡ (cid:3) It follows from Lemma 22 that we can write T N,χ in terms of the character sums X g = ∞ , ∞ g =0 χ ( g − ) and X g = ∞ , g = d χ ( g − ) . The next four lemmas show that these character sums can be written in terms of charactersums over finite fields for all χ ∈ { ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } . Lemma 23.
Let i be an element of F ∗ q \ F ∗ q such that i ∈ F ∗ q . Then, X g = ∞ , ∞ g =0 ψ − ( g − ) = φ ( − q − , X g = ∞ , ∞ g =0 ν γ ( g − ) = γ ( − q − for all γ ∈ Γ , X g = ∞ , ∞ g =0 η β ( g − ) = − β ( i )( q − for all β ∈ B. Proof.
The elements in
P GL (2 , q ) sending 0 to ∞ and ∞ to 0 are of the form, g λ := (cid:18) λ (cid:19) with λ ∈ F ∗ q . Note that the characteristic polynomial of g λ is p λ ( t ) := t − λ .To evaluate the character sums in this lemma we need to know to which conjugacy classesthe elements g λ belong.First, recall that the eigenvalues of g λ are defined up to multiplication by an element of F ∗ q . Now, if λ is a square in F ∗ q then p λ ( t ) is reducible and g λ has eigenvalues ±√ λ ∈ F ∗ q .This implies that g λ lies in the conjugacy class d − whenever λ is a square. On the otherhand, if λ is not a square the roots of p λ ( t ) lie on F ∗ q and they correspond to elements oforder 2 in F ∗ q / F ∗ q . Therefore, whenever λ is not a square we see that g λ lies on the conjugacyclass v i . ince there are equal number of squares and nonsquares in F ∗ q , the lemma follows from thecharacter table of P GL (2 , q ). (cid:3) Lemma 24.
For every γ ∈ Γ and d ∈ F ∗ q \ { } we have X g = ∞ , g = d ν γ ( g − ) = qP γ (2 d − . Proof.
The elements in
P GL (2 , q ) sending 0 to ∞ and 1 to d are of the form, g λ := (cid:18) αλα α ( d − λ ) (cid:19) with λ, α ∈ F ∗ q . To evaluate the sum in this lemma we need to know to which conjugacy classes theseelements belongs. However, we need to do this just for those elements which are not de-rangements because ν γ ( g ) = 0 if g is a derangement.Note that different values of α correspond to the same element g λ in P GL (2 , q ). Indeed,as was remarked earlier the eigenvalues of g λ are defined up to scalar multiplication.The characteristic polynomial of g λ is p λ ( t ) := t − α ( d − λ ) t − α λ and its eigenvaluesare, α ( d − λ ) ± p ( d − λ ) + 4 λ )2 ! . Thus, if p ( d − λ ) + 4 λ ∈ F ∗ q then there exists α ∈ F ∗ q such that the eigenvalues of g λ are { , x } for some x ∈ F ∗ q . This implies that g λ is contained in the same conjugacy classas d x (see Section 2.2). Here, we assume that d x with x = 1 corresponds to the element u ∈ P GL (2 , q ) defined in Section 2.2.For a fixed d ∈ F ∗ q \ { } and x ∈ F ∗ q we want to know for how many λ ∈ F ∗ q there existssome α such that g λ has eigenvalues { , x } . From the above analysis it is clear that d, x, α and λ must satisfy the equation below: p λ ( t ) = t − α ( d − λ ) t − α λ = ( t − x )( t −
1) = t − ( x + 1) t + x. This implies that α satisfies the following quadratic equation, dα − ( x + 1) α + x = 0 . Therefore, given x ∈ F ∗ q and d ∈ F ∗ q \ { } , the number of values of λ ∈ F ∗ q such that g λ isconjugate to d x is equal to1 + φ (( x + 1) − xd ) if x = − φ (( x + 1) − xd )) / x = − . Now using the above remarks and the character table of
P GL (2 , q ) we get X g = ∞ , g = d ν γ ( g ) = (1 + φ (1 − d )) γ (1) + (cid:18) φ ( d )2 (cid:19) (2 γ ( − X x =1 , − x ∈ F ∗ q (1 + φ (( x + 1) − xd ))( γ ( x ) + γ ( x − ))where the first two terms in the right hand side of Equation (16) corresponds to x = 1 and x = −
1. Furthermore, note that we have included a factor in front of the last expressionin Equation (16). This occurs because every element g λ having eigenvalues { , x } also has igenvalues { , x − } . Hence, given d ∈ F ∗ q \ { } , the elements x and x − are related to thesame values of λ . Simplifying the right hand side of Equation (16), X g = ∞ , g = d ν γ ( g ) = X x ∈ F ∗ q γ ( x ) φ ( x − d − x + 1)= qP γ (2 d − . Finally, applying basic properties of characters and Lemma 12 we obtain X g = ∞ , g = d ν γ ( g − ) = X g = ∞ , g = d ν γ ( g ) = qP γ − (2 d −
1) = qP γ (2 d − . The proof is now complete. (cid:3)
Lemma 25.
