Weierstrass semigroups on the Giulietti-Korchmáros curve
aa r X i v : . [ m a t h . AG ] A ug Weierstrass semigroups on the Giulietti–Korchm´aros curve
Peter Beelen and Maria Montanucci
Abstract
In this article we explicitly determine the structure of the Weierstrass semigroups H ( P ) for any point P of the Giulietti–Korchm´aros curve X . We show that as the point varies, exactly three possibilitiesarise: One for the F q -rational points (already known in the literature), one for the F q \ F q -rationalpoints, and one for all remaining points. As a result, we prove a conjecture concerning the structure of H ( P ) in case P is a F q \ F q -rational point. As a corollary we also obtain that the set of Weierstrasspoints of X is exactly its set of F q -rational points. Math. Subj. Class.:
Primary: 11G20. Secondary: 11R58, 14H05, 14H55.
Keywords:
Giulietti–Korchm´aros maximal curve, Weierstrass semigroup, Weierstrass points.
Let C be a nonsingular, projective algebraic curve of genus g defined over a field F . Let P be a rationalpoint on C . The Weierstrass semigroup H ( P ) is defined as the set of integers k such that there exists afunction on C having pole divisor exactly kP . More generally H ( P ) can be defined for any point P on C by considering C as an algebraic curve over the algebraic closure of F . It is clear that H ( P ) is a subset ofnatural numbers N = { , , , . . . } . The Weierstrass gap Theorem, see [10, Theorem 1.6.8], states that the set G ( P ) := N \ H ( P ) contains exactly g elements, which are called gaps . The structure of H ( P ) is not alwaysthe same for every point P of C . However, it is known that for generically the semigroup H ( P ) is the same,but there can exist finitely many points of C , called Weierstrass points , with a different gap set. These pointsare of intrinsic interest, for example in St¨ohr–Voloch theory [11], but in case F = F q , the finite field with q elements, they also occur in the study of algebraic geometry (AG) codes [12]. In this context, a commonlystudied class of curves are the so-called maximal curves , that is, algebraic curves defined over a finite field F q having as many rational points as possible according to the Hasse–Weil bound. More precisely, an algebraiccurve C with genus g ( C ) and defined over F q is said to be an F q -maximal curve if it has q + 1 + 2 g ( C ) √ q points defined over F q . Clearly, this can only be the case if the cardinality q of the finite field is a square.An important and well-studied example of an F q -maximal curve is given by the Hermitian curve H . Forfixed q , the curve H has the largest possible genus g ( H ) = q ( q − / F q -maximal curve can have.The Weierstrass points on H and the precise structure of the semigroups for P on H are known; see [3]. Bya result commonly attributed to Serre, see [9, Proposition 6], any F q -rational curve which is covered by an F q -maximal curve is also F q -maximal. Most of the known maximal curves are subcovers of the Hermitiancurve. The first known example of a maximal curve which is not a subcover of the Hermitian curve wasconstructed by Giulietti and Korchm´aros; see [4]. This curve is an F q -maximal curve and commonly calledthe Giulietti–Korchm´aros (GK) curve. The aim of this paper is to complete the description of the Weierstrasssemigroups occurring for this curve. 1he Weierstrass semigroup for any F q -rational point of X was computed in [4], but the structure of theWeierstrass semigroup H ( P ) where P
6∈ X ( F q ) is not known, except for q ≤
9, [2, 1]. Based on the availabledata for small q , a conjecture concerning the structure of H ( P ) was stated in [1] for P ∈ X ( F q ) \ X ( F q ).For P
6∈ X ( F q ) nothing specific is known about H ( P ) . In this article we determine settle the conjecturefrom [1] and also determine the structure of the generic semigroup for P on X . More precisely, we show thefollowing theorem. Theorem 1.1.
Let q be a prime power and let P be a point of the Giulietti–Korchm´aros curve X . TheWeierstrass semigroup H ( P ) is given by • H ( P ) = h q − q + q, q , q + 1 i , if P ∈ X ( F q ) ; • H ( P ) = h q − q + 1 , q + 1 , q + i ( q − q − q + q − | i = 0 , . . . , q − i , if P ∈ X ( F q ) \ X ( F q ) ; • H ( P ) = N \ G, if P
6∈ X ( F q ) , where G = ( iq + kq + m ( q + 1) + q − X s =1 n s (( s + 1) q ) + j + 1 | i, j, k, m, n , . . . , n q − ∈ Z ≥ , j ≤ q − and i + j + k + mq + q − X s =1 n s (( s + 1) q − s ) ≤ q − ) . As mentioned above, the case P ∈ X ( F q ) is already known and taken from [4]. As a bonus, we will alsoobtain the set of Weierstrass points of X . Corollary 1.2.
