There have been many researches about minimal surfaces in real and complex space forms^[1-3]. For quaternionic projective space, Bryant^[4] proved that the projections of horizontal holomorphic surfaces in $\mathbb{C} $P³ are minimal in $\mathbb{H}$P¹ through twistor map. Aithal^[5] described all the harmonic maps from S² to $\mathbb{H}$P². Considering $\mathbb{H}$Pⁿ as a totally geodesic submani-fold in the complex Grassmann manifold G(2, 2n+2), Bahy-El-Dien and Wood^[6] constructed explicitly all harmonic maps from the two-spheres to a quaternionic projective space. Based on Bahy-El-Dien and Wood's results, some classifications of minimal two-spheres of constant curvature in $\mathbb{H}$Pⁿ have been given^[7-9].

Recently, Chen and Jiao^[10] used twistor map to discuss conformal minimal surface in quaternionic projective space, based on the horizontal condition given by Yang^[11]. They got two polynomial equations as the horizontal conditions for the immersed surfaces in $\mathbb{C} $P²ⁿ⁺¹, and then they calculated Veronese surfaces in $\mathbb{C} $P⁵ and got a series of conformal minimal two-spheres with constant curvature in $\mathbb{H}$P².

In this paper we deal with the case $\mathbb{H}$P⁴.

In section 1, we review the knowledge of geometry of quaternionic projective space, minimal surfaces in complex projective space, and horizontal equation of immersed surfaces with respect to the twistor map.

In section 2, we try to improve the calculation of Veronese two-spheres to the case n. We will calculate explicitly the case n=4 and get a series of conformal minimal two-spheres ϕ_{4, p}=π○ (U₀·V_p⁹)(0≤p≤3) in $\mathbb{H}$P⁴.

In section 3, we talk about the interesting phenomenon that the middle terms, V₄⁹ and V₅⁹, of the Veronese sequences can not be rotated horizontally with respect to the twistor map. We will use the method of harmonic sequence to prove it.

1 Preliminaries 1.1 Geometry of quaternionic projective space

Let $\mathbb{H}$ be the quaternion algebra, which is regarded as a 4-dimensional real vector space with the basis 1, i, j, k. A ring multiplication is defined as

$ \begin{array}{*{20}{c}} {{{\rm{i}}^2} = {{\rm{j}}^2} = {{\rm{k}}^2} = - 1,}\\ {{\rm{ij}} = {\rm{k}} = - {\rm{ji}},{\rm{jk}} = {\rm{i}} = - {\rm{kj}},{\rm{ki}} = {\rm{j}} = - {\rm{ik}}{\rm{.}}} \end{array} $

(1)

In particular, $\mathbb{H}$ is a 2-dimensional ring module over $\mathbb{C} $ with the basis 1, j. For z=z₁+jz₂, w=w₁+jw₂∈$\mathbb{H}$, we have

$ zw = ({z_1}{w_1} - {\bar z_2}{w_2}) + {\rm{j}}({z_2}{w_1} + {\bar z_1}{w_2}). $

(2)

There is a natural conjugation in $\mathbb{H}$, denoted by

$ {({z_1} + {\rm{j}}{z_2})^*} = {z_1} - {\rm{j}}{z_2}. $

(3)

Let $\mathbb{H}$ⁿ be the set of n-tuples of quaternions written as column vectors. We can identify $\mathbb{H}$ⁿ with $\mathbb{C} $²ⁿ:

$ {z_1} + {\rm{j}}{z_2} \mapsto \left( {\begin{array}{*{20}{c}} {{z_1}}\\ {{z_2}} \end{array}} \right). $

(4)

The quaternionic projective space $\mathbb{H}$Pⁿ is the space of $\mathbb{H}$-lines in $\mathbb{H}$ⁿ⁺¹, that is, for [v₁]_$\mathbb{H}$ and [v₂]_$\mathbb{H}$∈$\mathbb{H}$Pⁿ, [v₁]_$\mathbb{H}$=[v₂]_$\mathbb{H}$ if and only if there is an x∈$\mathbb{H}$, such that v₂=v₁x.

The symplectic group is

$ Sp(n) = \{ \mathit{\boldsymbol{A}} \in {\rm{GL}} (n;\mathbb{H}){|^t}{\mathit{\boldsymbol{A}}^*} \cdot \mathit{\boldsymbol{A}} = {\mathit{\boldsymbol{I}}_n}\} , $

(5)

where I_n is the identity matrix of order n.

