Math. Z. DOI 10.1007/s00209-015-1452-5
Mathematische Zeitschrift
On conformal minimal immersions of constant curvature from S2 to HP n Ling He · Xiaoxiang Jiao
Received: 30 July 2014 / Accepted: 30 December 2014 © Springer-Verlag Berlin Heidelberg 2015
Abstract We generalize here our general characterization of the harmonic sequence generated by conformal minimal immersions from S 2 to the quaternionic projective space HPn . Then we give a classification theorem of linearly full unramified conformal minimal immersions of constant curvature from S 2 to HPn under some conditions. Keywords Conformal minimal immersion · Gauss curvature · Veronese curve · Classification · Quaternionic projective space Mathematics Subject Classification
Primary 53C42 · 53C55
1 Introduction Recently, we have obtained some characterization of the harmonic sequence generated by conformal minimal immersions from S 2 to the quaternionic projective plane HP2 and given a classification theorem of linearly full unramified conformal minimal immersions of constant curvature from S 2 to HP2 (cf. [6]). In this paper, we go further and study the case of the general quaternionic projective space HPn . Section 2 gives the similar preliminaries as that in [6]. Section 3 presents some
Project supported by the NSFC (Grant Nos. 11331002, 11471308). L. He (B) · X. Jiao School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 101408, People’s Republic of China e-mail:
[email protected] X. Jiao e-mail:
[email protected] Present address L. He Institute of Mathematics, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, People’s Republic of China
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geometric properties of reducible harmonic maps of constant curvature from S 2 to HPn . In Sect. 4, we use Bahy-El-Dien and Wood’s unique factorization theorem (cf. [1]) to study some properties of the harmonic sequence generated by an irreducible harmonic map from S 2 to HPn of finite isotropy order r = 2n − 2, and obtain the explicit characteristics of the corresponding harmonic map in HPn . Further we discuss geometric properties of conformal minimal immersions of constant curvature from S 2 to HPn and obtain a classification theorem of linearly full totally unramified conformal minimal immersions of S 2 in HPn with constant curvature (see Theorem 4.4).
2 Preliminaries 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+2 with Hn+1 . 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+2 → C2n+2 be the conjugate linear map given by left multiplication by j, i.e. J(z 1 , z 2 , . . . , z 2n+1 , z 2n+2 )T = (−¯z 2 , z¯ 1 , . . . , −¯z 2n+2 , z¯ 2n+1 )T . Then J2 = −id where id denotes the identity map on C2n+2 . In fact, for any v ∈ C2n+2 ,
where Jn+1
¯ Jv = Jn+1 v, 0 −1 0 −1 = diag ,..., . 1 0 1 0 n+1
Let G(2, 2n + 2) denote the Grassmann manifold of all complex 2-dimensional subspaces of C2n+2 with its standard Kähler structure. The quaternionic projective spaces HPn is the set of all one-dimensional quaternionic subspaces of Hn+1 . Throughout the above we shall regard HPn as the totally geodesic submanifold of G(2, 2n + 2) given by HPn = {V ∈ G(2, 2n + 2) : JV = V } . Let Sp(n + 1) = {g ∈ G L(n + 1; H), g ∗ g = In+1 } be the symplectic isometry group of HPn . The explicit description is that the following diagram commutes: i1
Sp(n + 1) −−−−→ U (2n + 2) ⏐ ⏐ ⏐ ⏐ π1 π2 HPn
i2
−−−−→ G(2, 2n + 2)
where i 1 , i 2 are inclusions and π1 , π2 are projections, and i 1 (g) = E, for 1 ≤ a, b ≤ n + 1 a 2a−1 2a−1 E 2b−1 = Aab , E 2b = −B b , a 2a 2a = A , E 2b−1 = Bba , E 2b b
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Conformal minimal immersions
where A = (Aab ), B = (Bba ) ∈ Mn+1 (C), g = A + B j ∈ Sp(n + 1); π1 (g) = g · [1, 0, . . . , 0]T ∈ HPn ;
T 1, 0, 0, . . . , 0 π2 (E) = E · ∈ G(2, 2n + 2); 0, 1, 0, . . . , 0 T z , z , . . . , z 2n+1 , z 2n+2 . i 2 [z 1 + z 2 j, . . . , z 2n+1 + z 2n+2 j]T = 1 2 −z 2 , z 1 , . . . , − z 2n+2 , z 2n+1 Here we suppose that the metric on G(2, 2n + 2) is given by section 2 of [8], then the metric induced by i 2 is twice as much as the standard metric on HPn . Thus we regard the harmonic map from S 2 to HPn as the one from S 2 to G(2, 2n + 2). For any g ∈ Sp(n + 1), the action of g on HPn induces an action of E on CP2n+1 , where J. Then E ∈ U (2n + 2) commutes with Sp(n + 1) = {E ∈ U (2n + 2), E ◦ J = J ◦ E} = E ∈ U (2n + 2), E Jn+1 E T = Jn+1 . In the following, we deal with the symplectic isometry of HPn through the corresponding symplectic isometry of CP2n+1 . Next, we simply introduce harmonic maps and harmonic sequences in G(k, N ) (cf. [3,4]) 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. [3]) are derived as follows: A ϕ
A ϕ
A ϕ
α−1
A ϕα
ϕ = ϕ 0 −→ ϕ 1 −→ · · · −→ ϕ α −→ · · · , 0
A ϕ 0
1
A ϕ −1
A ϕ −α+1
A ϕ−α
ϕ = ϕ 0 −→ ϕ −1 −→ · · · −→ ϕ −α −→ · · · ,
(2.1) (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 subbundles in (2.1) are not mutually orthogonal generally. So it is meaningful to define the isotropy order of ϕ (cf. [3], §3A) to be the greatest positive 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.
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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
A ϕα
α−1
A ϕ
A ϕ
A ϕα
α0 −1
A ϕβ
β0 −1
ϕ 0 −→ ϕ 1 −→ · · · −→ ϕ = ϕ α −→ · · · −→ ϕ α −→ · · · −→ ϕ β −→ 0, 0
1
0
0
0
0
(2.3)
where ϕ 0 , . . . , ϕ α are mutually orthogonal and k0 + · · · + kα0 = N (α0 ≤ β0 ) with kα = 0 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 = 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 ϕα = Wα Wα∗ ,
Wα∗ Wα
= Ikα ×kα ,
(2.4) Wα∗ Wα+1
= 0,
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 → CP N be a linearly full harmonic map. Eells and Wood’s result (cf. [5]) shows that the following sequence in CP 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 of
ψ0 (N ) ) , i.e. ∂ f (n) Let f 0 be a holomorphic section of ψ (N 0 0 ψ i(N ) such that (N )
fi
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(N )⊥
(N )
= ψi−1 (∂ f i−1 )
(N )
ψj
(2.9)
respectively (j = 0, . . . , N). (N )
= 0, and let f i
be a local section
Conformal minimal immersions
for i = 1, . . . , N . Then we have some formulas as follows (cf. [2]): (N )
= f i+1 + ∂ log | f i
(N )
= −li−1 f i−1 , i = 1, . . . , N ,
∂ fi ∂ fi ∂∂
(N )
(N ) log | f i |2 (N ) ∂∂ log li (N )
(N )
(N ) 2
(N )
| fi
, i = 0, . . . , N − 1,
(2.10)
(N ) (N )
(N ) li (N ) li+1
= =
(N ) − li−1 , (N ) − 2li
(2.11) (2.12)
(N ) + li−1 ,
(N )
i = 0, . . . , N − 1, (N )
(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 2.1 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 [2]. Now recall ([3], §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. [7]) ∂∂ 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. [7]) δα−1 − 2δα + δα+1 = −2kα ,
(2.15)
where δα = 2π √1 −1 S 2 L α dz ∧ dz. At last, we review the rigidity theorem of conformal minimal immersions with constant curvature from S 2 to CP N . T (N ) The Veronese sequence. Let f i = f i,0 , . . . , f i,N for each i = 0, . . . , N . Let f i, p be given for i, p = 0, 1, . . . , N as follows N p−i p N−p i! k z (zz)k . (−1) (2.16) f i, p = p i −k k (1 + zz)i k
(N )
(N )
Such a map ϕi = [ f i ] : S 2 → CP N is a conformal minimal immersion with constant (N ) 4 curvature N +2i(N given by −i) and constant Kähler angle θi tan (N )
(N )
θi
2
2 =
i(N − i + 1) . (i + 1)(N − i)
(N )
Such a harmonic sequence ϕ0 , . . . , ϕ N : S 2 → CP N is called the Veronese sequence. (N ) (N ) We always denote it by V0 , . . . , VN : S 2 → CP N .