For every β ∈ B and d ∈ F ∗ q \ { } we have, X g = ∞ , g = d η β ( g − ) = − qR β (2 d − . Proof.
Recall that all the elements in
P GL (2 , q ) sending 0 to ∞ and 1 to d take the form, g λ := (cid:18) αλα α ( d − λ ) (cid:19) with λ, α ∈ F ∗ q . To evaluate the sum in this lemma we have to know to which conjugacy classes theseelements belong. However, since η β ( g ) = 0 if g has two fixed points, we will pay attentionto derangements and the elements fixing one point only (see Section 2.2).We know that if r ∈ F ∗ q \ F ∗ q is an eigenvalue of g λ then g λ is a derangement with eigenvalues { r, r q } contained in the same conjugacy class as v r . On the other hand, if r ∈ F ∗ q is the onlyeigenvalue of g λ then this implies that g λ has exactly one fixed point and it is conjugated to u . In fact, when r ∈ F ∗ q every element of the form v r is conjugated to u .Fix r ∈ F ∗ q . We want to know for how many values of λ ∈ F ∗ q there exists α such that g λ has eigenvalues { r, r q } . From the characteristic polynomial of g λ the following equation isobtained t − α ( d − λ ) t − α λ = t − ( r + r q ) t + r q +1 , which implies that α ∈ F ∗ q must satisfy the quadratic equation below(17) dα − ( r + r q ) α + r q +1 = 0 . Distinct solutions of Equation (17) generate distinct values of λ unless r ∈ i F q where i isan element of F ∗ q \ F ∗ q such that i ∈ F ∗ q . Hence, given r ∈ F ∗ q and d ∈ F ∗ q \ { } , the numberof λ ∈ F ∗ q such that g λ is conjugated to v r is equal to:1 + φ (( r + r q ) − dr q +1 ) if r ∈ F ∗ q \ i F ∗ q and (1 + φ (( r + r q ) − dr q +1 )) / r ∈ i F ∗ q . Moreover, note that every element g λ having eigenvalues { r, r q } also has eigenvalues { ar, ( ar ) q } for any a ∈ F ∗ q . Thus, r and ar are related to the same values of λ for every ∈ F ∗ q . Therefore, X g = ∞ , g = d η β ( g − ) = 1 q − X r ∈ F ∗ q (1 + φ (( r + r q ) − dr q +1 ))( − β (1))+ 1 q − X r ∈ i F ∗ q (cid:18) φ (( r + r q ) − dr q +1 )2 (cid:19) ( − β ( i ))+ 12( q − X r ∈ F ∗ q \{ F ∗ q ,i F ∗ q } (1 + φ (( r + r q ) − dr q +1 ))( − β ( r ) − β ( r q ))= 12( q − X r ∈ F ∗ q φ (( r + r q ) − dr q +1 )( − β ( r ))= − q − X r ∈ F ∗ q φ (( r + r q ) − dr q +1 ) β ( r )Now, the lemma follows from Definition 8 and Lemma 12. (cid:3) Lemma 26.
For every d ∈ F ∗ q \ { } we have, X g = ∞ , g = d ψ − ( g ) = qP φ (2 d − . Proof.