For A∈Sp(n+1) and [v]_$\mathbb{H}$∈ $\mathbb{H}$Pⁿ, Sp(n+1) acts on $\mathbb{H}$Pⁿ by

$ \mathit{\boldsymbol{A}} \cdot {[\mathit{\boldsymbol{v}}]_\mathbb{H}} = {[\mathit{\boldsymbol{Av}}]_\mathbb{H}}. $

(6)

G₀=Sp(1)×Sp(n)is the isotropy group of this action at[(1, 0, …, 0)^T]_$\mathbb{H}$ and this action is transitive. Therefore,

$ \mathbb{H}{P^n} = Sp(n + 1)/Sp(1) \times Sp(n). $

(7)

There is the inclusion Sp(n)↪SU(2n) by $ \boldsymbol{X}+\text{j}\boldsymbol{Y}\mapsto \left( \begin{align} & \boldsymbol{X}\ \ \ -\boldsymbol{\bar{Y}} \\ & ~\boldsymbol{Y}\ \ \ \ \boldsymbol{\bar{X}} \\ \end{align} \right)$, where X, Y∈GL(n; C).

We have the following commutative diagram

$ \begin{matrix} {} & Sp(n+1) & {} \\ {{\pi }_{2}}\swarrow & {} & \searrow {{\pi }_{1}} \\ \mathbb{H}{{P}^{n}} & {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\leftarrow}$}}{\pi }} & \mathbb{C}{{P}^{2n+1}} \\ \end{matrix} $

(8)

where π:$\mathbb{C} $P²ⁿ⁺¹→$\mathbb{H}$Pⁿ, $\left[ \begin{align} & {{z}_{1}} \\ & {{z}_{2}} \\ \end{align} \right]\mapsto {{\left[ {{z}_{1}}+\text{j}{{z}_{2}} \right]}_{\mathbb{H}}} $, z₁, z₂∈$\mathbb{C} $ⁿ⁺¹, is the twistor map and it is a fibration, π₁ and π₂ are defined as follows. If A=X+jY∈Sp(n+1), write A=(A₀, …, A_n), X=(X₀, …, X_n), Y=(Y₀, …, Y_n). Then $ {\pi _1}(\mathit{\boldsymbol{A}}) = \left[ \begin{array}{l} {X_0}\\ {Y_0} \end{array} \right]$ and π₂(A)=[A₀]_$\mathbb{H}$ are the natural projections. Note that Sp(n+1)→$\mathbb{C} $P²ⁿ⁺¹ is a principal fiber bundle with structure group U(1)×Sp(n) and π₁ is a surjection.

Definition 1.1   The twistor map gives a fibration, and the horizontal distribution $\mathscr{H} $ on $\mathbb{C} $P²ⁿ⁺¹ is defined to be the orthogonal complement to the fiber of π with respect to the Fubini-Study metric on $\mathbb{C} $P²ⁿ⁺¹. In particular, if M is a Riemann surface, call a map f:M→ $\mathbb{C} $P²ⁿ⁺¹ horizontal surface if the image of f is tangent to $\mathscr{H} $.

As in Ref[11], for [v]∈$\mathbb{C} $P²ⁿ⁺¹, we write $\boldsymbol{v}=\left( \begin{align} & {{v}_{1}} \\ & {{v}_{2}} \\ \end{align} \right)\in {{\mathbb{C}}^{2n+2}} $, where v₁, v₂∈ $\mathbb{C} $ⁿ⁺¹, and the horizontal space $\mathscr{H} $ _[v] is

$ {\mathscr{H} _{[v]}} = \{ w \in {v^ \bot }|{\sigma _v}(w) = 0,{\sigma _v} = { - ^t}{z_2}{\rm{d}}{z_1}{ + ^t}{z_1}{\rm{d}}{z_2}\} . $

(9)

We have the following proposition^[10], which is a special case of Lemma 3.5 of Eells and Wood^[3].

Proposition 1.1   If ϕ=[f]:M→$\mathbb{C} $P²ⁿ⁺¹ is a horizontal conformal minimal surface, then π○ϕ:M→$\mathbb{H}$Pⁿ is conformal minimal, where π:$\mathbb{C} $P²ⁿ⁺¹→$\mathbb{H}$Pⁿ is the twistor map.

1.2 Minimal surface in a complex projective space

In this subsection, we introduce some knowledge about harmonic sequences and minimal surfaces in $\mathbb{C} $P^n[2-3].

Let $\mathbb{C} $Pⁿ denote the complex lines in $\mathbb{C} $ⁿ⁺¹, let T→$\mathbb{C} $Pⁿ denote the universal line bundle whose fibre at l∈$\mathbb{C} $Pⁿ is l itself.

Let M be a Riemann surface. We identify a smooth map ϕ:M²→$\mathbb{C} $Pⁿ with a subbundle ϕ of the trivial bundle $ \underline{{{\mathbb{C}}^{n+1}}}={{\boldsymbol{M}}^{2}}\times {{\mathbb{C}}^{n+1}}$ of rank one which has fibre at x∈M given by ϕ_x=ϕ(x). Thus ϕ=ϕ^-1T. Conversely any rank subbundle of $ \underline{{{\mathbb{C}}^{n+1}}}$ induces a map M²→$\mathbb{C} $Pⁿ.

