The Schwarz Lemma for Super-Conformal Maps

Abstract

A super-conformal map is a conformal map from a two-dimensional Riemannian manifold to the Euclidean four-space such that the ellipse of curvature is a circle. Quaternionic holomorphic geometry connects super-conformal maps with holomorphic maps. We report the Schwarz lemma for super-conformal maps and related results.

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1 Introduction

For a smooth manifold M, we denote the tangent bundle by TM and its fiber at $p\in M$ by $T_pM$. Let $\varSigma$ be a two-dimensional oriented Riemannian manifold and $f:\varSigma \rightarrow \mathbb {R}^4$ be an isometric immersion. We denote the Riemannian metric of $\varSigma$ by g. For a tangent vector $X\in T_p\varSigma$, we denote the norm with respect to the Riemannian metric by $\Vert X\Vert$. We denote the second fundamental form of f by h. Then

$$\begin{aligned}\begin{gathered} \mathscr {E}_p=\left\{ h(X,X):X\in T_p\varSigma ,\Vert X\Vert =1\right\} \end{gathered}\end{aligned}$$

is called the ellipse of curvature or the curvature ellipse of f at $p\in M$ [9]. It is indeed an ellipse in the normal space at p if it does not degenerate to a point or a line segment. If the ellipse of curvature is a circle or a point at any point p, then f is said to be super-conformal [2].

The author showed that a super-conformal map is a Bäcklund transform of a minimal surface [6]. Regarding f as an isometric immersion, the inequality

$$\begin{aligned} |\mathscr {H}|^2-K-|K^\perp |\ge 0 \end{aligned}$$

(1)

holds between the mean curvature vector $\mathscr {H}$, the Gaussian curvature K and the normal curvature $K^\perp$ [10]. The equality holds if and only if f is super-conformal. From this point of view, a super-conformal map is called a Wintgen ideal surface [8]. The integral of the left-hand side of (1) over $\varSigma$ is the Willmore energy of f. This implies that a super-conformal map is a Willmore surface with vanishing Willmore energy. Hence the super-conformality is invariant under Möbius transforms of $\mathbb {R}^4$.

We discuss the Möbius geometry of super-conformal immersion by exchanging a two-dimensional oriented Riemannian manifold with a Riemann surface and an isometric immersion with a conformal immersion. Regarding $\mathbb {C}$ as a subspace of $\mathbb {R}^4$ and a holomorphic function on $\varSigma$ as a map form $\varSigma$ to $\mathbb {R}^4$, a holomorphic function satisfies (1) and it is super-conformal. We may regard Möbius geometric theory of holomorphic functions on a Riemann surface as a special case of Möbius geometry of super-conformal immersion.

The author [7] discussed super-conformal maps as a higher codimensional analogue of holomorphic functions and meromorphic functions. In this paper, we report a part of the paper which discusses the Schwarz lemma for super-conformal maps.

For the discussion, we use quaternionic holomorphic geometry [3]. Quaternionic holomorphic geometry of surfaces in $\mathbb {R}^4$ connects theory of holomorphic functions with theory of surfaces in $\mathbb {R}^4$.

2 Classical Results

We begin with reviewing the classical results of super-conformal maps by Friedrich [4] and Wong [11].

Throughout this paper, all manifolds and maps are assumed to be smooth. We compute the ellipse of curvature. We denote the inner product of $\mathbb {R}^4$ by $\langle ,\rangle$. Let $e_1$, $e_2$, $e_3$, $e_4$ be an adapted orthonormal local frame of the pull-back bundle $f^*T\mathbb {R}^4$ and $\theta _1$, $\theta _2$, $\theta _3$, $\theta _4$ the dual frame. Assume that the second fundamental form is

$$\begin{aligned}\begin{gathered} h=\sum _{p=3}^4\sum _{i,j=1}^2h_{ijp}\,\theta _i\otimes \theta _j\otimes e_p. \end{gathered}\end{aligned}$$

