Results Math (2018) 73:93 Online First c 2018 Springer International Publishing AG,  part of Springer Nature https://doi.org/10.1007/s00025-018-0850-5

Results in Mathematics

Submanifolds with Parallel M¨ obius Second Fundamental Form in the Unit Sphere Zejun Hu and Shujie Zhai Abstract. In this paper, we establish a complete classification of umbilicfree submanifolds of the unit sphere with parallel M¨ obius second fundamental form and arbitrary codimension. Mathematics Subject Classification. Primary 53A30; Secondary 53B25. Keywords. parallel submanifold, M¨ obius second fundamental form, umbilicfree submanifold, Blaschke tensor.

1. Introduction In the theory of submanifolds, the first and second fundamental forms are fundamental invariants. Indeed, associated with the equations of Gauss–Codazzi– Ricci, the two fundamental invariants determine a submanifold up to an isometry of the ambient space. From this standpoint, the classification of parallel submanifolds (i.e. submanifolds with parallel second fundamental form) is a natural and important problem. Though usually very involved, a series of such achievements have been established. We would recall some of the important developments as below. First of all, the classification of parallel submanifolds of Euclidean space is due to Ferus [6–8] (cf. also [23]), he showed that all of them are products of a Euclidean factor and the standard minimal embeddings into hyperspheres of the symmetric R-spaces, which are orbits of special types of s-representations. The classification of parallel submanifolds in a sphere is an easy consequence of that in the Euclidean space, whereas the classification of parallel submanifolds This project was supported by NSF of China, Grant Numbers 11371330 and 11771404. 0123456789().: V,-vol

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of hyperbolic space was carried out independently by Backes and Reckziegel [2], and Takeuchi [24]. Next, in a series of papers [19–22], Naitoh and Takeuchi obtained the classification of parallel Lagrangian submanifolds of the complex projective space CPn . Recently, Dillen, Li, Vrancken and Wang [5] explicitly and geometrically give another classification of parallel Lagrangian submanifolds in CPn using a different method, which applies Calabi products to characterize parallel Lagrangian submanifolds. In the M¨ obius geometry of submanifolds in the unit sphere, established systematically by Wang [25], a totally similar classification problem can be raised. Let M n be an n-dimensional immersed umbilic-free submanifold in the (n + p)-dimensional unit sphere Sn+p (n ≥ 2, p ≥ 1), then M n is associated with the so called M¨ obius metric g, the M¨ obius second fundamental form B, the M¨ obius form Φ and the Blaschke tensor A, which are all invariants of obius transformation group of the ambient space Sn+p . M n M n under the M¨ is called a M¨ obius parallel submanifold means that, as an immersed submanifold it possesses parallel M¨obius second fundamental form, i.e. it satisfies the ¯ ≡ 0, where ∇ ¯ is the usual covariant differentiation with respect condition ∇B to the Levi-Civita connection of the M¨ obius metric g on T M n and the normal ⊥ n obius geometry of connection on T M . Obviously, from the standpoint of M¨ submanifolds, the classification of M¨ obius parallel submanifolds in Sn+p is a very natural and important problem. In fact, it has became a challenging problem in M¨ obius submanifolds geometry since the initial research of Li and the first author [10], where a complete classification for the hypersurface situation (i.e. p = 1) is established. For higher codimension case, Zhai, Hu and Wang [26] made an interesting breakthrough to the problem by showing that the assumption “parallel M¨ obius second fundamental form” implies that the Blaschke tensor is parallel (cf. Theorem 4.1 in [26]). This latter property plays a crucial role for us to know further about M¨ obius parallel submanifolds. In [26], by using the special character of p = 2, the authors classified all codimension two M¨ obius parallel submanifolds in the unit sphere. The aim of this paper is to work for a final solution of the problem for general codimensions. Indeed, using the method of moving frames and developing the algebraic techniques of Cecil and Jensen [4], we succeed in obtaining the complete classification of M¨ obius parallel submanifolds of arbitrary codimension. The main result can be stated as follows: Classification Theorem. Let x : M n → Sn+p be an n-dimensional umbilic-free submanifold with parallel M¨ obius second fundamental form. Then x is M¨ obius equivalent to an open part of one of the following submanifolds: (i) an umbilic-free submanifold of Sn+p with parallel second fundamental form,

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(ii) the image of σ of an umbilic-free submanifold of the Euclidean space Rn+p with parallel second fundamental form, (iii) the image of τ of an umbilic-free submanifold of the hyperbolic space Hn+p with parallel second fundamental form, (iv) the image of σ of a cone over some k-dimensional (k ≤ n−1) umbilic-free submanifold of Sk+p with parallel second fundamental form. Remark 1.1. In the above Classification Theorem, the two maps σ and τ denote the canonical conformal diffeomorphisms between the space forms whose definitions are stated in (2.1) and (2.2), respectively. Remark 1.2. After the completion of the present paper, Li and Song obtained an interesting complete classification of Blaschke parallel submanifolds in the unit sphere with vanishing M¨ obius form, see [16] for its published version. Although we have shown that parallel M¨ obius second fundamental form implies that the Blaschke tensor is parallel, comparing with the statement and proof of the above Classification Theorem with that of the main results in [16], one sees that both are quite different and are of independent meaning. This paper is organized as follows. In Sect. 2, we review some elementary facts of M¨ obius geometry of submanifolds in Sn+p . In Sect. 3, we establish the M¨ obius characterizations for two subclasses of submanifolds in the unit sphere with parallel M¨ obius second fundamental form. In Sect. 4, we made the preliminary discussions on submanifolds with parallel M¨ obius second fundamental form, and importantly we show that the number t of distinct Blaschke eigenvalues is at most p + 2. In Sects. 5 and 6, we settle two special cases t = 2 and t = p + 2, respectively. Then, in Sect. 7, we focus on the remaining much involved situation 3 ≤ t ≤ p + 1. Finally and as a summary, we complete the proof of Classification Theorem in Sect. 8.

2. M¨ obius Geometry of Submanifolds in Sn +p In this section, following Wang’s seminal article [25], we review the basic M¨ obius invariants and recall the structure equations for submanifolds in Sn+p . Let Hn+p (r) be a hyperbolic space of constant sectional curvature −r−2 defined by Hn+p (r) = {(y0 , y1 ) ∈ R+ × Rn+p | − y02 + y1 · y1 = −r2 }, where · denotes the canonical Euclidean inner product. Let Sn+p + (r) denote the open hemisphere in Sn+p (r) whose first coordinate is positive. For simplicity, we write Hn+p (1) = Hn+p , Sn+p (1) = Sn+p . Then we can define two conformal diffeomorphisms (cf. [18]) σ : Rn+p → Sn+p \{(−1, 0)} by

and τ : Hn+p → Sn+p +

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 σ(u) = τ (y0 , y1 ) =



1−|u|2 2u 1+|u|2 , 1+|u|2 1 y1 y0 , y0



,

 ,

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u ∈ Rn+p ,

(y0 , y1 ) ∈ Hn+p .

(2.1) (2.2)

By σ and τ , submanifolds of Rn+p and Hn+p can be regarded as ones of Sn+p . Let x : M n → Sn+p be an n-dimensional umbilic-free submanifold in n+p . We use the following range of indices throughout this paper, if not S stated otherwise: 1 ≤ i, j, k, l, m ≤ n,

n + 1 ≤ α, β, γ ≤ n + p.

Choose a local orthonormal basis {ei } for T M n with respect to the induced metric I = dx · dx with dual basis {θi }, and a local orthonormal normal basis form and the mean curvature vector {eα } of x, then the second fundamental   hα θi θj eα and H = n1 i,α hα eα := of x can be expressed as II = ij ii i,j,α  α H e , respectively. α α Let Rn+p+2 be the Lorentz space with standard inner product ·, · 1 . The 1 of x is defined by M¨ obius position vector Y : M n → Rn+p+2 1   n

II 2 − n H 2 > 0. (2.3) Y = ρ(1, x), with ρ2 = n−1 Then we have the following: Theorem 2.1 (cf. [25]). Two umbilic-free submanifolds x, x ˜ : M n → Sn+p are M¨ obius equivalent if and only if there exists T in the Lorentz group O(n + p + such that Y = Y˜ T . 1, 1) on Rn+p+2 1 obius It follows from Theorem 2.1 that g = dY, dY 1 = ρ2 dx · dx is a M¨ invariant, which is called the M¨ obius metric. Let Δ and R denote the Laplacian and the normalized scalar curvature of the M¨ obius metric g, respectively. We define N = − n1 ΔY −

1 2n2 (1

+ n2 R)Y.

(2.4)

Then we have Y, Y 1 = 0, N, Y 1 = 1, N, N 1 = 0. Put Ei = ρ−1 ei , ωi = ρθi , Eα = (H α , H α x + eα ).

(2.5)

Then {Ei } is a local orthonormal frame field with respect to g, with dual frame {ωi }, and {Eα } is a local orthonormal frame of the M¨ obius normal bundle of x. Denote Yi = Ei (Y ). Then we have Yi , Y 1 = Yi , N 1 = 0, Yi , Yj 1 = δij and {Y, N, Y1 , . . . , Yn , En+1 , . . . , En+p } forms a moving frame in Rn+p+2 1 along M n .

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We can write the structure equations as follows:    dY = Yi ωi , dN = Aij ωj Yi + Ciα ωi Eα , i

dYi = −



i,j

Aij ωj Y − ωi N +

j

dEα = −



Ciα ωi Y −

i





i,α

ωij Yj +

j α Bij ωj Yi +



i,j



α Bij ω j Eα ,

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(2.6) (2.7)

j,α

ωαβ Eβ ,

(2.8)

β

where {ωij } is the connection forms with respect to the M¨obius metric g, and {ωαβ } is the normal connection forms of x : M n → Sn+p . The M¨ obius invariants    α Bij ω i ω j Eα , A = Aij ωi ωj and Φ = Ciα ωi Eα B= i,j,α

i,j

i,α

are called the M¨ obius second fundamental form, the Blaschke tensor and the M¨ obius form of x, respectively. The relations between B, A and the associated Euclidean invariants of x are given by (cf. [25])   α α (2.9) Bij = ρ−1 hα ij − H δij ,   Aij = −ρ−2 Hessij (log ρ) − ei (log ρ)ej (log ρ) − (2.10) H α hα ij   − 12 ρ−2 ∇(log ρ) 2 − 1 + H 2 δij ,

α

where Hessij and ∇ are the Hessian matrix and the gradient with respect to dx · dx. The components of the covariant differentiation of B is defined by     β α α α α Bij,k ωk = dBij + Bik ωkj + Bkj ωki + Bij ωβα . (2.11) k

k

k

β

Among the integrability conditions for the structure Eqs. (2.6)–(2.8), we have the following (cf. [25]):   α α α Bik Rijkl = Bjl − Bilα Bjk α

  + δik Ajl + δjl Aik − δil Ajk − δjk Ail ,   α 2 α (Bij ) = n−1 Bii = 0, ∀ α, n , i,j,α

(2.12) (2.13)

i

where Rijkl denotes the components of the Riemannian curvature tensor of g.

3. M¨ obius Characterization of Typical Reducible Submanifolds To prove the Classification Theorem, a crucial step is to establish the M¨ obius characterizations for both (Euclidean) parallel submanifolds in space forms

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and cones over parallel submanifolds in spheres. This will be fulfilled in this section. First, the characterization of the (Euclidean) parallel submanifolds in space forms in terms of their M¨ obius invariants can be stated as follows. Theorem 3.1. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form. Then x is M¨ obius equivalent to an open part of one of the following submanifolds: (i) an umbilic-free submanifold of Sn+p with parallel second fundamental form, (ii) the image of σ of an umbilic-free submanifold of the Euclidean space Rn+p with parallel second fundamental form, (iii) the image of τ of an umbilic-free submanifold of the hyperbolic space Hn+p with parallel second fundamental form, if and only if there exist λ ∈ R and a parallel M¨ obius normal vector field ξ such that the Blaschke tensor A, the M¨ obius second fundamental form B, and the M¨ obius metric g of x satisfy the relation A + λg + B, ξ 1 = 0.

(3.1)

¯ = 0 implies that Φ = 0 (cf. Theorem 4.1 of [26]). Proof. The assumption ∇B Then, according to Theorem 2.2 of [17] (cf. also [13]), if there exist a constant λ and a parallel M¨ obius normal vector field ξ such that (3.1) holds, then x is M¨ obius equivalent to either a submanifold with constant scalar curvature and parallel mean curvature vector field in Sn+p , or the image of σ (or τ ) of a submanifold with constant scalar curvature and parallel mean curvature vector field in Rn+p (or Hn+p ). In each of the above possibilities, the facts that the mean curvature vector field being parallel imply the length of  it is2 constant.  n

II − n H 2 Therefore, by the Gauss equation, we can show that ρ2 = n−1 is constant. Then, according to Remark 3.2 of [26], x is M¨obius equivalent to a parallel submanifold. Conversely, if x is a parallel submanifold in Sn+p , then it follows from Proposition 3.1 of [26] that ρ is constant. Now, choosing  1 λ = − (1 + H 2 ), ξ = −ρ−1 H α Eα = −ρ−1 H, 2 α and taking account of the relations (2.9) and (2.10), we obtain A + λg +B, ξ 1 = 0. Similar discussions show that if x is M¨ obius equivalent to either the image of σ (or τ ) of a parallel submanifold of Rn+p (or Hn+p ), then (3.1) holds for some constant λ and some parallel M¨ obius normal vector field ξ.  Next, we consider the second construction of so called generalized cone, which is the extension of the usual notion of cone in Euclidean space.

