Large Automorphism Groups of Algebraic Curves in Positive Characteristic. BMS-LMS Conference

Transcription

1 Large Automorphism Groups of Algebraic Curves in Positive Characteristic Massimo Giulietti (Università degli Studi di Perugia) BMS-LMS Conference December 4-5, 2009 Leuven

2 Notation and Terminology K algebraically closed field of characteristic p. X projective, non-singular, geometrically irreducible, algebraic curve X of genus g.

3 Notation and Terminology K algebraically closed field of characteristic p. X projective, non-singular, geometrically irreducible, algebraic curve X of genus g. Aut(X ) = {φ : X X φ birational}

4 Notation and Terminology K algebraically closed field of characteristic p. X projective, non-singular, geometrically irreducible, algebraic curve X of genus g. Aut(X ) = {φ : X X φ birational} Any curve X is birationally equivalent to a plane curve F (X, Y ) = 0

5 Quotient Curves Let G < Aut(X ), G finite

6 Quotient Curves Let G < Aut(X ), G finite There is a curve X /G defined over K, and a separable morphism φ G : X X /G with φ G (P) = φ G (Q) Q Orb G (P) Definition X /G is the quotient curve of X by G.

7 How many automorphisms?

8 How many automorphisms? If g 2, then Aut(X ) is a finite group, [Schmid (1938), Iwasawa-Tamagawa (1951), Roquette (1952), Rosentlich (1955), Garcia (1993)].

9 How many automorphisms? If g 2, then Aut(X ) is a finite group, [Schmid (1938), Iwasawa-Tamagawa (1951), Roquette (1952), Rosentlich (1955), Garcia (1993)]. Hurwitz bound (1892): if p = 0 and g 2, Aut(X ) 84(g 1)

10 How many automorphisms? If g 2, then Aut(X ) is a finite group, [Schmid (1938), Iwasawa-Tamagawa (1951), Roquette (1952), Rosentlich (1955), Garcia (1993)]. Hurwitz bound (1892): if p = 0 and g 2, Aut(X ) 84(g 1) Example Klein quartic: K : X 3 + Y + XY 3 = 0. g = 3, Aut(K) = PSL(2, 7), Aut(K) = 168 = 84(3 1).

11 How many automorphisms? If g 2, then Aut(X ) is a finite group, [Schmid (1938), Iwasawa-Tamagawa (1951), Roquette (1952), Rosentlich (1955), Garcia (1993)]. Hurwitz bound (1892): if p = 0 and g 2, Aut(X ) 84(g 1) Example Klein quartic: K : X 3 + Y + XY 3 = 0. g = 3, Aut(K) = PSL(2, 7), Aut(K) = 168 = 84(3 1). If gcd(p, Aut(X ) ) = 1 then Aut(X ) 84(g 1).

12 What if p divides Aut(X )? Example Hermitian curve: H(n) : X n+1 = Y n + Y, n = p h. g = 1 2 n(n 1), Aut(H) = PGU(3, n) = (n3 + 1)(n 2 1)(n 3 )

13 What if p divides Aut(X )? Example Hermitian curve: H(n) : X n+1 = Y n + Y, n = p h. g = 1 2 n(n 1), Aut(H) = PGU(3, n) = (n3 + 1)(n 2 1)(n 3 ) Let G = Aut(X ). Let G Q be the stabilizer of a point Q X under the action of G. If p G Q, then G Q has a unique Sylow p-subgroup.