Then the ellipse of curvature is parametrized by the map

$$\begin{aligned}\begin{gathered} \varepsilon (u)=h(e_1\cos u+e_2\sin u,e_1\cos u+e_2\sin u)\\ =\mathscr {H} +\left( \frac{h_{113}-h_{223}}{2}e_3+\frac{h_{114}-h_{224}}{2}e_4\right) \cos 2u+\left( h_{123}e_3+h_{124}e_4\right) \sin 2u. \end{gathered}\end{aligned}$$

The map f is super-conformal map if and only if $\varepsilon (u)$ parametrizes a circle. The map f is minimal if and only if $\varepsilon (u)$ parametrize a curve of the linear transform of the circle centered at the origin. The linear transform is given by

$$\begin{aligned}\begin{gathered} P\begin{pmatrix}e_3&e_4\end{pmatrix}=\begin{pmatrix}e_3&e_4\end{pmatrix} \begin{pmatrix} h_{113}&{}h_{123}\\ h_{114}&{}h_{124} \end{pmatrix} \end{gathered}\end{aligned}$$

Hence f is super-conformal and minimal if and only if the ellipse of curvature is a circle centered at the origin.

We normalize the second fundamental form and the ellipse of curvature. Let $n(u)=e_3\cos u+e_4\sin u$. Because

$$\begin{aligned}\begin{gathered} \langle {h,n(u)}\rangle = \sum _{i,j=1}^2h_{ij3}\,\theta _i\otimes \theta _j \cos u +\sum _{i,j=1}^2h_{ij4}\,\theta _i\otimes \theta _j \sin u,\\ {{\mathrm{tr}}}\langle h,n(u)\rangle =h_{113}\cos u+h_{114}\sin u +h_{223}\cos u+h_{224}\sin u\\ =(h_{113}+h_{223})\cos u+(h_{114}+h_{224})\sin u, \end{gathered}\end{aligned}$$

we may assume that $h_{114}+h_{224}=0$. Let $A_{e_4}$ be the shape operator such that $\langle A_{e_4}(X),Y\rangle =\langle h(X,Y),e_4\rangle$ for any X, $Y\in T_p\varSigma$. Taking $e_1$ and $e_2$ as the eigenvectors of $A_{e_4}$, we may assume that $h_{124}=0$. The ellipse of curvature becomes

$$\begin{aligned}\begin{gathered} \varepsilon (u) =\frac{h_{113}+h_{223}}{2}e_3 +\left( \frac{h_{113}-h_{223}}{2}e_3+h_{114}e_4\right) \cos 2u+(h_{123}e_3)\sin 2u. \end{gathered}\end{aligned}$$

Then f is super-conformal if and only if

$$\begin{aligned}\begin{gathered} (h_{113}-h_{223})h_{123}=0,\left( \frac{h_{113}-h_{223}}{2}\right) ^2+h_{114}^2=h_{123}^2 \end{gathered}\end{aligned}$$

This is equivalent to

$$\begin{aligned}\begin{gathered} h_{123}=h_{114}=0,h_{113}=h_{223}\text { or } h_{113}=h_{223},h_{114}^2=h_{123}^2. \end{gathered}\end{aligned}$$

Hence the ellipse of curvature of a super-conformal map becomes

$$\begin{aligned}\begin{gathered} \varepsilon (u)=0 \text { or } \varepsilon (u) =h_{113} +(h_{114}e_4)\cos 2u+(\pm h_{114}e_3)\sin 2u. \end{gathered}\end{aligned}$$

If f is minimal, then the ellipse of curvature is

$$\begin{aligned}\begin{gathered} \varepsilon (u) =(h_{113}e_3+h_{114}e_4)\cos 2u+(h_{123}e_3)\sin 2u. \end{gathered}\end{aligned}$$

Hence f is super-conformal and minimal if and only if

$$\begin{aligned}\begin{gathered} \varepsilon (u) =(h_{114}e_4)\cos 2u+(\pm h_{114}e_3)\sin 2u. \end{gathered}\end{aligned}$$

Another notion is defined by the second fundamental form for surfaces in $\mathbb {R}^4$.

