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OF HIGHER ORDER OF A COMPACT RIEMANN SURFACE
Toshio. MATSUMOTO ,(Received(iCt・ber 1.・1988;Rgyi・ed・Nbvember 29,1988).../一 §1. Ilitroduction. . Let X be a comPact Riemann surface of genus g≧2, Ak be the complex vector space of analytic differentials of t)1)e k on X, Aut(X)be the group of auto− morphisms on X, and Fix〔h〕 be the fixed points of hεAut〔X〕. To each point PεX, there is a set of integers ・ . ’ 1=Y1<Y2<◆°’<Yd(k)≦2k(9−1)+1・u・h・h・tY・be…9・t・th・S蜘・n・e・f皿d・n・y・f・here ex・・tr・輌戦』Wi・h
a zero of order Y−l at P, where .d(k〕=1(、k.i〕〔、.、〕 1熟;B.
The sequence is called the k−gap sequence at P, the non−negative integer t d(k〕Wk〔P)=i=・〔Y」・う)・ .11 ・
i,ca11・d「 狽?Ek−th w・ight・t P. P・XiS call・d・k−W・ier・t・ass ’ o・iit dn X if「
wk(P)>0. It is known that 「 .『. ’
Wk〔9〕;鍵〔P)=t(;畿ご:1〕・ll㌫
for any given X of genus g. We shall denote by 〔k−WP)X the set of]ぐ一Weierstrass p.oints on X. Xwill be ca11・d k−a119・m・ip if wk〔P)=;f°「eveワPε〔k’WP〕X.・lhl f°ll°wing Tesultsare known. ’ t/
Theorem A. Let X be a 1−allge皿ein co㎎act Riemam surface of』genus g≧2, hεAut〔X)be of prime order N≠3,5. Then either Fix(h)⊂〔1−WP〕X or Fix(h〕n〔1−WP〕x =φ. .147
148
The・’・皿B・..L・ぱbe a 2−a119・皿・in・・mPact ki・壁㎜su・face・f、g・nu・g≧2・ hεAut(X〕be of prime order N≠3. Then either Fix(h)c(2−WP)X of’.Fi・〔h〕A:(2㊥X・φ∴. t. tt:.:二. ”
・・.E .../t
The fomer was proved by Lewittes in 1963, the latter by Duina in 1978. For a k−allgemein compact Riemann surface, k≧3, the analogous results were not known. Our reults are as follows. Theore皿1. Let X be a k−allgemeih cdmpact Rie血ann surface of genus g≧2, 3≦](≦7,hεAut(X) be of prime order N with t>O fixed points・ If Qne. of the Fix〔h)⊂〔k−wP)x or Fix〔h〕((k−wP)x一φ.following cases holds, then either (1〕. 〔2) 〔3)’ (4) 〔5) Theorem 2. k≧2,hεAut(X)either
k=3, ・N≠5. .』ik=4,N≠7.
k=5, N≠3,11.. k=6, N≠11, 13, 17. k;7, N≠11,13. Fix〔h)⊂〔k−WP)X §2.Proof of Theorem 1, Let X be a k−allge皿ein compact Riemann surface of prime order N with t>O fixed points.Weierstrass points
k−Weierstrass points that Pε〔k−WP)X, k=dN+k, When k≧2, Duma showed that only one of the .Case[1]case
case. 匹]四
we shall consider the Let X be a .k−qllgemein compact Ri㎝am sqrface of.genus g≧2, be of pri皿e order N with t>O fixed polnts. If N>4k二5, then or Fix(h)∧(k−WP〕X=φ・ 力 2. of gelms g.≧..2, hεAut(h)be Th・n・there are ・xacqy Wk〔9) ・n,x・㎝訓k〔9)−s.‡・a㎜1tip1・・f N.if・i・. th・n帥・「°f. in Fix(h〕. Assume that there are points P,QεFix(h〕 such Q¢(1(−WP〕x. Write ^‘@(2k−i)〔9−1)−qN・r,0≦食≦N−1,0≦・≦N二1. following cases is possible・ へ r=.0,』N+・1.=ぜ2k, N≧3.・ . . ’1≦食.Q.。≧N.、,、R、N∴2’≦t三・・.,”N’、.,.い.;
N+1−2k R・・.。≧N・・2,2・fe・〉.・N・1,2≦t・一禦一,.−N≧5.、. 2k−N−1 cases.2.1 case[1] L・t2k−・・P錐・P多・…幽,ぬer・’P、,P、,・r・,lh are・di・ti・r・ numbers and ∫↓1,∫↓2,.・・,父rm are positive integers. Then, Pζ・呼・….㎞・・2(dN・R)−1・2R−1〔鳳N)・ Since N…2h・・d・・N㎜・t b…ng・・{P、1・’P、 ’・…・lh}・Th・ref・re・ for k=3, 7 for k=4’, 3 for k=5, 11 for k=6, and 13.for k=7. 2.2 case [珂 odd prime N㎜st be 5 Now we shall show that t≠2. If t=2, then, r=N+1−2食, i.e., へ (2k.−1〕〔g−1) 三 1−2k (mod. N〕. ’ . It follows that A (2k−1〕g ≡ (2k−1〕+(1−2k) ≡.0 〔mod. N〕.