Each linearly full conformal minimal immersion φ from S² is obtained from a holomorphic curve φ₀ in $\mathbb{C} $Pⁿ, and φ₀ induces the harmonic sequence,

$ 0\overset{\partial }{\mathop{\to }}\,\underline{\varphi _{0}^{n}}\overset{\partial }{\mathop{\to }}\,\cdots \overset{\partial }{\mathop{\to }}\,\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{\varphi }=\underline{\varphi _{\alpha }^{n}}\overset{\partial }{\mathop{\to }}\,\cdots \overset{\partial }{\mathop{\to }}\,\underline{\varphi _{n}^{n}}\overset{\partial }{\mathop{\to }}\,0. $

(10)

Under local coordinate z, choose a local holomorphic section f₀ of φ₀ⁿ, such that $\frac{\partial }{\partial \bar{z}}{{f}_{0}}=0 $. We define local section f_i+1 of φ_i+1ⁿ by the formula

$ {f_{i + 1}} = \frac{\partial }{{\partial z}}{f_i} - \frac{{\left\langle {\frac{\partial }{{\partial z}}{f_i},{f_i}} \right\rangle }}{{|{f_i}{|^2}}}{f_i},0 \le i \le n - 1. $

(11)

Let $ {{\gamma }_{i}}=\frac{{{\left| {{f}_{i+1}} \right|}^{2}}}{{{\left| {{f}_{i}} \right|}^{2}}}$. Then we have the equations^[2]

$ \begin{array}{*{20}{c}} {\frac{\partial }{{\partial z}}{f_i} = - {\gamma _{i - 1}}{f_{i - 1}},1 \le i \le n,}\\ {\frac{{{\partial ^2}}}{{\partial z\partial \bar z}}{\rm{log}}|{f_j}{|^2} = {\gamma _j} - {\gamma _{j - 1}},}\\ {\frac{{{\partial ^2}}}{{\partial z\partial \bar z}}{\rm{log}}{\gamma _j} = {\gamma _{j + 1}} - 2{\gamma _j} + {\gamma _{j - 1}},0 \le j \le n - 1.} \end{array} $

(12)

For convenience, we set γ_-1=γ_n=0. Bolton et al.^[12] defined the Veronese sequence as follows: Let V_pⁿ:S²→$\mathbb{C} $Pⁿ be given by

$ V_p^n(z) = {[^t}({g_{p,0}}(z), \cdots ,{g_{p,n}}(z))], $

(13)

where z is a local holomorphic coordinate of S², p=0, 1, …, n. For j=0, 1, …, n, g_{p, j} are given by

$ {g_{p,j}}(z) = \frac{{p!}}{{{{(1 + z\bar z)}^p}}}\sqrt {C_n^j} {z^{j - p}}\sum\limits_k {{{( - 1)}^k}} C_j^{p - k}C_{n - j}^k{(z\bar z)^k}. $

(14)

Then, for each p, the Veronese map(or Veronese surface) V_pⁿ is a conformal minimal immersion with constant curvature $ \frac{4}{n+2p\left( n-p \right)}$, and {V₀ⁿ, …, V_nⁿ} is the Veronese sequence in $\mathbb{C} $Pⁿ.

Remark 1.1   Notice that g_{p, j}(z) have a common factor $ \frac{p!}{{{\left( 1+z\overline{z} \right)}^{p}}}$. We can omit it in the horizontal equations which are mentioned below.

The Veronese sequence is particularly important for the following rigidity theorem^[2].

Lemma 1.1   Let Ψ:S²→$\mathbb{C} $Pⁿ be a linearly full conformal minimal immersion with constant curvature, then, up to a holomorphic isometry of $\mathbb{C} $Pⁿ, the immersion Ψ is an element of the Veronese sequence.

1.3 Horizontal equations of immersed surfaces

Based on the above discussions, we can produce conformal minimal two-spheres from the Veronese two-spheres in $\mathbb{H}$Pⁿ by a unitary rotation. It is stated in the following proposition^[10].

Proposition 1.2   Let Ψ:S²→$\mathbb{C} $P²ⁿ⁺¹ be an element of the Veronese sequence in $\mathbb{C} $P²ⁿ⁺¹. If there exists a U∈U(2n+2) such that U·Ψ is horizontal in $\mathbb{C} $P²ⁿ⁺¹, then π○(U·Ψ):S²→$\mathbb{H}$Pⁿ is a conformal minimal two-spheres with constant curvature, where π:$\mathbb{C} $P²ⁿ⁺¹→$\mathbb{H}$Pⁿ is the twistor map.

