Algebraic equations of genus 2 Riemann surfaces with a hyperelliptic uniform dessin

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1 Algebraic equations of genus 2 Riemann surfaces with a hyperelliptic uniform dessin Ernesto Girondo (joint with David Torres) Automorphisms of Riemann Surfaces and related topics University Park, 2009

2 29 surfaces After David s talk:

3 29 surfaces After David s talk: Theorem.- There are exactly 29 isomorphy classes of genus 2 surfaces that contain a uniform dessin with only one face.

4 29 surfaces After David s talk: Theorem.- There are exactly 29 isomorphy classes of genus 2 surfaces that contain a uniform dessin with only one face. (Equivalently: genus 2 surfaces admitting a Belyĭ function with exactly one pole and constant ramification index on each fibre)

5 29 surfaces After David s talk: Theorem.- There are exactly 29 isomorphy classes of genus 2 surfaces that contain a uniform dessin with only one face. (Equivalently: genus 2 surfaces admitting a Belyĭ function with exactly one pole and constant ramification index on each fibre) A dream: equations?

6 29 Fuchsian groups waiting for equations

7 29 Fuchsian groups waiting for equations

8 29 Fuchsian groups waiting for equations

9 29 Fuchsian groups waiting for equations

10 29 Fuchsian groups waiting for equations

11 29 Fuchsian groups waiting for equations

12 Few immediate answers

13 Few immediate answers Easiest case: when the automorphism group uniquely determines the surface.

14 Few immediate answers For instance for S 1 :

15 Few immediate answers For instance for S 1 : The existence of an order 10 automorphism σ 10 with σ 5 10 = J guarantees (Schiller) that y 2 = x 5 1 is an algebraic model for S 1.

16 Few immediate answers Also for S 7 :

17 Few immediate answers Also for S 7 : There is now an order 8 automorphism σ 8 with σ8 4 = J and then y 2 = x(x 4 1) is an algebraic model for S 7.

18 Few immediate answers As for S 17 :

19 Few immediate answers As for S 17 : In this case the existence of an order 6 automorphism σ 6 with σ 3 6 J guarantees (Schiller) that y 2 = x 6 1 is an algebraic model for S 17.

20 Two more For another two of our surfaces an equation has been found in the literature.

21 Two more Fuertes and Mednykh gave recently the model y 2 = x x for S 2 :

22 Two more And the equation y 3 = (x 1)(x 3 1) for S 23 (as a cyclic 3-gonal curve) appears in a paper by Singerman and Syddall:

23 Two more And the equation y 3 = (x 1)(x 3 1) for S 23 (as a cyclic 3-gonal curve) appears in a paper by Singerman and Syddall: (By the way, we have found that y 2 = x 6 + 8x is a hyperelliptic model for this curve)

24 An affordable situation Let S be uniformized by K, where K < so that S D K [z] K β D Ĉ [z] is a Belyi function, and D = β 1 ([0, 1]) is a uniform dessin.

25 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. D K π D K, J β F D

26 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. S β π D F K, J D

27 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. S β π D F Ĉ K, J

28 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. S β π S J Ĉ F Ĉ

29 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. { y 2 = } 6 i=1 (x a i) β π S J Ĉ F Ĉ

30 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. { y 2 = } 6 i=1 (x a i) β π F x Ĉ

31 An affordable situation Let J be a lift of the hyperelliptic involution to the universal cover D, and assume J. Then the chain of inclusions K < K, J < induces the following commutative diagram of morphisms of Riemann surfaces. { y 2 = } 6 i=1 (x a i) β π F x F (x) = β(x, y)

32 An affordable situation The identity β = F π shows that F is a rational function with half the degree of β. The values a 1,..., a 6 are characterized as the branching values of π, therefore knowledge of the branching data of β (encoded in the dessin) can give information about them.

33 Example 1: S12

34 The case of S 12 (K < (2, 4, 12)) The order 4 rotation induces an automorphism τ and τ 2 = J.

35 The case of S 12 (K < (2, 4, 12)) A fundamental domain for K, J.

36 The case of S 12 (K < (2, 4, 12)) The six Weierstrass points.

37 The case of S 12 (K < (2, 4, 12)) Side-pairings in the fundamental domain of K, J.

38 The case of S 12 (K < (2, 4, 12)) 1 0 Normalization of three points of D/ K, J Ĉ.

39 The case of S 12 (K < (2, 4, 12)) 1 0 The real line in D/ K, J Ĉ.

40 The case of S 12 (K < (2, 4, 12)) 1 C b B a A 0 Further relevant points: 0 < A < a < B < b < C < 1. The equation of S 12 is thus y 2 = x(x 1)(x A)(x B)(x C).

41 The case of S 12 (K < (2, 4, 12)) π 1 C b B a A 0 F Ĉ From the local structure of β = F π we deduce: F (x) = kx(x 1)(x a) 2 (x b) 2 F (x) 1 = k(x A) 2 (x B) 2 (x C) 2.