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Bolton et al proved that (cf. [2]) if ψ is a linearly full conformal minimal 2-sphere of constant curvature immersed in CP N , then, up to a holomorphic isometry of CP N , ψ is an element of the Veronese sequence (i.e. a Veronese surface). 3 Reducible harmonic maps of constant curvature from S2 to HP n Let ϕ : S 2 → HPn be a linearly full harmonic map of isotropy order r . If ϕ has finite isotropy order, then r = 2s for 1 ≤ s ≤ n by ([1], Proposition 3.2 and Lemma 3.10); if ϕ is strongly isotropic, then r = ∞. If ϕ0 : S 2 → HPn is a reducible linearly full harmonic map, then by ([1], Proposition 3.7) we know that ϕ0 is a quaternionic mixed pair or a quaternionic Frenet pair. In the following we discuss the latter with ϕ0 of constant curvature, and the former with ϕ0 of constant curvature of isotropy order r = 2n − 2 or r = ∞. If ϕ0 is a linearly full quaternionic Frenet pair, then ⊕ f (2n+1) , ϕ 0 = f (2n+1) n n+1
(3.1)
where f (2n+1) , . . . , f (2n+1) : S 2 → CP2n+1 is a linearly full harmonic sequence with the 0 2n+1 . totally J-isotropic map f (2n+1) 0 Obviously ϕ0 belongs to the following harmonic sequence (cf. [3]) A 0
A 0
A n−1
A 1
ϕ
A 0 ϕ
A n+2
A 2n
A 2n+1
←− · · · ←− f (2n+1) ←− ϕ 0 −→ f (2n+1) −→ · · · −→ f (2n+1) −→ 0. 0 ←− f (2n+1) 0 n−1 n+2 2n+1 (3.2) Then we prove Proposition 3.1 Let ϕ0 : S 2 → HPn be a linearly full quaternionic Frenet pair of constant (2n+1) 2 curvature K 0 . Then K 0 = n(n+2) , and up to a symplectic isometry of HPn , ϕ 0 = U V n ⊕ (2n+1)
for some U ∈ G 2n+2 {U ∈ U (2n + 2), U Wn+1 U T = Jn+1 }, where Wn+1 = 0 −1 0 −1 . antidiag ,..., 1 0 1 0
U V n+1
n+1 2 and up to a holomorphic isometry of Proof By ([6], Lemma 4.1) we get that K 0 = n(n+2) is a Veronese surface. Then we can choose a complex coordinate z on C = CP2n+1 , f (2n+1) 0 (2n+1)
(2n+1)
(2n+1)
= U V0 , where U ∈ U (2n + 2) and V0 has the standard S 2 \ { pt} so that f 0 expression given in (2.16). So the condition of totally J-isotropic J f (2n+1) = f (2n+1) is 0 2n+1 equivalent to (2n+1) (2n+1) Jn+1 U V 0 = λU V2n+1 , (3.3) where λ is a parameter.
Set Wn+1 = antidiag
0 −1 0 −1 ,..., . From (2.16), we get 1 0 1 0 n+1
(2n+1) V0
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= 1,
T 2n + 1 2n + 1 n 2n + 1 2n 2n+1 , z, . . . , z ,..., z ,z 1 n 2n
Conformal minimal immersions
and (2n+1) V2n+1
(2n + 1)! = (1 + zz)2n+1 (2n+1)
which implies V2n+1
=
−z
2n+1
, . . . , (−1)
k
(2n+1) (2n+1)! W V . (1+zz)2n+1 n+1 0 (2n+1)
Jn+1 U V 0
T 2n + 1 k z ,...,1 , k
Then the condition (3.3) becomes (2n+1)
= U Wn+1 V 0
.
By differentiating with respect to z¯ in the above formula, the matrices Jn+1 U and U Wn+1 (2n+1) (2n+1) have the same effect on all derivatives of V 0 . Since V 0 is full, these span C2n+2 so that Jn+1 U = U Wn+1 , i.e. U Wn+1 U T = Jn+1 . (3.4) Define a set
G 2n+2 U ∈ U (2n + 2), U Wn+1 U T = Jn+1 .
Since Sp(n + 1) = checked
A ∈ U (2n + 2), A Jn+1 A T = Jn+1 , the following can be easily
(i) ∀ A ∈ Sp(n + 1), U ∈ G 2n+2 , we have that AU ∈ G 2n+2 ; (ii) ∀ U, V ∈ G 2n+2 , ∃ A = U V ∗ ∈ Sp(n + 1) s.t. U = AV .
So we get the conclusion. If ϕ0 is a linearly full quaternionic mixed pair, then ⊕ J f (m) , ϕ 0 = f (m) 0 0
(3.5)
: S 2 → CPm ⊆ CP2n+1 (n ≤ m ≤ 2n + 1) is holomorphic and f (m) ⊥ J f (m) . where f (m) 0 1 0 Obviously ϕ0 belongs to the following harmonic sequence A
A m
A 0
A
ϕ
A 0 ϕ
A 1
A m−1
A m
0 ←− J f (m) ←− · · · ←− J f (m) ←− ϕ 0 −→ f (m) −→ · · · −→ f (m) −→ 0. m 1 1 m m−1
1
(3.6)
So the induced metric by ϕ0 is given by (m)
ds02 = 2l0 dzdz,
(3.7)
(m) l0 dzdz
is the induced metric by the map f (m) . where 0 Then we prove Proposition 3.2 Let ϕ0 : S 2 → HPn be a linearly full quaternionic mixed pair of constant curvature K 0 . If its isotropy order is finite and r = 2n − 2, then up to a symplectic isometry (2n+1) (2n+1) 2 of HPn , ϕ 0 = U V 0 ⊕ U V 2n+1 with K 0 = 2n+1 for some U ∈ G 2n+2 , in fact in this case r = 2n. If it is strongly isotropic, then up to a symplectic isometry of HPn , ϕ 0 = (n)
(n)
JU V 0 ⊕ U V 0 with K 0 =
2 n
for some U ∈ G 2n+2 .
Proof By (3.7) we get that K 0 = m2 and up to a holomorphic isometry of CP2n+1 , f (m) 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 (2n + 2) and V0 has the standard expression given in (2.16) (m) (m) 2n+2 ). (adding zeros to V0 such that V0 ∈ C
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If the isotropy order of ϕ0 is r = 2n − 2, then it follows from the harmonic sequence (3.6) that 2n − 1 ≤ m ≤ 2n + 1. Thus f (m) ⊥ J f (m) holds if and only if the following equation 2n−2 0 (m) (m) T T U Jn+1 U = 0 (3.8) tr V2n−2 V0 holds. Set U T Jn+1 U = W , then we immediately get (m) (m) T W = 0(2n − 1 ≤ m ≤ 2n + 1), W T = −W, W ∗ W = I, tr V2n−2 V0
(3.9)
where I is the identity matrix. Define a set G W U ∈ U (2n + 2), U W U T = Jn+1 . For a given W , the following can be easily checked (i) ∀ A ∈ Sp(n + 1), U ∈ G W , we have that AU ∈ G W ; (ii) ∀ U, V ∈ G W , ∃ A = U V ∗ ∈ Sp(n + 1) s.t. U = AV . In the following we discuss W in the three cases m = 2n + 1, 2n, 2n − 1. (i) m = 2n + 1. K 0 = (2n+1)
(2n+1) T
2 2n+1 .