From the horizontal condition, if Ψ=[f] is horizontal, then ^tfBdf=0, where $\boldsymbol{B}=\left( \begin{align} & 0\ \ \ \ \ \ \boldsymbol{I} \\ & -\boldsymbol{I}\ \ \ \ 0 \\ \end{align} \right) $. More explictly, it is the following proposition^[10].

Proposition 1.3   For Ψ=[f]=[^t(f₁, f₂, …, f_2n+2)]:S²→$\mathbb{C} $P²ⁿ⁺¹, if U∈U(2n+2), then U·Ψ is horizontal if and only if

$ \left\{ {\begin{array}{*{20}{l}} {\sum\nolimits_{i,j = 1}^{2n + 2} {{A_{ij}}} {f_i}\partial {f_j} = 0,}\\ {\sum\nolimits_{i,j = 1}^{2n + 2} {{A_{ij}}} {f_i}\bar \partial {f_j} = 0,} \end{array}} \right. $

(15)

where A=(A_ij)=^tUBU, $\partial =\frac{\partial }{\partial z}, \overline{\partial }=\frac{\partial }{\partial \overline{z}} $, and z is a local coordinate of S².

2 Conformal minimal two-spheres in $ \mathbb{H}{{P}^{4}}$ 2.1 Reduced horizontal equation of Veronese maps

Definition 2.1   Let M be a Riemann surface, we call two surfaces Ψ, Φ:M²→$\mathbb{H}$Pⁿ are symplectic equivalent if there is an isometry, i.e., a symplectic matrix H, H·Ψ=Φ.

For Veronese sequence, V_p²ⁿ⁺¹ and V_2n+1-p²ⁿ⁺¹ have the following relation.

Lemma 2.1 V_2n+1-p²ⁿ⁺¹=J_2n+2· V_p²ⁿ⁺¹, here we choose V_p²ⁿ⁺¹=[^t(g_{p, 0}, …, g_{p, 2n+1})], where

$ {g_{p,j}} = \sqrt {C_{2n + 1}^j} {z^{j - p}}\sum\limits_k {{{( - 1)}^k}} C_j^{p - k}C_{2n + 1 - j}^k{(zz)^k}, $

(16)

$ {\mathit{\boldsymbol{J}}_m} = \left( {\begin{array}{*{20}{c}} 0&0& \cdots &0&{{{( - 1)}^{m - 1}}}\\ 0&0& \cdots &{{{( - 1)}^m}}&0\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 0&{{{( - 1)}^1}}& \cdots &0&0\\ {{{( - 1)}^0}}&0& \cdots &0&0 \end{array}} \right). $

(17)

for any integer m.

Proof

$ \begin{array}{*{20}{l}} {{g_{2n + 1 - p,2n + 1 - j}}}\\ { = \sqrt {C_{2n + 1}^{2n + 1 - j}} {z^{p - j}}\sum\limits_l {{{( - 1)}^l}} C_{2n + 1 - j}^{2n + 1 - p - l}C_j^l{{(zz)}^l}.} \end{array} $

(18)

Set k=p+l-j, then we easily see

$ \begin{array}{*{20}{l}} {{g_{2n + 1 - p,2n + 1 - j}} = {{( - 1)}^{j - p}}{{\bar g}_{p,j}}}\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = {{( - 1)}^p}{{( - 1)}^j}{{\bar g}_{p,j}}.} \end{array} $

(19)

Omit (-1)^p, then we get the conclusion.

Now, if V_p²ⁿ⁺¹ can be rotated horizontally, U·V_p²ⁿ⁺¹ satisfies the horizontal equations,

$ \left\{ {\begin{array}{*{20}{l}} {^tV{{_p^{2n + 1}}^t}\mathit{\boldsymbol{UBU}}V_{p + 1}^{2n + 1} = 0,}\\ {^tV{{_p^{2n + 1}}^t}\mathit{\boldsymbol{UBU}}V_{p - 1}^{2n + 1} = 0.} \end{array}} \right. $

(20)

By complex conjugate, we have

$ \left\{ {\begin{array}{*{20}{l}} {^tV{{_{2n + 1 - p}^{2n + 1}}^t}{J_{2n + 2}}^t\mathit{\boldsymbol{\bar UB\bar U}}{J_{2n + 2}}V_{2n - p}^{2n + 1} = 0,}\\ {^tV{{_{2n + 1 - p}^{2n + 1}}^t}{J_{2n + 2}}^t\mathit{\boldsymbol{\bar UB\bar U}}{J_{2n + 2}}V_{2n + 2 - p}^{2n + 1} = 0.} \end{array}} \right. $

(21)

That is to say, U·V_p²ⁿ⁺¹ is horizontal ⇔ UJ_2n+2·V_2n+1-p²ⁿ⁺¹ is horizontal.

