A new class of austere submanifolds

TL;DR

This paper introduces a new family of explicit austere submanifolds in Euclidean space that are non-Kaehler and of higher rank, expanding the known examples beyond minimal Kaehler cases.

Contribution

It provides the first explicit construction of higher-rank, non-Kaehler austere submanifolds using holomorphic data and a Weierstrass type parametrization.

Findings

First explicit examples of higher-rank, non-Kaehler austere submanifolds

Construction method based on holomorphic data

Uses a Weierstrass type parametrization

Abstract

Austere submanifolds of Euclidean space were introduced in 1982 by Harvey and Lawson in their foundational work on calibrated geometries. In general, the austerity condition is much stronger than minimality since it express that the nonzero eigenvalues of the shape operator of the submanifold appear in opposite pairs for any normal vector at any point. Thereafter, the challenging task of finding non-trivial explicit examples, other than minimal immersions of Kaehler manifolds, only turned out submanifolds of rank two, and these are of limited interest in the sense that in this special situation austerity is equivalent to minimality. In this paper, we present the first explicitly given family of austere non-Kaehler submanifolds of higher rank, and these are produced from holomorphic data by means of a Weierstrass type parametrization.

Equations

$$\psi (\xi (x))=(f(x),\xi (x))$$

$${\alpha }_1={\beta }_1\left(1-{\varphi }_0^2,i(1+{\varphi }_0^2),2{\varphi }_0\right)$$

$$\alpha ={\beta }_2\left(1-{\varphi }_1^2,i(1+{\varphi }_1^2),2{\varphi }_1\right)$$

$${\Lambda }_g(u,v)={\left(\mathrm{span}\{g_u,g_v,g_{uu},g_{uv}\}\right)}^{\perp }$$

$${\Lambda }_g=\alpha \wedge {\alpha }_z.$$

$$F_g(p,\xi )=g(p)+\xi ,p=\pi (\xi ).$$

$$G=(\cos \phi F_0,\sin \phi F_{\pi /2}):M^n\to {\mathbb{R}}^{n+2}\oplus {\mathbb{R}}^{n+2}\equiv {\mathbb{R}}^{2n+4},\phi \in [0,\pi ],$$

$$N_{g}L=N_1^g\oplus N_2^g\oplus \cdots \oplus N_m^g,m=[(n+1)/2],$$

$$N_k^g(p)=\text{mbox}span\left\{{\alpha }_g^{k+1}(e_1,\dots ,e_1,e_1)(p),{\alpha }_g^{k+1}(e_1,\dots ,e_1,e_2)(p)\right\}$$

$${\alpha }_g^s(X_1,\dots ,X_s)={\left({\nabla }_{X_s}^{\perp }\dots {\nabla }_{X_3}^{\perp }{\alpha }_g(X_2,X_1)\right)}^{\perp }$$

$${\mathcal{E}}_k^g(p)=\{{\alpha }_g^{k+1}(e(\theta ),\dots ,e(\theta ))(p):e(\theta )=\cos \theta e_1+\sin \theta e_2\text{mbox}and\theta \in [0,2\pi )\}.$$

$${\alpha }_g^{k+1}(e_1,\dots ,e_1,e_1)(p),{\alpha }_g^{k+1}(e_1,\dots ,e_1,e_2)(p)$$

$$TM=\mathcal{H}\oplus \mathcal{V}$$

$$\mathcal{V}={\mathcal{V}}^1\oplus {\mathcal{V}}^0$$

$${\omega }_{ij}^k=⟨{\tilde{\nabla }}_{e_k}{e}_i,e_j⟩,1\le k\le 2,1\le i,j\le n+2$$

$${\alpha }_{11}^2={\alpha }_g^2(e_1,e_1)={\kappa }_1{e}_3\text{mbox}and{\alpha }_{12}^2={\alpha }_g^2(e_1,e_2)={\kappa }_1{e}_4.$$

$${\alpha }_{111}^3=({\nabla }_{e_1}^{\perp }{\alpha }_{11}^2{)}_{N_2^g}={\kappa }_1({\nabla }_{e_1}^{\perp }{e}_3{)}_{N_2^g}$$

$${\alpha }_{112}^3=({\nabla }_{e_1}^{\perp }{\alpha }_{12}^2{)}_{N_2^g}={\kappa }_1({\nabla }_{e_1}^{\perp }{e}_4{)}_{N_2^g}$$