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Definition 3.1. For an immersed m-dimensional submanifold u : M m → Sm+p and an integer n ≥ m + 1, the immersed submanifold, defined by C : M m × R+ × Rn−m−1 → Rm+p+1 × Rn−m−1 = Rn+p with C(q, t, v) = (tu(q), v) for q ∈ M m , t ∈ R+ and v ∈ Rn−m−1 , is called a cone over u in Rn+p . A M¨ obius parallel submanifold may be not of (Euclidean) parallel second fundamental form. A typical example is that, by Definition 3.1, over a parallel submanifold of Sr+p (r ≤ n − 1) we have a “cone” in Rn+p . Proposition 3.4 of [26] shows that the image of σ of such cone is a M¨obius parallel submanifold in Sn+p . However, it is easy to check that it is not a (Euclidean) parallel submanifold. The following result, essentially due to Li, Ma and Wang, provides a nice criterion for judging whether a submanifold is M¨ obius equivalent to a cone in Rn+p . Proposition 3.1 (cf. Proposition 4 of [14]). Let x : M n → Sn+p be an immersed umbilic-free submanifold. If there exist an integer m ≤ n−1 and a submanifold ˜ m → Sm+p (r) such that the M¨ u:M obius position vector field Y of x can be written as ˜ m → Rn−m+1 × Sm+p (r) → Rn+p+2 , Y = (y, u) : Hn−m (r) × M 1 1 is the standard embedding as preceding where y : Hn−m (r) → Rn−m+1 1 stated. Then x is M¨ obius equivalent to the image of σ of the cone C over u in Rn+p . Proof. This is in fact a direct calculations and together with an application of Theorem 2.1. For reader’s convenience we include it with details. The standard embedding y : Hn−m (r) → Rn−m+1 can be parameterized 1 n−m+1 + n−m−1 → R1 , where by y : R × R  1 + r2 t2 + |v|2 1 − r2 t2 − |v|2 v  , , . y(t, v) = 2t 2t t Then, by writing  1 + r2 t2 + |v|2 1 − r2 t2 − |v|2 v Y = (y, u) = , , u, , (3.2) 2t 2t t ˜ m × R+ × Rn−m−1 , and x is parameterized we have the identification M n = M by

 1 − r2 t2 − |v|2 2tu(q) 2v x(q, t, v) = , , . (3.3) 1 + r2 t2 + |v|2 1 + r2 t2 + |v|2 1 + r2 t2 + |v|2 This implies that x(q, t, v) = σ((tu(q), v)), i.e. x is the image of σ of the cone ˜ m → Sm+p (r) in Rn+p .  over the submanifold u : M

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4. The Number of Distinct Blaschke Eigenvalues for M¨obius Parallel Submanifolds The main goal of this section is to prove that, for an n-dimensional M¨ obius parallel submanifold in Sn+p , its Blaschke tensor has at most (p + 2) distinct eigenvalues. To begin with, we first recall the following crucial result. Lemma 4.1 (cf. [26]). Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form. Then M n is of vanishing M¨ obius form Φ and parallel Blaschke tensor A. It follows that all eigenvalues of A are constant, we assume that t A has t distinct eigenvalues {A1 , . . . , At } and each As is of multiplicity ms , s=1 ms = n, and, around each point we can choose a local orthonormal frame field {Ei } such that A is diagonalized, without loss of generality, we assume that Aij = λi δij ,

(Aij ) = diag(A1 , . . . , A1 , A2 , . . . , A2 , . . . , At , . . . , At ).       m1

m2

(4.1)

mt

For each 1 ≤ s ≤ t, let Ds denote the eigen-distribution corresponding to As , then Ds is integrable. If we denote by Msms the ms -dimensional integral manifold of Ds , then (M n , g) can be locally decomposed as (set m0 := 0): (M n , g) = (M1m1 , g1 ) × · · · × (Mtmt , gt ), gs =

m1 +···+m  s

ωi2 .

i=m1 +···+ms−1 +1

Next, following the technique in [26] (the technique, which we originally learned from the paper of Cecil and Jensen [4], has been extended and successfully applied several times in other problems, cf. [9,11,12]), we introduce the set of vectors: vij =

n+p 

n+p n+1 n+2 α Bij Eα := (Bij , Bij , . . . , Bij ), 1 ≤ i, j ≤ n.

α=n+1

We will look { vij } as vectors in the p-dimensional Euclidean space Rp equipped with the canonical inner product “·”. Then (2.12) can be rewritten as   (4.2) Rijkl = vik · vjl − vil · vjk + (λi + λj ) δik δjl − δil δjk . For brevity, we denote [i] = {j | λj = λi } and sometimes we denote A[i] := λj for j ∈ [i]. We will write [i] = [j] if λi = λj , and [i] = [j] if otherwise. In [26], it was shown that the vectors { vij } have very nice properties.

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Lemma 4.2 ([26]). The vectors { vij } have the following properties: vij = 0,

if [i] = [j];

(4.3)

vii · vjj = −(λi + λj ),

if [i] = [j];

(4.4)

vii · vjl = 0,

if [i] = [j] = [l] and j = l;

(4.5)

vik · vjl = 0,

if [i] = [k] = [j] = [l], i = k and j = l.

(4.6)

For later’s convenience, we introduce further notations: Is = {i | λi = As }, Vs = { vii | i ∈ Is }, rank (Vs ) = rs , 1 ≤ s ≤ t. To explore the possible values attainable by t, we fix an arbitrary element s¯ ∈ Is for each 1 ≤ s ≤ t. For simplicity, we regard the vector vs¯s¯ as a point − → qs ∈ Rp , so that for O ∈ Rp the origin, we also have vs¯s¯ = Oq s , s = 1, 2, . . . , t. Now, we consider the set {q1 , . . . , qt }. According to (4.4), if t ≥ 4, then any four points qi , qj , qk , q ∈ {q1 , . . . , qt } satisfy (cf. also the proof of Lemma 5.1 in [26]): q−−→ q · q−−→ q = 0, − q−→ q ·− q−→ q = 0, − q−→ q ·− q−→ q = 0. (4.7) i j

k 

i k

j 

i 

j k

Based on the above fact, we can extend Lemma 5.1 of [26] to prove the following. Proposition 4.1. For an immersed umbilic-free submanifold x : M n → Sn+p with parallel M¨ obius second fundamental form, the number t of distinct Blaschke eigenvalues is at most p + 2. Proof. If p = 1, the assertion follows from Proposition 5.1 of [10]. If p = 2, according to Lemma 5.1 of [26], we know that t ≤ 4. Moreover, the proof of Lemma 5.1 in [26] shows that if t = 4, then any three points of {q1 , q2 , q3 , q4 } are non-collinear which form a triangular whose orthocenter is the fourth point. If p ≥ 3, we need only to consider the case t ≥ 5. Then, by use of (4.4), we see that the set {q1 , . . . , qt } consists of mutually distinct points in Rp . Obviously, the conclusion of Proposition 4.1 is a direct consequence of the following lemma.  Lemma 4.3. Let q1 , . . . , qt ∈ Rd (d ≥ 2) be t distinct points, if any four qi , qj , qk , ql of them satisfy the orthogonal conditions (4.7), then t ≤ d + 2. Moreover, if t = d + 2, then any d + 1 points of {q1 , q2 , . . . , qd+2 } span a d-dimensional simplex with the last point being the orthocenter. Proof of Lemma 4.3. We will prove the assertion by induction on d. From the proof of Lemma 5.1 in [26], the assertion holds for d = 2. Next, we assume that the assertion holds for d = s − 1 ≥ 2. To prove the assertion for the case d = s, we argue by assuming that there are s + 2 mutually distinct points {q1 , . . . , qs+2 } ⊂ Rs which satisfy (4.7).

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According to (4.7), it is easily seen that the points {q1 , . . . , qs+2 } ⊂ Rs satisfy the following relations: ⎧ −−−−→ → −−−−→ q− ⎪ qs+2 q1 ⊥ { − 2 q3 , . . . , q2 qs+1 } , ⎪ ⎨ −−−−→ − − → −−→ q− qs+2 q2 ⊥ { q1 q3 , . . . , − 1 qs+1 } , (4.8) ······ ······ ⎪ ⎪ ⎩ −−−−−→ − − → − − → ⊥ {q q , ..., q q }. q q s+2 s+1

1 2

Then we can state the following result.

1 s



Claim 4.1. The s + 1 points q1 , . . . , qs+1 span an s-dimensional simplex in Rs . In fact, if otherwise the points q1 , . . . , qs+1 lie in an (s − 1)-dimensional affine space As−1 ⊂ Rs , then, by the inductive assumption, there are at most s+1 points in As−1 which satisfy (4.7), and that any s points of {q1 , . . . , qs+1 } span an (s − 1)-dimensional simplex with the remaining unique point as its / As−1 . Thus by (4.8) we know that, starting from orthocenter. Moreover, qs+2 ∈ −−→ −−−−−→ q− the same point qs+2 , the s+1 vectors − s+2 q1 , . . ., qs+2 qs+1 are all perpendicular s−1 to the (s − 1)-dimensional affine space A . This is impossible, and the Claim is verified. Having shown that {q1 , q2 , . . . , qs+1 } span an s-dimensional simplex S in Rs , we deduce from (4.7) again that qs+2 is the orthocenter of S. Since the orthocenter (if it does exist) of any s-dimensional simplex in Rs is unique, there can exist at most s + 2 distinct points in Rs such that (4.7) is satisfied. This completes the proof of Lemma 4.3. We have completed the proof of Proposition 4.1.  The preceding discussions imply the following result, which is of independent meaning and will be used several times in later sections. Corollary 4.1. Assume that rank{ v¯1¯1 , . . . , vt¯t¯} = d. If there are distinct real numbers a1 , . . . , at such that for all i = j it holds v¯i¯i · v¯j¯j = ai + aj , then d ≥ t − 2. We recall that, in Theorem 4.2 of [26], it was shown that if an umbilic-free obius second fundamental form submanifold x : M n → Sn+p is of parallel M¨ with t = 1, then x is M¨ obius equivalent to an umbilic-free minimal submanifold of Sn+p with (Euclidean) parallel second fundamental form. Applying Proposition 4.1, we see that to deal with the remaining possibility of t ≥ 2, we may divide the discussions into three cases: Case I. t = 2. Case II. t = p + 2. Case III. 3 ≤ t ≤ p + 1. We will deal with each of these cases in the next three sections, respectively.

5. Case I: t = 2 In this section, we deal with the case t = 2. For simplicity, we separate it into two subcases:

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I-(ii). A1 + A2 = 0.

I-(i). A1 + A2 = 0;

In sequel of this section, we further make the following indices convention: 1 ≤ a, b, c ≤ m1 ; m1 + 1 ≤ μ, ν ≤ m1 + m2 = n.

(5.1)

From (4.3), we see that B=



α Bab ω a ω b Eα +



α Bμν ωμ ων Eα := B(1) + B(2) .

μ,ν,α

a,b,α

We first deal with case I-(i). Lemma 5.1. If t = 2 and A1 + A2 = 0, then we have either B(1) = 0 and A1 < 0, or B(2) = 0 and A2 < 0. Proof. Since B = 0, without loss of generality we assume B(1) = 0. Then, we claim that B(2) = 0. To verify the claim, we suppose on the contrary that B(2) = 0. By Lemma 4.2 and the condition A1 + A2 = 0, we have vab · vμν = 0, ∀ a, b ∈ I1 , μ, ν ∈ I2 .

(5.2)

Put p1 = rank { vab , a, b ∈ I1 }, p2 = rank { vμν , μ, ν ∈ I2 }. From the fact B(1) = 0 and B(2) = 0, we have 1 ≤ p1 , p2 ≤ p − 1. Moreover, because of (5.2), by making a transformation if necessary, we can choose the orthonormal basis {En+1 , . . . , En+p } of the normal bundle such that the following hold: 

n+p1 n+1 vab = (Bab , . . . , Bab ; 0, . . . , 0),

vμν = (0, . . . ,

n+p1 +1 0; Bμν ,

...,

n+p1 +p2 Bμν ; 0,

∀ a, b ∈ I1 , . . . , 0),

∀ μ, ν ∈ I2 .

(5.3)

Now, we put J1 = {n + 1, . . . , n + p1 },

J2 = {n + p1 + 1, . . . , n + p1 + p2 }.

¯ = 0 and (5.3), we have From (2.11), using ∇B 0=

 k

β Bab,k ωk =



α Bab ωαβ , ∀ a, b ∈ I1 , β ∈ J2 .

(5.4)

α∈J1

For each fixed β, we can look (5.4) as a system of linear equations with unknowns {ωαβ , α ∈ J1 }. Since the coefficients matrix is non-singular, we have ωαβ = 0 for all α ∈ J1 , β ∈ J2 . Then from the structure Eqs. (2.6)–(2.8) we

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have:     ⎧ dY = ωa Ya + ωμ Yμ , dN = A1 ωa Ya + A2 ωμ Yμ , ⎪ ⎪ ⎪ ⎪ a∈I1 μ∈I2 a∈I1 μ∈I2 ⎪ ⎪ ⎪   ⎪ ⎪ ⎪ Y + N ) = 2A ω Y , d(A Y + N ) = 2A ωμ Yμ , d(A ⎪ 1 1 a a 2 2 ⎪ ⎨ a∈I1 μ∈I2   α ⎪ ⎪ ωij Yj + Bij ωj Eα , i ∈ Is , s = 1, 2, dYi = −(As Y + N )ωi + ⎪ ⎪ ⎪ ⎪ j∈I j∈I ,α∈J s s s ⎪ ⎪   ⎪ ⎪ α ⎪ Bij ωj Yi + ωαγ Eγ , α ∈ Js , s = 1, 2. ⎪ ⎩ dEα = − i,j∈Is

γ∈Js

(5.5) The above equations imply that along M n the following distributions Vs = Span{As Y + N, Yi , Eα | i ∈ Is , α ∈ Js }, s = 1, 2 , which are orthogonal, i.e., V1 ⊥ V2 . define two parallel subspaces in Rn+p+2 1 Y +N A2 Y +N Now, we put u1 = A12A and u 2 = 2A2 . As A1 + A2 = 0, we have 1 to consider two possibilities: Case I-(i)-(1): A1 < 0;

Case I-(i)-(2): A1 > 0.

1 1 Case I-(i)-(1). In this case, u1 , u1 1 = 2A < 0 and u2 , u2 1 = − 2A > 0. It 1 1 follows that V2 is a spacelike subspace and V1 is a Lorentz subspace in Rn+p+2 . 1 m1 +p1 +1 ⊥ m2 +p2 +1 n We denote V1 = R1 and V2 = V1 = R . We see that, along M , D1 = Span{Y1 , . . . , Ym1 } and D2 = Span{Ym1 +1 , . . . , Yn } are both integrable distributions. Let M1m1 and M2m2 denote their maximal integral submanifolds, respectively. From (5.5), we get  ⎧ du1 = ωa Ya , ⎪ ⎪ ⎪ ⎪ a∈I1 ⎪ ⎪ ⎪   ⎨ α dYa = ωab Yb + Bab ωb Eα − 2A1 ωa u1 , a ∈ I1 , ⎪ b∈I b∈I ,α∈J 1 1 1 ⎪ ⎪   ⎪ ⎪ α ⎪ Bab ωb Ya + ωαγ Eγ , α ∈ J1 , ⎪ dEα = − ⎩ a,b∈I1

γ∈J1

which shows that u1 defines an immersion  1 +p1 +1 . u1 : M1m1 → Hm1 +p1 (1/ −2A1 ) → Rm 1  2 the second g¯1 = du1 , du1 1 = a ωa = g1 and h1 = B(1) are the firstα and  α = i Bii =0 fundamental forms of u1 , respectively. From the facts a Baa ¯ and ∇B = 0, we see that u1 is a minimal hypersurface with parallel second fundamental form. By Lemma 4.3 of [26] (or [24], or Lemma 3.7.10 of [3]) we get B(1) = 0, which contradicts the assumption B(1) = 0. Case I-(i)-(2). In this case, as A1 > 0, by (5.5) and similar arguments to the previous case we see that u2 defines an immersion of M2m2 into Hm2 +p2 (1/