Let W denote the set of Weierstrass points of the Giulietti–Korchm´aros curve X . Then W = X ( F q ) . The paper is organized as follows: In the next section we give the necessary background on the GK curveas well as some results on Weierstrass semigroups and their gaps that we will need later. In section three, wesettle the conjecture from [1] concerning H ( P ) for P ∈ X ( F q ) \ X ( F q ), while in section four, we computethe Weierstrass semigroup for P
6∈ X ( F q ). We finish with some concluding remarks and observations. Let q be a prime power and K = F q . The Giulietti–Korchm´aros (GK) curve X is a non-singular curve inPG(3 , K ) defined by the affine equations X : (cid:26) Y q +1 = X q + X,Z q − q +1 = Y q − Y. (1)This curve has genus g ( X ) = ( q − q + q ) / q − q + q + 1 F q -rational points. The curve X hasbeen introduced in [4], where it was proved that X is maximal over F q , that is, the number |X ( F q ) | of F q -rational points of X equals q +1+2 gq . Also, for q >
2, the curve X is not F q -covered by the Hermitiancurve maximal over F q ; X was the first maximal curve shown to have this property. Note that equation (1)implies that X is a cover of the Hermitian curve over F q given by the affine equation Y q +1 = X q + X . Wewill denote this curve by H . 2he automorphism group Aut( X ) of X is defined over F q and has order q ( q + 1)( q − q − q + 1).Moreover, it has a normal subgroup isomorphic to SU(3 , q), the automorphism group of the Hermitian curve H . The set X ( F q ) of the F q -rational points of X splits into two orbits under the action of Aut( X ): oneorbit O = X ( F q ) of size q + 1, which coincides with the intersection between X and the plane Z = 0;and another orbit O = X ( F q ) \ X ( F q ) of size q ( q + 1)( q − O and O are the short orbits of Aut( X ), that is, the unique orbits of points of X having a non-trivial stabilizer inAut( X ).Let x, y, z ∈ K ( X ) be the coordinate functions of the function field of X , which satisfy y q +1 = x q + x and z q − q +1 = y q − y . Then we denote by P ( a,b,c ) the affine point of X with coordinates ( a, b, c ) and by P ∞ the unique point at infinity. Similarly, we denote by Q ( a,b ) the affine point of the Hermitian curve H withcoordinates ( a, b ) and by Q ∞ its unique point at infinity.The Weierstrass semigroup at P ∞ , and hence at every F q -rational point of X (since they lie in the sameshort orbit O of Aut( X )) was computed in [4]. Proposition 2.1. [4, Proposition 6.2] The Weierstrass semigroup of X at P ∞ is generated by q − q + q , q , q + 1.Before describing what is known about H ( P ) for P
6∈ X ( F q ), we introduce several functions on X andgive their divisors. Some of these functions can be interpreted as functions on H as well and therefore have adivisor on H . To differentiate, we will write ( f ) H (resp. ( f ) X ) for divisors on the Hermitian curve H (resp.the GK curve X ). Given a point P = P ( a,b,c ) on X , we define the functions˜ x P = − a q − x + b q y, ˜ y P = y − b, ˜ z P = − a q − x + b q y + c q z. (2)Then it is not hard to show the following.(˜ x P ) X = q X ξ q − q +1 =1 P ( a,b,ξc ) + X ξ q − q +1 =1 P ( a q ,b q ,ξc q ) − ( q + 1) P ∞ , (3)(˜ y P ) X = X s q + s =0 , ξ q − q +1 =1 P ( a + s,b,ξc ) − ( q − q + q ) P ∞ , (4)(˜ z P ) X = q P ( a,b,c ) + P ( a q ,b q ,c q ) − ( q + 1) P ∞ , (5)( z ) X = X P ∈X ( F q ) ,P = P ∞ P − q P ∞ . (6)Now let P = P ( a,b,c ) be a fixed F q -rational point of X which is not F q -rational (implying c = 0). Inthis case equation (5) implies:(˜ z P ) X = ( q + 1)( P − P ∞ ) for P = P ( a,b,c ) ∈ X ( F q ) . (7)The Weierstrass semigroup H ( P ) is only completely known in finitely many cases if P ∈ X ( F q ) \ X ( F q ). Itwas computed for q = 2 and q = 3 in [2] and for 4 ≤ q ≤ q : Equations (3), (4) and (7) imply that the functions 1 / ˜ z P , ˜ y P / ˜ z P , ˜ x P / ˜ z P havepoles only in P of orders q + 1, q and q − q + 1 respectively. Hence h q − q + 1 , q , q + 1 i ⊆ H ( P ) for P ∈ X ( F q ) \ X ( F q ) . (8)Based on this and the results for q ≤
9, the following conjecture was stated in [1], which we will prove in thenext section. 3 onjecture 2.2.
The Weierstrass semigroup H ( P ) of X at P ∈ X ( F q ) \ X ( F q ) is given by H ( P ) = h q − q + 1 , q + 1 , q + i ( q − q − q + q − | i = 0 , . . . , q − i . Finally, for P
6∈ X ( F q ) nothing specific is known about the structure of semigroup H ( P ). We willcompletely determine its gap structure, but for now, we finish this section by stating some facts that wewill use to achieve this. We start with the following well-known lemma connecting regular differentials (i.e.,differential forms having no poles anywhere on X ) and gaps of H ( P ). Proposition 2.3. [13, Corollary 14.2.5]
Let X be an algebraic curve of genus g defined over K . Let P be apoint of X and ω be a regular differential on X . Then v P ( ω ) + 1 is a gap at P . This proposition has the following, for us very useful, consequence.
Corollary 2.4.
For any point P on the GK curve X distinct from P ∞ and for any f ∈ L ((2 g ( X ) − P ∞ ) , we have v P ( f ) + 1 ∈ N \ H ( P ) . Proof.
