A Factorization of a Super-Conformal Map

A factorization of a super-conformal map

Katsuhiro Moriya∗

Abstract A meromorphic function, a super-conformal map, and a minimal surface are factored into a product of two maps by modeling the Euclidean four-space and the complex Euclidean plane on the set of all quaternions. One of these two maps is a holomorphic map or a meromorphic map. These conformal maps adopt properties of a holomorphic function or a meromorphic function. Analogs of the Schwarz lemma, the Schwarz-Pick theorem, the Weierstrass factorization theorem, the Abel-Jacobi theorem, and a relation between zeros of a minimal surface and branch points of a super-conformal map are obtained.

1 Introduction

Pedit and Pinkall considered a conformal map from a Riemann surface to the Euclidean space of dimension four to be an analog of a holomorphic function or a meromorphic function in [13], by modeling R4 on the set of all quaternions H. Given a conformal map from a Riemann surface M to R4, there exists a complex structure of R4, parametrized by M, such that the conformal map is holomorphic about this complex structure at each point of M. The motivation for making this interpretation of a conformal map is to obtain its global properties. After a long history, the theory of meromorphic functions has been successful in obtaining global properties of meromorphic functions. Compared with the theory of meromorphic functions, the theory of conformal maps seems to be insufficiently developed in obtaining global properties. The above interpretation raises a problem whether a conformal map succeeds to a property of a meromorphic function. In [13], the order of a zero of a conformal map and the degree of a conformal map are defined (Theorem 3.2, Definition 3.2). The Riemann-Roch theorem for conformal maps is proved (Sec- tion 4). Quaternionic holomorphic curves in the quaternionic projective space in [7] succeeds various properties of meromorphic functions. This important achievement mainly arises for a compact Riemann surface without boundary. However, studies on open Riemann surfaces are still in their infancy. We will study whether a conformal map adopts properties of a meromorphic function on an open Riemann. ∗Division of Mathematics, Faculty of Pure and Applied Sciences, University of Tsukuba, 1-1-1 Tennodai, Tsukuba, Ibaraki 305-8571, Japan

1 Factoring a conformal map into a product of maps is an effective way to attack this problem. The example, in Example of [13], implies that a conformal map can be factored into a product of two conformal maps. We consider this factorization to be an analog of a factorization of a meromorphic function. The author has studied whether a Lagrangian conformal map becomes a product of two Lagrangian conformal maps in [12]. Restricting the scalar of H to C, we identify the quaternionic vector space H with a complex vector space C2. If one of the factors of a conformal map is a meromorphic map into C2, then the conformal map adopts a properties of a meromorphic function, such as zeros and poles, through the meromorphic factor. If the remaining factor reflects properties of a conformal map, this factorization would be a useful factorization in investigating a conformal map. In this paper, we provide this type of factorization of a meromorphic function, a super-conformal map, and a minimal surface. A super-conformal map is a conformal map whose curvature ellipse is a circle at each point ([2]). The curvature ellipse is a central topic of the study of surfaces in R4. There is comprehensive explanation for classical results in [18] and [16]. A super-conformal map is also called a Bor˚uvka’s surface after Bor˚uvka’s study ([1]) or a Wintgen ideal surface ([15]) because the equality holds in Wintgen’s inequality ([17]). A superminimal surface in [16] is a minimal super-conformal map. Rouxel [14] showed that a conformal transform of a super-conformal map is a super-conformal map. Castro [3] showed that the Whitney sphere is the only Lagrangian super-conformal map from a closed Riemann surface. Chen [4] classified all super-conformal maps such that the absolute value of the Gaussian curvature is equal to that of the normal curvature at each point. Friedrich [16] described superminimal surfaces in terms of the twistor space. In [2], it is shown that a super-conformal map is a stereographic projection of S4 followed by the twistor projection of a holomorphic map from M to CP 3, and that a super-conformal map is a Willmore surface with vanishing Willmore energy. In [11] and [5], it is shown that a holomorphic null curve is associated with a super-conformal map. Let S2 be the two-sphere with radius one centered at the origin with respect to all imaginary parts of the quaternions Im H. We model the complex projective line CP 1 on S2. Then, the following provides a factorization of a meromorphic function. Theorem 1. Let M be a Riemann surface. ∼ (a) A map N : M → CP 1 = S2 is a holomorphic map if and only if, for any point p ∈ M, there exist holomorphic functions λ0 and λ1 at p such that N = (λ + jλ )i(λ + jλ )−1. 0 1 0 1 ∼ (b) A map N : M → CP 1 = S2 is an anti-holomorphic map if and only if, for any point p ∈ M, there exist holomorphic functions λ0 and λ1 at p such that −1 N = −(λ0 + jλ1)i(λ0 + jλ1) .

When M is compact without boundary and λ0 and λ1 are meromorphic, we show that the degree of N is equal to the degrees of λ0 and of λ1 (Corollary 2).