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√ −2A ), whose first and second fundamental forms are g¯2 = du2 , du2 1 =  22   β (2) β , respectively. From the facts μ Bμμ = i Bii =0 μ ωμ = g2 and h2 = B ¯ = 0, we see that u2 is a minimal hypersurface with parallel second and ∇B fundamental form, and also by Lemma 4.3 of [26], we get B(2) = 0, which contradicts the assumption B(2) = 0. In conclusion, we have proved that if B(1) = 0, then B(2) = 0 holds. Next, for B(1) = 0 and B(2) = 0, we will show that A2 < 0. Suppose on the contrary that A1 < 0. Then similar to the preceding arguY +N defines an immersion from the integral maniments we see that u1 = A12A 1 √ m1 fold M1 of D1 := Span{Y1 , . . . , Ym1 } into Hm1 +p (1/ −2A1 ) → R1m1 +p+1 = V1 , which, in fact, is a minimal submanifold with parallel second fundamental form. Finally, by Lemma 4.3 of [26], we get B(1) = 0, which gives the desired contradiction. We have completed the proof of Lemma 5.1.  Proposition 5.1. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and that its Blaschke tensor has two distinct eigenvalues A1 and A2 of multiplicities m1 and m2 , respectively. obius equivalent √ to the image of σ of a cone over If A1 = −A2 > 0, then x is M¨ a minimal submanifold u1 : M1m1 → Sm1 +p (1/ 2A1 ) with parallel (Euclidean) second fundamental form. Proof. According to Lemma 5.1, we have B(1) = 0 and B(2) = 0. Then we can rewrite the structure equations as follows:   ⎧ d(A1 Y + N ) = 2A1 ωa Ya , d(A2 Y + N ) = 2A2 ωμ Yμ , ⎪ ⎪ ⎪ ⎪ a∈I1 μ∈I2 ⎪ ⎪ ⎪ ⎪ ⎪ dY = −(A Y + N )ω +  ω Y +  B α ω E , a ∈ I , ⎪ ⎪ a 1 a ab b 1 ab b α ⎪ ⎨ b∈I1 b∈I1 ,α∈J (5.6)  ⎪ ⎪ dY = −(A Y + N )ω + ω Y , μ ∈ I , μ 2 μ μν ν 2 ⎪ ⎪ ⎪ ⎪ ν∈I2 ⎪ ⎪   ⎪ ⎪ α ⎪ Bab ωb Ya + ωαγ Eγ , ∀ α. ⎪ ⎩ dEα = − a,b∈I1

γ

As A1 > 0, we can write, by the notations after (5.5), V1 = Rm1 +p+1 and 5.1 that V2 = V1⊥ = R1m2 +1 . Then, similar arguments as in the proof of Lemma √ m1 Y +N m1 +p defines an immersion of M into S (1/ 2A ) show that u1 := A12A 1 → 1 1 m1 +p+1 , which is a minimal submanifold with parallel second fundamental R form. Y +N From (5.6), we also see that u2 := A22A defines a totally geodesic 2 √ m2 m2 +1 the hyperbolic space H (1/ 2A1 ) → R1m2 +2 with immersion of M2 into  2 g¯2 := du2 , du2 1 = μ ωμ = g2 . This and (4.2) also imply that (M2m2 , g¯2 ) is of constant sectional curvature 2A2 = −2A1 < 0. Hence u2 defines a locally inclusion mapping

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u2 =

A2 Y +N 2A2

Results Math

 : M2m2 ∼ = Hm2 (1/ 2A1 ) → R1m2 +1 .

Y +N Y +N From u1 = A12A and u2 = A22A , we get Y = u1 + u2 . 1 2 According to the above discussions, we have shown that Y defines the following immersion:   Y = (u1 , u2 ) : M1m1 × Hm2 (1/ 2A1 ) → Sm1 +p (1/ 2A1 ) × R1m2 +1

→ Rn+p+2 . 1 From this and Proposition 3.1, we get the assertion of Proposition 5.1.



To settle Case I, we next deal with case I-(ii). Proposition 5.2. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and whose Blaschke tensor has obius equivalent two distinct eigenvalues A1 and A2 . If A1 +A2 = 0, then x is M¨ to an open part of one of the immersions as stated in (i)–(iii) of Classification Theorem. Proof. From (4.4) and that A1 + A2 = 0, we see that vii = 0 for any i. Notice that the ranks r1 and r2 of the sets V1 and V1 could depend on the choice of the orthonormal frame fields of the tangent bundle, nevertheless they are independent of the choice of the orthonormal frame field of the normal bundle. Obviously, we have 1 ≤ r1 , r2 ≤ p. Choose local orthonormal basis {Eα } of the normal bundle, such that n+1 n+r1 vaa = (Baa , . . . , Baa , 0, . . . , 0), ∀ a ∈ I1 .

(5.7)

From (4.4), we get ( vμμ − vnn ) · vaa = 0, ∀ μ ∈ I2 , a ∈ I1 , and therefore, by using (5.7) and rank { vaa , a ∈ I1 } = r1 , we deduce n+1 n+1 n+2 n+2 n+r1 n+r1 Bμμ = Bnn , Bμμ = Bnn , . . . , Bμμ = Bnn , ∀ μ ∈ I2 .

(5.8)

The fact vnn · vaa = −(A1 +A2 ) = 0, which implies that is not zero, allows us to make an orthogonal transformation of the normal frame such that n+1 n+r1 nn En+1 + · · · + Bnn En+r1 n+1 = B . E n+r1 2 n+1 2 (Bnn ) + · · · + (Bnn )

n+1 n+r1 (Bnn , . . . , Bnn )

It follows that, if necessary by making an orthogonal transformation of the normal frame, components of the tensor B can be assumed to satisfy n+1 n+2 n+r1 , Baa , . . . , Baa , 0, . . . , 0), vaa = (B11

∀ a ∈ I1 ,

vμμ =

∀ μ ∈ I2 .

n+1 (Bnn ,

0, . . . , 0,

n+r1 +1 Bμμ ,

...,

n+p Bμμ ),

(5.9)

n+1 Here, by vnn · vaa = −(A1 + A2 ) = 0 again, we have Bnn = 0; and for each n+1 n+1 a ∈ I1 it holds Baa = B11 , this is due to (4.4) and the fact n+1 n+1 n+1 0 = ( vaa − v11 ) · vnn = (Baa − B11 )Bnn , ∀ a ∈ I1 .

Submanifolds with Parallel M¨ obius

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Now the second equation of (2.13) becomes equivalently   vaa + vμμ = 0. a∈I1

(5.10)

μ∈I2

This fact, combining with the expressions in (5.9), shows that   α β Baa = 0, Bμμ = 0, n + 2 ≤ α, β ≤ n + p. a∈I1

93

(5.11)

μ∈I2

Thus, by making inner product between (5.10) and vnn , together with using (4.4), (5.9) and (5.11), we obtain  n+1 2 vμμ = −m1 (A1 + A2 ) + m2 (Bnn ) = 0. −m1 (A1 + A2 ) + vnn · μ∈I2

This implies that n+1 = constant. Bnn

(5.12)

Moreover, from the fact n+1 n+1 −(A1 + A2 ) = v11 · vnn = B11 Bnn = 0,

we obtain n+1 B11 = constant = 0. (5.13) n+1 n+1 n+1 n+1 The fact m1 B11 + m2 Bnn = 0 and B11 = 0 implies that B11 = n+1 . Hence we have two constants λ and ξ 1 , which satisfy the equations Bnn  n+1 1 A1 + λ + B11 ξ = 0, (5.14) n+1 1 ξ = 0. A2 + λ + Bnn

From (4.5), (5.9), (5.10) and (5.11), together with that rank (V1 ) = r1 , we get n+1 = 0, if a = b, Bab (5.15) n+1 n+2 n+r1 if μ = ν. Bμν = Bμν = · · · = Bμν = 0, Now, we define ξ := ξ 1 En+1 . Then from (5.14) and (5.15) it holds A + λg + B, ξ 1 = 0. Claim 5.1. En+1 is a parallel M¨ obius normal vector field, i.e. ⊥

∇ En+1 =

n+p 

ω(n+1)β Eβ = 0.

β=n+1

To verify the claim, we first notice from (5.15) that n+r1 +1 n+p , . . . , Bμν ), ∀ μ, ν ∈ I2 , μ = ν. vμν = (0, . . . , 0, Bμν

Next, we denote n+1 w μμ = vμμ − (Bnn , 0, . . . , 0), ∀μ ∈ I2 ,

and define p1 := rank {w μμ , vην | μ, η, ν ∈ I2 , η = ν}.

(5.16)

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For each vector of { vab | a, b ∈ I1 , a = b}, we make the decomposition vab = u ab + w ab ,

(5.17)

vaa , a ∈ I1 } and w ab ∈ V ⊥ . It is easily seen that such that u ab ∈ V = Span{ vμν . w ab ⊥w μμ and w ab ⊥ Define p2 := rank {w ab , a, b ∈ I1 , a = b}. Then, if necessary by changing the choice of {En+r1 +1 , . . . , En+p }, we can further assume ⎧ n+1 n+2 n+r1 vaa = (B11 , Baa , . . . , Baa , 0, . . . , 0), a ∈ I1 , ⎪ ⎪ ⎪ ⎪ n+1 n+r1 +1 n+r1 +p1 ⎪ ⎪ , . . . , Bμμ , 0, . . . , 0), μ ∈ I2 , ⎪ vμμ = (Bnn , 0, . . . , 0, Bμμ ⎨ n+r1 +1 n+r1 +p1 , . . . , Bμν , 0, . . . , 0), μ, ν ∈ I2 , μ = ν, vμν = (0, . . . , 0, Bμν ⎪ ⎪ n+r1 +p1 +1 n+r n+2 ⎪ 1 ⎪ vab = (0, Bab , . . . , Bab , 0, . . . , 0, Bab , ⎪ ⎪ ⎪ ⎩ n+r1 +p1 +p2 . . . , Bab , 0, . . . , 0), a, b ∈ I1 , a = b. (5.18) Now, we define J1 = {n + 1, . . . , n + r1 }, J2 = {n + r1 + 1, . . . , n + r1 + p1 }, J3 = {n + r1 + p1 + 1, . . . , n + r1 + p1 + p2 }, J4 = {n + r1 + p1 + p2 + 1, . . . , n + p}. ¯ ≡ 0, and using (5.12), (5.13) and (5.18), we get From the condition ∇B the following equations   n+1 α 0= Baa,k ωk = Baa ωα(n+1) , a ∈ I1 , (5.19) α∈J1

k

0=



n+1 Bab,k ωk

=

k

0=



n+1 Bμν,k ωk

=

 k

α Baa,k ωk =

β Bab ωβ(n+1) , a, b ∈ I1 , a = b,

β∈J3

n+1 (Bμμ





β Bab ωβ(n+1) +

β∈J1

k

0=



−

n+1 Bνν )ωμν

+



(5.20) β Bμν ωβ(n+1) ,

μ, ν ∈ I2 , (5.21)

β∈J2 β Baa ωβα , a ∈ I1 , α ∈ J4 .

(5.22)

β∈J1

From (5.18) and (5.19), we have ω(n+1)α = 0 for all α ∈ J1 . Combining this and (5.20), by the definition of p2 and (5.18), we have ω(n+1)α = 0 for all α ∈ J3 . From (5.21), by definition of p1 and (5.18), we get ω(n+1)α = 0 for all α ∈ J2 . Finally, from (5.22), by using (5.18) and that rank { vaa | a ∈ I1 } = r1 , we obtain ω(n+1)α = 0 for all α ∈ J4 . In summing up, we have completed the proof of Claim 5.1.

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Claim 5.1 implies that ξ = ξ 1 En+1 is a parallel M¨ obius normal vector field, and by (5.16) we can apply Theorem 3.1 to achieve the conclusion. This completes the proof of Proposition 5.2. 

6. Case II: t = p + 2 In this section, we will prove that if t = p+2 then x is locally M¨ obius equivalent to the immersion as stated in (iv) of the Classification Theorem. First of all, we prove three lemmas. Lemma 6.1. If t ≥ 3, then vii = 0 for all i. It follows that rs ≥ 1 for each s = 1, 2, . . . , t. Proof. If for some i it holds vii = 0, then for any j ∈ / [i], by (4.4) we have 0 = vii · vjj = −(A[i] + A[j] ). This implies that A[j] = −A[i] for all j satisfying [j] = [i], a contradiction to t ≥ 3.  Thus, vii = 0 for all i, and the assertion follows. Lemma 6.2. If t = p + 2, then r1 = r2 = · · · = rp+2 = 1, vij = 0 for all i = j. Proof. In fact, from the proof of Proposition 4.1, we know that any (p + 1) points of {q¯1 , q¯2 , . . . , qp+2 } span a p-dimensional simplex, thus any arbitrary p + 1 elements of { v¯1¯1 , v¯2¯2 , . . . , vp+2 p+2 } is of rank p. On the other hand, by (4.4) and (4.5), we get ( vaa − vbb ) · vs¯s¯ = 0, vab · vs¯s¯ = 0,

∀ a, b ∈ I1 , a = b, 2 ≤ s ≤ p + 2.

It follows that vaa = vbb = 0, vab = 0, ∀ a, b ∈ I1 , a = b. By similar arguments as above, we can verify the assertions that vii = vjj = 0, vij = 0, ∀ [i] = [j], i = j. This completes the proof of Lemma 6.2. Lemma 6.3. If t = p + 2, then ωαβ = 0,

α Bii

(6.1) 

= constant, ∀ α, β, i.

Proof. According to the proof of Lemma 6.2, we can choose the normal frame vs¯s¯}, such that {Eα }, if necessary by changing the ordering of { ⎧ n+1 v¯1¯1 = (B¯1¯1 , 0, . . . , 0), ⎪ ⎪ ⎪ ⎪ ⎨ v¯¯ = (B n+1 , B n+2 , 0, . . . , 0), 22 ¯ ¯ 2¯ 2 2¯ 2 (6.2) ⎪ ······ ······ ⎪ ⎪ ⎪ ⎩ n+p n+2 vp¯p¯ = (Bpn+1 ¯p¯ , Bp¯p¯ , . . . , Bp¯p¯ ), = 0 for all 1 ≤ s ≤ p. where Bs¯n+s s¯

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From (2.13) and (6.1), we get m1 v¯1¯1 + m2 v¯2¯2 + · · · + mp+2 vp+2 p+2 = 0.

(6.3)

Making inner product of the Eq. (6.3) with v¯1¯1 , and using (4.4), we obtain m1 | v¯1¯1 |2 − m2 (A1 + A2 ) − · · · − mp+2 (A1 + Ap+2 ) = 0. So we have | v¯1¯1 |2 = constant. In the same way, we can prove that | v¯2¯2 |2 = constant, . . . , | vp+2 p+2 |2 = constant. The above facts, together with (4.4) and (6.2), give that Bs¯n+1 = constant, . . . , Bs¯n+p = constant, s¯ s¯

1 ≤ s ≤ p + 2.

(6.4)

¯ = 0 and Lemma 6.2, we get From (6.2) and (6.4), combining with ∇B   β 0= Bs¯αs¯,k ωk = Bs¯s¯ωβα , 1 ≤ s ≤ p + 2. (6.5) k

β

This and rank{ v¯1¯1 , v¯2¯2 , . . . , vp+2 p+2 } = p then imply that ωαβ = 0 for all α, β.  The main result of this section is the following theorem. Theorem 6.1. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and that its Blaschke tensor has (p + 2) distinct eigenvalues of multiplicities m1 , . . . , mp+2 , respectively, then by changing the order of {ms } if necessary, we have 1 cs

:= 2As + | vs¯s¯|2 > 0, 1 ≤ s ≤ p + 1;

1 cp+2

:= 2Ap+2 + | vp+2 p+2 |2 < 0.

