of enneper type minimal surfaces - American Mathematical Society

proceedings of the american mathematical society Volume 108, Number 4, April 1990

A NEW FAMILY OF ENNEPER TYPE MINIMAL SURFACES YI FANG (Communicated by Jonathan M. Rosenberg)

Abstract. An Enneper type surface is a complete immersed minimal surface in R3 with only one end and finite total curvature. In this paper we construct a family of Enneper type surfaces of genus 1, total curvature -8(2n + \)n,n = 0,1,2, • ■• . We use the Weierstrass p elliptic function as a tool and also prove some results about p on a square torus.

1. Introduction An Enneper type minimal surface is a complete immersed minimal surface with finite total Gauss curvature and only one end; (i.e., conformally it is a closed genus k Riemannian surface with one puncture). The simplest example is Enneper's surface. It has genus 0 and total curvature -4it. There is also a family of genus 0 examples with total curvature -4nn, n = 1,2,3, • • • . In [2] Chen and Gackstätter constructed genus 1 and genus 2 examples with total curvature -87T and -12;:. In [6] Wohlgemuth constructed a family of genus 1 examples with total curvature -4n(2n + 1) for n > 1 . In this paper we will construct a family of genus 1 examples with total curvature -87t(2n + 1) for n > 0. Our main tools are Weierstrass representations for minimal surfaces and the Weierstrass elliptic function p associated to a lattice L —[1, t] in the complex plane C. The by-products of this study are some properties of the Weierstrass p function. Having not seen these properties in publication, we list them as a theorem in this paper.

2. Weierstrass

representation

A very important tool used in the construction of minimal surfaces is the Weierstrass representation formula. Here we state one version of it; for details

see [4] and [5]. Received by the editors January 23, 1989 and, in revised form, August 22, 1989.

1980 Mathematics Subject Classification (1985 Revision). Primary 53A10; Secondary 30C15. The research described in this paper was supported by research grant DE-FG02-86ER250125 of the Applied Mathematical Science subprogram of the Office of Energy Research, U. S. Department of Energy, and National Science Foundation grant DMS-8802858. © 1990 American Mathematical Society

0002-9939/90 $1.00+ $.25 per page

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994

YI FANG

Proposition 1. Let M be a compact Riemann surface and M = M-{px, ■■■,pn} . Suppose ~g: M-»Cu {00} is a meromorphic function and n is a meromorphic 1-form such that whenever g = ~g\M has a pole of order k, then r\ has a zero of order 2k and n has no other zeros on M. Let œx = ^(l-g2)n,

œ2 = ^(l+g2)n,

o)3 = gn.

If for any closed curve y in M,

(1)

Re /"«,. = (), for i =1,2,3, Jy

then the surface S, defined by X: M —>R3, is a regular minimal surface, where

X(z) = Re f / ft),, / co2, / a>i \J z0

J z0

J z0

Here, z0 is a fixed point of M. Moreover, if at any deleted point pt, one of oix, a>2, a>3 has a pole of order at least 2, then S is also complete. The total curvature of S is

C(S) = -4nm, where m is the degree of ~g.

Proof. See, for example, [4, pp. 112-113] and [5, p. 82], Theorem 9.2. a 3.Weierstrass

p elliptic

function

Let L = [co,, co2] be a lattice C. Associated to each L there is a doubly periodic meromorphic function, the Weierstrass p function. It is

z

Z?o\(z-œ)

w)

where œ = ma>x+ na>2 for all (m,n) eZxZ and (m, n) ^ (0,0). to see p is an even function. Moreover, we have

It is easy

Lemma 1. 1. p(2 ' is an even function, p{2 +1) is an odd function, where p(n) denotes

the nth derivative of p, n > 0. 2. p{2k)= Pk(p) and p(2k+X)= Qk(p)p . Where Pk and Qk are polyno-

mials. 3. If L = [1 ,t] and x e iR then pÇz + i) = p(z) and p(-~z+ 1) - p(z). Also, ,.i

f\P"'f(l +l)a,>0. for any n>0.