2 ∼ In [2], two maps from M to S2 = CP 1 are associated with a conformal map from M to H. These maps are called the left normal and the right normal of a conformal map. A super-conformal map has an anti-holomorphic left normal or an anti-holomorphic right normal. A minimal surface has a holomorphic left normal and a holomorphic right normal. For these conformal maps, we consider the case where the left normal is a non-constant holomorphic map or a non-constant anti-holomorphic map into S2 \{k}. Taking the stereographic projection from k, and the complex conjugate if necessary, we see that we assume the existence of a non-constant holomorphic function on M. Then, M is open. We contemplate a super-conformal map with anti-holomorphic left normal. Lemma 2 in Section 4 ensures the existence of a ∈ S2 such that Na + ai =6 0 on M for any map N : M → S2 \{k}. The following provides a factorization of a super-conformal map. ∼ Theorem 2. Let N : M → S2 \{k} = C be an anti-holomorphic map and a ∈ S2 such that ψ := Na + ai does not vanish on M. A map f : M → H is super-conformal with left normal N if and only if there exist holomorphic functions λ0 and λ1 on M such that f = ψ(λ0 + λ1j). This theorem is a variant of Theorem 3.1 in [13]. We factor the Whitney sphere by this theorem. We see that there exists a super-conformal map from M with anti-holomorphic left normal N, by taking λ as complex conjugate of the stereographic projection from k after N. Then, we prove an analog of the Schwartz lemma (Corollary 3) and the Schwarz-Pick theorem (Corollary 4). We use the notion of a divisor of a conformal map and prove an analog of the Weierstrass factorization theorem (Corollary 6) and that of the Abel-Jacobi theorem (Corollary 7). We assume that M is simply connected. Because we have assumed that M is open, M is bi-holomorphic to C or {z ∈ C | |z| < 1} by the uniformization theorem. Then, Theorem 1 in [11] and Theorem 1 in [10] imply the following factorization of a minimal surface, which is the real part of a holomorphic null curve (see [5], too). ∼ Theorem 3. Let N : M → S2 \{k} = C be a holomorphic map and a ∈ S2 such that ψ := −Na + ai does not vanish on M. For complex functions λ0 and { ∈ | } λ1 on M, set Qλ0,λ1 := p M (dN)p(λ0(p) + λ1(p)j) = 0 . (a) If Φ := f + ig : M → C ⊗ H is a holomorphic null curve and f and g are minimal surfaces with left normal N, then there exist holomorphic functions λ0 \ → H and λ1 on M, and µ: M Qλ0,λ1 with ψ d(λ0 + λ1j) = dN a(λ0 + λ1j)µ, such that f = a(λ0 + λ1j)(µ − 1) and g = −Na(λ0 + λ1j)µ + ai(λ0 + λ1j) up to constant addition to g. (b) Let λ0 and λ1 be holomorphic functions on M. Deﬁne a map µ: M \ → H Qλ0,λ1 by ψ d(λ0 + λ1j) = dN a(λ0 + λ1j)µ. Then, the maps f := a(λ0 + λ1j)(µ − 1) and g := −Na(λ0 + λ1j)µ + ai(λ0 + λ1j) are minimal surfaces with \ → H left normal N and Φ := f + ig : M Qλ0,λ1 is a holomorphic null curve. From this theorem, we have a relation between zeros of a minimal surface and

3 branch points of a super-conformal map (Corollary 8). We apply this relation to the Enneper surface. Acknowledgement. This work is supported by JSPS KAKENHI Grant number 22540064.

2 Conformal maps

We recall a conformal map from a Riemann surface to R4 and introduce the notion of a pole and a divisor of a conformal map. Let M be a Riemann surface with complex structure J M . For a one-form ω on M, we deﬁne a one-form ∗ ω on M by setting ∗ ω := ω ◦ J M . A one-form ω with values in the set of all complex numbers C is decomposed into the one-form of type (1, 0) and that of type (0, 1). We model R4 on the set of all quaternions H and R3 on the set of all purely imaginary quaternions Im H. For a quaternion a, we denote by a the quaternionic conjugate of a. Then, the inner product of a and b ∈ H is ha, bi = Re(ab) = 2−1(ab + ba) and the norm of a ∈ H is |a| := (aa)1/2. If a, b ∈ Im H, then ab = −ha, bi + a × b, where × is the cross product. Let S2 be the sphere of radius one centered at the origin in Im H. Then, S2 = {a ∈ Im H | a2 = −1}. Hence, S2 is the set of all square roots of −1 in Im H. Fix a map N : M → S2. We use N instead of i for the decomposition of a one-form with values in H. Because the multiplication in H is not commu- tative, two kinds of a one-form with values in H play a role of a one-form with values in C of type (1, 0) as follows. Let ω be a one-form with values in H on M. We deﬁne a one-form ωN and a one-form ωN by setting 1 1 ω := (ω − N ∗ ω), ωN := (ω − ∗ ω N). N 2 2

N −N Then, ω decomposes because ω = ωN + ω−N = ω + ω . We see that N ∗ ωN = N ωN and ∗ ω = ω N. Clearly, ω = ωN if and only if ω−N = 0. Similarly, ω = ωN if and only if ω−N = 0. The quaternionic conjugation −N provides an identity ωN = ω . We have the following decomposition of a two-form. Lemma 1. Let ω and η be one-forms with values in H on M. Then

N −N ω ∧ η = ω ∧ η−N + ω ∧ ηN .

N N N −N Proof. Because ∗ ω = ω N and ∗ ω = N ωN , we have ω ∧ηN = ω ∧η−N = 0 ([8], (35)). Then,

N −N N −N ω ∧ η = (ω + ω ) ∧ (ηN + η−N ) = ω ∧ η−N + ω ∧ ηN .

Hence, the lemma holds.

4 We regard C as a subset {a0 + a1i ∈ H | a0, a1 ∈ R} of H. Then, H is considered as a left complex vector space C ⊕ Cj or a right complex vector space C ⊕ jC. Let z be the standard holomorphic coordinate of C and (x, y) the real coordinate such that z = x + yi. Then, ∗ dx = −dy. For a map f : C → H, we have ( ) ∂f ∂f ∂f ∂f 2(df) = dx + dy − N ∗ dx + dy N ∂x ∂y ∂x ∂y ( ) ∂f ∂f ∂f ∂f = dx + dy − N − dy + dx ∂x ∂y ∂x ∂y ( ) ( ) ∂f ∂f ∂f ∂f = − N dx + N − N dy ∂x ∂y ∂x ∂y ( ) ( ) ∂f ∂f ∂f ∂f = (dx + N dy) − N = (dx) − N , ∂x ∂y N ∂x ∂y ( ) ∂f ∂f 2(df)N = − N (dx)N . ∂x ∂y

¯ −i A function h: M → C is holomorphic if and only if ∂h = (dh)−i = (dh) = 0. Let λ: M → H be a map. If (dλ)−i = 0, then λ = λ0 + λ1j with holomorphic −i functions λ0 : M → C and λ1 : M → C. If (dλ) = 0, then λ = λ0 + jλ1 with holomorphic functions λ0 : M → C and λ1 : M → C. A non-constant map f : M → H is called a conformal map with left normal 2 N : M → S if (df)−N = 0 ([13], Definition 2.1). Taking the quaternionic conjugate, we have (df)N = 0. A map f : M → H is called a conformal map with right normal N : M → S2 if (df)N = 0 ([2], Definition 2). A holomorphic function f : M → C is a conformal map with left normal i and right normal −i. Let N be a constant map. We define a complex structure J of H by Ja := Na for each a ∈ H. Then, df + J ∗ df = df + N ∗ df = 2(df)−N = 0. Hence f is holomorphic with respect to a complex structure J. Similarly, if R is constant, then f is holomorphic with respect to a complex structure of H. We recall a zero of a conformal map. Let p ∈ U and z a holomorphic coordinate on U centered at p. A map f : U → H vanishes to order at least n ≥ 0 at p if |f| ≤ C|z|n for some constant C > 0. If |f| ≤ C|z|n for some constant C > 0, but there does not exist a constant C˜ > 0 such that |f| ≤ C˜|z|n+1, then a map f vanishes to order n at p. Lemma 2. For a map N : U → N(U) ( S2, there exists a ∈ H with |a| = 1 such that ψ := Na + ai does not vanish on U and Nψ = ψi.