Moreover, x is M¨ obius equivalent to the image of σ of a cone over the nonminimal parallel submanifold √  √    Sm1 c1 × · · · × Smp+1 cp+1 → Sp+m1 +···+mp+1 −cp+2 in Rn+p . To prove Theorem 6.1, we first establish two auxiliary lemmas, where the first one is a necessary initial step to the inductive proof of the second, i.e. Lemma 6.5. Lemma 6.4. Let q1 , q2 , q3 be three non-collinear points in the Euclidean plane R2 , with origin O. Let q denote the orthocenter of the triangle q1 q2 q3 , if q1 , q2 , q3 , q are four mutually distinct points, then the following identities hold: 1 1 1 1 + −−→ −−→ + −−→ −→ + −→ −→ = 0, − → − −→ q− q q qq q · q q q · q q q · q q 1 2 1 3 2 1 2 3 3 2 3 3 · qq1 − → −−→ −−→ −−→ Oq Oq2 Oq3 Oq1 → − + + + = 0. − − → − − → − − → − → − − → − − → − → − → q1 q2 · q1 q3 q2 q1 · q2 q3 q3 q2 · q3 q qq3 · qq1

(6.6) (6.7)

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93

q3

c2 c1

q

q1

c3

q2

Figure 1. Proof. Since the discussion is totally similar for the case that q1 q2 q3 is an obtuse triangle, without loss of generality we will only consider the case that q1 q2 q3 is an acute triangle. This implies that the orthocenter q is located in the interior of q1 q2 q3 . Let c1 (resp., c2 and c3 ) be the foot point of q1 (resp., q2 and q3 ) on the edge q2 q3 (resp., q1 q3 and q1 q2 ). See Fig. 1. Then we have − → −−→ − → −−→ q− q− 1 q2 · q1 q3 = |q1 q2 | · |q1 c3 |, 2 q1 · q2 q3 = |q1 q2 | · |q2 c3 |, (6.8) − − →·− → = −|qc | · |qq |. q−→ q ·− q→q = |q c | · |q q|, qq qq 3 2

3

3 3

3

3

1

3

3

Substitute the above equations into the left side of (6.6), we get



 1 1 1 1 + − + |q1 q2 | · |q1 c3 | |q1 q2 | · |q2 c3 | |q3 c3 | · |q3 q| |qc3 | · |qq3 | (6.9) 1 1 − = 0, = |q1 c3 | · |q2 c3 | |q3 c3 | · |qc3 | where, the last equality holds because the two right triangles q1 qc3 and q3 q2 c3 are similar. From (6.8) and (6.9), we immediately get the identity (6.6). To prove (6.7), we notice in our case that −−→ −−→ −−→ |q2 c3 | −−→ |q1 c3 | −−→ Oc3 = Oq1 + q1 c3 = Oq1 + Oq2 . |q1 q2 | |q1 q2 | It follows that −−→ −−→ −−→ −−→ Oc3 Oq1 Oq2 Oc3 = = + . |q3 c3 | · |qc3 | |q1 c3 | · |q2 c3 | |q1 q2 | · |q1 c3 | |q1 q2 | · |q2 c3 | Similarly, we can get −−→ −−→ − → Oc3 Oq3 Oq = + . |qc3 | · |qq3 | |q3 c3 | · |qc3 | |q3 c3 | · |q3 q|

(6.10)

(6.11)

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From (6.10), (6.11) and (6.8), we obtain the assertion (6.7).

Results Math



The above arguments can be extended to get the following general result. Lemma 6.5. Let {q1 , q2 , . . . , qp+1 } (p ≥ 2) be (p + 1) points in Rp with origin / O, which span a p-dimensional simplex with an orthocenter cp+2 . If cp+2 ∈ {q1 , q2 , . . . , qp+1 }, then the following identities hold:  1 1 1 + − − → →·− −−−→ − − → − −→ + − −−−−−−→ q1 q2 · q1 q3 i=2 − qi− q q− q q q · i−1 i i+1 p+1 p qp+1 cp+2 p

1 + −−−−−−→ −−−−→ = 0, cp+2 qp+1 · cp+2 q1 −−−−→ −−→ −−→ p  Oqp+1 Oqi Oq1 + + −−−−→ −−−−−−→ − − → − − − → − − → − − − → q1 q2 · q1 q3 i=2 qi qi−1 · qi qi+1 qp+1 qp · qp+1 cp+2 −−−−→ Ocp+2 → − + −−−−−−→ −−−−→ = 0 . cp+2 qp+1 · cp+2 q1

(6.12)

(6.13)

Proof. We will verify the assertions by applying the method of induction on the integer p. As the first step, from Lemma 6.4, we have shown that Lemma 6.5 holds for p = 2. Next, assuming that Lemma 6.5 holds for p = d−1 (d−1 ≥ 2), we will show that it further holds if p = d. Let q1 q2 . . . qd+1 denote the d-dimensional simplex, spanned by the points q1 , q2 , . . . , qd+1 (d ≥ 3). Without loss of generality, we assume that the orthocenter cd+2 of q1 q2 . . . qd+1 is located in its interior. Let cd+1 be the foot point of qd+1 on the hyperplane determined by the simplex q1 q2 . . . qd , and similarly, cd be the foot point of qd on the hyperplane determined by the simplex q1 q2 . . . qd−1 . See Fig. 2 for an abridged general view corresponding to the special case d = 3. Obviously, the three points qd+1 , cd+2 , cd+1 are collinear, and cd+1 is the −−−−→ −−−−→ c− orthocenter of q1 q2 . . . qd which further satisfies the relation − d+2 cd+1 ⊥cd+1 cd . Since q1 q2 . . . qd+1 is a d-dimensional simplex which possesses an orthocenter, the point cd+1 we just defined above is the orthocenter of the (d − 1)dimensional simplex q1 q2 . . . qd . By the inductive assumption, we have  1 1 1 + + −−−−→ −−−−→ − − → − − − → − − → − − − → q1 q2 · q1 q3 i=2 qi qi−1 · qi qi+1 qd qd−1 · qd cd+1 d−1

1 + −−−−→ −−−−→ = 0; cd+1 qd · cd+1 q1 −−→ −−→ −−→ d−1  Oq1 Oqd Oqi + + − → − − −→ · − −q−−→ · q−−− − −→ − −→ − −→ q− q q q · q q q q q 1 2 1 3 i i+1 d d−1 d cd+1 i=2 i i−1 −−−−→ Ocd+1 → − + −−−−→ −−−−→ = 0 . cd+1 qd · cd+1 q1

(6.14)

(6.15)

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q4

c5

q1

q3 c4 c3 q2

Figure 2. −−−−→ However, from the fact − c− d+1 qd+1 ⊥q1 q2 . . . qd , we get − −−→ −−−−→ −−−−→ −−−−→ −−−−−−→ −−−−→ −−−−→ q− d qd−1 · qd cd+1 = qd qd−1 · (qd cd+1 + cd+1 qd+1 ) = qd qd−1 · qd qd+1 . Thus, we can rewrite (6.14) and (6.15) as follows: d  1 1 1 + − → − − −→ · − − −→ − −→ + c−−−−→ −−−−→ = 0, q− q q · q q q q q q 1 2 1 3 i i+1 d+1 d · cd+1 q1 i=2 i i−1 −−−−→ −−→ −−→ d  Ocd+1 Oqi Oq1 → − + + − − → − − − → − − − − → − − → − − − → −−→ = 0 . q1 q2 · q1 q3 i=2 qi qi−1 · qi qi+1 cd+1 qd · − c− d+1 q1

(6.16)

(6.17)

Furthermore, as cd+1 is the orthocenter of q1 q2 . . . qd , we have − −−→ −−−−−−→ −−−−−−→ −−−−−−→ q− d+1 qd · qd+1 cd+2 = |qd+1 cd+1 | · |qd+1 cd+2 |, − c−−−q−−→ · − c−−−→ q = −|− c−−−q−−→| · |− c−−−c−−→|, d+2 d+1

d+2 1

d+2 d+1

d+2 d+1

−−→ −−−−→ −−−−→ −−−−→ c−− d+1 qd · cd+1 q1 = −|cd+1 qd | · |cd+1 cd |.

(6.18)

On the other hand, cd+2 being the orthocenter of q1 q2 . . . qd+1 implies that − −−→ q− d qd+1 ⊥q1 q2 . . . qd−1 cd+2 . −−→ −−−−→ It follows that − q− d qd+1 ⊥cd cd+2 and ∠cd+1 cd cd+2 = ∠cd+1 qd+1 qd . This shows that the two right triangles cd+2 cd+1 cd and qd cd+1 qd+1 are similar, by which we have −−→ −−−−→ −−−−−−→ −−−−−−→ |c−− d+1 qd | · |cd+1 cd | = |qd+1 cd+1 | · |cd+2 cd+1 |.

(6.19)

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So, by using (6.18) and (6.19), the last term in (6.16) can be rewritten as 1 1 1 = −−−−→ −−−−−−→ + −−−−−−→ −−−−→ . − − − − → − − − − → cd+1 qd · cd+1 q1 qd+1 qd · qd+1 cd+2 cd+2 qd+1 · cd+2 q1 This and (6.16) show that (6.12) is true for p = d. To prove (6.13) holds for p = d, we notice in our case that −−−−→ −−−−→ −−−−→ Ocd+2 Ocd+1 Oqd+1 = + . |qd+1 cd+2 ||cd+1 cd+2 | |cd+1 cd+2 ||qd+1 cd+1 | |qd+1 cd+1 ||qd+1 cd+2 | Substitute (6.18) and (6.19) into the above formula, we get −−−−→ −−−−→ −−−−→ Ocd+1 Oqd+1 Ocd+2 = + −−→ −−−−→ − −−→ −−−−−−→ − −−−−→ −−−−→ . c−− q− c− d+1 qd · cd+1 q1 d+1 qd · qd+1 cd+2 d+2 qd+1 · cd+2 q1 Putting this into (6.17), we see that (6.13) holds for p = d. We have completed the proof of Lemma 6.5.

(6.20)



Proof of Theorem 6.1. By assuming (6.2), then from (2.6)–(2.8), together with the application of Lemmas 6.2 and 6.3, for each 1 ≤ s ≤ p + 2, we derive the following equations: ⎧   α 2 ⎪ d(A Y + N − B E ) = (2A + | v | ) ωk Yk , s α s s ¯ s ¯ ⎪ s ¯ s ¯ ⎪ ⎨ α k∈Is   (6.21)   α ⎪ ⎪ A , ∀ i ∈ I dY = ω Y − ω Y + N − B E , i ij j i s α s ⎪ s¯s¯ ⎩ α

j∈Is

Now, set m0 = 0, for 1 ≤ s ≤ p + 2, we define the following subspaces  Vs = Span{As Y + N − Bsαs Eα , Ys−1 mi +1 , . . . , Ysi=1 mi }. i=0

α

It is easily seen from (6.21) that, all the subspaces V1 , V2 , . . ., Vp+2 are parallel along M n in Rn+p+2 . Moreover, they are mutually orthogonal and 1 constitute a direct sum decomposition of Rn+p+2 : 1 . V1 ⊕ V2 ⊕ · · · ⊕ Vp+2 = Rn+p+2 1

(6.22)

From (6.21), we consider the following vector-valued functions in Rn+p+2 : 1 u ˜i := Ai Y + N −

 α

B¯iα¯i Eα , i = 1, . . . , p + 2.

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It is easily seen that ˜ ui , u ˜i 1 = 2Ai + | v¯i¯i |2 . On the other hand, by use of (4.4), and in notations of Sect. 4, we can write ⎧ → −−→ 2A + | v¯1¯1 |2 = − q− ⎪ 1 q2 · q1 q3 , ⎪ ⎪ 1 ⎪ ⎪ → −−→ ⎪ v¯2¯2 |2 = − q− 2 q1 · q2 q3 , ⎪ 2A2 + | ⎨ ······ ······ (6.23) ⎪ ⎪ − − − − → − − − − − − → 2 ⎪ ⎪ vp+1 p+1 | = qp+1 qp · qp+1 qp+2 , 2Ap+1 + | ⎪ ⎪ ⎪ ⎩ 2A −−−−→ −−−−→ vp+2 p+2 |2 = − q− p+2 + | p+2 qp+1 · qp+2 q1 . Analogously, we can deduce the relation → −−→ v¯i¯i |2 = ( v¯j¯j − v¯i¯i ) · ( vk¯k¯ − v¯i¯i ) = − q− 2Ai + | i qj · qi qk ,

(6.24)

for any i, j, k with [i] = [j] = [k] = [i]. For case t = p + 2 and any fixed i, j (i = j), by using Lemma 4.3 (or see the statement in the proof of Lemma 6.2), the vectors   v¯ ¯ − v¯¯ = − q−→ q | 1 ≤ k ≤ p + 2 and [k] = [i], [j] kk

ii

i k

constitute a basis of R . This fact and (6.24) allow us to make the next claim. p

Claim 6.1. If t = p + 2, then 2Ai + | v¯i¯i |2 = 0 for each i ∈ {1, 2, . . . , p + 2}. In fact, if 2Ai + | v¯i¯i |2 = 0 for some i, then for fixed [j] = [i], (6.24) implies vk¯k¯ − v¯i¯i ) = 0 for any [k] = [i], [j]. Hence v¯j¯j − v¯i¯i = 0, a that ( v¯j¯j − v¯i¯i ) · ( contradiction. , by Thus, we can further define ui : M n → Rn+p+2 1  α Ai Y + N − α B¯i¯i Eα ui := , i = 1, 2, . . . , p + 2. (6.25) 2Ai + | v¯i¯i |2 Then by (6.21) we have    du1 = ωa Ya , du2 = ωq Yq , . . . , dup+2 = ωs Ys , a∈I1

q∈I2

(6.26)

s∈Ip+2

and, direct calculation gives ui , uj 1 =

δij =: ci δij , 1 ≤ i, j ≤ p + 2. 2Ai + | v¯i¯i |2

(6.27)

Hence, by (6.25), (6.23) and Lemma 6.5, we can show that u1 + u2 + · · · + up+2 = Y.

(6.28)

In fact, by using (6.25), (6.23), (6.12) and (6.13), we easily verify that p+2

u1 + u2 + · · · + up+2 = cY,

(6.29)

Ai where c := i=1 2Ai +| v¯i¯i |2 is a constant. To see that c = 1, we exterior differentiate both sides of (6.29) and then compare (6.26) with (2.6), we have dY = c dY and the assertion c = 1 follows.

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Now, by (6.27) and the direct sum decomposition (6.22), we deduce that in the set {u1 , . . . , up+2 }, exact one element is a timelike vector, and all others are spacelike vectors. Without loss of generality, we may assume 2As + | vs¯s¯|2 > 0, s = 1, 2, . . . , p + 1; 2Ap+2 + | vp+2 p+2 |2 < 0.