So we only need to consider the case p≤n.

Denote g_{p, j}(0≤p, j≤n) by

$ {g_{p,j}} = \sqrt {C_n^j} {z^{j - p}}\sum\limits_k {{{( - 1)}^k}} C_j^{p - k}C_{n - j}^k{(z\bar z)^k}. $

(22)

By derivation, we get

$ \partial {g_{p,j}} = \sqrt {C_n^j} {z^{j - p - 1}}\sum\limits_k {{{( - 1)}^k}} C_j^{p - k}C_{n - j}^k(j - p + k){(z\bar z)^k}, $

(23)

$ \partial {g_{p,j}} = \sqrt {C_n^j} {z^{j - p + 1}}\sum\limits_k {{{( - 1)}^k}} C_j^{p - k}C_{n - j}^kk{(zz)^{k - 1}}. $

(24)

Set $ {{S}_{ij}}={{A}_{ij}}\sqrt{~C_{n}^{i}C_{n}^{j}}$, then we have

$ \left\{ {\begin{array}{*{20}{l}} {\sum\nolimits_{i + j = m} {{S_{ij}}} \sum\nolimits_{k + l = t} {C_i^{p - k}} C_{n - i}^kC_j^{p - l}C_{n - j}^l(j - p + l) = 0,}\\ {\sum\nolimits_{i + j = m} {{S_{ij}}} \sum\nolimits_{k + l = t} {C_i^{p - k}} C_{n - i}^kC_j^{p - l}C_{n - j}^ll = 0,} \end{array}} \right. $

(25)

for any nonnegative integers m, t.

If we change the positions of k and l in the second formula (notice that k+l=t=constant for some t and S_ij=-S_ji), then we get

$ \sum\limits_{i + j = m} {{S_{ij}}} \sum\limits_{k + l = t} {C_i^{p - k}} C_{n - i}^kC_j^{p - l}C_{n - j}^l = 0. $

(26)

Then, using i+j=m=constant for some m in the first equation, we finally get the equations

$ \left\{ {\begin{array}{*{20}{l}} {\sum\nolimits_{i + j = m} {{S_{ij}}} \sum\nolimits_{k + l = t} {C_i^{p - k}} C_{n - i}^kC_j^{p - l}C_{n - j}^l = 0,}\\ {\sum\nolimits_{i + j = m} {{S_{ij}}} \sum\nolimits_{k + l = t} {C_i^{p - k}} C_{n - i}^kC_j^{p - l}C_{n - j}^li = 0.} \end{array}} \right. $

(27)

Considering the projections of linearly full Veronese maps in $\mathbb{C} $P²ⁿ⁺¹, we only need to change n in the above equations into 2n+1, and then we have the following theorem.

Theorem 2.1   Let V_p²ⁿ⁺¹:S²→$\mathbb{C} $P²ⁿ⁺¹ be the Veronese map defined in (13) and (14). If U∈U(2n+2), U·V_p²ⁿ⁺¹ is horizontal with respect to the twistor map π:$\mathbb{C} $P²ⁿ⁺¹→$\mathbb{H}$Pⁿ, then it satisfies

$ \left\{ {\begin{array}{*{20}{l}} {\sum\nolimits_{i + j = m} {{S_{ij}}} \sum\nolimits_{k + l = t} {C_i^{p - k}} C_{2n + 1 - i}^kC_j^{p - l}C_{2n + 1 - j}^l = 0,}\\ {\sum\nolimits_{i + j = m} {{S_{ij}}} \sum\nolimits_{k + l = t} {C_i^{p - k}} C_{2n + 1 - i}^kC_j^{p - l}C_{2n + 1 - j}^li = 0,} \end{array}} \right. $

(28)

where S=QAQ, Q=diag{1, …, $\sqrt{C_{2n+1}^{i}} $, …, 1}, A=^tUBU is an anti-symmetric unitary matrix. The above equations should be right for any constant nonnegative integers m∈{0, …, 4n+2}, t∈{0, …, 2p}.

2.2 Projections of V_p⁹, p=0, 1, 2, 3, 4

As an application of (28), in this subsection, we calculate the Veronese maps V_p⁹, p=0, 1, 2, 3, 4.

First, we talk about the general V₀²ⁿ⁺¹, the equations (28) reduce to:

$ \left\{ {\begin{array}{*{20}{l}} {\sum\nolimits_{i + j = m} {{S_{ij}}} = 0,}\\ {\sum\nolimits_{i + j = m} {{S_{ij}}} i = 0.} \end{array}} \right. $

(29)

Notice that A is an anti-symmetric unitary matrix, so S is anti-symmetric too, the first equation is naturally satisfied, and we can reformulate the second to

$ \sum\limits_{i + j = m,j > i} {{S_{ij}}} (j - i) = 0, $

(30)

or, write

$ \sum\limits_{i + j = m,j > i} {{A_{ij}}} \sqrt {C_{2n + 1}^iC_{2n + 1}^j} (j - i) = 0. $

(31)

It is a linear combination of matrix elements in different lines of upper triangulation part of S. Now, we talk about the solution in different n(n≥1).