From the character table of
P GL (2 , q ) it follows that(18) ψ − ( g ) = , if g ∈ u, , if g ∈ d x and d x ⊂ P SL (2 , q ) , − , if g ∈ d x and d x ⊂ P GL (2 , q ) \ P SL (2 , q ) , − , if g ∈ v r and v r ⊂ P SL (2 , q ) , , if g ∈ v r and v r ⊂ P GL (2 , q ) \ P SL (2 , q ) . Thus, to evaluate the sum P g ψ − ( g ) we need to know: how many elements sending 0 to ∞ and 1 to d belong to each of the five categories considered in (18). In fact, these countingproblems follow from the proof of Case (4) of Lemma 15.For the sake of clarity, we recall some simple facts. There are q − P GL (2 , q )sending 0 to ∞ and 1 to d , and half of them are in P SL (2 , q ). It was proved by Meagher andSpiga [19] that if 1 − d is a square in F ∗ q then ( q − / − d is not a square then ( q + 1) / − d is a square. We can divide the ( q − / P SL (2 , q )sending 0 to ∞ and 1 to d into three categories: • • X x ∈ F ∗ q ,x =1 , − (1 + φ (( x + x − ) − d )) fix exactly two points. • q − − X x ∈ F ∗ q φ (( x + x − ) − d ) are derangements. similar analysis can be carried out when 1 − d is not a square. Specifically, from the( q − / P SL (2 , q ) sending 0 to ∞ and 1 to d , • There are no elements fixing exactly one point. • X x ∈ F ∗ q ,x =1 , − (1 + φ (( x + x − ) − d )) fix two points. • q − − X x ∈ F ∗ q φ (( x + x − ) − d ) are derangements.Putting all the above remarks together and assuming that 1 − d is a square we obtain, X g = ∞ , g = d ψ − ( g ) = 14 X x ∈ F ∗ q ,x =1 , − (1 + φ (( x + x − ) − d )) − q − − − X x ∈ F ∗ q ,x =1 , − (1 + φ (( x + x − ) − d )) − q − − X x ∈ F ∗ q φ (( x + x − ) − d ) + q − − q −
54 + 14 X x ∈ F ∗ q φ (( x + x − ) − d ) = 2 + X x ∈ F ∗ q φ (( x + x − ) − d )= 2 + X x ∈ F ∗ q φ ( x − d − x + 1)(1 + φ ( x ))= qP φ (2 d − . Here the last equality above follows from Equation (7).The case where (1 − d ) is not a square can be treated by similar computations. We omitthe details. (cid:3) The restriction of T N onto V ψ − , V ν γ and V η β In this section, we study the restriction of T N onto the irreducible constituents, V ψ − , { V ν γ } γ ∈ Γ and { V η β } β ∈ B , of V . We start with a technical lemma that will be useful forstudying the character sums T N,χ with χ ∈ { ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } . Lemma 27.
Let i be an element of F ∗ q \ F ∗ q such that i ∈ F ∗ q . Then for all γ ∈ Γ , T N,ν γ = ( q − q − q − ( q + 1) γ ( − φ ( − − q X b ∈ F ∗ q ,b =1 P γ (2 b h − P φ (2 b − . lso, for all β ∈ B , T N,η β = ( q − q + q + ( q + 1) β ( i ) φ ( −
1) + q X b ∈ F ∗ q ,b =1 R β (2 b h − P φ (2 b − , and T N,ψ − = ( q − q − q − − q X b ∈ F ∗ q ,b =1 P φ (2 b h − P φ (2 b − . Proof.
