中国科学院大学学报  2021, Vol. 38 Issue (6): 729-734   PDF    
Construction of totally real surfaces in complex Grassmannians
JIAO Xiaoxiang, XIN Jialin     
School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China
Abstract: We present a construction of the complex Grassmannian G(2, n+2) as a quotient of some minimal submanifold Qn+1 of $\mathbb{H}$Pn+1, then show that a surface in G(2, n+2) can be horizontally lifted to Qn+1 if and only if it is totally real.
Keywords: Grassmannian    totally real surface    horizontal lift    
复Grassmann流形中全实曲面的构造
焦晓祥, 辛嘉麟     
中国科学院大学数学科学学院, 北京 100049
摘要: 给出复Grassmann流形G(2,n+2)的全实曲面的一种构造方法,也就是把G(2,n+2)看作$\mathbb{H}$Pn+1中极小子流形Qn+1的商,并证明G(2,n+2)中的曲面可以水平提升到Qn+1中当且仅当它是全实的。
关键词: Grassmann流形    全实曲面    水平提升    

The theory of minimal surfaces is an important part of modern differential geometry. The theory is particularly fruitful when the ambient space is a symmetric space. Calabi[1] proved a rigidity theorem for minimal two-spheres of constant curvature in Sn. Bolton et al.[2] constructed all the minimal two-spheres of constant curvature in $\mathbb{C}$Pn, and showed that a totally real minimal two-sphere in $\mathbb{C}$Pn can be mapped, by a holomorphic isometry of $\mathbb{C}$Pn, into $\mathbb{R}$Pn$\mathbb{C}$Pn. Then He and Wang[3] proved a similar rigidity result for totally real minimal two-spheres in $\mathbb{H}$Pn.

In this paper, we present a construction of the complex Grassmannian G(2, n+2) due to Berndt[4], which considers G(2, n+2) as a quotient of some minimal submanifold Qn+1 of $\mathbb{H}$Pn+1. A Riemannian metric can be given on G(2, n+2) so that the projection π: Qn+1G(2, n+2) is a Riemannian submersion. Then we show that a surface in G(2, n+2) can be horizontally lifted to Qn+1 if and only if it is totally real.

Our result is a special case of Ref.[5], where the author considered a general Riemannian submersion NB, and characterized the existence of horizontal lifts of a submanifold of B using a family J of (1, 1)-tensors on B. In our paper, we make use of the fact that the projection Qn+1G(2, n+2) is a principal bundle, thus obtain a characterization by a first order PDE. Our method is largely inspired by Ref.[3], where the authors considered the Riemannian submersion S4n+3$\mathbb{H}$Pn.

1 Preliminaries

We denote by $\mathbb{H}$ the algebra of quaternions. This is a 4-dimensional vector space over $\mathbb{R}$ with basis {1, i, j, k}, and the multiplication is defined by

$ \mathrm{i}^{2}=\mathrm{j}^{2}=\mathrm{k}^{2}=-1, $
$ \mathrm{ij}=\mathrm{k}=-\mathrm{ji}, \mathrm{jk}=\mathrm{i}=-\mathrm{kj}, \mathrm{ki}=\mathrm{j}=-\mathrm{ik} . $

Thus $\mathbb{H}$ is associative but not commutative. $\mathbb{R}$ and $\mathbb{C}$ are naturally embedded into $\mathbb{H}$ as follows:

$ \mathbb{R}=\mathbb{R} \cdot 1 \subset \mathbb{H}, \quad \mathbb{C}=\mathbb{R} \cdot 1 \oplus \mathbb{R} \cdot \mathrm{i} \subset \mathbb{H}, $

and we sometimes express an element of $\mathbb{H}$ as q=z+wj, where z, w$\mathbb{C}$.

Conjugation is defined for quaternions:

$ \overline{a+b \mathrm{i}+c \mathrm{j}+d \mathrm{k}}=a-b \mathrm{i}-c \mathrm{j}-d \mathrm{k}(a, b, c, d \in \mathbb{R}). $

Or equivalently,

$ \overline{z+w \mathrm{j}}=\bar{z}-w \mathrm{j}(z, w \in \mathbb{C}). $

Then we have pq=qp, for any p, q$\mathbb{H}$.