42 The case of S 12 (K < (2, 4, 12)) π 1 C b B a A 0 F Ĉ The order four automorphism τ induces an involution ˆτ in S 12 / J sending {0,, a, A, B} to {1,, b, C, B} respectively. Thus ˆτ(x) = 1 x, B = 1/2, b = 1 a, C = 1 A.

43 The case of S 12 (K < (2, 4, 12)) π 1 A 1 1 a 1/2 a A 0 Ĉ The order four automorphism τ induces an involution ˆτ in S 12 / J sending {0,, a, A, B} to {1,, b, C, B} respectively. Thus ˆτ(x) = 1 x, B = 1/2, b = 1 a, C = 1 A. F

44 The case of S 12 (K < (2, 4, 12)) π 1 A 1 1 a 1/2 a A 0 F Ĉ We have therefore obtained the two expressions: F (x) = kx(x 1)(x a) 2 (x 1 + a) 2 F (x) 1 = k(x 1/2) 2 (x A) 2 (x 1 + A) 2.

45 The case of S 12 (K < (2, 4, 12)) (1) F (x) = kx(x 1)(x a) 2 (x 1 + a) 2 (2) F (x) 1 = k(x 1/2) 2 (x A) 2 (x 1 + A) 2

46 The case of S 12 (K < (2, 4, 12)) (1) F (x) = kx(x 1)(x a) 2 (x 1 + a) 2 (2) F (x) 1 = k(x 1/2) 2 (x A) 2 (x 1 + A) 2 Differentiating in (2) we get F (x) = 2k(x 1/2)(x A)(x 1 + A)(3x 2 3x + (A A )), where the roots of the last factor must be a and 1 a.

47 The case of S 12 (K < (2, 4, 12)) (1) F (x) = kx(x 1)(x a) 2 (x 1 + a) 2 (2) F (x) 1 = k(x 1/2) 2 (x A) 2 (x 1 + A) 2 Differentiating in (2) we get F (x) = 2k(x 1/2)(x A)(x 1 + A)(3x 2 3x + (A A )), where the roots of the last factor must be a and 1 a. Therefore (3x 2 3x + (A A )) = 3(x a)(x 1 + a), and replacing in (1) we get (1 )F (x) = kx(x 1)(x 2 x (A A ))2.

48 The case of S 12 (K < (2, 4, 12)) (1) F (x) = kx(x 1)(x a) 2 (x 1 + a) 2 (2) F (x) 1 = k(x 1/2) 2 (x A) 2 (x 1 + A) 2 Differentiating in (2) we get F (x) = 2k(x 1/2)(x A)(x 1 + A)(3x 2 3x + (A A )), where the roots of the last factor must be a and 1 a. Therefore (3x 2 3x + (A A )) = 3(x a)(x 1 + a), and replacing in (1) we get (1 )F (x) = kx(x 1)(x 2 x (A A ))2. Finally, comparing same degree terms in (1 ) and (2) we find 16A 2 16A + 1 = 0.

49 The case of S 12 (K < (2, 4, 12)) The roots of 16A 2 16A + 1 = 0 are A = and B = 1 A =

50 The case of S 12 (K < (2, 4, 12)) The roots of 16A 2 16A + 1 = 0 are A = and B = 1 A = We have therefore the following equation for S 12 : y 2 = x(x 1)(x 1 2 )(x )(x )

51 Example 2: S8

52 The case of S 8 (K < (2, 8, 8)) S 8 is constructed by the typical genus two identification in a regular 8-gon. a 1 b 1 a 1 1 b 1 1 a 2b 2 a 1 2 b 1 2 = 1

53 The case of S 8 (K < (2, 8, 8)) We shall nevertheless change to a more useful fundamental domain for K (for this we must deal with the whole tessellation by octogons).

54 The case of S 8 (K < (2, 8, 8)) Our new domain is determined by the index two inclusion (2, 8, 8) < (2, 4, 8).

55 The case of S 8 (K < (2, 8, 8)) Our new domain will be the reunion of two octogons with angle 2π/4.

56 The case of S 8 (K < (2, 8, 8)) Colours suggest the new side-pairings.

57 The case of S 8 (K < (2, 8, 8)) And we can forget about the group (2, 8, 8). The advantage is that J lies in (2, 4, 8).

58 The case of S 8 (K < (2, 8, 8)) The six Weierstrass points. One of the octogons serves as a fundamental domain of K, J.

59 The case of S 8 (K < (2, 8, 8)) Fundamental domain and side-pairings for the group K, J.

60 The case of S 8 (K < (2, 8, 8)) 0 π 1 Recall again that D/ K, J Ĉ and we can normalize three suitable points to be 0, 1,.

61 The case of S 8 (K < (2, 8, 8)) 0 π 1 There is again an anticonformal involution of S 8 that induces the mapping x x in Ĉ D/ K, J. We can detect where the real line lies.