14 What if p divides Aut(X )? Example Hermitian curve: H(n) : X n+1 = Y n + Y, n = p h. g = 1 2 n(n 1), Aut(H) = PGU(3, n) = (n3 + 1)(n 2 1)(n 3 ) Let G = Aut(X ). Let G Q be the stabilizer of a point Q X under the action of G. If p G Q, then G Q has a unique Sylow p-subgroup. Let { the Sylow p subgroup of GQ, if p G S Q = Q {1}, otherwise

15 What if p divides Aut(X )? Example Hermitian curve: H(n) : X n+1 = Y n + Y, n = p h. g = 1 2 n(n 1), Aut(H) = PGU(3, n) = (n3 + 1)(n 2 1)(n 3 ) Let G = Aut(X ). Let G Q be the stabilizer of a point Q X under the action of G. If p G Q, then G Q has a unique Sylow p-subgroup. Let { the Sylow p subgroup of GQ, if p G S Q = Q {1}, otherwise For H(n), Q = Y G Q = { X = ax + b Y = a n+1 Y + bx + c } a n2 1 = 1, b n2 = b c n + c = b n+1

16 The role of p-groups Theorem (Stichtenoth, 1973) If Aut(X ) 8g 3, then S Q > p p 1g for some Q X.

17 The role of p-groups Theorem (Stichtenoth, 1973) If Aut(X ) 8g 3, then S Q > p p 1g for some Q X. Definition Linear Stichtenoth bound S Q p p 1 g

18 Classification results Theorem (Stichtenoth, 1973) Aut(X ) 16g 4, unless X is a Hermitian curve.

19 Classification results Theorem (Stichtenoth, 1973) Aut(X ) 16g 4, unless X is a Hermitian curve. Theorem (Henn, 1978) If g 2 and Aut(X ) > 8g 3, then one of the following occurs for X :

20 Classification results Theorem (Stichtenoth, 1973) Aut(X ) 16g 4, unless X is a Hermitian curve. Theorem (Henn, 1978) If g 2 and Aut(X ) > 8g 3, then one of the following occurs for X : (A) p = 2, X non-singular model of Y 2 + Y = X 2k +1, k > 1

21 Classification results Theorem (Stichtenoth, 1973) Aut(X ) 16g 4, unless X is a Hermitian curve. Theorem (Henn, 1978) If g 2 and Aut(X ) > 8g 3, then one of the following occurs for X : (A) p = 2, X non-singular model of Y 2 + Y = X 2k +1, k > 1 (B) p > 2, X non-singular model of Y 2 = X n X, n = p h, h > 0

22 Classification results Theorem (Stichtenoth, 1973) Aut(X ) 16g 4, unless X is a Hermitian curve. Theorem (Henn, 1978) If g 2 and Aut(X ) > 8g 3, then one of the following occurs for X : (A) p = 2, X non-singular model of Y 2 + Y = X 2k +1, k > 1 (B) p > 2, X non-singular model of Y 2 = X n X, n = p h, h > 0 (C) Hermitian curve H(n)

23 Classification results Theorem (Stichtenoth, 1973) Aut(X ) 16g 4, unless X is a Hermitian curve. Theorem (Henn, 1978) If g 2 and Aut(X ) > 8g 3, then one of the following occurs for X : (A) p = 2, X non-singular model of Y 2 + Y = X 2k +1, k > 1 (B) p > 2, X non-singular model of Y 2 = X n X, n = p h, h > 0 (C) Hermitian curve H(n) (D) Suzuki curve S(n): p = 2, X non-singular model of X n 0 (X n + X ) = Y n + Y, n 0 = 2 r, r 1, n = 2n 2 0

24 Relationship with p-rank Definition The p-rank γ of a curve X is the integer such that p γ = {Q J(X ) [p]q = 0}.

25 Relationship with p-rank Definition The p-rank γ of a curve X is the integer such that p γ = {Q J(X ) [p]q = 0}. In general, γ g. If equality holds, then X is said to be ordinary.

26 Relationship with p-rank Definition The p-rank γ of a curve X is the integer such that p γ = {Q J(X ) [p]q = 0}. In general, γ g. If equality holds, then X is said to be ordinary. Theorem (Nakajima, 1987) If X is ordinary, then Aut(X ) 84(g 2 g).