Let $\mathbb{H}$n be the space of n-dimensional quaternion column vectors. We consider it as a right $\mathbb{H}$-module. If p=(p1, …, pn)T, q=(q1, …, qn)T$\mathbb{H}$n, two inner products of p, q are defined:

$ \langle\boldsymbol{p}, \boldsymbol{q}\rangle_{\mathbb{H}}=\sum\limits_{l} \bar{p}_{l} q_{l},\langle\boldsymbol{p}, \boldsymbol{q}\rangle_{\mathbb{R}}=\operatorname{Re}\langle\boldsymbol{p}, \boldsymbol{q}\rangle_{\mathbb{H}}. $

It is easily verified that 〈, 〉$\mathbb{R}$ is just the usual Euclidean inner product if $\mathbb{H}$n is identified as $\mathbb{R}$4n, and that the following properties hold:

$ \langle\boldsymbol{p} x, \boldsymbol{q} y\rangle_{\mathbb{H}} =\bar{x}\langle\boldsymbol{p}, \boldsymbol{q}\rangle_{\mathbb{H}} y, $
$ \langle\boldsymbol{p}, \boldsymbol{q}\rangle_{\mathbb{H}} =\overline{\langle\boldsymbol{q}, \boldsymbol{p}\rangle}_{\mathbb{H}}, $

where p, q$\mathbb{H}$n, x, y$\mathbb{H}$.

Similarly, for z=(z1, …, zn)t, w=(w1, …, wn)t$\mathbb{C}$n, we define their inner products:

$ \langle \boldsymbol{z}, \boldsymbol{w}\rangle_{\mathbb{C}}=\sum\limits_{1} \overline{\mathrm{z}}_{1} \mathrm{w}_{1},\langle \boldsymbol{z}, \boldsymbol{w}\rangle_{\mathbb{R}}=\operatorname{Re}\langle \boldsymbol{z}, \boldsymbol{w}\rangle_{\mathbb{C}}. $

We will often omit the subscripts $\mathbb{C}$and $\mathbb{H}$ for simplicity.

Next we consider the quaternion projective space $\mathbb{H}$Pn, the set of quaternionic lines in $\mathbb{H}$n+1. Equivalently, $\mathbb{H}$Pn=S4n+3/Sp(1), where S4n+3 is the unit sphere in $\mathbb{H}$n+1$\cong \mathbb{R}$4n+4, and Sp(1), the multiplicative group of unit quaternions, acts on S4n+3 by right multiplication. Since this is an isometric action, there is a unique Riemannian metric on $\mathbb{H}$Pn, called the Fubini-Study metric, such that the quotient map τ: S4n+3$\mathbb{H}$Pn is a Riemannian submersion. For any qS4n+3, let Hq be the horizontal space of τ at q, i.e. the normal space to the fibre τ-1(τ(q)). Then Hq={q′∈$\mathbb{H}$n+1|〈q, q′〉$\mathbb{H}$=0}. Let τq=dτq|Hq. By assumption, τq: HqTτ(q)$\mathbb{H}$Pn is a linear isometry.

2 The submanifold Qn+1$\mathbb{H}$Pn+1; the complex Grassmannian G(2, n+2)

We quote some results from Ref.[4].

SU(n+2) acts on S4n+7$\mathbb{H}$n+2 isometrically via

$ S U(n+2) \times S^{4 n+7} \rightarrow S^{4 n+7}, $
$ (\boldsymbol{A}, \boldsymbol{z}+\boldsymbol{v} \mathrm{j}) \mapsto \boldsymbol{A} \boldsymbol{z}+(\boldsymbol{A} \boldsymbol{v}) \mathrm{j}, $

where z, v$\mathbb{C}$n+2, with |z|2+|v|2=1. This action commutes with the Sp(1)-action on S4n+7 defined in the last section, hence descends to an isometric action on $\mathbb{H}$Pn+1.

By some straightforward calculations, we find that this SU(n+2)-action on $\mathbb{H}$Pn+1 has only two singular orbits, namely,

$ \mathbb{C} P^{n+1}=\left\{\boldsymbol{\tau}(\boldsymbol{z}+0 \cdot \mathrm{j}) \mid \boldsymbol{z} \in S^{2 n+3}\right\}, $ (1)

and

$ \begin{aligned} Q^{n+1} &=\{\boldsymbol{\tau}((1 / \sqrt{2})(\boldsymbol{z}+\boldsymbol{v} \mathrm{j})) \mid \boldsymbol{z},\\ \boldsymbol{v} &\left.\in S^{2 n+3},\langle \boldsymbol{z}, \boldsymbol{v}\rangle=0\right\}, \end{aligned} $ (2)

where S2n+3 is the unit sphere of $\mathbb{C}$n+2.

We have the following proposition from Ref.[4]:

Proposition 2.1  The singular orbits of the SU(n+2)-action on $\mathbb{H}$Pn+1 are $\mathbb{C}$Pn+1 and Qn+1.Qn+1 has codimension 3 in $\mathbb{H}$Pn+1, and is isometric to the homogeneous space SU(n+2)/SU(2)×SU(n) equipped with a suitable invariant metric. Furthermore, Qn+1 is a minimal submanifold of $\mathbb{H}$Pn+1.