(6.30)

From the structure Eq. (6.21), we have   ωk Yk ; dYj = ωjk Yk − ωj (2Ai + | v¯i¯i |2 )ui , j ∈ Ii . dui =

(6.31)

k∈Ii

k∈Ii

This shows that, each ui defines the standard totally umbilical imbedding √ ui : Smi ( ci ) → Rmi +1 := Vi , 1 ≤ i ≤ p + 1. On the other hand, up+2 satisfies up+2 , up+2 1 < 0. Thus, by (6.31) with i = p + 2, the mapping up+2 defines the standard totally umbilical imbedding   m +1 −cp+2 → R1 p+2 := Vp+2 . up+2 : Hmp+2 p+2 Therefore, by (6.22) and (6.28), noting that s=1 (ms + 1) = n + p + 2, the M¨ obius position vector Y of x can be expressed as: √ √ Y = (u1 , . . . , up+1 , up+2 ) : Sm1 ( c1 ) × · · · × Smp+1 ( cp+1 )  × Hmp+2 ( −cp+2 ) → Rn+p+2 . 1 According to Proposition 3.1 and that t = p + 2, x is M¨ obius equivalent to the image of σ of the cone over the non-minimal submanifold defined by √  √  c1 × · · · × Smp+1 cp+1 (u1 , . . . , up+1 ) : Sm1   −cp+2 , → Sp+m1 +···+mp+1 which is of parallel second fundamental form. We have completed the proof of Theorem 6.1.



7. Case III: 3 ≤ t ≤ p + 1 In this section, we assume that 3 ≤ t ≤ p + 1. We first prove a simple lemma. Lemma 7.1. If t ≥ 3 and for some s ∈ {1, 2, . . . , t} it holds rs = 1, then vaa = vbb , ∀ a, b ∈ Is . Proof. As rs = 1, for any a, b ∈ Is , we can assume that vaa = λ vbb . Then from (4.4), we get 0 = ( vaa − vbb ) · vjj = (λ − 1) vbb · vjj = (1 − λ)(As + A[j] ), if [j] = [a]. (7.1) Then, as t ≥ 3, (7.1) shows that it must be the case λ = 1.



From Lemma 6.1, we know that rs ≥ 1 for each s = 1, 2, . . . , t. Now, we are sufficient to consider the following two subcases:

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III-(i). r1 = r2 = · · · = rt = 1. III-(ii). ∃ s0 ≥ 1 such that r1 ≥ 2, . . . , rs0 ≥ 2, rs0 +1 = · · · = rt = 1. First, we consider case III-(i). In this case, from the fact m1 v¯1¯1 + m2 v¯2¯2 + · · · + mt vt¯t¯ = 0,

(7.2)

and Corollary 4.1, we have t − 2 ≤ rank (V1 ∪ · · · ∪ Vt ) ≤ t − 1. So we can divide the discussions of case III-(i) into the following two subcases. III-(i)-(a). r1 = r2 = · · · = rt = 1 and rank (V1 ∪ · · · ∪ Vt ) = t − 1. III-(i)-(b). r1 = r2 = · · · = rt = 1 and rank (V1 ∪ · · · ∪ Vt ) = t − 2. Corresponding to the case III-(i)-(a), we have the following result. Proposition 7.1. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and that its Blaschke tensor has t distinct eigenvalues. If 3 ≤ t ≤ p + 1 and that case III-(i)-(a) occurs, i.e. it holds r1 = r2 = · · · = rt = 1 and rank (V1 ∪ · · · ∪ Vt ) = t − 1, then x is M¨ obius equivalent to an open part of one of the immersions as stated in (i), (ii) and (iii) of the Classification Theorem. Proof. From (7.2) and the assumption that rank{ v¯1¯1 , . . . , vt¯t¯} = t − 1, we see that any (t − 1) vectors of the set { v¯1¯1 , . . . , vt¯t¯} are linearly independent. So we can choose an orthonormal normal frame {En+1 , . . . , En+p } such that ⎧ v¯1¯1 = (B¯1n+1 , 0, . . . , 0), ⎪ ¯ 1 ⎪ ⎪ ⎪ ⎨ v¯¯ = (B n+1 , B n+2 , 0, . . . , 0), 22 ¯ ¯ 2¯ 2 2¯ 2 (7.3) ⎪ ······ ······ ⎪ ⎪ ⎪ ⎩ n+1 n+t−1 vt−1 t−1 = (Bt−1 , . . . , Bt−1 , 0, . . . , 0), t−1 t−1 = 0 for each 1 ≤ s ≤ t − 1. It follows, by (4.5), that where Bs¯n+s s¯ n+1 n+t−1 Bij = · · · = Bij = 0, i = j.

(7.4)

Let us denote p¯ := rank { vij , i = j}. Then, if necessary by changing the choice of {En+t , . . . , En+p }, we can further assume that n+t+p−1 ¯ n+t , . . . , Bij , 0, . . . , 0), if i = j. vij = (0, . . . , 0, Bij

(7.5)

Put J1 = {n + 1, . . . , n + t − 1}, J2 = {n + t, . . . , n + t + p¯ − 1}, J3 = {n + t + p¯, . . . , n + p}. Because of (7.2), by using of (7.3), (4.4) and arguments similar to the α = constant for all i and α. proof of Lemma 6.3, we can derive that Bii

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Now we consider the system of equations with unknowns ξ1 , . . ., ξt−1 , λ: ⎧ n+1 B¯1¯1 ξ1 + λ + A1 = 0, ⎪ ⎪ ⎪ ⎪ ⎨ B n+1 ξ + B n+2 ξ + λ + A = 0, 1 2 2 ¯ ¯ 2¯ 2 2¯ 2 (7.6) ⎪ · · · · · · · · · · · · ⎪ ⎪ ⎪ ⎩ n+1 n+t−1 Bt¯t¯ ξ1 + Bt¯n+2 ξt−1 + λ + At = 0. t¯ ξ2 + · · · + Bt¯t¯ To solve the linear Eq. (7.6), we make use of (7.2) and (7.3) to deduce that    B¯1n+1 0 0 ··· 0 1  ¯ 1   n+1 n+2  B¯¯ B¯2¯2 0 ··· 0 1  22  n n+1  .. .. .. .. .. ..  n+t−1  . . . . . .  = m B¯1¯1 · · · Bt−1 t−1 = 0.  n+1 t n+2 n+t−1 B   t−1 t−1 Bt−1 t−1 · · · · · · Bt−1 t−1 1   B n+1 B n+2 · · · · · · B n+t−1 1 t¯t¯

t¯t¯

t¯t¯

It follows that (7.6) has unique constant solutions {ξ1 , ξ2 , . . . , ξt−1 , λ}. Thus ξ := ξ1 En+1 + ξ2 En+2 + · · · + ξt−1 En+t−1

(7.7)

defines a M¨ obius normal vector field. Moreover, from (7.6), it is easy to see that the tensors A, B and g are related to ξ by the following equation A + λg + B, ξ 1 = 0.

(7.8)

Importantly, we can made the following claim. Claim 7.1. The M¨ obius normal vector field ξ defined by (7.7) is parallel. ¯ = 0 and the fact B β To verify Claim 7.1, we use the assumption ∇B ii being constant for all i and β ∈ J1 . By (7.3) and (7.5), we have the following calculations:  β  α 0= Bii,k ωk = Bii ωαβ , ∀ i and β ∈ J1 , (7.9) α∈J1

k

0=



β Bij,k ωk

β β = (Bii − Bjj )ωij +

0=

k

J1 .

α Bij ωαβ , ∀ i = j and β ∈ J1 ,

α∈J2

k





α Bii,k ωk

=



(7.10) β Bii ωβα ,

∀ i and α ∈ J3 .

(7.11)

β∈J1

Then, (7.9) and rank (V1 ∪ · · · ∪ Vt ) = t − 1 shows that ωαβ = 0, α, β ∈

Next, for i = j and β ∈ J1 , by Lemma 7.1 and Lemma 4.1 of [26], we β β see that (Bii − Bjj )ωij = 0, Then, by using (7.10) and (7.5) and rank { vij | i = j} = p¯, we have ωαβ = 0, α ∈ J2 , β ∈ J1 . Finally, from (7.11) and (7.3), we get ωαβ = 0 for α ∈ J3 and β ∈ J1 .

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Summing up the above discussion, we have proved ωβα = 0, β ∈ J1 , α ∈ {n + 1, . . . , n + p}. Thus, each normal vector field of {En+1 , . . . , En+t−1 } is parallel. Accordingly, ξ defined by (7.7) is a parallel M¨ obius normal vector field. From (7.8) and Theorem 3.1, the proof of Proposition 7.1 is completed.  Corresponding to the case III-(i)-(b), the conclusion is as follows: Proposition 7.2. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and that its Blaschke tensor has t distinct eigenvalues. If 3 ≤ t ≤ p + 1 and that case III-(i)-(b) occurs, i.e. it holds r1 = r2 = · · · = rt = 1 and rank (V1 ∪ · · · ∪ Vt ) = t − 2, then x is M¨ obius equivalent to an open part of the immersion as stated in (iv) of the Classification Theorem. Proof. First, by assumption we know that the subspace V = Span{ v¯1¯1 , v¯2¯2 , . . . , vt¯t¯} of Rp is of dimension (t − 2). Then, adopting the same notations as in preceding sections, we see that any (t − 1) points of {q1 , q2 , . . . , qt } span a (t − 2)-dimensional simplex with the remaining one point as its orthocenter. From the assumption, and similar to the proof of Lemma 6.3, we can choose the normal frame {Eα }, such that ⎧ n+1 ⎪ ⎪ v¯1¯1 = (B¯1¯1 , 0, . . . , 0), ⎪ ⎪ ⎨ v¯¯ = (B n+1 , B n+2 , 0, . . . , 0), 22 ¯ ¯ 2¯ 2 2¯ 2 (7.12) ⎪ · · · · · · · · · · · · ⎪ ⎪ ⎪ ⎩ n+1 n+2 n+t−2 vt−2 t−2 = (Bt−2 , Bt−2 , . . . , Bt−2 , 0, . . . , 0), t−2 t−2 t−2 = constant for all 1 ≤ i ≤ j ≤ t − 2, and Bs¯n+s = 0 for all where B¯jn+i s¯ ¯ j 1 ≤ s ≤ t − 2. For the remaining cases, we can write n+2 n+t−2 vs¯ s¯ = (Bs¯n+1 , 0, . . . , 0), s¯ , Bs¯ s¯ , . . . , Bs¯ s¯

s = t − 1, t.

(7.13)

Next, if i = j, then by (4.3), (4.5) and (7.2) we see that vij · vs¯s¯ = 0 holds for all s = 1, 2, . . . , t. Hence, by using (7.12), we derive n+1 n+2 n+t−2 Bij = Bij = · · · = Bij = 0, ∀ i = j.

Accordingly, we can assume that n+p n+t−1 vij = (0, . . . , 0, Bij , . . . , Bij ),

i = j.

(7.14)

Similar to (but different from) that in the proof of Lemma 5.1, for each s ∈ {1, 2, . . . , t}, we define vab | a, b ∈ Is and a = b}. ps := rank {

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We point out that it may occur that ps = 0. For convenience of statement, we set p0 = 0. Then, by (4.6), we can further assume that, for a, b ∈ Is and a = b, n+t−1+

vab = (0, . . . , 0, Bab

s−1 i=0

pi

n+t−2+

, . . . , Bab

s

i=0

pi

, 0, . . . , 0).

(7.15)

Put J0 = {n + 1, . . . , n + t − 2}, s−1 s    Js = n + t − 1 + pi , . . . , n + t − 2 + pi , s = 1, 2, . . . , t, i=0

Jt+1

t   = n+t−1+ pi , . . . , n + p .

i=1

i=1

¯ = 0, (7.12) – (7.14), we obtain Now, using ∇B  β  α 0= Bii,k ωk = Bii ωαβ , ∀ i and β ∈ J0 , α∈J0

k

0=



β β β Bij,k ωk = (Bii − Bjj )ωij +

0=

α Bij,k ωk =

k

0=

 k

α Bij ωαβ , i, j ∈ Is , i = j, β ∈ J0 ,

α∈Js

k





(7.16)



(7.17) β Bij ωβα ,

α ∈ Jr , i, j ∈ Is , i = j, r = s,

(7.18)

β∈Js α Bij,k ωk =



β Bij ωβα , ∀ i, j and α ∈ Jt+1 .

(7.19)

β ∈J / t+1

Noticing that rank (V1 ∪ · · · ∪ Vt ) = t − 2, from (7.16) we can conclude that ωβα = 0 for all α, β ∈ J0 . Similar arguments as in the proof of Claim 7.1, by (7.17) and (7.15), we have ωβα = 0, α ∈ J0 , β ∈ Js , 1 ≤ s ≤ t.

(7.20)

It follows from (7.18) and the fact rank { vab | a, b ∈ Is and a = b} = ps that ωβα = 0, α ∈ Jr , β ∈ Js , r = s, 1 ≤ r, s ≤ t.

(7.21)

Finally, from (7.12) and (7.15), we can show that rank { vij | ∀ i, j} = t − 2 + p1 + · · · + pt , which, together with (7.19), implies that ωβα = 0 for all β ∈ J0 ∪ J1 ∪ · · · ∪ Jt and α ∈ Jt+1 .

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Using the above results, we can deduce from (2.6)–(2.8) the following equations: ⎧  n+t−2    ⎪ ⎪ ⎪ d As Y + N − Bs¯αs¯Eα = (2As + | vs¯s¯|2 ) ωi Yi , 1 ≤ s ≤ t; ⎪ ⎪ ⎪ ⎪ α=n+1 i∈Is ⎪ ⎪ ⎪ ⎪ n+t−2   ⎪    ⎨ β dYi = ωij Yj − ωi As Y + N − Bs¯αs¯Eα + Bij ω j Eβ , ⎪ α=n+1 j∈I j∈I ,β∈J s s s ⎪ ⎪ ⎪ ⎪ i ∈ I , 1 ≤ s ≤ t; ⎪ s ⎪ ⎪  β  ⎪ ⎪ ⎪ dEβ = − Bij ωj Yi + ωβα Eα , β ∈ Js , 1 ≤ s ≤ t. ⎪ ⎩ i,j∈Is

α∈Js

(7.22) Now, we define n+t−2   Vs := Span As Y + N − Bs¯αs¯Eα , Yi , Eβ | i ∈ Is , β ∈ Js , 1 ≤ s ≤ t. α=n+1

It is easily seen from (7.22) that, along M n , all the subspaces V1 , V2 , . Moreover, they are mutually orthogonal and . . ., Vt are parallel in Rn+p+2 1 constitute a direct sum decomposition of Rn+p+2 : 1 V1 ⊕ · · · ⊕ Vt = Rn+p+2 . 1

(7.23)

: From (7.22), we consider the following vector-valued functions in Rn+p+2 1 u ˜s := As Y + N −

n+t−2 

Bs¯αs¯Eα ,

1 ≤ s ≤ t.