$$\mathcal{H}=\text{mbox}span\{E_1,E_2\},{\mathcal{V}}^1=\text{mbox}span\{E_3,E_4\}\text{mbox}and{\mathcal{V}}^0=\text{mbox}span\{E_5,\dots ,E_n\}$$

$$F_g=g+\sum _{j=1}^{n-2}{t}_j{E}_{j+2}$$

$$A_{\xi }=\left[\begin{array}{ccccc}{\kappa }_1+h_1& h_2& r_1& s_1& \\ h_2& -{\kappa }_1-h_1& r_2& s_2& \\ r_1& r_2& 0& 0& \\ s_1& s_2& 0& 0& \end{array}\right],A_{\eta }=\left[\begin{array}{ccccc}{h}_2& {\kappa }_1-h_1& r_2& s_2& \\ {\kappa }_1-h_1& -h_2& -r_1& -s_1& \\ r_2& -r_1& 0& 0& \\ s_2& -s_1& 0& 0& \end{array}\right].$$

$$h_j=\frac{1}{{\Omega }^2}\left(t_1{D}_1^j+\cdots +t_4{D}_4^j\right),j=1,2,$$

$$2(r_1{r}_2+s_1{s}_2)h_2+(r_1^2+s_1^2-r_2^2-s_2^2)({\kappa }_1+h_1)=0$$

$$2(r_1{r}_2+s_1{s}_2)(h_1-{\kappa }_1)-(r_1^2+s_1^2-r_2^2-s_2^2)h_2=0.$$

$$r_1{r}_2+s_1{s}_2=0\text{mbox}and{r}_1^2+s_1^2=r_2^2+s_2^2$$

$$a_1{a}_2+b_1{b}_2=0\text{mbox}and{a}_1^2+b_1^2=a_2^2+b_2^2.$$

$$⟨{\alpha }_{111}^3,{\alpha }_{112}^3⟩={\kappa }_1^2(a_1{a}_2+b_1{b}_2)=0$$

$$\Vert {\alpha }_{111}^3{\Vert }^2={\kappa }_1^2(a_1^2+b_1^2)={\kappa }_1^2(a_2^2+b_2^2)=\Vert {\alpha }_{112}^3{\Vert }^2,$$

$${\alpha }_g^{k+1}(e_1,\dots ,e_1,e_1),{\alpha }_g^{k+1}(e_1,\dots ,e_1,e_2)$$

$$T{|}_{\mathcal{H}\oplus {\mathcal{V}}^1}=\left[\begin{array}{ccccc}0& 1& 0& 0& \\ -1& 0& 0& 0& \\ 0& 0& 0& -1& \\ 0& 0& 1& 0& \end{array}\right]$$

$$A_{\xi }\circ T=-T\circ A_{\xi }\text{mbox}and{A}_{\eta }\circ T=-T\circ A_{\eta }.$$

Full text

After the celebrated paper by Harvey and Lawson [8] on calibrated geometries the classification of austere Euclidean submanifolds became a rather challenging task in submanifold theory. An isometric immersion $f\colon M^{n}\to\mathbb{R}^{N}$ of a Riemannian manifold $M^{n}$, $n\geq 2$, into Euclidean space is called austere if the nonzero eigenvalues of the shape operator for any normal vector at any point appear in opposite pairs, or equivalently, if all odd degree symmetric polynomials on these eigenvalues vanish.

The notion of austerity was introduced by Harvey and Lawson [8] in connection with the class of special Lagrangian submanifolds in complex Euclidean space $\mathbb{C}^{N}$ that are not only minimal but absolutely area minimizing. Given an isometric immersion $f\colon M^{n}\to\mathbb{R}^{N}$, the embedding of its normal bundle $\psi\colon N_{f}M\to\mathbb{R}^{N}\oplus\mathbb{R}^{N}$ defined by

[TABLE]

is a Lagrangian submanifold of $\mathbb{C}^{N}\equiv\mathbb{R}^{N}\oplus\mathbb{R}^{N}$ with respect to the complex structure $J(X,Y)=(-Y,X)$. Then $\psi$ is special Lagrangian if and only if $f$ is austere.