Now consider an action of U(1) on Qn+1:

$ U(1) \times Q^{n+1} \rightarrow Q^{n+1},\left(\mathrm{e}^{\mathrm{i} t}, \boldsymbol{\tau}(\boldsymbol{q})\right) \mapsto \boldsymbol{\tau}\left(\mathrm{e}^{\mathrm{i} t} \boldsymbol{q}\right), $

where t$\mathbb{R}$. Again this is an isometric action. A vector field ξ on Qn+1 is defined:

$ \begin{aligned} \boldsymbol{\xi}_{\boldsymbol{\tau}(\boldsymbol{q})} &=\left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} \mathrm{e}^{\mathrm{i} t} \cdot \boldsymbol{\tau}(\boldsymbol{q})=\left.\frac{\mathrm{d}}{\mathrm{d} t}\right|_{t=0} \boldsymbol{\tau}\left(\mathrm{e}^{\mathrm{i} t} \boldsymbol{q}\right) \\ &=\mathrm{d} \boldsymbol{\tau}_{\boldsymbol{q}}(\mathrm{i} \boldsymbol{q})=\boldsymbol{\tau}_{\boldsymbol{q}}(\mathrm{i} \boldsymbol{q}). \end{aligned} $ (3)

Here $\boldsymbol{q}=\frac{1}{\sqrt{2}}(\boldsymbol{z}+\boldsymbol{v} \mathrm{j}) \in \boldsymbol{\tau}^{-1}\left(Q^{n+1}\right), \boldsymbol{z}, \boldsymbol{v} \in S^{2 n+3}$, 〈z, v〉=0. For the last equality, we note that 〈q, iq$\mathbb{H}$=0 for qτ-1(Qn+1), i.e., iqHq. Thus ξ is the field of tangent vectors to the orbits of the U(1)-action.

Let Bn+1=Qn+1/U(1). Since U(1) acts on Qn+1 isometrically, there is a unique Riemannian metric on Bn+1 such that the natural projection π: Qn+1Bn+1 is a Riemannian submersion.

For τ(q)∈Qn+1, let $\mathcal{H}$τ(q) be the orthogonal complement of ξτ(q) in Tτ(q)Qn+1, i.e. the horizontal space of the Riemannian submersion π: Qn+1Bn+1. Then the map $\pi_{\tau(\boldsymbol{q})}=\mathrm{d} \pi_{\tau(\boldsymbol{q})} \mid _{\mathcal{H}_{\tau(\boldsymbol{q})}}\;: \mathcal{H}_{\tau(\boldsymbol{q})}$Tπ(τ(q))Bn+1 is a linear isometry. By Ref.[4], the horizontal lift of $\mathcal{H}_{\tau(\boldsymbol{q})}$ through τ: S4n+7$\mathbb{H}$Pn+1 is

$ \boldsymbol{\tau}_{\boldsymbol{q}}^{-1} \mathcal{H}_{\boldsymbol{\tau}(\boldsymbol{q})}=\left\{\boldsymbol{X} \in \mathbb{H}^{n+2} \mid\langle\boldsymbol{X}, \boldsymbol{q}\rangle=\langle\boldsymbol{X}, \mathrm{i} \boldsymbol{q}\rangle=0\right\}. $ (4)

We define a (1, 1)-tensor φ on Qn+1 as φX=$-\nabla_{\boldsymbol{X}}^{Q^{n+1}} \boldsymbol{\xi}, \boldsymbol{X} \in T Q^{n+1}$, where $\nabla^{Q^{n+1}}$ is the Riemannian connection on Qn+1. Using the O'Neil formula for Riemannian submersions (see, for example, Proposition 4.5.1 of Ref.[6]), it can be shown that

$ \varphi X= \begin{cases}0, & \boldsymbol{X}=\boldsymbol{\xi}, \\ \boldsymbol{\tau}_{\boldsymbol{q}}\left(-\mathrm{i} \cdot \boldsymbol{\tau}_{\boldsymbol{q}}^{-1}(\boldsymbol{X})\right), & \boldsymbol{X} \in \mathcal{H}_{\boldsymbol{\tau}(\boldsymbol{q})} .\end{cases} $ (5)

Since by definition Tτ(q)Qn+1=$\mathbb{R}$·$\boldsymbol{\xi}_{\tau(\boldsymbol{q})} \oplus \mathcal{H}_{\tau(\boldsymbol{q})}$, this completely determines φ. In particular, $\varphi(\mathcal{H})\subset \mathcal{H}$.

