On the mean curvature of submanifolds with nullity

Abstract

In this paper, we investigate geometric conditions for isometric immersions with positive index of relative nullity to be cylinders. There is an abundance of noncylindrical n-dimensional minimal submanifolds with index of relative nullity $n-2$, fully described by Dajczer and Florit (Ill J Math 45:735–755, 2001) in terms of a certain class of elliptic surfaces. Opposed to this, we prove that nonminimal n-dimensional submanifolds in space forms of any codimension are locally cylinders provided that they carry a totally geodesic distribution of rank $n-2\ge 2,$ which is contained in the relative nullity distribution, such that the length of the mean curvature vector field is constant along each leaf. The case of dimension $n=3$ turns out to be special. We show that there exist elliptic three-dimensional submanifolds in spheres satisfying the above properties. In fact, we provide a parametrization of three-dimensional submanifolds as unit tangent bundles of minimal surfaces in the Euclidean space whose first curvature ellipse is nowhere a circle and its second one is everywhere a circle. Moreover, we provide several applications to submanifolds whose mean curvature vector field has constant length, a much weaker condition than being parallel.

1 Introduction

A fundamental concept in the theory of submanifolds is the index of relative nullity introduced by Chern and Kuiper [4]. At a point $x\in M^n$ the index of relative nullity$\nu (x)$ of an isometric immersion $f:M^n\rightarrow {\mathbb {Q}}_c^m$ is the dimension of the relative nullity tangent subspace ${\Delta }_f(x)$ of f at x, that is, the kernel of the second fundamental form $\alpha ^f$ at that point. Here, ${\mathbb {Q}}_c^m$ is the simply connected space form with curvature c, that is, the Euclidean space ${\mathbb {R}}^m,$ the sphere ${\mathbb {S}}^m$ or the hyperbolic space ${\mathbb {H}}^m$, according to whether $c=0,c=1$ or $c=-1,$ respectively. The kernels form an integrable distribution along any open subset where the index is constant and the images under f of the leaves of the foliation are totally geodesic submanifolds in the ambient space.

Cylinders are the simplest examples of submanifolds with positive index of relative nullity. An isometric immersion $f:M^n\rightarrow {\mathbb {R}}^m$ is said to be a k-cylinder if the manifold $M^n$ splits as a Riemannian product $M^n=M^{n-k}\times {\mathbb {R}}^k$ and there is an isometric immersion $g:M^{n-k}\rightarrow {\mathbb {R}}^{m-k}$ such that $f=g\times {\mathrm{id}}_{{\mathbb {R}}^k}$. A natural problem in submanifold theory is to find geometric conditions for an isometric immersion with index of relative nullity $\nu \ge k>0$ at any point to be a k-cylinder.

A fundamental result asserting that an isometric immersion $f:M^n\rightarrow {\mathbb {R}}^m$ of a complete Riemannian manifold with positive index of relative nullity must be a cylinder is Hartman’s theorem [21] that requires the Ricci curvature of $M^n$ to be nonnegative. Even for hypersurfaces, the same conclusion does not hold if instead we assume that the Ricci curvature is nonpositive. Notice that the latter is always the case if f is a minimal immersion. Counterexamples easy to construct are the complete irreducible ruled hypersurfaces of any dimension discussed in [7, p. 409].

The cylindricity of minimal submanifolds was studied in [8, 23] under global assumptions. These results are truly global in nature since there are plenty of (noncomplete) examples of minimal submanifolds of any dimension n with constant index $\nu =n-2$ that are not part of a cylinder on any open subset. They can be all locally parametrically described in terms of a certain class of elliptic surfaces (see [5, Th. 22]). Some of the many papers containing characterizations of submanifolds as cylinders without the requirement of minimality are [6, 20, 21, 26].

In this paper, we deal with nonminimal n-dimensional submanifolds of arbitrary codimension and index of relative nullity $\nu \ge n-2$ at any point. Our aim is to provide geometric conditions, in terms of the mean curvature, for an isometric immersion to be a cylinder. The choice of the geometric condition is inspired by the observation that cylinders are endowed with a totally geodesic distribution contained in the relative nullity distribution, such that the mean curvature is constant along each leaf. Throughout the paper, the mean curvature of an isometric immersion f is defined as the length $H=\Vert \mathcal {H}\Vert$ of the mean curvature vector field given by $\mathcal {H}=\mathrm {trace}(\alpha ^f)/n.$

The following result provides a characterization of cylinders of dimension $n\ge 4.$

Theorem 1

Let$f:M^n \rightarrow {\mathbb {Q}}^{n+p}_c, n\ge 4,$be an isometric immersion such that$M^n$carries a totally geodesic distributionDof rank$n-2$satisfying$D(x)\subseteq {\Delta }_f(x)$for any$x\in M^n$. If the mean curvature offis constant along each leaf ofD, then eitherfis minimal or$c=0$andfis locally a$(n-2)$-cylinder over a surface on the open subset where the mean curvature is positive. Moreover, the submanifold is globally a cylinder if the leaves ofDare complete.

It is interesting that the above theorem fails for substantial three-dimensional submanifolds of codimension $p\ge 2$. Being substantial means that the codimension cannot be reduced. We show that besides cylinders, there exist elliptic three-dimensional submanifolds in spheres satisfying the properties assumed in Theorem 1. Thus, the submanifolds being three-dimensional are special. The notion of elliptic submanifolds was introduced in [5]. In fact, the following result allows a parametrization of them in terms of minimal surfaces in the Euclidean space, the so-called bipolar parametrization, using the following construction.

Let $g:L^2 \rightarrow {\mathbb {R}}^{n+1}, n\ge 5,$ be a minimal surface. The map $\Phi _g:T^1L\rightarrow {\mathbb {S}}^n$ defined on the unit tangent bundle of $L^2$ and given by

$$\begin{aligned} \Phi _g(x,w)=g_{*_x}w \end{aligned}$$

(1)

parametrizes (outside singular points) an immersion with index of relative nullity at least one at any point.

Theorem 2

Let$f:M^3 \rightarrow {\mathbb {Q}}^{3+p}_c$be an isometric immersion such that$M^3$carries a totally geodesic distributionDof rank one satisfying$D(x)\subseteq {\Delta }_f(x)$for any$x\in M^3$. If the mean curvature offis constant along each integral curve ofD, then one of the following holds:

• (i) The immersionfis minimal.

• (ii) $c=0$andfis locally a cylinder over a surface.

• (iii) $c=1$and the immersionfis elliptic and locally parametrized by (1), where$g:L^2 \rightarrow {\mathbb {R}}^{n+1}, n\ge 5,$is a minimal surface whose first curvature ellipse is nowhere a circle and the second curvature ellipse is everywhere a circle.

Minimal surfaces satisfying the conditions in part (iii) of the above theorem can be constructed using the Weierstrass representation by choosing appropriately the holomorphic data. It is worth noticing that minimal surfaces in the Euclidean space that satisfy the Ricci condition, or equivalently are locally isometric to a minimal surface in ${\mathbb {R}}^3,$ fulfill these conditions (see Sect. 6 for details). These surfaces were classified by Lawson [25].

The above results allow us to provide applications to submanifolds with constant mean curvature and not necessarily constant positive index of relative nullity.

Having constant mean curvature is a much weaker restriction on the mean curvature vector field than being parallel in the normal bundle. One can check that three-dimensional elliptic submanifolds described in Theorem 2 do not have parallel mean curvature vector field along the totally geodesic distribution. Combining this with Theorem 1, it follows that a submanifold is locally a cylinder provided that it carries a totally geodesic distribution of rank $n-2\ge 1$ that is contained in the relative nullity distribution, along which the mean curvature vector field is parallel in the normal connection.

Opposed to the fact that there is an abundance of noncylindrical n-dimensional minimal submanifolds with index of relative nullity $n-2$ (see [5]), we prove the following result for submanifolds with constant positive mean curvature.

Theorem 3

Let$f:M^n \rightarrow {\mathbb {Q}}^{n+p}_c,n\ge 3,$be a nonminimal isometric immersion with index of relative nullity$\nu \ge n-2$at any point. If the mean curvature offis constant and either$n\ge 4$or$n=3$and$p=1$, then$c=0$. Moreover, there exists an open dense subset$V\subseteq M^n$such that every point has a neighborhood$U\subseteq V$so thatf(U) is an open subset of the image of a cylinder either over a surface in${\mathbb {R}}^{p+2}$, or over a curve in${\mathbb {R}}^{p+1}$with constant first Frenet curvature.

The following is an immediate consequence of the above result due to real analyticity of hypersurfaces with constant mean curvature.

Corollary 4

Let$f:M^n \rightarrow {\mathbb {Q}}^{n+1}_c, n\ge 3$, be a nonminimal isometric immersion with index of relative nullity$\nu \ge n-2$. If the mean curvature offis constant, then$c=0$andf(M) is an open subset of the image of a cylinder over a surface in${\mathbb {R}}^3$of constant mean curvature.

The next result extends Corollary 1 in [3] for hypersurfaces in every space form without any global assumption.

Corollary 5

Let$f:M^n \rightarrow {\mathbb {Q}}^{n+1}_c, n\ge 3$, be an isometric immersion with constant mean curvature. If$M^n$has sectional curvature$K\le c$, then eitherfis minimal or$c=0$andf(M) is an open subset of the image of a cylinder over a surface in${\mathbb {R}}^3$of constant mean curvature. In the latter case, fis a cylinder over a circle provided that$M^n$is complete.

The following rigidity result that was proved in [6] for $c=0$ is another consequence of our main results.

Corollary 6

Any nonminimal isometric immersion$f:M^n \rightarrow {\mathbb {Q}}^{n+1}_c,n\ge 3$, with constant mean curvature is rigid, unless$c=0$and f(M) is an open subset of the image of a cylinder over a surface in${\mathbb {R}}^3$of constant mean curvature.

Our next result extends to any dimension a well-known theorem for constant mean curvature surfaces due to Klotz and Osserman [24] (see [2] for another extension).

Theorem 7

Let$f:M^n \rightarrow {\mathbb {Q}}_c^{n+1},n\ge 3,$be an isometric immersion with constant mean curvature, where$c=0$ or $c=1$. If$M^n$is complete and its extrinsic curvature does not change sign, then eitherfis minimal or totally umbilical or a cylinder over a sphere of dimension$1\le k < n.$

For submanifolds with constant mean curvature of codimension two, we prove the following.

Theorem 8

Let$f:M^n \rightarrow {\mathbb {R}}^{n+2}, n\ge 3,$be a nonminimal isometric immersion with constant mean curvature. If the sectional curvature of$M^n$is nonpositive, then there exists an open dense subset$V\subseteq M^n$such that every point has a neighborhood$U\subseteq V$where one of the following holds:

• (i) The neighborhoodUsplits as a Riemannian product$U=M^2\times W^{n-2}$such that$f|_U=g\times j$is a product, where$g:M^2 \rightarrow {\mathbb {R}}^4$is a surface with constant mean curvature and$j:W^{n-2}\rightarrow {\mathbb {R}}^{n-2}$is the inclusion.

• (ii) The immersion onUis a composition$f|_U=h\circ F,$ where$h=\gamma \times id_{{\mathbb {R}}^{n-1}} :{\mathbb {R}}\times {\mathbb {R}}^n\rightarrow {\mathbb {R}}^{n+2}$is cylinder over a unit speed plane curve$\gamma (s)$with curvaturek(s) and$F:M^n \rightarrow {\mathbb {R}}^{n+1}$is a hypersurface. Moreover, the mean curvature$H_F$ofFis given by

  $$\begin{aligned} H_F^2=H_f^2-\frac{1}{n^2}k^2\circ F_a\left( 1-\langle \xi , a \rangle ^2\right) ^2, \end{aligned}$$

  where$F_a=\langle F,a \rangle$and$\langle \xi , a \rangle$are the height functions ofFand its Gauss map$\xi$relative to the unit vector$a=\partial /\partial s,$ respectively.

• (iii) The neighborhoodUsplits as a Riemannian product$U=M^2_1\times M_2^2\times W^{n-4}$such that$f|_U=g_1\times g_2\times j$is a product, where$g_i:M^2_i \rightarrow {\mathbb {R}}^3, i=1,2,$are surfaces with constant mean curvature and$j:W^{n-4}\rightarrow {\mathbb {R}}^{n-4}$is the inclusion.

For constant sectional curvature submanifolds with constant mean curvature of codimension two, we prove the following theorem that extends results in [11, 14].

Theorem 9

Let$f:M_{{\tilde{c}}}^n\rightarrow {\mathbb {Q}}_c^{n+2},n\ge 3,$be an isometric immersion of a Riemannian manifold of constant sectional curvature${\tilde{c}}$. If the mean curvature offis constant and either$n\ge 4$or$n=3$and$c=\tilde{c}$, then one of the following holds:

• (i) fis totally geodesic or totally umbilical.

