Annali di Matematica (2017) 196:1001–1023 DOI 10.1007/s10231-016-0605-4

Classification of conformal minimal immersions of constant curvature from S2 to Q n Xiaoxiang Jiao1 · Mingyan Li2

Received: 22 March 2016 / Accepted: 9 August 2016 / Published online: 23 August 2016 © Fondazione Annali di Matematica Pura ed Applicata and Springer-Verlag Berlin Heidelberg 2016

Abstract In this paper, we investigate geometry of conformal minimal two-spheres immersed in G(2, n; R) and give a classification theorem of linearly full conformal minimal immersions of constant curvature from S 2 to G(2, n; R), or equivalently, a complex hyperquadric Q n−2 under some conditions. Keywords Conformal minimal immersion · Gauss curvature · Second fundamental form · Complex hyperquadric · Classification Mathematics Subject Classification Primary 53C42 · 53C55

1 Introduction The classification of minimal surfaces M with constant curvature in various Riemannian spaces N is an important topic of differential geometry. In fact, these minimal surfaces have been investigated by several authors when the ambient spaces are real space forms and complex space forms (see [3,4,6,15]). When specialized to M = S 2 and N = C P n , Bolton et al. [3] proved that any linearly full conformal minimal immersion of constant curvature from S 2 to C P n belongs to the Veronese sequence, up to a rigid motion. It is well known that, when the ambient space N is general Riemannian symmetric space, for example, complex Grassmannian G(k, n; C), complex hyperquadric Q n and quaternionic projective space H P n and so on, Calabi’s rigidity fails for minimal 2-spheres of constant

B

Mingyan Li [email protected] Xiaoxiang Jiao [email protected]

1

School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, People’s Republic of China

2

School of Mathematics and Statistics, Zhengzhou University, Zhengzhou 450001, People’s Republic of China

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curvature immersed in them. When N is a complex Grassmannian manifold, the study of its harmonic (a generalization of minimal) two-sphere has been the topic of a sequence of papers: for N = G(2, 4; C), Chi and Zheng [8] classified all holomorphic curves of the Riemann sphere in G(2, 4; C) whose curvature is equal to 2 into two families, up to unitary equivalence, in which none of the curves are congruent; for N = G(2, 5; C), Aithal [1]; for N = G(2, n; C), Burstall and Wood [5], Chern and Wolfson [7], Uhlenbeck [17]. When N is a complex hyperquadric Q n or quaternionic projective space H P n , papers [10, 11,13,14,16,18] introduced conformal minimal immersions of 2-spheres in N and gave some geometric properties of them. In the two cases N = Q 3 , H P 2 , papers [10,16] gave classification theorems of linearly full totally unramified conformal minimal immersions of constant curvature from S 2 to Q 3 and H P 2 , respectively. Here, our interest is to investigate minimal 2-spheres with constant curvature in Q n . As is well known, G(2, n; R) may be identified with complex hyperquadric Q n−2 in C P n−1 (for detailed descriptions, see the Preliminaries below). In 1989, Bahy-El-Dien and Wood [2] gave the explicit construction of all harmonic two-spheres in G(2, n; R), which is considered as totally geodesic submanifolds in complex Grassmann manifolds G(2, n; C). They pointed out that any non-isotropic harmonic map S 2 → G(2, n; R) or Q n−2 can be obtained in a unique way from a holomorphic map f : S 2 → Q n−2 by certain flag transforms called forward and backward replacement ([2], Theorem 4.7). Using the method they gave, in this paper, we study classification of conformal minimal immersions of constant curvature from S 2 to G(2, n; R) by theory of harmonic maps. The paper is organized as follows. In Sect. 2, we identify Q n−2 and G(2, n; R), state some fundamental results concerning G(k, n; C) from the view of harmonic sequences, and then show some brief descriptions of Veronese sequence and the rigidity theorem in C P n . In Sect. 3, we characterize reducible harmonic maps of S 2 in G(2, n; R) with constant curvature (see Proposition 3.1). In Sect. 4, using Bahy-El-Dien and Wood’s results, we present some properties of the harmonic sequence generated by an irreducible harmonic map from S 2 to G(2, n; R) with finite isotropy order r ≥ n − 5, and obtain the explicit characteristics of the corresponding harmonic map in G(2, n; R). Further, we give a classification theorem of linearly full conformal minimal immersions of constant curvature from S 2 to G(2, n; R), or equivalently, a complex hyperquadric Q n−2 under some conditions (see Theorem 4.6).

2 Preliminaries (A) For 0 < k < n, let G(k, n; R) denote the Grassmannian of all real k-dimensional subspaces of Rn and σ : G(k, n; C) → G(k, n; C) denote the complex conjugation of G(k, n; C). It is easy to see that σ is an isometry with the standard Riemannian metric of G(k, n; C). Its fixed point set is G(k, n; R); thus, G(k, n; R) lies totally geodesically in G(k, n; C). Map Q n−2 → G(2, n; R) by √ −1 q → Z ∧ Z, 2

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where q ∈ Q n−2 and Z is a homogeneous coordinate vector of q. It is clear that the map is one-to-one and onto, and it is an isometry. Thus, we can identify Q n−2 and G(2, n; R) (for more details, see [19]). Here, suppose that the metric on G(2, n; R) is given by Section 2 of [12], then the metric is twice as much as the standard metric on Q n−2 induced by the inclusion τ : Q n−2 → C P n−1 , where this latter space is endowed with the Fubini-Study metric of constant holomorphic sectional curvature 4. (B) In this section we give general expression of some geometric quantities about conformal minimal immersions from S 2 to complex Grassmannian manifold G(k, n; C). 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

Consider complex Grassmann manifold G(k, n; C) as the set of Hermitian orthogonal projections from Cn onto a k-dimensional subspace in Cn . Then a map ϕ : M → G(k, n; C) is a Hermitian orthogonal projection onto a k-dimensional subbundle ϕ of the trivial bundle Cn = M × Cn given by setting fiber ϕ x = ϕ(x) for all x ∈ M. ϕ is called (a) harmonic ((sub-) bundle) whenever ϕ is a harmonic map (cf. [2]). Let ϕ : S 2 → G(k, n; C) be a harmonic map. Then from ϕ, two harmonic sequences are derived as follows: ∂

∂

∂

∂

ϕ = ϕ 0 −→ ϕ 1 −→ · · · −→ ϕ i −→ · · · , ∂ 

∂ 

∂ 

(2.1)

∂ 

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

(2.2)

where ϕ i = ∂  ϕ i−1 and ϕ −i = ∂  ϕ −i+1 are Hermitian orthogonal projections from S 2 × Cn  ⊥   ⊥  onto I m ϕi−1 ∂ϕi−1 and I m ϕ−i+1 ∂ϕ−i+1 , respectively, i = 1, 2, . . .. Now recall ([5], §3A) that a harmonic map ϕ : S 2 → G(k, n; C) 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. In particular, let ϕ be a harmonic map from    S 2 to G(2, n; R), then ϕ is ∂ -irreducible (resp. ∂ -reducible) if and only if ϕ is ∂ -irreducible  (resp. ∂ -reducible). In this case we simply say that ϕ is irreducible (resp. reducible). As in [9] call a harmonic map ϕ : S 2 → G(k, n; C) (strongly) isotropic if ϕi ⊥ϕ, ∀i ∈ Z, i  = 0. For an arbitrary harmonic map ϕ : S 2 → G(k, n; C), define its isotropy order (cf. [5]) to be the greatest integer r such that ϕi ⊥ϕ for all i with 1 ≤ i ≤ r ; if ϕ is isotropic, set r = ∞. Definition 2.1 Let ϕ : S 2 → G(k, n; C) be a map. ϕ is linearly full if ϕ cannot be contained in any proper trivial subbundle S 2 × Cm of S 2 × Cn , (m < n). In this paper, we always assume that ϕ is linearly full. Suppose that ϕ : S 2 → G(k, n; C) is a linearly full harmonic map and it belongs to the following harmonic sequence: ∂

∂

∂

∂

∂

∂

ϕ 0 −→ · · · −→ ϕ = ϕ i −→ ϕ i+1 −→ · · · −→ ϕ i −→ 0

(2.3)

0

(i)

(i)

(i)

for some i = 0, . . . , i 0 . We choose the local unit orthogonal frame e1 , e2 , . . . , eki such that they locally span subbundle ϕ i of S 2 × Cn , where ki = rank ϕ i .