Proof. For a ∈ H with |a| = 1, the map c 7→ aca−1 is a Euclidean motion in Im H. Hence, there exists a ∈ S2 such that N =6 −aia−1. Hence, Na + ai does not vanish. We see that Nψ = −a + Nai = ψi. We assume that f : U → H is conformal with left normal N such that N(U) =6 S2. By Theorem 3.5 in [13] and Lemma 3.9 in [7], there exist a

5 nowhere-vanishing map φf : U → H, a map ξ : U → H, and a non-negative integer n which depends only on f, such that | | n ξ(z) f(z) = ψ(z φf (z) + ξ(z)), lim < ∞. z→0 |z|n+1

The integer n is called the order of f at p and denoted by ordp f. As an analog of a zero, we introduce the notion of a pole of a conformal map as follows. Let f : U \{p} → H be a conformal map with left normal N such that N(U \{p}) ( S2 and |z|−n|f| ≤ C for some constant C > 0, but there does not exist a constant C˜ > 0 such that |z|−n−1|f| ≤ C˜ for a negative integer n.

Lemma 3. There exists a nowhere-vanishing map φf : U → H and a map ξ : U → H such that ( ) φf (z) | |−n−1| | ∞ f(z) = ψ − + ξ(z) , lim z ξ(z) < . (1) z n z→0 where n is a negative integer.

Proof. We give a proof which is parallel to the proof of Lemma 3.9 in [7]. Let −n us deﬁne λ := λ0 + jλ1 with λ0, λ1 : U → C by f = ψλz . Then, the equation (df)−N = 0 becomes

−n −n (dψ)−N λz + ψ(d(λz ))−i = 0.

Let α0 and α1 be complex one-forms on U such that (dψ)−N = ψ(α0 + jα1). Then, the above equation becomes

−n −n −n α0λ0z − α1λ1z + ∂(λ0z ) = 0, −n −n −n α0λ1z + α1λ1z + ∂(λ1z ) = 0.

Simplifying this system of equations, we have

α0λ0 − α1λ1 + ∂λ0 = 0, −1 α0λ1 + (α1 − nz )λ1 + ∂λ1 = 0.

This system of equations has the same form as equation (51) in [7]. Hence, there exists a nowhere-vanishing map φf : U → H and a map ξ : U → H such that (1) holds for a negative integer n. As the order of a zero, the negative integer n depends only on f.

Deﬁnition 1. We call a point p in Lemma 3 a pole of f and n the order of the pole p.

6 A pole p of f is of order at least n if and only if |z|n|f| ≤ C for some constant C > 0. If |z|n|f| ≤ C for some constant C > 0, but there does not exist a constant C˜ > 0 such that |z|n+1|f| ≤ C˜, then the point p is a pole of order n. If f is a meromorphic function on U, then N = i. Choosing a = −i/2, we have ψ = 1. Then, the order of f as a conformal map is equal to that as a meromorphic function. Recall that a divisor on M is a map D : M → Z such that, for any compact subset K of M, the set {p ∈ M | D(p) =6 0} ∩ K is a finite set (see [6]). The set {p ∈ M | D(p) =6 0} is called the support∑ of D and denoted by supp D. The degree of a divisor D is defined by deg D := p∈M D(p). We denote by Div(M) the set of all divisors on M. We introduce the notion of the divisor of a conformal map as follows. Let P be a subset of M such that, for any compact subset K of M, the set P ∩ K is a finite set. Let f : M \ P → H be a conformal map with left normal N and poles at P. We define a map (f): M → Z by (f)(p) := ordp f for each p ∈ M. Let us define nonnegative maps Z : M → Z and P : M → Z by (f) = Z − P . The map P is a divisor on M by the assumption. The map (f) is a divisor on M if and only if Z is a divisor on M. Definition 2. We assume that (f) is a divisor on M. We call (f) the divisor of f and the map f a conformal map with divisor (f). We call the divisors Z and P the zero divisor of f and the polar divisor of f respectively.

3 Meromorphic functions

We factor a meromorphic function on a Riemann surface. We model the conformal two-sphere S2 on the complex projective line CP 1. ∼ Lemma 4. A map N : M → S2 = CP 1 is holomorphic if and only if N satisfies −N (dN)N = (dN) = 0. Proof. The lemma follows from the formula between the differential of a conformal map, the mean curvature vector, and the left normal, and the fact that the Gauss map of a minimal surface is holomorphic ([2]). Here, we give a proof by direct calculation. 2 For a map N : M → S , we define real-valued function n1, n2, and n3 by N = n1i + n2j + n3k. Put M+ := M \{p ∈ M | N(p) = k} and M− := M \{p ∈ M | N(p) = −k}. −N We assume that (dN)N = (dN) = 0. This equation becomes

n1 ∗ dn1 + n2 ∗ dn2 − n3 ∗ dn3 = 0,

dn1 − n2 ∗ dn3 + n3 ∗ dn2 = 0,

dn2 − n3 ∗ dn1 + n1 ∗ dn3 = 0,

dn3 − n1 ∗ dn2 + n2 ∗ dn1 = 0.