• (ii) $\tilde{c}=c=0$and$f=g\times j,$where$g:M^2\rightarrow {\mathbb {R}}^4$is a flat surface with constant mean curvature and$j:W\rightarrow {\mathbb {R}}^{n-2}$is an inclusion.

• (iii) $\tilde{c}=0, c=-1$andfis a composition$f=i\circ F$, where$i:{\mathbb {R}}^{n+1}\rightarrow {\mathbb {H}}^{n+2}$is the inclusion as a horosphere and$F:M^n_{\tilde{c}} \rightarrow {\mathbb {R}}^{n+1}$is cylinder over a circle.

Cylinder theorems for complete minimal Kähler submanifolds were proved in [9, 19]. For Kähler submanifolds with constant mean curvature, we prove the following results.

Theorem 10

Let$f:M^n \rightarrow {\mathbb {R}}^{n+1},n\ge 4,$be an isometric immersion with constant mean curvature. If$M^n$is Kähler, then eitherfis minimal orf(M) is an open subset of the image of a cylinder over a surface in${\mathbb {R}}^3$with constant mean curvature.

Theorem 11

Let$f:M^n \rightarrow {\mathbb {R}}^{n+2}, n\ge 4$, be a nonminimal isometric immersion of a Kähler manifold$M^n$with constant mean curvature. If the Ricci curvature or the holomorphic curvature of$M^n$is nonnegative, then there exists an open dense subset$V\subseteq M^n$such that every point has a neighborhood$U\subseteq V$where$f|_U$is as in Theorem 8.

The paper is organized as follows: In Sect. 2, we recall well-known results about the relative nullity distribution, totally geodesic distributions that are contained in the relative nullity distribution, as well as results about their splitting tensor. In Sect. 3, we fix the notation, give some preliminaries and prove auxiliary results that will be used in the proofs of our main theorems. Section 4 is devoted to the proof of Theorem 1. In Sect. 5, we recall the notion of elliptic submanifolds, as well as the associated notion of higher curvature ellipses. We also discuss the polar and bipolar surfaces of elliptic submanifolds. In Sect. 6, we study the case of three-dimensional submanifolds. We provide a parametrization for these submanifolds in terms of certain elliptic surfaces, the so-called polar parametrization (see Theorem 21). Based on this, we give the proof of Theorem 2. We conclude this section by showing that minimal surfaces in the Euclidean space that are locally isometric to a minimal surface in ${\mathbb {R}}^3$ satisfy the conditions in part (iii) of Theorem 2. In Sect. 7, we prove Theorem 3 and the applications of our main results on submanifolds with constant mean curvature. In addition, we provide examples of submanifolds as in part (ii) of Theorems 8 and 9.

2 The relative nullity distribution

In this section, we recall some basic facts from the theory of isometric immersions that will be used throughout the paper.

Let $M^n,n\ge 3,$ be a Riemannian manifold and let $f:M^n\rightarrow {\mathbb {Q}}^m_c$ be an isometric immersion into a space form ${\mathbb {Q}}^m_c$. The relative nullity subspace $\Delta _f(x)$ of f at $x\in M^n$ is the kernel of its second fundamental form $\alpha ^f:TM\times TM\rightarrow N_fM$ with values in the normal bundle, that is,

$$\begin{aligned} \Delta _f(x)=\left\{ X\in T_xM:\alpha ^f(X,Y)=0\;\;\text {for all}\;\;Y\in T_xM\right\} . \end{aligned}$$

The dimension $\nu (x)$ of $\Delta _f(x)$ is called the index of relative nullity of f at $x\in M^n$.

A smooth distribution $D\subset TM$ on $M^n$ is totally geodesic if $\nabla _TS\in \Gamma (D)$ whenever $T,S\in \Gamma (D)$. Let D be a smooth distribution on $M^n$ and $D^{\perp }$ denote the distribution on $M^n$ that assigns to each $x\in M^n$ the orthogonal complement of D(x) in $T_xM$. We write $X=X^v+X^h$ according to the orthogonal splitting $TM=D\oplus D^{\perp }$ and denote ${\nabla }^{h}_XY = (\nabla _X Y)^h$ for all $X,Y\in TM$, where $\nabla$ is the Levi-Civitá connection on $M^n.$ The splitting tensor$C:D\times D^{\perp }\rightarrow D^{\perp }$ is given by

$$\begin{aligned} C(T,X)=-{\nabla }^{h}_XT \end{aligned}$$

for any $T\in D$ and $X\in D^{\perp }$.

When D is a totally geodesic distribution such that $D(x)\subseteq {\Delta }_f(x)$ for all $x\in M^n,$ the following differential equation for the tensor $\mathcal {C}_T=C(T,\cdot )$ is well-known to hold (cf. [7] or [12]):

$$\begin{aligned} \nabla ^{h}_S \mathcal {C}_T=\mathcal {C}_T\circ \mathcal {C}_S+\mathcal {C}_{\nabla _ST} +c\langle S,T \rangle I, \end{aligned}$$

(2)

where I is the identity endomorphism of $D^{\perp }$. Here $\nabla ^{h}_S \mathcal {C}_T \in \Gamma ({\mathrm{End}} (D^\perp ))$ is defined by

$$\begin{aligned} (\nabla ^{h}_S \mathcal {C}_T)X=\nabla ^{h}_S \mathcal {C}_TX-\mathcal {C}_T\nabla ^{h}_SX \end{aligned}$$

for all $T,S\in D$ and $X\in D^{\perp }$. The Codazzi equation gives

$$\begin{aligned} \nabla _TA_{\xi }=A_{\xi }\circ \mathcal {C}_T+ A_{\nabla ^{\perp }_T\xi } \end{aligned}$$

(3)

for any $T\in D,$ where the shape operator $A_{\xi }$ with respect to the normal direction $\xi$ is restricted to $D^{\perp }$ and $\nabla ^{\perp }$ stands for the normal connection of f. In particular, the endomorphism $A_\xi \circ \mathcal {C}_T$ of $D^\perp$ is symmetric, that is,

$$\begin{aligned} A_{\xi }\circ \mathcal {C}_T=\mathcal {C}_T^t\circ A_{\xi }. \end{aligned}$$

(4)

For later use, we recall the following known results.

Proposition 12

[12, Prop. 7.4] Let$f:M^n\rightarrow {\mathbb {Q}}^m_c$be an isometric immersion such that$M^n$carries a smooth totally geodesic distributionDof rank$0<k< n$satisfying$D(x)\subseteq {\Delta }_f(x)$for all$x\in M^n$. If the splitting tensorCvanishes, then$c=0$andfis locally ak-cylinder.

Proposition 13

[12, Prop. 1.18] For an isometric immersion$f:M^n\rightarrow {\mathbb {Q}}_c^m$, the following assertions hold:

• (i) The index of relative nullity$\nu$is upper semicontinuous. In particular, the subset

  $$\begin{aligned} M_0=\{x\in M^n:\nu (x)=\nu _0\}, \end{aligned}$$

  where$\nu$attains its minimum value$\nu _0$is open.

• (ii) The relative nullity distribution$x\mapsto \Delta _f(x)$is smooth on any subset of$M^n$where$\nu$is constant.

• (iii) If$U\subseteq M^n$is an open subset where$\nu$is constant, then${\Delta }_f$is a totally geodesic (hence integrable) distribution onUand the restriction offto each leaf is totally geodesic.

3 Auxiliary results

The aim of this section is to prove several lemmas that will be used in the proofs of our main results.

Throughout this section, we assume that $f:M^n \rightarrow {\mathbb {Q}}^{n+p}_c,n\ge 3,$ is a nonminimal isometric immersion such that $M^n$ carries a smooth totally geodesic distribution D of rank $n-2$ satisfying $D(x)\subseteq {\Delta }_f(x)$ for any $x\in M^n$. We also assume that the mean curvature of f is constant along each leaf of D.

Hereafter, we work on the open subset where the mean curvature is positive and choose a local orthonormal frame $\xi _{n+1}, \dots , \xi _{n+p}$ in the normal bundle $N_fM$, such that $\xi _{n+1}$ is collinear to the mean curvature vector field. We also choose a local orthonormal frame $e_1, \dots , e_n$ in the tangent bundle TM such that $e_1, e_2$ span $D^\perp$ and diagonalize $A_{\xi _{n+1}}|_{D^\perp }$, where $A_{\xi _{n+1}}$ denotes the shape operator of f with respect to $\xi _{n+1}$. Then, we have $A_{\xi _{n+1}}e_i=k_ie_i, \,\ i=1,2,$ and consequently the mean curvature is given by $nH=k_1 +k_2,$ where $k_1,k_2$ are the principal curvatures.

Since the mean curvature is positive, at least one of the principal curvatures $k_1$ and $k_2$ has to be different from zero. In the sequel, we assume without loss of generality, that $k_1\ne 0$ and define the function

$$\begin{aligned} \rho =-\frac{k_2}{k_1}. \end{aligned}$$

On the open subset where the mean curvature is positive we have

$$\begin{aligned} k_1=-\frac{nH}{\rho -1}\,\ \,\ \text { and }\,\ \,\ k_2=\frac{n\rho H}{\rho -1}. \end{aligned}$$

(5)

The above-mentioned notation is used throughout the paper.

The following lemma gives the form of the splitting tensor.

Lemma 14

On the open subset where the mean curvature is positive, the splitting tensor is given by

$$\begin{aligned} \mathcal {C}_T=\psi _1(T)L_1 + \psi _2(T)L_2 \end{aligned}$$

for any$T\in \Gamma (D),$where$\psi _1, \psi _2$ are 1-forms dual to the vector fields$\nabla _{e_2}e_2, \nabla _{e_1}e_2,$ respectively, and$L_1$, $L_2$$\in \Gamma ({\mathrm{End}}(D^\perp ))$are defined by$L_1e_1=\rho e_1=-L_2e_2$ and$L_1e_2=e_2 =L_2e_1$. Moreover, the following holds:

$$\begin{aligned}&T(k_1)=\rho k_1\psi _1(T)+\sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle , \end{aligned}$$

(6)

$$\begin{aligned}&T(k_2)=k_2\psi _1(T)-\sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle , \end{aligned}$$

(7)

$$\begin{aligned}&(k_1-k_2)\omega (T)=k_2\psi _2(T) + \sum _{\alpha ={n+2}}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_2 \rangle , \end{aligned}$$

(8)

$$\begin{aligned}&(k_1-k_2)\omega (T)=-\rho k_1\psi _2(T)+ \sum _{\alpha ={n+2}}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_2 \rangle \end{aligned}$$

(9)

for any$T\in \Gamma (D)$, where$\omega$denotes the connection form given by $\omega =\langle \nabla e_1,e_2 \rangle$.

Proof

From the Codazzi equation, we have

$$\begin{aligned} \big (\nabla _{T}A_{\xi _{n+1}}\big )e_i - \big (\nabla _{e_i}A_{\xi _{n+1}}\big )T=A_{\nabla ^\perp _T{\xi _{n+1}}}e_i - A_{\nabla ^\perp _{e_i}{\xi _{n+1}}}T \end{aligned}$$

for any $T\in \Gamma (D)$ and $i=1,2.$ The above is equivalent to the following:

$$\begin{aligned}&T(k_1)=k_1\langle \nabla _{e_1}e_1,T \rangle +\sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle , \\&T(k_2)=k_2\langle \nabla _{e_2}e_2,T \rangle -\sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle , \\&(k_1-k_2)\omega (T)=k_2\langle \nabla _{e_1}e_2,T \rangle + \sum _{\alpha ={n+2}}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_2 \rangle , \\&(k_1-k_2)\omega (T)=k_1\langle \nabla _{e_2}e_1,T \rangle + \sum _{\alpha ={n+2}}^{n+p}\langle \nabla ^\perp _{T}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_2 \rangle . \end{aligned}$$

Using the assumption that the mean curvature is constant along each leaf of the distribution D, the first two equations imply

$$\begin{aligned} \langle \nabla _{e_1}e_1,T \rangle =\rho \langle \nabla _{e_2}e_2,T \rangle \end{aligned}$$

for any $T\in \Gamma (D).$ Additionally, the last two equations yield

$$\begin{aligned} \langle \nabla _{e_2}e_1,T \rangle =-\rho \langle \nabla _{e_1}e_2,T \rangle . \end{aligned}$$

Now the structure of the splitting tensor and (6)-(9) follow easily from the above. $\square$

Lemma 15

Let$e_r, r\ge 3$, be an orthonormal frame of the distributionD. Then the functions$u_r:=\psi _1(e_r)$and$v_r:=\psi _2(e_r)$satisfy

$$\begin{aligned} 2\rho (u_ru_s + v_rv_s)-c\delta _{rs} = \frac{\rho -1}{nH}\sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{e_r}\xi _{n+1},\xi _{\alpha } \rangle \big (u_s\langle A_{\xi _\alpha }e_1,e_1 \rangle -v_s\langle A_{\xi _\alpha }e_1,e_2 \rangle \big ) \end{aligned}$$

(10)

for all$r, s\ge 3$, where$\delta _{rs}$is the Kronecker delta.