27 Relationship with p-rank Definition The p-rank γ of a curve X is the integer such that p γ = {Q J(X ) [p]q = 0}. In general, γ g. If equality holds, then X is said to be ordinary. Theorem (Nakajima, 1987) If X is ordinary, then Aut(X ) 84(g 2 g). Let S be a p-subgroup of Aut(X ). If then γ = 0. S > 2p p 1 g

28 Problems Problem (1) Classifying curves for which Stichtenoth/Nakajima bound on p-groups fails.

29 Problems Problem (1) Classifying curves for which Stichtenoth/Nakajima bound on p-groups fails. Problem (2) Classifying curves with γ = 0.

30 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action.

31 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and

32 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ

33 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ If S 4g 2 (p 1) 2, then

34 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ If S 4g 2 (p 1) 2, then S = Z/pZ

35 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ If S 4g 2 (p 1) 2, then S = Z/pZ X = X F : Y p Y = XF (X ) + cx, F additive

36 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ If S 4g 2 (p 1) 2, then S = Z/pZ X = X F : Y p Y = XF (X ) + cx, F additive If S 4g 2 (p 2 1) 2, then

37 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ If S 4g 2 (p 1) 2, then S = Z/pZ X = X F : Y p Y = XF (X ) + cx, F additive If S 4g 2 (p 2 1) 2, then S = (Z/pZ) n, 1 n 3

38 Problem (1): Big actions Definition (Lehr-Matignon, 2005) Let S be a p-subgroup of Aut(X ). If S > 2p p 1g then (X, S) is a big action. Theorem (Lehr-Matignon, Matignon-Rocher,Rocher, ) Let (X, S) be a big action. Then S S Q for some Q X, S S and S is not cyclic unless S = Z/pZ If S 4g 2 (p 1) 2, then S = Z/pZ X = X F : Y p Y = XF (X ) + cx, F additive If S 4g 2 (p 2 1) 2, then S = (Z/pZ) n, 1 n 3 Characterization given

39 Classification results Remark If Aut(X ) fixes a point, then Aut(X ) is the semidirect product of a p-group and a cyclic group of order m with (m, p) = 1.

40 Classification results Remark If Aut(X ) fixes a point, then Aut(X ) is the semidirect product of a p-group and a cyclic group of order m with (m, p) = 1. Theorem (G.-Korchmáros, 2009) p If g 2, S Q > p 1g for some Q X, and Aut(X ) fixes no point, then one of the following occurs for X : (B),(C),(D) in Henn s classification result (E) Ree curve R(n): p = 3, X non-singular model of the complete intersection of Y n Y = X n 0 (X n X ), Z n Z = X n 0 (Y n Y ), for n 0 = 3 r, r 0, n = 3n 2 0.

41 Sketch of the proof, I Let G := Aut(X ). Set Ω := {Q X S Q non-trivial}.

42 Sketch of the proof, I Let G := Aut(X ). Set Ω := {Q X S Q non-trivial}. Step 1: G acts on Ω as a 2-transitive permutation group.

43 Sketch of the proof, I Let G := Aut(X ). Set Ω := {Q X S Q non-trivial}. Step 1: G acts on Ω as a 2-transitive permutation group. Step 2: The 2-point stabilizer of the permutation group Ḡ induced by G on Ω is cyclic.

44 Sketch of the proof, I Let G := Aut(X ). Set Ω := {Q X S Q non-trivial}. Step 1: G acts on Ω as a 2-transitive permutation group. Step 2: The 2-point stabilizer of the permutation group Ḡ induced by G on Ω is cyclic. Step 3: The Kantor-O Nan-Seitz theorem: Let Ḡ be a finite 2-transitive permutation group whose 2-point stabiliser is cyclic. Then Ḡ has either a regular normal subgroup, or Ḡ is one of the following groups in their natural 2-transitive permutation representations: PSL(2, n), PGL(2, n), PSU(3, n), PGU(3, n), Sz(n), Ree(n).

45 Sketch of the proof, II Step 4: Actually Ḡ has no regular normal subgroup, this is shown using Huppert s classification theorem: Let Ḡ be a solvable 2-transitive permutation group of even degree n. Then n is a power of 2, and Ḡ is a subgroup of the affine semi-linear group AΓL(1, n).