Furthermore, _-i jfV-TG")"'%+•)«•

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A NEW FAMILY OF ENNEPER TYPE MINIMAL SURFACES

995

and

Si JO

\m(\

+ t)dt

are nonzero real numbers, for any n, 4. If L = [l,i],oj = Ij1, then p(co -p(a> + z), and p(w + z) = p(a> + double zero at w, and no other zeros

(2) (3)

^)(|

m e Z and n > 0. + iz) = p(co + z), p(co - iz) = z). p has a double pole at 0, a or poles. Furthermore,

+ ^_,-Vn)(i

+^

[P(n)f 0. Then P

(4n+l)

n

/v^

=P¿^a2,P j=0

2« 2/

>P

(4n+3)

ir»

n

2n+l

=P¿^a2j+lP j=0

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2j+l

■

YI FANG

998

We have that

P

(4n+4)

-P

/2v^/"i-

, i\

¿J2;

2"+1

2J ,

+ X)a2j+lP +P

"V1

2/1+1 2j+l

2^a2j+lP

j=0

j=0 n

n

= (4p3 - g2p) J2(2J + ! )a2;+l P2J + (6P2 - 82/2) J2 a2jtl P2J+1 ;=0

= ¿(4(2;

;=0

+ 1) + 6)a2;++1'^+3- f ¿(2(2;

j=0

+ 1) + \)a2£\p2i+\

j=0

Hence, (4/1+5)

/ I V^/i

• , -iiio ■ , im

2/i+l

= P \ 2^(2; + 3)(8; + I0)a2;+l p

2j+2

;=o

f^(27

+ l)(4; + 3)a2;++llp2^

,'j _ lfa]n+l + ¿(2; + I) [(8; + 2)a22;_+¡ - f (4; + 3)a22;++l'

I

y=i

/i ->\/o 2/1+1 2/1+2 +, {In +, 3)(8« +, in\ 10)a2„+1p

B+l

= P¿fl:

2/1+2 2;

j=0

Since there are only even terms and all the a2n+[ and ;' are real, so a"+ is real and claim 1 is true for k = 2« + 2 . Also we can see by the computation that a2"+2 = -\g2a2x +1. Similarly,

P

(4/1+6)

n+l ör^..

2/1+2 2;'-l

= P 2^2;a2;

p

,

n+l " V~»

2,!+2

+p 2^a2j

n+l

P

2>

n+l

n+l

E/o (8;■+, 6)a2;. ¿\ 2"+2 pJ2j'+2 -y2J8; 5? Wo j=0

• +, i\ 2"+2 P2> = V^ L 2>+2 l)a2, 2^è2j+2P

j=0

j=0

so

P

(4n+7)

n+l

'Wi-

= P ¿J2; j=0

, 1 . So we need only to prove that a0" ^ 0. Let o> = ^ . Since p(cw) — p'(to) = 0,

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A NEW FAMILY OF ENNEPER TYPE MINIMAL SURFACES

999

p"(co) ^ 0, so co is a double zero of p, a single zero of p . In (2) of Lemma 1 set t = 1/2 and n = 4k. We get p{4k](co) = -p(4k)(co), so p{4k)(co) = 0. Also by p{2k+l) = p'Qk(p) we know that p(2k+X)(co)= p(2k+])(co + ±) = 0. If a0" = 0, then co will be a zero of p "+ of order at least 3. So if we prove that p(4/I+ \co) t¿ 0, then p(4"+1) will have only a single zero at co and thus a2" ^ 0. We will count the number of zeros for p(4n+ }. Let / be the open interval (0,1/2). Then on co+I (for co+I we mean the interval l/2+i/2 + t,

0 < t < 1/2, similarly co+ H means the interval 1/2 + i/2 + it, 0 0, has these properties. If z —x + iy, then dRep{k)(co + z)

dy

y=0

on CO+ I. For a holomorphic function /,

/-2(Re/)z--^--z

dy

on co + / we have r

(fc+i).

,

dRep{k)(co + x)

(co + x) =-—-1-

dx

.8Rep(co + x)

dp(k)(co + x)

dy

dx

- i-

for all x G I. We claim that for any n > 0, p(4"+ ' has at least n + 1 different zeros on co + I. Since p'(co + 1/2) = p'(a>) = 0, by Rolle's theorem there is at least one xQ € / such that p"(co + x0) = 0. Hence for n = 0 the claim is true. We now apply induction. Suppose for n = k > 0, p has at least k + 1 different zero points on co + I. Then by Rolle's theorem, p