On conformal Gauss maps

TL;DR

This paper characterizes conformal Gauss maps of surfaces in spheres, providing an invariant formulation and a streamlined proof of their relation to harmonic maps of Willmore surfaces.

Contribution

It offers a new invariant approach and an efficient proof for identifying conformal Gauss maps of Willmore surfaces, extending prior characterizations.

Findings

Invariant formulation of conformal Gauss maps

Characterization of harmonic maps as conformal Gauss maps

Simplified proof of the Dorfmeister--Wang result

Abstract

We characterise the maps into the space of $2$-spheres in $S^n$ that are the conformal Gauss maps of conformal immersions of a surface. In particular, we give an invariant formulation and efficient proof of a characterisation, due to Dorfmeister--Wang \cites{DorWan13,DorWan}, of the harmonic maps that are conformal Gauss maps of Willmore surfaces.

Equations

$$f^{1,0}=\mathrm{span}\{\sigma ,d_Z\sigma |\sigma \in \Gamma f,Z\in T^{1.0}\Sigma \}.$$

$$V=\mathrm{span}\{\sigma ,d_Z\sigma ,d_{\bar{Z}}\sigma ,d_Z{d}_{\bar{Z}}\sigma |\sigma \in \Gamma f,Z\in \Gamma T^{1,0}\Sigma \}.$$

$$d=\mathcal{D}+\mathcal{N},$$

$$0$$

$${\mathcal{N}}_Z{V}^{\perp }=(\ker {\mathcal{N}}_Z{}_{|V}{)}^{\perp }\subseteq (f^{0,1}{)}^{\perp }\cap V=f^{0,1}\le \ker {\mathcal{N}}_Z{}_{|V}$$

$${\mathcal{N}}_Z{V}^{\perp }\subseteq f^{0,1}.$$

$$({\mathcal{D}}_{\bar{Z}}u,w)$$

$$({\mathcal{D}}_{\bar{Z}}u,u)$$

$${\tau }_V=2i\ast d^{\mathcal{D}}{\mathcal{N}}^{1,0}=-2i\ast d^{\mathcal{D}}{\mathcal{N}}^{0,1}.$$

$${\tau }_V{V}^{\perp }\subseteq f.$$

$$\mathcal{N}\circ {\tau }_V{}_{|V^{\perp }}$$

$$({\tau }_V{)}^2{}_{|V^{\perp }}$$

$$A:=\{p\in \Sigma |{\mathcal{N}}_Z{V}^{\perp }={\mathcal{N}}_{\bar{Z}}{V}^{\perp }\ne \{0\},Z\in T_p\Sigma \}$$

$$d_{\bar{Z}}\sigma ={\mathcal{D}}_{\bar{Z}}\sigma +{\mathcal{N}}_{\bar{Z}}\sigma ={\mathcal{D}}_{\bar{Z}}\sigma \in W,$$

$$\begin{gathered} {\mathcal{D}}_{\bar{Z}}w={\mathcal{D}}_{\bar{Z}}({\mathcal{N}}_Z\nu )=({\mathcal{D}}_{\bar{Z}}{\mathcal{N}}_Z)\nu +{\mathcal{N}}_Z({\mathcal{D}}_{\bar{Z}}\nu ) \\ ={\tau }_V\nu +{\mathcal{N}}_{[\bar{Z},Z]}^{1,0}\nu +{\mathcal{N}}_Z({\mathcal{D}}_{\bar{Z}}\nu )\in {\mathcal{N}}_Z{V}^{\perp }+{\tau }_V{V}^{\perp }. \end{gathered}$$

Full text

Introduction

A useful tool in conformal surface geometry is the central sphere congruence \citelist[Bla29]*§67[Tho24] or conformal Gauss map [Bry84]. Geometrically, the central sphere congruence of a surface in the conformal $n$-sphere attaches to each point of the surface a $2$-sphere, tangent to the surface at that point and having the same mean curvature vector as the surface at that point. The space of $2$-spheres in $S^{n}$ may be identified with the Grassmannian of $(3,1)$-planes in $\mathbb{R}^{n+1,1}$ and so the central sphere congruence may be viewed as a map, the conformal Gauss map, to this Grassmannian.