(2n+1)
By the standard expressions of V0
(2n+1)
and V2n−2 , we get
V2n−2 V0 is a polynomial matrix about z and z. But W is a constant matrix. Using the method of undeterminated coefficients by (3.9), we get that W = Wn+1 and G W = G 2n+2 . Then the proof of Proposition 3.1 gives (2n+1)
ϕ0 = U V 0
(2n+1)
⊕ U V 2n+1 ,
where U ∈ G 2n+2 . Obviously, in this case r = 2n. (ii) m = 2n. K 0 = n1 . Using (3.9), we find that there don’t exist such W . 2 (iii) m = 2n − 1. K 0 = 2n−1 . Using (3.9), we get
Wn √ 0 W = , 0 e −1θ W1 where θ is a real parameter. (2n−1)
In this case, for any U ∈ G W , we have JU V0
(2n−1)
ϕ0 = U V 0
(2n−1)
= −U V2n−1 , then (2n−1)
⊕ U V 2n−1 .
Obviously ϕ0 has image in HPn−1 , so it is not linearly full. If ϕ0 is strongly isotropic, then f (m) ⊥ J f (m) holds if and only if the following equation m 0 (m) T T tr Vm(m) V0 U Jn+1 U = 0 (3.10) holds. Similarly set U T Jn+1 U = W , then we immediately get (m) T W = 0(n ≤ m ≤ 2n − 2), W T = −W, W ∗ W = I, tr Vm(m) V0 where I is the identity matrix.
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(3.11)
Conformal minimal immersions
Using (3.11), we get
0 −B T , W = B 0
where B ∈ U (2n + 1 − m). In fact, from W ∗ W = I we know m = n, so K 0 = n2 . In this case, if we fix some B ∈ U (n + 1), then (n)
(n)
ϕ 0 = JU V 0 ⊕ U V 0 , where U ∈ G W . Now we claim ⎡ that for all B ∈ U⎤(n +1), the corresponding ϕ0 are symplecticly equivalent. b11 · · · b1,n+1 ⎥ ⎢ .. .. .. Set B = ⎣ . ⎦ ∈ U (n + 1), the corresponding . . bn+1,1 · · · bn+1,n+1 ⎤ ⎡ 1 0 ··· 0 0 ··· 0 ⎢0 0 · · · 0 b11 · · · bn+1,1 ⎥ ⎥ ⎢ ⎥ ⎢0 1 · · · 0 0 · · · 0 ⎥ ⎢ ⎢0 0 · · · 0 b12 · · · bn+1,2 ⎥ U =⎢ ⎥ ∈ GW . ⎥ ⎢. . . . .. .. .. ⎥ ⎢ .. .. . . .. . . . ⎥ ⎢ ⎦ ⎣0 0 · · · 1 0 · · · 0 0 0 · · · 0 b1,n+1 · · · bn+1,n+1 Then we have (n)
U V0
(n)
JU V0
= [(1, 0,
√
2z, 0, . . . , z n , 0)T ], √ = [(0, 1, 0, 2z, . . . , 0, z n )T ].
Thus, up to a symplectic isometry, ϕ0 is unique. For convenience, we choose ⎡ ⎤ 0 0 ··· 0 (−1)n ⎢0 0 · · · (−1)n−1 0 ⎥ ⎢ ⎥ ⎢ .. .. ⎥ , B = ⎢ ... ... . . . . . ⎥ ⎢ ⎥ ⎣0 −1 · · · 0 0 ⎦ 1 0 ··· 0 0 then G W = G 2n+2 . In summary we get the conclusion.
4 Irreducible harmonic maps of constant curvature from S2 to HP n Now, we consider the irreducible harmonic maps ϕ : S 2 → HPn of isotropy order r . In the following we consider the case r = 2n − 2. Here we use ([1], Theorem 4.7) to obtain the following lemma: Lemma 4.1 ([1]) Any non-isotropic irreducible harmonic map ϕ : S 2 → HPn of isotropy order 2n − 2 is of the form ϕ 2 where: (i) ϕ 0 = f (2n+1) ⊕ f (2n+1) , f (2n+1) : S 2 → CP2n+1 being a linearly full totally J0 2n+1 0 isotropic map,
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(ii) ϕ 1 is obtained from ϕ 0 by forward replacement of some holomorphic subbundle of ϕ 0 not equal to f (2n+1) , 2n+1
(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. Let ϕ0 : S 2 → HPn be an irreducible linearly full harmonic map of isotropy order 2n − 2. In the following we use Lemma 4.1 to characterize ϕ0 explicitly. In (i) of Lemma 4.1 ϕ 0 with isotropy order 2n belongs to the harmonic sequence as follows: A 0
A 0
A 2n
A 1
ϕ
A 0
A 2n+1
A 2n
A 1
ϕ
0 ←− f (2n+1) ←− · · · ←− f (2n+1) ←− ϕ 0 −→ f (2n+1) −→ · · · −→ f (2n+1) −→ 0, 0 2n 1 2n+1 (4.1) (2n+1) with a linearly full totally J-isotropic map f (2n+1) : S 2 → where ϕ 0 = f (2n+1) ⊕ f 0 2n+1 0
CP2n+1 . (2n+1) (2n+1) By (ii) of Lemma 4.1 and (4.1), there exists a local section V0 = x0 f 0 + J f0 0 1 0 such that V 0 is an antiholomorphic subbundle of ϕ , and ϕ is obtained from ϕ by forward replacement of V ⊥ 0 , i.e. , (4.2) ϕ 1 = V 0 ⊕ f (2n+1) 1 where x0 is a smooth function on S 2 expect some isolated points, and here V 0 denotes a line subbundle consists of V0 . Then ϕ 1 with isotropy order 2n − 1 belongs to the harmonic sequence as follows: A 0
f (2n+1) 0
0 ←−
A 2n−1
A 1
←− · · · ←−
f (2n+1) 2n−1
A 1
ϕ−1
←−
ϕ 1−1
A 1 ϕ0
←−
ϕ 10
A 1
A 2
ϕ0
−→ f (2n+1) −→ · · · 2
A 2n+1
A 2n
−→ f (2n+1) −→ 0, 2n+1
(4.3)
where ϕ 10 = ϕ 1 and ϕ 1−1 = Jϕ 1 .
(5)
By (iii) of Lemma 4.1 and (4.3), there exists a local section V = x1 f 1 + V0 such that = 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 , (4.4)
β ⊥ ∩ϕ 1
where X =
1
(2n+1) 2 |
| f1
(2n+1)
f1
− |Vx 1|2 V and x1 is a smooth function on S 2 expect some isolated
points. (2n+1) (2n+1) Obviously, X, V, f 2 , JX, JV, J f 2 are mutually orthogonal. Then ϕ0 with isotropy order 2n − 2 belongs to the harmonic sequence as follows: A 0
A 2n−2
A ϕ
A ϕ
−1
A ϕ
A ϕ
A 3
A 2n+1
0 ←− · · · ←− f (2n+1) ←− ϕ −1 ←− ϕ 0 −→ ϕ 1 −→ f (2n+1) −→ · · · −→ 0, 2n−2 3 0
0
1
(4.5)
where ϕ 0 = ϕ 2 , ϕ 1 = V ⊕ f (2n+1) and ϕ −1 = Jϕ 1 . 2 Since V 0 and V are antiholomorphic subbundles of ϕ 0 and ϕ 1 respectively, and A ϕ1 (ϕ 1 ) = ϕ 0 , we get πϕ 0 (∂ V0 ) ∈ V 0 , πϕ 1 (∂ V ) ∈ V ;
(4.6)
πϕ ⊥ (∂ V ) ∈ ϕ 0 .
