Math. Ann. DOI 10.1007/s00208-014-1013-y

Mathematische Annalen

Classification of conformal minimal immersions of constant curvature from S2 to HP 2 Ling He · Xiaoxiang Jiao

Received: 7 April 2013 / Revised: 22 October 2013 © Springer-Verlag Berlin Heidelberg 2014

Abstract In this paper, we study geometry of conformal minimal two-spheres immersed in quaternionic projective spaces. We firstly use Bahy-El-Dien and Wood’s results to obtain some characterizations of the harmonic sequences generated by conformal minimal immersions from S 2 to the quaternionic projective space HP2 . Then we give a classification theorem of linearly full totally unramified conformal minimal immersions of constant curvature from S 2 to the quaternionic projective space HP2 . Mathematics Subject Classification (2000)

Primary 53C42 · 53C55

1 Introduction It is well known that any linearly full conformal minimal immersion of constant curvature from S 2 to CPn belongs to the Veronese sequence, up to a rigid motion (cf.  In 1990 Ohnita (cf. [10]) gave a series of homogeneous minimal 2-spheres  [4]). ϕn,α of constant curvature in quaternionic projective spaces HPn , and conjectured   that ϕn,α exhaust all proper minimal isometric immersions of S 2 in HPn . In [7] He and Wang proved that the Veronese sequences in RP2m (2 ≤ 2m ≤ n) are the only totally real minimal 2-spheres with constant curvature in HPn .

Project supported by the NSFC (Grant No. 11071248, 11331002). L. He (B)· X. Jiao School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 101408, China e-mail: [email protected] X. Jiao e-mail: [email protected]

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L. He, X. Jiao

In this paper we mainly study classification of conformal minimal immersions of constant curvature from S 2 to the quaternionic projective space HP2 by theory of harmonic maps. In the paper [1], Aithal described all harmonic maps from the Riemann 2-sphere to HP2 . In 1991 Bahy-El-Dien and Wood gave the explicit construction of all harmonic two-spheres in quaternionic projective spaces (cf. [2]), which is considered as totally geodesic totally real submanifolds in complex Grassmann manifolds G(2, N ). Our arrangement is as follows. In the second section of this paper, firstly we give the definition of quaternionic projective space HPn−1 as the totally geodesic submanifold in G(2, 2n), then we give some fundamental results concerning G(k, N ) from the view of harmonic sequences, at last we give some brief description of Veronese sequence and the rigidity theorem in C P N . In the third section, we use their results to study some properties of the harmonic sequence generated by a harmonic map from S 2 to HP2 and obtain some characteristics of the corresponding harmonic map in HP2 . In the last section, we discuss geometric properties of conformal minimal immersions of constant curvature from S 2 to HP2 and obtain a classification theorem of linearly full totally unramified conformal minimal immersions of S 2 in HP2 with constant curvature (see Theorem 4.12). Furthermore our theorem verifies that Ohnita’ conjecture (cf. [10]) is true in the case n = 2. 2 Preliminaries (A) For any N = 1, 2, . . . , let ,  denote the standard Hermitian inner product on C N defined by z, w = z 1 w 1 + · · · + z N w N where z = (z 1 , . . . , z N )T , w = (w1 , . . . , w N )T ∈ C N and − denote complex conjugation. Let H denote the division ring of quaternions. Let j be a unit quaternion with j 2 = −1. Then we have an identification of C2 with H given by making (a, b) ∈ C2 correspond to a + bj ∈ H; let n ∈ {1, 2, . . .}, we have a corresponding identification of C2n with Hn . For any a + bj ∈ H, the left multiplication by j is given by j (a + bj) = −b + a j; the conjugation is given by a + bj = a − bj; the positive inner product is given by x, yH = Re(x y) for any x, y ∈ H. Let J : C2n → C2n be the conjugate linear map given by left multiplication by j, i.e. J(z 1 , z 2 , . . . , z 2n−1 , z 2n )T = (−¯z 2 , z¯ 1 , . . . , −¯z 2n , z¯ 2n−1 )T . Then J2 = −id where id denotes the identity map on C2n . In fact, for any v ∈ C2n , ¯ Jv = Jn v, 

   0 −1 0 −1 . where Jn = diag ,..., 1 0 1 0

  n

By the above, we immediately have the following lemma (cf. [2]).

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Classification of conformal minimal immersions

Lemma 2.1 The operator J has the following properties: (i) (ii) (iii) (iv)

Jv, Jw = w, v for all v, w ∈ C2n ; Jv, v = 0 for all v ∈ C2n ; ∂ ◦ J = J ◦ ∂, ∂ ◦ J = J ◦ ∂; J(λv) = λJv for any λ ∈ C, v ∈ C2n .

Let G(2, 2n) denote the Grassmann manifold of all complex 2-dimensional subspaces of C2n with its standard Kähler structure. The quaternionic projective spaces HPn−1 is the set of all one-dimensional quaternionic subspaces of Hn . Throughout the above we shall regard HPn−1 as the totally geodesic submanifold of G(2, 2n) given by H P n−1 = {V ∈ G(2, 2n) : JV = V } . Let Sp(n) = {g ∈ G L(n; H), g ∗ g = In } be the symplectic isometry group of HPn−1 . The explicit description is that the following diagram commutes: i1

Sp(n) −−−−→ U (2n) ⏐ ⏐ ⏐ ⏐ π1 π2 i2

H P n−1 −−−−→ G(2, 2n) where i 1 , i 2 are inclusions and π1 , π2 are projections, and i 1 (g) = E, for 1 ≤ a, b ≤ n 

a

2a−1 2a−1 = Aab , E 2b = −B b , E 2b−1 a 2a a 2a E 2b−1 = Bb , E 2b = Ab ,

where A = (Aab ), B = (Bba ) ∈ Mn (C), g = A + B j ∈ Sp(n); π1 (g) = g · [1, 0, . . . , 0]T ∈ H P n−1 ;  T 1, 0, 0, . . . , 0 ∈ G(2, 2n); π2 (E) = E · 0, 1, 0, . . . , 0  T   z 1 , z 2 , . . . , z 2n−1 , z 2n T . i 2 [z 1 + z 2 j, . . . , z 2n−1 + z 2n j] = −z 2 , z 1 , . . . , − z 2n , z 2n−1 Here we suppose that the metric on G(2, 2n) is given by Sect. 2 of [8], then the metric induced by i 2 is twice as much as the standard metric on HPn−1 . Thus we regard the harmonic map from S 2 to HPn−1 as the one from S 2 to G(2, 2n). For any g ∈ Sp(n), the action of g on HPn−1 induces an action of E which commutes with J. Then Sp(n) = on C P 2n−1 , where E ∈ U (2n)  {E ∈ U (2n), E ◦ J = J ◦ E} = E ∈ U (2n), E Jn E T = Jn . In the following, we deal with the symplectic isometry of HPn−1 through the corresponding symplectic isometry of C P 2n−1 .

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(B) In this section we simply introduce harmonic maps and harmonic sequences in G(k, N ) (cf. [5]) and calculate some corresponding geometric quantities. Let M be a simply connected domain in the unit sphere S 2 and let (z, z) be complex 2 = dzdz on M. Denote coordinates on M. We take the metric ds M ∂=

∂ ∂ , ∂= . ∂z ∂z

We consider the complex Grassmann manifold G(k, N ) as the set of Hermitian orthogonal projections from C N onto a k-dimensional subspace in C N . Then ϕ : S 2 → G(k, N ) is a Hermitian orthogonal projection onto a k-dimensional subbundle ϕ of the trivial bundle C N = M × C N given by setting the fibre of ϕ at x, ϕ x , equal to ϕ(x) for all x ∈ M. For any two orthogonal subbundles ϕ, ψ of C N , define vector bundle morphisms over M, A ϕ,ψ , A ϕ,ψ : ϕ → ψ called the ∂ - and ∂ -second fundamental forms of ϕ in ϕ ⊕ ψ by A ϕ,ψ (v) = πψ (∂v), A ϕ,ψ (v) = πψ (∂v) for v ∈ C ∞ (ϕ). Here πψ denotes orthogonal projection onto ψ and C ∞ (ϕ) denotes the vector space of smooth sections of ϕ. In particular A ϕ , A ϕ denote the second fundamental forms of ϕ in C N . Let ϕ : S 2 → G(k, N ) be a smooth harmonic map. Then from ϕ two harmonic sequences (cf. [5]) are derived as follows: A ϕ

A ϕ

A ϕ

α−1

A ϕα

ϕ = ϕ 0 −→ ϕ 1 −→ · · · −→ ϕ α −→ · · ·, 0

A ϕ

1

A ϕ −1

A ϕ −α+1

(2.1)

A ϕ−α

ϕ = ϕ 0 −→ ϕ −1 −→ · · · −→ ϕ −α −→ · · ·, 0

(2.2)

where ϕ α = I m A ϕα−1 and ϕ −α = I m A ϕ−α+1 are harmonic subbundles of C N (i.e., represent harmonic maps) respectively, α = 1, 2, . . .. We assume that ϕ is a linearly full harmonic map from S 2 to G(k, N ), here linearly full means that ϕ can not be contained in any proper trivial subbundle Cm of C N (m < N ). We know that several consecutive harmonic maps in (2.1) are not mutually orthogonal generally. So it is meaningful to define the isotropy order of ϕ (cf. [5], §3A) to be the greatest integer r such that ϕ i ⊥ ϕ j ∀i, j ∈ Z with 0 < |i − j| ≤ r ; if r = ∞, then ϕ is said to be strongly isotropic. Now we consider a special harmonic sequence. Suppose that ϕ : S 2 → G(k, N ) is a linearly full harmonic map and belongs to the following harmonic sequence: A ϕ

A ϕ

A ϕ

α−1

A ϕα

A ϕ

α0 −1

A ϕα

A ϕ

β0 −1

A ϕβ

ϕ 0 −→ ϕ 1 −→ · · · −→ ϕ = ϕ α −→ · · · −→ ϕ α −→ · · · −→ ϕ β −→ 0, 0 0 (2.3) where ϕ 0 , · · · , ϕ α are mutually orthogonal and k0 + . . . + kα0 = N (α0 ≤ β0 ) with 0 kα = rank ϕ α (α = 0, . . . , β0 ). For the harmonic sequence (2.3) we choose the local unitary frame e1 , e2 , . . . , e N of C N such that ek0 +···+kα−1 +1 , . . . , ek0 +···+kα−1 +kα locally span subbundle ϕ α (α = 0

123

1

0

0

Classification of conformal minimal immersions

1, . . . , α0 ). Let Wα = (ek0 +...+kα−1 +1 , . . . , ek0 +...+kα−1 +kα ) be an (N × kα )-matrix for α = 1, . . . , α0 and let W0 = (e1 , . . . , ek0 ) be an (N × k0 )-matrix. Then we have

Wα∗ Wα

ϕα = Wα Wα∗ , = Ikα ×kα ,

Wα∗ Wα+1

= 0,

(2.4) Wα∗ Wα−1

= 0.