46 Sketch of the proof, II Step 4: Actually Ḡ has no regular normal subgroup, this is shown using Huppert s classification theorem: Let Ḡ be a solvable 2-transitive permutation group of even degree n. Then n is a power of 2, and Ḡ is a subgroup of the affine semi-linear group AΓL(1, n). Step 5: For the groups PSL(2, n), PGL(2, n), PSU(3, n), PGU(3, n), Sz(n), Ree(n), the possible genera are computed. For each group at most one genus is consistent with S Q > p p 1g. Then Henn s classification applies.

47 Remark A similar classification under a bit weakened hypothesis than S Q > p p 1g may involve more curves, as suggested by the non-singular model of Y 3(p2 p+1) = X p3 + X (X p + X ) p2 p+1, p > 2, 3 p 2 for which g = p (p2 p) 1, S Q = p 3

48 Problem (2): zero p-rank curves Theorem (Deuring-Shafarevich formula) Let S be a p-subgroup of Aut(X ). Then γ 1 = S (γ 1) + where k ( S l i ) i=1 γ is the p-rank of X, γ is the p-rank of X /S, l i are the lengths of the short orbits of S.

49 Problem (2): zero p-rank curves Theorem (Deuring-Shafarevich formula) Let S be a p-subgroup of Aut(X ). Then where Corollary γ 1 = S (γ 1) + k ( S l i ) i=1 γ is the p-rank of X, γ is the p-rank of X /S, l i are the lengths of the short orbits of S. If γ = 0, then every Sylow p-subgroup of Aut(X ) fixes exactly one point of X,

50 Problem (2): zero p-rank curves Theorem (Deuring-Shafarevich formula) Let S be a p-subgroup of Aut(X ). Then where Corollary γ 1 = S (γ 1) + k ( S l i ) i=1 γ is the p-rank of X, γ is the p-rank of X /S, l i are the lengths of the short orbits of S. If γ = 0, then every Sylow p-subgroup of Aut(X ) fixes exactly one point of X, any two distinct Sylow p-subgroups of Aut(X ) have trivial intersection.

51 Trivial Intersection of Sylow p-subgroups Theorem (Burnside-Gow) Let G be a finite solvable group. If Sylow p-subgroups have trivial intersection, then one of the following occurs for any Sylow p-subgroup S P : S p is normal, S p is cyclic, p = 2 and S 2 is a generalized quaternion group.

52 Trivial Intersection of Sylow p-subgroups Theorem (Burnside-Gow) Let G be a finite solvable group. If Sylow p-subgroups have trivial intersection, then one of the following occurs for any Sylow p-subgroup S P : S p is normal, S p is cyclic, p = 2 and S 2 is a generalized quaternion group. Finite groups whose Sylow 2-subgroups have trivial intersection were classified by Suzuki, Shult and Hering

53 zero 2-rank curves From now on γ = 0, p = 2

54 zero 2-rank curves From now on γ = 0, p = 2 Lemma (G.-Korchmáros, 2009) If G = Aut(X ) does not fix a point, then one of the following cases occur (a) G is solvable. A Sylow 2-subgroup S Q of G is either a cyclic group or a generalized quaternion group. (b) The commutator subgroup G of G is isomorphic to one of the groups: with n = 2 r 4. PSL(2, n), PSU(3, n), SU(3, n), Sz(n)

55 Large automorphism groups of zero 2-rank curves Theorem (G.-Korchmáros, 2009) Let p = 2, g 2 and γ = 0. Assume that G = Aut(X ) fixes no point of X.

56 Large automorphism groups of zero 2-rank curves Theorem (G.-Korchmáros, 2009) Let p = 2, g 2 and γ = 0. Assume that G = Aut(X ) fixes no point of X. If G is solvable, then the Hurwitz bound holds for G; more precisely G 72(g 1).