Proof

Using Lemma 14, we have

$$\begin{aligned} (\nabla ^h_{e_r}\mathcal {C}_{e_s})=e_r(u_s)L_1+e_r(v_s)L_2+u_s\nabla ^h_{e_r}L_1+v_s\nabla ^h_{e_r}L_2 \end{aligned}$$

(11)

for any $r,s\ge 3$. A direct computation yields

$$\begin{aligned}&(\nabla ^h_{e_r}L_1)e_1=-(\nabla ^h_{e_r}L_2)e_2=e_r(\rho )e_1+(\rho -1)\omega (e_r)e_2, \end{aligned}$$

(12)

$$\begin{aligned}&(\nabla ^h_{e_r}L_1)e_2=(\nabla ^h_{e_r}L_2)e_1=(\rho -1)\omega (e_r)e_1. \end{aligned}$$

(13)

Then (6) and (7) imply that

$$\begin{aligned} e_r(\rho )=-\rho (\rho -1)u_r + \frac{(\rho -1)^2}{nH}\sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{e_r}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle . \end{aligned}$$

(14)

From (2), we know that the splitting tensor satisfies

$$\begin{aligned}&(\nabla _{e_r}^h\mathcal {C}_{e_s})e_i=\mathcal {C}_{e_s}\circ \mathcal {C}_{e_r}e_i + \mathcal {C}_{\nabla _{e_r}e_s}e_i +c\delta _{rs}e_i \end{aligned}$$

(15)

for any $r,s\ge 3$ and $i=1,2$.

Let $\omega _{rs}$ be the connection form given by $\omega _{rs}=\langle \nabla e_r,e_s \rangle$ for all $r,s\ge 3$. Using (11)-(14), we find that (15) for $i=1$ is equivalent to

$$\begin{aligned} \rho e_r(u_s)=&\rho (2\rho -1)u_ru_s-\rho v_rv_s -(\rho -1)v_s\omega (e_r)\qquad \nonumber \\&-u_s\frac{(\rho -1)^2}{nH} \sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{e_r}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle + \rho \sum _{t\ge 3}^n\omega _{st}(e_r)u_t+c\delta _{rs} \end{aligned}$$

(16)

and

$$\begin{aligned} e_r(v_s)=\rho u_r v_s +u_s v_r-(\rho -1)u_s\omega (e_r)+\sum _{t\ge 3}^n\omega _{st}(e_r)v_t \end{aligned}$$

(17)

for all $r,s\ge 3$. Moreover, (15) for $i=2$ implies that

$$\begin{aligned} e_r(u_s)=&u_ru_s-\rho v_rv_s+(\rho -1)v_s\omega (e_r)+\sum _{t\ge 3}^n\omega _{st}(e_r)u_t+c\delta _{rs} \end{aligned}$$

(18)

for all $r,s\ge 3$.

Combining (16) and (17), we obtain

$$\begin{aligned} 2\rho u_ru_s + \rho v_rv_s - c\delta _{rs}- v_s(\rho +1)\omega (e_r)=u_s\frac{\rho -1}{nH} \sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{e_r}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_1 \rangle . \end{aligned}$$

Using (5), it is easily seen that (9) is written as

$$\begin{aligned} (\rho +1)\omega (e_r)=-\rho v_r-\frac{\rho -1}{nH} \sum _{\alpha =n+2}^{n+p}\langle \nabla ^\perp _{e_r}\xi _{n+1},\xi _{\alpha } \rangle \langle A_{\xi _\alpha }e_1,e_2 \rangle , \end{aligned}$$

(19)

and now (10) follows directly from the above two equations. $\square$

We recall that the first normal space$N_1^f(x)$ of the immersion f at a point $x\in M^n$ is the subspace of its normal space $N_fM(x)$ spanned by the image of its second fundamental form $\alpha ^f$ at x, that is,

$$\begin{aligned} N_1^f(x)={\mathrm{span}}\left\{ \alpha ^f(X,Y): X,Y \in T_xM\right\} . \end{aligned}$$

The rank condition and the symmetry of the second fundamental form imply that $\dim N_1^f(x)\le 3$ for all $x\in M^n$.

Consider the open subset

$$\begin{aligned} M_3=\left\{ x\in M^n:\dim N_1^f(x)=3\right\} . \end{aligned}$$

Lemma 16

The splitting tensor vanishes on the open subset$M_3^*:=M_3\smallsetminus \{x\in M^n: H(x)=0\}$.

Proof

On the subset $M_3^*,$ we consider the orthogonal splitting $N_1^f=\hat{N}^f_1 \oplus {\mathrm{span}}\{\mathcal {H}\}.$ Choose the local frame such that $\xi _{n+1}$ is collinear to the mean curvature vector field $\mathcal {H}$, and $\xi _{n+2},\xi _{n+3}$ span the plane bundle $\hat{N}^f_1$. Then, we have

$$\begin{aligned} \text {trace}A_{\xi _{n+2}}|_{D^\perp }=0=\text {trace}A_{\xi _{n+3}}|_{D^\perp }. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} A_{\xi _{n+2}}|_{D^\perp }\circ J=J^t\circ A_{\xi _{n+2}}|_{D^\perp } \,\ \text {and} \,\ A_{\xi _{n+3}}|_{D^\perp }\circ J=J^t\circ A_{\xi _{n+3}}|_{D^\perp }, \end{aligned}$$

where J denotes the unique, up to a sign, almost complex structure acting on the plane bundle $D^\perp$.

It follows using (4) that

$$\begin{aligned} A_{\xi _{n+2}}|_{D^\perp }\circ \mathcal {C}_T=\mathcal {C}_T^t\circ A_{\xi _{n+2}}|_{D^\perp } \,\ \text {and} \,\ A_{\xi _{n+3}}|_{D^\perp }\circ \mathcal {C}_T=\mathcal {C}_T^t\circ A_{\xi _{n+3}}|_{D^\perp } \end{aligned}$$

for any $T\in \Gamma (D)$. Since $\hat{N}^f_1$ is a plane bundle, the above imply that $\mathcal {C}_T\in {\mathrm{span}}\{I,J\}\subseteq {\mathrm{End}}(D^\perp )$. This, combined with Lemma 14, yields

$$\begin{aligned} (\rho -1)\psi _1(T)=0 \,\ \text {and} \,\ (\rho -1)\psi _2(T)=0 \end{aligned}$$

for any $T\in \Gamma (D)$. Thus, the splitting tensor vanishes identically on $M_3^*$. $\square$

Hereafter, we assume that $M_3$ is not dense on $M^n$ and consider the open subset

$$\begin{aligned} M_2=\left\{ x\in M^n\smallsetminus {\overline{M}}_3:\dim N_1^f(x)=2\right\} . \end{aligned}$$

In the sequel, we assume that the open subset $M_2^*:=M_2\smallsetminus \{x\in M^n: H(x)=0\}$ is nonempty. Choose a local orthonormal frame such that $\xi _{n+1}$ and $\xi _{n+2}$ span the plane bundle $N_1^f$ on this subset and $\xi _{n+1}$ is collinear to the mean curvature vector field. Thus, there exist smooth functions $\lambda ,\mu$ such that

$$\begin{aligned} A_{\xi _{n+2}}e_1=\lambda e_1 + \mu e_2, \,\ A_{\xi _{n+2}}e_2=\mu e_1 - \lambda e_2 \,\,\,\, \text {and}\,\,\,\, \lambda ^2+\mu ^2>0. \end{aligned}$$

We proceed with some auxiliary lemmas.

Lemma 17

The plane bundle$N_1^f$is parallel in the normal connection along the distributionDon the subset$M_2^*$. Moreover, the following holds:

$$\begin{aligned}&\mu \psi _1(T)=-\lambda \psi _2(T), \end{aligned}$$

(20)

$$\begin{aligned}&\mu \phi (T)=-(\lambda ^2+\mu ^2)\frac{\rho -1}{nH}\psi _2(T),\end{aligned}$$

(21)

$$\begin{aligned}&T(\mu )+2\lambda \omega (T)+(\rho +1)\lambda \psi _2(T)=0, \end{aligned}$$

(22)

$$\begin{aligned}&T(\lambda )-2\mu \omega (T)-\mu \rho \psi _2(T)-\lambda \psi _1(T)=\frac{n\rho H}{\rho -1}\phi (T), \end{aligned}$$

(23)

$$\begin{aligned}&T(\lambda )-2\mu \omega (T)-\mu \psi _2(T)-\lambda \rho \psi _1(T)=\frac{nH}{\rho -1}\phi (T) \end{aligned}$$

(24)

for any$T\in \Gamma (D)$, where$\phi$is the normal connection form given by$\phi =\langle \nabla ^\perp \xi _{n+1},\xi _{n+2} \rangle .$

Proof

It follows from (3) that

$$\begin{aligned} \langle \nabla ^\perp _T\xi _{\alpha }, \xi \rangle =0 \;\;\text {if}\;\; \alpha =n+1,n+2 \end{aligned}$$

for any $T\in \Gamma (D)$ and any $\xi \in \Gamma ({N^f_1}^{\perp })$. Thus, the subbundle $N_1^f$ is parallel in the normal connection along the distribution D.

Moreover, from (3) we have

$$\begin{aligned} (\nabla _TA_{\xi _{n+2}})e_i=A_{\xi _{n+2}}\circ \mathcal {C}_Te_i+ A_{\nabla ^{\perp }_T\xi _{n+2}}e_i, \;\;i=1,2, \end{aligned}$$

for any $T\in \Gamma (D)$. Bearing in mind the form of the splitting tensor given in Lemma 14, the above equations yield directly (23), (24) and the following

$$\begin{aligned}&T(\mu )+2\lambda \omega (T)+\lambda \rho \psi _2(T)-\mu \psi _1(T)=0, \\&T(\mu )+2\lambda \omega (T)-\mu \rho \psi _1(T)+\lambda \psi _2(T)=0 \end{aligned}$$

for any $T\in \Gamma (D)$. Subtracting the above equations, we obtain (20). Similarly, (21) follows by subtracting (23), (24) and using (20). Finally, plugging (20) into the first of the above equations, we obtain (22). $\square$

Now suppose that the subset ${M_3\cup M_2}$ is not dense on $M^n$ and consider the open subset

$$\begin{aligned} M_1=\left\{ x\in M^n\smallsetminus \overline{M_3\cup M_2}: \dim N_1^f(x)=1\right\} . \end{aligned}$$

Lemma 18

If the subset$M_1^*:=M_1\smallsetminus \{x\in M^n: H(x)=0\}$ is nonempty, then$c=0$and$f|_{M_1^*}$is locally a cylinder either over a surface in${\mathbb {R}}^{p+2}$or over a curve in${\mathbb {R}}^{p+1}$.

Proof

On the subset $M_1^*$ we choose a local orthonormal frame $\xi _{n+1},\dots ,\xi _{n+p}$ in the normal bundle such that $\xi _{n+1}$ is collinear to the mean curvature vector field. Then we have $A_{\xi _\alpha }=0$ for all $\alpha \ge n+2.$ The Codazzi equation yields

$$\begin{aligned} A_{\nabla ^\perp _{e_i}\xi _\alpha }e_r=A_{\nabla ^\perp _{e_r}\xi _\alpha }e_i \end{aligned}$$

for all $\alpha \ge n+2,$$i=1,2,$ and $r\ge 3$. Thus, we obtain $\nabla ^\perp _{e_r}\xi _{n+1}=0$ and Lemma 15 gives

$$\begin{aligned} 2\rho (u_ru_s + v_rv_s)=c\delta _{rs} \end{aligned}$$

(25)

for all $r\ge 3$. Moreover, (14) becomes

$$\begin{aligned} e_r(\rho )=-\rho (\rho -1)u_r. \end{aligned}$$

Differentiating (25) with respect to $e_r$ and using the above along with (17) and (18), we obtain

$$\begin{aligned} \rho u_r(\rho -3)(u_r^2+v_r^2)-2c\rho u_r + 2\rho \sum _{s\ge 3}^n\omega _{rs}(e_r)\big (u_su_r +v_sv_r\big )=0 \end{aligned}$$

for all $r\ge 3$. In view of (25), the above equation simplifies to the following

$$\begin{aligned} c(\rho +1)u_r=0. \end{aligned}$$

Now we prove that $c=0$. Arguing indirectly, we suppose that $c\ne 0.$ Assume that the open set of points where $\rho \ne -1$ is nonempty. On this subset, we have $u_r=0$ for all $r\ge 3.$ Thus, (25) becomes $2\rho v_r^2=c$ for all $r\ge 3$. Using (19), (18) yields $2\rho ^2 v_r^2=c(\rho +1),$ which is a contradiction. Assume now that the set of points where $\rho =-1$ has nonempty interior. On this subset, (6) yields $u_r=0$ and (8) implies that $v_r=0,$ which contradicts the assumption that $c\ne 0.$

Hence, $c=0$ and (25) becomes

$$\begin{aligned} \rho (u_r^2+v_r^2)=0 \end{aligned}$$

for all $r\ge 3.$ If $\rho \ne 0,$ then the splitting tensor vanishes and Proposition 12 implies that f is locally a cylinder over a surface. If the subset of points where $\rho =0$ has nonempty interior, then the Codazzi equation implies that the tangent bundle splits as an orthogonal sum of two parallel distributions one of which has rank $n-1$. Thus, the manifold splits locally as a Riemannian product by the De Rham decomposition theorem. Since the second fundamental form is adapted to this splitting, the result follows from [12, Th. 8.4]. $\square$

4 Submanifolds of dimension $n\ge 4$

We are now ready to give the proof of our first main result.