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  (i) (i) (i) Let Wi = e1 , e2 , . . . , eki be (n × ki )-matrix. Then we have ϕi = Wi Wi∗ , Wi∗ Wi = Iki ×ki , Wi∗ Wi+1 = 0, Wi∗ Wi−1 = 0. By (2.4), a straightforward computation shows that  ∂ Wi = Wi+1 i + Wi i , ∗ − W  ∗, ∂ Wi = −Wi−1 i−1 i i

(2.4)

(2.5)

where i is a (ki+1 × ki )-matrix and i is a (ki × ki )-matrix for i = 0, 1, 2, . . . , i 0 and i0 = 0. It is very evident that integrability conditions for (2.5) are ∗ ∂i = i+1 i − i i∗ , ∗ ∂i + ∂i∗ = i∗ i + i∗ i − i−1 i−1 − i i∗ .

Set L i = tr(i i∗ ), the metric induced by ϕi is given in the form dsi2 = (L i−1 + L i )dzdz.

(2.6)

From (2.6) and Section 2 of [13], suppose ds 2 = λ2 dzdz is the induced metric by ϕ, K and B are its Gauss curvature and second fundamental form, respectively. Then we have ⎧ 2 ⎪ ⎨ λ = tr∂ϕ∂ϕ, (2.7) K = − λ22 ∂∂ log λ2 , ⎪ ⎩ 2 ∗ B = 4trP P ,   where P = ∂ λA2z with A z = (2ϕ − I )∂ϕ, A z = (2ϕ − I )∂ϕ, and I is the identity matrix,   then P ∗ = −∂ λA2z . In the following, we give a definition of the unramified harmonic map as follows: Definition 2.2 ([12]) If det(i i∗ )dz ki+1 dz ki+1  = 0 everywhere on S 2 in (2.3) for some i, we say that ϕi : S 2 → G(ki , n; C) is unramified. If det(i i∗ )dz ki+1 dz ki+1  = 0 everywhere on S 2 in (2.1) [resp. (2.2)] for each i = 0, 1, 2, . . ., we say that the harmonic sequence (2.1) [resp. (2.2)] is totally unramified. If (2.1) and (2.2) are both totally unramified, we say that ϕ is totally unramified. Set δi =

1 √ 2π −1

 S2

L i dz ∧ dz.

(2.8)



For some i, if ϕi is ∂ -irreducible and unramified in (2.3), then | det i |2 dz ki dz ki is a welldefined invariant and has no zeros on S 2 . It can be checked that (cf. [12])  1 ∂∂ log | det i |2 dz ∧ dz = −2ki . (2.9) √ 2π −1 S 2 (C) In this section, we review the rigidity theorem of conformal minimal immersions of constant curvature from S 2 to C P n .

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Let ψ : S 2 → C P n be a linearly full conformal minimal immersion, a harmonic sequence is derived as follows ∂

∂

∂

∂

∂

∂

0 −→ ψ 0 −→ · · · −→ ψ = ψ i −→ · · · −→ ψ n −→ 0,

(2.10)

for some i = 0, 1, . . . , n. We define a sequence f 0 , . . . , f n of local sections of ψ 0 , . . . , ψ n inductively such that f 0 is a nowhere zero local section of ψ 0 (without loss of generality, assume that ∂ f 0 ≡ 0 ) and f i+1 = ψi⊥ (∂ f i ) for i = 0, . . . , n − 1. Then we have some formulae as follows: ∂ f i = f i+1 +

∂ f i , f i  f i , i = 0, . . . , n, | f i |2

(2.11)

| f i |2 f i−1 , i = 1, . . . , n. (2.12) | f i−1 |2 (n) Let li = | f i+1 |2 /| f i |2 and δi = 2π √1 −1 S 2 li dz ∧ dz, i = 0, . . . , n − 1, l−1 = ln = 0. It is easy to check that they are in accordance with L i and δi , respectively, in the case k = 1. Moreover, if (2.10) is a totally unramified harmonic sequence (i.e., ψi is unramified, i=0,…, n), Bolton et al. proved (cf. [3]) ∂ fi = −

(n)

δi

= (i + 1)(n − i).

(2.13)

Consider the Veronese sequence ∂

(n)

(n)

∂

∂

∂

0−→V 0 −→ V 1 −→ · · · −→ V (n) n −→ 0. (n)

For each i = 0, . . . , n, let Vi for z ∈ S 2 and j = 0, . . . , n,

i! vi, j (z) = (1 + zz)i (n)

Each map V i

(n)

: S 2 → C P n be given by Vi

= (vi,0 , . . . , vi,n )T , where,

     n j−i  j n− j (−1)k z (zz)k . j i −k k

has induced metric dsi2 =

(2.14)

k

n+2i(n−i) dzdz, (1+zz)2

the corresponding constant cur-

vature K i is given by K i = By Calabi’s rigidity theorem, Bolton et al. proved the following rigidity result (cf. [3]). 4 n+2i(n−i) .

Lemma 2.3 ([3]) Let ψ : S 2 → C P n be a linearly full conformal minimal immersion of constant curvature. Then, up to a holomorphic isometry of C P n , the harmonic sequence determined by ψ is the Veronese sequence.

3 Reducible harmonic maps of constant curvature from S2 to G(2, n; R) Recall that an immersion of S 2 in Q n is conformal and minimal if and only if it is harmonic. Thus, we shall analyze harmonic maps from S 2 to G(2, n; R) by reducible and irreducible cases, respectively. In this section, we first consider reducible harmonic maps of S 2 in G(2, n; R) with constant curvature, and for the irreducible ones, we will discuss in detail in Sect. 4 below. Let ϕ : S 2 → G(2, n; R) be a linearly full reducible harmonic map with isotropy order r and constant curvature K . By ([2], Proposition 2.12) we know that, if ϕ has finite isotropy

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order, then it is a real mixed pair; if ϕ is (strongly) isotropic, then r = ∞. To characterize ϕ, we only need to consider the (strongly) isotropic ones. In this case, ϕ has isotropy order ∞ and belongs to the following harmonic sequence: ∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

0 ←− ϕ k ←− · · · ←− ϕ 1 ←− ϕ = ϕ 0 −→ ϕ 1 −→ · · · −→ ϕ k −→ 0

(3.1)

for some k ≥ 1. Here, ϕ 1 , . . . , ϕ k , ϕ 0 , ϕ 1 , . . . , ϕ k are mutually orthogonal, and they are all harmonic maps from S 2 to C P m ⊂ C P n−1 (m < n) from the fact that ϕ is harmonic and reducible. Let p = m − k, choose f p+1 , . . . , f m such that they are local sections of ϕ 1 , . . . , ϕ k , respectively, and satisfy equations (2.11) and (2.12) (here, the derived f 0 is holomorphic). From (3.1), we know that f p and f p are both local sections of ϕ, where f p is obtained from f p+1 by equation (2.12). Then we prove Proposition 3.1 Let ϕ : S 2 → G(2, n; R) be a linearly full reducible harmonic map with Gauss curvature K . Suppose that K is constant. Then ϕ belongs to one of the following cases. (a) ϕ = f p ⊕ c0 , where c0 is a constant vector in Cn , f p = f p : S 2 → C P n−2 and 2 p = n − 2; (b) ϕ is a real mixed pair, i.e., ϕ = f 0 ⊕ f 0 , where f 0 : S 2 → Q n−2 is holomorphic. Further, the curvature of f p in (a) and f 0 in (b) are both constants. Proof We first study the easier case f p = f p . In this case, it is easy to see that f p+i = f p−i for 1 ≤ i ≤ k and 2 p = m. Let ρ be a section of ϕ such that ϕ = f p ⊕ ρ, then we have ρ, f i  = 0 for 0 ≤ i ≤ m since ϕ is (strongly) isotropic. Using the theory of harmonic maps, ρ is a constant vector in Cn , this implies that ϕ belongs to case (a), completing the proof in the case f p = f p . From now on, we assume that f p  = f p . In this case, we claim p = 0 holds, which implies that  f 0 , f 0  = 0, i.e., ϕ is a real mixed pair. Suppose by contrary that p ≥ 1 holds. Let ρ = | f p |2 f p −  f p , f p  f p ,

(3.2)

thus, we have ϕ = f p ⊕ ρ and |ρ|2 = | f p |6 − | f p |2  f p , f p  f p , f p . Choose local frame e1 =

f p+1 f p+2 fp f p+1 f p+2 ρ , e2 = , e3 = , e4 = , e5 = , e6 = . | f p| |ρ| | f p+1 | | f p+1 | | f p+2 | | f p+2 |

Set W0 = (e1 , e2 ), W1 = (e3 ), W−1 = (e4 ), W2 = (e5 ), W−2 = (e6 ), then by (2.5), we obtain  0 =

⎞ ⎛ |f |   f , f  − | fp+1 p | f p+1 | 3 p p| ⎠. , 0 , −1 = ⎝ |f ||ρ| | f p| − p+1 4

(3.3)

| f p|

From (3.3), applying the equation L i = tr(i i∗ ), a straightforward computation shows L 0 = L −1 =

| f p+1 |2 . | f p |2

Then the metric induced by ϕ is as follows ds 2 = 2l p dzdz.