57 Large automorphism groups of zero 2-rank curves Theorem (G.-Korchmáros, 2009) Let p = 2, g 2 and γ = 0. Assume that G = Aut(X ) fixes no point of X. If G is solvable, then the Hurwitz bound holds for G; more precisely G 72(g 1). If G is not solvable and G 24g(g 1), then G is known and the possible genera of X are computed from the order of G.

58 Large automorphism groups of zero 2-rank curves Theorem (G.-Korchmáros, 2009) Let p = 2, g 2 and γ = 0. Assume that G = Aut(X ) fixes no point of X. Remark If G is solvable, then the Hurwitz bound holds for G; more precisely G 72(g 1). If G is not solvable and G 24g(g 1), then G is known and the possible genera of X are computed from the order of G. In case (b), for each group G the situation is investigated carefully by using both ramification theory and classification of subgroups of those linear groups.

59 List of g when G 24g(g 1) (i) G = PSL(2, n) with n = 2 h, h 3 and g = 1 2 (t 1)(n 1) with t (n + 1). (ii) G = PSU(3, n); and either g = 1 2 (n 1)(t(n + 1)2 (n 2 + n + 1)) with t (n 2 n + 1)/µ, and µ = gcd(3, n + 1), or with t (n + 1); g = 1 2 (n 1) ( t(n 3 + 1) µ ) (n 2 + n + 1), Remark in the former case, t = 1 only occurs when X is isomorphic to H(n).

60 List of g when G 24g(g 1), cont. (iii) G = Sz(n), n = 2n0 2, n 0 = 2 h, h 1; either g = 1 2 [(t 1)(n2 1) 2tn 0 (n 1)] with t (n + 2n 0 + 1), or g = 1 2 [(t 1)(n2 1) + 2tn 0 (n 1)] with t (n 2n 0 + 1); Remark in the latter case, t = 1 only occurs when X is isomorphic to S(n).

61 List of g when G 24g(g 1), cont. (iv) G = SU(3, n) with 3 (n + 1) and either g = 1 2 (n 1)[3t(n + 1)2 (n 2 + n + 1)] with t (n 2 n + 1)/3, or g = 1 2 (n 1) ( t(n 3 + 1) (n 2 + n + 1) ), with t (n + 1);

62 Open problems Find examples for the following cases: (iii), t > 1 (iv)

63 Open problems Find examples for the following cases: (iii), t > 1 (iv) solved!

64 Open problems Find examples for the following cases: (iii), t > 1 (iv) solved! Find similar results in odd characteristic.

65 Open problems Find examples for the following cases: (iii), t > 1 (iv) solved! Theorem Find similar results in odd characteristic. Let g 2 and γ = 0. Assume that G = Aut(X ) fixes no point of X. If G is solvable and p 2 G, then G 24g(g 1).

66 Work in progress for γ > 0 Nakajima s bound for p = 2: S 4(g 1)

67 Work in progress for γ > 0 Nakajima s bound for p = 2: S 4(g 1) Case under investigation 2g 1 < S 4(g 1)

68 Theorem (G.-Korchmáros) Let 8 2g 1 < S 4(g 1). If S does not fix a point, then one of the following cases occurs:

69 Theorem (G.-Korchmáros) Let 8 2g 1 < S 4(g 1). If S does not fix a point, then one of the following cases occurs: (i) For every central involution u S, the genus ḡ of the quotient curve X = X / u satisfies the condition S > 2(ḡ 1) where S = S/ u is the subgroup of Aut( X ) induced by S on X, and S has no fixed point on X.

70 Theorem (G.-Korchmáros) Let 8 2g 1 < S 4(g 1). If S does not fix a point, then one of the following cases occurs: (i) For every central involution u S, the genus ḡ of the quotient curve X = X / u satisfies the condition S > 2(ḡ 1) where S = S/ u is the subgroup of Aut( X ) induced by S on X, and S has no fixed point on X. (ii) S = 4(g 1) and X is an ordinary, bielliptic curve. Furthermore, either (iia) S is dihedral and has no inductive central involution; or (iib) S = (E U) V where E is cyclic group of order g 1 and U = V = 2. The factor group S/U is a dihedral, and the involution of E is the unique central inductive involution of S.