Proof of Theorem 1

If the open subset $M_3^*$ is nonempty, then Lemma 16 implies that the splitting tensor vanishes identically on it. Then, by Proposition 12 the immersion f is locally a cylinder over a surface on $M_3^*$.

Now assume that the subset $M_3$ is not dense on $M^n$ and suppose that $M_2^*$ is nonempty. Hereafter, we work on $M_2^*$. Due to the choice of the local orthonormal frame $\xi _{n+1}, \xi _{n+2}$ in the normal subbundle $N_1^f$, and using (20) and (21), (10) of Lemma 15 takes the following form

$$\begin{aligned} v_rv_s\left( \lambda ^2+\mu ^2\right) \Big (2\rho -\left( \lambda ^2+\mu ^2\right) \frac{(\rho -1)^2}{n^2H^2}\Big )=c\mu ^2 \delta _{rs} \end{aligned}$$

(26)

for any $r,s\ge 3$.

We claim that $v_r=0$ for any $r\ge 3$. In fact, at points where

$$\begin{aligned} 2\rho -\left( \lambda ^2+\mu ^2\right) \frac{(\rho -1)^2}{n^2H^2}\ne 0, \end{aligned}$$

it follows from (26) that

$$\begin{aligned} v_r^2=\frac{c\mu ^2}{\left( \lambda ^2+\mu ^2\right) \left( 2\rho -\left( \lambda ^2+\mu ^2\right) \frac{(\rho -1)^2}{n^2H^2}\right) } \end{aligned}$$

for any $r\ge 3$ and $v_rv_s=0$ for $r\ne s\ge 3$. Thus, $v_r=0$ for any $r\ge 3$ at those points.

It remains to prove that the same holds on the subset $U\subseteq M_2^*$ of points where

$$\begin{aligned} 2\rho -\left( \lambda ^2+\mu ^2\right) \frac{(\rho -1)^2}{n^2H^2}=0. \end{aligned}$$

Notice that because of (5), the subset U is the set of points where

$$\begin{aligned} \lambda ^2+\mu ^2=-2k_1k_2. \end{aligned}$$

(27)

In order to prove that $v_r=0$ for any $r\ge 3$ on U, we suppose that the interior of U is nonempty. Suppose to the contrary that there exists $r_0\ge 3$ such that $v_{r_0}\ne 0$ on an open subset of U. Differentiating (27) with respect to $e_{r_0}$ and using (5), (6), (7), (22) and (23), we obtain

$$\begin{aligned} \lambda ^2u_{r_0}-\lambda \mu v_{r_0} + (\rho +1)k_1k_2 u_{r_0}=\lambda (k_1-2k_2)\phi (e_{r_0}). \end{aligned}$$

Multiplying by $\mu$ the above and using (21), we find that

$$\begin{aligned} \mu u_{r_0}\left( \lambda ^2+ (\rho +1)k_1k_2\right) = \lambda v_{r_0}\Big (\mu ^2- (\lambda ^2+\mu ^2)(k_1-2k_2)\frac{\rho -1}{nH}\Big ). \end{aligned}$$

Taking into account (5), (20) and (27), the above yields

$$\begin{aligned} \lambda v_{r_0}(\rho +1)(\lambda ^2+\mu ^2)=0. \end{aligned}$$

Due to (27), we conclude that $\lambda =0$ and consequently $\mu \ne 0$. Then, it follows from (20) that $u_s=0$ for any $s\ge 3$. It is easily seen from (16) and (18) for $s= r_0$ that

$$\begin{aligned} \rho v^2_{r_0}+(\rho -1)v_{r_0}\omega (e_{r_0})-c=0 \;\;{\text {and}}\;\; \rho v^2_{r_0}-(\rho -1)v_{r_0}\omega (e_{r_0})-c=0. \end{aligned}$$

Hence, $\omega (e_{r_0})=0$, and consequently (19) yields

$$\begin{aligned} \rho v_{r_0}+\frac{\rho -1}{nH}\mu \phi (e_{r_0})=0. \end{aligned}$$

Using (5), (21) and (27), we find that $\rho =0$, which contradicts (27). Thus, we have proved the claim that $v_r=0$ for any $r\ge 3$.

Now, we claim that $u_r=0$ for any $r\ge 3$. It follows using (20) that $\mu u_r=0$ for any $r\ge 3$. Obviously, the function $u_r$ vanishes at points where $\mu \ne 0$.

Assume that the set of points where $\mu =0$ has nonempty interior and argue on this subset. Since $\lambda \ne 0$ on this subset, it follows from (22) that $\omega (e_r)=0$ for any $r\ge 3,$ and consequently (23) and (24) yield

$$\begin{aligned} \phi (e_r)=\frac{\rho -1}{n H}\lambda u_r\,\,\, \text {and} \,\,\, e_r(\lambda )=(\rho +1)\lambda u_r \end{aligned}$$

(28)

for all $r\ge 3$. Using the first of the above equations, (10) is written equivalently as

$$\begin{aligned} u_r^2\Big (2\rho -\frac{\lambda ^2(\rho -1)^2}{n^2H^2}\Big )=c \end{aligned}$$

(29)

for all $r\ge 3$.

Since we already proved that $v_r=0$ for all $r\ge 3$, Lemma 14 implies that the image of the splitting tensor $C:D\rightarrow {\mathrm{End}}(D^\perp )$ satisfies $\dim {{\mathrm{Im}}}\, C\le 1.$ Thus, $\dim \ker \, C\ge n-3.$

Now suppose that $\dim \ker \, C=n-3$. Then, there exists a unique $r_0\ge 3$ such that $u_{r_0}\ne 0$ and $u_s=0$ for any $s\ne r_0$. Thus, (29) implies that $c=0$ and

$$\begin{aligned} 2\rho =\frac{\lambda ^2(\rho -1)^2}{n^2H^2}. \end{aligned}$$

On account of (5), the above equation becomes $\lambda ^2=-2k_1k_2>0.$ Differentiating this equation with respect to $e_{r_0}$ and using (5), (6), (7) and the second of (28), we obtain $2\lambda ^2 + k_1k_2=0,$ which contradicts the previous equation. Thus, the splitting tensor vanishes identically on the subset $M_2^*$ and consequently, by Proposition 12, the immersion f is locally a cylinder over a surface.

If the open subset $M_1^*$ is nonempty, then Lemma 18 implies that f is locally a cylinder over a surface or over a curve. $\square$

5 Elliptic submanifolds

In this section, we recall from [5] the notion of elliptic submanifolds of a space form as well as several of their basic properties.

Let $f:M^n\rightarrow {\mathbb {Q}}_c^m$ be an isometric immersion. The $\ell ^{th}$-normal space$N^f_\ell (x)$ of f at $x\in M^n$ for $\ell \ge 1$ is defined as

$$\begin{aligned} N^f_\ell (x)=\hbox {span}\left\{ \alpha ^f_{\ell +1}\left( X_1,\ldots ,X_{\ell +1}\right) : X_1,\ldots ,X_{\ell +1}\in T_xM\right\} . \end{aligned}$$

Here $\alpha ^f_2=\alpha ^f$ and for $s\ge 3$ the so-called $s^{th}$-fundamental form is the symmetric tensor $\alpha ^f_s:TM\times \cdots \times TM\rightarrow N_fM$ defined inductively by

$$\begin{aligned} \alpha ^f_s(X_1,\ldots ,X_s)=\pi ^{s-1}\left( \nabla ^\perp _{X_s}\cdots \nabla ^\perp _{X_3}\alpha ^f(X_2,X_1)\right) , \end{aligned}$$

where $\pi ^k$ stands for the projection onto $(N_1^f\oplus \cdots \oplus N_{k-1}^f)^{\perp }$.

An isometric immersion $f:M^n\rightarrow {\mathbb {Q}}_c^m$ is called elliptic if $M^n$ carries a totally geodesic distribution D of rank $n-2$ satisfying $D(x)\subseteq {\Delta }_f(x)$ for any $x\in M^n$ and there exists an (necessary unique up to a sign) almost complex structure $J:D^\perp \rightarrow D^\perp$ such that the second fundamental form satisfies

$$\begin{aligned} \alpha ^f(X,X)+\alpha ^f(JX,JX)=0 \end{aligned}$$

for all $X\in D^\perp$. Notice that J is orthogonal if and only f is minimal.

Assume that $f:M^n\rightarrow {\mathbb {Q}}_c^m$ is substantial and elliptic. Assume also that f is nicely curved which means that for any $\ell \ge 1$ all subspaces $N^f_\ell (x)$ have constant dimension and thus form subbundles of the normal bundle. Notice that any f is nicely curved along connected components of an open dense subset of $M^n$. Then, along that subset the normal bundle splits orthogonally and smoothly as

$$\begin{aligned} N_fM=N^f_1\oplus \cdots \oplus N^f_{\tau _f}, \end{aligned}$$

(30)

where all $N^f_\ell$’s have rank two, except possibly the last one that has rank one in case the codimension is odd. Thus, the induced bundle $f^*T{\mathbb {Q}}_c^m$ splits as

$$\begin{aligned} f^*T{\mathbb {Q}}_c^m=f_*D\oplus N^f_0\oplus N^f_1\oplus \cdots \oplus N^f_{\tau _f}, \end{aligned}$$

where $N^f_0=f_*D^\perp$. Setting

$$\begin{aligned} \tau ^o_f = \left\{ \begin{array}{l} \tau _f\;\;\;\;\;\;\;\;\;\;\;\hbox {if}\;\;m-n\;\;\; \hbox {is even}\\ \tau _f-1\;\;\;\;\; \hbox {if}\;\;m-n\;\;\; \hbox {is odd} \end{array} \right. \end{aligned}$$

it turns out that the almost complex structure J on $D^\perp$ induces an almost complex structure $J_\ell$ on each $N_\ell ^f$, $0\le \ell \le \tau ^o_f$, defined by

$$\begin{aligned} J_\ell \alpha _{\ell +1}^f\left( X_1,\ldots ,X_\ell ,X_{\ell +1}\right) =\alpha _{\ell +1}^f\left( X_1,\ldots ,X_\ell ,J X_{\ell +1}\right) , \end{aligned}$$

where $\alpha _1^f=f_*$.

The $\ell ^{th}$-order curvature ellipse$\mathcal {E}_\ell ^f(x)\subset N^f_\ell (x)$ of f at $x\in M^n$ for $0\le \ell \le \tau ^o_f$ is

$$\begin{aligned} \mathcal {E}_\ell ^f(x)=\big \{\alpha ^f_{\ell +1}(Z_{\theta },\dots ,Z_{\theta }): Z_{\theta }=\cos \theta Z+\sin \theta J Z\;\;\hbox {and}\;\;\theta \in [0,\pi ) \big \}, \end{aligned}$$

where $Z\in D^\perp (x)$ has unit length and satisfies $\langle Z,JZ \rangle =0$. From ellipticity, such a Z always exists and $\mathcal {E}_\ell ^f(x)$ is indeed an ellipse.

We say that the curvature ellipse $\mathcal {E}_\ell ^f$ of an elliptic submanifold f is a circle for some $0\le \ell \le \tau ^o_f$ if all ellipses $\mathcal {E}_\ell ^f(x)$ are circles. That the curvature ellipse $\mathcal {E}_\ell ^f$ in a circle is equivalent to the almost complex structure $J_\ell$ being orthogonal. Notice that $\mathcal {E}_0^f$ is a circle if and only if f is minimal.

Let $f:M^n\rightarrow {\mathbb {Q}}_c^{m-c}, c\in \{0,1\},$ be a substantial nicely curved elliptic submanifold. Assume that $M^n$ is the saturation of a fixed cross section $L^2\subset M^n$ to the foliation of the distribution D. The subbundles in the orthogonal splitting (30) are parallel in the normal connection (and thus in ${\mathbb {Q}}_c^{m-c}$) along D. Hence, each $N^f_\ell$ can be seen as a vector bundle along the surface $L^2$.

A polar surface to f is an immersion h of $L^2$ defined as follows:

• (a) If $m-n-c$ is odd, then the polar surface $h:L^2\rightarrow {\mathbb {S}}^{m-1}$ is the spherical image of the unit normal field spanning $N^f_{\tau _f}$.