We will prove that the expression for T N,ν γ holds for every γ ∈ Γ. The proofs for thecharacters sums T N,η β and T N,ψ − are similar; we omit those details.First, assume that q ≡ T N,ν γ = ( q − − q − X g = ∞ , ∞ g =0 ν γ ( g − ) + ( q − X b ∈ F ∗ q b =1 X g = ∞ , g = b h ν γ ( g − ) N (0 , ∞ ) , (1 ,b ) = ( q − − ( q − γ ( −
1) + ( q − X b ∈ F ∗ q b =1 X g = ∞ , g = b h ν γ ( g − ) N (0 , ∞ ) , (1 ,b ) , where for the last equality we have applied Lemma 23. Also, recall that h ∈ P GL (2 , q ) isthe unique element sending 0 to 0, 1 to ∞ and ∞ to 1.Let us define S := X b ∈ F ∗ q b =1 X g = ∞ , g = b h ν γ ( g − ) N (0 , ∞ ) , (1 ,b ) . Applying Corollary 16 and Lemma 24 we obtain S = X b ∈ F ∗ q b =1 qP γ (2 b h − (cid:18) q − − φ (1 − b )2 − P φ (2 b − (cid:19) = q ( q − X b ∈ F ∗ q b =1 P γ (2 b h − − q X b ∈ F ∗ q b =1 φ (1 − b ) P γ (2 b h − − q X b ∈ F ∗ q b =1 P γ (2 b h − P φ (2 b − . We now simplify the first two character sums in the above expression for S .The following computation uses the connection between Legendre sums and hypergeomet-ric sums given by Lemma 13. We have X b ∈ F ∗ q b =1 P γ (2 b h −
1) = X a ∈ F q a = ± P γ ( a )= X a ∈ F q a = ± F (cid:20) γ γ − ǫ ; 1 − a q (cid:21) . ow, using Greene’s definition of hypergeometric sums given in Equation (21) we get X b ∈ F ∗ q b =1 P γ (2 b h −
1) = γ − ( − q X a ∈ F q a = ± X x ∈ F q γ − ( x ) γ (1 − x ) γ − (cid:18) −
12 (1 − a ) x (cid:19) = γ − ( − q X x ∈ F ∗ q γ − ( x ) γ (1 − x ) X a ∈ F q a = ± γ − (cid:18) −
12 (1 − a ) x (cid:19) = γ − ( − q X x ∈ F ∗ q γ − ( x ) γ (1 − x )( − − γ − (1 − x ))= 1 q (1 + γ ( − . On the other hand, to compute the second sum we use the definition of Legendre sumsgiven in Definition 7 and noting that φ ( −
1) = 1 when q ≡ X b ∈ F ∗ q b =1 φ (1 − b ) P γ (2 b h −
1) = 1 q X b ∈ F ∗ q b =1 φ (1 − b ) X x ∈ F ∗ q γ ( x ) φ (1 + (2 − b h ) x + x )= 1 q X x ∈ F ∗ q γ ( x ) X b ∈ F ∗ q b =1 φ (( x + 1) − b h x ) φ ( b − q X x ∈ F ∗ q γ ( x ) X b ∈ F ∗ q b =1 φ (( x − b − ( x + 1) )= 1 q X x ∈ F ∗ q ,x =1 γ ( x ) X b ∈ F ∗ q b =1 φ (( x − b − ( x + 1) ) + 1 q X b ∈ F ∗ q b =1 φ ( − q X x ∈ F ∗ q ,x =1 γ ( x )( − φ ( − x ) − φ ( − ( x + 1) )) + q − q = 1 + 1 q γ ( − . Putting all the above results together we have S = − ( q − q − γ ( − − q X b ∈ F ∗ q b =1 P γ (2 b h − P φ (2 b − , and plugging in S into the expression for T N,ν γ we obtain T N,ν γ = q − q − q − ( q − γ ( − − q X b ∈ F ∗ q b =1 P γ (2 b h − P φ (2 b − . he computations for the case q ≡ T N,ν γ assuming that q ≡ T N,ν γ = q − q − q + ( q − γ ( − − q X b ∈ F ∗ q b =1 P γ (2 b h − P φ (2 b − . Finally, note that φ ( −
1) = 1 when q ≡ φ ( −
1) = − q ≡ (cid:3) From Schur’s Lemma we know that the restriction of T N onto any irreducible module isan isomorphism or the zero map. The next theorem shows that the restriction of T N onto V η β is a P GL (2 , q )-module isomorphism for every β ∈ B .For the proofs below, we will need the following function in ℓ ( F q , m ), f : F q → C x φ (1 − x ) P φ ( x )Note that the norm of f is closely related to the norm of P φ , k f k = X x ∈ F q f ( x ) m ( x ) = X x ∈ F q x =1 P φ ( x ) m ( x ) = k P φ k − q + 1 q = 1 − q − q , where we have used Lemma 9 in the last equality. Theorem 28.
For every β ∈ B we have T N ( V η β ) ∼ = V η β . Proof.