First note that ( dy ) H = ( q − q − Q ∞ . The set of points that ramify in the covering of X by H isexactly H ( F q ), the set of F q -rational points of the Hermitian curve, all with ramification index q − q + 1.Moreover, the points of X above H ( F q ) are precisely the F q -rational points of X . Therefore, we immediatelyobtain that ( dy ) X = ( q − q + q − P ∞ + ( q − q ) X P ∈X ( F q ) ,P = P ∞ P. Thus, from z q − q +1 = y q − y and equation (6),( dz ) X = ( − dy/z q − q ) X = ( q − q + q − P ∞ . In particular a differential f dz is regular if and only if f ∈ L (( q − q + q − P ∞ ) = L ((2 g ( X ) − P ∞ ).The corollary now follows by applying Proposition 2.3. H ( P ) for P ∈ X ( F q ) \ X ( F q ) This section is devoted to the proof of Conjecture 2.2 for any prime power q . In particular in this section P = P ( a,b,c ) will always denote a point in X ( F q ) \ X ( F q ). Further we define the semigroup T := h q − q + 1 , q + 1 , q + i ( q − q − q + q − | i = 0 , . . . , q − i . Conjecture 2.2 then simply states that H ( P ) = T . Our proof of the conjecture consists of two main steps.In the first step, we will show that T ⊂ H ( P ) by showing that the generators of T are in H ( P ). In thesecond step, we show that the number of gaps of the semigroup T (also known as the genus of T ) is exactlyequal to the genus of X . Once this has been established, the equality H ( P ) = T will follow immediately,proving Conjecture 2 . T ⊂ H ( P ) As before we use the function ˜ x P defined in equation (2) and its divisor in equation (3). Moreover, for k ∈ Z ,we define the k -th Frobenius twist of ˜ x P as the follows:˜ x ( k ) P := − a q k +1 − x + b q k +1 y for P = P ( a,b,c ) . (9)4ince we assume that P ∈ X ( F q ) \ X ( F q ) , equation (3) implies that(˜ x (1) P ) X = q X ξ q − q +1 =1 P ( a q ,b q ,ξc q ) + X ξ q − q +1 =1 P ( a q ,b q ,ξc q ) − ( q + 1) P ∞ , (˜ x (2) P ) X = q X ξ q − q +1 =1 P ( a q ,b q ,ξc q ) + X ξ q − q +1 =1 P ( a,b,ξc ) − ( q + 1) P ∞ . (10) Lemma 3.1.
Let P = P ( a,b,c ) ∈ X ( F q ) \ X ( F q ) and let ˜ f i = f i / ˜ z iq − i +1 P where f i := (˜ x P ) qi · ˜ x (2) P (˜ x (1) P ) i , for i = 1 , . . . , q − . Then ( ˜ f i ) ∞ = ( q + i ( q − q − q + q − P and in particular q + i ( q − q − q + q − ∈ H ( P ) for i = 1 , . . . , q − .Proof. Using equations (3) and (10), we directly obtain that( f i ) X = ( iq + 1) X ξ q − q +1 =1 P ( a,b,ξc ) + ( q − i ) X ξ q − q +1 =1 P ( a q ,b q ,ξc q ) − ( q + 1)( iq − i + 1) P ∞ . Now using the divisor of ˜ z P given in equation (7), we find that( ˜ f i ) X = − ( q + i ( q − q − q + q − P + ( iq + 1) X ξ q − q +1 =1 ,ξ =1 P ( a,b,ξc ) + ( q − i ) X ξ q − q +1 =1 P ( a q ,b q ,ξc q ) . The lemma now follows.Note that the lemma is also true for i = 0. Considering the corresponding function ˜ f = ˜ x (2) P / ˜ z P , givesa way to show that q ∈ H ( P ). However, this is already known, see equation (8). Proposition 3.2.
Let P ∈ X ( F q ) \ X ( F q ) . Then T ⊂ H ( P ) . Proof.
Equation (8) and Lemma 3.1 imply that { q − q +1 , q +1 , q + i ( q − q − q + q − | i = 0 , . . . , q − } ⊂ H ( P ). Since by definition these numbers generate T , the proposition follows. T equals g ( X ) We now show that the genus g ( T ) of the numerical semigroup T = h q − q +1 , q +1 , q + i ( q − q − q + q − | i = 0 , . . . , q − i is equal to g ( X ) = ( q − q + q ) /
2. In this way, since we already know that T ⊆ H ( P ( a,b,c ) )from Proposition 3.2, Conjecture 2.2 will be completely proved. We recall that a numerical semigroup iscalled telescopic if it is generated by a telescopic sequence, that is by a sequence ( a , . . . , a k ) such that • gcd( a , . . . , a k ) = 1; • for each i = 2 , . . . , k , a i /d i ∈ h a /d i − , . . . , a i − /d i − i , where d i = gcd( a , . . . , a i ) and d = 0;5ee [8]. From [7, Proposition 5.35], the genus of a semigroup Γ generated by a telescopic sequence ( a , . . . , a k )is g (Γ) = 12 (cid:18) k X i =1 (cid:18) d i − d i − (cid:19) a i (cid:19) . (11)For the semigroup S defined by S := h q − q + 1 , q + 1 i we obtain the following: Lemma 3.3.
The numerical semigroup S = h q − q + 1 , q + 1 i is telescopic. Its genus g ( S ) is given by g ( S ) = q ( q − q )2 . Proof.
Let a = q − q + 1 and a = q + 1. Then gcd ( a , a ) = 1 and, using the same notation as above, d = a and d = 1. Since a /d ∈ h i = h a /d i , S is telescopic. Thus from equation (11), g ( S ) = 12 (cid:18) − a + ( a − a (cid:19) = q ( q − q )2 . Now the idea is to compute the number of gaps of T by identifying the elements of T that are gaps of S .The following observation is trivial, but will be very useful. Observation 3.4.