Finally, notice that φ commutes with the U(1)-action on Qn+1. In other words, if Lt denotes the map Qn+1Qn+1, $\tau(\boldsymbol{q}) \mapsto \tau\left(\mathrm{e}^{\mathrm{i} t} \boldsymbol{q}\right)$, then $\mathrm{d} L_{t} \circ \varphi=\varphi \circ \mathrm{d} L_{t}$ for all t$\mathbb{R}$. Therefore, there exists a (1, 1)-tensor J on Bn+1 satisfying *=π*φ. As φ2X=-X for all $\boldsymbol{X} \in \mathcal{H}$, it follows that J is an almost Hermitian structure on Bn+1. Actually, as is proved in Ref.[4], J turns out to be Kähler, and Bn+1 is holomorphically isometric to the complex Grass-mannian

$ G(2, n+2)=U(n+2) / U(2) \times U(n), $

where the metric on G(2, n+2) is induced by the following bi-invariant metric on U(n+2):

$ \langle\boldsymbol{X}, \boldsymbol{Y}\rangle=-\frac{1}{4} \operatorname{tr}(\boldsymbol{X} \boldsymbol{Y}),(\boldsymbol{X}, \boldsymbol{Y} \in \mathfrak{U}(n+2)). $

Thus, for example, B2 is isometric to G(2, 3)=$\mathbb{C}$P2, with the Fubini-Study metric of constant holomorphic sectional curvature 8.

Remark  The isometry between G(2, n+2) and Bn+1 can be explicitly given as

$ G(2, n+2) \rightarrow B^{n+1}, $
$ \mathbb{C} \boldsymbol{z} \oplus \mathbb{C} \boldsymbol{v} \mapsto \boldsymbol{\pi}\left(\boldsymbol{\tau}\left(\frac{1}{\sqrt{2}}(\boldsymbol{z}+\boldsymbol{v} \mathrm{j})\right)\right), $

where z, v$\mathbb{C}$n+2, |z|=|v|=1, 〈z, v$\mathbb{C}$=0.

3 The main theorem

Definition 3.1  Suppose N is a Hermitian manifold, J is its complex structure, f: MN is an immersion from a surface M to N. Then f is called totally real if J Im f*p⊥Imf*p for all pM.

If we choose a local frame X, Y for M, then fis totally real if and only if Jf*Xf*Y everywhere. This follows easily from the Hermitian condition 〈Ju, Jv〉=〈u, v〉, J2=-1, where 〈, 〉 is the Riemannian metric on N.

Now we can state our main result.

Theorem 3.1  Suppose M is a surface, ψ: MBn+1 an immersion, then the following are equivalent:

1) ψ is totally real;

2) ψ has local horizontal lifts to Qn+1, that is, for any pM, there is a neighborhood U of p, and an immersion η: UQn+1, such that $\pi \circ \eta=\psi$, and Im $\eta_{*} \subset \mathcal{H}$.

Furthermore, η is minimal in Qn+1 if and only if ψ is minimal in Bn+1.

We prove the theorem step by step.

Step 1  Let U be an open subset of M, η: UQn+1 an immersion, we shall find a sufficient and necessary condition for η to be horizontal.

First, since τ: S4n+7$\mathbb{H}$Pn+1 is a submersion, η can be lifted to S4n+7, that is, there is an immersion $\boldsymbol{q}=\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\boldsymbol{V} \mathrm{j}): U \rightarrow \tau^{-1}\left(Q^{n+1}\right) \subset S^{4 n+7}$ such that $\eta=\tau \circ q$, where Z, V: U$\mathbb{C}$n+2, |Z|=|V|=1, 〈Z, V〉=0. Now

$ \begin{aligned} \mathrm{d} \eta &=\mathrm{d} \boldsymbol{\tau} \mathrm{d} \boldsymbol{q} \\ &=\mathrm{d} \boldsymbol{\tau}(\mathrm{d} \boldsymbol{q}-\boldsymbol{q}\langle\boldsymbol{q}, \mathrm{d} \boldsymbol{q}\rangle), \end{aligned} $ (6)

so the horizontal lift of dη to S4n+7 is τq-1dη=dq-qq, dq〉, namely the orthogonal projection of dq onto Hq, the horizontal space of τ at q.

Recall from the last section that

η is horizontal with respect to π

$\Leftrightarrow \operatorname{Im}(\mathrm{d} \eta) \subset \mathcal{H}$

$\Leftrightarrow$τq-1dη, q〉=〈τq-1dη, iq〉=0

$\Leftrightarrow$〈dq-qq, dq〉, iq〉=0

$\Leftrightarrow$〈dq, iq〉=0.

For the last equivalence note that qτ-1(Qn+1) implies 〈q, iq〉=0.