In the special case of a submanifold $M^{n}$ in $\mathbb{R}^{N}$ of rank $\rho=2$, that is, when the kernel of the second fundamental form (called the relative nullity subspace) of the submanifold has constant dimension (called the index of relative nullity) $n-2$, we have that austerity and minimality are equivalent. Notice that $\rho$ is the rank of the Gauss map with values in the Grassmannian $G_{n,N}$ of oriented subspaces. But for submanifolds for higher rank, the austerity condition is much more demanding than minimality. This makes it rather hard to find examples of austere submanifolds other than the obvious examples of holomorphic isometric immersions of Kaehler manifolds into $\mathbb{C}^{N}$. In fact, we know from [6] that for an isometric immersion of a Kaehler manifold into $\mathbb{R}^{N}$ to be austere it suffices to be minimal, but these immersions are always the “real part” of a holomorphic one in $\mathbb{C}^{N}$.

The quest to construct new examples of austere submanifolds was initiated by Bryant [1] who classified the rank two submanifolds of dimension three as well as a quite simple family of examples of higher dimension called generalized helicoids. Bryant showed that the interesting examples of dimension three are “twisted cones” over minimal surfaces in spheres. As for dimension four, he provided a careful full pointwise description of the structures of all possible second fundamental forms. In a somehow dual parametric form, Bryant’s construction in the three dimensional case was extended by Dajczer and Florit [3] to submanifolds of rank two of any dimension. Roughly speaking, they showed that these submanifolds are subbundles of the normal bundles of a class of Euclidean or spherical surfaces called elliptic that, in addition, satisfy that the ellipses of curvature of a certain order are circles. But outside special cases, it is not known how to generate these surfaces. Finally, the four dimensional case was intensively studied by Ionel and Ivey [9], [10] building on Bryant’s algebraic results. In particular, they obtained a non-parametric classification in the special case of the submanifolds ruled by planes.

In this paper, we take advantage of our results in [7] in order to characterize in an explicit parametric form a class of austere submanifolds $M^{n}$ in $\mathbb{R}^{n+2}$ of dimension $n\geq 4$ and rank $\rho=4$. Besides being the first non-trivial known examples, other than minimal Kaehler submanifolds, having any possible dimension and rank $\rho>2$, what makes this new class of particular interest is that they are given in terms of a Weierstrass type parametrization depending on $n$ holomorphic functions on a domain. Consequently, the same is true for the special Lagrangian submanifolds that can be constructed from them as shown above.

Before stating our results, we first briefly recall some facts that can be seen exposed with many details in [7]. In fact, in the sequel we will make systematic use of results in that paper, sometimes without further referrence.

A substantial minimal surface $g\colon L^{2}\to\mathbb{R}^{N}$ is called $m$-isotropic, $m\geq 1$, if at any point of $L^{2}$ all ellipses of curvature (defined below) until order $m$ are circles. Being substantial means that the surface is not contained in any proper affine subspace of $\mathbb{R}^{N}$, in fact, not even locally since $g$ is real analytic. It is well-known that $g\colon L^{2}\to\mathbb{C}^{N/2}\cong\mathbb{R}^{N}$ for $N$ even is a holomorphic curve if and only if the ellipses of curvature of any order at any point are circles; for instance see [2].

Any simply connected $m$-isotropic surface admits a Weierstrass type representation given in [5] based on results in [2]. In particular, any simply connected $2$-isotropic surface is obtained as follows: Start with a nonzero holomorphic map $\alpha_{0}\colon U\to\mathbb{C}^{N-4}$ on a domain $U\subset\mathbb{C}$ and define $\alpha_{1}\colon U\to\mathbb{C}^{N-2}$ by

[TABLE]

where $\phi_{0}=\int_{U}^{z}\alpha_{0}dz$ and $\beta_{1}\neq 0$ is any holomorphic function. Define $\alpha\colon U\to\mathbb{C}^{N}$ by

[TABLE]

where $\phi_{1}=\int_{U}^{z}\alpha_{1}dz$ and $\beta_{2}\neq 0$ is any holomorphic function. If $\phi=\int_{U}^{z}\alpha dz$ then $g=\mbox{Re}\,\phi$ is a $2$-isotropic surface in $\mathbb{R}^{N}$.

It is easy to see that the above procedure yields examples of $2$-isotropic surfaces with complete metrics. For instance, see the construction at the final part of [5].

Let $g\colon L^{2}\to\mathbb{R}^{n+2}$ be a $1$-isotropic oriented surface. Then let $\Lambda_{g}\subset N_{g}L$ be the vector subbundle of the normal bundle of $g$ with $(n-2)$-dimensional fibers

[TABLE]

where $g=g(u,v)$ is parametrized in local isothermal coordinates. If $g=\mbox{Re}\,\phi$ is as above, then

[TABLE]

It was shown in [7] that the dimension of $\Lambda_{g}(u,v)$ may fail to be $n-2$ only at isolated points and that the vector bundle extends smoothly to these points. Hence, from now on $\pi\colon\Lambda_{g}\to L^{2}$ denotes the extended vector bundle.