It suffices to show that T N,η β = 0 for all β ∈ B . From Lemma 27 it follows that(19) T N,η β = ( q − q + q + ( q + 1) β ( i ) φ ( −
1) + q X b ∈ F ∗ q ,b =1 R β (2 b h − P φ (2 b − , where i ∈ F ∗ q \ F ∗ q such that i ∈ F ∗ q . We will show that the expression on the right handside of Equation (19) is not equal to zero.We claim that the character sum(20) X b ∈ F ∗ q ,b =1 R β (2 b h − P φ (2 b − f . Recall that h is the unique element in P GL (2 , q )sending 0 to 0, 1 to ∞ and ∞ to 1. Hence, if b ∈ F ∗ q and b = 1 then b h = 0 , , ∞ . Moreover,we have the following formula for b h when b ∈ F ∗ q and b = 1, b h = bb − b h ) h = b for any b ∈ F q . Thus, we can rewrite the sum in (20) as, X b ∈ F ∗ q ,b =1 R β (2 b h − P φ (2 b −
1) = X b ∈ F ∗ q ,b =1 P φ (2 b h − R β (2 b − . sing the relation between Legendre sums and hypergeometric sums given by Lemma 13and the transformation formula in Lemma 5, the following expression for P φ (2 b h −
1) isobtained P φ (2 b h −
1) = F (cid:20) φ φǫ ; 11 − b ; q (cid:21) = φ (1 − b ) F (cid:20) φ φǫ ; 1 − b ; q (cid:21) = φ (1 − b ) P φ (2 b − , for b ∈ F q , b = 0 ,
1. Putting all the above remarks together we conclude that X b ∈ F ∗ q ,b =1 R β (2 b h − P φ (2 b −
1) = X b ∈ F ∗ q ,b =1 φ (1 − b ) P φ (2 b − R β (2 b − φ (2) X x ∈ F q ,x = ± φ (1 − x ) P φ ( x ) R β ( x )= φ (2) (cid:18) q (cid:19) / h f, R ′ β i − ( q + 1) β ( i ) φ ( − q where i is an element of F ∗ q \ F ∗ q such that i ∈ F ∗ q . Therefore, plugging in the above expressioninto Equation (19), we can also express T N,η β in terms of the function f ,(21) T N,η β = q ( q − " q + φ (2) (cid:18) q (cid:19) / h f, R ′ β i . Note that Equation (21) implies that if |h f, R ′ β i| ≤ T N,η β = 0. We claim that |h f, R ′ β i| ≤ β ∈ B ; note that the theorem follows from the validity of this claim.Recall that { P ′ ǫ , P ′ φ , P ′ γ , R ′ β : γ ∈ Γ , β ∈ B } is an orthonormal basis of ℓ ( F q , m ). Thus,we can express f in terms of this orthonormal basis, f = h f, P ′ ǫ i P ′ ǫ + h f, P ′ φ i P ′ φ + X γ h f, P ′ γ i P ′ γ + X β h f, R ′ β i R ′ β . Analogously, the squared norm of f can also be expressed in terms of this orthonormal basis, k f k = h f, P ′ ǫ i + h f, P ′ φ i + X γ h f, P ′ γ i + X β h f, R ′ β i , where we have used the fact the coefficients in the expansion of f are all real (cf. Lemma 12).On the other hand, we know that the squared norm of f is 1 − /q − /q . This impliesthat the square of every coefficient of the form h f, g i is less than 1 for all g ∈ { P ′ ǫ , P ′ φ , P ′ γ , R ′ β : γ ∈ Γ , β ∈ B } . In particular, h f, R ′ β i ≤ − /q − /q for all β ∈ B . Thus, our claim isproved. (cid:3) Unfortunately, the argument used in the proof of Theorem 28 cannot be applied to showthat the restriction of T N onto the irreducible module V ψ − is a P GL (2 , q )-module isomor-phism. To deal with this case we exploit the connection between Legendre sums and Hyper-geometric sums shown by Kable in [14]. Lemma 29.
Let γ be a nontrivial multiplicative character of F q . Then φ (2) q h f, P γ i = q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) + φ ( − γ ( − q. roof. Applying Lemmas 4 and 13 we obtain, φ (2) q h f, P γ i = φ (2) q X x ∈ F q x = ± φ (1 − x ) P φ ( x ) P γ ( x ) + q P φ ( − P γ ( − m ( − q X y ∈ F ∗ q y =1 φ ( y ) F (cid:20) φ φǫ ; y ; q (cid:21) F (cid:20) γ γ − ǫ ; y ; q (cid:21) + φ ( − γ ( − q + 1)= q X y ∈ F q φ ( y ) F (cid:20) φ φǫ ; y ; q (cid:21) F (cid:20) γ γ − ǫ ; y ; q (cid:21) + φ ( − γ ( − q = q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) + φ ( − γ ( − q. (cid:3) Theorem 30. If q ≥ then, T N ( V ψ − ) ∼ = V ψ − . Proof.