Write $\boldsymbol{q}=\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\boldsymbol{V} \mathrm{j}), \mathrm{d} \boldsymbol{q}=\frac{1}{\sqrt{2}}(\mathrm{~d} \boldsymbol{Z}+\mathrm{d} \boldsymbol{V} \cdot \mathrm{j})$. Differentiating 〈V, V〉=1, 〈Z, V〉=0 gives

$ \left\{\begin{array}{l} \langle\mathrm{d} \boldsymbol{V}, \boldsymbol{V}\rangle+\langle\boldsymbol{V}, \mathrm{d} \boldsymbol{V}\rangle=0, \\ \langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{V}\rangle+\langle\boldsymbol{Z}, \mathrm{d} \boldsymbol{V}\rangle=0. \end{array}\right. $

Then

$ \begin{aligned} \langle&\mathrm{d} \boldsymbol{q}, \mathrm{i} \boldsymbol{q}\rangle=0 \\ \Leftrightarrow 0=&\langle\mathrm{d} \boldsymbol{Z}+\mathrm{d} \boldsymbol{V} \cdot \mathrm{j}, \boldsymbol{Z} \mathrm{i}+\boldsymbol{V} \boldsymbol{\mathrm { k }}\rangle \\ =&(\langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{Z}\rangle-\langle\boldsymbol{V}, \mathrm{d} \boldsymbol{V}\rangle) \mathrm{i}+\\ &(\langle\boldsymbol{Z}, \mathrm{d} \boldsymbol{V}\rangle+\langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{V}\rangle) \mathrm{k} \\ =&(\langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{Z}\rangle+\langle\mathrm{d} \boldsymbol{V}, \boldsymbol{V}\rangle) \mathrm{i} . \end{aligned} $

In summary, we have proved

Lemma 3.1  Suppose $\eta=\tau\left(\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\boldsymbol{V} \mathrm{j})\right)$ : UQn+1 is an immersion. Then η is horizontal with respect to π: Qn+1Bn+1 if and only if

$ \langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{Z}\rangle+\langle\mathrm{d} \boldsymbol{V}, \boldsymbol{V}\rangle=0. $ (7)

Step 2  Let ψ: MBn+1 be an immersion of a surface M into Bn+1. We look for the condition under which ψ has a local horizontal lift to Qn+1.

Let $\eta=\tau \circ \boldsymbol{q}=\tau\left(\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\boldsymbol{V} \mathrm{j})\right): U \rightarrow Q^{n+1}$ be any local lift of ψ, and η0 a horizontal lift of ψ. Recall that Bn+1 is defined as the quotient of Qn+1 under the U(1)-action eit·τ(q)=τ(eitq), then for any pU, η(p) and η0(p) lie in the same orbit. It follows that there is a map λ: UU(1) such that η0(p)=λ(pη(p) for all pU. In short,

$ \eta_{0}=\lambda \cdot \eta=\boldsymbol{\tau}\left(\frac{1}{\sqrt{2}}(\lambda \boldsymbol{Z}+\lambda \boldsymbol{V} \mathrm{j})\right). $ (8)

Since η0 is horizontal, we apply Lemma 1 to obtain

$ \begin{aligned} 0 &=\langle\mathrm{d}(\lambda \boldsymbol{Z}), \lambda \boldsymbol{Z}\rangle+\langle\mathrm{d}(\lambda \boldsymbol{V}), \lambda \boldsymbol{V}\rangle \\ &=\langle\mathrm{d} \lambda \cdot \boldsymbol{Z}+\lambda \mathrm{d} \boldsymbol{Z}, \lambda \boldsymbol{Z}\rangle+\langle\mathrm{d} \lambda \cdot \boldsymbol{V}+\lambda \mathrm{d} \boldsymbol{V}, \lambda \boldsymbol{V}\rangle \\ &=\lambda \mathrm{d} \bar{\lambda}(\langle\boldsymbol{Z}, \boldsymbol{Z}\rangle+\langle\boldsymbol{V}, \boldsymbol{V}\rangle)+\lambda \bar{\lambda}(\langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{Z}\rangle+\langle\mathrm{d} \boldsymbol{V}, \boldsymbol{V}\rangle) \\ &=-2 \bar{\lambda} \mathrm{d} \lambda+\langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{Z}\rangle+\langle\mathrm{d} \boldsymbol{V}, \boldsymbol{V}\rangle. \end{aligned} $

Here we have used λλ=1 and λdλ+λdλ=0. Since λdλ=λ-1dλ=d(logλ) we get

$ 2 \mathrm{~d}(\log \lambda)=\langle\mathrm{d} \boldsymbol{Z}, \boldsymbol{Z}\rangle+\langle\mathrm{d} \boldsymbol{V}, \boldsymbol{V}\rangle. $ (9)