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Under the assumption that ϕ is of constant curvature, the curvature of f p is also a constant. By Lemma 2.3, up to a holomorphic isometric of C P n−1 , f p is a Veronese surface. Since ϕ k and ϕ k in (3.1) are orthogonal, so  f m , f m  = 0, which implies that  f 0 , f 0  = 0.

(3.4)

Using (3.1) and (3.2) we have f p−i =

 f p−i , f p+1   f p−i , f p+i−1   f p−i , ρ ρ+ f p+1 + · · · + f p+i−1 2 2 |ρ| | f p+1 | | f p+i−1 |2 +

 f p−i , f p+i  | f p+i |2

f p+i

(3.5)

for 1 ≤ i ≤ p. In particular, take i = p, then (3.5) becomes f0 =

 f 0 , f p+1   f 0 , f 2 p−1   f0 , f 2 p   f 0 , ρ ρ+ f p+1 + · · · + f 2 p−1 + f 2p. |ρ|2 | f p+1 |2 | f 2 p−1 |2 | f 2 p |2

(3.6)

Substituting (3.6) into (3.4), using the fact that f p+1 , . . . , f 2 p , ρ, f p+1 , . . . , f 2 p are mutually orthogonal, we have ( f 0 , ρ)2 ρ, ρ = 0, which can be rewritten as  f p , f p  f 0 , f p  = 0, by making use of  f 0 , ρ = | f p |2  f 0 , f p , ρ, ρ = − ||ρ|  f p , f p . We shall divide our f |2 2

p

discussion into two cases, according as  f p , f p  = 0 or  f 0 , f p  = 0. If  f p , f p  = 0, under the assumption that p ≥ 1, from (3.1) we get f p−1 = f p+1 , which implies that f p = f p . This contradicts our supposition f p  = f p . Thus, we obtain  f 0 , f p  = 0, i.e.,  f 0 , ρ = 0 and  f p , f p   = 0. From (3.5) we have  f p−i , f p+i+1  = 0, which gives  f p−i , f p+i  = (−1)i

l p l p+1 . . . l p+i−1  f p , f p , 1 ≤ i ≤ p. l p−1 l p−2 . . . l p−i

This shows that  f p−i , f p+i   = 0 for 0 ≤ i ≤ p. Since p ≥ 1, in the following we write equation (3.5) in the form ⎧  f p−1 , f p+1   f p−1 ,ρ ⎪ ⎪ ⎪ f p−1 = |ρ|2 ρ + | f p+1 |2 f p+1 , ⎪ ⎪ ⎪  f p−2 , f   f p−2 , f  f ,ρ ⎪ ⎪ ρ + | f |p+1 f p+1 + | f |p+2 f p+2 , f p−2 = p−2 2 2 ⎪ |ρ|2 ⎪ p+1 p+2 ⎪ ⎪  f p−3 , f   f p−3 , f   f p−3 , f  f ,ρ ⎪ ⎨ f p−3 = p−3 ρ + | f |p+1 f p+1 + | f |p+2 f p+2 + | f |p+3 f p+3 , 2 2 2 |ρ|2 p+1 p+2 p+3 (3.7) . ⎪ .. ⎪ ⎪ ⎪ ⎪ ⎪  f 1 , f p+1   f 1 , f 2 p−1  ⎪ f 1 ,ρ ⎪ f 1 =  |ρ| ⎪ 2 ρ + |f 2 f p+1 + · · · + | f 2 f 2 p−1 , ⎪ p+1 | 2 p−1 | ⎪ ⎪ ⎪  f 0 , f 2 p−1   f0 , f  ⎩ f =  f0 , f p+1  f f 2 p−1 + | f |22p f 2 p . 0 p+1 + · · · + | f |f |2 |2 p+1

2 p−1

2p

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Hence, it is possible to have the equations  f 0 , f p+1  = 0,  f 0 , f p+2  = 0, . . . ,  f 0 , f 2 p−1  = 0, by substituting (3.7) into the following equations  f p−1 , f 0  = 0,  f p−2 , f 0  = 0,  f p−3 , f 0  = 0, . . . ,  f 1 , f 0  = 0. We can thus arrive at the relation f 0 = f 2 p , 2 p = m. Moreover, we have f p = f p , which contradicts the supposition f p  = f p . Summarizing the above results, if f p  = f p , then we get p = 0 and  f 0 , f 0  = 0, i.e., ϕ is a real mixed pair. Since the curvature K of ϕ is a constant, then the curvature of f p in (a) and f 0 in (b) are both constants.   By Proposition 3.1, in the following we discuss case (b) with ϕ of finite isotropy order r ≥ n − 4, and case (a) and case (b) with ϕ of r = ∞.

3.1 Reducible harmonic maps from S2 to G(2, n; R) with constant curvature and finite isotropy order r ≥ n − 4 Let ϕ : S 2 → G(2, n; R) be a linearly full reducible harmonic map with constant curvature and finite isotropy order r ≥ n −4. It follows from [2] that r is odd and ϕ can be characterized by harmonic maps from S 2 to C P n−1 , in fact, ϕ = f 0 ⊕ f 0, where f 0 : S 2 → Q n−2 is holomorphic. By using ϕ, a harmonic sequence is derived as follows ∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

0 ←− f m ←− · · · ←− f 1 ←− ϕ −→ f 1 −→ · · · −→ f m −→ 0, ∂

∂

∂

∂

∂

where 0 −→ f 0 −→ f 1 −→ · · · −→ f m −→ 0 is a linearly full harmonic sequence in C P m ⊂ C P n−1 satisfying   f 0 , f i  = 0 (0 ≤ i ≤ r ), (3.8)  f 0 , fr +1   = 0, and r + 1 ≤ m ≤ n − 1. The induced metric of ϕ is given by ds 2 = 2l0 dzdz,

(3.9)

where l0 dzdz is the induced metric of f 0 . Since ϕ is of constant curvature, using (3.9) we get that the curvature K of ϕ satisfies K = m2 . By Lemma 2.3, up to a holomorphic isometry of C P n−1 , f 0 is a Veronese surface. We can choose a complex coordinate z on C = S 2 \{ pt} (m) (m) so that f 0 = U V0 , where U ∈ U (n) and V0 has the standard expression given in part (m) (m) (C) of Sect. 2 (adding zeros to V0 such that V0 ∈ Cn ). Then (3.8) becomes  (m) (m) U V0 , U V r  = 0, (m) (m) U V0 , U V r +1   = 0,

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which is equivalent to



(m)

1009

(m)T

trW V0 Vr = 0, (m) (m)T trW V0 Vr +1  = 0,

(3.10)

where W = U T U , it satisfies W ∈ U (n) and W T = W . Define a set G W  {U ∈ U (n)|U T U = W }.

(3.11)

For a given W , the following can be easily checked (1) ∀ U ∈ G W , A ∈ S O(n), we have AU ∈ G W ; (2) ∀ U, V ∈ G W , ∃ A ∈ S O(n), s.t. U = AV. To determine ϕ, firstly we need to determine the isotropy order r of it. If n is even, since the isotropy order r of ϕ is greater than or equal to n − 4 and it is odd, then r = n − 3. If n is odd, r = n − 2 or n − 4. In the following we discuss W in cases r = n − 3, n − 2, n − 4, respectively. Case I, r = n − 3. In this case, m = n − 2 or n − 1 and ϕ belongs to the following two harmonic sequences, respectively: ∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

(Ia) ϕ −→ f 1 −→ · · · −→ f n−3 −→ f n−2 −→ 0;

∂

(Ib) ϕ −→ f 1 −→ · · · −→ f n−3 −→ f n−2 −→ f n−1 −→ 0. Case II, r = n − 2. In this case, m = n − 1 and ϕ belongs to the following harmonic sequence: (IIa) ϕ −→ f 1 −→ · · · −→ f n−2 −→ f n−1 −→ 0. Case III, r = n − 4. In this case, m = n − 3, n − 2 or n − 1, and ϕ belongs to the following three harmonic sequences, respectively: (IIIa) ϕ −→ f 1 −→ · · · −→ f n−4 −→ f n−3 −→ 0; ∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

∂

(IIIb) ϕ −→ f 1 −→ · · · −→ f n−4 −→ f n−3 −→ f n−2 −→ 0;

∂

(IIIc) ϕ −→ f 1 −→ · · · −→ f n−4 −→ f n−3 −→ f n−2 −→ f n−1 −→ 0. To determine ϕ, we first should consider the type of W or U . From (3.10), the matrices W in cases (Ia), (IIa) and (IIIa) satisfy (m)