Definition 1

([5, 11]) The set

$$\begin{aligned}\begin{gathered} \mathscr {I}_p=\left\{ \langle h,n\rangle : n\in T_p\varSigma ^\perp ,\Vert n\Vert =1 \right\} . \end{gathered}\end{aligned}$$

is called the indicatrix of the normal curvature or the Kommerell conic of f.

The indicatrix is parametrized by

$$\begin{aligned}\begin{gathered} \iota (u)=\langle h,n(u)\rangle = \sum _{i,j=1}^2(h_{ij3}\cos u+h_{ij4}\sin u)\theta _i\otimes \theta _j. \end{gathered}\end{aligned}$$

By the normalization, we may assume that

$$\begin{aligned}\begin{gathered} \iota (u) = (h_{113}\cos u+h_{114}\sin t)\theta _1\otimes \theta _1+h_{123}\cos u\,\theta _1\otimes \theta _2\\ +h_{213}\cos u\,\theta _2\otimes \theta _1+(h_{223}\cos u-h_{114}\sin u)\theta _2\otimes \theta _2. \end{gathered}\end{aligned}$$

We regard $\langle h,n(u)\rangle$ as the shape operator which is is a symmetric (1, 1)-tensor. With the standard inner product of symmetric (1, 1)-tensors, the curve $\langle h,n(u)\rangle$ is isometrically the curve parametrized by

$$\begin{aligned}\begin{gathered} \iota (u)= \left( \frac{1}{\sqrt{2}}(h_{113}\cos u+h_{114}\sin u),\sqrt{2}h_{123}\cos u,\frac{1}{\sqrt{2}}(h_{223}\cos u-h_{114}\sin u)\right) \end{gathered}\end{aligned}$$

in $\mathbb {R}^3$. Hence f is super-conformal if and only if the indicatrix is parametrized by

$$\begin{aligned}\begin{gathered} \iota (u)=\left( \frac{1}{\sqrt{2}}(h_{113}\cos u),0,\frac{1}{\sqrt{2}}(h_{113}\cos u)\right) \end{gathered}\end{aligned}$$

or

$$\begin{aligned}\begin{gathered} \iota (u)=\left( \frac{1}{\sqrt{2}}(h_{113}\cos u+h_{114}\sin u),\pm \sqrt{2}h_{114}\cos u,\frac{1}{\sqrt{2}}(h_{113}\cos u-h_{114}\sin u)\right) . \end{gathered}\end{aligned}$$

We see that f is minimal if and only if the indicatrix is parametrized by

$$\begin{aligned}\begin{gathered} \iota (u)=\left( \frac{1}{\sqrt{2}}(h_{113}\cos u+h_{114}\sin u),\sqrt{2}h_{123}\cos u,\frac{1}{\sqrt{2}}(-h_{113}\cos u-h_{114}\sin u)\right) . \end{gathered}\end{aligned}$$

Moreover, f is super-conformal and minimal if and only if

$$\begin{aligned}\begin{gathered} \iota (u)=0\text { or }\iota (u)=\left( \frac{1}{\sqrt{2}}(h_{114}\sin u),\pm \sqrt{2}h_{114}\cos u,\frac{1}{\sqrt{2}}(-h_{114}\sin u)\right) . \end{gathered}\end{aligned}$$

Set

$$\begin{aligned}\begin{gathered} \mathscr {I}_p(X)=\left\{ \langle h(X,),n\rangle : n\in T_p\varSigma ^\perp ,\Vert n\Vert =1 \right\} \subset (T_p\varSigma )^*. \end{gathered}\end{aligned}$$

Definition 2

([4]) An immersion f is called superminimal if $\mathscr {I}_p(X)$ is a circle centered at 0 in $(T_p\varSigma )^*$.

The following lemma explains the relation among the super-conformal maps, minimal surfaces and superminimal surfaces.