• (b) If $m-n-c$ is even, then the polar surface $h:L^2\rightarrow {\mathbb {R}}^m$ is any surface such that $h_*T_xL=N^f_{\tau _f}(x)$ up to parallel identification in ${\mathbb {R}}^m$.

Polar surfaces always exist since in case $\mathrm{(b)}$ any elliptic submanifold admits locally many polar surfaces.

The almost complex structure J on $D^\perp$ induces an almost complex structure ${\tilde{J}}$ on TL defined by $P\circ {\tilde{J}}=J\circ P$, where $P:TL \rightarrow D^\perp$ is the orthogonal projection. It turns out that a polar surface to an elliptic submanifold is necessarily elliptic. Moreover, if the elliptic submanifold has a circular curvature ellipse then its polar surface has the same property at the “corresponding” normal bundle. As a matter of fact, up to parallel identification it holds that

$$\begin{aligned} N_s^h=N_{\tau ^o_f-s}^f\;\;\hbox {and}\;\; J^h_s=\big (J^f_{\tau ^o_f-s}\big )^t,\;\; 0\le s\le \tau ^o_f. \end{aligned}$$

(31)

In particular, the polar surface is nicely curved.

A bipolar surface to f is any polar surface to a polar surface to f. In particular, if we are in case $f:M^3\rightarrow {\mathbb {S}}^{m-1}$, then a bipolar surface to f is a nicely curved elliptic surface $g:L^2\rightarrow {\mathbb {R}}^m$.

6 Three-dimensional submanifolds

In this section, we study the case of three-dimensional submanifolds and we provide the proof of Theorem 2. To this purpose, we need the following results.

Proposition 19

Let$f:M^3 \rightarrow {\mathbb {Q}}^{3+p}_c$be an isometric immersion such that$M^3$carries a totally geodesic distributionDof rank one satisfying$D(x)\subseteq {\Delta }_f(x)$for any$x\in M^3$. If the mean curvature offis constant along each integral curve ofDand the normal bundle offis flat, thenfis minimal or$c=0$andfis locally a cylinder.

Proof

Assume that f is nonminimal. If the open subset $M_3^*$ is nonempty, then Lemma 16 and Proposition 12 imply that the immersion f is locally a cylinder over a surface.

Now suppose that the open subset $M_2^*$ is nonempty and argue on it. Having flat normal bundle implies that $\mu =0$ and according to (20), we obtain $v_3=0$. Consequently, (18) is written as

$$\begin{aligned} e_3(u_3)=u_3^2+c. \end{aligned}$$

(32)

Comparing (23) and (24), we obtain

$$\begin{aligned} \phi (e_3)=\frac{\rho -1}{nH}\lambda u_3. \end{aligned}$$

Thus,

$$\begin{aligned} e_3(\lambda )=(\rho +1)\lambda u_3 \end{aligned}$$

(33)

and consequently (14) becomes

$$\begin{aligned} e_3(\rho )=u_3(\rho -1)(\tau -\rho ), \end{aligned}$$

(34)

where $\tau$ is the function given by

$$\begin{aligned} \tau =\frac{\lambda ^2(\rho -1)^2}{n^2H^2}. \end{aligned}$$

Moreover, (10) is written as $u_3^2(2\rho -\tau )=c.$ Differentiating with respect to $e_3$ and using (32)-(34), we derive that

$$\begin{aligned} u_3^2(\rho +1)(\rho -\tau )=0. \end{aligned}$$

Now we claim that $u_3=0.$ Arguing indirectly, we suppose that $u_3\ne 0$ on an open subset. Observe that $\rho \ne -1$ due to our assumption and (6). Hence, $\rho =\tau ,$ or equivalently $\rho n^2H^2=\lambda ^2(\rho -1)^2$ and $e_3(\rho )=0$ by (34). Thus, $e_3(\lambda )=0$, which contradicts (33) since $\lambda \ne 0.$ This proves the claim that $u_3=0$ and consequently the splitting tensor vanishes. That the immersion f is locally a cylinder on $M_2^*$ follows from Proposition 12.

If the open subset $M_1^*$ is nonempty, then Lemma 18 implies that the immersion f is locally a cylinder over a surface or over a curve. $\square$

Proposition 20

Let$f:M^3 \rightarrow {\mathbb {Q}}^{3+p}_c$be a nonminimal isometric immersion such that$M^3$carries a totally geodesic distributionDof rank one satisfying$D(x)\subseteq {\Delta }_f(x)$for any$x\in M^3$. If the mean curvature offis constant along each integral curve ofDandfis not locally a cylinder, then the splitting tensor offis an almost complex structure on$D^\perp .$Moreover, fis a spherical elliptic submanifold with respect to this almost complex structure and its first curvature ellipse is a circle.

Proof

Since by assumption the immersion f is not a cylinder on any open subset, it follows from Proposition 12, Lemmas 16 and  18 that the open subsets $M_3^*$ and $M_1^*$ are both empty.

Proposition 19 implies that the immersion f has nonflat normal bundle on $M_2^*.$ Thus, we have $\mu \ne 0$ and $\rho \ne -1.$ Using (20) and (21), it is easily seen that (10), (14), (17), (19), (22) and (23) are written as

$$\begin{aligned}&\omega (e_3)=-\frac{\rho -\tau }{\rho +1}v_3,\nonumber \\&e_3(\rho )=\frac{\lambda }{\mu }(\rho -1)(\rho -\tau )v_3, \end{aligned}$$

(35)

$$\begin{aligned}&e_3(\mu )=-\frac{\lambda }{\rho +1}\left( 2\tau +\rho ^2+1\right) v_3,\nonumber \\&e_3(\lambda )=\Big (\frac{2\mu }{\rho +1}\tau -\frac{2\mu \rho }{\rho +1}-\frac{\lambda ^2}{\mu }(\rho +1)\Big )v_3, \end{aligned}$$

(36)

$$\begin{aligned}&e_3(v_3)=\frac{\lambda }{\mu (\rho +1)}\left( (\rho -1)\tau -(2\rho ^2+\rho +1)\right) v_3^2,\nonumber \\&(\lambda ^2+\mu ^2)(2\rho -\tau )v_3^2=c\mu ^2, \end{aligned}$$

(37)

where $\tau$ is the function given by

$$\begin{aligned} \tau =(\lambda ^2+\mu ^2)\frac{(\rho -1)^2}{n^2H^2}. \end{aligned}$$

By differentiating (37) and using all the above equations, we obtain

$$\begin{aligned} \lambda (\lambda ^2+\mu ^2)\Big (\rho (5\rho ^2+6\rho +5)-(4\rho ^2+2\rho +4)\tau - 2\tau ^2\Big )v_3^3=c\lambda \mu ^2v_3. \end{aligned}$$

We claim that $\lambda v_3=0.$ Arguing indirectly, we assume that the open subset where $\lambda v_3\ne 0$ is nonempty. Thus, comparing the above equation with (37), we derive that $\tau =\rho .$ This along with (35) implies that $e_3(\tau )=e_3(\rho )=0.$ By the definition of $\tau ,$ it follows that $e_3(\lambda ^2+\mu ^2)=0.$ Using the above equations, it is easy to see that

$$\begin{aligned} e_3(\lambda ^2+\mu ^2)=-2\frac{\lambda }{\mu }(\lambda ^2+\mu ^2)(\rho +1)v_3, \end{aligned}$$

which is a contradiction and this proves our claim.

Now we claim that $v_3$ cannot vanish on any open subset. Arguing indirectly, we suppose that $v_3=0$ on an open subset. Then (20) implies that $u_3=0.$ By Lemma 14, the splitting tensor vanishes and consequently the immersion f would be a cylinder by Proposition 12. This contradicts our assumption.

Since we already proved that $\lambda v_3=0$, we obtain $\lambda =0$ and (20) implies that $u_3=0.$ It follows from (36) that

$$\begin{aligned} \mu ^2=\frac{\rho n^2H^2}{(\rho -1)^2}. \end{aligned}$$

(38)

In particular, we have $\rho >0.$ This, along with (37) yields

$$\begin{aligned} \rho v_3^2=c. \end{aligned}$$

(39)

Hence, $c=1.$ Now observe that the splitting tensor satisfies $\mathcal {C}_3^2=-I,$ where I is the identity endomorphism of $D^{\perp },$ that is, $\mathcal {C}_3$ is an almost complex structure $J:D^\perp \rightarrow D^\perp .$ Using (39) and the fact that the shape operator $A_{\xi _5}$ satisfies $A_{\xi _5}e_i=\mu e_j$ for $i\ne j=1,2,$ we easily verify that the second fundamental form of f satisfies $\alpha ^f(Je_1,e_2)=\alpha ^f(e_1,Je_2).$ This is equivalent to the ellipticity of the immersion f.

In order to prove that the first curvature ellipse of f is a circle, it is equivalent to prove that the vector fields $\alpha ^f(e_1,e_1)$ and $\alpha ^f(e_1,Je_1)$ are of the same length and perpendicular. Obviously, they are perpendicular since

$$\begin{aligned} \alpha ^f(e_1,e_1)=k_1\xi _4 \;\; \text {and} \;\; \alpha ^f(e_1,Je_1)=\mu v_3\xi _5. \end{aligned}$$

Using (5) and (38), we obtain

$$\begin{aligned} \frac{\Vert \alpha ^f(e_1,Je_1)\Vert ^2}{\Vert \alpha ^f(e_1,e_1)\Vert ^2}=\rho v_3^2. \end{aligned}$$

Bearing in mind (39), we conclude that the first curvature ellipse is a circle. $\square$

The following result parametrizes all three-dimensional submanifolds in spheres that carry a totally geodesic distribution of rank one, contained in the relative nullity distribution, such that the mean curvature is constant along each integral curve. This parametrization, given in terms of their polar surfaces, was introduced in [5] as the polar parametrization.

Theorem 21

Let$h:L^2\rightarrow {\mathbb {Q}}_c^{N+1}, c\in \{0,1\},N\ge 5,$be a nicely curved elliptic surface of substantial even codimension, such that the curvature ellipses$\mathcal {E}_{\tau _h-2}^h, \mathcal {E}_{\tau _h}^h$are circles and$\mathcal {E}_{\tau _h-1}^h$is nowhere a circle. Then, the map$\Psi _h:M^3\rightarrow {\mathbb {S}}^{N+c}$defined on the circle bundle$M^3=UN^h_{\tau _h}=\{(x,w)\in N^h_{\tau _h}: \Vert w\Vert =1\}$ by $\Psi _h(x,w)=w$is a nonminimal elliptic isometric immersion with polar surfaceh. Moreover, $M^3$carries a totally geodesic distributionDof rank one satisfying$D(p)\subseteq {\Delta }_{\Psi _h}(p)$for any$p\in M^3$such that the mean curvature of$\Psi _h$is constant along each integral curve ofD.

Conversely, let$f:M^3 \rightarrow {\mathbb {S}}^{3+p}, p\ge 2,$be a substantial nonminimal isometric immersion such that$M^3$carries a totally geodesic distributionDof rank one satisfying$D(x)\subseteq {\Delta }_f(x)$for any$x\in M^3$. If the mean curvature offis constant along each integral curve ofD, then f is elliptic and there exists an open dense subset of$M^3$such that for each point there exist a neighborhoodU, and a local isometry$F:U\rightarrow UN^h_{\tau _h}$such that$f=\Psi _h\circ F$, wherehis a polar surface tofwith curvature ellipses as above.

Proof

Let $h:L^2\rightarrow {\mathbb {Q}}_c^{N+1}, c\in \{0,1\},$ be a substantial elliptic surface, where $N=2m+3, m\ge 1$. Choose a local orthonormal frame $e_1, e_2$ in the tangent bundle of $L^2$ such that the almost complex structure J of the elliptic surface is given by

$$\begin{aligned} Je_1=be_2 \,\ \, \text {and} \,\ \, Je_2=-\frac{1}{b}e_1, \end{aligned}$$

where b is a positive smooth function.