(2.5)

By (2.5), a straightforward computation shows 

∂ Wα = Wα+1 α + Wα α , ∂ Wα = −Wα−1 ∗α−1 − Wα α∗ ,

(2.6)

where α is a (kα+1 × kα )-matrix and α is a (kα × kα )-matrix for α = 0, . . . , α0 . It is very evident that integrability conditions for (2.6) are ∗ − ∗, ∂ α = α+1 α α α

∂ α + ∂ α∗ = ∗α α + α∗ α − α−1 ∗α−1 − α α∗ . 2  2   Set L α = tr( α ∗α ) =  A ϕα  =  A ϕα+1  . Then the metric induced by ϕα is given by dsα2 = (L α−1 + L α )dzdz.

(2.7)

The Laplacian α and the curvature K α of dsα2 are given by α =

4 ∂∂, L α−1 + L α

Kα = −

2 ∂∂ log(L α−1 + L α ). L α−1 + L α

(2.8)

Especially, let ψ : S 2 → C P N be a linearly full harmonic map. Eells and Wood’s result (cf. [6]) shows that the following sequence in C P N is uniquely determined by ψ A 0

A 0

Ai−1

A N −1

Ai

A N

) ) −→ · · · −→ ψ = ψ i(N ) −→ · · · −→ ψ (N −→ 0, 0 ←− ψ (N 0 N

for some i = 0, 1, . . . , N , and here A 0 , A j denote A (N ) , A ψ0

(N )

ψj

(2.9)

respectively (j = 0,

. . ., N). ) , i.e. ∂ f (n) = 0, and let f (N ) be a local Let f 0(N ) be a holomorphic section of ψ (N 0 i 0 section of ψ i(N ) such that (N )

fi

(N )⊥

(N )

= ψi−1 (∂ f i−1 )

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L. He, X. Jiao

for i = 1, . . . , N . Then we have some formulas as follows (cf. [4]): (N )

∂ fi

∂∂ (N )

(N ) 2 (N ) | f i , i = 0, . . . , N − 1, (N ) (N ) ∂ f i(N ) = −li−1 f i−1 , i = 1, . . . , N , (N ) 2 (N ) (N ) ∂∂ log | f i | = li − li−1 , (N ) (N ) (N ) (N ) log li = li+1 − 2li + li−1 , i = 0, . . . , N − 1,

(N )

(N )

= f i+1 + ∂ log | f i

(N )

(N )

(2.10) (2.11) (2.12) (2.13)

(N )

where li = | f i+1 |2 /| f i |2 for i = 0, . . . , N , and l−1 = l N = 0. For convenience, we denote f i(N ) = ψ i(N ) for i = 0, 1, . . . , N . In the following, we give a definition of the unramified harmonic map as follows: Definition 2.2 If det( α ∗α )dz kα+1 dz kα+1 = 0 everywhere on S 2 in (2.3) for some α = 0, 1, . . . , β0 − 1, we say that ϕα : S 2 → G(kα , N ) is unramified. If det( α ∗α )dz kα+1 dz kα+1 = 0 everywhere on S 2 in (2.3) for each α = 0, 1, . . . , β0 − 1 , we say that the harmonic sequence (2.3) is totally unramified. In this case we also say that each map ϕα in (2.3) is totally unramified. In the case k = 1, the above definition is in accordance with that in §3 of [4]. Now recall ([5], §3A) that a harmonic map ϕ : S 2 → G(k, N ) in (2.1) (resp. (2.2)) is said to be ∂ -irreducible (resp. ∂ -irreducible) if rank ϕ = rank ϕ 1 (resp. rank ϕ = rank ϕ −1 ) and ∂ -reducible (resp. ∂ -reducible) otherwise. We assume that ϕα

in (2.3) is ∂ -irreducible, then | det α |2 dz kα dz kα is a well-defined invariant and has only isolated zeros on S 2 . Under this condition, it can be checked that (cf. [9]) ∂∂ log | det α |2 = L α−1 − 2L α + L α+1 ,

(2.14)

which is in accordance with (2.13) in the case k = 1. Furthermore if ϕα is ∂ -irreducible and unramified, then | det α |2 dz kα dz kα is a well-defined invariant and has no zeros on S 2 . It follows from (2.14) that (cf. [9]) δα−1 − 2δα + δα+1 = −2kα ,

(2.15)

 where δα = 2π √1 −1 S 2 L α dz ∧ dz. (C) In this section we review the rigidity theorem of conformal minimal immersions with constant curvature from S 2 to C P N . T (N ) = f i,0 , . . . , f i,N for each i = 0, . . . , N . Let The Veronese sequence. Let f i f i, p be given for i, p = 0, 1, . . . , N as follows

f i, p

123

i! = (1 + zz)i

     N p−i  p N−p z (zz)k . (−1)k p i −k k k

Classification of conformal minimal immersions (N )

(N )

Such a map ϕi = [ f i ] : S 2 → C P N is a conformal minimal immersion with (N ) 4 given by constant curvature N +2i(N −i) and constant Kähler angle θi 

θ (N ) tan i 2

2 =

i(N − i + 1) . (i + 1)(N − i)

) : S 2 → C P N is called the Veronese Such a harmonic sequence ϕ0(N ) , . . . , ϕ (N N (N ) (N ) sequence. We always denote it by V0 , . . . , VN : S 2 → C P N . Bolton et al proved that (cf. [4]) if ψ is a linearly full conformal minimal 2-sphere of constant curvature immersed in C P N , then, up to a holomorphic isometry of C P N , ψ is an element of the Veronese sequence (i.e. a Veronese surface). This rigidity theorem is our main tool to obtain the classification theorem in HP2 .

3 Characterization of harmonic maps from S2 to HP 2 We analyze the harmonic maps from S 2 to HP2 by the reducible and irreducible case respectively. It follows from ([2], Definition 3.4, 3.5 and Proposition 3.7) that all reducible harmonic maps from S 2 to HP2 have been characterized by the harmonic maps from S 2 to C P 5 . Now, we consider the irreducible harmonic maps ϕ : S 2 → H P 2 of isotropy order r . If ϕ has finite isotropy order, then r = 2 by ([2], Proposition 3.2 and Lemma 3.10); if ϕ is strongly isotropic, then r = ∞. So we study the following two cases respevtively. (a) r = 2. Here we state one of Bahy-El-Dien and Wood’s results (cf. [2], §6B) (equivalent to that of Aithal [1]) as follows: Lemma 3.1 ([2]) Any non-isotropic irreducible harmonic map ϕ : S 2 → H P 2 is of the form ϕ 2 where: (i) ϕ 0 = f ⊕ f 5 , f : S 2 → C P 5 being a full totally J-isotropic map, (ii) ϕ 1 is obtained from ϕ 0 by forward replacement of some holomorphic subbundle of ϕ 0 not equal to f 5 , (iii) ϕ 2 is obtained from ϕ 1 by backward replacement of β ⊥ ∩ ϕ 1 where β is the unique holomorphic subbundle of ϕ 1 not equal to the image of the first ∂ -return map of ϕ 1 such that I m(A ϕ 1 |β ) ⊥ Jβ. Firstly we recall ([2], §3) that a full holomorphic map f : S 2 → C P 2n−1 in (3.1) = J f is said to be totally J-isotropic, satisfying f (2n−1) 2n−1 A 0

A 0

A n−2

A n−1

A n

A 2n−2

A 2n−1

0 ←− f (2n−1) = f −→ · · · −→ f (2n−1) −→ f (2n−1) −→ · · · −→ f (2n−1) −→ 0. 0 n−1 n 2n−1 (3.1) From the definition of totally J-isotropic and (3.1), we can immediately get (2n−1)

lj

(2n−1)

= l2n−2− j ,

(3.2)

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L. He, X. Jiao (2n−1)

and set J f 0

(2n−1)

= τ0 f 2n−1

|τ0 |2 =

(2n−1) 2 |

| f0

(2n−1)

| f 2n−1 |2

, then (2n−1)

, J fj

(2n−1)

= (−1) j τ0

| f 2n−1 |2

(2n−1)

(2n−1)

| f 2n−1− j |2

f 2n−1− j ,

(3.3)

where j = 0, . . . , n − 1. Let ϕ0 : S 2 → H P 2 be an irreducible linearly full harmonic map of isotropy order 2. In the following we use Lemma 3.1 to characterize ϕ0 explicitly. In (i) of Lemma 3.1 ϕ 0 with isotropy order 4 belongs to the harmonic sequence as follows: A 0

A 1

A 0

A 4

ϕ

A 0

A 1

ϕ

A 5

A 4

←− · · · ←− f (5) ←− ϕ 0 −→ f (5) −→ · · · −→ f (5) −→ 0, 0 ←− f (5) 0 4 1 5 (5)

where ϕ 0 = f (5) ⊕ f (5) with a full totally J-isotropic map f 0 0 5

(3.4)

: S2 → C P 5. (5)

(5)

By (ii) of Lemma 3.1 and (3.4), there exists a local section V0 = x0 f 0 + J f 0 such that V 0 is an antiholomorphic subbundle of ϕ 0 , and ϕ 1 is obtained from ϕ 0 by forward replacement of V ⊥ 0 , i.e. ϕ 1 = V 0 ⊕ f (5) , 1 where x0 is a smooth function on S 2 expect some isolated points, and here V 0 denotes the line subbundle spanned by V0 . Then ϕ 1 with isotropy order 3 belongs to the harmonic sequence as follows: A 0

0 ←−

f (5) 0

A 1

A 3

f (5) 3

←− · · · ←−

A 1

ϕ−1

←−

ϕ 1−1

A 1 ϕ0

←−

ϕ 10

A 1

A 2

ϕ0

A 5

A 4

−→ f (5) −→ · · · −→ f (5) −→ 0, 2 5 (3.5)

where ϕ 10 = ϕ 1 and ϕ 1−1 = Jϕ 1 .