7 The map st: S2 \{k} → C deﬁned by

x1 x2 st(x1i + x2j + x3k) = + i (x1, x2, x3 ∈ R) 1 − x3 1 − x3 is the stereographic projection from k. We deﬁne functions l1 and l2 with values in R on M+ by

n1 n2 l1 + l2i := st ◦N = + i. 1 − n3 1 − n3 Then,

dn1(1 − n3) − n1 d(1 − n3) dn1 − dn1 n3 + n1 dn3 dn1 + ∗ dn2 dl1 = 2 = 2 = 2 , (1 − n3) (1 − n3) (1 − n3) dn2 − dn2 n3 + n2 dn3 dn2 − ∗ dn1 dl2 = 2 = 2 , (1 − n3) (1 − n3) Hence,

(d(l1 + l2i))−i = 0. | Then, l1 + l2i is a holomorphic function. Hence, N M+ is a holomorphic map. The map

−iNi = n1i − n2j − n3k is a rotation of N centered at the origin. We have M \{p ∈ M | − i(N(p))i = } − | k = M−. In an analogous discussion as above, we show that iNi M− is a | holomorphic map. This is equivalent to having N M− as a holomorphic map. Therefore, N is a holomorphic map on M. Conversely, we assume that N is holomorphic. Then, (d(l1 + l2i))−i = 0. Because 2 2 − 2l1 2l2 l1 + l2 1 N = 2 2 i + 2 2 j + 2 2 k, l1 + l2 + 1 l1 + l2 + 1 l1 + l2 + 1 we have ( ) 2(−l2 + l2 + 1) 4l l dN = 1 2 dl − 1 2 dl i (l2 + l2 + 1)2 1 (l2 + l2 + 1)2 2 ( 1 2 1 2 ) 2(l2 − l2 + 1) 4l l + 1 2 dl − 1 2 dl j (l2 + l2 + 1)2 1 (l2 + l2 + 1)2 2 ( 1 2 1 2 ) 4l 4l + 1 dl + 2 dl k, (l2 + l2 + 1)2 1 (l2 + l2 + 1)2 2 (1 2 1 2 ) 4l l 2(−l2 + l2 + 1) N dN = 1 2 dl + 1 2 dl i (l2 + l2 + 1)2 1 (l2 + l2 + 1)2 2 ( 1 2 1 2 ) 2 − 2 4l1l2 2(l1 l2 + 1) + 2 2 2 dl1 + 2 2 2 dl2 j (l1 + l2 + 1) (l1 + l2 + 1)

8 ( ) − 4l2 4l1 + 2 2 2 dl1 + 2 2 2 dl2 k. (l1 + l2 + 1) (l1 + l2 + 1) | − | Hence, (dN)N M+ = 0. A similar discussion for iNi shows that (dN)N M− = −N 0. Hence, (dN)N = (dN) = 0 over M. ∼ Corollary 1. A map N : M → S2 = CP 1 is anti-holomorphic if and only if −N is holomorphic. In other words, a map N : M → S2 is anti-holomorphic if N and only if (dN)−N = (dN) = 0. Proof. We have

−(n + n i) −(n2 + n2) st ◦(−N) = 1 2 = 1 2 1 + n3 (1 + n3)(n1 − n2i) −(1 − n2) −(1 − n ) 1 = 3 = 3 = − . (1 + n3)(n1 − n2i) n1 − n2i st ◦N

Hence, N is anti-holomorphic if and only if −N is holomorphic; that is (dN)−N = (dN)N = 0.

Proof of Theorem 1. (b) follows from (a) and Corollary 1. We show (a). We assume that λ0 and λ1 are holomorphic functions at p and set λ := −i −1 −1 λ0 + jλ1. Then (dλ) = 0. Because |λiλ | = 1, the map N := λiλ is deﬁned on a zero of λ. Then

dN = dλ iλ−1 − λiλ−1 dλ λ−1, N ∗ dN = λiλ−1 ∗ dλ iλ−1 + ∗ dλ λ−1 = −λiλ−1dλ λ−1 + dλ iλ−1.

Hence, (dN)N = 0. Then, N is holomorphic at p. Conversely, we assume that N is holomorphic at p. Then, (dN)N = 0. For any a ∈ H with |a| = 1, the quaternion aia−1 is a rotation of i centered at the origin in Im H. Hence, there exists a map ξ with |ξ| = 1 such that N = ξiξ−1. The equation (dN)N = 0 becomes

dξ iξ−1 − ξiξ−1 dξ ξ−1 = ξiξ−1 ∗ dξ iξ−1 + ∗ dξ ξ−1.

Simplifying this equation, we have

iξ−1(dξ)−i = ξ−1(dξ)−i i.

−1 −i Hence ω := ξ (dξ) is a complex-valued (0, 1)-form. Let ξ0 and ξ1 be complex- valued functions such that ξ = ξ0 + jξ1. Then, ¯ ¯ ∂ξ0 + j ∂ξ1 = ξ0ω + jξ1ω.

Hence, ¯ ¯ ω = ∂ log ξ0 = ∂ log ξ1.

9 Then, there exist holomorphic functions λ0 and λ1 at p such that λ0ξ0 = λ1ξ1. Then

−1 −1 N = ξiξ = (ξ0 + jξ1)i(ξ0 + jξ1) −1 −1 = (λ0ξ0 + jλ0ξ1)λ0 iλ0(λ0ξ0 + jλ0ξ1) −1 = (λ1ξ1 + jλ0ξ1)i(λ1ξ1 + jλ0ξ1) −1 = (λ1 + jλ0)i(λ1 + jλ0) . Thus, the theorem holds.

−1 If N = (λ0 + jλ1)i(λ0 + jλ1) is a holomorphic map with holomorphic functions λ0 and λ1, then N = ΛiΛ with Λ := (λ0 + jλ1)/|λ|. From this factorization, we have a relation between the degree of N and the degree of λ when M is closed.

Corollary 2. Let M be a closed Riemann surface, λ0 and λ1 are meromorphic −1 functions on M, λ := λ0 + jλ1, and N := λiλ . The degree of a holomorphic map N is m if and only if the degree of λ0 and λ1 are m. Proof. We assume that the degree of N is m. The equation N = i has m solutions counting multiplicities. This equation is equivalent to the equation iλ = λi. Rewriting this equation, we have λ0i = λ0i and −λ1i = λ1i. The former equation is trivial; the latter is equivalent to λ1 = 0. Hence the equation λ1 = 0 has m solutions counting multiplicities. Then, λ1 is a meromorphic function of degree m. Next, we consider the equation N = −i. This equation is equivalent to λ0 = 0. Hence λ0 is a meromorphic function of degree m. The converse is trivial.