If we take a local coordinate (x, y) on M, this amounts to

$ \left\{\begin{array}{l} 2 \frac{\partial \log {\lambda}}{\partial x}=\left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}\right\rangle+\left\langle\boldsymbol{V}_{x}, \boldsymbol{V}\right\rangle, \\ 2 \frac{\partial \log \lambda}{\partial y}=\left\langle\boldsymbol{Z}_{y}, \boldsymbol{Z}\right\rangle+\left\langle\boldsymbol{V}_{y}, \boldsymbol{V}\right\rangle, \end{array}\right. $ (10)

where $\boldsymbol{Z}_{x}=\frac{\partial \boldsymbol{Z}}{\partial x}, \boldsymbol{Z}_{y}=\frac{\partial \boldsymbol{Z}}{\partial y}$, etc. This is a system of first-order PDEs inλ. By the Frobenius theorem for PDEs, an initial value problem of such a system is solvable if and only if the integrability condition

$ \frac{\partial}{\partial y}\left(\frac{\partial \log \lambda}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\partial \log \lambda}{\partial y}\right), $

that is,

$ \frac{\partial}{\partial y}\left(\left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}\right\rangle+\left\langle\boldsymbol{V}_{x}, \boldsymbol{V}\right\rangle\right)=\frac{\partial}{\partial x}\left(\left\langle\boldsymbol{Z}_{y}, \boldsymbol{Z}\right\rangle+\left\langle\boldsymbol{V}_{y}, \boldsymbol{V}\right\rangle\right) $

holds. This equation simplifies to

$ \left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}_{y}\right\rangle+\left\langle\boldsymbol{V}_{x}, \boldsymbol{V}_{y}\right\rangle=\left\langle\boldsymbol{Z}_{y}, \boldsymbol{Z}_{x}\right\rangle+\left\langle\boldsymbol{V}_{y}, \boldsymbol{V}_{x}\right\rangle . $ (11)

Thus we obtain

Lemma 3.2  Suppose $ \psi=\pi \circ \tau\left(\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\right.\boldsymbol{V}\text{ j) ) }$ : MBn+1 is an immersion. Then ψ has local horizontal lifts to Qn+1 if and only if (11) holds.

Step 3  Let $\psi=\pi \circ \tau \circ q: M \rightarrow B^{n+1}$, where q=$\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\boldsymbol{V} \mathrm{j}): M \rightarrow \tau^{-1}\left(Q^{n+1}\right) \subset S^{4 n+7}$. We shall find out the equation for ψ to be totally real.

We have

$ \begin{aligned} \mathrm{d} \psi &=\mathrm{d} \pi \mathrm{d} \boldsymbol{\tau} \mathrm{d} \boldsymbol{q} \\ &=\mathrm{d} \pi \mathrm{d} \boldsymbol{\tau}(\mathrm{d} \boldsymbol{q}-\boldsymbol{q}\langle\boldsymbol{q}, \mathrm{d} \boldsymbol{q}\rangle-\mathrm{i} \boldsymbol{q}\langle\mathrm{i} \boldsymbol{q}, \mathrm{d} \boldsymbol{q}\rangle) \\ &=\mathrm{d} \pi \mathrm{d} \boldsymbol{\tau}\left(\mathrm{d} \boldsymbol{q}^{\mathcal{H}}\right), \end{aligned} $ (12)

where $\mathrm{d} \boldsymbol{q}^{\mathcal{H}}=\mathrm{d} \boldsymbol{q}-\boldsymbol{q}\langle\boldsymbol{q}, \mathrm{d} \boldsymbol{q}\rangle-\mathrm{i} \boldsymbol{q}\langle\mathrm{i} \boldsymbol{q}, \mathrm{d} \boldsymbol{q}\rangle$ is the orthogonal projection of dq onto $\tau_{\boldsymbol{q}}^{-1} \mathcal{H}_{\tau(\boldsymbol{q})}$. In other words, $\mathrm{d} \boldsymbol{q}^{\mathcal{H}}=\tau_{\boldsymbol{q}}^{-1} {\pi}_{\tau(\boldsymbol{q})}^{-1}(\mathrm{~d} \psi)$.