(m)T

(m)

Vm(m)T  = 0,

(3.12)

where m is even. Similarly, the matrices W in cases (Ib) and (IIIb) satisfy (m) (m)T (m) (m)T trW V0 Vm−2 = 0, trW V0 Vm−1  = 0,

(3.13)

trW V0

Vm−1 = 0, trW V0

where m is odd, and the matrix W in case (IIIc) satisfies (m)

trW V0

(m)T

(m)

Vm−3 = 0, trW V0

(m)T

Vm−2  = 0,

(3.14)

where m is even. (m) (m) We first consider the type of W in (3.12). The standard expressions of V0 and Vm−1 (m)

(m)T

show that V0 Vm−1 is a polynomial matrix in z and z. Assume W = (ai j ), 0 ≤ i, j ≤ n − 1, which is a constant matrix. From (2.14), we have

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  m j m (m − 1)! v0, j (z) = z , vm−1, j (z) = (−z)m− j−1 (m − j − j zz), m−1 j j (1 + zz) for 0 ≤ j ≤ m, and v0, j (z) = vm−1, j (z) = 0, m + 1 ≤ j ≤ n − 1. (m) (m)T Then equation trW V0 Vm−1 = 0 can be rewritten in the form

   m  m m ai j (−1)m− j−1 (z)m−i− j−1 [(m − j)(zz)i − j (zz)i+1 ] = 0. (3.15) i j

i, j=0

For any fixed integer k (0 ≤ k ≤ 2m), the formula (3.15) yields a polynomial equation for zz given by

    m m ai j (3.16) (−1) j [(m − j)(zz)i − j (zz)i+1 ] = 0. i j i+ j=k

Using the method of indeterminate coefficients by (3.16), we get  a0k = 0,      m  m  m m (k − i + 1) i−1 k−i+1 ai−1,k−i+1 + (m − k + i) i k−i ai,k−i = 0 for 0 ≤ k ≤ m − 1, 1 ≤ i ≤ k;  am,k−m = 0,     m  m m  (m − k + i) mi k−i ai,k−i + (k − i + 1) i−1 k−i+1 ai−1,k−i+1 = 0

(3.17)

(3.18)

for m + 1 ≤ k ≤ 2m, m + 1 ≤ i ≤ k. Rearranging (3.16) (3.17) and (3.18), it is easy to see that W in (3.12) satisfies  ai j = 0, 0 ≤ i, j ≤ m, i + j  = m, (3.19) ai−1,m−i+1 + ai,m−i = 0, 1 ≤ i ≤ m. Similarly, W in (3.13) satisfies  ai j = 0, 0 ≤ i, j ≤ m, i + j ≥ m + 2 or i + j ≤ m − 2, 3ai,i+1 + ai−1,i+2 = 0, 2i + 1 = m,

(3.20)

and W in (3.14) satisfies ai j = 0, 0 ≤ i, j ≤ m, i + j ≥ m + 3 or i + j ≤ m − 3.

(3.21)

Therefore, we obtain Proposition 3.2 Let ϕ : S 2 → G(2, n; R) be a linearly full reducible harmonic map with constant curvature K and finite isotropy order r . Suppose that r ≥ n − 4. Then n must be (n−1) (n−1) 2 odd, and up to an isometry of G(2, n; R), ϕ = U V 0 ⊕UV0 with K = n−1 for some T U ∈ G  {U ∈ U (n)|U U = W0 }, where W0 = antidiag{1, −1, 1, −1, . . . , 1}, which is an (n × n)-matrix. Proof At first, we discuss W in cases r = n − 3, n − 2, n − 4, respectively. (I) r = n − 3. In this case, m = n − 2 or n − 1. If m =  n − 2, ϕ belongs to (Ia). Using 0 W1 , where W1 = (3.19) and the property of unitary matrix, we get W = 0 am+1,m+1 antidiag{a0m , a1,m−1 , · · · , am0 }. Obviously, ϕ is not linearly full in this condition. If m = n − 1, using (3.20) and the property of unitary matrix, we find that there doesn’t exist such W .

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(II) r = n − 2. In this case, m = n − 1 and f 0 = f m , which is equivalent to (m)

UV0

= μU Vm(m) ,

(3.22)

where μ is a parameter. Set W0 = antidiag{1, −1, 1, −1, · · · , 1}, which is an (n × n)matrix. Then condition (3.22) becomes U = U W0 .

(3.23)

G  {U ∈ U (n)|U T U = W0 }.

(3.24)

Define a set Then the following can be easily checked (1) ∀ U ∈ G, A ∈ S O(n), we have AU ∈ G; (2) ∀ U, V ∈ G, ∃ A ∈ S O(n), s.t. U = AV. (III) r = n − 4. In this case, m = n − 3, n − 2 or n − 1. If m = n − 3, then ϕ belongs  W1 0 to (IIIa). Using (3.19) and the property of unitary matrix, we get W = , 0 W2   am+1,m+1 am+1,m+2 . where W1 = antidiag{a0m , a1,m−1 , · · · , am0 } and W2 = am+1,m+2 am+2,m+2 Obviously ϕ is not linearly full in this condition. If m = n − 2 or m = n − 1, using (3.20) and (3.21), we find that there doesn’t exist such W .  

In summary we get the conclusion. Remark 3.3 In Proposition 3.2, G  = ∅. Since n is odd, let n = 2 p + 1. If p is even, choose ⎡

√1 √2 √−1 2

⎢ ⎢ ⎢ ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢ 0 U0 = ⎢ ⎢ ⎢. ⎢. ⎢. ⎢ ⎢0 ⎢ ⎢ ⎣0 0 ⎡ 1

.. .

√ √2 √−1 2

⎢ ⎢ ⎢ ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢ 0 If p is odd, choose U0 = ⎢ ⎢ ⎢. ⎢. ⎢. ⎢ ⎢0 ⎢ ⎢ ⎣0 0

.. .

0

···

0

0

0

···

0

0

···

0

0

0

···

0

√1 2 √ √−1 2

··· ··· .. .

0

···

0 0

··· ···

0 0 √1 2 √ √−1 2

0

.. . .. . .. .

.. . .. . .. .

.. . .. . .. .

···

√1 2√ − √−1 2

− √1

0

2 √ √−1 2

··· .. .. . . ··· 0

0 .. .

0

− √1

0

0 1

0

···

0

0

0

···

0

···

0

0

0

···

0

√1 √2 √−1 2

. · · · .. . · · · .. .. .. . . · · · √1

0

···

0

···

√2 √−1 2

0

√ 2 √−1 2

.. . .. . .. .

.. . .. . .. . 0 0 √

−1

··· ···

0 0

···

√1 2√ − √−1 2

0

0 0 0

··· .. .. . . ··· 0

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥; ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

√1 2√ − √−1 2

− √1

2 √ −1 √ 2

⎤

0 0 .. . 0

···

0

0

···

0

0

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

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Let Ci denote the i-th column of U0 . For 1 ≤ i ≤ p, all the√ elements of Ci are zero expect the (2i − 1)-th and (2i)-th elements, which are √1 and √−1 respectively. Moreover, the 2

2

column vectors of U0 satisfy equation C i + (−1)i Cn−i+1 = 0, 1 ≤ i ≤ n. Then we have U0 ∈ G and ∀U ∈ G can be obtained from U0 by an S O(n)-motion. As to the second fundamental form B of ϕ, we have the following corollary. Corollary 3.4 Let ϕ : S 2 → G(2, n; R) be a linearly full reducible harmonic map with constant curvature K and finite isotropy order r . Suppose that r ≥ n − 4. Then its second fundamental form B satisfies B2 = 4(n−2) n−1 , n ≥ 4. Proof Let n = m + 1, and take U0 ∈ U (n), which is the one shown in Remark 3.3. For any integer i (0 ≤ i ≤ m), let (m) f i = U0 Vi . (3.25) Then f 0 = f m , and ϕ can be rewritten as ϕ = f 0 ⊕ f 0 , where f 0 is of constant curvature. By (2.7), a series of calculations give ⎧ ∂ϕ = | f1|2 [ f 0 ( f 1 )∗ + f 1 f 0∗ ], ⎪ ⎪ 0 ⎨ 1 A z = | f |2 [ f 0 ( f 1 )∗ − f 1 f 0∗ ], 0 ⎪ ⎪ ⎩ P = 2| f1 |2 [ f 0 ( f 2 )∗ − f 2 f 0∗ ]. 1

Hence, we get 

 l1 | f 0 , f 2 |2 − B = 2 . l0 | f 1 |4 2

Since f 0 is of constant curvature and m + 1 = n, so 

l1 l0

=

(m)

δ1

(m)

δ0

=

 2(n − 2) | f 0 , f 2 |2 B = 2 − . n−1 | f 1 |4 2

2(n−2) n−1 ,

which implies that

(3.26)

From the fact that n is odd, we discuss it as follows: (1) If n = 3. Equation f 0 = f 2 follows immediately from (3.25). Substituting it into (3.26), we obtain B2 = 0, i.e., ϕ is totally geodesic. In this case, ϕ is a conformal minimal immersion from S 2 to G(2, 3; R) (or equivalently Q 1 ), which is trivial, we don’t consider it in this paper; (2) If n ≥ 5. Since f 0 = f m , then we have  f 0 , f 2  = 0. Equation (3.26) becomes

B2 =

4(n−2) n−1 .