First, when n=1, the number of elements in the lines which are parallel to antidiagonal in the upper triangulation part of S (respect A) is less than 2, then they should be all zeros, so with respect to the twistor map π:$\mathbb{C} $P³→$\mathbb{H}$P¹, there only have the linear relation in the antidiagonal:

$ {\sum\limits_{i + j = 3,j > i} {{S_{ij}}} (j - i) = 0,} $

(32)

or, write

$ {\sum\limits_{i + j = 3,j > i} {{A_{ij}}} C_3^i(j - i) = 0.} $

(33)

The solution is

$ \left( {\begin{array}{*{20}{c}} 0&0&0&{{A_{03}}}\\ 0&0&{ - {A_{03}}}&0\\ 0&{{A_{03}}}&0&0\\ { - {A_{03}}}&0&0&0 \end{array}} \right) $

(34)

So, when n=1, A=λJ₄, λ∈$\mathbb{C} $, |λ|=1.

In general, we have

Lemma 2.2   $\sum\nolimits_{i+j=2n+1, j>i}{{{A}_{ij}}C_{2n+1}^{i}\left( j-i \right)=0} $ or $\sum\nolimits_{i+j=2n+1}{{{A}_{ij}}C_{2n+1}^{i}i=0} $ has a solution:

$ \left\{ {\begin{array}{*{20}{l}} {{A_{ij}} = 0,i + j \ne 2n + 1,}\\ {{A_{ij}} = {{( - 1)}^i}{A_{0,2n + 1}},i + j = 2n + 1.} \end{array}} \right. $

(35)

or, equivalently, A=λJ_2n+2, λ∈$\mathbb{C} $, |λ|=1.

Proof   For (1+x)²ⁿ⁺¹= $ \sum\nolimits_{i=0}^{2n+1}{C_{2n+1}^{i}{{x}^{i}}}$, derivative both sides.

Remark 2.1   Readers could see $ \sum\nolimits_{i+j=2n+1}{{{A}_{ij}}C_{2n+1}^{i}{{i}^{t}}=0}$ also have this solution for any nonnegative integer t.

Lemma 2.3   Choose two solutions of ^tUBU=A, denote them by M, N, then π○(M·V_p²ⁿ⁺¹) is symplectic equivalent to π○(N·V_p²ⁿ⁺¹).

Proof  It's easy to see MN^-1 are symplectic matrix with the form $\boldsymbol{M}{{\boldsymbol{N}}^{-1}}=\left( \begin{align} & \boldsymbol{X}\ \ \ \ \ \ -\overline{\boldsymbol{Y}} \\ & \boldsymbol{Y}\ \ \ \ \ \ \ \ \overline{\boldsymbol{X}} \\ \end{align} \right) $, denote N·V_p²ⁿ⁺¹ by $\left[ \begin{align} & {{v}_{1}} \\ & {{v}_{2}} \\ \end{align} \right] $, then π○(M·V_p²ⁿ⁺¹)=[Xv₁-Yv₂+j(Yv₁+Xv₂)]_$\mathbb{H}$=(X+jY)·[v₁+jv₂]_$\mathbb{H}$=(X+jY)·(π○(N·V_p²ⁿ⁺¹)).

From the above Lemma, the key point of "rotate horizontally" is to decide the existence of A, about U, we only need to search for the solution of special form, here we consider that U possess this form: $\mathit{\boldsymbol{U}} = \left( \begin{array}{l} {\mathit{\boldsymbol{U}}_1}\;\;\;0\\ 0\;\;\;\;\;\mathit{\boldsymbol{I}} \end{array} \right) $, U₁∈U(n+1).

Then $ \mathit{\boldsymbol{A}} = \left( \begin{array}{l} \;\;0\;\;\;\;\;\;\;{\;^t}{\mathit{\boldsymbol{U}}_1}\\ - {\mathit{\boldsymbol{U}}_1}\;\;\;\;\;\;0 \end{array} \right)$, so, when p=0, for any n≥1, U has one solution $ \mathit{\boldsymbol{U}} = {\mathit{\boldsymbol{U}}_0} = \left( \begin{array}{l} {\mathit{\boldsymbol{J}}_{n + 1}}\;\;\;0\\ 0\;\;\;\;\;\;\;\mathit{\boldsymbol{I}} \end{array} \right)$.

In the rest part of this article, we will always denote the above matrix by U₀.