(4.7)
1
123
Conformal minimal immersions
Through a direct computation, the conditions (4.6) and (4.7) are equivalent to the equations ⎧ (2n+1) 2 ⎪ | = 0, ⎨∂ x0 + x0 ∂ log | f 0 (2n+1) 2 (4.8) ∂ x1 + x0 + x1 ∂ log | f 1 | = 0, ⎪ ⎩ (2n+1) (2n+1) 2 ∂ x0 − 2x1 l0 − x0 ∂ log | f 0 | =0 hold. Thus we have given the property of the harmonic sequence (4.5). Then we have Proposition 4.2 The map ϕ0 : S 2 → HPn is an irreducible linearly full harmonic map of (2n+1) 1 finite isotropy order 2n − 2 if and only if ϕ 0 = X ⊕ JX , where X = (2n+1) f − |Vx 1|2 V 2 1 (2n+1)
(2n+1)
| f1
(2n+1)
|
+ x0 f 0 + J f0 (where f (2n+1) is a linearly full totally J with V = x1 f 1 0 isotropic map), and the corresponding coefficients x 0 , x1 satisfy the Eq. (4.8). Proof Through the above construction, the necessity is obvious. Conversely, Eq. (4.8) yield that 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. Then, we consider the case of constant curvature and prove Proposition 4.3 Let ϕ0 : S 2 → HPn be an irreducible linearly full totally unramified harmonic map of finite isotropy order 2n − 2 with constant curvature K 0 . Then up to a (2n+1) (2n+1) 2 symplectic isometry of HPn , ϕ 0 = U V 1 ⊕ U V 2n with K 0 = 6n+1 for some U ∈ G 2n+2 . Proof Let ϕ0 : S 2 → HPn be an irreducible linearly full harmonic map of finite isotropy order 2n − 2. By Proposition 4.2, we choose the local unitary frame ⎧ (2n+1) (2n+1) f β−2 ⎪ f2 JX X V ⎪ ⎨e1 = |X | , e2 = |X | , e3 = |V | , e4 = | f (2n+1) | , eβ = | f (2n+1) | , β = 5, . . . , 2n, 2 β−2 (4.9) (2n+1) ⎪ J f2 JV ⎪ , e2n+2 = |V , ⎩e2n+1 = (2n+1) | | f2
where X =
1
|
(2n+1) 2 |
| f1
(2n+1)
f1
−
x1 V, V |V |2
(2n+1)
= x1 f 1
(2n+1)
+ x0 f 0
(2n+1)
+ J f0
and x0 , x1
are smooth functions on S 2 except some isolated points. Set W0 = (e1 , e2 ), W1 = (e3 , e4 ), Wα = eα+3 (α = 2, . . . , 2n − 3) and W−1 = (e2n+1 , e2n+2 ), then by (2.6), we get ⎧ ' (
(2n+1) (2n+1) ⎪ λ0 t0 |f | | f3 | ⎪ ⎪ α = 2, . . . , 2n − 3, , 1 = 0 (2n+1) , α = α+2 ⎪0 = (2n+1) , ⎨ | f2 | | f α+1 | 0 μ0 ' ( (4.10) ⎪ −μ0 −t0 ⎪ ⎪ , ⎪ ⎩−1 = 0 λ 0
where λ0 = − J
(2n+1) . f0
(2n+1)
| f1 | |V0 | , μ0
=
(2n+1)
|V0 || f 2
|
(2n+1) |V || f 1 |
(2n+1)
| x1 , and t0 = − |V |X | ∂ |V |2 with V0 = x 0 f 0
+
123
L. He, X. Jiao
A straightforward computation shows (2n+1)
|det0 |2 dz 2 dz 2 =
|2 (2n+1) (2n+1) 2 2 | f0 l0 l1 dz dz , 2 |V | (2n+1)
det (α ∗α )dzdz = lα+1
dzdz, α = 1, 2, . . . , 2n − 3,
L 0 = L −1 = λ0 λ0 + μ0 μ0 + t0 t 0 , Lα =
(2n+1) lα+1 ,
α = 1, 2, . . . , 2n − 3.
(4.11) (4.12) (4.13) (4.14)
We assume that ϕ0 is totally unramified, then |det0 |2 dz 2 dz 2 = 0 and det (α ∗α ) (2n+1) dzdz = dzdz = 0 everywhere on S 2 . It follows from (4.11) and (4.12) that li 2 0(i = 0, . . . , 2n) everywhere on S . Then we obtain that the harmonic sequence f (2n+1) , . . . , f (2n+1) : S 2 → CP2n+1 is totally unramified, from ([2], §3) we get 0 2n+1 (2n+1)
δi
= (i + 1)(2n + 1 − i), i = 0, . . . , 2n.
(4.15)
By (2.15), we have where δα =
δ0 = δ−1 and δ1 =
δ1 − 2δ0 + δ−1 = −4,
S 2 L α dz ∧ (2n+1) δ2 = 6n
1 √ 2π −1
(4.16)
dz(α = −1, 0, 1). It follows from (4.13) and (4.14) that − 3, so that δ0 = 6n + 1.
(4.17)
→ ⊂ G(2, 2n + 2) is a harmonic map with constant We assume that ϕ0 : 2 curvature K 0 , then it follows from (4.17) that K 0 = 6n+1 , and we can choose a complex 2 coordinate z on C = S \ { pt} so that the induced metric ds02 = 2L 0 dzdz by ϕ0 is given by S2
HPn
ds02 =
12n + 2 dzdz, (1 + zz)2
which implies L 0 = λ0 λ0 + μ0 μ0 + t0 t 0 =
(4.18)
6n + 1 . (1 + zz)2
(4.19)
(2n+1) (2n+1) = f0 ∧ ... ∧ Consider a local lift of the ith osculating curve (cf. [2]) Fi (2n+1) 2n+2 fi (i = 0, . . . , 2n + 1). We choose a nowhere zero holomorphic C -valued function (2n+1) (2n+1) f0 , then Fi is a nowhere zero holomorphic curve and is a polynomial function on (2n+1) (2n+1) 2 (2n+1) C of degree δi satisfying ∂∂ log |Fi | = li . So using (4.11), (4.13), (4.14)
(4.19), and (2.14), we obtain
∂∂ log By (4.11) we know that
|V |2
(2n+1) 2 (2n+1) 2 | |F1 | (2n+1) 2 6n+1 | f0 | (1 + zz)
|V |2 |F0
= 0.
(4.20)
is a globally defined function without zeros on S 2 . Then
(2n+1) 2 | f0 | (2n+1) 2 (2n+1) 2 |V |2 |F0 | |F1 | (2n+1) 2 | f0 | (1+zz)6n+1
it follows from (4.15) that
is globally defined on C and has a positive
constant limit c1 as z → ∞. Thus from (4.20)
(2n+1) 2 (2n+1) 2 | |F1 | (2n+1) 2 6n+1 | f0 | (1 + zz)
|V |2 |F0
123
= c1 ,
(4.21)
Conformal minimal immersions
i.e.
|V |2
(2n+1)
= |x0 |2 + 1 + |x1 |2 l0
(2n+1) 2 |
| f0
=
c1 (1 + zz)6n+1 (2n+1) 2 (2n+1) 2 | |F1 |
|F0
.