Choose a local coordinate (x, y) on M. Then, using the definitions of the tensors φ, J (see (5)), and the fact that τ, π are Riemannian submersions, we obtain

ψ is totally real

$ \begin{aligned} \Leftrightarrow 0 &=\left\langle\psi_{x}, J \psi_{y}\right\rangle_{B^{n+1}} \\ &=\left\langle\pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{x}, \pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} J \psi_{y}\right\rangle_{Q^{n+1}} \\ &=\left\langle{\pi}_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{x}, {\varphi} \pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{y}\right\rangle_{Q^{n+1}} \\ &=\left\langle\boldsymbol{\tau}_{\boldsymbol{q}}^{-1} \pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{x}, \boldsymbol{\tau}_{\boldsymbol{q}}^{-1} \varphi \pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{y}\right\rangle_{\mathbb{R}} \\ &=\left\langle\boldsymbol{\tau}_{\boldsymbol{q}}^{-1} \pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{x},-\mathrm{i} \cdot \boldsymbol{\tau}_{\boldsymbol{q}}^{-1} \pi_{\boldsymbol{\tau}(\boldsymbol{q})}^{-1} \psi_{y}\right\rangle_{\mathbb{R}} \\ &=\left\langle\boldsymbol{q}_{x}^{\mathcal{H}},-\mathrm{i} \boldsymbol{q}_{y}^{\mathcal{H}}\right\rangle_{\mathbb{R}} . \end{aligned} $ (13)

Since $\left\langle\boldsymbol{q}_{x}^{\mathcal{H}}, \mathrm{i} \boldsymbol{q}_{y}^{\mathcal{H}}\right\rangle_{\mathbb{R}}=\operatorname{Re}\left\langle\boldsymbol{q}_{x}^{\mathcal{H}}, \mathrm{i} \boldsymbol{q}_{y}^{\mathcal{H}}\right\rangle$, let us calculate $\left\langle\boldsymbol{q}_{x}^{\mathcal{H}}, \mathrm{i} \boldsymbol{q}_{y}^{\mathcal{H}}\right\rangle$ first. Now

For the second step note that qτ-1(Qn+1) implies 〈q, iq〉=0. Differentiating〈q, q〉=1 yields

$ \begin{aligned} 0 &=\left\langle\boldsymbol{q}_{x}, \boldsymbol{q}\right\rangle+\left\langle\boldsymbol{q}, \boldsymbol{q}_{x}\right\rangle \\ &=\left\langle\boldsymbol{q}_{x}, \boldsymbol{q}\right\rangle+\overline{\left\langle\boldsymbol{q}_{x}, \boldsymbol{q}\right\rangle}, \end{aligned} $

i.e., 〈qx, q〉∈Im$\mathbb{H}$. Similarly, differentiating 〈q, iq〉=0 yields

$ \begin{aligned} 0 &=\left\langle\boldsymbol{q}_{y}, \mathrm{i} \boldsymbol{q}\right\rangle+\left\langle\boldsymbol{q}, \mathrm{i} \boldsymbol{q}_{y}\right\rangle \\ &=-\left\langle\mathrm{i} \boldsymbol{q}_{y}, \boldsymbol{q}\right\rangle+\left\langle\boldsymbol{q}, \mathrm{i} \boldsymbol{q}_{y}\right\rangle \\ &=-\overline{\left\langle\boldsymbol{q}, \mathrm{i} \boldsymbol{q}_{y}\right\rangle}+\left\langle\boldsymbol{q}, \mathrm{i} \boldsymbol{q}_{y}\right\rangle, \end{aligned} $

i.e., 〈q, iqy〉∈$\mathbb{R}$. Therefore〈qx, q〉〈q, iqy〉∈Im$\mathbb{H}$. Similarly 〈qx, iq〉〈iq, iqy〉∈Im$\mathbb{H}$. Thus we get

$ \begin{aligned} & 2\left\langle\boldsymbol{q}_{x}^{\mathcal{H}}, \mathrm{i} \boldsymbol{q}_{y}^{\mathcal{H}}\right\rangle_{\mathbb{R}} \\ =& 2 \operatorname{Re}\left\langle\boldsymbol{q}_{x}^{\mathcal{H}}, \mathrm{i} \boldsymbol{q}_{y}^{\mathcal{H}}\right\rangle \\ =& 2 \operatorname{Re}\left\langle\boldsymbol{q}_{x}, \mathrm{i} \boldsymbol{q}_{y}\right\rangle \\ =& \operatorname{Re}\left\langle\boldsymbol{Z}_{x}+\boldsymbol{V}_{x} \mathrm{j}, \boldsymbol{Z}_{y} \mathrm{i}+\boldsymbol{V}_{y} \mathrm{k}\right\rangle \\ =& \operatorname{Re}\left(\left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}_{y}\right\rangle \mathrm{i}-\left\langle\boldsymbol{V}_{y}, \boldsymbol{V}_{x}\right\rangle \mathrm{i}\right) \\ =& \operatorname{Im}\left(\left\langle\boldsymbol{V}_{y}, \boldsymbol{V}_{x}\right\rangle-\left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}_{y}\right\rangle\right) . \end{aligned} $ (14)