α=n+1

Then we have ˜ us , u ˜s 1 = 2As + | vs¯s¯|2 , 1 ≤ s ≤ t. Totally similar to the proof of Claim 6.1, we can show that vs¯s¯|2 = 0, s = 1, . . . , t. 2As + | Thus, we can further define us : M n → Rn+p+2 , by 1 n+t−2 α As Y + N − α=n+1 Bs¯s¯Eα us := , s = 1, . . . , t. 2As + | vs¯s¯|2 Then by (7.22) we have dus =



(7.24)

ωi Yi , s = 1, . . . , t,

(7.25)

δij =: ci δij , 1 ≤ i, j ≤ t. 2Ai + | v¯i¯i |2

(7.26)

i∈Is

and, direct calculation gives ui , uj 1 =

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Hence, similar to that in Sect. 6, by (7.25) and Lemma 6.5, we can show that [cf. (6.28)] u1 + u2 + · · · + ut = Y.

(7.27)

Similar to the proof of (6.30), we can infer that in the set {u1 , . . . , ut }, exact one element is a timelike vector, and all other (t − 1) elements are spacelike vectors. Thus, without loss of generality, we can assume u1 , u1 1 > 0, . . . , ut−1 , ut−1 1 > 0; ut , ut 1 < 0.

(7.28)

From the structure Eqs. (7.22), we have  ⎧ dui = ωj Yj , ⎪ ⎪ ⎪ ⎪ j∈I ⎪ i ⎪ ⎪   ⎨ β ωjk Yk + Bjk ωk Eβ − ωj (2Ai + | v¯i¯i |2 )ui , j ∈ Ii , dYj = ⎪ k∈Ii k∈Ii ,β∈Ji ⎪ ⎪  β  ⎪ ⎪ ⎪ dE = − B ω Y + ωβα Eα , β ∈ Ji . ⎪ β k j jk ⎩ j,k∈Ii

α∈Ji

(7.29) This shows that, for each i = 1, . . . , t − 1, the mapping ui defines an mi dimensional immersed submanifold √  ui : Mimi → Smi +pi ci → Rmi +pi +1 := Vi √  in Smi +pi ci of codimension pi (pi ≥ 0), with (Euclidean) parallel second √  fundamental form. Here, if pi = 0 then Mimi = Smi ci . On the other hand, the mapping ut satisfies ut , ut 1 < 0 and that  ⎧ dut = ωj Yj , ⎪ ⎪ ⎪ ⎪ j∈It ⎪ ⎪ ⎪   ⎨ β ωjk Yk + Bjk ωk Eβ − ωj (2At + | vt¯t¯|2 )ut , j ∈ It , dYj = ⎪ k∈It k∈It ,β∈Jt ⎪ ⎪  β  ⎪ ⎪ ⎪ Bjk ωk Yj + ωβα Eα , β ∈ Jt . ⎪ ⎩ dEβ = − j,k∈It

α∈Jt

(7.30) Therefore, the mapping ut defines an mt -dimensional immersed submanifold √  ut : Mtmt → Hmt +pt −ct → R1mt +pt +1 := Vt √  in Hmt +pt −ct of codimension pt (pt ≥ 0), with (Euclidean) parallel second  β = 0 for all fundamental form. Moreover, due to the easily seen fact i∈It Bii √  mt mt +pt β ∈ Jt , we further derive from (7.30) that ut : Mt → H −ct is minimal. Thus by [24] (cf. Lemma 3.7.10 of [3]) the immersion √ ut is in fact totally geodesic and pt = 0, so we locally have Mtmt = Hmt −ct .

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t Therefore, by (7.23) and (7.27), noting that s=1 (ms +ps +1) = n+p+2, the M¨ obius position vector Y of x can be expressed as: √  mt−1 × Hmt −ct Y = (u1 , . . . , ut−1 , ut ) : M1m1 × · · · × Mt−1 √  √  → Sm1 +p1 c1 × · · · × Smt−1 +pt−1 ct−1 × R1mt +1 → Rn+p+2 . 1 According to Proposition 3.1, x is M¨ obius equivalent to the image of σ of the cone over the (Euclidean) parallel submanifold defined by √  mt−1 → Sm1 +p1 c1 × · · · (u1 , . . . , ut−1 ) : M1m1 × · · · × Mt−1 √  × Smt−1 +pt−1 ct−1 t−1 √  → S s=1 (ms +ps )+t−2 −ct . This finally completes the proof of Proposition 7.2.



Next, we consider the second subcase. III-(ii): ∃ s0 ≥ 1 such that r1 ≥ 2, . . . , rs0 ≥ 2, rs0 +1 = · · · = rt = 1. To deal with this case, similar to that in Sect. 4, we will look vii (1 ≤ i ≤ n) as vectors in the Euclidean space Rp with origin O, so that we have points −−→ H1 , H2 , . . . , Hn such that OH i = vii (1 ≤ i ≤ n). Our first result is the following crucial observation. Lemma 7.2. For 3 ≤ t ≤ p + 1, if for some s ∈ {1, 2, . . . , t} it holds rs ≥ 2, then there exists a unique (rs − 1)-dimensional hyperplane Πs in Rp such that O∈ / Πs and {Hi | i ∈ Is } ⊂ Πs . Proof. Without loss of generality, we assume that s = 1. Since t ≥ 3, there exists s ∈ {2, 3, . . . , t} such that A1 + As = 0, so we can assume A1 + A2 = 0. Now, by fixing an arbitrary vector v¯2¯2 ∈ V2 , we can decompose each vector vaa (a ∈ I1 ) such that vaa = λa v¯2¯2 + w aa , a ∈ I1

(7.31)

and w aa · v¯2¯2 = 0. Then, by (4.4), we get − (A1 + A2 ) = vaa · v¯2¯2 = λa | v¯2¯2 |2 = 0, a ∈ I1 ,

(7.32)

v¯2¯2 | , a, b ∈ I1 . 0 = ( vaa − vbb ) · v¯2¯2 = (λa − λb )|

(7.33)

2

This implies that λa = λb := λ = 0 for any a, b ∈ I1 , a = b. Thus we can rewrite (7.31) as vaa = λ v¯2¯2 + w aa , a ∈ I1 . (7.34) In V1 = { vaa | a ∈ I1 }, we choose a maximum linearly independent group, say (for simplicity), { v11 , v22 , . . . , vr1 r1 }. Then, for each a ∈ I1 , there are unique ordered numbers (κ1 , κ2 , . . . , κr1 ), which may depend on a, such that vaa = κ1 v11 + · · · + κr1 vr1 r1 , a ∈ I1 . (7.35) Substitute (7.34) into (7.35), we can get v¯2¯2 + κ1 w 11 + · · · + κr1 w r1 r1 , a ∈ I1 . (7.36) λ v¯2¯2 + w aa = λ(κ1 + · · · + κr1 )

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This, together with w aa · v¯2¯2 = 0, imply that κ1 + · · · + κr1 = 1, w aa = κ1 w 11 + · · · + κr1 w r1 r1 .

(7.37) (7.38)

From (7.35) and (7.37), we have vaa − v11 = κ2 ( v22 − v11 ) + · · · + κr1 ( vr1 r1 − v11 ), a ∈ I1 .

(7.39)

Since the linear independence of the r1 vectors { v11 , v22 , . . . , vr1 r1 } implies v22 − v11 , . . . , vr1 r1 − v11 } are linear independent, (7.39) that the r1 −1 vectors { shows that all points {Ha , a ∈ I1 } are located in a (r1 −1)-dimensional hyperplane Π. Here, Π is uniquely determined such that it passes through the point −−−→ −−−−→ H1 and is spanned by the (r1 − 1) vectors {H1 H2 , . . . , H1 Hr1 }. Next, we show that O ∈ / Π. In fact, if otherwise, there exist numbers {h2 , . . . , hr1 } such that −−→ −−−→ −−−−→ H1 O = h 2 H1 H2 + · · · + h r 1 H1 Hr 1 , which is equivalent to − v11 = h2 ( v22 − v11 ) + · · · + hr1 ( vr1 r1 − v11 ). Thus (h2 + · · · + hr1 − 1) v11 = h2 v22 + · · · + hr1 vr1 r1 . Since v11 , v22 , . . . , vr1 r1 are linear independent, we should have h2 = · · · = hr1 = 0, h2 + · · · + hr1 − 1 = 0, which is impossible. Thus, we complete the proof of Lemma 7.2.



According to Lemma 7.2, in the case t ≥ 3, for each s ∈ {1, 2, . . . , t} with rs ≥ 2, the set Vs = { vaa | a ∈ Is } determines a (rs − 1)-dimensional hyperplane Πs such that O ∈ / Πs . Hence, there exists a unique point Fs ∈ Πs −−→ −−→ such that OFs ⊥ Πs . Put u s¯s¯ := OFs = 0. Then, for each a ∈ Is , the vector vaa has a unique decomposition vaa = u s¯s¯ + zaa , a ∈ Is ,

(7.40)

such that zaa · u s¯s¯ = 0. Definition 7.1. For 3 ≤ t ≤ p + 1 and rs ≥ 2, the above decomposition (7.40) of the vectors vaa (a ∈ Is ) is called the canonical decomposition. Lemma 7.3. Assume that 3 ≤ t ≤ p + 1 and rs ≥ 2. Then, associated with the zaa | a ∈ Is } canonical decomposition (7.40) and any fixed s¯ ∈ Is , the vectors { satisfy zaa − zs¯s¯ | a ∈ Is }. (7.41) Span{ zaa | a ∈ Is } = Span{

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Proof. We carry the proof with s = 1, i.e. we assume that r1 ≥ 2. By Lemma 7.2 and preceding statements, we have the (r1 − 1)-dimensional hyperplane Π1 as well as the canonical orthogonal decomposition (7.40), such that (7.42) vaa = u ¯1¯1 + zaa , a ∈ I1 , −−→ −−→ where u ¯1¯1 = OF1 , F1 ∈ Π1 and OF1 ⊥ Π1 . Without loss of generality, we may assume s¯ = 1. According to the proof of Lemma 7.2, we may further assume that { v11 , v22 , . . . , vr1 r1 } is a maximum vaa | a ∈ I1 }. Thus, for each linearly independent group of the set V1 = { a ∈ I1 , there are unique ordered numbers {κ1 , κ2 , . . . , κr1 }, which may depend on a, such that (7.43) vaa = κ1 v11 + · · · + κr1 vr1 r1 , a ∈ I1 . Substitute (7.42) into (7.43), we can get κ1 + · · · + κr1 = 1, zaa = κ1 z11 + · · · + κr1 zr1 r1 , a ∈ I1 .

(7.44) (7.45)

Using the relation (7.44), we can rewrite (7.45) into the following two equations: z22 − z11 ) + · · · + κr1 ( zr1 r1 − z11 ), a ∈ I1 , zaa − z11 = κ2 ( zaa − z11 = (κ1 − 1) z11 + κ2 z22 + · · · + κr1 zr1 r1 , a ∈ I1 .

(7.46) (7.47)

Now, (7.45) and (7.47) imply that zaa | a ∈ I1 }, Span{ z11 , . . . , zr1 r1 } = Span{

(7.48)

Span{ zaa − z11 | a ∈ I1 } ⊂ Span{ z11 . . . . , zr1 r1 },

(7.49)

From (7.46), we see that zaa − z11 , a ∈ I1 }. Span{ z22 − z11 , . . . , zr1 r1 − z11 } = Span{

(7.50)

Moreover, it is obvious seen that vss − v11 | 2 ≤ s ≤ r1 } = r1 − 1. (7.51) rank { zss − z11 | 2 ≤ s ≤ r1 } = rank { −−→ By the fact u ¯1¯1 = OF1 and F1 ∈ Π1 , there exists unique ordered numbers {h2 , . . . , hr1 }, such that v22 − v11 ) + · · · + hr1 ( vr1 r1 − v11 ). u ¯1¯1 − v11 = h2 (

(7.52)

This, together with (7.42), implies that z22 − z11 ) + · · · + hr1 ( zr1 r1 − z11 ), − z11 = h2 (

(7.53)

z11 = h2 z22 + · · · + hr1 zr1 r1 . (h2 + · · · + hr1 − 1)

(7.54)

or equivalently, This shows that the vectors { z11 , . . . , zr1 r1 } are in fact linear dependent, and by (7.48) we have z11 , . . . , zr1 r1 } ≤ r1 − 1. rank { zaa | a ∈ I1 } = rank {

(7.55)

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On the other hand, by (7.49), (7.50) and (7.51), we know that r1 − 1 = rank { zaa − z11 } ≤ rank { zaa | a ∈ I1 }.

(7.56)

From (7.49), (7.55) and (7.56), we immediately get the desired conclusion Span{ zaa | a ∈ I1 } = Span{ zaa − z11 | a ∈ I1 }. This completes the proof of Lemma 7.3.

(7.57) 

The another crucial observation towards the final solution of case III-(ii) is the next lemma. Lemma 7.4. If case III-(ii) occurs, i.e. for 3 ≤ t ≤ p + 1, ∃ s0 ≥ 1 such that it holds r1 ≥ 2, . . . , rs0 ≥ 2 and rs0 +1 = · · · = rt = 1, then we have Span{ zii | i ∈ Is , 1 ≤ s ≤ s0 } ⊥ Span{ u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯}, (7.58) t − 2 ≤ r := rank { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯} ≤ t − 1.

(7.59)

Proof. To prove (7.58), we first consider the case s0 = 1. In this case, from (4.4) and (7.42), we obtain zii − z¯1¯1 ) · vkk , ∀ i ∈ I1 , k ∈ / I1 . 0 = ( vii − v¯1¯1 ) · vkk = (

(7.60)

/ I1 , it Combining (7.60) and Lemma 7.3 we obtain that, for all i ∈ I1 and k ∈ holds zii · vkk = 0. It follows that zii · v¯j¯j = 0 holds for all i ∈ I1 and 2 ≤ j ≤ t. Furthermore, according to the decomposition (7.42), the relation zii · u ¯1¯1 = 0 holds for all i ∈ I1 . Thus, (7.58) holds true if s0 = 1. Next, we consider the case s0 ≥ 2. Let us take any μ, ν ≤ s0 with μ = ν, then (4.4) and (7.40) imply that 0 = ( vii − vμ¯μ¯ ) · vkk = ( zii − zμ¯μ¯ ) · vkk , i ∈ Iμ , k ∈ / Iμ , (7.61) 0 = ( vii − vμ¯μ¯ ) · ( vjj − vν¯ν¯ ) = ( zii − zμ¯μ¯ ) · ( zjj − zμ¯μ¯ ), i ∈ Iμ , j ∈ Iν . (7.62) Combining (7.61), (7.62) and Lemma 7.3, we immediately get zii · vkk = 0, i ∈ Iμ , μ ≤ s0 , k ∈ / Iμ , zii · zjj = 0, i ∈ Iμ , μ ≤ s0 , j ∈ Iν , ν ≤ s0 , μ = ν.

(7.63) (7.64)

These further imply that zii · u ν¯ν¯ = zii · vjj = 0,

i ∈ Iμ , μ ≤ s0 , j ∈ Iν , ν ≤ s0 , μ = ν.

(7.65)

We also observe that by definition (7.40) we direct have zii · u μ¯μ¯ = 0, i ∈ Iμ .

(7.66)

From (7.63), (7.65) and (7.66), we have proved that (7.58) is true also for s0 ≥ 2. Now, we come to prove (7.59). To begin with we use (7.58) to obtain u μ¯μ¯ · vς¯ς¯ = vii · vς¯ς¯ = −(Aμ + Aς ), i ∈ Iμ , μ ≤ s0 , ς ≥ s0 + 1.