(5)

By (iii) of Lemma 3.1 and (3.5), there exists a local section V = x1 f 1 + V0 such that β ⊥ ∩ ϕ 1 = V is an antiholomorphic subbundle of ϕ 1 , and ϕ 2 is obtained from ϕ 1 by backward replacement of V , i.e. ϕ 2 = X ⊕ JX , where X =

(5) 1 f (5) | f 1 |2 1

x1 V |V |2

−

and x1 is a smooth function on S 2 expect at some

isolated points. (5) (5) Obviously, X, V, f 2 , JX, JV, J f 2 are mutually orthogonal. Then ϕ0 with isotropy order 2 belongs to the harmonic sequence as follows: A 0

A 1

A 2

A ϕ

−1

A ϕ

←− f (5) ←− f (5) ←− ϕ −1 ←− 0 ←− f (5) 0 1 2 A ϕ

A ϕ

A 3

A 4

0

A 5

−→ f (5) −→ f (5) −→ 0, ϕ 0 −→ ϕ 1 −→ f (5) 3 4 5 0

1

and ϕ −1 = Jϕ 1 . where ϕ 0 = ϕ 2 , ϕ 1 = V ⊕ f (5) 2

123

(3.6)

Classification of conformal minimal immersions

Since V 0 and V are antiholomorphic subbundles of ϕ 0 and ϕ 1 respectively, and = ϕ 0 , we get

A ϕ1 (ϕ 1 )

πϕ 0 (∂ V0 ) ∈ V 0 , πϕ 1 (∂ V ) ∈ V ;

(3.7)

πϕ ⊥ (∂ V ) ∈ ϕ 0 .

(3.8)

1

Through a direct computation, the conditions (3.7) and (3.8) are equivalent to the equations ⎧ (5) 2 ⎪ ⎨ ∂ x0 + x0 ∂ log | f 0 | = 0, (5) (3.9) ∂ x1 + x0 + x1 ∂ log | f 1 |2 = 0, ⎪ ⎩ (5) (5) ∂ x0 − 2x1 l0 − x0 ∂ log | f 0 |2 = 0 hold. Thus we give the property of the harmonic sequence (3.6). Then we have Proposition 3.2 The map ϕ0 : S 2 → H P 2 is an irreducible linearly full harmonic (5) 1 f − |Vx 1|2 V map of finite isotropy order if and only if ϕ 0 = X ⊕JX , where X = (5) 2 1 (5)

(5)

(5)

| f1 |

(5)

with V = x1 f 1 + x0 f 0 + J f 0 (where f 0 is a full totally J -isotropic map), and the corresponding coefficients x0 , x1 satisfy the equations (3.9). Proof Through the above construction, the necessity is obvious. From (3.9), we know V 0 and V are antiholomorphic subbundles of ϕ 0 and ϕ 1 respectively. It follows from the above that ϕ 2 is obtained by forward and then backward replacement of V ⊥ 0 , then   V , so that ϕ 2 is harmonic. Thus we get the sufficiency. (b) r = ∞. At first we give a lemma which is the totally isotropic case of ([2], Lemma 4.3). Lemma 3.3 Let ϕ 1 : S 2 → G(2, 2n) be a ∂ -reducible totally isotropic harmonic map with ϕ 1−1 = Jϕ 1 . Let β be the holomorphic line subbundle of ϕ 1 defined by K er A ϕ 1 . Let ϕ be obtained from ϕ 1 by backward replacement of α = β ⊥ ∩ ϕ 1 . Then ϕ is quaternionic. Furthermore, ϕ is a quaternionic mixed pair or a quaternionic Frenet pair. Proof Let 2 ≤ m ≤ 2n − 1 and 1 ≤ r ≤ m, then ϕ 1 belongs to the following harmonic sequence: A m

0 ←− J

(m) fm

A m−1

A

m−1

Ar

←− · · · ←− J

f r(m)

A 1

ϕ−1

←−

A m

(m) · · · −→ f m −→ 0,

ϕ 1−1

A 1 ϕ0

←−

ϕ 10

A 1 ϕ0

Ar

−→ f r(m) −→ (3.10)

(m)

where ϕ 10 = ϕ 1 , ϕ 1−1 = Jϕ 1 with f i denotes A (m) (i = 0, . . . , m).

(m)

, J fi

: S 2 → C P m ⊆ C P 2n−1 and Ai

J fi

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L. He, X. Jiao

Through observing (3.10), we get α = f r(m) . Set α −1 = I m(A ϕ 1 |α ) and β −1 = −1 0

1 α⊥ −1 ∩ ϕ −1 , then ϕ = β ⊕ α −1 . With ϕ 1−1 = Jϕ 1 and a simple computation, we have α −1 = Jβ and β −1 =

Jα = J f r(m) . So ϕ = β ⊕ Jβ is quaternionic and belongs to the following harmonic −1 sequence: A m

A

A

A ϕ

r −1

A ϕ

Ar −1

A m−1

A m

(m) ←− · · · ←− J f r(m) ←− ϕ −→ f r(m) −→ · · · −→ f m −→ 0. −1 −1 (3.11) From ([2], Proposition 3.7) and (3.11), we immediately have that ϕ is a quaternionic mixed pair or a quaternionic Frenet pair.  

0 ←− J

(m) fm

m−1

Let ϕ : S 2 → H P 2 be an irreducible linearly full strongly isotropic harmonic map, then ϕ belongs to the following harmonic sequence: A ϕ

A ϕ

−1

A ϕ

A ϕ

0 ←− ϕ −1 ←− ϕ 0 = ϕ −→ ϕ 1 −→ 0, 0

0

1

(3.12)

where ϕ −1 = Jϕ 1 and ϕ−1 , ϕ1 : S 2 → G(2, 6) are holomorphic and antiholomorphic respectively. Let α 0 be any holomorphic line subbundle of ϕ 0 , then ϕ 0 = α 0 ⊕ Jα 0 , using ([2], Lemma 4.1) we get a new harmonic sequence from (3.12) as follows: A 1

ϕ−2

0 ←−

A 1

ϕ 1−2

ϕ−1

←−

ϕ 1−1

A 1 ϕ0

←−

ϕ 10

A 1 ϕ0

−→

ϕ 11

A 1 ϕ1

−→ 0,

(3.13)

where ϕ 10 = Jα 0 ⊕ α 1 , ϕ 11 = β 1 , ϕ 1−1 = Jϕ 10 and ϕ 1−2 = Jϕ 11 with that α 1 = I m(A ϕ0 |α0 ) and β 1 = α ⊥ 1 ∩ ϕ 1 are both line subbundles of ϕ 1 . 1 : S 2 → G(2, 6) are harmonic and From (3.13), we immediately see that ϕ01 , ϕ−1 1 1 2 m 5 ϕ1 , ϕ−2 : S → C P ⊆ C P (m = 2, 3, 4, 5) are linearly full antiholomorphic and holomorphic respectively. Then in (3.13) we set (m) , ϕ 11 = f m (m) : S 2 → C P m ⊆ C P 5 is a harmonic sequence. where f 0(m) , . . . , f m By Lemma 3.3 and (3.13) we obtain a new harmonic map ϕ 2 : S 2 → H P 2 which belongs to the following harmonic sequence: A m

0 ←− J

(m) fm

A

m−1

←− J

(m) f m−1

A 2 ϕ

A 2 ϕ

A m−1

where ϕ 2 = β ⊕ Jβ with β = K er A ϕ 1 . 0

With ([2], Proposition 3.7), we can express ϕ 2 as follows: (m) (m) ⊕ J f m−2 . ϕ 2 = f m−2

123

A m

(m) (m) ←− ϕ −→ f m−1 −→ f m −→ 0, 2

(3.14)

Classification of conformal minimal immersions

In the following, we discuss the harmonic sequence (3.14) in cases m = 2, 3, 4, 5 respectively, and characterize the corresponding ϕ explicitly. (b.I) m = 5. Obviously ϕ 2 is a quaternionic Frenet pair by ([2], Definition 3.5). (5) Thus we get that f 0 is a full totally J-isotropic map and the harmonic sequence (3.14) becomes A 0

0 ←−

f (5) 0

A 1

←−

f (5) 1

A 2 ϕ

←− ϕ = 2

f (5) 2

⊕ f (5) . where ϕ 2 = f (5) 2 3

⊕

A 2

A 5

A 4

ϕ

f (5) 3

−→ f (5) −→ f (5) −→ 0, 4 5

(5)

In (3.12) we choose a local section V = x2 f 2

(5)

+ x3 f 3

(5)

+ f4

(3.15)

such that

, ϕ 1 = V ⊕ f (5) 5 where x2 , x3 are smooth functions on S 2 expect some isolated points and x2 , x3 = 0. (5) 1 Set X = (5) f − |V1|2 V , then in (3.12) we have 2 4 | f4 |

ϕ 0 = X ⊕ JX . (5)

(5)

Obviously, X, V, f 5 , JX, JV, J f 5 and A ϕ1 (ϕ 1 ) = ϕ 0 , we get

are mutually orthogonal. Since A ϕ1 (ϕ 1 ) = 0

πϕ ⊥ (∂ V ) = 0;

(3.16)

πϕ ⊥ (∂ V ) ∈ ϕ 0 .

(3.17)

1 1

Through a direct computation, the conditions (3.16) and (3.17) are equivalent to the equations ⎧ (5) 2 (5) 2 ⎪ ⎨ ∂ x2 − x2 x3 − x2 ∂ log | f 4 | + x2 ∂ log | f 2 | = 0, (5) 2 (3.18) ∂ x3 + x2 − x3 − x3 ∂ log l3 = 0, ⎪ ⎩ (5) (5) 2 x3 ∂ x2 − x2 ∂ x3 + 2x2 l1 − x3 l2 = 0 hold. Thus we give a property of the harmonic sequence (3.12) in the case m = 5. Then we prove Proposition 3.4 ϕ0 is an irreducible linearly full strongly isotropic harmonic map in : S2 → C P 5 (3.12) corresponding to the harmonic sequence (3.13) with ϕ 11 = f (5) 5 if and only if x2 , x3 satisfy the equations (3.18). Proof Through the above construction, the necessity is obvious. From (3.18), we know ϕ1 is antiholomorphic. It follows from A ϕ1 (ϕ 1 ) = ϕ 0 that ϕ0 is harmonic. Thus we get the sufficiency.  