(4.22)
(2n+1)
In view of the first equation of (4.8), we have ∂(x 0 | f 0 |2 ) = 0. Observing (4.22), we (2n+1) 2 | is analytic on C at most with the pole z = ∞. So it is a polynomial find that x 0 | f 0 function about z and from (4.15), (4.22) we know that its degree can’t exceed 4n +2. Without loss of generality, set (2n+1) 2
x 0 | f0
| = h(z) =
4n+2
αi z i ,
i=0
where αi are complex coefficients, then x0 =
h (2n+1) 2 |
| f0
.
(4.23)
Substitution of (4.23) in the third equation of (4.8) yields x1 = (2n+1)
g (2n+1) 2 2|F1 |
,
(4.24)
(2n+1)
where g = | f 0 |2 ∂h − 2h∂| f 0 |2 . (2n+1) 4 2 | . Substituting (4.23) and (4.24) into (4.22), we get Set t = |h| + | f 0 (2n+1) 2
4|F1 (2n+1)
(2n+1)
(2n+1) 2
| t + |g|2 = 4c1 (1 + zz)6n+1 | f 0 (2n+1)
(2n+1)
| ,
(4.25)
(2n+1)
|2 = | f 0 |2 ∂∂| f 0 |2 − ∂| f 0 |2 ∂| f 0 |2 . where |F1 From (4.10), (4.19), (4.21), (4.23) and (4.24), we get
1 (2n+1) 2 2 (2n+1) 2 (2n+1) 4 | t = c1 (1 + zz)6n−1 (6n + 1)|F1 | t − (1 + zz)2 |F1 | − |G|2 , |F2 4 (4.26) where G = (1 + zz)∂g − (6n + 1)gz. (2n+1) 2 | = (1 + zz)d Q, here 0 ≤ d ≤ 6n − 3(d ∈ Z) and Observing (4.26), we know |F2 (2n+1) 2 | divided by (1 + zz)d , which is a general polynomial Q denotes the quotient of |F2 about z, z (in the following, we also use this notation). In the following, we discuss the three cases of d = 2n − 3, 0 ≤ d ≤ 2n − 4 and 2n − 2 ≤ d ≤ 6n − 3 respectively to prove x0 = x1 = 0. In fact, we only need to prove the first case, because other two cases can be transformed into the first case. Now we give the proof of the first case. (2n+1) 2 (1) If d = 2n − 3, i.e. |F2 | = (1 + zz)2n−3 Q. Then by (4.26), we have t = 2n+1 2 (1 + zz) Q. Since |g| must be this form (1 + zz)2k Q (k ∈ Z+ ), by observing (4.25), we have g = (1 + zz)n+1 Q, so G = (1 + zz)n+1 Q. Thus using (4.26) again, we have t = (1 + zz)2n+2 Q. Now, we apply t = (1 + zz)2n+2 Q, g = (1 + zz)n+1 Q and (4.25) to prove h = 0. Our main idea is to prove αi = 0 (i = 0, . . . , 4n + 2) respectively by induction. |V |2 From (4.11), (2n+1) is invariant as z → 1z . By (4.22), we have 2 | f0
|
(2n+1) 2 1 (2n+1) 2 1 (2n+1) 2 (2n+1) 2 |F1 z 6n+1 z 6n+1 = | f 0 | f0 | | | (z)|F1 | (z). z z
(4.27)
123
L. He, X. Jiao (2n+1) 1 (z)
Because l0
(2n+1)
= l0
(z)z 2 z 2 , we have (2n+1) 2 1 (2n+1) 2 | z |F1 | (z) 2 2 |F1 z z . = (2n+1) 4 1 (2n+1) 4 | f0 | z | f0 | (z)
By (4.27) and (4.28), we get (2n+1) 2
| f0
|
(4.28)
1 2n+1 2n+1 (2n+1) 2 z z = | f0 | (z). z
So set (2n+1) 2
| =
| f0
2n+1
ai j z i z j ,
(4.29)
(4.30)
i, j=0
where a2n+1−i,2n+1− j = ai j , ai j = a ji with a00 = a2n+1,2n+1 = 1. Since (2n+1) 4
t = |h|2 + | f 0
| =
4n+2
ti j z i z j ,
(4.31)
i, j=0 (2n+1) 2
(2n+1) 2
| ∂h − 2h∂| f 0
g = | f0
| =
6n+1 2n+1
gi j z i z j ,
(4.32)
i=0 j=0
where ti j , gi j are the combinations of αi , ai j , α i , a i j as follows, akl ai−k, j−l (i, j = 0, . . . , 4n + 2), ti j = αi α j +
(4.33)
k,l
gi j =
(i + 1 − 3k)αi+1−k ak j (i = 0, . . . , 6n + 1,
j = 0, . . . , 2n + 1). (4.34)
k
we set t = (1 + zz)2n+2 λ = (1 + zz)2n+2
2n
λi j z i z j ,
(4.35)
i, j=0
g = (1 + zz)n+1 μ = (1 + zz)n+1
n 5n
μi j z i z j ,
(4.36)
i=0 j=0
where λi j , μi j are complex coefficients and λi j = λ ji . Then by comparing the coefficients of (4.31) and (4.35), we get 2n + 2 λi−k, j−k (i, j = 0, . . . , 4n + 2), ti j = k
(4.37)
k
which implies t2n+1,0 = · · · = t4n+2,0 = t2n+2,1 = · · · = t4n+2,1 = · · · = t4n+2,2n+1 = 0, and for p = 0, . . . , 2n, we have
t p0 , t p+1,1 , . . . , t4n+2,4n+2− p
123
T
(4.38)
⎡
⎤ A T = ⎣ B ⎦ λ p0 , λ p+1,1 , . . . , λ2n,2n− p , AT
(4.39)
Conformal minimal immersions
where A, B are (2n − p + 1) × (2n − p + 1), ( p + 1) × (2n − p + 1) matrix respectively given by ⎤ ⎡ 2n+2 ⎤ ⎡ 2n+2 2n+2 · · · 2n+2 2n+1− p 2n− p 1 0 ⎢ 2n+2 2n+2 ⎥ ⎢ 2n+2 2n+2 · · · 2n+2⎥ ⎢ 1 ⎥ ⎢ 0 p 2n+1− p 2 ⎥ ⎥ , B = ⎢ 2n+2− A=⎢ .. .. .. .. .. ⎥ ⎥. ⎢ .. ⎥ ⎢ . .. ⎦ ⎣ . . . ⎦ ⎣ . . . 2n+2 2n+2 2n+2 2n+2 2n+2 2n+2 · · · ··· 2n− p 2n−1− p 0 2n+1