This solution is special, now U₀V₀²ⁿ⁺¹ is horizontal, since $ {\mathit{\boldsymbol{U}}_{\rm{0}}}{\mathit{\boldsymbol{J}}_{{\rm{2}}n + 2}} = \left( \begin{array}{l} {\rm{0}}\;\;\;\;\;\;\;\;{\rm{ - }}\mathit{\boldsymbol{I}}\\ {\mathit{\boldsymbol{J}}_{n + 1}}\;\;\;\;{\rm{0}} \end{array} \right)$, then

$ {\mathit{\boldsymbol{U}}_0}V_{2n + 1}^{2n + 1} = {\mathit{\boldsymbol{U}}_0}{\mathit{\boldsymbol{J}}_{2n + 2}}\bar V_0^{2n + 1} = {\bf{J}} \circ {\mathit{\boldsymbol{U}}_0}V_0^{2n + 1} $

(36)

is also horizontal.

From the above discussion, we have

Theorem 2.2   V₀⁹ can be rotated horizontally, and π○(U₀·V₀⁹) is conformal minimal two-sphere with constant curvature 4/9 in $\mathbb{H}$P⁴.

For V₁⁹, If we set ${F_t} = \sum\nolimits_{i + j = m} {{S_{ij}}{i^t}} $ for some nonnegative integer m, t, then the first equation of (28) reduce to

$ \left\{ {\begin{array}{*{20}{l}} {m{F_1} - {F_2} = 0,}\\ {9m{F_0} - 2(m{F_1} - {F_2}) = 0,}\\ {9(9 - m){F_0} + m{F_1} - {F_2} = 0,} \end{array}} \right. $

(37)

the second equation of (28) reduce to

$ \left\{ {\begin{array}{*{20}{l}} {m{F_2} - {F_3} = 0,}\\ {9m{F_1} - 2(m{F_2} - {F_3}) = 0,}\\ {9(9 - m){F_1} + m{F_2} - {F_3} = 0,} \end{array}} \right. $

(38)

notice F₀=0 is naturally satisfied, then we have

$ {F_1} = 0,{F_2} = 0,{F_3} = 0. $

(39)

The equations F_t=0 which t is even are degenerate to those t is odd, since

$ {F_{2k}} = 0 \Leftrightarrow \sum\limits_{i + j = m,j > i} {{S_{ij}}} ({(m - i)^{2k}} - {i^{2k}}) = 0, $

(40)

and

$ \sum\limits_{i + j = m,j > i} {{S_{ij}}} ({(m - i)^{2k}} - {i^{2k}}) = 0 \Leftarrow {F_t} = 0 $

(41)

by every 0≤t≤2k-1.

So, finally, the equations reduce to

$ {F_1} = 0,{F_3} = 0. $

(42)

The other discussions are just like V₀⁹, we give the following result:

Theorem 2.3   For V₁⁹ can be rotated horizontally, and π○(U₀·V₁⁹) is conformal minimal two-sphere with constant curvature 4/25 in $\mathbb{H}$P⁴.

For V₂⁹, (28) reduce to

$ \begin{array}{*{20}{l}} {\left[ {\begin{array}{*{20}{l}} {m{F_1} - {F_2} = 0,}\\ {m{F_2} - {F_3} = 0,}\\ {{F_4} - 2m{F_3} + {m^3}{F_1} = 0,}\\ {{F_5} - 2m{F_4} + ({m^2} + m - 1){F_3} + (m - {m^2}){F_2} = 0,}\\ {{F_3} - m{F_2} + \frac{{72m(m - 1)}}{{48m - 208}}{F_1} = 0,} \end{array}} \right.} \end{array} $

(43)

then we have

$ {F_1} = 0,{F_2} = 0,{F_3} = 0,{F_4} = 0,{F_5} = 0. $

(44)

So, finally, the equations reduce to

$ {F_1} = 0,{F_3} = 0,{F_5} = 0. $

(45)

The other discussions are just like V₀⁹, we give the following result:

Theorem 2.4   V₂⁹ can be rotated horizontally, and π○(U₀·V₂⁹) is conformal minimal two-sphere with constant curvature 4/37 in $\mathbb{H}$P⁴.

For V₃⁹, (28) reduce to

$ \mathit{\boldsymbol{B}}3{ * ^t}({F_7},{F_6},{F_5},{F_4},{F_3},{F_2},{F_1},{F_0}) = 0, $

(46)

where B3 is a 14×8 matrix, it is a big matrix, by using the software Matlab, we calculate rank(B3)=8, so we have

$ {F_i} = 0 $

(47)

for 0≤i≤7.

The equations F_t=0 which t is even are degenerate to those t is odd.