123

L. He, X. Jiao

(b.II) m = 4. Obviously ϕ 2 is a quaternionic Frenet pair. Thus the harmonic sequence (3.14) becomes A

A 2

A

A 2

ϕ

ϕ

A 3

A 4

0 ←− J f (4) ←− J f (4) ←− ϕ 2 −→ f (4) −→ f (4) −→ 0, 4 3 3 4 4

3

(4)

where ϕ 2 = f (4) ⊕ J f (4) and we see f i 2 2 Then we have

(3.19)

as the vector in C6 naturally.

Proposition 3.5 m = 4 if ϕ0 is an irreducible linearly full strongly isotropic harmonic map in (3.12) corresponding to the harmonic sequence (3.13). (4)

Proof Because f i is linearly full as the vector in C5 , the trivial bundle S 2 × C6 over S 2 has a corresponding decomposition S 2 ×C6 = S 2 ×C5 ⊕ S 2 ×C. Then from (3.19) we see that one of the six orthogonal linear bundles is in S 2 × C, what is impossible. So there doesn’t exist this case.   (b.III) m = 3. Similarly we have the following conclusion Proposition 3.6 m = 3 if ϕ0 is an irreducible linearly full strongly isotropic harmonic map in (3.12) corresponding to the harmonic sequence (3.13). (b.IV) m = 2. In this case ϕ 2 is a quaternionic mixed pair by ([2], Definition 3.4). Thus the harmonic sequence (3.14) becomes A

A

A ϕ

A ϕ

A 1

A 2

←− J f (2) ←− ϕ 2 −→ f (2) −→ f (2) −→ 0, 0 ←− J f (2) 2 1 1 2 2

1

(2)

(2)

(2)

(2)

(2)

(3.20)

(2)

where ϕ 2 = J f (2) ⊕ f (2) and f 0 , f 1 , f 2 , J f 0 , J f 1 , J f 2 are mutually orthog0 0 onal. (2) (2) (2) In (3.12) we choose a local section V = x0 f 0 + x1 f 1 + J f 0 such that ϕ 1 = V ⊕ f (2) , 2 where x0 , x1 are smooth functions on S 2 expect some isolated points. 1 Set X = (2) f (2) − |Vx 1|2 V , then in (3.12) we have 2 1 | f1 |

ϕ 0 = X ⊕ JX . (2)

(2)

Obviously, X, V, f 2 , JX, JV, J f 2 are mutually orthogonal. Through a direct computation, the condition (3.16) and (3.17) are equivalent to the equations ⎧ (2) 2 ⎪ ⎨ ∂ x0 + x0 ∂ log | f 0 | = 0, (3.21) ∂ x1 + x0 + x1 ∂ log | f 1(2) |2 = 0, ⎪ ⎩ (2) (2) 2 ∂ x0 − 2x1 l0 − x0 ∂ log | f 0 | = 0 hold.

123

Classification of conformal minimal immersions

Thus we give a property of the harmonic sequence (3.12) in the case m = 2. Then we have Proposition 3.7 The map ϕ0 : S 2 → H P 2 is an irreducible linearly full strongly isotropic harmonic map in (3.12) corresponding to the harmonic sequence (3.13) : S 2 → C P 2 ⊂ C P 5 , if and only if ϕ 0 = X ⊕ JX ,where X = with ϕ 11 = f (2) 2 (2) 1 f (2) | f 1 |2 1

−

x1 V |V |2

(2)

(2)

(2)

(2)

with V = x1 f 1 + x0 f 0 + J f 0 (where f 0

is a full totally J -

isotropic map), and the corresponding coefficients x0 , x1 satisfy the equations (3.21).  

Proof The proof is similar to that of Proposition 3.2. 4 Conformal minimal immersions of constant curvature from S2 to HP 2

In this section, we regard the harmonic maps from S 2 to HP2 as the conformal minimal immersions of S 2 in HP2 . Then we consider the harmonic maps of constant curvature from S 2 to HP2 by the reducible case and irreducible case. So we divide the two cases into the following two subsections. 4.1 Reducible harmonic maps of constant curvature from S 2 to HP2 Let ϕ0 : S 2 → H P 2 be a reducible linearly full harmonic map, then by ([2], Proposition 3.7) we know that ϕ0 is a quaternionic mixed pair or a quaternionic Frenet pair. In the following we discuss the two cases with ϕ0 of constant curvature respectively. (I) If ϕ0 is a linearly full quaternionic Frenet pair, then (5)

i 2 ◦ ϕ0 = f 2

(5)

⊕ f3 ,

(4.1)

where f (5) , f (5) , f (5) , f (5) , f (5) , f (5) : S 2 → C P 5 is the harmonic sequence gener0 1 2 3 4 5 (5)

ated by the full totally J-isotropic map f 0 . Obviously ϕ0 belongs to the following harmonic sequence A 0

0 ←−

f (5) 0

A 1

←−

f (5) 1

A 0 ϕ

A 0 ϕ

A 4

A 5

←− ϕ 0 −→ f (5) −→ f (5) −→ 0. 4 5

(4.2)

Here we state one of Shen Yibing’s results as follows. Lemma 4.1 ([11]) Let ψ : S 2 → C P N be a full nondegenerate holomorphic curve ) = ψ, . . . , f (N ) : S 2 → C P N . For any integer p with the harmonic sequence f (N 0 N −1 (N ) ] : S2 → (0 ≤ p ≤ N ), the associated map y p of ψ p is defined by y p = [ f p(N ) ∧ f p+1 G(2, N ). If there is an integer p such that y p induces a metric of constant curvature on S 2 , then up to a holomorphic isometry of CP N , ψ is a Veronese surface.

Then we prove that

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L. He, X. Jiao

Proposition 4.2 Let ϕ0 : S 2 → H P 2 be a linearly full quaternionic Frenet pair of constant curvature K 0 . Then K 0 = 41 and, up to a symplectic isometry of HP2 , i 2 ◦   (5) (5) ϕ0 = U V2 ⊕ U V3 for U∈ G   U ∈ U (6), U W3 U T = J3 , where  some  0 −1 0 −1 0 −1 W3 = antidiag , , . 1 0 1 0 1 0 Proof By Lemma 4.1 we get that K 0 = 41 and up to a holomorphic isometry of C P 5 , f 0(5) is a Veronese surface. We can choose a complex coordinate z on C = (5) (5) (5) S 2 \{ pt} so that f 0 = U V0 , where U ∈ U (6) and V0 has the standard expression (5) (5) given in part (C) of Sect. 2. Then J f 0 = f 5 is equivalent to (5)

(5)

J3 U V 0 = λU V5 , where λ is a parameter.  Set W3 = antidiag

(4.3)

     0 −1 0 −1 0 −1 , , . From part (C) of Sect. 2, 1 0 1 0 1 0

we get (5)

V0

= (1,

√

5z,

√

10z 2 ,

√

10z 3 ,

√

5z 4 , z 5 )T ,

and V5(5) =

√ √ √ √ 5! (−z 5 , 5z 4 , − 10z 3 , 10z 2 , − 5z, 1)T , 5 (1 + zz)

(5)

which implies V5

=

(5) 5! W V . (1+zz)5 3 0

Then the condition (4.3) becomes

(5)

(5)

J3 U V 0 = U W3 V 0 . By differentiating with respect to z¯ in the above formula, the matrices J3 U and U W3 have the same effect on all derivatives of V¯0(5) . Since V0(5) is full, these span C 6 so that J3 U = U W3 , i.e. (4.4) U W3 U T = J3 . Define a set   G  U ∈ U (6), U W3 U T = J3 .   With Sp(3) = A ∈ U (6), A J3 A T = J3 , the following can be easily checked (i) ∀ A ∈ Sp(3), U ∈ G, we have that AU ∈ G; (ii) ∀ U, V ∈ G, ∃ A = U V ∗ ∈ Sp(3) s.t. U = AV . So we get the conclusion.

123

 

Classification of conformal minimal immersions

⎡

1 ⎢0 ⎢ ⎢0 Remark 4.3 (i) G = ∅. Simply we choose U0 = ⎢ ⎢0 ⎢ ⎣0 0

0 0 1 0 0 0

0 0 0 0 1 0

0 0 0 0 0 1

⎤ 0 0 0 1⎥ ⎥ 0 0⎥ ⎥, then −1 0⎥ ⎥ 0 0⎦ 0 0

i 2 ◦ ϕ0 = f 2(5) ⊕ f 3(5) ∈ G(2, 6) √ √ √ √ √ √ (5) with f 2 = [( 10z 2 , 10z 3 , 3 2zz 2 − 2 2z, 2 2z 3 z − 3 2z 2 , 3z 2 z 2 − √ √ √ (5) 6zz + 1, z 2 z 2 − 6zz + 3)T ] and f 3 = [(− 10z 3 , 10z 2 , −2 2zz 3 + √ √ √ 3 2z 2 , 3 2z 2 z − 2 2z, −z 2 z 2 + 6zz − 3, 3z 2 z 2 − 6zz +√ 1)T ]. √ Further, from inclusion map i 2 that ϕ0 = [( 10z 2 + 10z 3 j, √ √ the √ √ 2 it follows 3 2zz −2 2z +(2 2zz 3 −3 2z 2 ) j, 3z 2 z 2 −6zz +1+(z 2 z 2 −6zz +3) j)T ] ∈ H P 2 . Here j is the unit quaternion appeared in (A) of Sect. 2. (ii) From (4.2), ϕ0 is strongly isotropic. (II) If ϕ0 is a full quaternionic mixed pair, then (m)

i 2 ◦ ϕ0 = f 0

(m)

⊕ J f0

,

(4.5)

(m)

where f 0 : S 2 → C P m ⊆ C P 5 (m = 2, 3, 4, 5) is holomorphic and f 1(m) ⊥ J f 0(m) . Obviously ϕ0 belongs to the following harmonic sequence A m

A

A 0

A

A 0

ϕ

ϕ

A 1

A m−1

A m

(m) 0 ←− J f m ←− · · · ←− J f 1(m) ←− ϕ 0 −→ f (m) −→ · · · −→ f (m) −→ 0. 1 m (4.6) So the induced metric by ϕ0 is given by m−1

1

(m)

ds02 = 2l0 dzdz, (m)

(m)

where l0 dzdz is the induced metric by f 0 Then we prove

(4.7)

.