Therefore, .
Wk〔9)一(2k−i)9〔2k−・〕〔9−・)2・0.〔・・d. N).. On the other hand, since there is only.二〇ne krWeierstrass points in Fix(h), Wk〔g〕−1must be a multiple of N,砲ich is a contradictioh. Hence t≠2, and we may assume that t≧3. Then, A A 〕皇3≦t− 2「^≦2〔N−4,〕・2・2k−4^, .・・.
N+1−2k N+1−2k N+1−2k .1 ハ and this implies that N≦4k−5. So we can replace the ass㎜ption in case[田by.〔2−2−1〕3≦t−2「.,
. 、 N+.1−2k 、 、 ム (2−2−2〕 0<r≦N−1−k, . tt (2−2−3〕 max{5,2R}≦N≦4食一5. 』 Suppose now that k=3. SinCe N≧5, we have 食=3. Then, from 〔2−2−3), it must be N=7. Therefore, we obtain (i)t=r=3 from〔2−2−1〕and 〔2’2−2). へ Suppose now that ](=4. Since N≧5, we have k=4. Then, from (2−2−3), it must be N=.ll. Therefore, we obtain (ii〕 t=3,r=6 fro皿 〔2−2−1〕and(2−2−2). へ SupPose.now that k=5・ If N=5,、then k〒0, so 〔2−2−3〕 ca皿Qt hold・ ・ If N≧7, then喪=5. From (2・−2「3〕, it rnust. be N=11・or N〒13, When N=11, we obtain 〔i亘〕 t=r=3, (iv〕 t『r〒4, and 〔v〕t=rFs from〔2−2−!) apd 〔2−2−2). Similarly, we obtain 〔vi)t=3,r=6 when N=13.150
.T・. MATSUMOTO ’ へ SupPose now that k;6. If N=5, then k=1, so 〔2−2−3) cannot hold・’1・If N『≧7, ・k・n㊧・6・F・㎝〔2−.2−5).i・㎜・a・N=,13・rNロ・rN=19・F・・IF.、〔.2’2−1〕・nd (2−2−2),.we oもtain 6亘〕’t=T」3i(殖)t=r=4, 〔iX)t=r」5,』(x)t=rゴ6 when N=13,(Xi.)t・3,。・9ぬ。。 N・17,㈹t・3,。・12寵e・’N三19.1.・・ ご:’
SupP・se n・w th・t k・7.1.f N・’5, then㊧一2,・・〔2二2」3〕ca・⑳t’h・1q.>If N・7, ・h・n食・0,・・.(2−2−3)ca皿・ゆlq,. t・g・・f N≧1・,・・th・n葦・7ぽ・m.,(2.−Z−、3)・ it皿st b・N・.17・r N・19・r N・23・F・・m、〔2.ご2三1)・and〔Z−2−2)・wr.・bj aip. .・ 顧)t=3,r=6, 〔㎞)t=4,r=8曲en N=17, (>ctt) t=3,r=9 when N=19, (麺) t=3,r=15 when N=23. 、 ・ ・ .ω㈲㈹旬ω回㈲㈹㈱ω㈲⑰⑪⑰㈲圃
k
3455556666667777
N
7111133333797793
111111111111112
Table 1.・ t3334533456333433
. 一 −r
3634563456926895
...−9(mod;N〕
3、 5 5 . 10 4 6 6 . 12 5 ! 11 8 9 『 、 8 16 . 9 11 Wk・〔9) 6 4 ・ 1 6 1 8 2』 10 8‘ 6 2 ’>@ 4
16 4 7 14 〔m6d. N)・ It is shown in Table l that the cases (i),(ii〕,〔t〕,(W〕,匝),〔iX),〔x),{海), 6匝),6⑩,(W〕,6畦) camot occur, since none of the numbers珊(〔9)−s,s=1,2,… ,t−1, is a multiple of N in each case of th㎝. ・ ’ . 2.3 case [町 N・・W・・haU…w・h・tt’1・・.,・ft・・,中・nN−r〒2R−N・・,1i・・i・.、’... (2k−1)〔g−1) ≡1−2食 .(mod. N)... . . 、.. 、、 . It follows that(.2k−1)g三〇・〔mod. N). Therefore, Wl((g〕』三〇 〔md. N).・一・..・ On the other hand, since there is㎝1y one k−Weierstrass pointS in Fix(h),閲 Wk(9)−1must be a multiple of N,曲ich is a conltradiction二 ・ ・』幽 Hence t≠.2, and we may ass血ne that t≧3. .Then, ’ :.』 ・−. ’3≦t・器・器・
and this i∬plies that 2(§−2)≧3〔2食一N−1),i:e.,.・3N≧4食+1. So we can replace the assumption in gaSe.1珂.by 、. ・ .、 ., , 、、 ,.