Let $F_{g}\colon\Lambda_{g}\to\mathbb{R}^{n+2}$ be the immersion associated to $g$ defined on $\pi\colon\Lambda_{g}\to L^{2}$ by

[TABLE]

In the sequel, we denote by $M^{n}$ the manifold $\Lambda_{g}$ when endowed with the metric induced by $F_{g}$ and by $j\colon L^{2}\to M^{n}$ the immersion in $M^{n}$ of the zero-section of $\Lambda_{g}$. We have by construction that $F_{g}$ is a $(n-2)$-ruled submanifold, and it is easily seen that $j(L)$ is a totally geodesic cross section that is orthogonal to the rulings.

Conversely, let $F\colon M^{n}\to\mathbb{R}^{n+2}$, $n\geq 4$, be an austere $(n-2)$-ruled isometric immersion that has rank $\rho=4$ on an open dense subset of $M^{n}$. If there exists a totally geodesic global cross section $j\colon L^{2}\to M^{n}$ orthogonal to the rulings, then the surface $g=F\circ j\colon L^{2}\to\mathbb{R}^{n+2}$ is $2$-isotropic and $F$ can be parametrized by $F_{g}$.

Assume that $L^{2}$ is simply connected. By Theorem $4$ in [7] there is a one-parameter family of minimal isometric immersions $F_{\theta}$ for $\theta\in[0,\pi)$ with $F_{0}=F_{g}$ such that each $F_{\theta}$ is also austere carrying the same rulings and relative nullity subspaces as $F_{g}$. Consequently, we have the isometric immersions in higher codimension

[TABLE]

that are also austere with the same rulings and relative nullity subspaces.

The following result analyzes when the submanifold $M^{n}$ above is Kaehler, which turns out to be always the case for $n=4$. On the other hand, we see that the property of being Kaehler is exceptional for higher even dimensions.

If $M^{n}$ above is Kaehler and simply-connected, being $F_{g}$ not holomorphic it follows from a result in [4] that $F_{g}$ admits an non-trivial associated one-parameter family of isometric minimal immersions. It can be shown that this family coincides with the one discussed after Theorem 1.

1 The proofs

Let $g\colon L^{2}\to\mathbb{R}^{n+2}$, $n\geq 4$, be a substantial $1$-isotropic isometric immersion. Hence the surface is minimal and the first ellipse of curvature is a circle at all points. The minimality condition yields that the normal bundle of $g$ splits along an open dense subset of $L^{2}$ as the orthogonal sum

[TABLE]

of the higher normal bundles and these have rank two except possible the last one that has rank one if $n$ is odd. Given an orthonormal tangent frame $\{e_{1},e_{2}\}$ we have

[TABLE]

at $p\in L^{2}$. Here $\alpha_{g}^{2}=\alpha_{g}\colon TL\times TL\to N_{g}L$ is the second fundamental form of $g$ and $\alpha_{g}^{s}\colon TL\times\cdots\times TL\to N_{g}L$, $s\geq 3$, is the $s^{th}$-fundamental form defined inductively by

[TABLE]

where $(\;\;)^{\perp}$ denotes the projection onto the normal complement of $N^{g}_{1}\oplus\cdots\oplus N^{g}_{s-2}$.

The $k^{th}$-order ellipse of curvature ${\cal E}_{k}^{g}(p)\subset N_{k}^{g}(p)$ at $p\in L^{2}$ is

[TABLE]

Then ${\cal E}_{k}^{g}(p)$ is indeed an ellipse and is a circle if and only if the vectors

[TABLE]

are orthogonal with equal norm.

Proof of Theorem 1: The minimal submanifold $F_{g}\colon M^{n}\to\mathbb{R}^{n+2}$ parametrized by (1) is $(n-2)$-ruled of rank four. Its tangent bundle splits orthogonally as

[TABLE]

where ${\cal H}$ is the tangent distribution orthogonal to the rulings and $\mathcal{V}=\ker\pi_{*}$ is the vertical bundle of the submersion $\pi$. Then $j_{*}T_{p}L={\cal H}(j(p))$ at every point $p\in L^{2}$ and the fibers of $\mathcal{V}$ form the distribution tangent to the rulings. We also have the orthogonal splitting

[TABLE]

where $\mathcal{V}^{1}$ is identified with the fibers of $N_{2}^{g}$ and $\mathcal{V}^{0}$ is identified with $N_{3}^{g}\oplus\cdots\oplus N_{m}^{g}$ and are the relative nullity subspaces of $F_{g}$.