It suffices to show that T N,ψ − = 0. It follows from Lemma 27 that T N,ψ − = ( q − q − q − − q X b ∈ F ∗ q ,b =1 P φ (2 b h − P φ (2 b − . Let f be the function in ℓ ( F q , m ) defined before the statement of Theorem 28. By Lemmas5 and 13 we see that the sum X b ∈ F ∗ q ,b =1 P φ (2 b h − P φ (2 b − f . In particular, X b ∈ F ∗ q ,b =1 P φ (2 b h − P φ (2 b −
1) = X b ∈ F ∗ q ,b =1 φ (1 − b ) P φ (2 b − P φ (2 b − φ (2) X x ∈ F q ,x = ± φ (1 − x ) P φ ( x ) P φ ( x )= φ (2) h f, P φ i − q + 1 q . Thus, T N,ψ − can be expressed in terms of f :(22) T N,ψ − = ( q − (cid:2) q − q − − φ (2) q h f, P φ i (cid:3) . We claim that φ (2) q h f, P φ i ≤ q / . This claim together with Equation (22) immediatelyimplies that T N,ψ − = 0 for every q ≥ φ (2) q h f, P φ i can be written in termsof a hypergeometric sum F . Letting γ = φ in Lemma 29, φ (2) q h f, P φ i = q F (cid:20) φ φ φ φǫ ǫ ǫ ; 1; q (cid:21) + q. herefore, our claim follows directly from the final conclusion of Proposition 6. (cid:3) To study the restriction of T N onto V ν γ we consider two cases. First, if γ is a characterwhose order is not equal to three, four or six then we can apply arguments similar to theones used in the proof of Theorem 28 to prove that the restriction is an isomorphism. Onthe other hand, different ideas have to be used to show that the same result holds when γ has order three, four or six. The next theorem deals with these cases. Theorem 31.
Assume that q ≥ . If γ ∈ Γ then T N ( V ν γ ) ∼ = V ν γ . Proof.
We proceed as we did in the proof of Theorem 28. Thus, to prove this theorem it isenough to show that T N,ν γ = 0. It follows from Lemma 27 that T N,ν γ = ( q − q − q − ( q + 1) γ ( − φ ( − − q X b ∈ F ∗ q ,b =1 P γ (2 b h − P φ (2 b − . Applying Lemmas 5 and 13 it is possible to write the sum of products of Legendre sumsin terms of the function f . In fact, X b ∈ F ∗ q ,b =1 P γ (2 b h − P φ (2 b −
1) = φ (2) (cid:18) − q (cid:19) / h f, P ′ γ i − ( q + 1) γ ( − φ ( − q . Therefore, for every γ ∈ Γ we have(23) T N,ν γ = q ( q − " − q − φ (2) (cid:18) − q (cid:19) / h f, P ′ γ i . Recall that(24) k f k = h f, P ′ ǫ i + h f, P ′ φ i + X γ h f, P γ i + X β h f, R ′ β i = 1 − q − q , where { P ′ ǫ , P ′ φ , P ′ γ , R ′ β : γ ∈ Γ , β ∈ B } is an orthonormal basis of ℓ ( F q , m ). Equation (24)implies that at most one of the coefficients h f, g i with g ∈ { P ′ ǫ , P ′ φ , P ′ γ , R ′ β : γ ∈ Γ , β ∈ B } can be close to 1. On the other hand, it is clear from (23) that T N,ν γ = 0 if and only if thecoefficient h f, P ′ γ i is close to 1.To prove the theorem we proceed by contradiction. Assume that there exists γ ∈ Γ suchthat T N,ν γ = 0. Hence, it follows from equation (23) that(25) h f, P ′ γ i = 1 − q + 4 q ( q − . Let Gal( Q ( ζ q − ) / Q ) be the Galois group where ζ q − is a primitive ( q − γ is a nontrivial character whose order is not equal to three, four or six, thereexists σ ∈ Gal( Q ( ζ q − ) / Q ) such that γ σ = γ and γ σ = γ − . Now, applying the Galois utomorphism σ to both sides of (25) we conclude that σ (cid:0) h f, P ′ γ i (cid:1) = σ (cid:18) − q + 4 q ( q − (cid:19) h f, P ′ γ σ i = 1 − q + 4 q ( q − . Thus, h f, P ′ γ i and h f, P ′ γ σ i are equal to 1 − q + q ( q − which is a contradiction because atmost one of the coefficients h f, g i with g ∈ { P ′ ǫ , P ′ φ , P ′ γ , R ′ β : γ ∈ Γ , β ∈ B } can be close to 1.Assume now γ ∈ Γ is a character of order 3, 4 or 6. From equation (23) we get the followingexpression for T N,ν γ , T N,ν γ = ( q − (cid:2) q − q − φ (2) q h f, P γ i (cid:3) . By Lemma 29, φ (2) q h f, P γ i = q F (cid:20) γ γ − φ φǫ ǫ ǫ ; 1; q (cid:21) + φ ( − γ ( − q. Now applying Proposition 6, we conclude that T Nν γ = 0. (cid:3) Finally, we are ready to prove Theorem 3.