Proposition 4.4 Let ϕ0 : S 2 → H P 2 be a linearly full quaternionic mixed pair of (5) constant curvature K 0 . Then up to a symplectic isometry of HP2 , i 2 ◦ ϕ0 = U V0 ⊕ U V5(5) with K 0 = 25 for some U ∈ G or i 2 ◦ ϕ0 = JU V0(2) ⊕ U V0(2) with K 0 = 1 for some U ∈ G. (m)

Proof By (4.7) we get that K 0 = m2 and up to a holomorphic isometry of C P 5 , f 0 is a Veronese surface. We can choose a complex coordinate z on C = S 2 \{ pt} so that (m) (m) (m) f 0 = U V0 , where U ∈ U (6) and V0 has the standard expression given in part (m) (m) (C) of Sect. 2 (adding zeros to V0 such that V0 ∈ C6 ). Then f 1(m) ⊥ J f 0(m) is equivalent to % & (m)

tr V1

(m) T

V0

U T J3 U = 0.

(4.8)

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L. He, X. Jiao

Set U T J3 U = W , then we immediately get & % (m) (m) T W = 0, W T = −W, W ∗ W = I, tr V1 V0

(4.9)

where I is the identity matrix. Define a set   G W  U ∈ U (6), U W U T = J3 . For a given W , the following can be easily checked (i) ∀ A ∈ Sp(3), U ∈ G W , we have that AU ∈ G W ; (ii) ∀ U, V ∈ G W , ∃ A = U V ∗ ∈ Sp(3) s.t. U = AV . In the following we discuss W in the cases m = 2, 3, 4, 5. (5)

(5)

(5)

(5) T

(i) m = 5. K 0 = 25 . By the standard expression of V0 and V1 , we get V1 V0 is a polynomial matrix in z and z. But W is a constant matrix. Using the method of undeterminated coefficients by (4.9), we get that W = W3 and G W = G. Then by the proof of Proposition 4.2, we have (5)

i 2 ◦ ϕ0 = U V0

(5)

⊕ U V5

for some U ∈ G. (ii) m = 4. K 0 = 21 . Using (4.9), we find that there don’t exist such W . (iii) m = 3. K 0 = 23 . Using (4.9), we get ⎡

0 ⎢0 ⎢ ⎢0 W =⎢ ⎢−1 ⎢ ⎣0 0

0 0 1 0 0 0

0 −1 0 0 0 0

1 0 0 0 0 0

0 0 0 0 0 √ e

−1θ

0 0 0 0 √

⎤

⎥ ⎥ ⎥ ⎥, ⎥ ⎥ −e −1θ ⎦ 0

where θ is a real parameter. (3) (3) In this case, for any U ∈ G W , we have JU V0 = −U V3 , then i 2 ◦ ϕ0 = U V0(3) ⊕ U V3(3) . Obviously ϕ0 is included in HP1 , so it is not linearly full. (iv) m = 2. K 0 = 1. Using (4.9), we get  W =

123

0 B

 −B T , 0

Classification of conformal minimal immersions

where B ∈ U (3). In this case if we fix some B ∈ U (3), then (2)

i 2 ◦ ϕ0 = JU V0

(2)

⊕ U V0

for some U ∈ G W . Now we claim that for all B ∈ U (3), the corresponding ϕ0 are symplecticly equivalent. ⎡ ⎤ b11 b12 b13 Set B = ⎣b21 b22 b23 ⎦ ∈ U (3), the corresponding b31 b32 b33 ⎡

1 ⎢0 ⎢ ⎢0 U =⎢ ⎢0 ⎢ ⎣0 0

0 0 1 0 0 0

0 0 0 0 1 0

⎤ 0 0 b21 b31 ⎥ ⎥ 0 0 ⎥ ⎥ ∈ GW . b22 b32 ⎥ ⎥ 0 0 ⎦ b23 b33

0 b11 0 b12 0 b13

Then we have U V0(2) = [(1, 0, (2)

JU V0

√

2z, 0, z 2 , 0)T ], √ = [(0, 1, 0, 2z, 0, z 2 )T ].

Thus, up ⎤to a symplectic isometry, ϕ0 is unique. For convenience, we choose B = ⎡ 0 0 1 ⎣0 −1 0⎦, then G W = G. 1 0 0 In summary we get the conclusion.   Remark 4.5 (i) In the case m = 5, we choose the same U0 as in Remark 4.3 (i), then ⊕ f (5) ∈ G(2, 6) i 2 ◦ ϕ0 = f (5) 0 5 √ √ √ √ (5) (5) with f 0 = [(1, z 5 , 5z, − 5z 4 , 10z 2 , 10z 3 )T ] and f 5 = [(−z 5 , 1, √ √ √ 4 √ √ 3 √ 2 T √ − 10z , 10z ) ]. Further ϕ0 = [(1+z 5 j, 5z− 5z 4 j, 10z 2 + √5z ,3 5z, 10z j)T ] ∈ H P 2 . From (4.6), √ ϕ0 has isotropy order 4. (ii) In the case m = 2, ϕ0 = [(1, 2z, z 2 )T ] is the Veronese surface in C P 2 . From (4.6), ϕ0 is strongly isotropic. 4.2 Irreducible harmonic maps of constant curvature from S 2 to HP2 In the following, we discuss the cases (a), (b.I) and (b.IV) with ϕ0 of constant curvature in Sect. 3 respectively. Here we give a simple lemma that we need in the following proofs.

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Lemma 4.6 Let v = (v1 , . . . , vn )T and w = (w1 , . . . , wn )T be two n-dimensional complex vectors. If there exists a real (n × n)-matrix H , every element of which is not zero, satisfying H 2 = In where In denotes the (n × n)-identity matrix such that v = H w, and for any 1 ≤ i ≤ n, v1 = w1 , . . . , vi−1 = wi−1 , vi+1 = wi+1 , . . . , vn = w n hold, then vi = wi holds. Proof It follows from the known conditions that (w 1 , . . . , wi−1 , vi , wi+1 , . . . , w n )T = H (w1 , . . . , wi−1 , wi , wi+1 , . . . , wn )T , which implies (0, . . . , 0, v i − wi , 0, . . . , 0)T = H (0, . . . , 0, wi − vi , 0, . . . , 0)T . Since every element of H is not zero, the above formula verifies the conclusion.

 

(a) In this case, we prove Proposition 4.7 Let ϕ0 : S 2 → H P 2 be an irreducible linearly full totally unramified harmonic map of finite isotropy order with constant curvature K 0 . Then up to a 2 for some U ∈ G. symplectic isometry of HP2 , i 2 ◦ ϕ0 = U V1(5) ⊕U V4(5) with K 0 = 13 Proof Let ϕ0 : S 2 → H P 2 be an irreducible linearly full harmonic map of finite isotropy order. By Proposition 3.2, we choose the local unitary frame f (5) J f 2(5) JX X V JV , e2 = , e3 = , e4 = 2(5) , e5 = (5) , , e6 = |X | |X | |V | |V | | f2 | | f2 |

e1 =

1 f (5) − |Vx 1|2 V, V = x1 f 1(5) + x0 f 0(5) (5) | f 1 |2 1 on S 2 except some isolated points.

where X =

(4.10)

+ J f 0(5) and x0 , x1 are smooth

functions Set W0 = (e1 , e2 ), W1 = (e3 , e4 ) and W−1 = (e5 , e6 ), then by (2.6), we get    λ 0 t0 , 1 = 0 0 = 0 μ0

(5)

| f3 | (5)

| f2 |



 −μ0 −t0 , = 0 λ0 

, −1

where ( (5) JX, ∂ V |f | λ0 = ∂e1 , e3  = − =− 1 , |X ||V | |V0 | ) * (5) (5) X, ∂ f 2 |V0 || f 2 | μ0 = ∂e2 , e4  = − = , (5) (5) |X || f 2 | |V || f 1 | '

and t0 = ∂e2 , e3  =

123

∂ X, V  |V | x 1 =− ∂ , |X ||V | |X | |V |2

(4.11)

Classification of conformal minimal immersions

' ( (5) (5) with V0 = x0 f 0 + J f 0 and JX, ∂ V = 1 by the third equation of (3.9). A straightforward computation shows (5)

| f 0 |2 (5) (5) 2 2 l l dz dz , |V |2 0 1 (5) det ( 1 ∗1 )dzdz = l2 dzdz,

|det 0 |2 dz 2 dz 2 =

(4.12) (4.13)

L 0 = L −1 = λ0 λ0 + μ0 μ0 + t0 t 0 , L1 =

(4.14)

(5) l2 .

(4.15)

If ϕ0 is totally unramified, then |det 0 |2 dz 2 dz 2 = 0 and det ( 1 ∗1 )dzdz = 0 (5) everywhere on S 2 . It follows from (4.12) and (4.13) that li dzdz = 0(i = 0, 1, 2) (5) (5) (5) (5) (5) everywhere on S 2 . By (3.2), we have that l0 = l4 and l1 = l3 . So li dzdz = 0(i = 0, 1, 2, 3, 4) everywhere on S 2 . Then we obtain that the harmonic sequence , . . . , f (5) : S 2 → C P 5 is totally unramified, from ([4], §3) we get f (5) 0 5 (5)

(5)

(5)

(5)

(5)

δ0 = δ4 = 5, δ1 = δ3 = 8, δ2 = 9.

(4.16)

By (2.15), we have where δα =

1 √ 2π −1

δ1 − 2δ0 + δ−1 = −4,

 S2

(4.17)

L α dz ∧ dz (α = −1, 0, 1). It follows from (4.14) and (4.15) (5)

that δ0 = δ−1 and δ1 = δ2 = 9, so that δ0 = 13.

(4.18)

If ϕ0 : S 2 → H P 2 ⊂ G(2, 6) is a harmonic map with constant curvature K 0 , then 2 , and we can choose a complex coordinate z on it follows from (4.18) that K 0 = 13 2 C = S \{ pt} so that the induced metric ds02 = 2L 0 dzdz by ϕ0 is given by ds02 =

26 dzdz, (1 + zz)2

which implies L 0 = λ0 λ0 + μ0 μ0 + t0 t 0 =

13 . (1 + zz)2 (5)

(4.19) (5)

(5)

Consider a local lift of the i-th osculating curve (cf. [4]) Fi = f 0 ∧· · ·∧ f i (i = (5) 0, 1, 2, 3, 4, 5). We choose a nowhere zero holomorphic C6 -valued function f 0 , then (5) Fi is a nowhere zero holomorphic curve and is a polynomial function on C of degree (5) (5) (5) δi satisfying ∂∂ log |Fi |2 = li . So using (2.14), (4.12), (4.14), (4.15) and (4.19), we obtain   (5) (5) |V |2 |F0 |2 |F1 |2 ∂∂ log = 0. (4.20) | f 0(5) |2 (1 + zz)13

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L. He, X. Jiao

By (4.12) we know that

|V |2 (5) | f 0 |2

it follows from (4.16) that

is a globally defined function without zeros on S 2 . Then (5)

(5)

|V |2 |F0 |2 |F1 |2 (5) | f 0 |2 (1+zz)13

is globally defined on C and has a positive

constant limit c1 as z → ∞. Thus from (4.20) (5)

(5)

|V |2 |F0 |2 |F1 |2 (5)

| f 0 |2 (1 + zz)13 i.e.

|V |2 (5) | f 0 |2

= c1 ,

(5)

= |x0 |2 + 1 + |x1 |2 l0 =

c1 (1 + zz)13 (5)

(5)

|F0 |2 |F1 |2

.