We argue for the case where $m\ge 2.$ The case where $m=1$ can be handled in a similar manner. We know from (30) that the normal bundle splits orthogonally as

$$\begin{aligned} N_hL=N^h_1\oplus \cdots \oplus N^h_{m-1}\oplus N^h_m\oplus N^h_{m+1}. \end{aligned}$$

Let $\zeta _3,\dots ,\zeta _{2m+4}$ be an orthonormal frame in the normal bundle, defined on an open subset $V\subseteq L^2,$ such that $\zeta _{2s+1}, \zeta _{2s+2}$ span the plane subbundle $N^h_s$ for any $1\le s\le m+1.$ The corresponding normal connection forms $\omega _{\alpha \beta }$ are given by $\omega _{\alpha \beta }=\langle \nabla ^\perp \zeta _\alpha ,\zeta _\beta \rangle , \alpha ,\beta =3,\dots ,2m+4.$

Due to our hypothesis, we may choose the frame such that

$$\begin{aligned} \alpha ^h_m(e_1,\dots ,e_1)=\kappa _{m-1}\zeta _{2m-1}, \,\ \,\ \alpha ^h_m(e_1,\dots ,e_1,e_2)=\frac{\kappa _{m-1}}{b}\zeta _{2m} \end{aligned}$$

and

$$\begin{aligned} \alpha ^h_{m+2}(e_1,\dots ,e_1)=\kappa _{m+1}\zeta _{2m+3}, \,\ \,\ \alpha ^h_{m+2}(e_1,\dots ,e_1,e_2)=\frac{\kappa _{m+1}}{b}\zeta _{2m+4}, \end{aligned}$$

where $\kappa _{m-1}, \kappa _{m+1}$ denote the radii of the circular curvature ellipses $\mathcal {E}^h_{m-1}, \mathcal {E}^h_{m+1},$ respectively. Since the curvature ellipse $\mathcal {E}^h_{m}$ is nowhere a circle, we may choose $\zeta _{2m+1}, \zeta _{2m+2}$ to be collinear to the major and minor axes of this ellipse, respectively. Thus, we may write

$$\begin{aligned} \alpha ^h_{m+1}(e_1,\dots ,e_1)=v_{11}\zeta _{2m+1}+v_{12}\zeta _{2m+2} \,\ \,\ \text {and} \,\ \,\ \alpha ^h_{m+1}(e_1,\dots ,e_1,e_2)=v_{21}\zeta _{2m+1}+v_{22}\zeta _{2m+2}, \end{aligned}$$

where $v_{ij}$ are smooth functions such that

$$\begin{aligned} b^2v_{21}v_{22}+v_{11}v_{12}=0, \,\ \,\ \kappa _m=\left( v_{11}^2+b^2v_{21}^2\right) ^{1/2}, \,\ \,\ \mu _{m}=\left( v_{12}^2+b^2v_{22}^2\right) ^{1/2} \end{aligned}$$

(40)

and $\kappa _m,\mu _m$ denote the lengths of the semi-axes of the curvature ellipse $\mathcal {E}^h_{m}$.

Bearing in mind the definition of the higher fundamental forms, their symmetry and the ellipticity of the surface h,  we have

$$\begin{aligned} \alpha _{s+1}^h\left( e_1,\dots ,e_1,e_2\right) =\left( \nabla ^\perp _{e_2}\alpha _s^h\left( e_1,\dots ,e_1\right) \right) ^{N^h_s}=\left( \nabla ^\perp _{e_1}\alpha _s^h\left( e_1,\dots ,e_1,e_2\right) \right) ^{N^h_s},\\ \alpha _{s+1}^h\left( e_1,\dots ,e_1\right) =-b^2\left( \nabla ^\perp _{e_2}\alpha _s^h\left( e_1,\dots ,e_1, e_2\right) \right) ^{N^h_s}=\left( \nabla ^\perp _{e_1}\alpha _s^h\left( e_1,\dots ,e_1\right) \right) ^{N^h_s} \end{aligned}$$

for $s=m,m+1,$ where $( \cdot )^{N^h_s}$ denotes taking the projection onto the normal subbundle $N^h_s$. From these, we obtain

$$\begin{aligned}&\omega _{2m-1,2m+1}(e_1)=\frac{v_{11}}{\kappa _{m-1}}, \,\ \,\ \omega _{2m-1,2m+2}(e_1)=\frac{v_{12}}{\kappa _{m-1}}, \end{aligned}$$

(41)

$$\begin{aligned}&\omega _{2m-1,2m+1}(e_2)=\frac{v_{21}}{\kappa _{m-1}}, \,\ \,\ \omega _{2m-1,2m+2}(e_2)=\frac{v_{22}}{\kappa _{m-1}},\end{aligned}$$

(42)

$$\begin{aligned}&\omega _{2m,2m+1}(e_1)=\frac{b v_{21}}{\kappa _{m-1}}, \,\ \,\ \omega _{2m,2m+2}(e_1)=\frac{b v_{22}}{\kappa _{m-1}},\end{aligned}$$

(43)

$$\begin{aligned}&\omega _{2m,2m+1}(e_2)=-\frac{v_{11}}{b \kappa _{m-1}}, \,\ \,\ \omega _{2m,2m+2}(e_2)=-\frac{v_{12}}{b \kappa _{m-1}}, \end{aligned}$$

(44)

$$\begin{aligned}&\omega _{2m+1,2m+3}(e_1)=\frac{b \kappa _{m+1}}{\kappa _m \mu _m}v_{22}, \,\ \,\ \omega _{2m+1,2m+3}(e_2)=\frac{\kappa _{m+1}}{b \kappa _m \mu _m}v_{12}, \end{aligned}$$

(45)

$$\begin{aligned}&\omega _{2m+1,2m+4}(e_1)=-\frac{\kappa _{m+1}}{\kappa _m \mu _m}v_{12}, \,\ \,\ \omega _{2m+1,2m+4}(e_2)=\frac{\kappa _{m+1}}{\kappa _m \mu _m}v_{22},\end{aligned}$$

(46)

$$\begin{aligned}&\omega _{2m+2,2m+3}(e_1)=-\frac{b \kappa _{m+1}}{\kappa _m \mu _m}v_{21}, \,\ \,\ \omega _{2m+2,2m+3}(e_2)=-\frac{\kappa _{m+1}}{b \kappa _m \mu _m}v_{11},\end{aligned}$$

(47)

$$\begin{aligned}&\omega _{2m+2,2m+4}(e_1)=\frac{\kappa _{m+1}}{\kappa _m \mu _m}v_{11}, \,\ \,\ \omega _{2m+2,2m+4}(e_2)=-\frac{\kappa _{m+1}}{\kappa _m \mu _m}v_{21}. \end{aligned}$$

(48)

Let $\Pi :M^3\rightarrow L^2$ the natural projection of the circle bundle

$$\begin{aligned} M^3=UN^h_{\tau _h}=\left\{ (x,\delta )\in N^h_{m+1}:\Vert \delta \Vert =1, x\in L^2\right\} . \end{aligned}$$

We parametrize $\Pi ^{-1}(V)$ by $V\times {\mathbb {R}}$ via the map

$$\begin{aligned} (x,\theta )\mapsto \left( x,\cos \theta \zeta _{2m+3}(x)+\sin \theta \zeta _{2m+4}(x)\right) \end{aligned}$$

and consequently, we have

$$\begin{aligned} \Psi _h(x,\theta )=\cos \theta \zeta _{2m+3}+\sin \theta \zeta _{2m+4}. \end{aligned}$$

Notice that $\nabla ^\perp N^h_{m+1}\subseteq N_m^h\oplus N_{m+1}^h.$ It is easily seen that

$$\begin{aligned} \Psi _{h_*}E_i&=\big (\cos \theta \omega _{2m+3,2m+1}(e_i)+\sin \theta \omega _{2m+4,2m+1}(e_i)\big )\zeta _{2m+1}\\&+\big (\cos \theta \omega _{2m+3,2m+2}(e_i)+\sin \theta \omega _{2m+4,2m+2}(e_i)\big )\zeta _{2m+2}, \end{aligned}$$

where the vector fields $E_i\in TM,i=1,2,$ are given by

$$\begin{aligned} E_i=e_i-\omega _{2m+3,2m+4}(e_i)\frac{\partial }{\partial \theta }. \end{aligned}$$

Using (45)-(48), we obtain

$$\begin{aligned} \Psi _{h_*}E_1=\frac{\kappa _{m+1}}{\kappa _m\mu _m}\Big (\big (- b v_{22}\cos \theta +v_{12}\sin \theta \big )\zeta _{2m+1} +\big (b v_{21}\cos \theta - v_{11}\sin \theta \big )\zeta _{2m+2}\Big ) \end{aligned}$$

(49)

and

$$\begin{aligned} \Psi _{h_*}E_2=\frac{\kappa _{m+1}}{\kappa _m\mu _m}\Big (-\big (\frac{v_{12}}{b}\cos \theta + v_{22}\sin \theta \big )\zeta _{2m+1}+\big (\frac{v_{11}}{b}\cos \theta + v_{21}\sin \theta \big )\zeta _{2m+2}\Big ). \end{aligned}$$

(50)

Additionally, we have

$$\begin{aligned} \Psi _{h_*}(\partial /\partial \theta )=-\sin \theta \zeta _{2m+3}+\cos \theta \zeta _{2m+4}. \end{aligned}$$

(51)

It follows that the normal bundle of the isometric immersion $\Psi _h$ is given by

$$\begin{aligned} N_{\Psi _h}M=c\ {\mathrm{span}} \{h\}\oplus N^h_1\oplus \cdots \oplus N^h_{m-2}\oplus N^h_{m-1}. \end{aligned}$$

It is easy to see that the first normal bundle of $\Psi _h$ is $N_1^{\Psi _h}=N^h_{m-1}.$ Moreover, it follows easily that the distribution $D={\mathrm{span}}\{\partial /\partial \theta \}$ is contained in the nullity distribution ${\Delta }_{\Psi _h}$ of $\Psi _h.$ In particular, from (51) and the Gauss formula we derive that $\nabla _{\partial /\partial \theta }\partial /\partial \theta =0.$ This implies that the distribution D is totally geodesic.

It remains to show that the mean curvature of the immersion $\Psi _h$ is constant along each integral curve of D. The shape operator $A_{\zeta _{2m-j}}$ of $\Psi _h$ with respect to the normal direction $\zeta _{2m-j}, j=0,1,$ is given by the Weingarten formula as

$$\begin{aligned} -\Psi _{h_*}\left( A_{\zeta _{2m-j}}E_i\right) =\nabla ^\perp _{e_i}\zeta _{2m-j}-\Big ({\tilde{\nabla }}_{e_i}\zeta _{2m-j}\Big )^{N^h_{m-2}\oplus N^h_{m-1}} =\left( \nabla ^\perp _{e_i}\zeta _{2m-j}\right) ^{N^h_m},\;\;i=1,2, \end{aligned}$$

(52)

since $\zeta _{2m-1},\zeta _{2m}\in N_{m-1}^h$. Here, ${\tilde{\nabla }}$ stands for the induced connection of the induced bundle $h^*T{\mathbb {Q}}_c^{N+1}.$ It follows from (52) using (41)-(44) that

$$\begin{aligned}&\Psi _{h_*}\left( A_{\zeta _{2m-1}}E_1\right) =-\frac{1}{\kappa _{m-1}}\left( v_{11}\zeta _{2m+1}+v_{12}\zeta _{2m+2}\right) , \end{aligned}$$

(53)

$$\begin{aligned}&\Psi _{h_*}\left( A_{\zeta _{2m-1}}E_2\right) =-\frac{1}{\kappa _{m-1}}\left( v_{21}\zeta _{2m+1}+v_{22}\zeta _{2m+2}\right) ,\end{aligned}$$

(54)

$$\begin{aligned}&\Psi _{h_*}\left( A_{\zeta _{2m}}E_1\right) =-\frac{b}{\kappa _{m-1}}\left( v_{21}\zeta _{2m+1}+v_{22}\zeta _{2m+2}\right) ,\end{aligned}$$

(55)

$$\begin{aligned}&\Psi _{h_*}\left( A_{\zeta _{2m}}E_2\right) =\frac{1}{b \kappa _{m-1}}\left( v_{11}\zeta _{2m+1}+v_{12}\zeta _{2m+2}\right) . \end{aligned}$$

(56)

We may set

$$\begin{aligned} A_{\zeta _{2m-1}}E_i=\lambda _{i1}E_1+\lambda _{i2}E_2 \,\ \,\ \text {and} \,\ \,\ A_{\zeta _{2m}}E_i=\gamma _{i1}E_1+\gamma _{i2}E_2, \,\ i=1,2, \end{aligned}$$

(57)

where $\lambda _{ij}$ and $\gamma _{ij}$ are smooth functions on the manifold $M^3$. From (49), (53), (54) and the first one of (57), we obtain

$$\begin{aligned} \lambda _{11}=\frac{1}{\kappa _{m-1}\kappa _{m+1}}\left( \left( v_{11}^2+v_{12}^2\right) \cos \theta + b\left( v_{11}v_{21}+v_{12}v_{22}\right) \sin \theta \right) \end{aligned}$$

and

$$\begin{aligned} \lambda _{22}=\frac{1}{\kappa _{m-1}\kappa _{m+1}}\left( - b^2\left( v_{21}^2+v_{22}^2\right) \cos \theta + b\left( v_{11}v_{21}+v_{12}v_{22}\right) \sin \theta \right) . \end{aligned}$$

Hence

$$\begin{aligned} {\mathrm{trace}} A_{\zeta _{2m-1}}=\frac{1}{\kappa _{m-1}\kappa _{m+1}}\left( \left( v^2_{11}+v^2_{12}- b^2v_{21}^2-b^2v_{22}^2\right) \cos \theta + 2b\left( v_{11}v_{21}+v_{12}v_{22}\right) \sin \theta \right) . \end{aligned}$$