 

3.2 Reducible (strongly) isotropic harmonic maps from S2 to G(2, n; R) with constant curvature Above we determine all harmonic maps of S 2 in G(2, n; R) with constant curvature and finite isotropy order r ≥ n − 4. In the following, using Proposition 3.1, we shall give a classification of reducible (strongly) isotropic harmonic maps from S 2 to G(2, n; R) with constant curvature.

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Proposition 3.5 Let ϕ : S 2 → G(2, n; R) be a linearly full reducible harmonic map with constant curvature K and second fundamental form B. Suppose that ϕ is (strongly) isotropic. Then n must be even, and up to an isometry of G(2, n; R), ϕ belongs to one of the following cases. (2 p)

8 and B2 = 2(n−4)(n+2) for some U ∈ U (n−1), n ≥ (a) ϕ = U V p ⊕c0 with K = n(n−2) n(n−2) 4, where c0 is a constant vector in Cn and 2 p = n − 2; (m)

(m)

(b) ϕ = U V 0 ⊕ U V 0 n = 2m + 2.

with K =

and B2 =

4 n−2

4(n−4) n−2

for some U ∈ U (n),

Proof It follows from Proposition 3.1 that, ϕ belongs to case (a) or case (b). In the following we discuss the two cases of ϕ, respectively. (a) ϕ = f p ⊕ c0 , where c0 is a constant vector in Cn , f p : S 2 → C P n−2 with constant curvature satisfies f p = f p and 2 p = n − 2. In this case, we have l p−1 = l p . By (2.7), direct calculations give ⎧ 2 λ = 2l p , ⎪ ⎪ ⎪ 2 8 ⎨ K = p( p+1) = n(n−2) ,

(3.27)

l

⎪ B2 = 2 p+1 ⎪ l p , p ≥ 2, ⎪ ⎩ B2 = 0, p = 1.

Since f p is of constant curvature, then the map f p : S 2 → C P 2 p is totally unramified. When p ≥ 2, it follows from (2.13) and (3.27) that (2 p)

B2 = 2

δ p+1 l p+1 2(n − 4)(n + 2) . = 2 (2 p) = lp n(n − 2) δp

Moreover, harmonic sequence f 0 , f 1 , . . . , f 2 p is the Veronese sequence in C P 2 p ⊂ C P n−2 , up to U (n − 1). Then we can choose a complex coordinate z on C = S 2 \{ pt} so (2 p)

(2 p)

that f i = U Vi , 0 ≤ i ≤ 2 p, where U ∈ U (n − 1) and Vi given in part (C) of Sect. 2. Equation f p = f p is equivalent to (2 p)

UV p

(2 p)

= μU V p

has the standard expression

,

(3.28)

where μ is a parameter. Set W0 = antidiag{1, −1, 1, −1, · · · , 1}, which is an (2 p + 1) × (2 p + 1)-matrix. Then condition (3.28) becomes U = U W0 . Thus, choose U = U0 as the one shown in Remark 3.3, up to an isometry of G(2, n; R), p) ⊕ c0 . ϕ = U 0 V (2 p

(b) ϕ is a real mixed pair. In this case, ϕ belongs to the following harmonic sequence: ∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

0 ←− f m ←− · · · ←− f 1 ←− ϕ = f 0 ⊕ f 0 −→ f 1 −→ · · · −→ f m −→ 0, (3.29)

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where f 0 , . . . , f m : S 2 → C P m is a linearly full harmonic sequence in C P m ⊂ C P n−1 . By (2.7) and a series of calculations , we obtain ⎧ 2 ⎪ ⎨ λ = 2l0 , K = m2 , (3.30) ⎪ ⎩ l1 2 B = 2 l0 . Since f 0 is of constant curvature, by virtue of Lemma 2.3, f 0 , f 1 , . . . , f m is the Veronese sequence in C P m ⊂ C P n−1 , up to U (n), and they are all totally unramified. It follows from (2.13) and (3.30) that (m)

B2 = 2

δ l1 4(m − 1) = 2 1(m) = . l0 m δ0

From (3.29), by the assumption that ϕ is (strongly) isotropic, we get  f 0 , f m  = 0, which is equivalent to

(m)

trU T U V0 (m)

Vm(m)T = 0,

(3.31)

(m)T

(m)

here, V0 Vm is a polynomial matrix in z and z by the standard expressions of V0 and (m) (m) Vm given in Part (C) of Sect. 2, and U ∈ U (n) satisfying U V0 = f 0 is a constant matrix. Using the method of indeterminate coefficients by (3.31), assume U T U = (ai j ), 0 ≤ i, j ≤ n − 1, it has the form ⎤ ⎡ a0,m+1 · · · a0,n−1 .. .. ⎥ ⎢ ⎥ ⎢ . . ⎥ ⎢ ⎥ ⎢ · · · a a m,m+1 m,n−1 T ⎥. ⎢ U U =⎢ ⎥ a · · · a a · · · a m+1,0 m+1,m m+1,m+1 m+1,n−1 ⎥ ⎢ ⎥ ⎢. .. .. .. ⎦ ⎣ .. . . .

0

an−1,0 Choose

⎡

· · · an−1,m

√1 √2 √−1 2

⎢ ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢0 ⎢ ⎢ ⎢ U1 = ⎢ 0 ⎢ ⎢0 ⎢ ⎢ ⎢ .. ⎢. ⎢ ⎢0 ⎣ 0

· · · an−1,n−1

an−1,m+1

0

0

···

0

0

0

···

0

√1 √2 √−1 2

0

···

0

0

···

0

0

√1 √2 √−1 2

···

0

0

0

··· 0 . .. . .. · · · √1

0

0

···

0 .. .

.. .

⎤

*

√2 √−1 2

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(3.32)

Let Ci denote the i-th column of U1 . For 1 ≤ i ≤ m + 1, all√the elements of Ci are zero expect the (2i − 1)-th and (2i)-th elements, which are √1 and √−1 , respectively. 2

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From the above discussion, we have n = 2m + 2, and up to an isometry of G(2, n; R), (m) (m) ϕ = U 1 V 0 ⊕ U 1 V 0 . Its curvature K and the second fundamental form B of ϕ are as follows K =

4 4(n − 4) , B2 = . n−2 n−2  

This finishes the proof.

4 Irreducible harmonic maps of constant curvature from S2 to G(2, n; R) In Sect. 3, we consider conformal minimal immersions from S 2 to G(2, n; R) with constant curvature under the assumption that they are reducible. To complete the classification of all conformal minimal immersions of S 2 in G(2, n; R) with constant curvature, we must discuss the irreducible ones. For a linearly full irreducible harmonic map ϕ : S 2 → G(2, n; R) with isotropy order r , in this section we consider the case that ϕ is of finite isotropy order r ≥ n − 5, where n ≥ 5 (it is easy to check that, if n = 4 and ϕ is of finite isotropy order, then ϕ must be reducible). Here, we state one of Bahy-El-Dien and Wood’ results as follows: Lemma 4.1 (Special case of [2], Theorem 4.7) Let ϕ : S 2 → G(2, n; R) be an irreducible harmonic map of finite isotropy order r ( here n ≥ 5). Suppose that r ≥ n − 5. Then, there is a unique sequence of harmonic maps ϕ i : S 2 → G(2, n; C), (i = 0, 1, 2) such that (i) ϕ 0 is a real mixed pair, in fact ϕ 0 = f 0 ⊕ f 0 has finite isotropy order r + 2 and f 0 is a holomorphic map; (ii) ϕ = ϕ 2 ; (iii) ϕ 1 is obtained from ϕ 0 by forward replacement of f 0 ; (iv) ϕ 2 is obtained from ϕ 1 by backward replacement of V ⊥ ∩ ϕ 1 , where V is a holomorphic  line subbundle of ϕ 1 not equal to the image of the first ∂ -return map of ϕ 1 . Let ϕ : S 2 → G(2, n; R) be a linearly full irreducible harmonic map of finite isotropy order r ≥ n − 5 (here we know that n ≥ 5). In the following we characterize ϕ explicitly by virtue of Lemma 4.1. First recall ([2], Proposition 2.8 and Lemma 2.15) that the isotropy order r of ϕ is n − 4 if n is odd and n − 5 if n is even. In (i) of Lemma 4.1, ϕ 0 with isotropy order r + 2 belongs to the harmonic sequence as follows: ∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

∂

∂

0 ←− f m ←− · · · ←− f 1 ←− ϕ 0 −→ f 1 −→ · · · −→ f m −→ 0, ∂

∂

(4.1)

∂

where ϕ 0 = f 0 ⊕ f 0 and 0 −→ f 0 −→ f 1 −→ · · · −→ f m −→ 0 is a linearly full harmonic sequence in C P m ⊂ C P n−1 . Here, m = n − 1 if n is odd, and n − 2 ≤ m ≤ n − 1 if n is even. By (iii) of Lemma 4.1, ϕ 1 is obtained from ϕ 0 by forward replacement of f 0 , using (4.1) we have ϕ1 = f 0 ⊕ f 1.