2n
p+1
A straightforward computation shows ⎡ ⎤ (−1)0 2n+1 0 ⎢ (−1)1 2n+2 ⎥ (−1)0 2n+1 ⎢ ⎥ 1 0 ⎥. A−1 = ⎢ .. .. .. ⎢ ⎥ . . . ⎣ ⎦ p 2n−1− p 4n− p 0 2n+1 (−1) · · · (−1) (−1)2n− p 4n+1− 2n− p 2n−1− p 0
(4.40)
From (4.39) and (4.40), we get T T t2n+1,2n+1− p , t2n+2,2n+2− p , . . . , t4n+2,4n+2− p = D p t p0 , t p+1,1 , . . . , t2n,2n− p , (4.41) where D p is a (2n + 2) × (2n − p + 1) matrix given by ⎡
p 2n− p p 2n−1− p (−1)2n−1− p 4n+1− (−1)2n− p 4n+2− 2n+1 0 2n+1 0 ⎢(−1)2n− p 4n+2− p 2n+1− p (−1)2n−1− p 4n+1− p 2n− p ⎢ 2n 1 2n 1 Dp = ⎢ .. .. ⎢ ⎣ . . p 4n+1− p 2n−1− p 4n+1− p 4n− p (−1) (−1)2n− p 4n+2− 0 2n+1 0 2n+1
0 ⎤ · · · (−1)0 2n+2 2n+10 1 ⎥ · · · (−1)0 2n+2 2n 1 ⎥ ⎥. .. .. ⎥ . ⎦ . 2n+22n+1 0 · · · (−1) 0 2n+1
From (4.41) we have T T t4n+2,4n+2− p , t4n+1,4n+1− p , . . . , t2n+2+ p,2n+2 = H p t p0 , t p+1,1 , . . . , t2n,2n− p , (4.42) where H p is a (2n − p + 1) × (2n − p + 1) real matrix satisfying H p2 = I2n− p+1 . Similarly, from (4.32) and (4.36), we have n + 1 gi j = (4.43) μi−k, j−k (i = 0, . . . , 6n + 1, j = 0, . . . , 2n + 1), k k
which implies g5n+1,0 = · · · = g6n+1,0 = g5n+2,1 = · · · = g6n+1,1 = · · · = g6n+1,n = 0, g0,n+1 = · · · = g0,2n+1 = g1,n+2 = · · · = g1,2n+1 = · · · = gn,2n+1 = 0,
(4.44)
and for l = 0, . . . , 4n, we have T T gl+n+1,n+1 , gl+n+2+ p,n+2 , . . . , gl+2n+1,2n+1 = E gl0 , gl+1,1 , . . . , gl+n,n , (4.45) where E is a (n + 1) × (n + 1) matrix given by ⎡ n ⎤ n−1 2n n−1 · · · (−1)0 n+1 0 (−1)n 2n+1 (−1) n 0 2n+1 n2n 0n n 0 ⎢ n+1 0 n+1 1 ⎥ (−1)n−1 n−1 ⎢(−1)n n−1 1 1 · · · (−1) n−1 1 ⎥ ⎢ ⎥, E =⎢ .. .. .. .. ⎥ . ⎣ ⎦ . . . 2n+1 2n 2n 2n−1 n+1 n n n−1 0 (−1) · · · (−1) 0 n (−1) 0 n 0 n
123
L. He, X. Jiao
and for q = 4n + 1, . . . , 5n, we have
T T g5n+1,5n+1−q , g5n+2,5n+2−q , . . . , g6n+1,6n+1−q = Fq gq0 , gq+1,1 , . . . , g5n,5n−q , T T g5n+1−q,n+1 , g5n+2−q,n+2 , . . . , g6n+1−q,2n+1 = Fq g0,q−4n , g1,q−4n+1 , . . . , g5n−q,n .
(4.46) where Fq is a (n + 1) × (5n − q + 1) matrix given by ⎡ 5n−q 5n−1−q (−1)5n−q 6n+1−q (−1)5n−1−q 6n−q n 0 n 0 ⎢ 6n+1−q 5n+1−q 5n−1−q 6n−q 5n−q (−1) ⎢(−1)5n−q n−1 1 n−1 1 Fq = ⎢ .. .. ⎢ ⎣ . . 6n−q 6n−1−q (−1)5n−1−q 6n−q (−1)5n−q 6n+1−q 0 n 0 n
0⎤ · · · (−1)0 n+1 n 01⎥ · · · (−1)0 n+1 n−1 1 ⎥ ⎥. .. .. ⎥ . ⎦ . n+1 n 0 · · · (−1) 0 n
Using (4.31) and (4.32) again, we get ti j = αi α j + akl ai−k, j−l (i, j = 0, . . . , 4n + 2),
(4.47)
k,l
gi j =
(i + 1 − 3k)αi+1−k ak j (i = 0, . . . , 6n + 1,
j = 0, . . . , 2n + 1). (4.48)
k
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 α0 = 0.
(4.49)
Otherwise if α0 = 0, then firstly we conclude a2n+1,0 = 0.
(4.50)
In fact, if a2n+1,0 = 0, then by (4.38) and (4.47), we have 2 t4n+2,0 = α4n+2 α 0 + a2n+1,0 = 0,
so α4n+2 = 0. For any 0 ≤ ρ ≤ n(ρ ∈ Z) we prove an+ρ+1,0 = 0 by induction on ρ. If ρ = n then there is nothing to prove. Assume now that the assertion is correct for ρ + 1, . . . , n. Consider the case of ρ. In view of (4.38) and (4.47) again, we have t4n+2,2n−2ρ = α4n+2 α 2n−2ρ + 2
n+ρ
a i0 a 2n+2ρ+2−i,0 + a 2n+ρ+1,0 = 0,
i=2ρ+1
by induction hypotheses we get an+ρ+1,0 = 0. Thus it follows from t2n+2ρ+1,0 = t2n+2ρ+2,0 = 0 that α2n+2ρ+1 = α2n+2ρ+2 = 0 by (4.47). For any 0 ≤ ρ ≤ 2n(ρ ∈ Z) we prove αρ = 0 by induction on ρ. If ρ = 2n then substitution of p = 2n in (4.41) yields t3n+1,n+1 = (2n + 2)t2n,0 = (2n + 2)t4n+2,2n+2 . Since α3n+1 = 0, we find t3n+1,n+1 = t 3n+1,n+1 by (4.47). Then we get t4n+2,2n+2 = t 2n,0 , which implies α4n+2 α 2n+2 = α 2n α0 by (4.47), hence α2n = 0 holds. Assume now that the assertion is correct for ρ + 1, . . . , 2n. Consider the case of ρ. Let p = ρ in (4.42), by induction hypotheses and ([6], Lemma 4.6) we get t4n+2,4n+2−ρ = t ρ0 , which implies αρ = 0 by (4.47). Obviously it contradicts our supposition α0 = 0 and verifies (4.50).
