By J. Lucas M. Barbosa, A. Gervasio Colares (auth.)
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P k exists of This C5]). M and of N: M * $ 2 ( I ) , which N: In M 4 S;2(1). M and M representation) all would in R 3, Since values be pj , as of But N cur- surface in R3 the metric assume of M R3 theorem. extends is with would pj the g imply M. M. has complete in t h e s e curvature after mapping Indeed, we have (great) g hence, total have by at it that of the theorem, times, the function an essential to h y p o t h e s i s also no poles were many a pole; We total function that is c o n t r a r y at m o s t finite to a m e r o m o r p h i c infinitely to presented the G a u s s $2(I) to of p o i n t s Furthermore, by Picardts form that with then, which number M - {Pl .....
Of that FI a_~t 8 = ~I z = 8. occurs if and only if Then 2n l-z 2n = o2n-z 2n = (0-z) Z 82n-j j=l z j-I , n zn+l-i = zn+l-@ n+l = (z-e) E j=O On-j zj l z and 2n Z Fl(Z) 1 (_(_~ - ~ : 8 2 n - j z j-I j:l ) = n ( Z 2 z - e Gj_( 2 en-Jz j) z) . j=O Now, z = O, if %(e) we obtain 2n E j=l : O2 n - j 8 j - i n ( e~-J z . e J) 2 ( j=0 Hence, z = e is 2nd case: a pole e ~ ±i. 4) 2n E j=l n We of order one of F 1 = - -~1 lim ze+l may then ( (n 2n)+l2z) rewrite i F1 F1 . F1 Thus, - ±n (n+l) 2 " as l-z 2n = 2(z_e)2 n ( ~.
A complete viewed to perpendicular this grows that k is e q u i v a l e n t "gets passing $2(I), is h o m e o - The origin. corresponds means M M. the planes points that in through finite __in however~ pj "gets with theorem; of of If to i n f i n i t y . corresponding pj quotient of neighborhood is a c o n s e q u e n c e k M at M M. F 1 .... ,Yk of this the G a u s s one the R3 k as to dif- compact to curves the h y p o t h e s i s normal of R 3, it. e. compact). 7). [i]). with k Z lj j=l 2m = where following inequality (Jorge-Meeks immersed the From multiplicity if a n d the only previous X(M) of the if e a c h theorem, ~ k end - X(M), E end is for each corresponding 0 to P j- embedded.