Similarly, from (50), (55), (56) and the second of (57), we find that

$$\begin{aligned} \gamma _{11}=\frac{1}{\kappa _{m-1}\kappa _{m+1}}\left( b\left( v_{11}v_{21}+v_{12}v_{22}\right) \cos \theta + b^2\left( v_{21}^2+v_{22}^2\right) \sin \theta \right) \end{aligned}$$

and

$$\begin{aligned} \gamma _{22}=\frac{1}{\kappa _{m-1}\kappa _{m+1}}\left( b\left( v_{11}v_{21}+v_{12}v_{22}\right) \cos \theta -\left( v_{11}^2+v_{12}^2\right) \sin \theta \right) . \end{aligned}$$

Then, it follows that

$$\begin{aligned} {\mathrm{trace}} A_{\zeta _{2m}}=\frac{1}{\kappa _{m-1}\kappa _{m+1}}\left( 2b\left( v_{11}v_{21}+v_{12}v_{22}\right) \cos \theta -\left( v^2_{11}+v^2_{12}- b^2v_{21}^2-b^2v_{22}^2\right) \sin \theta \right) . \end{aligned}$$

Thus, the mean curvature of the isometric immersion $\Psi _h$ is given by

$$\begin{aligned} \Vert \mathcal {H}_{\Psi _h}\Vert ^2=\frac{1}{(3\kappa _{m-1}\kappa _{m+1})^2}\Big (\big (v_{11}^2+v_{12}^2+b^2v_{21}^2+b^2v_{22}^2\big )^2-4\big (v_{11}^2+b^2v_{21}^2\big )^2\big (v_{12}^2+b^2v_{22}^2\big )^2\Big ). \end{aligned}$$

Using (40), the above equation becomes

$$\begin{aligned} \Vert \mathcal {H}_{\Psi _h}\Vert =\frac{|\kappa ^2_m-\mu ^2_m|}{3\kappa _{m-1}\kappa _{m+1}}. \end{aligned}$$

It is clear that the mean curvature of the isometric immersion $\Psi _h$ is constant along each integral curve of the distribution D. This completes the proof of the direct statement of the theorem for $m\ge 2.$ The case $m=1$ can be treated in a similar manner. In this case, the mean curvature of $\Psi _h$ is given by

$$\begin{aligned} \Vert \mathcal {H}_{\Psi _h}\Vert =\frac{|\kappa ^2_1-\mu ^2_1|}{3\kappa _{2}^2}. \end{aligned}$$

Conversely, let $f:M^3\rightarrow {\mathbb {S}}^{3+p}$ be a nonminimal isometric immersion. Suppose that $M^3$ carries a totally geodesic distribution D of rank one satisfying $D(x)\subseteq {\Delta }_f(x)$ for any $x\in M^3$ such that the mean curvature is constant along each integral curve of D. From Proposition 20, we know that f is an elliptic submanifold and its first curvature ellipse is a circle. Hereafter, we work on a connected component of an open dense subset where f is nicely curved.

Consider a polar surface $h:L^2\rightarrow {\mathbb {Q}}^{p-c+4}_c$ to the immersion f,  where $c=0$ if p is even and $c=1$ if p is odd. Notice that $\tau ^0_f=\tau _h-1.$ Using (31), we conclude that the curvature ellipse $\mathcal {E}^h_{\tau _h-2}$ of the surface h is a circle and the curvature ellipse $\mathcal {E}^h_{\tau _h-1}$ is nowhere a circle.

We claim that the last curvature ellipse $\mathcal {E}^h_{\tau _h}$ is a circle. Observe that $N^h_{\tau _h}={\mathrm{span}}\{\xi ,\eta \},$ where the sections $\xi ,\eta$ of the normal bundle $N_hL$ are given by $\xi =f\circ \pi \;\;\text {and}\;\; \eta =f_*e_3\circ \pi .$ Here $\pi$ denotes the natural projection $\pi :M^3\rightarrow L^2$ onto the fixed cross section $L^2\subset M^3$ to the foliation generated by the distribution D.

Let $X_1,\dots ,X_{\tau _h}\in TL$ be arbitrary vector fields. By (31), we have $N^h_{\tau _h-1}=N^f_0=f_*D^\perp .$ Thus, there exists $X\in \Gamma (D^\perp )$ such that

$$\begin{aligned} \alpha ^h_{\tau _h}\left( X_1,\dots ,X_{\tau _h}\right) =f_*X. \end{aligned}$$

For every vector field $Y\in TL$ there exists a vector field $Z\in \Gamma (D^\perp )$ such that $Y=\pi _*Z.$ Then we have

$$\begin{aligned} \alpha ^h_{\tau _h+1}(X_1,\dots ,X_{\tau _h},Y)&=\left( \nabla ^\perp _Y\alpha ^h_{\tau _h}(X_1,\dots ,X_{\tau _h})\right) ^{N^h_{\tau _h}}\nonumber \\&= -\langle f_*X,f_*Z \rangle \xi -\langle f_*X, {\tilde{\nabla }}_Zf_*e_3 \rangle \eta . \end{aligned}$$

Using the Gauss formula and the definition of the splitting tensor, the above equation becomes

$$\begin{aligned} \alpha ^h_{\tau _h+1}(X_1,\dots ,X_{\tau _h},Y)=-\langle X,Z \rangle \xi +\langle X,\mathcal {C}_3Z \rangle \eta . \end{aligned}$$

From Proposition 20, we know that the splitting tensor in the direction of $e_3$ is the almost complex structure $J_0^f:D^\perp \rightarrow D^\perp$ of f. Hence, we obtain

$$\begin{aligned} \alpha ^h_{\tau _h+1}(X_1,\dots ,X_{\tau _h},Y)=-\langle X,Z \rangle \xi +\langle X,J^f_0Z \rangle \eta . \end{aligned}$$

On account of $\pi _*\circ J^f_0=J^h_0\circ \pi _*,$ we have $J^h_0Y=\pi _*J^f_0Z.$ Thus, it follows that

$$\begin{aligned} \alpha ^h_{\tau _h+1}(X_1,\dots ,X_{\tau _h},J^h_0Y)=-\langle X,J_0^hZ \rangle \xi -\langle X,Z \rangle \eta . \end{aligned}$$

Since $\xi ,\eta$ is an orthonormal frame of the subbundle $N^h_{\tau _h},$ it is now obvious that the normal vector fields $\alpha ^h_{\tau _h+1}(X_1,\dots ,X_{\tau _h+1},Y)$ and $\alpha ^h_{\tau _h+1}(X_1,\dots ,X_{\tau _h},J_0^hY)$ are of the same length and perpendicular. Hence, the last curvature ellipse of the polar surface h is a circle.

Finally, observe that the isometric immersion f is written as the composition $f=\Psi _h\circ F,$ where $F:U\rightarrow UN_{\tau _h}^h$ is the local isometry given by $F(x)=(\pi (x),f(x)), \, x\in U,$ and U is the saturation of the cross section $L^2\subset M^3$. $\square$

Remark 22

It follows from the computation of the mean curvature of the submanifold $\Psi _h$ in the proof of Theorem 21, that the mean curvature is constant by properly choosing the elliptic surface h. Ejiri [13] proved that tubes in the direction of the second normal bundle of a pseudoholomorphic curve in the nearly Kähler sphere ${\mathbb {S}}^6$ have constant mean curvature. Opposed to our case, the index of relative nullity of these tubes is zero.

Proof of Theorem 2

Assume that the isometric immersion f is neither minimal nor locally a cylinder. Proposition 20 implies that f is spherical. Thus, from Theorem 21 we know that for each point on an open dense subset there exist an elliptic surface $h:L^2\rightarrow {\mathbb {Q}}^{p-c+4}_c,$ where $c=0$ if p is even and $c=1$ if p is odd, a neighborhood U and a local isometry $F:U\rightarrow UN^h_{\tau _h}$ such that $f=\Psi _h\circ F.$ In fact, the elliptic surface h is a polar to f. Moreover, we know that the curvature ellipses $\mathcal {E}^h_{\tau _h-2}$ and $\mathcal {E}^h_{\tau _h}$ are circles, while the curvature ellipse $\mathcal {E}^h_{\tau _h-1}$ is nowhere a circle.

Now consider a bipolar surface g to f,  that is, a polar surface to the elliptic surface h. Then it follows from (31) that the curvature ellipse $\mathcal {E}^g_0$ of g is a circle. This means that the bipolar surface is minimal. Furthermore, its first curvature ellipse is nowhere a circle and the second one is a circle. That the isometric immersion f is locally parametrized by (1) follows from the fact that $f=\Psi _h\circ F$ and $N_0^g=N_{\tau _h}^h$. $\square$

6.1 Minimal surfaces

The following proposition provides a way of constructing minimal surfaces in ${\mathbb {R}}^6$ that satisfy the properties that are required in part (iii) of Theorem 2.

Proposition 23

Let$\hat{g}:M^2\rightarrow {\mathbb {R}}^6$be the minimal surface defined by

$$\begin{aligned} \hat{g}=\cos \varphi g_\theta \oplus \sin \varphi g_{\theta +\pi /2}, \end{aligned}$$

where$g_\theta , \theta \in [0,\pi )$, is the associated family of a simply connected minimal surface$g:M^2\rightarrow {\mathbb {R}}^3$with negative Gaussian curvature, and$\oplus$denotes the orthogonal sum with respect to an orthogonal decomposition of${\mathbb {R}}^6$. If$\varphi \ne \pi /4$, then its first curvature ellipse is nowhere a circle and its second curvature ellipse is a circle.

Let $g:M\rightarrow {\mathbb {R}}^n$ be an oriented minimal surface. The complexified tangent bundle $TM\otimes {\mathbb {C}}$ is decomposed into the eigenspaces $T^{\prime }M$ and $T^{\prime \prime }M$ of the complex structure J, corresponding to the eigenvalues i and $-i.$ The r-th fundamental form $\alpha ^g_r$, which takes values in the normal subbundle $N_{r-1}^g$, can be complex linearly extended to $TM\otimes {\mathbb {C}}$ with values in the complexified vector bundle $N_{r-1}^g\otimes {\mathbb {C}}$ and then decomposed into its (p, q)-components, $p+q=r,$ which are tensor products of p differential 1-forms vanishing on $T^{\prime \prime }M$ and q differential 1-forms vanishing on $T^{\prime }M.$ The minimality of g is equivalent to the vanishing of the (1, 1)-component of the second fundamental form. Hence, the (p, q)-components of $\alpha ^g_r$ vanish unless $p=r$ or $p=0$.

It is known (see [30, Lem. 3.1]) that the curvature ellipse of order $r-1$ is a circle if and only if the (r, 0)-component of $\alpha ^g_r$ is isotropic, that is

$$\begin{aligned} \langle \alpha ^g_r(X,\dots ,X),\alpha ^g_r(X,\dots ,X) \rangle =0 \end{aligned}$$

for any $X\in T^{\prime }M,$ where $\langle \cdot ,\cdot \rangle$ denotes the bilinear extension over the complex numbers of the Euclidean metric.

Proof of Proposition 23

Choose a local tangent orthonormal frame $e_1,e_2$ such that the shape operator A of g satisfies $AE=k{\bar{E}},$ where $E=e_1+ie_2$ and k is a positive smooth function. The associated family satisfies $g_{\theta _*}=g_{*}\circ J_\theta ,$ where $J_\theta =\cos \theta I+\sin \theta J$ and I is the identity endomorphism of the tangent bundle. Then we have

$$\begin{aligned} {\hat{g}}_*E=e^{-i{\theta }}\left( \cos \varphi g_*E,-i\sin \varphi g_*E\right) . \end{aligned}$$

(58)

Using the Gauss formula and the fact that the shape operator $A_\theta$ of $g_{\theta }$ is given by $A_\theta =A\circ J_\theta$, we find that the second fundamental form ${\hat{\alpha }}$ of ${\hat{g}}$ satisfies

$$\begin{aligned} {\hat{\alpha }}(E,E)=2ke^{-i{\theta }}\left( \cos \varphi N, -i\sin \varphi N\right) , \end{aligned}$$

(59)

where N is the Gauss map of g. It is obvious that ${\hat{\alpha }}(E,E)$ is not isotropic if $\varphi \ne \pi /4$, which implies that the first curvature ellipse of ${\hat{g}}$ is nowhere a circle.