71 Theorem (G.-Korchmáros) Let 8 2g 1 < S 4(g 1). If S does not fix a point, then one of the following cases occurs: (i) For every central involution u S, the genus ḡ of the quotient curve X = X / u satisfies the condition S > 2(ḡ 1) where S = S/ u is the subgroup of Aut( X ) induced by S on X, and S has no fixed point on X. (ii) S = 4(g 1) and X is an ordinary, bielliptic curve. Furthermore, either (iia) S is dihedral and has no inductive central involution; or (iib) S = (E U) V where E is cyclic group of order g 1 and U = V = 2. The factor group S/U is a dihedral, and the involution of E is the unique central inductive involution of S. (iii) S = 2g + 2 and X is ordinary. Also, S = D E, with D elementary abelian of index 2 and E = 2. If S is abelian, then it is elementary abelian and X is hyperelliptic.

72 Maximal curves Theorem (Hasse-Weil bound) Let X be a curve defined over F q. The number N of F q -rational points of X satisfies N q g q. Definition X is F q -maximal if equality holds

73 Maximal curves Theorem (Hasse-Weil bound) Let X be a curve defined over F q. The number N of F q -rational points of X satisfies N q g q. Definition X is F q -maximal if equality holds Example H(n) is F q -maximal for q = n 2m, m 1 odd S(n) is F q -maximal for q = n 4m, m 1 odd R(n) is F q -maximal for q = n 6m, m 1 odd

74 Maximal curves Theorem (Hasse-Weil bound) Let X be a curve defined over F q. The number N of F q -rational points of X satisfies N q g q. Definition X is F q -maximal if equality holds Example H(n) is F q -maximal for q = n 2m, m 1 odd S(n) is F q -maximal for q = n 4m, m 1 odd R(n) is F q -maximal for q = n 6m, m 1 odd Theorem An F q -maximal curve has zero p-rank.

75 Classification results largest genus: 1 2 q( q 1) (Ihara, 1981)

76 Classification results largest genus: 1 2 q( q 1) (Ihara, 1981) if g = 1 2 q( q 1), then X = H( q) (Rück-Stichtenoth, 1994)

77 Classification results largest genus: 1 2 q( q 1) (Ihara, 1981) if g = 1 2 q( q 1), then X = H( q) (Rück-Stichtenoth, 1994) second largest genus: 1 4 ( q 1) 2 (Fuhrmann-Torres, 1996)

78 Classification results largest genus: 1 2 q( q 1) (Ihara, 1981) if g = 1 2 q( q 1), then X = H( q) (Rück-Stichtenoth, 1994) second largest genus: 1 4 ( q 1) 2 (Fuhrmann-Torres, 1996) if g(x ) = 1 4 ( q 1) 2, then

79 Classification results largest genus: 1 2 q( q 1) (Ihara, 1981) if g = 1 2 q( q 1), then X = H( q) (Rück-Stichtenoth, 1994) second largest genus: 1 4 ( q 1) 2 (Fuhrmann-Torres, 1996) if g(x ) = 1 4 ( q 1) 2, then for q odd X is birationally equivalent over F q to Y (Fuhrmann-Garcia-Torres, 1997) q+1 2 = X q + X

80 Classification results largest genus: 1 2 q( q 1) (Ihara, 1981) if g = 1 2 q( q 1), then X = H( q) (Rück-Stichtenoth, 1994) second largest genus: 1 4 ( q 1) 2 (Fuhrmann-Torres, 1996) if g(x ) = 1 4 ( q 1) 2, then for q odd X is birationally equivalent over F q to Y q+1 2 = X q + X (Fuhrmann-Garcia-Torres, 1997) for q 16 even X is birationally equivalent over F q to Y q+1 = X + X 2 + X 4 q q X 4 + X 2 (Abdón-Torres 1999 for q = 16, Korchmáros-Torres 2002 for q > 16)

81 Classification results g = 0

82 Classification results g = 0 g = 1 Number of isomorphism classes ( p (Schoof, 1987) 1 12 ( 3 p ) 3 ( )) 4 p

83 Classification results g = 0 g = 1 Number of isomorphism classes ( p (Schoof, 1987) 1 12 ( 3 p ) 3 ( )) 4 p Problem Determine the spectrum of genera g for which an F q -maximal curve of genus g exists.