Finally, from (13) and (14) we obtain

Lemma 3.3  $\psi=\pi \circ \tau\left(\frac{1}{\sqrt{2}}(\boldsymbol{Z}+\boldsymbol{V} \mathrm{j})\right): \boldsymbol{M}$Bn+1 is totally real if and only if

$ \operatorname{Im}\left(\left\langle\boldsymbol{V}_{y}, \boldsymbol{V}_{x}\right\rangle-\left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}_{y}\right\rangle\right)=0, $

or equivalently,

$ \left\langle\boldsymbol{V}_{y}, \boldsymbol{V}_{x}\right\rangle-\left\langle\boldsymbol{Z}_{x}, \boldsymbol{Z}_{y}\right\rangle=\left\langle\boldsymbol{V}_{x}, \boldsymbol{V}_{y}\right\rangle-\left\langle\boldsymbol{Z}_{y}, \boldsymbol{Z}_{x}\right\rangle $ (15)

Comparing with Lemma 3.2, we find that ψ have a local horizontal lift to Qn+1 if and only if it is totally real.

Step 4  We need a simple lemma.

Lemma 3.4  Suppose $\pi: \bar{N} \rightarrow N$ is a Riemannian submersion, $\bar{M} \subset \bar{N}$ is a horizontal submanifold, and $M=\pi(\bar{M}) \subset N$. Then

$ \boldsymbol{H}_{M}(\pi(p))=\pi_{*}\left(\boldsymbol{H}_{\bar{M}}(p)\right) $

for any pM. Furthermore, HM is horizontal. Here HM, HM are the mean curvature vectors of M, M, respectively.

Proof  Let e1, …, em be an orthonormal frame on M, then, since $\left.\pi\right|_{\bar{M}}: \bar{M} \rightarrow M$ is an isometry, e1=π*(e1), , em=π*(em) is an orthonormal frame on M. By O'Neil's formula, $\nabla_{\bar{\boldsymbol{e}}_{i}}^{\bar{N}} \bar{\boldsymbol{e}}_{i}$ is the horizontal lift of $\nabla_{\boldsymbol{e}_{i}}^{N} \boldsymbol{e}_{i}$, hence horizontal, and $\boldsymbol{B}_{\bar{M}}\left(\overline{\boldsymbol{e}}_{i}, \overline{\boldsymbol{e}}_{i}\right)$=$\nabla_{\overline{\boldsymbol{e}}_{i}}^{\bar{N}} \overline{\boldsymbol{e}}_{i}-\nabla_{\overline{\boldsymbol{e}}_{i}}^{\bar{M}} \overline{\boldsymbol{e}}_{i}$ is also horizontal. Thus HM=$\sum\limits_{i} \boldsymbol{B}_{\bar{M}}\left(\overline{\boldsymbol{e}}_{i}, \overline{\boldsymbol{e}}_{i}\right)$ is horizontal. On the other hand,

$ \begin{aligned} \nabla_{\boldsymbol{e}_{i}}^{N} \boldsymbol{e}_{i} &=\pi_{*}\left(\nabla_{\bar{\boldsymbol{e}}_{i}}^{\bar{N}} \overline{\boldsymbol{e}}_{i}\right) \\ &=\pi_{*}\left(\nabla_{\bar{\boldsymbol{e}}}^{\bar{M}} \overline{\boldsymbol{e}}_{i}+\boldsymbol{B}_{\bar{M}}\left(\overline{\boldsymbol{e}}_{i}, \overline{\boldsymbol{e}}_{i}\right)\right) \\ &=\nabla_{\boldsymbol{e}_{i}}^{M} \boldsymbol{e}_{i}+\pi_{*}\left(\boldsymbol{B}_{\bar{M}}\left(\overline{\boldsymbol{e}}_{i}, \overline{\boldsymbol{e}}_{i}\right)\right). \end{aligned} $ (16)

Comparing with the Gauss equation in N, we find that

$ \boldsymbol{B}_{M}\left(\boldsymbol{e}_{i}, \boldsymbol{e}_{i}\right)=\pi_{*}\left(\boldsymbol{B}_{\bar{M}}\left(\overline{\boldsymbol{e}}_{i}, \overline{\boldsymbol{e}}_{i}\right)\right). $ (17)

The conclusion follows immediately.

From the above lemma, we see that HM=$0 \Leftrightarrow \boldsymbol{H}_{\bar{M}}=0$. That is, $M {\rm { minimal }} \Leftrightarrow \bar{M} \rm { minimal }$. This applies to our situation and the main theorem is fully proved.

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