Let $e_{1},\ldots,e_{n+2}$ be an orthonormal frame such that $N_{r}^{g}=\mbox{span}\{e_{2r+1},e_{2r+2}\}$. Denote

[TABLE]

where $\tilde{\nabla}$ stands for the connection in $\mathbb{R}^{n+2}$. By assumption

[TABLE]

Thus

[TABLE]

As shown in [7] there is an orthonormal tangent frame $E_{i}$, $1\leq i\leq n$, such that

[TABLE]

where $E_{3},E_{4}$ are taken constant in each ruling and $F_{*}E_{j}=e_{j+2}$, $3\leq j\leq n$. Then the submanifold can be parametrized as

[TABLE]

where $t_{1},\ldots,t_{n-2}\in\mathbb{R}$. Moreover, there is an orthogonal normal frame $\xi,\eta$ satisfying $\|\xi\|=\Omega=\|\eta\|$ with $\Omega\in C^{\infty}(M)$ such that the shape operators of $F_{g}$ vanish on $\mathcal{V}_{0}$ and restricted to $\mathcal{H}\oplus\mathcal{V}^{1}$ have the form

[TABLE]

Moreover, $r_{j}=-a_{j}/\Omega$, $s_{j}=-b_{j}/\Omega$, $j=1,2$, with $\kappa_{1},a_{1},a_{2},b_{1},b_{1}\in C^{\infty}(L)$ whereas

[TABLE]

where $D_{i}^{j}\in C^{\infty}(L)$, $1\leq i\leq 4$, and $t_{1},\ldots,t_{4}\in\mathbb{R}$ are independent parameters.

We obtain from (5) that if $F_{g}$ is austere then the coefficients of the terms of third order of the characteristic polynomials of both shape operators have to vanish. From this it turns out that austerity implies that

[TABLE]

and

[TABLE]

It follows that austerity yields

[TABLE]

that is equivalent to

[TABLE]

Using (3) it follows that

[TABLE]

and

[TABLE]

hence $g$ is $2$-isotropic.

To prove the converse, we have to verify that if (6) holds then the coefficient of the term of third order of the characteristic polynomial of $A_{\cos\varphi\xi+\sin\varphi\eta}$ vanishes for any $\varphi\in[0,2\pi]$. In this case, since (8) and (9) hold we can choose $e_{5}$ and $e_{6}$ collinear with $\alpha_{111}^{3}$ and $\alpha_{112}^{3}$, respectively, and the remaining of the proof is just a long but straightforward computation.

In the sequel, we will be dealing with the case when $M^{n}$ is a Kaehler manifold.

Let $F=F_{g}\colon M^{n}\to\mathbb{R}^{n+2}$, $n=2m\geq 4$, be a minimal $(n-2)$-ruled submanifold associated to a $1$-isotropic oriented surface $g\colon L^{2}\to\mathbb{R}^{n+2}$. The orientation of $L^{2}$ induces an orientation on each plane vector bundle $N_{k}^{g}$ in (2) given by the ordered pair

[TABLE]

where $\{e_{1},e_{2}\}$ is a positively oriented tangent frame. Then let the orthonormal frame $e_{1},\ldots,e_{n+2}$ be such that the pairs $e_{2r+1},\;e_{2r+2}$ spanning $N_{r}^{g}$ are positively oriented. Now define $T\colon TM\to TM$ with respect to the orthonormal frame $E_{1},\ldots,E_{n}$ as in (4) by

[TABLE]

and $T|_{\mathcal{V}^{0}}=I$. Thus $T$ leaves invariant the distributions tangent to the rulings.

Lemma 3

**. **The following facts are equivalent:

• (i)

$\alpha_{F}(TX,Y)=\alpha_{F}(X,TY)$* for all $X,Y\in TM$.*

• (ii)

$g$* is $2$-isotropic.*

*Proof: *We have that $(i)$ is equivalent to

[TABLE]

It is straightforward to verify that the above is equivalent to

[TABLE]

Thus (7) holds and $g$ is $2$-isotropic. Conversely, if $g$ is $2$-isotropic then the pair of orthogonal vectors with the same norm $\alpha_{111}^{3},\alpha_{112}^{3}$ is positively oriented. Hence, we can take $\alpha_{111}^{3}=\kappa e_{5}$ and $\alpha_{112}^{3}=\kappa e_{6}$ and (10) holds.