84 Problem Theorem (Serre, 1987) If X is F q -maximal and φ : X Y is a non-constant morphism defined over F q, then Y is F q -maximal Question Does there exist an F q -maximal curve not covered by an Hermitian, Suzuki or Ree curve?

85 A nice curve (iv) G = SU(3, n) with 3 (n + 1) and either g = 1 2 (n 1)[3t(n + 1)2 (n 2 + n + 1)] with t (n 2 n + 1)/3, or g = 1 2 (n 1) ( t(n 3 + 1) (n 2 + n + 1) ), with t (n + 1);

86 A nice curve (iv) G = SU(3, n) with 3 (n + 1) and either g = 1 2 (n 1)[3t(n + 1)2 (n 2 + n + 1)] with t (n 2 n + 1)/3, or with t (n + 1); Proposition g = 1 2 (n 1) ( t(n 3 + 1) (n 2 + n + 1) ), In the latter case, t = n + 1 occurs when X is birationally equivalent to the curve X : Y n3 +1 (X n3 + X (X n + X ) n2 n+1 ) = 0.

87 Properties X is F n 6-maximal, also for p > 2 (G.-Korchmáros)

88 Properties X is F n 6-maximal, also for p > 2 (G.-Korchmáros) For n > 2, X is the first known example of a maximal curve which is not covered by the Hermitian, Suzuki or Ree curve (G.-Korchmáros)

89 Properties X is F n 6-maximal, also for p > 2 (G.-Korchmáros) For n > 2, X is the first known example of a maximal curve which is not covered by the Hermitian, Suzuki or Ree curve (G.-Korchmáros) X has a large variety of quotient curves producing many new genera of maximal curves (G.-Fanali)

90 Properties X is F n 6-maximal, also for p > 2 (G.-Korchmáros) For n > 2, X is the first known example of a maximal curve which is not covered by the Hermitian, Suzuki or Ree curve (G.-Korchmáros) X has a large variety of quotient curves producing many new genera of maximal curves (G.-Fanali) The linear codes arising from X sometimes have better parameters than previously known ones (G.-Fanali)

91 A(nother) characterization of the Hermitian curve Remark Aut(H(n)) is transitive on the set of its F n 2-rational points.

92 A(nother) characterization of the Hermitian curve Remark Aut(H(n)) is transitive on the set of its F n 2-rational points. Theorem (G.-Korchmáros) Let p = 2. Let X be an F q -maximal curve of genus g 2. Then Aut(X ) acts on the set of its F q -rational points as a transitive permutation group if and only if X is the Hermitian curve H( q).

93 References M. G. and S. Fanali, Quotient curves of the GK curve, arxiv: M. G. and G. Korchmáros, Algebraic curves with a large non-tame automorphism group fixing no point, Trans. Am. Math. Soc., to appear. M. G. and G. Korchmáros, Automorphism groups of algebraic curves with p-rank zero, J. London Math. Soc., to appear. M. G. and G. Korchmáros, A new family of maximal curves over a finite field, Math. Ann., 343 (2009), C. Lehr and M. Matignon, Automorphism groups for p-cyclic covers of the affine line, Compositio Math. 141 (2005), M. Matignon and M. Rocher, On smooth curves endowed with a large automorphism p-group in characteristic p > 0, Algebra & Number Theory 2 (2008), M. Rocher, Large p-groups actions with a p-elementary abelian second ramification group, J. Alg. 321 (2009), M. Rocher, Large p-groups actions with G /g 2 > 4/(p 2 1) 2, arxiv: