Automorphism group of plane curve computed by Galois points, II
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(2) 60. T. HARUI, K. MIURA and A. OHBUCHI. G be a group of automorphisms of C. Then, it is well-known that G is considered as a subgroup of PGLð3; CÞ ¼ AutðP2 Þ. Let Fd be the Fermat curve X d þ Y d þ Z d ¼ 0. We denote by Kd a smooth curve defined by XY d1 þ Y Z d1 þ ZX d1 ¼ 0 (In [3], Kd is called Klein curve of degree d). For a non-zero monomial cX i Y j Z k with c 2 C n f0g, we define its exponent as maxfi; j; kg. For a homogeneous polynomial F ðX; Y ; ZÞ, the core of F ðX; Y ; ZÞ is defined as the sum of all terms of F ðX; Y ; ZÞ with the greatest exponent. Definition 2. Let C0 be a smooth plane curve with defining equation F0 ðX; Y ; ZÞ ¼ 0. Then a pair ðC; GÞ of a smooth plane curve C and a subgroup G AutðCÞ is said to be a descendent of C0 if C is defined by a homogeneous polynomial whose core coincides with F0 ðX; Y ; ZÞ and G acts on C0 in a suitable coordinate system. Definition 3. We denote by PBDð2; 1Þ the following subgroup of PGLð3; CÞ: 8 9, 1 0 a11 a12 0 > > < = C B PBDð2; 1Þ :¼ A ¼ @ a21 a22 0 A 2 GLð3; CÞ C : > > : ; 0 0 . We remark that there exists a natural group homomorphism : PBDð2; 1Þ ! PGLð2; CÞ, i.e., A 7! ðaij Þ. Using these concepts, the first author proved the following theorem. Theorem 4 ([3]). Let C be a smooth plane curve of degree d 4, G a subgroup of AutðCÞ. Then one of the following holds: (a-i) G fixes a point on C and G is a cyclic group whose order is at most dðd 1Þ. Furthermore, if d 5 and jGj ¼ dðd 1Þ, then C is projectively equivalent to the curve Y Z d1 þ X d þ Y d ¼ 0 and AutðCÞ ¼ Zdðd1Þ . (a-ii) G fixes a point not lying on C and there exists a commutative diagram 1. 1. C. N. PBD(2, 1). G. ρ. PGL(2, C). 1. G. 1,. where N is a cyclic group whose order is a factor of d and G0 is a subgroup of PGLð2; CÞ, i.e., a cyclic group Zm , a dihedral group D2m , the tetrahedral group A4 , the octahedral group S4 or the icosahedral group A5 . Furthermore, m d 1 and if G0 ¼ D2m then. [Vol. 94(A),. m j d 2 or N is trivial. In particular, jGj maxf2dðd 2Þ; 60dg. (b-i) ðC; GÞ is a descendant of the Fermat curve Fd : X d þ Y d þ Z d ¼ 0. In this case jGj 6d2 . (b-ii) ðC; GÞ is a descendant of the Klein curve Kd : XY d1 þ Y Z d1 þ ZX d1 ¼ 0. In this case jGj 3ðd2 3d þ 3Þ if d 5. (c) G is conjugate to a finite primitive subgroup of PGLð3; CÞ. Namely, the icosahedral group A5 , the Klein group of order 168, the alternating group A6 , the Hessian group H216 or its subgroup of order 36 or 72. In particular, jGj 360. 2. Remark on (a-i). Let P1 ; ; Pm be all inner and outer Galois points for C and GðCÞ denote the group generated by GPi ði ¼ 1; 2; . . . ; mÞ. The group GðCÞ is called the group generated by automorphisms belonging to all Galois points for C. In [5], we have studied the difference between AutðCÞ and GðCÞ. Referring to [2], if ðCÞ 1 and 0 ðCÞ 1, then C is projectively equivalent to the curve as in Theorem 4 (a-i). We denote the curve by CðdÞ, i.e., CðdÞ : Y Z d1 þ X d þ Y d ¼ 0. If d 5, then P ¼ ð0 : 0 : 1Þ is the only inner Galois point and Q ¼ ð1 : 0 : 0Þ is the only outer Galois point for CðdÞ. We put GP ¼ hi and GQ ¼ hi. Then GðCðdÞÞ ¼ h; i. In [5], we obtain AutðCðdÞÞ ¼ GðCðdÞÞ. Thus Galois points play an important role in studying the automorphism groups of smooth plane curves. 3. Main results. In this section, we first remark on Theorem 4 (a-ii). In general, we have 2dðd 2Þ > 60d. However, clearly 2dðd 2Þ < 60d if d < 32. Hence we consider the case d < 32, and try to construct C with jAutðCÞj ¼ 60d. Let Fi ðX; Y Þ (i ¼ 1; 2; 3) be the homogeneous polynomials of X and Y defined by F30 ¼ X 30 þ 522 ðX 25 Y 5 X 5 Y 25 Þ 10005ðX 20 Y 10 þ X 10 Y 20 Þ þ Y 30 , F20 ¼ X 20 228 ðX 15 Y 5 X 5 Y 15 Þ þ 494X 10 Y 10 þ Y 20 and F12 ¼ XY ðX 10 þ 11X 5 Y 5 Y 10 Þ. For these polynomials, we have well-known facts as follows: Fact 1. Let 5 be a primitive 5th root of unity and put ! 0 1 53 0 ¼ ; ¼ ; 1 0 0 52.
(3) No. 6]. Automorphism group computed by Galois points, II. 1. ¼ 2 5 53. 5 þ 51. 1. 1. ð5 þ 51 Þ. !. 0. and I ¼ h; ; i. Then C½X; Y I ¼ C½F30 ; F20 ; F12 . Note that I ¼ SLð2; 5Þ: the binary icosahedral subgroup of SLð2; CÞ. Under the situation above, our main results are stated as follows: Theorem 5. Let C30 , C20 and C12 be the plane curves defined by C30 : Z 30 þ F30 ðX; Y Þ ¼ 0, C20 : Z 20 þ F20 ðX; Y Þ ¼ 0 and C12 : Z 12 þ F12 ðX; Y Þ ¼ 0. Then jAutðCd Þj ¼ 60d (d ¼ 30; 20; 12). Furthermore, the following hold: Z15 SLð2; 5Þ, AutðC30 Þ ¼ AutðC20 Þ ¼ Z5 ðSLð2; 5Þ o Z2 Þ and AutðC12 Þ ¼ Z3 ðSLð2; 5Þ o Z2 Þ. 4. Proofs of Theorem 5. First of all, we review Theorem 4 (a-ii) from the viewpoint of Galois points. Let C be a smooth plane curve of degree d 4 with a unique Galois point P , G a subgroup of AutðCÞ. Then by Proposition 3, we may assume that the defining equation of C is given by Z d þ Fd ðX; Y Þ ¼ 0 for some homogeneous polynomial Fd ðX; Y Þ of degree d and P ¼ ð0 : 0 : 1Þ. Let P : P2 ! P1 be the projection with the center P . Then P is represented as P ððX : Y : ZÞÞ ¼ ðX : Y Þ. The Galois group GP is represented by 0 1 * 1 0 0 + B C GP ¼ @ 0 1 0 A ; 0 0 d where d is a primitive d-th root of unity. We denote by d this matrix generating GP . Then we get the following commutative diagram as in Theorem 4 (a-ii): 1. 1. C. PBD(2, 1). N. ρ. G. PGL(2, C). 1. G. 1.. In this case N ¼ GP . Thus we get the exact sequence ð]Þ. . 1 ! GP ! G ! G0 ! 1;. where G0 PGLð2; CÞ. Now, we put. B ¼B @. 5 54 pffiffi 5. 53 52 pffiffi 5. 53 52 pffiffi 5. 5pffiffi5 5. 0. 0. 0. 61. 0. 1. C C 0 A;. 4. 1. 53 B ¼@ 0. 0 52. 1 0 C 0A. 0. 0. 1. and. 5. B ’¼@ 0 0. B. 0 ¼ @. 0. 1. 1 0. C 0 A; 1. 0. 0. 1. 0. 1 0. 0. 1. C 0 A; ¼ @. C A;. 1 B. 0 ¼ @. 5. B ¼@ 0 0 0. where
(4) 12 ¼ 5 . We also put 0 1. 0. 0. B. C 0 A;
(5). 1 1. C A 2 GLð3; CÞ: 1. Referring to [1], we see that the image of I under the natural homomorphism SLð2; CÞ ! PGLð2; CÞ is isomorphic to A5 . Further, we define Sð2; 1Þ :¼ h0 ; 0 ; 0 i ¼ I. f0 ¼ h; ; 30 i First we deal with C30 . Put G GLð3; CÞ and H ¼ h; i. Then we can check that 0 1 0 12 5 0 0 5 0 0 ð 4 Þ2 ¼ @ 0 5 0 A and ¼ @ 0 5 0 A . 0 0 1 0 0 1 So we have 2 H. Furthermore, since 0 ¼ ð2 Þ2 , 0 ¼ ð2 Þ2 and 0 ¼ 2 , we obtain H Sð2; 1Þ. 0 1 1 0 0 We also remark that @ 0 1 0 A 2 Sð2; 1Þ 0 0 1 0 1 1 0 0 and ¼ @ 0 1 0 A 0 0 , ¼ 0 0 1 0 1 0 12 1 0 0 5 0 0 @ 0 1 0 A0 @ 0 5 0 A . Thus we ob0 0 1 0 0 1 0 1+ * 5 0 0 tain Sð2; 1Þ; @ 0 5 0 A ¼ Sð2; 1Þ 0 0 1 1+ *0 5 0 0 @ 0 5 0 A ¼ H. 0 0 1 Therefore, we have that.
(6) 62. T. HARUI, K. MIURA and A. OHBUCHI. 0 * 1 f0 ¼ H B G @0. 0 1. 0 0 0 * 5 B ¼ Sð2; 1Þ @ 0 0 Put. Z. 1 0 + C 0 A 30. we obtain *. 1 0 + C 0 A : 0 1 0 0 30 80 9 1 < 0 0 = f0 \ @ 0 0 A 2 C ¼ :¼ G : ; 0 0 1 0 1+ 0 5 0 0 0 A; @ 0 5 0 A . Then 1 0 0 5 0 5. *0 1 0 @ 0 1 0 0 f0 =Z G and G0 :¼ G 0 * 1 B Sð2; 1Þ @ 0. 0 1 0 + * 1 B C 0A @0. 0 1. 1 + 0 C 0 A 0 0 1 1 G0 ¼ *0 + 1 0 0 B C @ 0 1 0 A 0 0 1 0 1 0 1 + * 5 0 0 + * 1 0 0 B C B C @ 0 5 0 A @ 0 1 0 A 0 0 1 0 0 15 0 1 : * 5 0 0 + B C @ 0 5 0 A 0 0 5 Hence G0 ¼ SLð2; 5Þ Z15 . In particular, jG0 j ¼ 120 15 ¼ 1800. On the other hand, we see that jGj ¼ 30 60 ¼ 1800 by ð]Þ. Hence G0 ¼ G, which completes the proof of this case. By a similar argument to the above, we can prove the other cases. So, we give the proofs in brief. f0 ¼ h; ; 20 i For the curve C20 , we put G GLð3; CÞ. We see that 0 1 * 1 0 + 0 B C f0 ¼ Sð2; 1Þ @ 0 1 0 A G pffiffiffiffiffiffiffi 0 0 1 0 1 0 1 * 5 0 0 + * 1 0 0 + B C B C @ 0 5 0 A @ 0 1 0 A : 0. 0. 0 1. 1. Since its center Z is 0 1 0 * 1 0 0 5 B C B 0 1 0 ; @ A @ 0 0 0 1 0. [Vol. 94(A),. 0 0 5 0. 0 5 1. + C 0 A ; 5 0. 0. 1. + B C Sð2; 1Þ @ 0 1 0 A pffiffiffiffiffiffiffi 0 0 1 0 1 G0 ¼ * 1 0 + 0 B C @ 0 1 0 A 0 0 1 0 0 1 * 5 0 0 + * 1 0 B B C @ 0 5 0 A @ 0 1 0 0 1 0 0 0 1 * 5 0 0 + B C @ 0 5 0 A 1. 0. 0. 0. 0. f0 =Z G. where G0 ¼ G Further, since. 0. 1. C 0 A 5. +. ;. 5. 1 + 1 0 0 B C Sð2; 1Þ @ 0 1 0 A 0 0 1 0 1 SLð2; 5Þ ¼ * 1 0 + ; 0 B C @ 0 1 0 A 0 0 1 *. 0. we have the following exact sequence: 0 * 1 0 B Sð2; 1Þ @ 0 1. 1 + 0 C 0 A pffiffiffiffiffiffiffi 0 0 1 0 1 * 1 0 + 0 B C @ 0 1 0 A 0 0 1. 1 ! SLð2; 5Þ !. . ! f
(7) 1g ! 1; 0 B where : @ 0. 1. A. 0 by 1 7! @ 1 0. C A 7! 2 . The sequence is split 1 0 0. 1 0 A p0ffiffiffiffiffiffiffi . 1. Hence G0 ¼ ðSLð2; 5Þ o Z2 Þ Z5 . In particular, jG0 j ¼ 120 2 5 ¼ 1200. On the other hand, we see that jGj ¼ 20 60 ¼ 1200 by ð]Þ. Hence G0 ¼ G, which completes the proof of this case. f0 ¼ h; ’i, Finally, for the curve C12 , we put G and K ¼ h; i. We can check that K ¼ Sð2; 1Þ..
(8) No. 6]. Automorphism group computed by Galois points, II. 0. 5 0 Putting " :¼ ð’4 Þ3 ¼ @ 0 5 0 0 f0 ¼ hK; 12 ; "i. Furthermore we G f0 ¼ hKi h 12 i h"i G 0 * 1 0 B ¼ Sð2; 1Þ @ 0 1 * . 0. 1 B @0 0. 1 0 0 A, we obtain 5 get 1. 0 1. + 0 C 0 A pffiffiffiffiffiffiffi 0 0 1 1 0 + * 0 5 0 C B 0 A @ 0 5. 0. !. 0. 0. 1 0 + C 0 A ; 5. where ! is a cubic root of unity. ðSLð2; 5Þ o Z2 Þ Z3 . In particular, Hence G0 ¼ jG0 j ¼ 120 2 3 ¼ 720. On the other hand, we see that jGj ¼ 12 60 ¼ 720 by ð]Þ. Hence G0 ¼ G, which completes the proof of this case. Acknowledgments. The second author was partially supported by JSPS KAKENHI Grant Number JP26400057. The third author was partial-. 63. ly supported by JSPS KAKENHI Grant Number JP15K04822. References [ 1 ] H. F. Blichfeldt, Finite collineation groups, with an introduction to the theory of groups of operators and substitution groups, Univ. of Chicago Press, Chicago, Ill., 1917. [ 2 ] S. Fukasawa, On the number of Galois points for a plane curve in positive characteristic. III, Geom. Dedicata 146 (2010), 9–20. [ 3 ] T. Harui, Automorphism groups of smooth plane curves, arXiv:1306.5842v2. [ 4 ] K. Miura, Field theory for function fields of singular plane quartic curves, Bull. Austral. Math. Soc. 62 (2000), no. 2, 193–204. [ 5 ] K. Miura and A. Ohbuchi, Automorphism group of plane curve computed by Galois points, Beitr. Algebra Geom. 56 (2015), no. 2, 695–702. [ 6 ] K. Miura and H. Yoshihara, Field theory for function fields of plane quartic curves, J. Algebra 226 (2000), no. 1, 283–294. [ 7 ] H. Yoshihara, Function field theory of plane curves by dual curves, J. Algebra 239 (2001), no. 1, 340–355..
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