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The isotropy order of ϕ 1 is r + 1, and a harmonic sequence is derived as follows: ∂ 

∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

0 ←− f m ←− · · · ←− f 2 ←− ϕ 1−1 ←− ϕ 1 −→ f 2 −→ · · · −→ f m −→ 0, where ϕ 1−1 = ϕ 1 = f 1 ⊕ f 0 .

(4.2)



From (4.2), the image of the first ∂ -return map of ϕ 1 is f 0 . By (iv) of Lemma 4.1, let V = f 1 + x0 f 0 , where x0 is a smooth function on S 2 expect some isolated points. Let X = −| f 0 |2 x 0 f 1 + | f 1 |2 f 0 , it satisfies X = V ⊥ ∩ ϕ 1 . Since ϕ 2 is obtained from ϕ 1 by backward replacement of X , so ϕ 2 = V ⊕ W with W = πϕ 1⊥ (∂ X ), and it belongs to the following harmonic sequence ∂ 

∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

∂

0 ←− f m ←− · · · ←− f 3 ←− Y ⊕ f 2 ←− ϕ 2 −→ Y ⊕ f 2 −→ f 3 −→ · · · −→ f m −→ 0,

(4.3)

where Y = W ⊥ ∩ ϕ 1−1 .

Applying the equation W = πϕ 1⊥ (∂ X ) we obtain W = | f 1 |2 V ,

(4.4)

this implies that W = V , Y = X. Obviously, X , X , V and V are mutually orthogonal. Then ϕ = ϕ 2 = V ⊕ V , and (4.3) reduces to ∂ 

∂ 

∂ 

∂ 

∂ 

∂

∂

∂

∂

∂

0 ←− f m ←− · · · ←− f 3 ←− X ⊕ f 2 ←− ϕ −→ X ⊕ f 2 −→ f 3 −→ · · · −→ f m −→ 0.

Since V is a holomorphic line subbundle of ϕ 1 , we get πϕ 1 (∂ V ) ∈ V .

(4.5)

Through a direct computation, condition (4.5) is equivalent to the following equation ∂ x 0 + x 0 ∂ log | f 0 |2 = 0,

(4.6)

which can also be obtained from the fact that πϕ ⊥ (∂ϕ1 ) = f 3 , where ϕ1 = X ⊕ f 2 . Then 1 we have Proposition 4.2 The map ϕ : S 2 → G(2, n; R) is a linearly full irreducible harmonic map with finite isotropy order r ≥ n − 5 if and only if ϕ = V ⊕ V with V = f 1 + x0 f 0 , where   f 0 , fr +2  = 0 f 0 is a holomorphic map satisfying , and the corresponding coefficient x0  f 0 , fr +3   = 0 satisfies equation (4.6). Proof Through the construction  of ϕ as shown above, the necessity is obvious. Since f 0 is a  f 0 , fr +2  = 0 holomorphic map satisfying , using (4.6), direct calculation gives ∂ A z =  f 0 , fr +3   = 0   [A z , A z ], which implies that ϕ is harmonic. Thus, we get the sufficiency. Proposition 4.2 gives a characterization of linearly full irreducible harmonic maps from S 2 to G(2, n; R) with finite isotropy order r ≥ n − 5. In the following we consider the case of constant curvature.

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Let ϕ : S 2 → G(2, n; R) be a linearly full irreducible harmonic map of constant curvature K and finite isotropy order r . Suppose that r ≥ n − 5. By Proposition 4.2, we choose local frame e1 =

V X f2 f3 V X f , e2 = , e3 = , e4 = , e5 = , e6 = 2 , e7 = , |V | |V | |X | | f2 | |X | | f2 | | f3 |

where V = f 1 + x0 f 0 and x0 is a smooth function on S 2 except some isolate points. Here, the local frame we choose is not unitary frame except r ≥ 3. Set W0 = (e1 , e2 ), W1 = (e3 , e4 ), W−1 = (e5 , e6 ) and W2 = (e7 ), then by (2.5), we obtain  | f | ∂ V,X   ∂ V,X  | f |   2 − | f01 | |X ||V | | f3 | |X ||V | |V | 0 = , 1 = 0, . (4.7) , −1 = − | f2 | | f2 | − || ff01 || 0 0 |V | This together with equation L i = tr(i i∗ ) implies that ∂ V, X X, ∂ V  | f 2 |2 + + l0 , |X |2 |V |2 |V |2 L 1 = l2 , 2 | f | 0 dz 2 dz 2 , | det 0 |2 dz 2 dz 2 = l02 l1 |V |2 det 1 ∗1 dzdz = l2 dzdz.

L 0 = L −1 =

(4.8) (4.9) (4.10) (4.11)

Using (4.8) (4.9) and (4.10), direct calculations give ∂∂ log | det 0 |2 = L −1 − 2L 0 + L 1 .

(4.12)

Suppose that ϕ is totally unramified, then | det 0 |2 dz 2 dz 2  = 0 and det 1 ∗1 dzdz  = 0 everywhere on S 2 and f i : S 2 → C P m ⊂ C P n−1 (3 ≤ i ≤ m) are all unramified. It follows from (4.11) that l2 dzdz  = 0 everywhere on S 2 . And, we claim l1 dzdz  = 0 everywhere 2 2 on S 2 holds. Otherwise, if l1 dzdz = 0 at some point z = z 0 , since ||Vf2||2 = ||Vf1||2 l1 and 2 2 | f 1 |2 = |x |2 | |f f 1|2| +| f |2 ≤ 1 < +∞, then ||Vf2||2 = 0 at the point z = z 0 . It follows from |V |2 0 0 1 0 in (4.7) that ϕ is degenerate at the point z = z 0 , which contradicts to the fact that ϕ is non-degenerate. Thus, we have l1 dzdz  = 0 everywhere on S 2 . Also from the expression of 0 in (4.7), we get l0 dzdz  = 0 everywhere on S 2 .

From the above discussion, we conclude li dzdz  = 0 (i = 0, 1, 2, . . . , m) everywhere on S 2 , i.e., the harmonic sequence f 0 , f 1 , . . . , f m : S 2 → C P m ⊂ C P n−1 is totally unramified. In this case, we prove Proposition 4.3 Let ϕ : S 2 → G(2, n; R) be a linearly full irreducible totally unramified harmonic map of constant curvature K and finite isotropy order r . Suppose that r ≥ n − 5. (n−1) (n−1) Then n must be odd, and up to an isometry of G(2, n; R), ϕ = U V 1 ⊕ UV1 with 2 for some U ∈ G. K = 3n−5 Proof Since the harmonic sequence f 0 , . . . , f m : S 2 → C P m ⊂ C P n−1 is totally unramified, it follows from (2.13), (4.8) and (4.9) that (m)

δ−1 = δ0 , δ1 = δ2

(m)

= 3(m − 2), δ0

(m)

= m, δ1

= 2(m − 1).

(4.13)

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On the other hand, by (2.8) (2.9) and (4.12) we have

where δi =

1 √ 2π −1

δ1 − 2δ0 + δ−1 = −4,

S2

(4.14)

L i dz ∧ dz, i = −1, 0, 1. Hence, δ0 = 3m − 2.