For any integer n , there exist unique integers a and b such that n = a ( q − q +1)+ b ( q +1) and ≤ b ≤ q − q. An integer n is an element of the semigroup S = h q − q + 1 , q + 1 i if and only if thereexist integers a and b such that n = a ( q − q + 1) + b ( q + 1) , a ≥ and ≤ b ≤ q − q. In the following lemma, we identify several elements of T \ S that turn out to play an important role. Lemma 3.5.
For any i = 0 , . . . , q − and j = 1 , . . . , q − , define the set S i,j := { ( iq − jq + k )( q − q + 1) + ( jq − i + k )( q + 1) | k = 0 , . . . , q − , k = 0 , . . . , q − q − jq + i } . Then we have:1. S i,j ⊂ T \ S. S i,j ∩ S i ′ j ′ = ∅ if ( i ′ , j ′ ) = ( i, j ) , ≤ i ′ ≤ q − and ≤ j ′ ≤ q − . | S i,j | = q ( q − q − jq + i + 1) . Proof.
First of all note that jq + i ( q − q − q + q −
1) = ( − jq + iq )( q − q + 1) + ( jq − i )( q + 1) . Using this, it is clear from Proposition 3.2, that ( iq − jq + k )( q − q + 1) + ( jq − i + k )( q + 1) ∈ T forany i, j, k , k in the given range. To show that these elements are not in S , observe that iq − jq + k ≤ ( q − q − q + q − < ≤ jq − i + k ≤ q − q. (12)Observation 3.4 now implies that ( iq − jq + k )( q − q + 1) + ( jq − i + k )( q + 1) S. This completes theproof of the first item. 6ow suppose that S i,j ∩ S i ′ j ′ = ∅ . Then there exist integers k , k ′ , k , k ′ satisfying the defining require-ments of S i,j and S i ′ j ′ such that( iq − jq + k )( q − q + 1) + ( jq − i + k )( q + 1) = ( i ′ q − j ′ q + k ′ )( q − q + 1) + ( j ′ q − i ′ + k ′ )( q + 1) . As above, we have equation (12) as well as the similar equation i ′ q − j ′ q + k ′ < ≤ j ′ q − i ′ + k ′ ≤ q − q. Observation 3.4 therefore implies that iq − jq + k = i ′ q − j ′ q + k ′ and jq − i + k = j ′ q − i ′ + k ′ , and in particular ( i − i ′ ) q − ( j − j ′ ) q + ( k − k ′ ) = 0 . Considering this equation modulo q and modulo q ,we see that k = k ′ and i = i ′ , implying that j = j ′ as well. Then it is also clear that k = k ′ . This impliesthe second item.As for the third item: if( iq − jq + k )( q − q + 1) + ( jq − i + k )( q + 1) = ( iq − jq + k ′ )( q − q + 1) + ( jq − i + k ′ )( q + 1) , with integers k , k ′ , k , k ′ satisfying the defining requirements of S i,j , then the same reasoning as in aboveproof of the second item, shows that k = k ′ and k = k ′ . Hence the cardinality of S i,j is simply the numberof possibilities for k times that for k .Picture 3.2 describes the sets S i,j for q = 3. In this picture a point of coordinates ( a, b ) is used torepresent the element a ( q − q + 1) + b ( q + 1). Black dots represent elements of the numerical semigroup S , while white dots represent the elements contained in S i,j for some i and j . ab SS , S , S , S , S , S , − − − − − − − S i,j and S for q = 3 ◦ : Elements in S i,j • : Elements in S
7e are now ready to prove Conjecture 2.2.
Theorem 3.6.
We have g ( T ) = g ( X ) and in particular H ( P ) = T. Proof.
Proposition 3.2 implies that g ( T ) ≥ g ( X ). Hence the theorem follows once we show that g ( T ) ≤ g ( X ).However, using the first two items of Lemma 3.5, we see that g ( T ) ≤ g ( S ) − q − X i =0 q − X j =1 | S i,j | . Using Lemma 3.3 and item three of Lemma 3.5 we obtain g ( T ) ≤ q − q − q − X i =0 q − X j =1 q ( q − q + 1 − jq + i )= q − q − q − X i =0 q − X j =1 q ( q − q + 1) + q − X i =0 q − X j =1 jq − q − X i =0 q − X j =1 iq = q − q − q ( q − q − q + 1) + q ( q − − q ( q − q − q + q g ( X ) . A direct consequence of the above theorem is that H ( P ) = (cid:16)S i,j S i,j (cid:17) ∪ S . It is not hard to obtain moreinformation about H ( P ) from the above calculations. For example, it is clear that the multiplicity of H ( P )(i.e., the smallest positive element in H ( P )) is equal to q − q + 1, while its conductor (i.e., the largest gap)is 2 g ( X ) −
1. This means in particular that like H ( P ∞ ), the semigroup H ( P ) is symmetric. Since H ( P ∞ )has multiplicity q − q + q , we also see that H ( P ) = H ( P ∞ ) . H ( P ) for P
6∈ X ( F q ) In this section we determine the Weierstrass semigroup H ( P ) for P
6∈ X ( F q ). In particular in this section P = P ( a,b,c ) will always denote a point on X not in X ( F q ). For future reference, note that as in the previoussection, this means that c = 0. As we will see, the semigroup H ( P ) is the same for all P
6∈ X ( F q ) andhence the ‘generic’ semigroup for a point on X . Our approach is use Corollary 2.4 to construct gaps of H ( P ) by computing the valuation at P of functions f ∈ L ((2 g ( X ) − P ∞ ) . It is very easy to find a basisof the Riemann–Roch space L ((2 g ( X ) − P ∞ ). For example the functions x i y j z k where i ≥
0, 0 ≤ j ≤ q ,0 ≤ k ≤ q + q and i ( q + 1) + j ( q − q + q ) + kq ≤ g ( X ) − P . Therefore an effort must bemade to construct functions in L ((2 g ( X ) − P ∞ ) having distinct valuations at P . In the next subsection,we construct functions with various valuations at P . After that we will combine these functions and obtaina set G of several explicitly described gaps of H ( P ) using Corollary 2.4. The remainder of the section willthen be a somewhat lengthy calculation showing that the set G in fact contains g ( X ), and hence all, gaps of H ( P ). 8 .1 Construction of functions. We start by constructing a function g with small, but positive, valuation at P = P ( a,b,c ) . It will beconvenient to define β = b q − b . Note that b q − b = c q − q +1 = 0, since P
6∈ X ( F q ) (and therefore a fortiori P
6∈ X ( F q )). We define g := ( β q − − x qP + β q + q + β q (cid:0) (˜ y P − β )(˜ x P + β q (˜ y P − β )) q − (cid:1) . The functions ˜ x P and ˜ y P are as in equation (2). This definition may seen ad hoc, but it arises naturallywhen constructing functions of low pole order at P ∞ and large vanishing order at P . More precisely, wehave the following lemma. Lemma 4.1.