(4.21)

By the first equation of (3.9), we get ∂(x 0 | f 0(5) |2 ) = 0. Observing (4.21), we find (5) that x 0 | f 0 |2 is a holomorphic function on C at most with the pole z = ∞. So it is a polynomial function about z and from (4.16), (4.21) we know that its degree can’t +10 (5) exceed 10. Without loss of generality, we set x 0 | f 0 |2 = h(z) = i=0 αi z i , where αi are complex coefficients, then x0 =

h | f 0(5) |2

.

(4.22)

Substituting (4.22) into the third equation of (3.9), we get x1 = (5)

g (5)

2|F1 |2

,

(4.23)

(5)

where g = | f 0 |2 ∂h − 2h∂| f 0 |2 . (5) Set t = |h|2 + | f 0 |4 . Substituting (4.22) and (4.23) into (4.21), we get (5)

(5)

4|F1 |2 t + |g|2 = 4c1 (1 + zz)13 | f 0 |2 ,

(4.24)

where |F1(5) |2 = | f 0(5) |2 ∂∂| f 0(5) |2 − ∂| f 0(5) |2 ∂| f 0(5) |2 . From (4.11), (4.19), (4.21), (4.22) and (4.23), we get   1 (5) (5) (5) |F2 |2 t 2 = c1 (1 + zz)11 13|F1 |2 t − (1 + zz)2 |F1 |4 − |G|2 , 4

(4.25)

where G = (1 + zz)∂g − 13gz. (5) Observing (4.25), we know |F2 |2 = (1 + zz)d Q, here 0 ≤ d ≤ 9(k ∈ Z) and Q (5) 2 denotes the quotient of |F2 | divided by (1 + zz)d , which is a general polynomial of z, z (in the following, we also use this notation). In the following, we discuss the ten cases respectively and prove x0 = x1 = 0. In fact, we only need to prove the first case, because other cases can be transformed into the first case. Now we give the proof of the first case.

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Classification of conformal minimal immersions

(1) If d = 0, since t 2 must be this form (1 + zz)2n Q (n ∈ Z+ ), then we have t = (1 + zz)6 Q by (4.25). By observing (4.24), we have g = (1 + zz)3 Q. Now, we apply t = (1 + zz)6 Q, g = (1 + zz)3 Q and (4.24) to prove h = 0. Our main idea is to prove αi = 0 (i = 0, . . . , 10) respectively by induction. 2 From (4.12), |V(5)| 2 is invariant as z → 1z . By (4.23), we have | f0 |

(5) | f 0 |2

(5)

    1 (5) 2 1 (5) (5) |F1 | z 13 z 13 = | f 0 |2 (z)|F1 |2 (z). z z

(4.26)

(5)

Because l0 ( 1z ) = l0 (z)z 2 z 2 , we have |F1(5) |2 | f 0(5) |4

% & 1 z

% & =

(5)

|F1 |2 (z)

z2 z2.

(4.27)

  1 5 5 (5) z z = | f 0 |2 (z). z

(4.28)

1 z

(5) | f 0 |4 (z)

By (4.26) and (4.27), we get (5) | f 0 |2

By (4.28), we set (5)

| f 0 |2 =

5 

ai j z i z j ,

(4.29)

i, j=0

where a5−i,5− j = ai j , ai j = a ji with a00 = a55 = 1. Since

t = |h|2 + | f 0(5) |4 =

10 

ti j z i z j ,

(4.30)

i, j=0 (5)

(5)

g = | f 0 |2 ∂h − 2h∂| f 0 |2 =

13  5 

gi j z i z j ,

(4.31)

i=0 j=0

where ti j , gi j are given by + ti j = αi α j + k,l akl ai−k, j−l (i, j = 0, . . . , 10), (4.32) + gi j = k (i + 1 − 3k)αi+1−k ak j (i = 0, . . . , 13, j = 0, . . . , 5), (4.33)

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L. He, X. Jiao

we set 4 

t = (1 + zz)6 λ = (1 + zz)6

λi j z i z j ,

(4.34)

i, j=0

g = (1 + zz)3 μ = (1 + zz)3

10  2 

μi j z i z j ,

(4.35)

i=0 j=0

where λi j , μi j are complex coefficients and λi j = λ ji . Then by comparing the coefficients of (4.30) and (4.34), we get ti j =

  6 λi−k, j−k (i, j = 0, . . . , 10), k k

which implies t50 = · · · = t10,0 = t61 = · · · = t10,1 = · · · = t10,5 = 0,

(4.36)

and for p = 0, 1, 2, 3, 4, we have ⎡ ⎤ A  T T  t p0 , t p+1,1 , . . . , t10,10− p = ⎣ B ⎦ λ p0 , λ p+1,1 , . . . , λ4,4− p , AT

(4.37)

% & % & where A, B are (5 − p) × (5 − p) -, ( p + 1) × (5 − p) -matrix respectively given by ⎤

⎡ 6 0 6 1

⎥ ⎢ 6 ⎥ ⎢ ⎥ ⎢ 0 A=⎢ . ⎥, .. . . ⎥ ⎢ . ⎣ .   .  .  ⎦ 6 6 6 4− p 3− p . . . 0

⎡

6  6  5− p 4− p ⎢    6 ⎢ 6 ⎢ 6− p 5− p

B=⎢ ⎢ ⎣

.. . 6 5

...

6 ⎤ 1

 ⎥ . . . 26 ⎥ ⎥ . .. . . .. ⎥ ⎥ . . ⎦ .  6  6  . . . p+1 4

A straightforward computation shows ⎡ A

−1

(−1)0

5

⎤

0 6 1 (−1) 1

⎢ ⎥  ⎢ ⎥ (−1)0 05 ⎢ ⎥ =⎢ ⎥. .. .. .. ⎢ ⎥ . . . ⎣ ⎦       9− p 3− p 8− p . . . (−1)0 5 (−1) (−1)4− p 4− p 3− p 0

(4.38)

From (4.37) and (4.38), we get  T  T t5,5− p , t6,6− p , . . . , t10,10− p = D p t p0 , t p+1,1 , . . . , t4,4− p ,

123

(4.39)

Classification of conformal minimal immersions

% & where D p is a 6 × (5 − p) -matrix given by  p4− p       3− p 9− p 3− p . . . (−1)0 6 0 ⎤ (−1)4− p 10− 5  0  (−1) 5  0    50 ⎢(−1)4− p 10− p 5− p (−1)3− p 9− p 4− p . . . (−1)0 6 1 ⎥ ⎢ 4 1 4 1 4 1 ⎥ Dp = ⎢ ⎥. .. .. .. .. ⎣ ⎦ . . . . 10− p9− p 9− p8− p 65 4− p 3− p 0 (−1) . . . (−1) 0 5 (−1) 0 0 5 5 ⎡

From (4.39) we have  T  T t10,10− p , t9,9− p , . . . , t6+ p,6 = H p t p0 , t p+1,1 , . . . , t4,4− p ,

(4.40)

% & where H p is a real (5 − p) × (5 − p) -matrix satisfying H p2 = I5− p . Similarly, from (4.31) and (4.35), we have  3  μi−k, j−k (i = 0, . . . , 13, j = 0, . . . , 5), gi j = k k

which implies g11,0 = g12,0 = g13,0 = g12,1 = g13,1 = g13,2 = g03 = g04 = g05 = g14 = g15 = g25 = 0,

(4.41) and g35 = g02 .

(4.42)

In order to achieve our objective, the proof is composed of the following 3 steps. Note that the proof of every step later is built on the bases of all the steps before it. Step 1, we claim (4.43) α0 = 0. Otherwise if α0 = 0, then firstly we conclude a50 = 0.

(4.44)

In fact, if a50 = 0, then by (4.32) and (4.36), we have 2 t10,0 = α10 α 0 + a50 = 0,

so α10 = 0. Through (4.32) and (4.36) again, we have t10,2 = α10 α 2 + 2a 30 a 50 + a 240 = 0, which implies a40 = 0. By the same way we have t10,4 = α10 α 4 + 2a 10 a 50 + 2a 20 a 40 + a 230 = 0, which implies a30 = 0.

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L. He, X. Jiao

Hence for any 5 ≤ s ≤ 9 (s ∈ Z), it follows from ts0 = 0 that αs = 0 by (4.32). For any 0 ≤ ρ ≤ 4 (ρ ∈ Z) we prove αρ = 0 by induction on ρ. When ρ = 4 let p = 4 in (4.39), we have t73 = 6t40 = 6t10,6 . Since α7 = 0, we find t73 = t 73 by (4.32). Then we get t10,6 = t 40 , which means α10 α 6 = α 4 α0 by (4.32), hence α4 = 0 holds. Suppose the conclusion is true for ρ +1, . . . , 4. Consider the case of ρ. Let p = ρ in (4.40), by induction hypotheses and Lemma 4.6 we get t10,10−ρ = t ρ0 , which implies αρ = 0. Obviously it contradicts our supposition α0 = 0 and verifies (4.44). Secondly, we prove 2 . (4.45) |α0 |2 = a50 In fact, by t50 = t10,0 = 0 and (4.32), we can get the expressions of α5 and α10 as follows a2 2a50 + 2a10 a40 + 2a20 a30 α5 = − , α10 = − 50 . (4.46) α0 α0 Then substituting (4.46) into t10,5 = 0, we get by (4.32) % & 2 |α0 |2 + a50 (2a50 + 2a10 a40 + 2a20 a30 ) = 0, which implies that a50 + a10 a40 + a20 a30 = 0, i.e. α5 = 0. Then by (4.32) we find t95 = t 51 . Let p = 4 in (4.39), we have t95 = t51 = 6t40 and t10,6 = t40 . So we get t10,6 = t 40 , which means α10 α 6 = α 4 α0 by (4.32). Similarly by t60 = t10,4 = 0 we can get the expressions of α6 and α4 , thus we have &% & % 4 2 2a10 a50 + 2a20 a40 + a30 = 0, |α0 |4 − a50