4 Super-conformal maps

We factor a super-conformal map. We recall the deﬁnition and basic properties of a super-conformal map (see [2]). A conformal map f : M → H with left normal N : M → S2 and right normal R: M → S2 is called a super-conformal map if its curvature ellipse is a circle. A conformal map f is super-conformal if and only if N or R is anti-holomorphic. That is, f is super-conformal if and only if (dN)−N = 0 or (dR)−R = 0 by Corollary 1. ∼ Lemma 5. A holomorphic map and an anti-holomorphic map from M to S2 = CP 1 are super-conformal.

2 Proof. Let N : M → S be a holomorphic map. By Lemma 4, (dN)N = 0. Then, N is a conformal map with left normal −N. The map −N is anti- holomorphic by Corollary 1. Hence, N is super-conformal. ∼ Let N : M → CP 1 = S2 be an anti-holomorphic map. By Corollary 1, (dN)−N = 0. Then, N is a conformal map with left normal N. Hence, N is super-conformal.

10 Proof of Theorem 2. Because ψ is nowhere-vanishing, any map f : M → H is factored by the product ψ(λ0 + jλ1) with complex functions λ0 and λ1 on M. Let λ := λ0 + λ1j. The functions λ0 and λ1 are holomorphic if and only if (dλ)−i = 0. We have

d(ψλ) = dN aλ + ψ dλ,

2(d(ψλ))−N = dN aλ + ψ dλ + N ∗ (dN aλ + ψ dλ) = 2ψ(dλ)−i.

Hence, f = ψλ with (dλ)−i = 0 if and only if f is super-conformal with anti- holomorphic left normal N.

By the above theorem, the set of super-conformal maps from M to H with ∼ anti-holomorphic left normal N : M → S2 \{k} = C is parametrized by two holomorphic functions. Let z be the standard holomorphic coordinate of C,(x, y) the real coordinate such that z = x + yi, and set D := {z ∈ C | |z| < 1}. The following is an analog of the Schwarz lemma (see [9]). Corollary 3. Let f : D → H be a super-conformal map with anti-holomorphic ∼ left normal N : D → S2 \{k} = C. We assume that f(0) = 0 and f is bounded. Then, there exists C > 0 such that

∂f ∂f |f(z)| ≤ 2C|z|, (0) − N (0) ≤ 2C. ∂x ∂y

−1 θ0i θ1i The equality holds if and only if N = aia , f(z) = ψ(C0e z + C1e zj) and 2 2 1/2 ≥ C = (C0 + C1 ) (C0,C1 0) with real-valued functions θ0 and θ1.

Proof. Let f = ψ(λ0+λ1j) be the factorization by Theorem 2. Because f(0) = 0 and ψ is nowhere-vanishing, we have λ0(0) = λ1(0) = 0. Also, because |f| is bounded and |ψ| = |Na + ai| ≤ |Na| + |ai| ≤ 2, the map λ := λ0 + λ1j is bounded. The equality |ψ| = 2 holds if and only if Na = ai. Then N is a constant map. Because |λn| ≤ |λ|, the functions λ0 and λ1 are bounded. Let |λn(z)| ≤ Cn (n = 0, 1). By the Schwarz lemma, we have |λn(0)| ≤ Cn|z| θni and |(∂λn/∂z)(0)| ≤ Cn. The equality holds if and only if λn(z) = Cne z with real-valued function θn. Then

|f(z)| = |ψ(z)||λ(z)| ≤ 2(|λ (z)|2 + |λ (z)|2)1/2 ≤ 2(C2 + C2)1/2, ( 0 )1 ( 0 1 )

∂f ∂f ∂ψ ∂ψ ∂λ ∂λ (0) − N (0) = (0) − N (0) λ(0) + ψ(0) (0) − i (0) ∂x ∂y ∂x ∂y ∂x ∂y ( )

∂λ ∂λ ∂λ0 ∂λ1 = ψ(0) (0) − i (0) = |ψ(0)| (0) + (0)j ∂x ∂y ∂z ∂z ( ) 2 2 1/2 ∂λ0 ∂λ1 2 2 1/2 ≤ 2 (0) + (0) ≤ 2(C + C ) . ∂z ∂z 0 1

Hence the corollary holds.

11 The following is an analog of the Schwarz-Pick theorem (see [9]). Let B3 := {a ∈ H | |a| < 1}. Corollary 4. Let f : D → B3 be a super-conformal map with anti-holomorphic → 2 ∈ → left normal N : D S . For z1 D, deﬁne M¨obiustransforms τz1 : D D 3 → 3 and Θz1 : B B by ( ) − −1 z z1 − − τz1 (z) = , Θz1 (a) = (a f(z1)) 1 f(z1)a . 1 − z1z ◦ ◦ We assume that the image of the left normal of a conformal map g := Θz1 f τ −1 : D → B3 is contained in S2 \{k}. Then for all z ∈ D, we have z1 2

| − | − f(z1) f(z2) z1 z2 ≤ 2C , 1 − z1z2 1 − f(z1)f(z2)

∂f ∂f (x1 + y1i) (x1 + y1i) ∂x ∂y ≤ 2C 2 = 2 2 . 1 − |f(x1 + y1i)| 1 − |f(x1 + y1i)| 1 − |x1 + y1i| Proof. Rouxel [14] showed that a conformal transform of a super-conformal map is a super-conformal map. Hence g is a super-conformal map with g(0) = 0. Then the image of the map Θ ◦ N ◦ τ −1 is contained in a round sphere in H. z1 z1 By the assumption, the image of the left normal of g is contained in S2 \{k}. By Cororllary 3, there exists C > 0 such that |g(z)| ≤ 2C |z|. Hence ( ) −1 − z z1 (f(z) − f(z1)) 1 − f(z1)f(z) ≤ 2C . 1 − z1z Then

| − | − f(z2) f(z1) z2 z1 ≤ 2C 1 − z1z2 1 − f(z1)f(z2) for any z2 ∈ D. Let z1 = x1 + y1i and z2 = x2 + y1i,(x1, x2, y1 ∈ R). Then

|f(x + y i) − f(x + y i)| x − x 2 1 1 1 ≤ 2C 2 1 . − 1 − f(x1 + y1i)f(x2 + y1i) 1 (x1 + y1i)(x2 + y1i)

Hence

|f(x + y i) − f(x + y i)| 1 2 1 1 1 ≤ 2C . − |x2 − x1| 1 − f(x1 + y1i)f(x2 + y1i) 1 (x1 + y1i)(x2 + y1i)

Let x2 tend to x1. Then

∂f (x1 + y1i) ∂x ≤ 2C 2 2 . 1 − |f(x1 + y1i)| 1 − |x1 + y1i|

12 Because f is confomal, we have

∂f ∂f (x1 + y1i) = (x1 + y1i) . ∂x ∂y

Then the corollary holds. 2 − 2 Let ds be the Poincar´emetric on D with curvature 1 and dsB3 be the Poincar´emetric on B3 with curvature −1. Then 4 ds2 = (dx ⊗ dx + dy ⊗ dy), D (1 − (x2 + y2))2 ∑3 2 4 ds 3 = ∑ (dan ⊗ dan). B − 3 2 2 (1 n=0 an) n=0 The following is a geometric interpretation of Corollay 4

Corollary 5. Let f : D → B3 be a super-conformal map with anti-holomorphic 2 left normal N : D → S . We denote by D˜ the set of all points z1 ∈ D such ◦ ◦ −1 → 3 that the image of the left normal of a conformal map Θz1 f τz1 : D B is 2 \{ } ∗ 2 ≤ 2 2 ˜ contained in S k . Then f dsB3 4C dsD on D. Proof. A straightforward calculation prove the corollary. Recalling the deﬁnition of a pole of a conformal map, the map f := ψλ with λ = λ0 + jλ1 for meromorphic functions λ0 and λ1 is a super-conformal map with poles. Hence, a super-conformal map with poles adopts properties of a meromorphic function. For example, the following is an analog of the Weierstrass factorization theorem (see [6]).

Corollary 6. Any divisor on M is a divisor of a super-conformal map with poles.

Proof. By the Weierstrass factorization theorem, any divisor on M is a divisor of a meromorphic function. Let D be a divisor on M and λ a meromorphic function on M with divisor D. Then, f := ψλ: M \ supp D → H is a super- conformal map by Theorem 6 and (f) = D by the deﬁnition of a divisor of a conformal map.

We assume that M is a connected open subset of a closed Riemann surface M˜ . We denote by C1(M) the set of all one-chains in M. We deﬁne a map δ : C1(M) → Div(M) by, for c: [0, 1] → M,  1 (p = c(1)), (δ(c))(p) := −1 (p = c(−1)),  0 (otherwise).

The following is an analog of the Abel-Jacobi theorem (see [6]).

13 Corollary 7. Let D be a divisor on M with deg D = 0. Then, D is the divisor of a super-conformal map from M with poles and left normal N : M → S2 \{k}, if and only if there exists c ∈ C1(M˜ ) such that δ(c) = D and ∫ ω = 0 c for every holomorphic one-form ω on M˜ . Proof. By the Abel-Jacobi theorem, the divisor D is a divisor of a meromorphic function λ on M˜ . Hence f := ψλ: M \ supp D → H is a super-conformal map by Theorem 6. We see that (f) = D by the deﬁnition of a divisor of a conformal map.

5 A conformal maps with induced metric

We connect a conformal map with a classical surface. R Let f : M → H be a conformal map with (df)−N = (df) = 0. We induce a (singular) metric on M from H by f. Consequently, the Gauss curvature K, the normal curvature K⊥ and the mean curvature vector H of f can be deﬁned. We have

df H = −N(dN)N , H df = R(dR)R, 1 K|df|2 = (h∗ dR, R dRi + h∗ dN, N dNi), 2 1 K⊥|df|2 = (h∗ dR, R dRi − h∗ dN, N dNi). 2 ([2], Proposition 8, Proposition 9). A conformal map f is minimal if and only if N is holomorphic or, equivalently, R is holomorphic. Hence, if f is super- conformal and minimal, then N or R is a constant map. Then, f is a holomorphic map with respect to a complex structure of H. We consider the class of surfaces with |K| = |K⊥|. Chen [4] completely classiﬁed Wintgen ideal surfaces in this class. Wintgen [17] showed that K + |K⊥| ≤ |H|2 for any conformal map and K + |K⊥| = |H|2 if and only if a conformal map is super-conformal. A super-conformal map is called a Wintgen ideal surface in [15]. We denote by σ the area element of the two sphere with radius one. R Lemma 6. Let f : M → H be a conformal map with (df)−N = (df) = 0. If |K| = |K⊥|, then N ∗σ = 0 or R∗σ = 0. Proof. From the assumption, we have K = K⊥. Then h∗ dN, N dNi = 0 or h∗ dR, R dRi = 0. It is known that h∗ dN, N dNi = N ∗σ and h∗ dR, R dRi = R∗σ ([2], Proposition 10). Hence the lemma holds. If N ∗σ = 0, then N is not anti-holomorphic. Hence, if f is super-conformal R ⊥ ∗ with (df)−N = (df) = 0 and |K| = |K |, then (a) N σ = 0 and R is anti- holomorphic or (b) R∗σ = 0 and N is anti-holomorphic.

14 Because H = C ⊕ jC, we can identify H with C2. A conformal map f : M → 2 ∼ R θi C = H with (df)−N = (df) = 0 is Lagrangian if and only if R = je where θ is a real-valued function ([12]). The map R = jeθi is not anti-holomorphic. Hence if f is super-conformal and Lagrangian, then N is an anti-holomorphic map.

6 The Whitney sphere

We apply Theorem 2 to the Whitney sphere for an example of the previous sections. Let (u, v) be the standard coordinate of R2. The map f : R2 → H deﬁned by sin u f(u, v) = (sin v − (cos u sin v)i − (cos v)j − (cos u cos v)k) 1 + (cos u)2

4 3 is a parametrization of the Whitney sphere (see [4]). We deﬁne ρabc : R → R by 2 ρabc(x0, x1, x2, x3) = (xa, xb, xc). Figures 1, 2, 3, and 4 are plots of (ρabc◦f)(R ).

Castro [3] showed that the Whitney sphere is the only Lagrangian super- conformal map from a closed Riemann surface. Chen [4] showed that the Whit- ney sphere is a super-conformal map with |K| + |K⊥| = |H|2 and |K| = |K⊥|. By the discussion in the previous section, we can assume that the left normal of f is anti-holomorphic and the right normal is jeβi for a real-valued function β. We have that

fu fv df = fu du + fv dv = |fu| du + |fv| dv, |fu| |fv| 1 ( f (u, v) = cos u sin v ((sin u)2 + 2) + sin v (3(sin u)2 − 2)i u (1 + (cos u)2)2 ) + cos u cos v ((cos u)2 − 3)j + cos v (3(sin u)2 − 2)k , sin u f (u, v) = (cos v + (− cos u cos v)i + (sin v)j + (cos u sin v)k) , v 1 + (cos u)2 1 (sin u)2 |f |2 = , hf , f i = 0, |f |2 = . u 1 + (cos u)2 u v v 1 + (cos u)2

Hence f is not immersion at (nπ, v) for any integer n and any v ∈ R. Deﬁne a 2 complex structure J of R by J(fu/|fu|) = fv/|fv|. We have

fv fu ∗ df = |fu| du − |fv| dv. |fv| |fu| Then the left normal N and the right normal R are respectively ( ) −1 ( fv fu sin u − 2 N(u, v) = = 2 ( 2(sin u) cos v sin v)i |fv| |fu| | sin u|(1 + (cos u) )

15 0.3 x3 0.8 -0.3 x2 -0.3 -0.8 0 0.3 x1

2 Figure 1: (ρ123 ◦ f)(R )

16 0.3 x 3 0 -0.3 0.8

-0.8 0 x 0 2 x0 0.8 -0.8

2 Figure 2: (ρ023 ◦ f)(R )

17 0.3 x 3 0 -0.3

-0.8 0 0.3 x0 0 0.8-0.3 x1

2 Figure 3: (ρ013 ◦ f)(R )

18 0.8 x 2 0

-0.8

-0.8 0 0.3 x0 0 0.8-0.3 x1

2 Figure 4: (ρ012 ◦ f)(R )

Let V := {(u, v) ∈ R2 | N(u, v) = −i} and deﬁne a map λ: R2 \ V → H by f = (N + i)λ. Then,

cos u sin u(sin v + sin u cos v) sin u(sin v − sin u cos v) λ(u, v) = + i D D cos u sin v(sin u sin v + cos v) sin u(sin u sin v − cos v) + j + k, D D D = 4(sin u)3 cos v sin v + (sin u)4 − (sin u)2 − 2 satisﬁes the equation (dλ)−i = 0 with respect to J. Since f has no zero on R2 \ V , the map λ has no zero on R2 \ V .

7 Minimal surfaces

We give a factorization of a minimal surface. R Let f : M → H be a minimal surface with (df)−N = (df) = 0. A minimal surface g : M → H such that dg = − ∗ df is called a conjugate minimal surface of f. There exists a conjugate minimal surface of f if and only if ∗ df is exact. A conjugate minimal surface g shares the same left normal and the same right normal with the original minimal surface f. If there exists a conjugate minimal ∼ surface g : M → H, then the holomorphic map f + ig : M → C ⊗ H = C4 is called a holomorphic null curve. For a factorization of a minimal surface, we assume that M is simply connected and induce a map µ as follows. ∼ Let N : M → S2 \{k} = C be a holomorphic map. By Lemma 2, there exists a ∈ H with |a| = 1 such that ψ := −Na + ai does not vanish on M. Let λ0 and λ1 be holomorphic functions on M and λ := λ0 + λ1j. Then, (dλ)−i = 0. Put { ∈ | } Qλ0,λ1 := p M (dN)pλ(p) = 0 . Because (ψ dλ)N = 0 and (dN aλ)N = 0, the equation ψ dλ = dN aλµ deﬁnes a map µ: M \ Q → H. ∼ λ0,λ1 If N : M → S2 \{k} = C is holomorphic, then −N is anti-holomorphic by Corollary 1. The map ψλ with (dλ)−i = 0 is super-conformal with left normal − N by Theorem 2. Because N is holomorphic and (dλ)−i = 0, the set Qλ0,λ1 is discrete.

Proof of Theorem 3. (a) We assume that Φ := f + ig : M → C ⊗ H is a holomorphic null curve such that f and g : M → H are minimal surfaces with left N normal N. We have d(dN f)) = −dN ∧ df = −(dN) ∧ (df)N = 0. Then, the one-form dN f on M is exact. Hence, there exists a function Λ: M → H such

20 that dN f = dΛ. We deﬁne λ: M → H by λ := ψ−1Λ. Then,

dN f = −dN aλ + ψ dλ = d(ψλ).

Because (dN f)N = 0 and (dN aλ)N = 0, we have (ψ dλ)N = 0. Then,

(ψ dλ) − N ∗ (ψ dλ) = ψ(dλ + i ∗ dλ) = 0.

Hence, (dλ)−i = 0. Then,

dN f = dN aλ(µ − 1)

\ − on M Qλ0,λ1 . Then, f = aλ(µ 1). Because the left hand side is deﬁned on M, the right hand side is extended to M. Then,

− ∗ df = −N df = −d(Nf) + dN f = −d(Nf) − dN aλ + (−Na + ai) dλ = d(−Naλ(µ − 1) + (−Na + ai)λ) = d(−Naλµ + aiλ).

Hence g = −Naλµ + aiλ up to an additive constant. (b) We have

dN f = dN (aλ(µ − 1)) = −dN aλ + dN aλµ = −dN aλ + ψ dλ = d(ψλ).

Diﬀerentiating the above equation, we have

N −dN ∧ df = (dN) ∧ (df)−N = 0.

Hence (df)−N = 0. Then f is a minimal surface with left normal N. Then,

− ∗ df = −N df = −d(Nf) + dN f = −d(Nf) − dN aλ + (−Na + ai) dλ = d(−Naλ(µ − 1) + (−Na + ai)λ) = d(−Naλµ + aiλ)

Hence, g is a minimal surface with left normal N and Φ := f+ig is a holomorphic null curve.

The equation f = aλ(µ − 1) is a factorization of a minimal surface which has a conjugate minimal surface g and Φ := f + gi is a holomorphic null curve. The arrangement of the zeros of f is unclear because that of µ − 1 is unclear. However, we have the following property.

Corollary 8. Let f = aλ(µ − 1): M → H be a minimal surface factored by Theorem 3. A point on M is a branch point of a super-conformal map ψλ if and only if it is a zero of f or a branch point of N. Proof. From the proof of Theorem 3, we have dN f = d(ψλ). Hence the corollary holds.

21 8 The Enneper surface

We apply Corollary 8 to the Enneper surface for an example of the previous section. Let z be the standard holomorphic coordinate of C and (x, y) the real coordinate such that z = x + iy. The map f : C → Im H deﬁned by ( ) ( ) 1 1 ( ) f(z) = x − x3 + xy2 i + −y − x2y + y3 j + x2 − y2 k 3 3 is a parametrization of the Enneper surface and g : C → Im H deﬁned by ( ) ( ) 1 1 g(z) = y − x2y + y3 i + x − xy2 + x3 j + (2xy) k 3 3 is its conjugate surface. The map f has only one zero at the origin. The left normal is 2x 2y x2 + y2 − 1 N = i + j + k. x2 + y2 + 1 x2 + y2 + 1 x2 + y2 + 1

The stereographic√ projection of N from k is z. Hence, N is unbranched. Let a := (i + k)/ 2, then −Na + ai does not vanishes on C. Diﬀerentiating N, we have ( ) 2(y2 − x2 + 1) −4xy 4x dN = i + j + k dx (x2 + y2 + 1)2 (x2 + y2 + 1)2 (x2 + y2 + 1)2 ( ) −4xy 2(x2 − y2 + 1) 4y + i + j + k dy. (x2 + y2 + 1)2 (x2 + y2 + 1)2 (x2 + y2 + 1)2

Then, [ ( ) 2x y4 + 2x2y2 + 6y2 + x4 + 2x2 + 3 dN f = − 3(x2 + y2 + 1)2 ( ) 4xy 2y2 + 3 + i 2 2 2 ( 3(x + x + 1) ) 2 3y4 + 3y2 + x4 + 3x2 + j 3(x2 + y2 + 1)2 ( ) ] 2 y5 + 2x2y3 − 2y3 + x4y + 6x2y − 3y + k dx 3(x2 + y2 + 1)2 [ ( ) 2 y5 + 2x2y3 + 2y3 + x4y + 6x2y + 3y + 3(x2 + y2 + 1)2 ( ) 2 y4 + 3y2 + 3x4 + 3x2 + i 3(x2 + y2 + 1)2

22 ( ) 4xy 2x2 + 3 + j 3(x2 + y2 + 1)2 ( ) ] 2x y4 + 2x2 y2 + 6y2 + x4 − 2x2 − 3 + k dy 3(x2 + y2 + 1)2 ( y4 + 5y2 − x4 − x2 + 2 2y(y2 + 3x2) = d + i 3 (x2 + y2 + 1) 3 (x2 + y2 + 1) ( ) ( ) ) 2x 3y2 + x2 2x y3 + x2y − 3y + j + k . 3 (x2 + y2 + 1) 3 (x2 + y2 + 1)

Hence y4 + 5y2 − x4 − x2 + 2 2y(y2 + 3x2) ψλ := 2 2 + 2 2 i ( 3 (x + y) + 1) ( 3 (x + y )+ 1) 2x 3y2 + x2 2x y3 + x2y − 3y + j + k 3 (x2 + y2 + 1) 3 (x2 + y2 + 1) is a super-conformal map with left normal −N. By Corollary 8, the map ψλ has only a single branch point at the origin. ∼ Let (r, t) be the polar coordinate of R2 = C with radial coordinate r, and the angular coordinate t. We ﬁx a set U := {(r, t) | 0 ≤ r ≤ 3, 0 ≤ t ≤ 2π} ⊂ R2. 4 3 We deﬁne ρabc : R → R by ρabc(x0, x1, x2, x3) = (xa, xb, xc). Figures 5, 6, 7, and 8 are plots of (ρabc ◦ ψλ)(U). We have λ = (−Na + ai)−1 ( y4 + 5y2 − x4 − x2 + 2 2y(y2 + 3x2) × + i 3 (x2 + y2 + 1) 3 (x2 + y2 + 1) ( ) ( ) ) 2x 3y2 + x2 2x y3 + x2y − 3y + j + k 3 (x2 + y2 + 1) 3 (x2 + y2 + 1) 3xy2 − 3y2 − x3 + 3x2 + 2x − 2 = √ 6 2 y(y2 − 3x2 + 6x + 2) + √ i 6 2 3xy2 + 3y2 − x3 − 3x2 + 2x + 2 − √ j 6 2 y(y2 − 3x2 − 6x + 2) − √ k. 6 2 The map µ with (−Na + ai) dλ = dN aλµ becomes 3y6 + 9x2y4 + 29y4 + 9x4y2 + 18x2y2 + 16y2 + 3x6 + 13x4 − 8x2 + 4 µ = ( D ) −2y 18x2y2 − 13y2 − 6x4 + 27x2 − 6 + i D

23 1.5 x 3 0

-1.5

2 1 -2 0 -1 x2 0 -1 1 x1 2 -2

Figure 5: (ρ123 ◦ ψλ)(U)

24 1.5 x 3 0 -1.5

2 -2 -1 1 0 0 x x 1 2 -1 2 0 3 4 -2

Figure 6: (ρ023 ◦ ψλ)(U)

25 1.5 x 3 0 -1.5

2 -2 -1 1 0 0 x x 1 2 -1 1 0 3 4 -2

Figure 7: (ρ013 ◦ ψλ)(U)

26 2 1 x 2 0 -1 -2

2 -2 -1 1 0 0 x x 1 2 -1 1 0 3 4 -2

Figure 8: (ρ012 ◦ ψλ)(U)

27 ( ) −2x 6y4 − 18x2y2 − 27y2 + 5x2 − 6 + j ( D ) 4xy 17y2 − 13x2 + 6 + k, D D = y6 + 3x2y4 + 13y4 + 3x4y2 + 18x2y2 + 16y2 + x6 + 5x4 − 8x2 + 4.

Hence, ( ) 2 y6 + 3x2y4 + 8y4 + 3x4y2 + x6 + 4x4 µ − 1 = ( D ) −2y 18x2y2 − 13y2 − 6x4 + 27x2 − 6 + i ( D ) −2x 6y4 − 18x2y2 − 27y2 + 5x2 − 6 + j ( D ) 4xy 17y2 − 13x2 + 6 + k. D Then f = aλ(µ − 1) by Theorem 3.

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