123
Conformal minimal immersions
Secondly, we have the conclusion 2 . |α0 |2 = a2n+1,0
(4.51)
In fact, by t2n+1,0 = t4n+2,0 = 0, we get the expressions of α2n+1 and α4n+2 as follows α2n+1 = −
n 2 a2n+1,0 2 ai0 a2n+1−i,0 , α4n+2 = − . α0 α0
(4.52)
i=0
In view of (4.47), substitution of (4.52) in t4n+2,2n+1 = 0 yields n 2 2 |α0 | + a2n+1,0 a i0 a 2n+1−i,0 = 0, i=0
)n
which implies that i=0 a i0 a 2n+1−i,0 = 0, i.e. α2n+1 = 0. By (4.46) we find t4n+1,2n+1 = t 2n+1,1 . By (4.41) we have that t4n+1,2n+1 = t2n+1,1 = (2n + 2)t2n,0 and t4n+2,2n+2 = t2n,0 . So we see that t4n+2,2n+2 = t 2n,0 , which means that α4n+2 α 2n+2 = α 2n α0 by (4.46). Similarly by t2n+2,0 = t4n+2,2n = 0 we get the expressions of α2n+2 and α2n , thus we have n 4 4 2 a i0 a 2n+2−i,0 + a n+1,0 = 0, |α0 | − a2n+1,0 2 i=1
)n then 2 i=1 a i0 a 2n+2−i,0 + a 2n+1,0 = 0, i.e. α2n+2 = which implies that if |α0 = α2n = 0. Using the same way and ([6], Lemma 4.6), by induction we get that α2n+3 = α2n−1 = 0, . . . , α4n+1 = α1 = 0 respectively. At last we find t4n+2,4n+2 = t 00 by induction, i.e. |α4n+2 |2 = |α0 |2 , it follows from (4.52) that 2 2 |α0 |2 + a2n+1,0 = 0, (4.53) |α0 |2 − a2n+1,0 |2
2 a2n+1,0 ,
which verifies (4.51). Finally, it follows from (4.38), (4.47), (4.51) and induction that a p1 =
a p0 a 2n,0 ( p = 1, . . . , 2n). a2n+1,0
(4.54)
) (5) Set |F1 |2 = i,4nj=0 bi j z i z j . By (4.30), (4.41), (4.47), (4.48), (4.54) and a series of calculations, we get that ⎧ b4n,0 = · · · = b2n+1,0 = 0, b2n,0 = (2n + 1) a 2n,0 − a 10 a2n+1,0 , ⎪ ⎪ ⎪ ⎪ a10 ⎪ ⎪ ⎨b00 = a2n+1,0 a 2n,0 − a 10 a2n+1,0 ; 1 2 (4.55) t2n,0 = − n+1 a 2n,0 − a 10 a2n+1,0 , t00 = a2n+1,0 + 1; ⎪ ⎪ ⎪ = · · · = g = 0, g ⎪ 0n 01 ⎪ ⎪ 2a2n+1,0 ⎩g 2α0 a2n,0 − a10 a2n+1,0 . a 2n,0 − a 10 a2n+1,0 , g00 = a2n+1,0 4n,0 = α0 It follows from (4.25), (4.38), (4.44) and (4.55) that 4b00 t00 + |g00 |2 = 4c1 , 4b2n,0 t2n,0 + g2n,0 g 00 = 0, which implies that 1 a2n+1,0
2 a 2n,0 − a 10 a2n+1,0 a10 + a2n,0 a2n+1,0 = c1 , a 2n,0 − a 10 a2n+1,0 = 0.
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Then we clearly see that c1 = 0, which contradicts the fact that c1 is a positive constant and verifies (4.49). Furthermore, by (4.38) and (4.47) we have an+s+1,0 = 0 (s = 0, . . . , n).
(4.56)
α1 = 0.
(4.57)
Step 2, we prove Suppose α1 = 0, then applying (4.48), (4.56) and g1,2n+1 = · · · = g1,n+2 = 0 we get a2n+1,1 = · · · = an+2,1 = 0.
(4.58)
By (4.48), (4.56) and t4n+2,1 = · · · = t2n+2,1 = 0 we get α4n+2 = · · · = α2n+2 = 0. T Let q = 5n in (4.46), we have g1,n+1 , . . . , gn+1,2n+1 = F5n g0n , which implies an0 = 0, then t2n,0 = 0 by (4.47) and (4.56). Let p = 2n in (4.41), we have t2n+1,1 = 0, which implies α2n+1 = 0 by (4.47), (4.56) and (4.58). For any 1 ≤ ρ ≤ 2n(ρ ∈ Z) we prove αρ = 0 by induction on ρ. If ρ = 2n then let p = 2n − 1 in (4.42) and it follows from ([6], Lemma 4.6) that t4n+1,2n+2 = t 2n,1 , which implies α2n = 0 by (4.47). Assume now that the assertion is correct for ρ + 1, . . . , 2n. Consider the case of ρ. Let p = ρ−1 in (4.42), then by induction hypotheses and ([6], Lemma 4.6) we get t4n+1,4n+2−ρ = t ρ,1 , which implies αρ = 0 by (4.47). Obviously it contradicts our supposition α1 = 0 and verifies (4.57). Moreover for any 1 ≤ ρ ≤ n +1(ρ ∈ Z) we prove aρ0 = 0 by induction on ρ. If ρ = n +1 then there is nothing to prove by (4.56). Assume now that the assertion is correct for ρ + 1, . . . , n + 1. Consider the case of ρ. If aρ0 = 0, by induction hypotheses we have ⎧ ⎪ ⎨b4n,0 = · · · = b2n+ρ,0 = 0, b2n+ρ−1,0 = (2n − ρ)aρ0 a2n,1 ; 2 ; (4.59) t2n,0 = · · · = t2ρ+1,0 = 0, t2ρ,0 = aρ0 ⎪ ⎩ g0n = · · · = g00 = 0.
It follows from (4.25), (4.38), (4.44), (4.56) and (4.59) that 4b2n+ρ−1,0 t2ρ,0 = 0, which implies b2n+ρ−1,0 = 0. Similarly by induction we obtain that b2n+ρ−2,0 = · · · = b00 = 0, which shows |F12n+1 |2 has zero point z = 0, what is impossible. So aρ0 = 0. Then for any 2 ≤ ρ ≤ 2n + 1(ρ ∈ Z) we prove aρ1 = 0 by induction on ρ. If ρ = 2n + 1 then there is nothing to prove by a2n+1,1 = a0,2n = 0. Assume now that the assertion is correct for ρ + 1, . . . , 2n + 1. Consider the case of ρ. By induction hypotheses we have ⎧ ⎪ ⎨b4n,0 = · · · = bρ,0 = 0, bρ−1,0 = ρaρ1 ; (4.60) t2n,0 = · · · = t10 = 0, t00 = 1; ⎪ ⎩ g0n = · · · = g00 = 0. It follows from (4.25), (4.38), (4.44), (4.56) and (4.60) that 4bρ−1,0 t00 = 0,
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which implies aρ1 = 0. Step 3, For any 1 ≤ δ ≤ 2n we prove αδ = 0, aρδ = 0 (δ + 1 ≤ ρ ≤ 2n + 1),
(4.61)
by induction on δ. If δ = 1 then α1 = 0, aρ1 = 0 (2 ≤ ρ ≤ 2n + 1), by step 2. Assume now that the assertion is correct for 1, . . . , δ −1. Consider the case of δ. If αδ = 0, then it follows from (4.47) and t4n+2,δ = · · · = t2n+1+δ,δ = 0 that α4n+2 = · · · = α2n+1+δ = 0. For any 2n − δ + 1 ≤ p ≤ 2n, by (4.41), (4.47) and induction hypotheses we have T = 0, which implies t p+δ,δ = 0. Then t2n+1,2n+1− p , t2n+2,2n+2− p , . . . , t4n+2,4n+2− p it follows from (4.47) that α p+δ = 0 (2n + 1 ≤ p + δ ≤ 2n + δ). Now for any δ ≤ ρ ≤ 2n + 1(ρ ∈ Z) we prove αρ = 0 by induction on ρ. If ρ = 2n + 1 then there is nothing to prove by the above formula. Assume now that the assertion is correct for ρ + 1, . . . , 2n + 1. Consider the case of ρ. Let p = ρ − δ in (4.42), then by induction hypotheses and ([6], Lemma 4.6) we get t4n+2−δ,4n+2−ρ = t ρ,δ , which implies αρ = 0 by (4.47). So αδ = 0. In the following, for any δ + 1 ≤ ρ ≤ 2n + 1(ρ ∈ Z) we prove aρδ = 0 by induction on ρ. If ρ = 2n + 1 then there is nothing to prove by a2n+1,δ = a0,2n+1−δ = 0. Assume now that the assertion is correct for ρ + 1, . . . , 2n + 1. Consider the case of ρ. By induction hypotheses we have ⎧ b4n,ζ = · · · = bζ +1,ζ = 0, bζ,ζ = 0 (0 ≤ ζ ≤ δ − 2), ⎪ ⎪ ⎪ ⎨b 4n,δ−1 = · · · = bρ,δ−1 = 0, bρ−1,δ−1 = ρδaρδ ; )ζ ⎪ ⎪ ⎪t2n+ζ,ζ = · · · = tζ +1,ζ = 0, tζ ζ = k=0 akk aζ −k,ζ −k = 0 (0 ≤ ζ ≤ δ − 1); ⎩ gζ 0 = · · · = gζ,2n+1 = 0 (0 ≤ ζ ≤ δ − 1). (4.62) It follows from (4.25) that 4bρ−1,δ−1 t00 = 0, which implies aρδ = 0. At last we end this step through proving that for any 2n + 1 ≤ δ ≤ 4n + 2 αδ = 0,
(4.63)
by induction on δ. If δ = 2n +1 then α2n+1 = 0. Otherwise if α2n+1 = 0 then it follows from (4.47) and t4n+2,2n+1 = 0 that α4n+2 = 0. For any 1 ≤ p ≤ 2n, by (4.41) and (4.47) we have T t2n+1,2n+1− p , t2n+2,2n+2− p , . . . , t4n+2,4n+2− p = 0, which implies t2n+1+ p,2n+1 = 0. Then (4.47) yields α2n+1+ p = 0. Then let p = 0 in (4.41) we get t00 +t22 +· · ·+t4n+2,4n+2 = t11 + t33 + · · · + t4n+1,4n+1 , which implies α2n+1 = 0.