(7.67)

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Moreover, if s0 ≥ 2, then, for any μ, ν ≤ s0 with μ = ν, we have u μ¯μ¯ · u ν¯ν¯ = vii · vjj = −(Aμ + Aν ), i ∈ Iμ , j ∈ Iν .

(7.68)

From (7.67) and (7.68), we can apply Corollary 4.1 to the following set of vectors { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯} to obtain that t − 2 ≤ rank { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯}.

(7.69)

On the other hand, by (2.13), we have 

vii + · · · +

i∈I1



t 

vii +

mj v¯j¯j = 0.

(7.70)

j=s0 +1

i∈Is0

Substitute (7.40) into (7.70), and use (7.58), we get s0 

mi u¯i¯i +

i=1

t 

mj v¯j¯j = 0.

(7.71)

j=s0 +1

This shows that rank { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯} ≤ t − 1. Combining (7.69) and (7.72), we get (7.59). In summary, we have completed the proof of Lemma 7.4.

(7.72) 

Remark 7.1. For the number r defined by (7.59), the following two statements are obvious: (1) if r = t − 2, then s0 ≤ min{p + 2 − t, t}; (2) if r = t − 1, then s0 ≤ min{p + 1 − t, t}. The following result is an immediate consequence of Lemma 7.4. Lemma 7.5. If case III-(ii) occurs, i.e. for 3 ≤ t ≤ p + 1, ∃ s0 ≥ 1 such that it holds r1 ≥ 2, . . . , rs0 ≥ 2 and rs0 +1 = · · · = rt = 1, then we can choose the normal frame {En+1 , En+2 , . . . , En+p }, such that the following expressions hold: ⎧ n+p n+r+1 ⎪ z = (0, . . . , 0, Bii , . . . , Bii ), i ∈ Is , 1 ≤ s ≤ s0 , ⎪ ⎨ ii n+1 n+r (7.73) 1 ≤ s ≤ s0 , u s¯s¯ = (Bs¯s¯ , . . . , Bs¯s¯ , 0, . . . , 0), ⎪ ⎪ ⎩ v = (B n+1 , . . . , B n+r , 0, . . . , 0), s + 1 ≤ s ≤ t. s¯s¯

s¯s¯

s¯s¯

0

Proof. According to (7.58), there is an orthonormal frame {En+1 , En+2 , . . . , obius normal bundle of x, such that En+p } of the M¨ n+p n+r+1 zii = (0, . . . , 0, zii , . . . , zii ),

i ∈ Is , 1 ≤ s ≤ s0 ,

n+r u s¯s¯ = (un+1 s¯s¯ , . . . , us¯s¯ , 0, . . . , 0), n+r vs¯s¯ = (Bs¯n+1 s¯ , . . . , Bs¯s¯ , 0, . . . , 0),

1 ≤ s ≤ s0 , s0 + 1 ≤ s ≤ t.

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On the other hand, from (7.40), we have n+p n+1 vii = u s¯s¯ + zii = (Bii , . . . , Bii ), i ∈ Is , 1 ≤ s ≤ s0 ,

which immediately implies the first two expressions in (7.73).



Now, according to Lemma 7.4 and adopting the preceding notations, we see that Case III-(ii) is reduced to the following two subcases: ∃ s0 ≥ 1 such that r1 ≥ 2, . . . , rs0 ≥ 2, rs0 +1 = · · · = rt = 1 and r = rank { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯} = t − 1. III-(ii)-(b). ∃ s0 ≥ 1 such that r1 ≥ 2, . . . , rs0 ≥ 2, rs0 +1 = · · · = rt = 1 and r = rank { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯} = t − 2. III-(ii)-(a).

Corresponding to case III-(ii)-(a), we have one of the main results in this section as follows. Proposition 7.3. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and its Blaschke tensor has t distinct eigenvalues. If 3 ≤ t ≤ p + 1 and that there exists s0 ≥ 1 such that r1 ≥ 2, . . . , rs0 ≥ 2, rs0 +1 = · · · = rt = 1 and r = t − 1, then x is M¨ obius equivalent to an open part of one of the immersions as stated in (i), (ii) and (iii) of the Classification Theorem. Proof. From the assumption r = t − 1 and (7.71), we see that any (t − 1) vectors of the set { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯} are linearly independent. This fact and Lemma 7.5 imply that we can choose the normal frame vs¯s¯}, {En+1 , En+2 , . . . , En+p }, if necessary re-numbering the subscripts of { such that ⎧ u ¯1¯1 = (B¯1n+1 , 0, . . . , 0), ⎪ ¯ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ... ⎪ ⎪ ⎪ ⎪ ⎪ n+s0 n+1 ⎪ ⎪ ⎪ u s¯0 s¯0 = (Bs¯0 s¯0 , . . . , Bs¯0 s¯0 , 0, . . . , 0), ⎨ 0 +1 , . . . , Bsn+s , 0, . . . , 0), vs0 +1 s0 +1 = (Bsn+1 (7.74) 0 +1 s0 +1 0 +1 s0 +1 ⎪ ⎪ ⎪ .. ⎪ ⎪ ⎪ . ⎪ ⎪ ⎪ ⎪ n+1 n+t−1 ⎪ ⎪ vt−1 t−1 = (Bt−1 , . . . , Bt−1 , 0, . . . , 0), ⎪ t−1 t−1 ⎪ ⎪ ⎩ n+t−1 vt¯t¯ = (Bt¯n+1 , 0, . . . , 0), t¯ , . . . , Bt¯t¯ where Bs¯n+s = 0 for all 1 ≤ s ≤ t − 1, and moreover s¯ n+p n+t zii = (0, . . . , 0, Bii , . . . , Bii ), i ∈ Is , 1 ≤ s ≤ s0 .

(7.75)

Similar to the proof of Lemma 6.3, making use of (4.4), (7.71) and (7.74), α = constant for all i and n + 1 ≤ α ≤ n + t − 1. we can show that Bii

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For 1 ≤ s ≤ s0 and i, j ∈ / Is , by (4.5) we get vkk − vs¯s¯) = vij · ( zkk − zs¯s¯), k ∈ Is , 0 = vij · (

(7.76)

0 = vij · vkk = vij · ( us¯s¯ + zkk ), k ∈ Is , i = j.

(7.77)

(7.76) and Lemma 7.3 imply that / Is . vij · zkk = 0, k ∈ Is , i, j ∈ This together with (7.77) shows that vij · u s¯s¯ = 0, i, j ∈ / Is , i = j.

(7.78)

Hence, by (4.3), (4.5), (7.71) and (7.78), we obtain u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯}, ∀i = j. vij ⊥Span {

(7.79)

This, together with (7.74), implies that we can write n+p n+t , . . . , Bij ), i = j. vij = (0, . . . , 0, Bij

(7.80)

By denoting p¯ := rank { vij , zkk | i = j, k ∈ Is , 1 ≤ s ≤ s0 } and noticing that p¯ ≥ 1, we can adjust the choice of {En+t , . . . , En+p } so that the following expressions hold  n+t+p−1 ¯ n+t , . . . , Bij , 0, . . . , 0), i = j, vij = (0, . . . , 0, Bij n+t+p−1 ¯ n+t zkk = (0, . . . , 0, Bkk , . . . , Bkk , 0, . . . , 0), k ∈ Is , 1 ≤ s ≤ s0 . (7.81) Now we put J1 = {n + 1, . . . , n + t − 1},

J2 = {n + t, . . . , n + t + p¯ − 1}, J3 = {n + t + p¯, . . . , n + p}, and consider the following system of linear equations with unknowns ξ1 , . . . , ξt−1 , λ: ⎧ n+1 B¯1¯1 ξ1 + λ + A1 = 0, ⎪ ⎪ ⎪ ⎪ ⎪ B¯2n+1 ξ1 + B¯2n+2 ξ2 + λ + A2 = 0, ⎪ ¯ ¯ ⎪ 2 2 ⎪ ⎨ .. (7.82) . ⎪ ⎪ ⎪ n+1 n+t−1 ⎪B ⎪ ξ + · · · + Bt−1 t−1 ξt−1 + λ + At−1 = 0, ⎪ t−1 t−1 1 ⎪ ⎪ ⎩ n+1 Bt¯t¯ ξ1 + · · · + Bt¯n+t−1 ξt−1 + λ + At = 0. t¯ Similar to (7.6), by making use of (7.71) and (7.74), it is direct to check that the coefficient matrix of (7.82) is nondegenerate. Thus (7.82) has unique constant solution {ξ1 , ξ2 , . . . , ξt−1 , λ}. Now we define ξ := ξ1 En+1 + ξ2 En+2 + · · · + ξt−1 En+t−1 .

(7.83)

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Claim 7.2. The M¨ obius normal vector field ξ defined by (7.83) is parallel. ¯ ≡ 0 and the fact B β being constant To verify Claim 7.2, we use ∇B ii for all i and β ∈ J1 . Then, by using (2.11), (7.74) and (7.81), we have the following derivation:   β α (Bii,k − Bs¯βs¯,k )ωk = (Bii − Bs¯αs¯)ωαβ , i ∈ Is , 1 ≤ s ≤ s0 , β ∈ J1 , 0= α∈J2

k

0=



β Bij,k ωk

=



[i] = [j], i = j, β ∈ J1 ,

α Bij ωαβ ,

α∈J2

k

0=

(7.84)



β Bii,k ωk =



α Bii ωαβ +

α∈J1

k



α Bii ωαβ , ∀ i and β ∈ J1 .

k

0=

 k

(7.87)

β∈J2

α Bij,k ωk

=

k

0=

(7.86)

α∈J2

Analogously, we have   β α 0= (Bii,k − Bs¯αs¯,k )ωk = (Bii − Bs¯βs¯)ωβα , 

(7.85)



i ∈ Is , 1 ≤ s ≤ s0 , α ∈ J3 ; β Bij ωβα ,

β∈J2 α Bii,k ωk =



β∈J1

β Bii ωβα +

[i] = [j], i = j, α ∈ J3 , 

β Bii ωβα , ∀ i and α ∈ J3 .

(7.88) (7.89)

β∈J2

According to (7.41), we know that rank { vij , zkk − zs¯s¯ | i = j, k ∈ Is , 1 ≤ s ≤ s0 } = rank { vij , zkk | i = j, k ∈ Is , 1 ≤ s ≤ s0 } = p¯, Then, from (7.84), (7.85) and (7.81), we easily see that ωαβ = 0 for all β ∈ J1 and α ∈ J2 . From this and (7.86) and (7.74), we easily get ωαβ = 0 for all α, β ∈ J1 . Similarly, by (7.87), (7.88) and (7.41), (7.81) we can show that ωαβ = 0 for all β ∈ J2 and α ∈ J3 , which together with (7.89) and (7.74) further implies that ωαβ = 0 for all β ∈ J1 and α ∈ J3 . In conclusion, we have proved that ωβα = 0 for β ∈ J1 and for all α. Thus, Claim 7.2 immediately follows. On the other hand, from (7.80) and (7.82), we see that the tensors A, B and g satisfy the equation (3.1). Then, due to that ξ is parallel, applying Theorem 3.1, we have completed the proof of Proposition 7.3.  Finally, for case III-(ii)-(b), we have the last main result in this section as follows. Proposition 7.4. Let x : M n → Sn+p be an immersed umbilic-free submanifold with parallel M¨ obius second fundamental form and its Blaschke tensor has t

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distinct eigenvalues. If 3 ≤ t ≤ p + 1 and that there exists s0 ≥ 1 such that r1 ≥ 2, . . . , rs0 ≥ 2, rs0 +1 = · · · = rt = 1 and r = t − 2, then x is M¨ obius equivalent to an open part of the immersion as stated in (iv) of the Classification Theorem. Proof. The assumption r = t−2 implies that there are (t−2) linearly independent vectors in { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt¯t¯}, without loss of generality, we assume that { u¯1¯1 , . . . , u s¯0 s¯0 , vs0 +1 s0 +1 , . . . , vt−2 t−2 } are linearly independent. Then, according to Lemma 7.5, we can adjust the choice of the normal frame {Eα }, such that ⎧ u = (B¯1n+1 , 0, . . . , 0), ¯ ⎪ 1 ⎪ ¯1¯1 ⎪ ⎪ ⎪ ······ ······ ⎪ ⎪ ⎪ ⎪ n+s0 ⎪ u s¯ s¯ = (Bs¯n+1 ⎪ ¯0 , . . . , Bs¯0 s¯0 , 0, . . . , 0), 0s ⎪ ⎨ 0 0 0 +1 , . . . , Bsn+s , 0, . . . , 0), vs0 +1 s0 +1 = (Bsn+1 (7.90) 0 +1 s0 +1 0 +1 s0 +1 ⎪ ⎪ ⎪ · · · · · · · · · · · · ⎪ ⎪ ⎪ ⎪ ⎪ v n+1 n+t−2 ⎪ ⎪ t−1 t−1 = (Bt−1 t−1 , . . . , Bt−1 t−1 , 0, . . . , 0), ⎪ ⎪ ⎩ n+t−2 vt¯t¯ = (Bt¯n+1 , 0, . . . , 0), t¯ , . . . , Bt¯t¯ = 0 for all 1 ≤ s ≤ t − 2, and moreover where Bs¯n+s s¯ n+p n+t−1 , . . . , Bii ), i ∈ Is , 1 ≤ s ≤ s0 . zii = (0, . . . , 0, Bii

(7.91)

Similar as the proof of Lemma 6.3, making use of (4.4), (7.71) and (7.90), α we can show that Bii = constant for all i and n + 1 ≤ α ≤ n + t − 2. From (7.79) and (7.90), we can further write n+p n+t−1 , . . . , Bij ), i = j. vij = (0, . . . , 0, Bij

Now, we put Ds =

(7.92)



 zkk , vij | i, j, k ∈ Is , i = j , 1 ≤ s ≤ s0 ,   vij | i, j ∈ Is , i = j , s0 + 1 ≤ s ≤ t,

and assume that rank Ds = ps , 1 ≤ s ≤ t; p0 = 0.

(7.93)

Then, by (7.64), (7.77) and (4.6), we can easily check the following perpendicularity (7.94) Ds ⊥ Ds , ∀ s = s . It follows that by further changing the choice of {En+t−1 , . . . , En+p }, we can assume     n+t−2+ si=1 pi n+t−1+ s−1 i=1 pi , . . . , Bkk , 0, . . . , 0 , (7.95) zkk = 0, . . . , 0, Bkk k ∈ Is , 1 ≤ s ≤ s0 ;

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    n+t−2+ si=1 pi n+t−1+ s−1 i=1 pi vij = 0, . . . , 0, Bij , . . . , Bij , 0, . . . , 0 ,

(7.96)

i, j ∈ Is , i = j, 1 ≤ s ≤ t. Now we denote J = {n + 1, . . . , n + p}, J0 = {n + 1, . . . , n + t − 2}, J1 = {n + t − 1, . . . , n + t − 2 + p1 }, s−1 s    pi , . . . , n + t − 2 + pi , s = 2, . . . , t, Js = n + t − 1 + i=1

Jt+1

t   = n+t−1+ pi , . . . , n + p .

i=1

i=1

¯ ≡ 0 and the fact B β being constant for all i and From the condition ∇B ii β ∈ J0 , in view of (7.90), (7.95) and (7.96), we have  β  α 0= (Bii,k − Bs¯βs¯,k )ωk = (Bii − Bs¯αs¯)ωαβ , (7.97) α∈Js

k

i ∈ Is , 1 ≤ s ≤ s0 , β ∈ J0 ,   β α Bij,k ωk = Bij ωαβ , i, j ∈ Is , i = j, β ∈ J0 , 0= α∈Js

k

0=

 k

β Bii,k ωk =



α Bii ωαβ +

α∈J0



α Bii ωαβ , i ∈ Is , β ∈ J0 .