Lemma 1

A map f is superminimal if and only if f is super-conformal and minimal.

Proof

For $X=X_1e_1+X_2e_2$ and the normalization, we have

$$\begin{aligned}\begin{gathered} \langle h(X,),n(u)\rangle = \sum _{i,j=1}^2X_ih_{ij3}\theta _j \cos u +\sum _{i,j=1}^2X_ih_{ij4}\theta _j \sin u\\ =((X_1h_{113}+X_2h_{123})\theta _1+(X_1h_{123}+X_2h_{223})\theta _2 )\cos u\\ +\,(X_1h_{114}\theta _1-X_2h_{114}\theta _2 )\sin u. \end{gathered}\end{aligned}$$

Hence f is superminimal if and only if

$$\begin{aligned}\begin{gathered} (X_1^2h_{113}-X_2^2h_{223})h_{114}=0,\\ (X_1h_{113}+X_2h_{213})^2+(X_1h_{123}+X_2h_{223})^2=(X_1^2+X_2^2)h_{114}^2 \end{gathered}\end{aligned}$$

Hence

$$\begin{aligned}\begin{gathered} h_{114}=X_1h_{113}+X_2h_{213}=X_1h_{123}+X_2h_{223}=0 \end{gathered}\end{aligned}$$

or

$$\begin{aligned}\begin{gathered} X_1^2h_{113}-X_2^2h_{223}=0,\\\ (X_1h_{113}+X_2h_{123})^2+(X_1h_{123}+X_2h_{223})^2=(X_1^2+X_2^2)h_{114}^2 \end{gathered}\end{aligned}$$

Because $X_1$ and $X_2$ is arbitrary under $X_1^2+X_2^2\ne 0$, we have $h=0$, or $h_{113}=h_{223}=0$ and $h_{123}^2=h_{114}^2$. Hence the lemme holds. $\square$

For a holomorphic function g(z) on $\mathbb {C}$, define a map $\tilde{g}:\mathbb {C}\rightarrow \mathbb {C}^2\cong \mathbb {R}^4$ by $\tilde{g}(z)=(z,g(z))$. Then $\tilde{g}$ is called an R-surface [5]. Kommerell showed that an R-surface is superminimal.

3 Quaternionic Holomorphic Geometry

We review super-conformal maps by quaternionic holomorphic geometry of surfaces in $\mathbb {R}^4$ [2]. We identify $\mathbb {R}^4$ with the set of all quaternions $\mathbb {H}$. The inner product of $\mathbb {R}^4$ becomes

$$\begin{aligned}\begin{gathered} \langle a,b\rangle ={{\mathrm{Re}}}(\overline{a}b)={{\mathrm{Re}}}(\overline{b}a)=\frac{1}{2}(\overline{a}b+\overline{b}a). \end{gathered}\end{aligned}$$

We identify $\mathbb {R}^3$ with the set of all imaginary parts of quaternions ${{\mathrm{Im}}}\,\mathbb {H}$. Then two-dimensional sphere with radius one centered at the origin in $\mathbb {R}^3$ is $S^2=\{a\in {{\mathrm{Im}}}\,\mathbb {H}:a^2=-1\}$.

Let $\varSigma$ be a Riemann surface with complex structure $J_\varSigma$. For a one-form $\omega$ on $\varSigma$, we define a one-form $*\,\omega$ by $*\,\omega =\omega \circ J_\varSigma$. A map $f:\varSigma \rightarrow \mathbb {H}$ is called a conformal map if $\langle df\circ J_\varSigma ,df\rangle =0$. This is equivalent to that $*\,df=N\,df=-df\,R$ with maps N, $R:\varSigma \rightarrow S^2$. Each point where f fails to be an immersion is isolated. This means that a conformal map is a branched immersion. The second fundamental form of f is

$$\begin{aligned}\begin{gathered} h(X,Y)=\frac{1}{2}(*\,df(X)\,dR(Y)-dN(X)*df(Y)). \end{gathered}\end{aligned}$$