Proof of Theorem 3.
Recall that in Section 3.3 we proved the following lower and upperbounds on the rank of the derangement matrix M of P SL (2 , q ) acting on P G (1 , q ),(26) X { χ : T N,χ =0 } dim( V χ ) ≤ rank( M ) ≤ q ( q − . These bounds imply that if T N,χ is not zero for every χ ∈ { λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } thenthe rank of M is q ( q − q ≥
11 then it follows from Theorems 28, 30 and 31 that T N,χ = 0 for all χ ∈{ λ , ψ − , { η β } β ∈ B , { ν γ } γ ∈ Γ } . Furthermore, for each odd prime power q , 3 < q <
11, weuse a computer to check that the rank of M is exactly q ( q − (cid:3) Conclusions
In this paper we consider the natural right action of
P SL (2 , q ) on P G (1 , q ), where q is an odd prime power. Using the eigenvalue method, it was proved in [19, 2] that themaximum size of an intersecting family in P SL (2 , q ) is q ( q − /
2. Meagher and Spiga [19]conjectured that the cosets of point stabilizers are the only intersecting families of maximumsize in
P SL (2 , q ), when q > P GL (2 , q ) and deep results fromnumber theory.For future research, one could consider the stability problem concerning intersecting fam-ilies of P SL (2 , q ). To present this problem we introduce the notion of stability.Let X be a finite set and G a finite group acting on X . Recall that a subset S of G is said to be an intersecting family if for any g , g ∈ S there exists an element x ∈ X such that x g = x g . We will refer to intersecting families of maximum size as extremalfamilies . Moreover, intersecting families whose sizes are close to the maximum are called lmost extremal families . We say that the extremal families of a group G acting on X are stable if almost extremal families are similar in structure to the extremal ones.The stability of intersecting families has been studied during the past few years (cf. [7,8, 24]). Consider the action of S n on [ n ]. As was remarked in the introduction, the size ofextremal families in S n is ( n − S n was established by Ellis [7], who provedthat for any ǫ > n > N ( ǫ ), any intersecting family of size at least (1 − /e + ǫ )( n − P GL (2 , q ) acting on P G (1 , q ). In fact, the size of extremal families in P GL (2 , q ) is q ( q −
1) and every extremal family is a coset of a point stabilizer. Recently, in[24] it was proved that the extremal families in
P GL (2 , q ) are stable.We conjecture that the extremal families in P SL (2 , q ) are also stable. The precise state-ment is given below. Conjecture 32.
Let S be an intersecting family in P SL (2 , q ) with q > δ > | S | ≥ (1 − δ ) q ( q − / S is containedwithin a coset of a point stabilizer. Acknowledgment
The authors would like to thank the reviewers for their helpful comments.
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E-mail address : [email protected] Rafael Plaza, Department of Mathematical Sciences, University of Delaware, Newark,DE 19716, USA
E-mail address : [email protected] Peter Sin, Department of Mathematics, University of Florida, Gainesville, FL 32611,USA
E-mail address : [email protected] Qing Xiang, Department of Mathematical Sciences, University of Delaware, Newark, DE19716, USA
E-mail address : [email protected]@udel.edu