(7.98) (7.99)

α∈Js

For 1 ≤ s ≤ s0 , Lemma 7.3 implies that rank {vij , zkk − zs¯s¯ | i, j, k ∈ Is , i = j} = rank {vij , zkk | i, j, k ∈ Is , i = j}. (7.100) The above fact, together with (7.97), (7.98), (7.95) and (7.96), gives α ∈ Js , 1 ≤ s ≤ s0 , β ∈ J0 .

0 = ωαβ ,

Also, from (7.98) and (7.96), we derive 0 = ωαβ ,

α ∈ Js , s0 + 1 ≤ s ≤ t, β ∈ J0 .

Hence, we have the conclusion ωαβ = 0,

α ∈ J1 ∪ · · · ∪ Jt , β ∈ J0 .

(7.101)

Then, from (7.99) and (7.101), and in view of (7.90), we easily get ωαβ = 0,

α, β ∈ J0 .

(7.102)

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Analogously, for each α ∈ Jt+1 , we can derive 0=



α (Bii,k − Bs¯αs¯,k )ωk =

k



β (Bii − Bs¯βs¯)ωβα ,

(7.103)

β∈Js

i ∈ Is , 1 ≤ s ≤ s0 , α ∈ Jt+1 ;  β  α Bij,k ωk = Bij ωβα , i, j ∈ Is , i = j, α ∈ Jt+1 , 0= k

0=



β∈Js α Bii,k ωk =

k



β Bii ωβα +

β∈J0



β Bii ωβα ,

i ∈ Is , α ∈ Jt+1 .

(7.104) (7.105)

β∈Js

From (7.100), together with (7.103), (7.104), (7.95) and (7.96), we obtain α ∈ Jt+1 , β ∈ Js , 1 ≤ s ≤ s0 .

ωαβ = 0,

Similarly, from (7.104) and (7.96), we can also get α ∈ Jt+1 , β ∈ Js , s0 + 1 ≤ s ≤ t.

ωαβ = 0,

Inserting the above result into (7.105), for any i, we get 0=



α Bii,k ωk =

k



β Bii ωβα , α ∈ Jt+1 .

β∈J0

This together with (7.90) gives ωαβ = 0,

α ∈ Jt+1 , β ∈ J0 .

In summary, we have obtained that ωαβ = 0,

α ∈ Jt+1 , β ∈ J0 ∪ J1 ∪ · · · ∪ Jt .

(7.106)

Furthermore, for s, s ∈ {1, 2, . . . , t} with s = s , using (7.101), we have 0=

 k

α Bij,k ωk =



β Bij ωβα , i, j ∈ Is , α ∈ Js , s = s .

(7.107)

β∈Js

This combining with (7.93), (7.95) and (7.96) gives that ωαβ = 0,

α ∈ Js , β ∈ Js , s = s .

(7.108)

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Applying the above results on the normal connection forms to the structure equations (2.6) – (2.8), we can derive the following ⎧   α 2 ⎪ d(A Y + N − B E ) = (2A + | u | ) ωi Yi , 1 ≤ s ≤ s0 , s α s s ¯ s ¯ ⎪ s ¯ s ¯ ⎪ ⎪ ⎪ α∈J i∈I 0 s ⎪ ⎪   ⎪ ⎪ α 2 ⎪ d(A Y + N − B E ) = (2A + | v | ) ωi Yi , s0 + 1 ≤ s ≤ t, ⎪ s s s¯s¯ s¯s¯ α ⎪ ⎪ ⎪ α∈J i∈I ⎪ 0 s ⎨      β α A + dY = ω Y − ω Y + N − B E Bij ω j Eβ , i ij j i s α s ¯ s ¯ ⎪ ⎪ ⎪ j∈Is α∈J0 j∈Is ,β∈Js ⎪ ⎪ ⎪ ⎪ ⎪ i ∈ Is , 1 ≤ s ≤ t; ⎪ ⎪ ⎪  β  ⎪ ⎪ ⎪ Bij ωi Yj + ωβα Eα , β ∈ Js , 1 ≤ s ≤ t. dE = − ⎪ ⎩ β i,j∈Is

α∈Js

(7.109) If we define n+t−2   Bs¯βs¯Eβ , Ya , Eα | a ∈ Is , α ∈ Js , 1 ≤ s ≤ t. Vs = Span As Y + N − β=n+1

Then, by (7.109), we can easily check that, V1 , V2 , . . . , Vt are parallel subspaces , and Vi ⊥Vj for i = j. Thus we have along M n in Rn+p+2 1 . V1 ⊕ V2 ⊕ · · · ⊕ Vt = Rn+p+2 1 Now, similar as in the proof of Claim 6.1, we can prove that 2As + | us¯s¯|2 = 0, 1 ≤ s ≤ s0 , vs¯s¯|2 = 0, s0 + 1 ≤ s ≤ t. 2As + | Thus, we can further define ⎧ n+s As Y + N − α=n+1 Bs¯αs¯Eα ⎪ ⎪ ⎪ us := , s = 1, . . . , s0 , ⎨ 2As + | us¯s¯|2 n+t−2 ⎪ Aς Y + N − α=n+1 Bς¯ας¯Eα ⎪ ⎪ ⎩ uς := , ς = s0 + 1, . . . , t. 2Aς + | vς¯ς¯|2 Then direct calculation gives ⎧ 1 =: ci , ⎪ ⎨ 2Ai +|u¯i¯i |2 1 ui , uj 1 = 2Ai +|v¯¯|2 =: ci , ii ⎪ ⎩ 0,

(7.110)

1 ≤ i = j ≤ s0 , s0 + 1 ≤ i = j ≤ t,

(7.111)

otherwise.

Moreover, similar to the proof of (6.28), we can verify that u1 + u2 + · · · + ut = Y.

(7.112)

Similar as in the proof of Proposition 7.2, we can get the assertion that in the set {u1 , . . . , ut }, exact one element is a timelike vector, and all other (t − 1) elements are spacelike vectors. Due to the assumption that rs ≥ 2

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for 1 ≤ s ≤ s0 , we see that the timelike vector can only be some vector in {us | s0 + 1 ≤ s ≤ t}. Thus, without loss of generality, we may assume that u1 , u1 1 > 0, . . . , ut−1 , ut−1 1 > 0; ut , ut 1 < 0. From the structure equations (7.109), we further have ⎧  ⎪ ωi Yi , 1 ≤ s ≤ t, ⎪ dus = ⎪ ⎪ ⎪ i∈I s ⎪ ⎪   ⎪ ⎪ β ⎪ ωij Yj − (cs )−1 ωi us + Bij ω j Eβ , ⎨ dYi = j∈Is

j∈Is ,β∈Js

⎪ ⎪ ⎪ i ∈ Is , 1 ≤ s ≤ t; ⎪ ⎪ ⎪  β  ⎪ ⎪ ⎪ Bij ωi Yj + ωβα Eα , β ∈ Js , 1 ≤ s ≤ t. ⎪ ⎩ dEβ = − i,j∈Is

(7.113)

(7.114)

α∈Js

This shows that, for each i = 1, 2, . . . , t − 1, ui defines an mi -dimensional immersed submanifold √ ui : Mimi → Smi +pi ( ci ) → Vi =: Rmi +pi +1 √ in Smi +pi ( ci ) of codimension pi , with (Euclidean) parallel second fundamental form, whereas ut defines an mt -dimensional immersed submanifold √ ut : Mtmt → Hmt +pt ( −ct ) → Vt =: R1mt +pt +1 √ in Hmt +pt ( −ct ) of codimension pt , with parallel second fundamental form.  β Moreover, due to the fact i∈It Bii = 0 for all β ∈ Jt , we further see from √ mt mt +pt ( −ct ) is minimal. Thus by [24] (cf. Lemma (7.114) that ut : Mt → H √ 3.7.10 of [3]) the immersion ut : Mtmt → √Hmt +pt ( −ct ) is in fact totally geodesic and pt = 0, so locally Mtmt = Hmt ( −ct ). Therefore, by (7.112), the M¨ obius position vector Y of x is expressed as: √ mt−1 Y =(u1 , . . . , ut−1 , ut ) : M1m1 × · · · × Mt−1 × Hmt ( −ct ) → √ √ Sm1 +p1 ( c1 ) × · · · × Smt−1 +pt−1 ( ct−1 ) × R1mt +1 → Rn+p+2 . 1 According to Proposition 3.1, x is M¨ obius equivalent to the image of σ of a cone over the submanifold defined by √ mt−1 (u1 , . . . , ut−1 ) : M1m1 × · · · × Mt−1 → Sm1 +p1 ( c1 ) √ × · · · × Smt−1 +pt−1 ( ct−1 ) t−1 √ → S s=1 (ms +ps )+t−2 ( −ct ), which is of parallel second fundamental form. We have completed the proof of Proposition 7.4.



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8. Completion Proof of Classification Theorem Summing up the preceding discussions from Sect. 4 up to Sect. 7, more specifically, unifying all the results of Propositions 5.1 and 5.2 (corresponding to t = 2), Theorem 6.1 (corresponding to t = p + 2), Propositions 7.1, 7.2, 7.3 and 7.4 (corresponding to 3 ≤ t ≤ p + 1), and Theorem 4.2 of [26] (corresponding to t = 1), we finally obtain a complete proof of the Classification Theorem. 

Acknowledgements The authors would like to express their heartfelt thanks to the referee for helpful comments and suggestions. We would also thank Professors H. Li and C.P. Wang for fruitful cooperations and their longstanding encouragement and support with our working on the present challenging project. We especially express our thanks to Prof. X. Ma for very enlightening discussions with the topic and particularly on establishing the result of Proposition 4.1. Finally, we express our thanks also to Prof. T.Z. Li for his kindly informing us his joint paper with J. Qing and C.P. Wang [15] soon after its preliminary version was completed. The important ideas of [15] are very helpful for our finally completion of this project.

References [1] Akivis, M.A., Goldberg, V.V.: A conformal differential invariant and the conformal rigidity of hypersurfaces. Proc. Am. Math. Soc. 125, 2415–2424 (1997) [2] Backes, E., Reckziegel, H.: On symmetric submanifolds of spaces of constant curvature. Math. Ann. 263, 419–433 (1983) [3] Berndt, J., Console, S., Olmos, C.: Submanifolds and Holonomy, Research Notes in Mathematics. Chapman & Hall 434 (2003) [4] Cecil, T.E., Jensen, G.R.: Dupin hypersurfaces with three principal curvatures. Invent. Math. 132, 121–178 (1998) [5] Dillen, F., Li, H., Vrancken, L., Wang, X.: Lagrangian submanifolds in complex projective space with parallel second fundamental form. Pac. J. Math. 255, 79– 115 (2012) [6] Ferus, D.: Immersions with parallel second fundamental form. Math. Z. 140, 87–92 (1974) [7] Ferus, D.: Immersionen mit paralleler zweiter Fundamentalform Beispiele und Nicht-Beispiele. Manuscripta Math. 12, 153–162 (1974) [8] Ferus, D.: Symmetric submanifolds of Euclidean space. Math. Ann. 247, 81–93 (1980) [9] Hu, Z.J., Li, D.Y.: M¨ obius isoparametric hypersurfaces with three distinct principal curvatures. Pac. J. Math. 232, 289–311 (2007)

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[10] Hu, Z.J., Li, H.Z.: Classification of hypersurfaces with parallel M¨ obius second fundamental form in S n+1 . Sci. China Ser. A Math. 47, 417–430 (2004) [11] Hu, Z.J., Li, X.X., Zhai, S.J.: On the Blaschke isoparametric hypersurfaces in the unit sphere with three distinct Blaschke eigenvalues. Sci. China Math. 54, 2171–2194 (2011) [12] Hu, Z.J., Zhai, S.J.: M¨ obius isoparametric hypersurfaces with three distinct principal curvatures, II. Pac. J. Math. 249, 343–370 (2011) [13] Li, H.Z., Wang, C.P.: M¨ obius geometry of hypersurfaces with constant mean curvature and scalar curvature. Manuscripta Math. 112, 1–13 (2003) [14] Li, T.Z., Ma, X., Wang, C.P.: Wintgen ideal submanifolds with a low-dimensional integrable distribution. Front. Math. China 10, 111–136 (2015) [15] Li, T.Z., Qing, J., Wang, C.P.: M¨ obius curvature, Laguerre curvature and Dupin hypersurface. Adv. Math. 311, 249–294 (2017) [16] Li, X.X., Song, H.R.: A complete classification of Blaschke parallel submanifolds with vanishing M¨ obius form. Sci. China Math. 60, 1281–1310 (2017) [17] Li, X.X., Zhang, F.Y.: A M¨ obius characterization of submanifolds in real space forms with parallel mean curvature and constant scalar curvature. Manuscripta Math. 117, 135–152 (2005) [18] Liu, H.L., Wang, C.P., Zhao, G.S.: M¨ obius isotropic submanifolds in Sn . Tohoku Math. J. 53, 553–569 (2001) [19] Naitoh, H.: Totally real parallel submanifolds in P n (c). Tokyo J. Math. 4, 279– 306 (1981) [20] Naitoh, H.: Parallel submanifolds of complex space forms, I. Nagoya Math. J. 90, 85–117 (1983) [21] Naitoh, H.: Parallel submanifolds of complex space forms, II. Nagoya Math. J. 91, 119–149 (1983) [22] Naitoh, H., Takeuchi, M.: Totally real submanifolds and symmetric bounded domains. Osaka J. Math. 19, 717–731 (1982) [23] Str¨ ubing, W.: Symmetric submanifolds of Riemannian manifolds. Math. Ann. 245, 37–44 (1979) [24] Takeuchi, M.: Parallel submanifolds of space forms, Manifolds and Lie groups, In honor of Y. Matsushima (J. Hano et al., eds.), Birkh¨ auser, Boston (1981), pp. 429–447 [25] Wang, C.P.: M¨ obius geometry of submanifolds in Sn . Manuscripta Math. 96, 517–534 (1998) [26] Zhai, S.J., Hu, Z.J., Wang, C.P.: On submanifolds with parallel M¨ obius second fundamental form in the unit sphere. Int. J. Math. 25, 1450062 (2014)

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Zejun Hu and Shujie Zhai School of Mathematics and Statistics Zhengzhou University Zhengzhou 450001 People’s Republic of China email: [email protected] Shujie Zhai email: [email protected] Received: January 1, 2018. Accepted: May 28, 2018.

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Results Math