(4.15)

and This formula and the fact that ϕ is of constant curvature enable us to set K = 2 2 complex coordinate z on C = S \{ pt} can be chosen so that the induced metric ds = 2L 0 dzdz of ϕ is given by 2 3m−2 ,

ds 2 =

2(3m − 2) dzdz, (1 + zz)2

where L0 =

3m − 2 . (1 + zz)2

(4.16)

Consider local lift of the i-th osculating curve Fi = f 0 ∧. . .∧ f i (i = 0, . . . , m). We choose a nowhere zero holomorphic Cn -valued function f 0 , then Fi is a nowhere zero holomorphic (m) curve and it is a polynomial function on C of degree δi satisfying ∂∂ log |Fi |2 = li . So using (4.8) (4.9) (4.10) (4.12) and (4.16), we obtain ∂∂ log From (4.10), we know that

| f 0 |2 l |V |2 0

(1 + zz)δ0 | f 0 |2 = 0. |F0 |6 |V |2

(4.17)

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

it follows from (4.13) and (4.15) that C and has a positive constant limit

1 c

(1+zz)δ0 | f 0 |2 |F0 |6 |V |2

=

(1+zz)δ0 |F0 |2 |F1 |2

·

| f 0 |2 l |V |2 0

is globally defined on

as z → ∞. Thus, (4.17) gives us that

(1 + zz)δ0 | f 0 |2 1 = , |F0 |6 |V |2 c i.e., |V |2 =

c(1 + zz)δ0 . | f 0 |4

(4.18)

Applying the equation V = f 1 + x0 f 0 , equation (4.18) reduces to |x0 |2 | f 0 |4 + |F1 |2 =

c(1 + zz)δ0 . | f 0 |2

(4.19)

By equation (4.6) we get ∂(x0 | f 0 |2 ) = 0. Observing (4.19), from the fact that both |F1 |2 δ0 and (1+zz) have no singular points except z = ∞, we have x0 | f 0 |2 is a holomorphic function | f 0 |2 on C at most with the pole z = ∞. So, it is a polynomial function about z. Without loss of generality, set x0 | f 0 |2 = h(z), (4.20) the formula (4.19) is rewritten as |h|2 + |F1 |2 =

123

c(1 + zz)δ0 . | f 0 |2

(4.21)

Classification of conformal minimal immersions of constant…

1019 (m)

Since both sides of (4.21) are polynomial functions and δ0

= m, then we have

| f 0 |2 = μ(1 + zz)m ,

(4.22)

where μ is a real parameter. If h  = 0, then 1 + zz is a factor of it, which contracts the fact that h is holomorphic. Thus, we have h = 0, which implies that x0 = 0. Then V = f1 , ϕ = f 1 ⊕ f 1 . From (4.22), by Lemma 2.3, up to a holomorphic isometry of C P n−1 , f 0 is a Veronese (m) surface. We can choose a complex coordinate z on C = S 2 \{ pt} so that f 0 = U V0 , where (m) U ∈ U (n) and V0 has the standard expression given in part (C) of Sect. 2 (adding zeros to (m)

V0

(m)

such that V0

(m)

∈ Cn ). Thus, we have ϕ = U V 1

(m)

⊕ U V 1 . To determine ϕ, we just (m)

(m)

need to determine the matrix U . To do this, consider the reducible map U V 0 ⊕ U V 0 : S 2 → G(2, n; R), which is of constant curvature and finite isotropy order r + 2. Thus, Proposition 4.3 immediately follows from Proposition 3.2.   Remark 4.4 In Proposition 4.3, we have m = n − 1. Choose the same U0 as the one shown (n−1) (n−1) in Remark 3.3, then, up to an isometry of G(2, n; R), ϕ = U 0 V 1 ⊕ U 0V 1 with 2 K = 3n−5 . As to the second fundamental form B of ϕ, we have the following corollary. Corollary 4.5 Let ϕ : S 2 → G(2, n; R) be a linearly full irreducible totally unramified harmonic map of constant curvature K and finite isotropy order r . Suppose that r ≥ n − 5.  B2 = 0, n = 5, Then the second fundamental form B of ϕ satisfies , n ≥ 7. B2 = 12(n−2)(n−3) (3n−5)2 Proof For any integer i, 0 ≤ i ≤ m, let (m)

f i = U0 Vi

,

(4.23)

U0 ∈ U (n) is the one shown in Remark 3.3. Up to an isometry of G(2, n; R), ϕ can be expressed as ϕ = f 1 ⊕ f 1 , where f 1 is of constant curvature. Then by (2.7) and a series of calculations, we obtain ⎧ ∂ϕ = | f1|2 [ f 1 ( f 2 )∗ + f 2 f 1∗ ] − | f1|2 [ f 0 ( f 1 )∗ + f 1 f 0∗ ], ⎪ ⎪ 1 0 ⎪ ⎪ ⎨ A = 1 [ f ( f )∗ − f f ∗ ] + 1 [ f ( f )∗ − f f ∗ ], z 2 1 1 0 1 2 0 1 | f 1 |2 | f 0 |2 (4.24) ⎪ P = 0, m = 2, ⎪ ⎪ ⎪ 2 ⎩ P = 2(| f |2 || ff 0||2 +| f |4 ) [ f 1 ( f 3 )∗ − f 3 f 1∗ ], m ≥ 3. 0

2

1

If m = 2, it is easy to see that B2 = 0, i.e., ϕ is totally geodesic. Otherwise if m ≥ 3, we get   l2 2 | f 1 , f 3 |2 2 B = − . | f 2 |4 (1 + l0 )2 l1 l1

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1020

X. Jiao, M. Li

Since f 1 is of constant curvature and m + 1 = n, we have l2 l1

=

(m) δ2 (m) δ1

=

3(n−3) 2(n−2) ,

l0 l1

=

(m)

δ0

(m)

δ1

=

n−1 2(n−2)

and

which implies that 8(n − 2)2 B = (3n − 5)2



2

 3(n − 3) | f 1 , f 3 |2 − . 2(n − 2) | f 2 |4

(4.25)

From the fact that n is odd, we discuss it as follows: (1) If n = 3. ϕ is totally geodesic by (4.24). In this case, ϕ is a conformal minimal immersion from S 2 to G(2, 3; R) (or equivalently Q 1 ), which is trivial, we don’t consider it in this paper; (2) If n = 5. It follows from (4.23) that f 1 = f 3 . Substituting it into (4.25), we obtain B2 = 0, i.e., ϕ is totally geodesic; (3) If n ≥ 7. It should be remarked that r ≥ n − 5 and r is odd, then we have r ≥ 3, which implies that  f 1 , f 3  = 0. Hence, (4.25) reduces to B2 = 12(n−2)(n−3) .   (3n−5)2 By Proposition 3.2, 3.5, 4.3, we conclude a classification of conformal minimal immersions of constant curvature from S 2 to G(2, n; R) as follows: Theorem 4.6 Let ϕ : S 2 → G(2, n; R) be a linearly full conformal minimal immersion with constant curvature K and isotropy order r . Then, up to an isometry of G(2, n; R), (2 p)

(i) If ϕ is reducible and (strongly) isotropic, then, ϕ = U 0 V p 2 p = n − 2, or ϕ =

(m) U 1V 0

(m)

⊕ U 1V 0

with K =

4 n−2 ,

⊕ c0 with K =

8 n(n−2) ,

n = 2m + 2;

(n−1)

(n−1)

(ii) If ϕ is reducible and has finite isotropy order r ≥ n−4, then, ϕ = U 0 V 0 ⊕U 0 V 0 2 with K = n−1 . (iii) If ϕ is totally unramified irreducible and has finite isotropy order r ≥ n − 5, then, (n−1) (n−1) 2 ϕ = U 0V 1 ⊕ U 0V 1 with K = 3n−5 . Theorem 4.6 shows that, up to an isometry of G(2, n; R), conformal minimal immersions of constant curvature from S 2 to G(2, n; R), or equivalently, a complex hyperquadric Q n−2 can be presented by the Veronese surfaces in C P n−1 under three conditions, respectively: the immersion is reducible with finite isotropy order r ≥ n − 4; the immersion is totally unramified irreducible with finite isotropy order r ≥ n − 5; the immersion is reducible and (strongly) isotropic. It is easy to check that these minimal immersions are all homogeneous. By Theorem 4.6, we obtain classifications of linearly full conformal minimal immersions of constant curvature from S 2 to Q 2 and Q 3 , respectively as follows: Corollary 4.7 Let ϕ : S 2 → G(2, 4; R) be a linearly full conformal minimal immersion of constant curvature K . Then, ϕ is totally geodesic and up to an isometry of G(2, 4; R), √ √ (1) ϕ = f 0 ⊕ f 0 with K = 2, f 0 = [(1, −1, z, −1z)T ]; √ √ (2) ϕ = f 1 ⊕ c0 with K = 1, c0 = (0, 0, 0, 1), f 1 = [(z − z, − −1(z + z), −1(1 − zz))T ]. Corollary 4.8 ([16], Theorem 4.9) Let ϕ : S 2 → G(2, 5; R) be a linearly full conformal minimal immersion of constant curvature K . Then, up to an isometry of G(2, 5; R),