2 , then α = α = 0. Next let p = 3, 2, 1 in (4.40) which implies that if |α0 |2 = a50 6 4 respectively, by Lemma 4.6 and simple induction we get t10,7 = t 30 , t10,8 = t 20 and t10,9 = t 10 respectively. Using the same method as the above we have α7 = α3 = 0, α8 = α2 = 0 and α9 = α1 = 0 respectively. At last let p = 0 in (4.40), by Lemma 4.6 we find t10,10 = t 00 , which implies |α10 |2 = |α0 |2 by (4.32). So it follows from (4.46) that &% & % 2 2 |α0 |2 + a50 = 0, (4.47) |α0 |2 − a50

which verifies (4.45). Finally, since t10,1 = · · · = t61 = 0, it follows from (4.32) and simple induction that as0 a 40 as1 = (s = 1, . . . , 4). (4.48) a50

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Classification of conformal minimal immersions

+ (5) Set |F1 |2 = i,8 j=0 bi j z i z j . Then by (4.29), (4.32), (4.33), (4.39), (4.48) and a series of calculations, we get ⎧ a10 ⎪ ⎨ b80 = b70 = b60 = b50 = 0, b40 = 5 (a 40 − a 10 a50 ) , b00 = a50 (a 40 − a 10 a50 ) ; 2 + 1; t40 = − 13 (a 40 − a 10 a50 ) , t00 = a50 ⎪ ⎩ g01 = g02 = 0, g80 = 2a50 (a 40 − a 10 a50 ) , g00 = 2α0 (a40 − a10 a50 ) . α0 a50 (4.49) It follows from (4.24), (4.36), (4.41) and (4.49) that 4b00 t00 + |g00 |2 = 4c1 , 4b40 t40 + g40 g 00 = 0, which implies 1 (a 40 − a 10 a50 ) (a10 + a40 a50 ) = c1 , (a 40 − a 10 a50 )2 = 0. a50 Then we clearly see c1 = 0, which contradicts the fact that c1 is a positive constant and verifies (4.43). Furthermore, by t10,0 = t80 = t60 = 0 and (4.32) we have a50 = a40 = a30 = 0.

(4.50)

α1 = 0.

(4.51)

Step 2, we prove Suppose α1 = 0, then applying g14 = 0 we get from (4.33) a41 = 0.

(4.52)

By t10,1 = · · · = t61 = 0 we get from (4.32) α10 = · · · = α6 = 0. Since g35 = g02 , we have a20 = 0 from (4.33), then t40 = 0 by (4.32). Let p = 4 in (4.39), we have t51 = 0, which implies α5 = 0 by (4.32). Then let p = 3, 2, 1, 0 in (4.40) respectively, by Lemma 4.6 and simple induction we get t96 = t 41 , t97 = t 31 , t98 = t 21 and t99 = t 11 respectively, which implies α4 = α3 = α2 = α1 = 0 by (4.32) respectively. Obviously it contradicts our supposition α1 = 0 and verifies (4.51). Furthermore, we have by simple induction a20 = a10 = 0.

(4.53)

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L. He, X. Jiao

In fact, if a20 = 0, then a straightforward calculation shows ⎧ ⎨ b80 = b70 = b60 = 0, b50 = 2a20 a41 ; 2 ; t = a20 ⎩ 40 g02 = g01 = g00 = 0.

(4.54)

It follows from (4.24), (4.36), (4.41) and (4.54) that 4b50 t40 = 0, which implies b50 = 0. Similarly by induction we obtain b40 = · · · = b00 = 0, which shows |F1(5) |2 is zero at z = 0, what is impossible. So a20 = 0. Then by the same way we get a10 = 0 at once. So it verifies (4.53). At this time, a straightforward calculation shows ⎧ ⎨ b80 = · · · = b40 = 0, bs0 = (s + 1)as+1,1 (s = 3, 2, 1); t40 = · · · = t10 = 0, t00 = 1; (4.55) ⎩ g02 = g01 = g00 = 0. It follows from (4.24), (4.36), (4.41), (4.55) and simple induction that a41 = a31 = a21 = 0.

(4.56)

Step 3, for any 1 ≤ δ ≤ 4 we prove 

αδ = 0, aρδ = 0 (δ + 1 ≤ ρ ≤ 5),

(4.57)

by induction on δ. When δ = 1 we have 

α1 = 0, aρ1 = 0 (2 ≤ ρ ≤ 5),

by step 2. Suppose the conclusion is true for 1, . . . , δ − 1. Consider the case of δ. If αδ = 0, then it follows from t10,δ = · · · = t5+δ,δ = 0 and (4.32) that α10 = · · · = α5+δ = 0. For any 4 − δ + 1 ≤ p ≤ 4, by (4.32), (4.39) and induction hypotheses we have t p+δ,δ = 0. Then it follows from (4.32) that α p+δ = 0 (5 ≤ p + δ ≤ 4 + δ). Now let p = 4 − δ, . . . , 0 in (4.40) respectively, by Lemma 4.6 and simple induction we get t10−δ,6 = t 4,δ , . . . , t10−δ,10−δ = t δ,δ respectively, which implies α4 = · · · = αδ = 0 by (4.32) respectively.

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Classification of conformal minimal immersions

In the following, for any δ + 1 ≤ ρ ≤ 5(ρ ∈ Z) we prove aρδ = 0 by induction on ρ. When ρ = 5 obviously a5δ = a0,5−δ = 0. Suppose the conclusion is true for ρ +1, . . . , 5. Consider the case of ρ. By induction hypotheses we have ⎧ b8ζ = · · · = bζ +1,ζ = 0, bζ,ζ = 0 (0 ≤ ζ ≤ δ − 2), ⎪ ⎪ ⎨b 8,δ−1 = · · · = bρ,δ−1 = 0, bρ−1,δ−1 = ρδaρδ ; +ζ ⎪ t = · · · = tζ +1,ζ = 0, tζ ζ = k=0 akk aζ −k,ζ −k = 0 (0 ≤ ζ ≤ δ − 1); ⎪ ⎩ 4+ζ,ζ gζ 0 = · · · = gζ 5 = 0 (0 ≤ ζ ≤ δ − 1). (4.58) It follows from (4.24), (4.36), (4.41) and (4.58) that 4bρ−1,δ−1 t00 = 0, which implies aρδ = 0. So it verifies (4.57). At last we end this step through proving by induction on δ that for any 5 ≤ δ ≤ 10 αδ = 0.

(4.59)

When δ = 5 if α5 = 0 then it follows from t10,5 = 0 and (4.23) that α10 = 0. For any 1 ≤ p ≤ 4, by (4.32) and (4.39) we have t5+ p,5 = 0, which implies α5+ p = 0. Then let p = 0 in (4.39) we get t00 + t22 + · · · + t10,10 = t11 + t33 + · · · + t99 , which implies α5 = 0 by (4.32). Suppose the conclusion is true for 5, . . . , δ − 1. Consider the case of δ. If αδ = 0, then for any 1 ≤ p ≤ 4 it follows from (4.32) and (4.39) that t p+δ,δ = 0, which implies α p+δ = 0 (δ + 1 ≤ p + δ ≤ 10). Let p = 0 in (4.40). By induction hypotheses and Lemma 4.6 we have tδδ = t 10−δ,10−δ , which verifies (4.59) by (4.32). In summary, we have made h = 0 true by (4.43), (4.57) and (4.59). Then from (4.22) and (4.23), we immediately get x0 = x1 = 0.

(4.60)

Then we transform all other cases into the first case and finish our proof. (5) (2) If d = 1, i.e. |F2 |2 = (1 + zz)Q, then by (4.25) we have t = (1 + zz)5 Q. 2 Since |g| must be this form (1 + zz)2n Q (n ∈ Z+ ), we have g = (1 + zz)3 Q by (4.24), so G = (1 + zz)3 Q. Thus using (4.25) again, we have t = (1 + zz)6 Q. Now it belongs to the case (1). (5) (3) If d = 2, i.e. |F2 |2 = (1 + zz)2 Q, by (4.25) we have t = (1 + zz)5 Q. Then it is the same as the case (2) and belongs to the case (1). (4) If d = 3, i.e. |F2(5) |2 = (1 + zz)3 Q, then by (4.25) we have t = (1 + zz)4 Q. (5) (5) (5) (5) (5) Since f 0 is a full totally J-isotropic map, we obtain | f 0 |2 | f 5 |2 = | f 1 |2 | f 4 |2 =

123

L. He, X. Jiao (5)

(5)

(5)

(5)

| f 2 |2 | f 3 |2 = c2 , where c2 is a positive constant. So |F3 |2 = c2 |F1 |2 . Then we have (5) (5) (5) (5) (5) (4.61) |F2 |2 ∂∂|F2 |2 − ∂|F2 |2 ∂|F2 |2 = c2 |F1 |4 . (5)

From (4.61), we get |F1 |2 = (1 + zz)2 Q. By (4.24) we have g = (1 + zz)3 Q, so G = (1 + zz)3 Q. Thus using (4.25), we have t = (1 + zz)7 Q. Now it belongs to the case (1). (5) (5) If d = 4, i.e. |F2 |2 = (1 + zz)4 Q, then by (4.25) we have t = (1 + zz)4 Q. Now it is similar to the case (4) and belongs to the case (1). (5) (6) If d = 5, i.e. |F2 |2 = (1 + zz)5 Q, then by (4.25) we have t = (1 + zz)3 Q. From (4.61), we get |F1(5) |2 = (1 + zz)4 Q. By(4.24), we have g = (1 + zz)4 Q, so G = (1 + zz)4 Q. Thus using (4.25), we have t = (1 + zz)7 Q. Now it belongs to the case (1). (7) If d = 6, i.e. |F2(5) |2 = (1 + zz)6 Q, then by (4.25) we have t = (1 + zz)3 Q. Now it is similar to the case (6) and belongs to the case (1). (8) If d = 7, i.e. |F2(5) |2 = (1 + zz)7 Q, then by (4.25) we have t = (1 + zz)2 Q. (5) From (4.61), we get |F1 |2 = (1 + zz)6 Q. By (4.24), we have g = (1 + zz)4 Q, so 4 G = (1 + zz) Q. Thus using (4.25), we have t = (1 + zz)6 Q. Now it belongs to the case (1). (5) (9) If d = 8, i.e. |F2 |2 = (1 + zz)8 Q, then by (4.25) we have t = (1 + zz)2 Q. Now it is similar to the case (8) and belongs to the case (1). (5) (5) (10) If d = 9, i.e. |F2 |2 = (1 + zz)9 Q, but |F2 |2 is a polynomial function of degree 9, then it can be easily checked that x0 = x1 = 0. Consequently, in the harmonic sequence (3.6) we have ϕ −1 = f (5) ⊕ f (5) , ϕ 0 = f (5) ⊕ f (5) , ϕ 1 = f (5) ⊕ f (5) , 0 3 1 4 2 5 (5)