The function g is an element of L ((2 g ( X ) − P ∞ ) . Moreover v P ∞ ( g ) ≥ − q ( q + 1) and v P ( g ) = q + 1 .Proof. It is clear that g only can have a pole at P ∞ . Moreover, from equations (3) and (4) imply that ˜ x P (resp. ˜ y P ) has a pole at P ∞ of order q + 1 (resp. q − q + q ). Therefore, the triangle inequality impliesthat v P ∞ ( g ) ≥ v P ∞ (˜ x qP ) = − q ( q + 1) , which is what we want to show.From equation (4), we see that the function ˜ y P is a local parameter for the point P = P ( a,b,c ) . Thedefining equation for H q directly implies that ˜ x qP + ˜ x P = β ˜ y qP − ˜ y q +1 P . Hence we easily can obtain the powerseries development of ˜ x P in terms of ˜ y P . More precisely, we obtain that˜ x P = β ˜ y qP − ˜ y q +1 P − ˜ x qP = β ˜ y qP − ˜ y q +1 P − β q ˜ y q P + ˜ y q + qP + · · · = (˜ y P − β )( − ˜ y qP + (˜ y P − β ) q − ˜ y q P ) + · · · (13)Using this, we also obtain that(˜ y P − β ) (˜ x P + β q (˜ y P − β )) q − = (˜ y P − β ) (cid:16) (˜ y P − β )( − ˜ y qP + (˜ y P − β ) q − ˜ y q P ) + β q (˜ y P − β ) (cid:17) q − + · · · = (˜ y P − β ) q (cid:16) − (˜ y P − β ) q + (˜ y P − β ) q − ˜ y q P (cid:17) q − + · · · = (˜ y P − β ) q − q +1 (cid:16) − (˜ y P − β ) + ˜ y q P (cid:17) q − + · · · = (˜ y P − β ) q − (˜ y P − β ) q − ˜ y q P + · · · = − β q + (1 − β q − )˜ y q P + β q − ˜ y q +1 P + · · · . (14)Combining equations (13) and (14), we see that g = ( β q − − β q ˜ y q P + β q + q + β q ( − β q + (1 − β q − )˜ y q P + β q − ˜ y q +1 P ) + · · · = β q + q − ˜ y q +1 P + · · · This implies that v P ( g ) = q + 1, which is what we wanted to show.The next functions are inspired by the previous section in the sense that we again use the functions ˜ x ( k ) P introduced in equation (9), but now for P = P ( a,b,c )
6∈ X ( F q ). For s = 1 , . . . , q − h s := ˜ x qP ˜ x (1) P ! s +1 · ˜ x (2) P .
9e have the following lemma about these functions.
Lemma 4.2.
Let s = 1 , . . . , q − . The function h s is an element of L ((2 g ( X ) − P ∞ ) . Moreover v P ∞ ( h s ) = − ( q ( s + 1) − s )( q + 1) and v P ( h s ) = ( s + 1) q .Proof. Using equations (3) and (10), we see that v P ∞ ( h s ) = − ( q ( s + 1) − s )( q + 1) and that h s has noother poles. Further it is well known that H q ( F q ) = H q ( F q ) . Since any point in H q ( F q ) ramifies totally inthe cover X → H , this means that also X ( F q ) = X ( F q ) . Therefore v P (˜ x (2) P ) = 0, since P
6∈ X ( F q ). Thisimplies that v P ( h s ) = ( s + 1) (cid:16) qv P (˜ x P ) − v P (˜ x (1) P ) (cid:17) = ( s + 1) q , as claimed.Now we able to determine several gaps of H ( P ). Proposition 4.3.