123

Classification of conformal minimal immersions of constant…

1021

(1) If ϕ is reducible, then ϕ = f 0 ⊕ f 0 with K = 21 , B2 = 3, √ √ √ f 0 = [(1 + z 4 , −1(1 − z 4 ), 2(z − z 3 ), 2 −1(z + z 3 ), 2 3z 2 )T ]; (2) If ϕ is totally unramified irreducible, then, ϕ = f 1 ⊕ f 1 is totally geodesic with K = 15 , √ 3 − z), −2 −1(z 3 + z), 1 − 3zz − z 2 (3 − zz), f 1 = [((2(z √ √ −1[1 − 3zz + z 2 (3 − zz)], 2 3z(1 − zz))T ]. Irreducible examples. In the following, we give some examples about linearly full irreducible harmonic map ϕ : S 2 → G(2, n; R), which is (strongly) isotropic with constant curvature K and second fundamental form B. Let n = 2m + 2, and take U1 ∈ U (n), which is of the same type as (3.32). Then for any integer i (0 < i < m), (m) (m) ϕ = U 1V i ⊕ U 1V i (4.26) are all linearly full irreducible (strongly) isotropic harmonic maps from S 2 to G(2, n; R). (m) Let f i = U1 Vi , ϕ can be rewritten as ϕ = f i ⊕ f i , where f i is of constant curvature and f 0 , f 1 , . . . , f m , f 0 , f 1 , . . . , f m are mutually orthogonal. Then by (2.7), a series of calculations give ⎧ ∗ ], ∂ϕ = | f1|2 [ f i ( f i+1 )∗ + f i+1 f i∗ ] − | f 1 |2 [ f i−1 ( f i )∗ + f i f i−1 ⎪ ⎪ i i−1 ⎪ ⎪ ⎪ ∗ ], ⎨ A z = 1 2 [ f i ( f i+1 )∗ − f i+1 f i∗ ] + 1 2 [ f i−1 ( f i )∗ − f i f i−1 |f | |f | i

⎪ P= ⎪ ⎪ ⎪ ⎪ ⎩ P=

i−1

1

2| f i

|2 (l

i−1

[ f i ( f i+2 )∗ − f i+2 f i∗ ] − +l ) i

1 [ f 1 ( f 3 )∗ 2| f 1 |2 (l0 +l1 )

and

1 [ f i−2 ( f i )∗ 2(li−1 +li )| f i−2 |2

∗ ], i ≥ 2, − f i f i−2

− f 3 f 1∗ ], i = 1,

⎧ 2 λ = 2(li−1 + li ), ⎪ ⎪ ⎪ ⎨K = 2 4 m+2i(m−i) = n−2+2i(n−2−2i) , ⎪ (m) (m) (m) (m) ⎪ ⎪ i+1 +δi−1 δi−2 ) ⎩ B2 = 2(δi δ(m) . (m) 2 (δi−1 +δi

)

(m)

(m)

Thus, for any integer i (0 < i < m), ϕ = U 1 V i ⊕ U 1 V i is a linearly full irreducible (strongly) isotropic harmonic map from S 2 to G(2, n; R) with constant curvature K = 4 n−2+2i(n−2−2i)

and B2 = 2

(m) (m) (m) (m) δi+1 +δi−1 δi−2 (m) (m) 2 (δi−1 +δi )

δi

(m)

, where δi

= (i + 1)(m − i).

Remark 4.9 Let ϕ be a linearly full conformal minimal immersion of constant curvature from S 2 to G(2, n; R), and f ϕ be the corresponding map of ϕ from S 2 to Q n−2 . Summing the maps ϕ given in Sects. 3 and 4, they can be divided into two classes: (m)

(m)

(1) ϕ = U V i ⊕ U V i for any integer i (0 ≤ i ≤ m < n) and some matrix U ∈ U (n). In this case, the corresponding map of ϕ from S 2 to Q n−2 is as follows (m) T

f ϕ = [(U Vi

) ] : S 2 → Q n−2 ⊂ C P n−1 , (m)

(m)

which is also minimal from S 2 to C P n−1 . In particular, when ϕ = U 1 V i0 ⊕ U 1 V i0 , where U1 is of the one shown in (3.32) and 4i 0 + 2 = 2m + 2 = n, the corresponding f ϕ : S 2 → Q n−2 is also totally real.

123

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X. Jiao, M. Li (n−2)

(2) ϕ = U 0 V p ⊕ c0 , where c0 is a constant vector in Cn and 2 p = n − 2, U0 is of the same type as the matrix shown in Remark 3.3. In this case, the corresponding map of ϕ from S 2 to Q n−2 is as follows √  (n−2) T (n−2) (n−2) , −1 U0 V p , U 0V p ) ] : S 2 → Q n−2 ⊂ C P n−1 . f ϕ = [(U0 V p By a simple test, we can check that f ϕ is a totally real and minimal map from S 2 to Q n−2 , but it is not minimal of S 2 in C P n−1 . To completely classify conformal minimal immersions of constant curvature from S 2 to G(2, n; R), or equivalently, a complex hyperquadric Q n−2 , there are many problems deserving further consideration. (Q1) When the immersion is of finite isotropy order r , our classification theorem is true under assumptions that the immersion is reducible with r ≥ n − 4 and that the immersion is totally unramified irreducible with r ≥ n −5. Then whether it is also true after omitting these assumptions. (Q2) For the (strongly) isotropic case, we determine all reducible (strongly) isotropic conformal minimal immersions of two-spheres in complex hyperquadric with constant curvature. Then what about the irreducible (strongly) isotropic ones? How to determine them? Acknowledgements Project supported by the NSFC (Grant No. 11331002).

References 1. Aithal, A.R.: Harmonic maps from S 2 to G 2 (C 5 ). J. Lond. Math. Soc. 32(2), 572–576 (1985) 2. Bahy-El-Dien, A., Wood, J.C.: The explicit construction of all harmonic two-spheres in G 2 (R n ). J. Reine Angew. Math. 398, 36–66 (1989) 3. Bolton, J., Jensen, G.R., Rigoli, M., Woodward, L.M.: On conformal minimal immersions of S 2 into C P n . Math. Ann. 279, 599–620 (1988) 4. Bryant, R.L.: Minimal surfaces of constant curvature in S n . Trans. Am. Math. Soc. 290, 259–271 (1985) 5. Burstall, F.E., Wood, J.C.: The construction of harmonic maps into complex Grassmannians. J. Diff. Geom. 23, 255–297 (1986) 6. Calabi, E.: Isometric embedding of complex manifolds. Ann. Math. 58, 1–23 (1953) 7. Chern, S.S., Wolfson, J.G.: Harmonic maps of the two-spheres in a complex Grassmann manifold I I . Ann. Math. 125, 301–335 (1987) 8. Chi, Q.S., Zheng, Y.B.: Rigidity of pseudo-holomorphic curves of constant curvature in Grassmann manifolds. Trans. Am. Math. Soc. 313, 393–406 (1989) 9. Erdem, S., Wood, J.C.: On the construction of harmonic maps into a Grassmannian. J. Lond. Math. Soc. 28(2), 161–174 (1983) 10. He, L., Jiao, X.X.: Classification of conformal minimal immersions of constant curvature from S 2 to H P 2 . Math. Ann. 359(3–4), 663–694 (2014) 11. Hoffman, D.A., Osserman, R.: The geometry of the generalized Gauss map. Mem. Am. Math. Soc. 28(236)(1980) 12. Jiao, X.X.: Pseudo-holomorphic curves of constant curvature in complex Grassmannians. Isr. J. Math. 163(1), 45–63 (2008) 13. Jiao, X.X., Li, M.Y.: On conformal minimal immersions of two-spheres in a complex hyperquadric with parallel second fundamental form. J. Geom. Anal. 26, 185–205 (2016) 14. Jiao, X.X., Wang, J.: Minimal surfaces in a complex hyperquadric Q 2 . Manus. Math. 140, 597–611 (2013) 15. Kenmotsu, K., Masuda, K.: On minimal surfaces of constant curvature in two-dimensional complex space form. J. Reine Angew. Math. 523, 182–191 (2000) 16. Li, M.Y., Jiao, X.X., He, L.: Classification of conformal minimal immersions of constant curvature from S 2 to Q 3 . J. Math. Soc. Jpn. 68(2), 863–883 (2016)

123

Classification of conformal minimal immersions of constant…

1023

17. Uhlenbeck, K.: Harmonic maps into Lie groups (classical solutions of the chiral model). J. Diff. Geom. 30, 1–50 (1989) 18. Wang, J., Jiao, X.X.: Totally real minimal surfaces in the complex hyperquadrics. Diff. Geom. Appl. 31, 540–555 (2013) 19. Wolfson, J.G.: Harmonic maps of the two-sphere into the complex hyperquadric. J. Diff. Geom. 24, 141–152 (1986)

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