where f 0 : S 2 → C P 5 is a full totally J-isotropic map. (5) (5) Then the metric induced by ϕ0 is dsϕ20 = 2(l0 + l1 )dzdz. Because ϕ0 is a 2 , up to a holomorphic isometry of harmonic map with constant curvature K 0 = 13 (5) 5 C P , f 0 is a Veronese surface. Using the proof of Proposition 4.2, we have the conclusion.   Remark 4.8 We choose the same U0 with the one in Remark 4.3 (i), then i 2 ◦ ϕ0 = f 1(5) ⊕ f 4(5) ∈ G(2, 6) √ √ √ √ √ with f 1(5) = [( 5z, − 5z 4 , 4zz − 1, −z 4 z + 4z 3 , 3 2z 2 z − 2 2z, 2 2z 3 z − √ √ √ √ √ 2 T (5) 3 2z ) ] and f 4 = [( 5z 4 , 5z, zz 4 − 4z 3 , 4zz − 1, −2 2zz 3 + 3 2z 2 , √ √ √ √ √ ϕ0 = [( 5z − 5z 4 j, 4zz −1−(zz 4 −4z 3 ) j, 3 2z 2 z − 3√2zz 2 −2√ 2z)T ]. Further √ 2 2z + (2 2zz 3 − 3 2z 2 ) j)T ] ∈ H P 2 . From (3.6), ϕ0 has isotropy order 2. (b.I) In this case, we prove

123

Classification of conformal minimal immersions

Proposition 4.9 Let ϕ0 be an irreducible linearly full unramified strongly isotropic harmonic map of constant curvature; let m be as defined just after (3.13). Then m = 5. Proof Let ϕ0 : S 2 → H P 2 be an irreducible linearly full strongly isotropic harmonic map. We suppose that m = 5. By Proposition 3.4, we choose the local unitary frame (5)

e1 =

(5)

f J f5 JX X V JV , e2 = , e3 = , e4 = 5(5) , e5 = (5) , , e6 = |X | |X | |V | |V | | f5 | | f5 | (5) (5) 1 f − |V1|2 V, V = x2 f 2 (5) | f 4 |2 4 on S 2 except some isolated points.

where X =

(5)

+ x3 f 3

(5)

+ f4

(4.62)

and x2 , x3 are smooth

functions Set W0 = (e1 , e2 ), W1 = (e3 , e4 ) and W−1 = (e5 , e6 ), then by (2.6), we get     λ 0 t0 −μ0 −t0 , −1 = , 0 = 0 μ0 0 λ0 where λ0 = (5)

| f 0 |2 (5)

| f 5 |2

(5)

(5)

x 2 τ0 | f 4 || f 5 |2 (5) |V0 || f 3 |2

(5)

|V0 || f 5 |

, μ0 =

(5) |V || f 4 |

(4.63)

λ 2 = , and t0 = − |X ||V | with |τ0 |

, V0 = x2 f 2(5) + x3 f 3(5) , λ = x3 − ∂ log |V |2 + ∂ log | f 4(5) |2 .

It follows from (4.63) that |det 0 |2 dz 2 dz 2 =

(5)

|x2 |2 | f 2 |2 (5) (5) 2 2 l0 l1 dz dz , |V |2

L 0 = L −1 = λ0 λ0 + μ0 μ0 + t0 t 0 .

(4.64) (4.65)

Let ϕ0 be totally unramified, then |det 0 |2 dz 2 dz 2 = 0 everywhere on S 2 . It follows from (4.64) that li(5) dzdz = 0(i = 0, 1, 3, 4) everywhere on S 2 . By ([4], §3), we get 1 3 (5) (5) (5) (5) (5) δ0 = δ4 = 5 + r (∂2 ), δ1 = δ3 = 8 + r (∂2 ), δ2 = 9 + r (∂2 ). (4.66) 2 2 Then by (2.15) and (4.65), we have δ0 = δ−1 = 4.

(4.67)

If ϕ0 : S 2 → H P 2 ⊂ G(2, 6) is a harmonic map with constant curvature K 0 , then it follows from (4.67) that K 0 = 21 , and we can choose a complex coordinate z on C = S 2 \{ pt} such that the induced metric ds02 = 2L 0 dzdz by ϕ0 is given by ds02 =

8 dzdz, (1 + zz)2

which implies L 0 = λ0 λ0 + μ0 μ0 + t0 t 0 =

4 . (1 + zz)2

(4.68)

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L. He, X. Jiao (5)

Similarly, we choose a nowhere zero holomorphic C6 -valued function f 0 , then   (5) ,(5) , where ρi (z) is the greatest common divisor of the 4 we write Fi = ρi (z) F i i+1 (5)

(5)

components of Fi . Set ρ0 (z) = ρ1 (z) = ρ2 (z) = 1. We obtain that Fi (i = ,(5) , F ,(5) are nowhere zero holomorphic curves and are polynomial functions 0, 1, 2), F 3

4

(5)

(5)

(5)

(5)

on C of degrees δi , δ3 , δ4 respectively. And they satisfy that ∂∂ log |Fi |2 = ,(5) |2 = l (5) . ∂∂ log | F i i By the first equation of (3.18), we know that x 2 is an analytic type function, then x 2 has only isolated zeros or is identical to 0. Now we suppose that x 2 has only isolated zeros Z 2 , then on S 2 \ Z 2 we use (2.14), (4.64), (4.65) and (4.68) to obtain   |V |2 |F0(5) |2 |F1(5) |2 ∂∂ log = 0. (4.69) (5) (5) |x2 |2 | f 2 |2 |F2 |2 (1 + zz)4 By (4.64) we know that

|V |2 (5) |x2 |2 | f 2 |2

Then it follows from (4.66) that

is a globally defined function without zeros on S 2 . (5) (5) |V |2 |F0 |2 |F1 |2 (5) 2 (5) 2 2 |x2 | | f 2 | |F2 | (1+zz)4

is globally defined on C and

has a positive constant limit c1 as z → ∞. Thus from (4.69) (5)

(5)

|V |2 |F0 |2 |F1 |2 (5)

(5)

|x2 |2 | f 2 |2 |F2 |2 (1 + zz)4

= c1 ,

i.e. (5)

|V |2 (5)

|x2 |2 | f 2 |2

=1+

|x3 |2 | f 3 |2 (5)

|x2 |2 | f 2 |2

(5)

+

| f 4 |2

(5)

|x2 |2 | f 2 |2

(5)

=

c1 |F2 |2 (1 + zz)4 (5)

(5)

|F0 |2 |F1 |2

.

(4.70)

Here we assume that x2 , x3 are the rational functions of z, z by ([3], Proposition 5.1). A straightforward calculation shows (5) 2

2 1 + |x3 | | f3(5) |  2 | f (5) |2   |x | 1 |x2 |2 | f 2 |2 2 2 t0 t 0 = ∂ log + ∂ log  . (5) 2 | f 4 |2 |V | x2 2

(4.71)

(5)

|x2 |2 | f 2 |2

By analyzing (4.68), (4.70) and (4.71), we conclude that x2 have no zeros and poles on S 2 . In fact if z i is a zero or pole of x2 , then at z = z i , t0 t 0 → ∞, which contradicts (4.68). But from the first and second equations of (3.18) we find that x2 = 0 at z → ∞, what is impossible. Thus x2 = 0. Through substituting it into the third equation of (3.18), we have x3 = 0. But in this case x2 , x3 = 0, we have a contradiction. So we get the conclusion.   (b.IV) In this case, we prove

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Classification of conformal minimal immersions

Proposition 4.10 Let ϕ0 : S 2 → H P 2 be an irreducible linearly full strongly isotropic harmonic map of constant curvature K 0 . Then up to a symplectic isome(2) (2) try of HP2 , i 2 ◦ ϕ0 = JU V1 ⊕ U V1 with K 0 = 21 for some U ∈ G. Proof The proof is completely similar to that of Proposition 4.7 and even simpler.   Remark 4.11 We choose the same U0 with the one in Remark 4.3 (i), then (2)

i 2 ◦ ϕ0 = J f 1

(2)

⊕ f1

∈ G(2, 6)

√ √ √ (2) (2) with f 1 = [(−2z, 0, 2 − 2zz, 0, 2z, 0)T ] and J f 1 = [(0, 2z, 0, − 2 + √ √ √ 2zz, 0, −2z)T ]. Further ϕ0 = [(−2z, 2 − 2zz, 2z)T ] is the Veronese surface in C P 2 . From (3.12), ϕ0 is strongly isotropic. By Proposition 4.2, 4.4, 4.7, 4.9 and 4.10, we obtain a classification of conformal minimal immersions of constant curvature from S 2 to HP2 as follows: Theorem 4.12 Let ϕ : S 2 → H P 2 be a linearly full conformal minimal immersion of constant curvature. Then, (i) If ϕ is reducible, then, up to a symplectic isometry of HP2 , i 2 ◦ ϕ is U V0(5) ⊕ (5) (5) (5) U V5 with constant curvature 25 , or U V2 ⊕ U V3 with constant curvature 41 , (2) (2) or JU V0 ⊕ U V0 with constant curvature 1, for some U ∈ G; (ii) If ϕ is totally unramified irreducible, then, up to a symplectic isometry of HP2 , i 2 ◦ϕ (5) (5) (2) (2) 2 , or JU V1 ⊕ U V1 with is either U V1 ⊕ U V4 with constant curvature 13 1 constant curvature 2 , for some U ∈ G. Theorem 4.12 shows that all totally unramified conformal minimal immersions of constant curvature from S 2 to HP2 are presented by the Veronese surfaces in C P 2 and C P 5 , up to a symplectic isometry of HP2 . It is easy to check that these minimal immersions are all homogeneous (i.e. SU (2)-equivariant) (cf. [10]). Of course there is a problem deserving further consideration. When ϕ is irreducible, our classification theorem is true under the condition of totally unramified. Then whether it is also true after omitting this condition. Acknowledgments

The authors would like to express gratitude for the referee’s comments.

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