The utility of this construction is that it links the (parabolic) conformal geometry of the sphere to the (reductive) pseudo-Riemannian geometry of the Grassmannian. For example, a surface is Willmore if and only if the conformal Gauss map is harmonic \citelist[Bla29]*§81[Bry84][Eji88][Rig87]. In another direction, away from umbilic points, the metric induced by the conformal Gauss map, which is in the conformal class of the surface, is invariant by conformal diffeomorphisms of $S^{n}$ and even arbitrary rescalings of the ambient metric [Fia44, Wan98].

The purpose of this short note is to characterise those maps into the Grassmannian which are the conformal Gauss map of a conformal immersion. In so doing, we build on a result of Dorfmeister–Wang [DorWan13, DorWan] which treats the case where the map is harmonic. As a by-product of our analysis, we give an invariant formulation and efficient proof of their result.

It is a pleasure to thank David Calderbank, Udo Hertrich-Jeromin and Franz Pedit for their careful reading of and helpful comments on a previous draft of this paper.

1. The conformal Gauss map

We view the conformal $n$-sphere $S^{n}$ as the projective lightcone $\mathbb{P}(\mathcal{L})$ of $\mathbb{R}^{n+1,1}$ \citelist[Dar72a]*Livre II, Chapitre VI[Her03]*Chapter 1. Here $\mathcal{L}=\{v\in\mathbb{R}^{n+1,1}_{\times}\mathrel{|}(v,v)=0\}$ and $(\,,\,)$ is the signature $(n+1,1)$ inner product.

Let $f:\Sigma\to S^{n}=\mathbb{P}(\mathcal{L})$ be a conformal immersion of a Riemann surface into the conformal $n$-sphere. Equivalently, $f$ is a null line subbundle of the trivial bundle $\underline{\mathbb{R}}^{n+1,1}:=\Sigma\times\mathbb{R}^{n+1,1}$.

Define $f^{1,0}\leq\underline{\mathbb{C}}^{n+2}$ by

[TABLE]

Here the notation $U\leq V$ means $U$ is a subbundle of $V$. That $f$ is a conformal immersion is equivalent to $f^{1,0}$ being a rank $2$ isotropic subbundle of $\underline{\mathbb{C}}^{n+2}$. Set $f^{0,1}:=\overline{f^{1,0}}$ and note that111We make no notational distinction between a real bundle and its complexification $f^{1,0}\cap f^{0,1}=f$.

The conformal Gauss map of $f$ is the bundle of $(3,1)$-planes $V\leq\underline{\mathbb{R}}^{n+1,1}$ given by

[TABLE]

We have a decomposition $\underline{\mathbb{R}}^{n+1,1}=V\oplus V^{\perp}$ which induces a decomposition of the flat connection $\operatorname{d}$:

[TABLE]

where $\mathcal{D}$ is a metric connection preserving $V$ and $V^{\perp}$ while $\mathcal{N}$ is a $1$-form taking values in skew-endomorphisms of $\underline{\mathbb{R}}^{n+1,1}$ which permute $V$ and $V^{\perp}$.

Remark*.*

We may view $V$ as a map from $\Sigma$ into the Grassmannian of $(3,1)$-planes in $\mathbb{R}^{n+1,1}$ and then $\mathcal{N}$ can be identified with its differential.

The flatness of $\operatorname{d}$ yields the structure equations of the situation:

[TABLE]

Here $\operatorname{d}^{\mathcal{D}}$ is the exterior derivative on bundle-valued forms with $\mathcal{D}$ used to differentiate coefficients.

The conformal Gauss map $V$ is defined by the following properties:

$f^{0,1}\leq V$; 2. 2.

$f^{0,1}\leq\ker\mathcal{N}^{1,0}$.

Now, for $Z$ a local section of $T^{1,0}\Sigma$, $\mathcal{N}_{Z}$ is skew while $f^{0,1}$ is maximal isotropic in $V$ so that

[TABLE]

and we conclude:

Lemma 1.1** (c.f. [DorWan13]*Proposition 2.2).**

If $V$ is the conformal Gauss map of a conformal immersion then $(\mathcal{N}^{1,0})^{2}{}_{|V^{\perp}}=0$.

Following [DorWan13], we say that $V$ with the property of 1.1 is strongly conformal.

The conformal Gauss map also satisfies a second order condition. First note that (1.2) tells us that

[TABLE]

Moreover, $f^{0,1}$ is $\mathcal{D}^{0,1}$-stable thanks to the following lemma which will see further use in Section 2:

Lemma 1.2**.**

Let $W\leq V$ be maximal isotropic in $V$ with a never-vanishing section $w$ such that $\mathcal{D}_{\bar{Z}}w\in W$, for $\bar{Z}\in T^{0,1}\Sigma$. Then $W$ is $\mathcal{D}^{0,1}$-stable.

Proof.

Let $u\in\Gamma W$ be a local section so that $u,w$ locally frame $W$. It suffices to show that $\mathcal{D}_{\bar{Z}}u\in W$. However,

[TABLE]

since $W$ is isotropic. Thus $\mathcal{D}_{\bar{Z}}u\in W^{\perp}\cap V=W$ since $W$ is maximal isotropic in $V$. ∎

In the case at hand, for $\sigma\in\Gamma f$, we have $\mathcal{D}_{\bar{Z}}\sigma\in f^{0,1}$ by definition so 1.2 applies to show that $f^{0,1}$ is $\mathcal{D}^{0,1}$-stable.

Now contemplate the tension field $\tau_{V}:=*\operatorname{d}^{\mathcal{D}}*\mathcal{N}$ of $V$. Since $*\mathcal{N}=i(\mathcal{N}^{1,0}-\mathcal{N}^{0,1})$, (1.1b) yields

[TABLE]

In view of the last paragraph, $\tau_{V}V^{\perp}$ takes values in $f^{0,1}$ since $*\operatorname{d}^{\mathcal{D}}\mathcal{N}^{1,0}V^{\perp}$ does. However, $\tau_{V}$ is real so that $\tau_{V}V^{\perp}$ takes values in $f^{0,1}\cap f^{1,0}=f$:

[TABLE]

In particular, since $f$ is a null line subbundle on which $\mathcal{N}$ vanishes, we conclude:

Proposition 1.3**.**

If $V$ is the conformal Gauss map of a conformal immersion with tension field $\tau_{V}$. Then:

[TABLE]

This line of argument additionally give us some control on the rank of $\mathcal{N}:T\Sigma\otimes V^{\perp}\to V$:

Lemma 1.4**.**

Let $V$ be the conformal Gauss map of a conformal immersion $f$, then the set

[TABLE]

is nowhere dense.

Proof.

Any open set in the closure of $A$ must contain an open set where $\mathcal{N}_{Z}V^{\perp}=\mathcal{N}_{\bar{Z}}V^{\perp}\neq\{0\}$. On this latter set, we immediately see from (1.3) that $\mathcal{N}_{Z}V^{\perp}=f$. Since $\tau_{V}V^{\perp}\leq f$ also, by (1.4), we rapidly conclude (c.f. 2.2 below) that $f$ is $\mathcal{D}^{0,1}$-stable and so $\mathcal{D}$-stable. Since $\mathcal{N}f=0$ also, $f$ is constant: a contradiction. ∎

In the next section, we will establish a generic converse to these results.

2. Reconstruction of $f$ from $V$

Suppose now that we have a bundle $V\leq\underline{\mathbb{R}}^{n+1,1}$ of $(3,1)$-planes and ask whether $V$ is the conformal Gauss map of some conformal immersion $f$. Our task is then to construct $f\leq V$ but, in fact, it will be more convenient to construct $f^{0,1}$:

Proposition 2.1**.**

Let $W\leq V$ be a maximal isotropic subbundle of $V$ such that:

$W$* is $\mathcal{D}^{0,1}$-stable;* 2. 2.

$\mathcal{N}^{1,0}W=0$, or, equivalently (c.f. (1.2)), $\mathcal{N}_{Z}V^{\perp}\subseteq W$, for all $Z\in T^{1,0}\Sigma$.

Then $f:=W\cap\overline{W}$ is a real, null, line subbundle which, on the open set where it immerses, is conformal with $W=f^{0,1}$ and conformal Gauss map $V$.

Proof.

Since $V$ has signature $(3,1)$, $W$ and $\overline{W}$ must intersect in a line bundle, necessarily null and real. Since $f$ is real, $f\leq\ker\mathcal{N}^{1,0}\cap\ker\mathcal{N}^{0,1}$ so that, for $\sigma\in\Gamma f$, $\bar{Z}\in T^{0,1}\Sigma$,

[TABLE]

since $f\leq W$ and $W$ is $\mathcal{D}^{0,1}$-stable. Thus $W=f^{0,1}$ on the set where $f$ immerses. We conclude that, on that set, $f$ is conformal, since $f^{0,1}$ is isotropic and $V$ is the conformal Gauss map of $f$ since $f^{0,1}\leq\ker\mathcal{N}^{1,0}$. ∎

For our main result, we need the following simple observation:

Lemma 2.2**.**

Let $V\leq\underline{\mathbb{R}}^{n+1,1}$ be a bundle of $(3,1)$-planes with tension field $\tau_{V}$. Let $w=\mathcal{N}_{Z}\nu$, for $\nu\in\Gamma V^{\perp}$ and $Z\in T\Sigma$. Then $\mathcal{D}_{\bar{Z}}w\in\mathcal{N}_{Z}V^{\perp}+\tau_{V}V^{\perp}$.

Proof.

For suitable $Z\in T^{1,0}\Sigma$, $\tau_{V}=\mathcal{D}_{\bar{Z}}\mathcal{N}_{Z}-\mathcal{N}^{1,0}_{[\bar{Z},Z]}$ so that

[TABLE]

∎

With all this in hand, we have:

Theorem 2.3**.**

Let $V\leq\underline{\mathbb{R}}^{n+1,1}$ be a bundle of $(3,1)$-planes with tension field $\tau_{V}$. Suppose that:

$V$* is strongly conformal.* 2. 2.

Equations (1.5) hold. 3. 3.

$\{p\in\Sigma\mathrel{|}\mathcal{N}_{Z}V^{\perp}=\mathcal{N}_{\bar{Z}}V^{\perp}\neq\{0\},Z\in T_{p}\Sigma\}$* is empty.*

Set $U:=\mathcal{N}_{Z}V^{\perp}+\tau_{V}V^{\perp}$ and restrict attention to the open dense subset of $\Sigma$ where $U$ has fibres of locally constant dimension and so is a vector bundle.

Then $\operatorname{rank}U\leq 2$ and we have:

1. (a)

Where $\operatorname{rank}U=2$, there is a unique real, null line subbundle $f\leq V$ which, where it immerses, is a conformal immersion with conformal Gauss map $V$. 2. (b)

Where $\operatorname{rank}U=1$, there are exactly two real, null line subbundles $f,\hat{f}\leq V$, which, where they immerse, are conformal immersions with conformal Gauss map $V$. In this case, $V$ is harmonic and $f,\hat{f}$ are a dual pair of Willmore, thus $S$-Willmore [Eji88], surfaces. 3. (c)

Where $\operatorname{rank}U=0$, $V$ is constant and there are infinitely many real, null line subbundles $f\leq V$ defining conformal immersions with conformal Gauss map $V$.

Proof.

First note that hypotheses 1 and 2 amount to the assertion that $U\leq V$ is isotropic so that $\operatorname{rank}U\leq 2$.

We now consider each possibility for $\operatorname{rank}U$ in turn.

First suppose that $\operatorname{rank}U=2$. Then $U$ is maximal isotropic in $V$ and is $\mathcal{D}^{0,1}$-stable by 1.2 in view of 2.2. By construction $\mathcal{N}_{Z}V^{\perp}\subseteq U$ so that we may take $U=W$ in 2.1 to learn that $V$ is the conformal Gauss map of $f=U\cap\overline{U}$ where the latter immerses.

Now suppose that $\operatorname{rank}U=1$. We claim that $U=\mathcal{N}_{Z}V^{\perp}$: first this holds on a dense open set $\Omega$, (if $\mathcal{N}_{Z}V^{\perp}$ vanishes on an open set, so does $\tau_{V}$) so that, by hypothesis 3, we have $U\cap\overline{U}=\{0\}$ on $\Omega$. Since $\tau_{V}$ is real, we must have $\tau_{V}=0$ on $\Omega$ and hence everywhere so that the claim follows and $V$ is a harmonic map. It is now immediate that $U$ is $\mathcal{D}^{0,1}$-stable. By hypothesis 3, we have that $U\cap\overline{U}=\{0\}$ everywhere so that there are exactly two real, null line subbundles $f_{1},f_{2}\leq V$ orthogonal to $U\oplus\overline{U}$ and we set $W_{i}=f_{i}\oplus U$, $i=1,2$. 1.2, applied to a section $w$ of $U$ assures us that each $W_{i}$ is $\mathcal{D}^{0,1}$-stable so that 2.1 gives that each $f_{i}$ is conformal where it immerses with conformal Gauss map $V$. In this case, the $f_{i}$ are dual Willmore surfaces.

Finally, if $\mathcal{N}^{1,0}=0$ then $\mathcal{N}$ vanishes also so that $V$ is $\operatorname{d}$-stable and so constant. Thus $S^{2}:=\mathbb{P}(\mathcal{L}\cap V)$ is a conformal $2$-sphere and any conformal immersion $f:\Sigma\to S^{2}$ (in particular, any meromorphic function on $\Sigma$, off its branch locus) has $V$ as conformal Gauss map. ∎

Remarks*.*

The caveat that $f$ immerse is not vacuous: one can readily construct $V$ satisfying the hypotheses of 2.3 for which $f$ we find is constant. Indeed, given constant $f\in\mathbb{P}(\mathcal{L})$, let $W\leq\underline{\mathbb{C}}^{n+2}$ be a non-constant rank $2$ isotropic subbundle containing $f$ with $\overline{W}$ holomorphic with respect to the trivial holomorphic structure of $\underline{\mathbb{C}}^{n+2}$ and choose $V^{\perp}$ to be a complement to $W+\overline{W}$ in $f^{\perp}$. Then it is not difficult to show that $W$ is $\mathcal{D}^{0,1}$-stable and $\mathcal{N}^{1,0}W=\{0\}$. 2. 2.

For strongly conformal $V$, equations (1.5) are not independent. Indeed, when $\operatorname{rank}\mathcal{N}^{1,0}{}_{|V^{\perp}}=2$, $\mathcal{N}_{Z}V^{\perp}$ is maximal isotropic in $V$ so that (1.5a) forces $\tau_{V}V^{\perp}\leq\mathcal{N}_{Z}V^{\perp}$. Thus $\tau_{V}V^{\perp}$ is isotropic and (1.5b) holds. Again, when $\operatorname{rank}\mathcal{N}^{1,0}{}_{|V^{\perp}}=1$, it is easy to see that (1.5a) holds automatically.

In the interesting case of harmonic $V$ (so that $\tau_{V}=0$), matters simplify considerably. Here, of course, hypothesis 2 of 2.3 is vacuous. Moreover, $\mathcal{N}^{1,0}$ is a holomorphic $1$-form with respect to the Koszul–Malgrange holomorphic structure of $\underline{\mathbb{C}}^{n+2}$ with $\bar{\partial}$-operator $\mathcal{D}^{0,1}$. It follows that $\mathcal{N}_{Z}{}_{|V^{\perp}}$ has constant rank off a divisor and, moreover, that there is a $\mathcal{D}^{0,1}$-holomorphic subbundle of $\underline{\mathbb{C}}^{n+2}$ that coincides with $\mathcal{N}_{Z}V^{\perp}$ away from that divisor. In this setting, we conclude with Dorfmester–Wang:

Corollary 2.4** (c.f. \citelist[DorWan13]*Theorem 3.11[DorWan]*Theorem 3.11).**

Let $V\leq\underline{\mathbb{R}}^{n+1,1}$ be a strongly conformal harmonic bundle of $(3,1)$-planes.

Let $U\leq V$ be the $\mathcal{D}^{0,1}$-holomorphic, isotropic bundle that coincides with $\mathcal{N}_{Z}V^{\perp}$ off a divisor.

1. (a)

if $\operatorname{rank}U=2$, there is a unique real, null line subbundle $f\leq V$ which, where it immerses, is a Willmore, non $S$-Willmore, surface with conformal Gauss map $V$. 2. (b)

if $\operatorname{rank}U=1$ and $U\cap\overline{U}=\{0\}$, there are exactly two real, null line subbundles $f,\hat{f}\leq V$, which, where they immerse, are a dual pair of $S$-Willmore surfaces.

Remark*.*

In the notation of Dorfmeister–Wang [DorWan13, DorWan], after a gauge transformation that renders $V,V^{\perp}$ constant, $\mathcal{N}^{1,0}$ is represented by the matrix $B_{1}$.