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Assume now that the assertion is correct for 2n + 1, . . . , δ − 1. Consider the case of δ. If αδ = 0, then for any 1 ≤ p ≤ 2n it follows from (4.41) and (4.47) that t p+δ,δ = 0, which implies α p+δ = 0 (δ + 1 ≤ p + δ ≤ 4n + 2). Let p = 0 in (4.42). By induction hypotheses and ([6], Lemma 4.6) we have tδδ = t 4n+2−δ,4n+2−δ , which verifies (4.63) by (4.47). At present, we have made h = 0 true by (4.49), (4.61) and (4.63). Then from (4.23) and (4.24), we immediately get x0 = x1 = 0. (4.64) Then we transform the other two cases into the first case and finish our proof. (2n+1) 2 | = (1 + zz)d Q, then by (4.26), we have t 2 = (2) If 0 ≤ d ≤ 2n − 4, i.e. |F2 6n−1−d (1 + zz) Q. Since 4n + 3 ≤ 6n − 1 − d ≤ 6n − 1, we have t = (1 + zz)2n+2 Q. Then the formula (4.25) yields g = (1 + zz)n+1 Q. Now it belongs to the case (1). (2n+1) 2 (3) If 2n − 2 ≤ d ≤ 6n − 3, i.e. |F2 | = (1 + zz)d Q, then by (4.26), we have 2 6n−1−d t = (1 + zz) Q, where 2 ≤ 6n − 1 − d ≤ 4n + 1. In view of the properties of harmonic sequence, we have (2n+1) 2
|Fk
(2n+1) 2
| ∂∂|Fk
Since f (2n+1) 0
(2n+1) 2
| − ∂|Fk
(2n+1) 2
| ∂|Fk
(2n+1) 2
(2n+1) 2
| , 1 ≤ k ≤ 2n. (4.65) (2n+1) (2n+1) is a linearly full totally J-isotropic map, we have ln+k = ln−k and
(2n+1) 2 (2n+1) 2 | | f n+1 | = c2 , where (2n+1) 2 (2n+1) |Fn+1 | = c2 |Fn−1 |2 , further a
| fn
simple induction shows (2n+1) 2
| = c2k |Fn−k
(2n+1)
| |Fk+1
1 ≤ k ≤ n and c2 is a positive constant. Hence
(2n+1) 2
|Fn+k
| = |Fk−1
| , 1 ≤ k ≤ n.
(4.66)
(2n+1)
|2 |F1 |2 = (1+zz)2d−2 Q. Without loss of generality, Let k = 2 in (4.65), we get |F3 (2n+1) 2 (2n+1) 2 | = (1 + zz)b Q (0 ≤ b ≤ 2d − 2, b ∈ Z), then we have |F3 | = set |F1 2d−b−2 (1 + zz) Q. It follows from (4.65) that (2n+1) 2
|Fk+1
| = (1 + zz)kd−(k−1)b−k(k−1) Q, 2 ≤ k ≤ 2n − 1.
(4.67)
Let k = 2n − 3 in (4.67), we have (2n+1)
|F2n−2 |2 = (1 + zz)(2n−3)d−(2n−4)b−(2n−3)(2n−4) Q. (2n+1) |F2n−2 |2
(4.68)
(2n+1) 2 c2n−2 |F2 | ,
= which implies (2n − 3)d − Let k = n − 2 in (4.66), we have (2n − 4)b − (2n − 3)(2n − 4) ≤ d. If n ≥ 3 then b ≥ d − 2n + 3, i.e. (2n+1) 2
|F1
| = (1 + zz)d−2n+3 Q, 1 ≤ d − 2n + 3 ≤ 4n. (5)
(5)
(4.69) (5)
(5)
In fact if n = 2 then from (4.65) and (4.66) we have |F2 |2 ∂∂|F2 |2 − ∂|F2 |2 ∂|F2 |2 = (5) (5) c2 |F1 |4 , i.e. |F1 |2 = (1 + zz)d−1 Q, which satisfies (4.69) obviously. At this time, using (4.25), (4.26) and (4.69), we get t 2 = (1 + zz)4n+3 Q, i.e. t = (1 + zz)2n+2 Q. Using (4.25) again, we have g = (1 + zz)n+1 Q. Hence it also belongs to the case (1). Consequently, in the harmonic sequence (4.5) we have ⊕ f (2n+1) , ϕ 0 = f (2n+1) ⊕ f (2n+1) , ϕ 1 = f (2n+1) ⊕ f (2n+1) , (4.70) ϕ −1 = f (2n+1) 0 2n−1 1 2n 2 2n+1 where f (2n+1) : S 2 → CP2n+1 is a linearly full totally J-isotropic map. 0
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Conformal minimal immersions (2n+1)
(2n+1)
Then the metric induced by ϕ0 is dsϕ20 = 2(l0 + l1 )dzdz. Because ϕ0 is 2 a harmonic map with constant curvature K 0 = 6n+1 , up to a holomorphic isometry of CP2n+1 , f (2n+1) is a Veronese surface. Using the proof of Proposition 3.1, we have the 0 conclusion. By Proposition 3.1, 3.2 and 4.3, we obtain a classification of conformal minimal immersions of constant curvature from S 2 to HPn as follows: Theorem 4.4 Let ϕ : S 2 → HPn be a linearly full conformal minimal immersion of constant curvature. Then, (i) If ϕ is reducible and strongly isotropic, then, up to a symplectic isometry of HPn , ϕ (2n+1)
(2n+1)
(n)
(n)
2 is U V n ⊕ U V n+1 with constant curvature n(n+2) , or JU V 0 ⊕ U V 0 with 2 constant curvature n , for some U ∈ G 2n+2 ; (ii) If ϕ is reducible and its isotropy order is r = 2n − 2, then, up to a symplectic isometry (2n+1) (2n+1) 2 of HPn , ϕ is U V 0 ⊕U V 2n+1 with constant curvature 2n+1 for some U ∈ G 2n+2 , in fact in this case r = 2n; (iii) If ϕ is unramified irreducible and its isotropy order is r = 2n − 2, then, up to a (2n+1) (2n+1) 2 symplectic isometry of HPn , ϕ is U V 1 ⊕ U V 2n with constant curvature 6n+1 for some U ∈ G 2n+2 .
Theorem 4.4 shows that all unramified conformal minimal immersions of S 2 in HPn of isotropy order 2n − 2 and 2n with constant curvature are presented by the Veronese surfaces in CP2n+1 , up to a symplectic isometry of HPn . It is easy to check that these minimal immersions are all homogeneous (i.e. SU(2)-equivariant) (cf. [9]).
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