Let $\mathscr {H}:\varSigma \rightarrow \mathbb {H}$ be the mean curvature vector of f. Then

$$\begin{aligned}\begin{gathered} df\,\overline{\mathscr {H}}=-\frac{1}{2}(*\,dN+N\,dN),\overline{\mathscr {H}}\,df=\frac{1}{2}(*\,dR+R\,dR). \end{gathered}\end{aligned}$$

The ellipse of curvature is

$$\begin{aligned}\begin{gathered} \mathscr {E}_p= \left\{ \mathscr {H}|df(e_1)|^2+\frac{1}{4}\cos 2\theta (a-b)(e_1)+\frac{1}{4}\sin 2\theta N(a+b)(e_1):\theta \in \mathbb {R}\right\} ,\\ a=df\,(*\,dR-R\,dR),b=(*\,dN-N\,dN)\,df. \end{gathered}\end{aligned}$$

Then f is super-conformal if and only one of the following equations holds.

$$\begin{aligned}\begin{gathered} *\,dR-R\,dR=0,*\,dN-N\,dN=0 \end{gathered}\end{aligned}$$

at any point $p\in \varSigma$.

In the following, we restrict ourselves to super-conformal maps with $*\,dN=N\,dN$.

Lemma 2

A super-conformal map $f:\varSigma \rightarrow \mathbb {H}$ with $*\,df=N\,df$ and $*\,dN=N\,dN$ is superminimal if and only if f is holomorphic with respect to a right quaternionic linear complex structure of $\mathbb {H}$.

Proof

A super-conformal map $f:\varSigma \rightarrow \mathbb {H}$ with $*\,df=N\,df$ and $*\,dN=N\,dN$ satisfies the equation

$$\begin{aligned}\begin{gathered} df\,\overline{\mathscr {H}}=-N\,dN. \end{gathered}\end{aligned}$$

Hence f is minimal if and only if N is a constant map. Define $J :\mathbb {H}\rightarrow \mathbb {H}$ by $Jv=Nv$ for any $v\in \mathbb {H}$. Then J is a right quaternionic linear complex structure of $\mathbb {H}$. Because $*\,df=J\,df$, the map f is holomorphic with respect to J.

By the above lemma, we see that a holomorphic map g from $\varSigma$ to $\mathbb {C}^2\cong \mathbb {C}\oplus \mathbb {C}j\cong \mathbb {H}$ is superminimal because $*\,dg=i\,dg$. A holomorphic function and an R-surface are special cases of this superminimal surface.

4 The Schwarz Lemma

Because a holomorphic function is a super-conformal map, we may expect that a super-conformal map is an analogue of a holomorphic function. A factorization of super-conformal map given in Theorem 4.3 in [7] may support this idea. The following is a variant of the theorem.

Theorem 1

([7], Theorem 4.3) Let $\phi :\varSigma \rightarrow \mathbb {H}$ be a super-conformal map with $*\,d\phi =N\,d\phi$, $*\,dN=N\,dN$ and $N\phi =\phi i$ and $h:\varSigma \rightarrow \mathbb {C}^2\cong \mathbb {C}\oplus \mathbb {C}j\cong \mathbb {H}$ be a holomorphic map. Then, a map $f=\phi h$ is a super-conformal map with $*\,df=N\,df$.

This theorem shows that a holomorphic section of a complex vector bundle of rank two, trivialized by the super-conformal map f is a super-conformal map. We see that $N+i$ is a super-conformal map with $N(N+i)=(N+i)i$. The condition $*\,dN=N\,dN$ implies N is the inverse of the stereographic projection followed by an anti-holomorphic function ([7], Corollary 3.2). Hence if $\varSigma$ is an open Riemann surface and $N:\varSigma \rightarrow S^2$ is the inverse of the stereographic projection of an anti-holomorphic function with $N(\varSigma )\subset S^2\setminus \{-i\}$, then $N+i$ is a global super-conformal trivializing section. A super-conformal map $f:\varSigma \rightarrow \mathbb {H}$ with $*\,df=N\,df$, $*\,dN=N\,dN$ always factors $f=(N+i)h$ with a holomorphic map $h:\varSigma \rightarrow \mathbb {C}\oplus \mathbb {C}j$. We don’t need to see $-i$ in a special light. If $a\in S^2$ and $a\not \in N(\varSigma )$, then we can rotate f so that $-i\not \in N(\varSigma )$. The condition that N fails to be surjective is necessary.