So, finally, the equations reduce to

$ {F_1} = 0,{F_3} = 0,{F_5} = 0,{F_7} = 0. $

(48)

It is still not determinate for the elements of matrix S, so we still have a solution U=U₀, then:

Theorem 2.5   V₃⁹ can be rotated horizontally, and π○(U₀·V₃⁹) is conformal minimal two-sphere with constant curvature 4/45 in $\mathbb{H}$P⁴.

We continue to talk about V₄⁹, (28) reduce to

$ \mathit{\boldsymbol{B}}4{ * ^t}({F_9},{F_8},{F_7},{F_6},{F_5},{F_4},{F_3},{F_2},{F_1},{F_0}) = 0, $

(49)

where B4 is a 18×10 matrix, we calculate rank(B4)=10, so we have

$ {F_i} = 0 $

(50)

for 0≤i≤9.

The equations F_t=0 which t is even are degenerate to those t is odd.

So, finally, the equations reduce to

$ {F_1} = 0,{F_3} = 0,{F_5} = 0,{F_7} = 0,{F_9} = 0. $

(51)

Different like the former cases, this is a over-determined system for the elements of matrix S, since (51) is equivalent to:

$ \left\{ {\begin{array}{*{20}{l}} {\sum\nolimits_{i + j = m,j > i} {{S_{ij}}} (j - i) = 0,}\\ {\sum\nolimits_{i + j = m,j > i} {{S_{ij}}} ({j^3} - {i^3}) = 0,}\\ {\sum\nolimits_{i + j = m,j > i} {{S_{ij}}} ({j^5} - {i^5}) = 0,}\\ {\sum\nolimits_{i + j = m,j > i} {{S_{ij}}} ({j^7} - {i^7}) = 0,}\\ {\sum\nolimits_{i + j = m,j > i} {{S_{ij}}} ({j^9} - {i^9}) = 0.} \end{array}} \right. $

(52)

By checking rank of the coefficient matrix of any m, S must be zero matrix, then

Theorem 2.6   V₄⁹ can not be rotated horizontally.

3 The middle terms of harmonic sequence

In this section, we give another proof for the phenomenon that V₄⁹ and V₅⁹ can not be rotated horizontally. In fact, it is also right for any harmonic sequence in $\mathbb{C} $P⁹.

Theorem 3.1   In the following harmonic sequence associated to a linearly full holomorphic map [f₀]=f:M²→$\mathbb{C} $P⁹,

$ 0\mathop \to \limits^\partial \underline {{f_0}} \mathop \to \limits^\partial \cdots \mathop \to \limits^\partial \underline {{f_4}} \mathop \to \limits^\partial \underline {{f_5}} \mathop \to \limits^\partial \cdots \mathop \to \limits^\partial \underline {{f_9}} \mathop \to \limits^\partial 0. $

(53)

f₄, f₅ can not be rotated horizontally.

Proof  Assume f₄ can be rotated horizontally, ^tf₄Bdf₄=0, from(11), (12), we have

$ \begin{array}{*{20}{l}} {^t{f_4}\mathit{\boldsymbol{B}}{\rm{d}}{f_4}{ = ^t}{f_4}\mathit{\boldsymbol{B}}(\partial {f_4}{\rm{d}}z + \bar \partial {f_4}{\rm{d}}\bar z)}\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} { = ^t}{f_4}\mathit{\boldsymbol{B}}({f_5} + \partial {\rm{log}}|{f_4}{|^2}{f_4}){\rm{d}}z + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{\kern 1pt} ^t}{f_4}\mathit{\boldsymbol{B}}( - {\gamma _3}{f_3}){\rm{d}}z}\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} = 0,} \end{array} $

(54)

Since B is antisymmetric, ^tf₄Bf₄=0, then it is equivalent to

$ \left\{ {\begin{array}{*{20}{l}} {^t{f_4}\mathit{\boldsymbol{B}}{f_5} = 0,}\\ {^t{f_4}\mathit{\boldsymbol{B}}{f_3} = 0,} \end{array}} \right. $

(55)

then we have

$ ^t{f_4}\mathit{\boldsymbol{B}}{f_5} = 0{\mathop \to \limits^\partial {^t}}{f_4}\mathit{\boldsymbol{B}}{f_6} = 0{\mathop \to \limits^{\bar \partial } {^t}}{f_3}\mathit{\boldsymbol{B}}{f_6} = 0. $

(56)

Continue this process, we will get ^tf₀Bf₉=0, then do ∂ continuously for it, we will get

$ ^t{f_0}\mathit{\boldsymbol{B}}{f_i} = 0,i = 0,1,2,3,4,5,6,7,8,9, $

(57)

then it is easy to check that f₀ must be a local zero section of the trivial bundle M²×$\mathbb{C} $¹⁰, this is a contradiction.

f₅ is similar, we only need to show f₉ is a local zero section.