Proof of Theorem 2: Assume that $g$ is holomorphic. Then

[TABLE]

Moreover, from [7] the connection forms $\omega_{\alpha,\beta}=\langle\nabla^{\perp}e_{\alpha},e_{\beta}\rangle$ satisfy

[TABLE]

and

[TABLE]

where $\omega_{1},\omega_{2}$ are dual to $e_{1},e_{2}$, respectively, and $\tau_{s}=\kappa_{s}/\kappa_{s-1}$ with $\kappa_{0}=1$, $1\leq s\leq n/2$.

Let $E_{1},\ldots,E_{n}$ be an orthonormal frame as in (4). We have to show that the almost complex structure $J$ defined as $J|_{\mathcal{H}\oplus\mathcal{V}^{1}}=T|_{\mathcal{H}\oplus\mathcal{V}^{1}}$ and $JE_{2i+1}=E_{2i+2}$, $JE_{2i+2}=-E_{2i+1}$, $i\geq 2$, is parallel. That is,

[TABLE]

or equivalently,

[TABLE]

Since $g$ holomorphic, we have from [7] that

[TABLE]

We only argue for nontrivial cases:

Let $i=1$ and $j=2s+1,\;s\geq 2$. Then

[TABLE]

[TABLE]

[TABLE]

The last equality holds trivially for $s\geq 2$ and by (11) and (12) for $s=1$. The proof for the cases $i=1,2$ and $j=2s+1,2s+2$, $s\geq 1$, is similar.

Let $i=2s+1$ and $j=2r+1$ with $r\neq s$. Then

[TABLE]

where the last equality either holds trivially or follows from (11). The proof for the remaining cases is similar.

Now let us assume that $M^{n}$ is Kaehler. Being $F_{g}$ is austere we have that $g$ is $2$-isotropic. Being $F_{g}$ minimal we have

[TABLE]

for any $X,Y\in TM$. It follows easily that the three subspaces in the decomposition $TM=\mathcal{H}\oplus\mathcal{V}^{1}\oplus\mathcal{V}^{0}$ are $J$-invariant. Hence

[TABLE]

where $\epsilon=\pm 1$.

Since $g$ is $2$-isotropic we have from [7] that

[TABLE]

on $\mathcal{H}\oplus\mathcal{V}_{1}$. That $A_{\xi}J+JA_{\xi}=0$ and $A_{\eta}J+JA_{\eta}=0$ hold is equivalent to $\epsilon=1$.

We define an isometry $J^{\perp}\colon N_{g}L\to N_{g}L$ by

[TABLE]

Then $J^{\perp}$ is an almost complex structure since

[TABLE]

and

[TABLE]

We claim that $J^{\perp}$ is parallel with respect to the normal connection of $g$. Since $J$ is parallel, we have

[TABLE]

which is equivalent to

[TABLE]

for any $X\in TM$.

If $i,j\geq 3$, we have

[TABLE]

which gives

[TABLE]

If $i\geq 3$ we have

[TABLE]

Since $\nabla^{\perp}_{X}e_{3},\nabla^{\perp}_{X}e_{4}\in N_{1}^{g}\oplus N_{2}^{g}$, we obtain

[TABLE]

and

[TABLE]

Similarly, we obtain

[TABLE]

If follows from (14), (15) and (16) that

[TABLE]

and in the same way that

[TABLE]

It follows from (13), (17) and (18) that

[TABLE]

The same type of arguments yield

[TABLE]

and this proves the claim.

We have

[TABLE]

and

[TABLE]

Hence

[TABLE]

Let $\tilde{J}\colon g^{*}T\mathbb{R}^{n+2}\to g^{*}T\mathbb{R}^{n+2}$ be defined as

[TABLE]

It is now straightforward to verify that $\tilde{\nabla}\tilde{J}=0$, that is, that $\tilde{J}$ is a complex structure in $\mathbb{R}^{n+2}$ that satisfies $\tilde{J}\circ g_{*}=g_{*}\circ J,$ hence $g$ is holomorphic.

For the last statement, observe that if we had that $F_{g}$ is holomorphic then we would have in (5) that $A_{\eta}=\pm J\circ A_{\xi}$, and it is easy to verify that this cannot be the case.