This fact suggests that we should distinguish the case where the Riemann surface $\varSigma$ is parabolic or hyperbolic. In the case where $\varSigma =\mathbb {C}$, we have an analogue of Liouville’s theorem.

Theorem 2

([7], Theorem 4.4) Let $\phi :\mathbb {C}\rightarrow \mathbb {H}$ be a super-conformal map with $*\,d\phi =N\,d\phi$, $*\,dN=N\,dN$ and $N\phi =\phi i$. Assume that $N(\mathbb {C})\subset S^2\setminus \{-i\}$ and $|\phi |^{-1}$ is bounded above. If $f:\mathbb {C}\rightarrow \mathbb {H}$ is a super-conformal map with $*\,df=N\,df$ and |f| is bounded above, then $f=\phi C$ for some constant $C\in \mathbb {H}$.

In the case where $\varSigma =B^2=\{z\in \mathbb {C}:|z|<1\}$, we have an analogue of the Schwarz lemma.

Theorem 3

([7], Theorem 4.5) Let $\phi :B^2\rightarrow \mathbb {H}$ be a super-conformal map with $*\,d\phi =N\,d\phi$, $*\,dN=N\,dN$ and $N\phi =\phi i$. Assume that $N(B^2)\subset S^2\setminus \{-i\}$ and $|\phi |<c$ and $|\phi |^{-1}<\tilde{c}$. If $f:B^2\rightarrow \mathbb {H}$ is a super-conformal map with $*\,df=N\,df$ and $f(0)=0$, then there exists a constant C such that

$$\begin{aligned}\begin{gathered} |f(z)|\le C|z|. \end{gathered}\end{aligned}$$

Moreover, if $f=\phi (\lambda _0+\lambda _1j)$ for holomorphic functions $\lambda _0$ and $\lambda _1$, then there exist constants $C_0$, $C_1>0$ such that

$$\begin{aligned}\begin{gathered} |f(z)|\le c(C_0^2+C_1^2)^{1/2}|z|. \end{gathered}\end{aligned}$$

The equality holds if and only if $\phi =c$ and there exists $z_0\in B^2$ such that $|\lambda _n(z)|=C_n|z_0|$ $(n=0,1)$. We also have

$$\begin{aligned}\begin{gathered} \left| f_x(0)-N(0)f_y(0)\right| \le c(C_0^2+C_1^2)^{1/2}. \end{gathered}\end{aligned}$$

The equality holds if and only if $f=c$ and there exists $z_0\in B^2$ such that $|\lambda _n(z)|=C_n|z_0|$ $(n=0,1)$.

Assume that $f(B^2)\subset B^4=\{a\in \mathbb {H}:|a|<1\}$. It is known that

$$\begin{aligned}\begin{gathered} T(a)=\frac{(1-|a_1|^2)(a-a_1)-|a-a_1|^2a_1}{1+|a|^2|a_1|^2-2\langle a,a_1\rangle } \end{gathered}\end{aligned}$$

is a Möbius transform of $\mathbb {R}^4$ with $T(a_1)=0$ [1]. The transform T is

$$\begin{aligned}\begin{gathered} T(a)=\frac{a-a_1-|a_1|^2a+|a_1|^2a_1-|a|^2a_1+a|a_1|^2+a_1\overline{a}a_1-|a_1|^2a_1}{|1-\overline{a}_1a|^2}\\ =\frac{a-a_1-|a|^2a_1+a_1\overline{a}a_1}{|1-\overline{a}_1a|^2} =\frac{(1-a_1\overline{a})(a-a_1)}{|1-\overline{a}_1a|^2}\\ =(1-a\overline{a}_1)^{-1}(a-a_1) \end{gathered}\end{aligned}$$