Let P
6∈ X ( F q ) be a point on X . Then G := { iq + j + kq + m ( q + 1) + q − X s =1 n s (( s + 1) q ) + 1 | i, j, k, m, n , . . . , n q − ∈ Z ≥ , and i ( q + 1) + jq + k ( q + 1) + mq ( q + 1) + q − X s =1 n s (( s + 1) q − s )( q + 1) ≤ ( q + 1)( q − } , is a set of gaps at P .Proof. Let i, j, k, m, n , . . . , n q − be nonnegative integers and write f = ˜ z iP ˜ y jP ˜ x kP g m Q q − s =1 h n s s . Equations(3), (4), (5) combined with Lemmas 4.1 and 4.2 imply that f ∈ L ((2 g ( X ) − P ∞ ) if i ( q + 1) + j ( q − q + q ) + k ( q + 1) + m ( q + q ) + q − X s =1 n s (( s + 1) q − s )( q + 1) ≤ q − q + q − , which is equivalent to i ( q + 1) + jq + k ( q + 1) + mq ( q + 1) + q − X s =1 n s (( s + 1) q − s )( q + 1) ≤ ( q + 1)( q − . (15)On the other hand we have v P ( f ) = iq + j + kq + m ( q + 1) + q − X s =1 n s (( s + 1) q ) . Hence the claim follows from Lemma 2.4.
Observation 4.4.
Inequality (16) implies in particular that i ≤ q − , j ≤ q + q − , k ≤ q − , m ≤ q − and n s ≤ ⌊ ( q + 1) / ( s + 1) ⌋ . This implies directly that the largest gap of H ( P ) that is contained in G isobtained by putting i = q − and all other remaining variables to . In other words: the largest element in G is q − q + 1 = 2 g ( X ) − q + 1 . bservation 4.5. If j ≥ q and the tuple ( i, j, k, m, n , . . . , n q − ) satisfies inequality (16) , then the tuple ( i, j − q, k + 1 , m, n , . . . , n s ) will also satisfy inequality (16) . This implies that when calculating the set G ,we may assume that j ≤ q − . Moreover, inequality (15) is equivalent to i + j + k + mq + q − X s =1 n s (( s + 1) q − s ) ≤ q − jq + 1 , which for j ≤ q − is equivalent to i + j + k + mq + q − X s =1 n s (( s + 1) q − s ) ≤ q − , (16) since all variables involved are integers. | G | = g ( X ) . We now prove that G is exactly the set of gaps G at P = P ( a,b,c )
6∈ X ( F q ), that is | G | = g ( X ). Since wealready know that G contains gaps of H ( P ), it is sufficient to show that | G | ≥ g ( X ). This will require adetailed study of the elements of G . To this end we consider the following map ϕ : Z q +2 ≥ → Z ≥ , with ϕ ( i, j, k, m, n , . . . , n q − ) = iq + j + kq + m ( q + 1) + q − X s =1 n s (( s + 1) q ) + 1 , and consider the set G = { ( i, j, k, m, n , . . . , n q − ) ∈ Z q +2 ≥ | j ≤ q − , inequality (16) holds } . Then by Observation 4.5 we have G = ϕ ( G ). The main difficulty is that ϕ (cid:12)(cid:12) G , the restriction of the map ϕ to G , is not injective. This makes estimating the cardinality of G somewhat tricky. We proceed by studyingthe image of ϕ on the following three subsets of G . G := { ( i, , k, m, , . . . , ∈ G} , G := { ( i, j, k, m, , . . . , ∈ G | ≤ j ≤ q − , k ≤ q − , j + m ≤ q − }G := { ( i, j, k, , . . . , , n s , , . . . , ∈ G | k ≤ q − , ≤ s ≤ q − , n s = 1 , i + k + ( s + 1) q ≥ q − } . Further, we write G = ϕ ( G ), G = ϕ ( G ) and G = ϕ ( G ). We will show that these sets are mutuallydisjoint and that their cardinalities add up to | G | in a series of lemmas. Lemma 4.6.
Let G and G = ϕ ( G ) be as above. Then ϕ restricted to G is injective and | G | = 12 q ( q − (cid:18) q + 56 q + 12 (cid:19) . Proof.
If ( i, , k, m, , . . . , ∈ G , then ϕ ( i, , k, m, , . . . ,
0) = iq + kq + m ( q + 1) + 1 and by inequality(16) i + k + mq ≤ q − . This implies in particular that0 ≤ m ≤ q − ≤ kq + m ( q + 1) ≤ ( k + mq ) q + q − ≤ ( q − q + q − < q . i , , k , m , , . . . , , ( i , , k , m , , . . . , ∈ G and i q + k q + m ( q + 1) = i q + k q + m ( q + 1) . Calculating modulo q and using that 0 ≤ m ≤ q − ≤ m ≤ q − m = m . Further, since 0 ≤ k q + m ( q + 1) < q and 0 ≤ k q + m ( q + 1) < q , wesee that k q + m ( q + 1) = k q + m ( q + 1) and i q = i q . Combining these equalities, we see that( i , , k , m , , . . . ,
0) = ( i , , k , m , , . . . , | G | . First of all, from the above we see that | G | = |G | . Further we have |G | = q − X m =0 q − − mq X i =0 q − − mq − i X k =0 q − X m =0 q − − mq X i =0 ( q − − mq − i )= q − X m =0 ( q − − mq )( q − mq )2 = ( q − q q − X m =0 − q − q + q m + (cid:18) m + 12 (cid:19) q = ( q − q − q − q + q (cid:18) q (cid:19) + (cid:18) q + 13 (cid:19) q . In the last equality we used summation on the upper index to evaluate the summation P m (cid:0) m +12 (cid:1) ; see [5,Eqn. (5.10)]. The desired equality for | G | now follows. Lemma 4.7.