62 The case of S 8 (K < (2, 8, 8)) a 0 b a π 1 b The remaining Weierstrass points are two pairs of complex conjugate numbers. The equation of S 8 is then y 2 = x(x 1)(x a)(x a)(x b)(x b)

63 The case of S 8 (K < (2, 8, 8)) a A 0 C b a π B 1 b We introduce also three more relevant points. Here 0 < A < C < B.

64 The case of S 8 (K < (2, 8, 8)) a A 0 C b a π B 1 b F Ĉ From the local structure of β = F π we find: (1) F (x) = kx(x 1)(x a)(x a)(x b)(x b)(x C) 2 (2) F (x) 1 = k(x A) 4 (x B) 4

65 The case of S 8 (K < (2, 8, 8)) (1) F (x) = kx(x 1)(x a)(x a)(x b)(x b)(x C) 2 (2) F (x) 1 = k(x A) 4 (x B) 4

66 The case of S 8 (K < (2, 8, 8)) (1) F (x) = kx(x 1)(x a)(x a)(x b)(x b)(x C) 2 (2) F (x) 1 = k(x A) 4 (x B) 4 (i) Evaluating in x = 0 we find k = 1/(AB) 4.

67 The case of S 8 (K < (2, 8, 8)) (1) F (x) = kx(x 1)(x a)(x a)(x b)(x b)(x C) 2 (2) F (x) 1 = k(x A) 4 (x B) 4 (i) Evaluating in x = 0 we find k = 1/(AB) 4. (ii) Evaluating in x = 1 and using the value of k above, we find A + B = 1.

68 The case of S 8 (K < (2, 8, 8)) (1) F (x) = kx(x 1)(x a)(x a)(x b)(x b)(x C) 2 (2) F (x) 1 = k(x A) 4 (x B) 4 (i) Evaluating in x = 0 we find k = 1/(AB) 4. (ii) Evaluating in x = 1 and using the value of k above, we find A + B = 1. (iii) Differentiating in (2) we get F (x) = 4k(x A) 3 (x B) 3 (2x A B). The root of the last factor must be C, hence C = 1/2.

69 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 1 (AB) 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) 1 = 1 (AB) 4 (x A) 4 (x B) 4 (i) Evaluating in x = 0 we find k = 1/(AB) 4. (ii) Evaluating in x = 1 and using the value of k above, we find A + B = 1. (iii) Differentiating in (2) we get F (x) = 4k(x A) 3 (x B) 3 (2x A B). The root of the last factor must be C, hence C = 1/2.

70 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 1 (AB) 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) 1 = 1 (AB) 4 (x A) 4 (x B) 4

71 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 1 (AB) 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) 1 = 1 (AB) 4 (x A) 4 (x B) 4 (iv) We also can write (2) as F (x) 1 = 1 (AB) 4 (x 2 (A+B)x +AB) 4 = 1 (AB) 4 (x 2 x +AB) 4.

72 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 1 (AB) 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) 1 = 1 (AB) 4 (x 2 x + AB) 4 (iv) We also can write (2) as F (x) 1 = 1 (AB) 4 (x 2 (A+B)x +AB) 4 = 1 (AB) 4 (x 2 x +AB) 4.

73 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 1 (AB) 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) 1 = 1 (AB) 4 (x 2 x + AB) 4 (v) Evaluating this last expression in x = 1/2 we find and we deduce AB = = F (1/2) 1 = 1 (AB) 4 ( AB)4

74 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 8 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) 1 = 8 4 (x 2 x )4 (v) Evaluating this last expression in x = 1/2 we find and we deduce AB = = F (1/2) 1 = 1 (AB) 4 ( AB)4

75 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 8 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) = 8 4 (x 2 x )4 + 1 (vi) A factorization of this last expression for F in (2) is F (x) = 32x(x 1)(32x 4 64x x 2 8x + 1)( 1 + 2x) 2 and we deduce by comparison with (1) that 32(x a)(x a)(x b)(x b) = (32x 4 64x x 2 8x + 1).

76 The case of S 8 (K < (2, 8, 8)) (1) F (x) = 8 4 x(x 1)(x a)(x a)(x b)(x b)(x 1 2 )2 (2) F (x) = 8 4 (x 2 x )4 + 1 (vi) A factorization of this last expression for F in (2) is F (x) = 32x(x 1)(32x 4 64x x 2 8x + 1)( 1 + 2x) 2 and we deduce by comparison with (1) that 32(x a)(x a)(x b)(x b) = (32x 4 64x x 2 8x + 1). The equation of S 8 results y 2 = x(x 1)(x 4 2x x x )

A FUCHSIAN GROUP PROOF OF THE HYPERELLIPTICITY OF RIEMANN SURFACES OF GENUS 2

Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 28, 2003, 69 74 A FUCHSIAN GROUP PROOF OF THE HYPERELLIPTICITY OF RIEMANN SURFACES OF GENUS 2 Yolanda Fuertes and Gabino González-Diez Universidad