and T preserves $B^4$. If $f:B^2\rightarrow B^4$ is a super-conformal map with $*\,df=N\,df$ and $*\,dN=N\,dN$, then

$$\begin{aligned}\begin{gathered} *\,d(T\circ f)=(1-f\overline{a}_1)^{-1}(*\,df\,\overline{a})(1-f\overline{a}_1)^{-1}(1-f)-(1-f\overline{a}_1)^{-1}*\,df\\ =(1-f\overline{a}_1)^{-1}N(1-f\overline{a}_1)\,d(T\circ f). \end{gathered}\end{aligned}$$

It is known that a Möbius transform of a super-conformal map is super-conformal. Then we have an analogue of the Schwarz-Pick theorem.

Theorem 4

([7], Theorem 4.7) Let $\phi :B^2\rightarrow B^4$ be a super-conformal map with $*\,d\phi =N\,d\phi$, $*\,dN=N\,dN$ and $N\phi =\phi i$. Assume that $|\phi |$ and $|\phi |^{-1}$ are bounded. Let $f:\varSigma \rightarrow \mathbb {H}$ be a super-conformal map with $*\,df=N\,df$. Assume that the map

$$\begin{aligned}\begin{gathered} \tilde{N}:=(1-\tilde{f}(z)\overline{\tilde{f}(z_1)})^{-1}N(1-\tilde{f}(z)\overline{\tilde{f}(z_1)}):\varSigma \rightarrow S^2 \end{gathered}\end{aligned}$$

satisfies $\tilde{N}(B^2)\subset S^2\setminus \{-i\}$. Then there exists a constant $C>0$ such that

$$\begin{aligned}\begin{gathered} \frac{\left| f(z)-f(z_1)\right| }{\left| 1-\overline{f(z_1)}f(z)\right| } \le C\frac{\left| z-z_1\right| }{\left| 1-\overline{z_1}z\right| } \end{gathered}\end{aligned}$$

Moreover,

$$\begin{aligned}\begin{gathered} \frac{\left| f_x(z_1)\right| }{1-|f(z_1)|^2}=\frac{\left| f_y(z_1)\right| }{1-|f(z_1)|^2}\le \frac{C}{1-|z_1|^2}. \end{gathered}\end{aligned}$$

We fix Riemannian metrics $ds_{B^2}^2$ on $B^2$ and $ds_{B^4}^2$ on $B^4$ as

$$\begin{aligned}\begin{gathered} ds^2_{B^2}=\frac{4}{(1-(x^2+y^2))^2}(dx\otimes dx+dy\otimes dy),\\ ds^2_{B^4}=\frac{4}{(1-\sum _{n=0}^3a_n^2)^2}(\sum _{n=0}^3da_n\otimes da_n). \end{gathered}\end{aligned}$$

Then a geometric version of the Schwarz-Pick theorem becomes as follows.

Theorem 5

Let $\phi :B^2\rightarrow B^4$ be a super-conformal map with $*\,d\phi =N\,d\phi$, $*\,dN=N\,dN$ and $N\phi =\phi i$. Assume that $|\phi |$ and $|\phi |^{-1}$ are bounded. Let $f:\varSigma \rightarrow \mathbb {H}$ be a super-conformal map with $*\,df=N\,df$. Assume that the map

$$\begin{aligned}\begin{gathered} \tilde{N}:=(1-f(z)\overline{f(z_1)})^{-1}N(1-f(z)\overline{f(z_1)}):\varSigma \rightarrow S^2 \end{gathered}\end{aligned}$$

satisfies $\tilde{N}(B^2)\subset S^2\setminus \{-i\}$. Then there exists a constant $C>0$ such that $f^*ds_{B^4}^2\le C ds_{B^2}^2$.