Let G and G = ϕ ( G ) be as above. Then ϕ restricted to G is injective and | G | = 12 q ( q − (cid:18) q − q − (cid:19) . Proof.
If ( i, j, k, m, , . . . , ∈ G , then ϕ ( i, j, k, m, , . . . ,
0) = iq + j + kq + m ( q + 1) + 1 and by definitionwe have 1 ≤ j ≤ q −
1, 1 ≤ j + m ≤ q − ≤ k ≤ q −
1. Moreover, inequality (16) gives that i + j + k + mq ≤ q − . Similarly as in the previous lemma, we obtain that0 ≤ m ≤ q − ≤ j + kq + m ( q + 1) ≤ ( k + mq ) q + q − ≤ ( q − q + q − < q . Now suppose ( i , j , k , m , , . . . , , ( i , j , k , m , , . . . , ∈ G and i q + j + k q + m ( q + 1) = i q + j + k q + m ( q + 1) . Reasoning exactly as in the previous lemma, we obtain that j + m = j + m , j + k q + m ( q + 1) = j + k q + m ( q +1) and i = i . Combining the first two equations, we deduce that k q + m q = k q + m q .Since 0 ≤ k ≤ q − ≤ k ≤ q −
1, we see k = k , which now implies that ( i , j , k , m , , . . . ,
0) =( i , j , k , m , , . . . , . Now we compute | G | . First note that k ≤ q −
1, but for a given j and m , we also have k ≤ q − − j − mq .However, since j ≥ ≤ j + m ≤ q −
1, we see that m ≤ q −
2. Hence q − − j − mq ≥ q − − − ( q − q ≥ − , implying that the condition k ≤ q − − j − mq is trivially satisfied. Hence |G | = q − X j =1 q − − j X m =0 q − X k =0 q − − j − k − mq X i =0 q − X j =1 q − − j X m =0 q − X k =0 ( q − − j − k − mq )= q − X j =1 q − − j X m =0 ( q − − j − mq ) q − (cid:18) q (cid:19) = q − X j =1 (cid:18) ( q − − j ) q − (cid:18) q (cid:19)(cid:19) ( q − j ) − q (cid:18) q − j (cid:19) = q − X j =1 (cid:18) ( q − q ) q − (cid:18) q (cid:19)(cid:19) ( q − j ) − ( q − q ) (cid:18) q − j (cid:19) = (cid:18) ( q − q ) q − (cid:18) q (cid:19)(cid:19) (cid:18) q (cid:19) − ( q − q ) (cid:18) q (cid:19) . The desired equality now follows.
Lemma 4.8.
Let G and G = ϕ ( G ) be as above. Then ϕ restricted to G is injective and | G | = 12 q ( q − (cid:18) q − (cid:19) . Proof.
If ( i, j, k, , , . . . , , n s , , . . . , ∈ G , then ϕ ( i, j, k, , , . . . , , n s , , . . . ,
0) = iq + j + kq +( s +1) q +1and by definition we have n s = 1, 1 ≤ s ≤ q −
2, 0 ≤ j ≤ q −
1, 0 ≤ k ≤ q − i + k + ( s + 1) q ≥ q − i + k + sq ≥ q − q − i + j + k + s ( q − ≤ q − q − . Note that the inequalities i + k + sq ≥ q − q − i + j + k + s ( q − ≤ q − q − j ≤ s −
1, so we may assume this as well in the remainder of the proof.Now suppose ( i , j , k , , , . . . , , n s , , . . . , , ( i , j , k, , , . . . , , , , . . . , ∈ G and i q + j + k q + ( s + 1) q = i q + j + k q + ( s + 1) q . Since the q -ary expansion of a number is unique, we immediately obtain that j = j , k = k and s = s ,since all variables involved at between 0 and q −
1. Hence i = i as well and the first part of the lemmafollows.Now we compute | G | . Recall that we may assume j ≤ s −
1. Hence |G | = q − X s =1 s − X j =0 q − X k =0 q − q − − j − k − s ( q − X i = q − q − − k − sq q − X s =1 s − X j =0 q − X k =0 ( s − j )= q q − X s =1 s − X j =0 ( s − j ) = q q − X s =1 (cid:18) s + 12 (cid:19) = q (cid:18) q (cid:19) . The desired equality now follows.Finally to obtain an estimate for | G | , we need to study the intersections of the sets G , G and G . Itturns out that they are disjoint, as we will now show. Lemma 4.9.
The sets G , G and G defined above are mutually disjoint. roof. Part 1. G ∩ G = ∅ . Let ( i , , k , m , , . . . , ∈ G , ( i , j , k , m , , . . . , ∈ G and suppose that i q + k q + m ( q + 1) = i q + j + k q + m ( q + 1) . Since 0 ≤ m ≤ q − ≤ j + m ≤ q −
1, we see that m = j + m and hence that i q + k + m q = i q + k + m q. Note that m − m = j ≥
0, where the inequality follows from the definition of G . Inequality(16) implies that k + m q < q as well as k + m q < q . Hence we obtain i = i and k + m q = k + m q ,whence ( m − m ) q = k − k . This implies that k ≡ k (mod q ), but since k ≥ ≤ k ≤ q − k − k ≥
0. On the other hand we already have seen that m − m = j ≥
1, but then we arriveat a contradiction, since 0 < ( m − m ) q = k − k ≤ Part 2. G ∩ G = ∅ . Let ( i , , k , m , , . . . , ∈ G , ( i , j , k , , , . . . , , , , . . . , ∈ G and supposethat i q + k q + m ( q + 1) = i q + j + k q + ( s + 1) q . Similarly as in part 1 above, we obtain that m = j , whence i q + k + m q = i q + k + ( s + 1) q , as wellas the inequality k + m q < q . However, since k ≤ q − s + 1 ≤ q −
1, we also have k + ( s + 1) q < q .Therefore we obtain that i = i as well as k + m q = k + ( s + 1) q . This implies that i + k + ( s + 1) q = i + k + m q ≤ q − , where we have used inequality (16) to obtain the inequality. On the other hand i + k + ( s + 1) q ≥ q − G and we arrive at a contradiction. Part 3. G ∩ G = ∅ . Let ( i , j , k , m , , . . . , ∈ G , ( i , j , k , , , . . . , , , , . . . , ∈ G and supposethat i q + j + k q + m ( q + 1) = i q + j + k q + ( s + 1) q . Reasoning very similarly as in Part 1 and Part 2, we obtain j + m = j , i = i and i + k + ( s + 1) q = i + k + m q ≤ q − . Again we arrive at a constriction.We are now ready to prove the main theorem of this section.
Theorem 4.10.
Let P be a point of X with P
6∈ X ( F q ) . Then the set of gaps of H ( P ) is given by, G = { iq + kq + m ( q + 1) + q − X s =1 n s (( s + 1) q ) + j + 1 | i, j, k, m, n , . . . , n q − ∈ Z ≥ , j ≤ q − , and i + j + k + mq + q − X s =1 n s (( s + 1) q − s ) ≤ q − } . Moreover, the set of Weierstrass points W on X coincides with X ( F q ) .Proof. Combing Lemmas 4.6, 4.7, 4.8, and 4.9 we see that | G | ≥ | G | + | G | + | G | = 12 q ( q − q + q −
1) = g ( X ) . Since we know that H ( P ) has exactly g ( X ) gaps, Proposition 4.3 then implies that H ( P ) = N \ G . FromObservation 4.4, we deduce that the largest gap in H ( P ) is 2 g ( X ) − q + 1, while we already know that forany P ∈ X ( F q ), the largest gap is 2 g ( X ) −
1. This implies the last statement in the theorem.14he proof also shows that the gaps of H ( P ) are precisely G ∪ G ∪ G , which is convenient when checkingif a particular number is a gap or not. For example, this allows us to compute the multiplicity (smallestpositive element) of H ( P ) fairly easily. Corollary 4.11.
Let P be a point of X with P
6∈ X ( F q ) . The multiplicity of H ( P ) is equal to q − .Proof. From St¨ohr-Voloch Theory we know that q − q are non-gaps at P , since P is not a Weierstrasspoint; see [6, Proposition 10.9]. It is also not difficult to verify this directly. On the other hand, let1 ≤ a ≤ q − a − c + c q + c q with 0 ≤ c t ≤ q − t = 1 , ,
3. Then wedistinguish three cases.
Case 1. c ≥ c and ( c , c ) = ( q − , q − . In this case a direct verification shows that a = ϕ (0 , , c +( c − c ) q, c , . . . ,
0) and that (0 , , c + ( c − c ) q, c , . . . , ∈ G . Case 2. c < c . We have a = ϕ (0 , c − c , c , c , . . . ,
0) and (0 , c − c , c , c , . . . , ∈ G in this case. Case 3. ( c , c ) = ( q − , q −
1) Note that in this case c ≤ q − , since a − c +( q − q +( q − q ≤ q − a = ϕ (0 , c , q − , , , . . . , ,
1) and that (0 , c , q − , , , . . . , , ∈ G .At this point seems to be reasonable to ask for the generators of the Weierstrass semigroup H ( P ) for P
6∈ X ( F q ). Their explicit determination seems to be a challenging task as the following examples show. Inparticular the number of generators of H ( P ) seems to grow quickly with respect to q . Example 4.12.
Let P ∈ X such that P
6∈ X ( F q ) . • If q = 2 then g = 10 and G = { , , , , , , , , , } . Clearly and must be generators of H ( P ) and since
6∈ h , i and
6∈ h , , i we obtain thatalso and are generators. Note that h , , , i ∩ { , . . . , } = { , , , , , , } and hencealso is a generator. In fact H ( P ) = h , , , , i . Moreover, if P ∈ X then H ( P ) = { , , , , , , , , , , , . . . } , if P ∈ X ( F ) , { , , , , , , , , , , , . . . } , if P ∈ X ( F ) \ X ( F ) , { , , , , , , , , , , . . . } , otherwise . • If q = 3 then g = 99 and G = { , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , } . Arguing as for the previous case, one can prove that H ( P ) = h , , , , , , , , , , , , , , i . It is unclear what the number of generators for general q is. For q = 4 the semigroup turns out to have generators. X in a table, leaving a questionmark for the minimal number of generators in the case P
6∈ X ( F q ). Determining this number could beinteresting future work. P multiplicity conductor number of generators P ∈ X ( F q ) q − q + q g ( X ) − P ∈ X ( F q ) \ X ( F q ) q − q + 1 2 g ( X ) − q + 2 P
6∈ X ( F q ) q − g ( X ) − q + 1 ? Acknowledgments
The first author gratefully acknowledges the support from The Danish Council for Independent Research(Grant No. DFF–4002-00367). The second author would like to thank the Italian Ministry MIUR, StruttureGeometriche, Combinatoria e loro Applicazioni, Prin 2012 prot. 2012XZE22K and GNSAGA of the ItalianINDAM.
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