Differentiating (59) with respect to E and using the Weingarten formula, we obtain

$$\begin{aligned} {\tilde{\nabla }}_E{\hat{\alpha }}(E,E)=2e^{-i{\theta }}E(k)\left( \cos \varphi N,-i\sin \varphi N\right) -2k^2e^{-i{\theta }}\left( \cos \varphi g_*{\bar{E}},-i\sin \varphi g_*{\bar{E}}\right) , \end{aligned}$$

where ${\tilde{\nabla }}$ is the connection of the induced bundle of ${\hat{g}}$. Since ${\hat{g}}_*E$ and ${\hat{g}}_*{\bar{E}}$ span $N_0^{\hat{g}}\otimes {\mathbb {C}},$ the above equation along with (58) yield

$$\begin{aligned} \big ({\tilde{\nabla }}_E{\hat{\alpha }}(E,E)\big )^{N_0^{\hat{g}}\otimes {\mathbb {C}}}=-2k^2e^{-2i{\theta }}\cos 2\varphi {\hat{g}}_*{\bar{E}}. \end{aligned}$$

It follows using (59) that $N_1^{{\hat{g}}}\otimes {\mathbb {C}}={\mathrm{span}}_{{\mathbb {C}}}\{\xi ,\eta \},$ where $\xi =(N,0)$ and $\eta =(0,iN)$. Then, we find that

$$\begin{aligned} \big ({\tilde{\nabla }}_E{\hat{\alpha }}(E,E)\big )^{N_1^{\hat{g}}\otimes {\mathbb {C}}}=2e^{-i{\theta }}E(k)\left( \cos \varphi N,-i\sin \varphi N\right) . \end{aligned}$$

Using the above and since the (3, 0)-component of the third fundamental form of ${\hat{g}}$ is given by

$$\begin{aligned} {\hat{\alpha }}_3(E,E,E)=\big ({\tilde{\nabla }}_E{\hat{\alpha }}(E,E)\big )^{\left( N^{{\hat{g}}}_0\otimes {\mathbb {C}}\oplus N^{{\hat{g}}}_1\otimes {\mathbb {C}}\right) ^{\perp }}, \end{aligned}$$

we obtain

$$\begin{aligned} {\hat{\alpha }}_3(E,E,E)=k^2e^{-i{\theta }}\sin 2\varphi \left( -\sin \varphi g_*{\bar{E}},i\cos \varphi g_*{\bar{E}}\right) . \end{aligned}$$

Thus, the (3,0)-component of the third fundamental form of ${\hat{g}}$ is isotropic, and consequently the second curvature ellipse is a circle. $\square$

7 Submanifolds with constant mean curvature

In this section, we provide the proofs of the applications of our main results to submanifolds with constant mean curvature.

Proof of Theorem 3

The manifold $M^n$ is the disjoint union of the subsets

$$\begin{aligned} M_{n-i}=\left\{ x\in M^n:\nu (x)=n-i\right\} , \,\ \,\ i=1,2. \end{aligned}$$

Assume that the subset $M_{n-2}$ is nonempty. Then, using Proposition 13 it follows from Theorem 1 for $n\ge 4$, or Theorem 2 for $n=3$ and $p=1,$ that the isometric immersion f is locally a cylinder over a surface on $M_{n-2}$.

Suppose that the interior ${\mathrm{int}}(M_{n-1})$ of the subset $M_{n-1}$ is nonempty. It follows from the Codazzi equation that the relative nullity distribution is parallel in the tangent bundle along ${\mathrm{int}}(M_{n-1})$. Thus, the tangent bundle splits as an orthogonal sum of two parallel orthogonal distributions of rank one and $n-1$ on ${\mathrm{int}}(M_{n-1})$. By the De Rham decomposition theorem, ${\mathrm{int}}(M_{n-1})$ splits locally as a Riemannian product of two manifolds of dimension one and $n-1.$ Then, the Gauss equation yields $c=0$. Since the second fundamental form is adapted to the orthogonal decomposition of the tangent bundle, it follows that f is a cylinder over a curve in ${\mathbb {R}}^{p+1}$ with constant first Frenet curvature (see [12, Th. 8.4]).

Finally, observe that the open subset $V={\mathrm{int}}(M_{n-1})\cup M_{n-2}$ is dense on $M^n$. $\square$

In order to proceed to the proofs of the applications of our main results, we need to recall Florit’s estimate of the index of relative nullity for isometric immersions with nonpositive extrinsic curvature. The extrinsic curvature of an isometric immersion $f:M^n\rightarrow \tilde{M}^{n+p}$ for any point $x\in M^n$ and any plane $\sigma \subset T_xM$ is given by

$$\begin{aligned} K_f(\sigma )=K_M(\sigma )-K_{\tilde{M}}(f_*\sigma ), \end{aligned}$$

where $K_M$ and $K_{\tilde{M}}$ are the sectional curvatures of $M^n$ and $\tilde{M}^{n+p},$ respectively. Florit [15] proved that the index of relative nullity satisfies $\nu \ge n-2p$ at points where the extrinsic curvature of f is nonpositive.

Proof of Corollary 5

We have that the index of relative nullity of f satisfies $\nu \ge n-2.$ Theorem 3 implies that $c=0$ and, on an open dense subset, f splits locally as a cylinder over a surface in ${\mathbb {R}}^3$ of constant mean curvature. By real analyticity, the splitting is global. If $M^n$ is complete, then the surface is also complete with nonnegative Gaussian curvature. That the surface is a cylinder over a circle follows from [24]. $\square$

Proof of Corollary 6

Assume that the hypersurface is nonrigid. Then, the well-known Beez-Killing Theorem (see [12]) implies that the index of relative nullity satisfies $\nu \ge n-2.$ The result follows from Corollary 4. $\square$

Proof of Theorem 7

Suppose that the hypersurface is nonminimal.

At first assume that the extrinsic curvature is nonnegative. If $c=0$, a result of Hartman [22] asserts that $f(M^n)={\mathbb {S}}_R^{k}\times {\mathbb {R}}^{n-k},$ where $1\le k\le n.$ If $c=1,$ then $M^n$ is compact by the Bonnet-Myers theorem. According to [28, Th. 2], f is totally umbilical.

In the case of nonpositive extrinsic curvature, the result follows from Corollary 5. $\square$

Proof of Theorem 8

According to the aforementioned result due to Florit [15], we have $\nu \ge n-4.$ Clearly the manifold $M^n$ is the disjoint union of the subsets

$$\begin{aligned} M_{n-i}=\left\{ x\in M^n:\nu (x)=n-i\right\} , \,\ \,\ i=1,2,3. \end{aligned}$$

We distinguish the following cases.

Case I: Suppose that the subset $M_{n-4}$ is nonempty. According to Proposition 13, this subset is open. Using [16, Th. 1], we have that on an open dense subset of $M_{n-4}$ the immersion f is locally a product $f=f_1\times f_2$ of two hypersurfaces $f_i:M^{n_i}\rightarrow {\mathbb {R}}^{n_i+1},i=1,2,$ of nonpositive sectional curvature. The assumption that f has constant mean curvature implies that both hypersurfaces have constant mean curvature as well. Each hypersurface $f_i,i=1,2,$ has index of relative nullity $n_i-2.$ Then, it follows from Corollary 4 that the submanifold is locally as in part (iii) of the theorem.

Case II: Suppose that the interior of the subset $M_{n-3}$ is nonempty. Due to [17, Th. 1], on an open dense subset of ${\mathrm{int}}(M_{n-3})$, f is written locally as a composition $f=h\circ F$, where $h=\gamma \times id_{{\mathbb {R}}^{n-1}} :{\mathbb {R}}\times {\mathbb {R}}^n\rightarrow {\mathbb {R}}^{n+2}$ is cylinder over a unit speed plane curve $\gamma (s)$ with nonvanishing curvature k(s) and $F:M^n \rightarrow {\mathbb {R}}^{n+1}$ is a hypersurface. The second fundamental form of f is given by

$$\begin{aligned} \alpha ^f(X,Y)=h_*\alpha ^F(X,Y)+\alpha ^h\left( F_*X,F_*Y\right) ,\;\;X,Y\in TM. \end{aligned}$$

From this, we obtain $k\langle F_*T,\partial /\partial s \rangle ^2=0$ for any $T \in \Delta _f.$ This implies that the height function $F_a=\langle F, a \rangle$ relative to $a=\partial /\partial s$ is constant along the leaves of $\Delta _f.$ Then, the mean curvature vector field of f is given by

$$\begin{aligned} n\mathcal {H}_f=nH_Fh_*\xi +k\circ F_a\Vert \hbox {grad}F_a\Vert ^2\eta , \end{aligned}$$

where $\xi ,\eta$ stand for the Gauss maps of F and h, respectively. Using that

$$\begin{aligned} \Vert \hbox {grad}F_a\Vert ^2=1-\langle \xi , a \rangle ^2, \end{aligned}$$

it follows that the mean curvature of F is given as in part (ii) of the theorem.

Case III: Suppose that the subset $M_{n-2}\cup M_{n-1}$ has nonempty interior. Then Theorem 3 implies that the submanifold is locally as in part (i) of the theorem. $\square$

Proof of Theorem 9

It follows from [12, Th. 5.1] that $\tilde{c}\ge c$ if $n\ge 4$. We distinguish the following cases.

Case I: Suppose that $\tilde{c}>c.$ From [10, Prop. 9] or [29, Lem. 8], we have that the second fundamental form splits orthogonally and smoothly as

$$\begin{aligned} \alpha ^f(\cdot ,\cdot )=\beta (\cdot ,\cdot )+\sqrt{\tilde{c}-c}\ \langle \cdot ,\cdot \rangle \eta , \end{aligned}$$

where $\eta$ is a unit normal vector field and $\beta$ is a flat bilinear form. Thus, the shape operator $A_\xi ,$ associated to a unit normal vector field $\xi$ perpendicular to $\eta ,$ has $\mathrm {rank}A_\xi \le 1.$ The mean curvature H of f is given by

$$\begin{aligned} H^2=\frac{k^2}{n^2} + \frac{\tilde{c}-c}{n}, \end{aligned}$$

where $k=\mathrm {trace}A_\xi .$ Obviously, the function k is constant. If $k=0,$ then f is totally umbilical.

Assume now that $k\ne 0.$ Let X be a unit vector field such that $A_\xi X=kX$. The Codazzi equation

$$\begin{aligned} (\nabla _XA_{\eta })T-(\nabla _TA_{\eta })X=A_{\nabla _X^\perp \eta }T-A_{\nabla _T^\perp \eta }X \end{aligned}$$

implies that

$$\begin{aligned} \nabla _T^\perp \xi =\nabla _T^\perp \eta =0 \end{aligned}$$

for any $T\in \ker A_\xi$. Moreover, from the Codazzi equation

$$\begin{aligned} (\nabla _XA_{\xi })T-(\nabla _TA_{\xi })X=A_{\nabla _X^\perp \xi }T-A_{\nabla _T^\perp \xi }X \end{aligned}$$

it follows that

$$\begin{aligned} \nabla _X X=0 \;\;\text {and}\;\;\langle \nabla _S T, X \rangle =\frac {1}{k}{\sqrt{\tilde{c}-c}}\langle \nabla ^\perp _T\xi , \eta \rangle \end{aligned} \langle T, S \rangle$$

for any $T,S \in \ker A_\xi$. Hence, the distributions $D^1=\mathrm {span}\{X\}$ is totally geodesic and $D^{n-1}=\ker A_\xi$ is umbilical. The flatness of the normal bundle implies that $D^{n-1}$ is spherical. Thus the manifold splits locally as a warped product $M_{\tilde{c}}^n=M^1\times_\rho M^{n-1}$ and f is a warped product of a curve and an umbilical submanifold (see [12, Th. 10.4 and Th. 10.21]). This implies that $\rho$ is constant and the manifold splits locally as a Riemannian product $M_{\tilde{c}}^n=M^1\times M^{n-1}$. Consequently, we have $\tilde{c}=0$ and $c=-1$. Clearly $M^{n-1}$ is flat and the second fundamental form is adapted to this decomposition. Then it follows that f is a composition $f=i\circ F$, where $i:{\mathbb {R}}^{n+1}\rightarrow {\mathbb {H}}^{n+2}$ is the inclusion as a horosphere and $F:M^n_{\tilde{c}} \rightarrow {\mathbb {R}}^{n+1}$ is the cylinder over a circle (see [12, Th. 8.4]).

Case II: Suppose that $c=\tilde{c}.$ It is known that $\nu \ge n-2$ (see Example 1 and Corollary 1 in [27]). Then, the result follows from Theorem 3.

If $n=3,$ then Theorem 2 implies that either $c=0$ and f(M) is an open subset of a cylinder over a flat surface $g:M^2\rightarrow {\mathbb {R}}^4$ of constant mean curvature, or $c=1$ and f is parametrized by (1). In the latter case, it follows from Proposition 20 that f is either totally geodesic or elliptic. However, the ellipticity of f implies that the sectional curvature cannot be equal to one. $\square$

Proof of Theorem 10

Assume that f is nonminimal. According to Abe [1], the index of relative nullity satisfies $\nu \ge n-2.$ Corollary 4 implies that the hypersurface is a cylinder over a surface with constant mean curvature. $\square$

Proof

Using [18, Cor. 2], it follows that $\nu \ge n-4.$ The rest of the proof is omitted since it is similar to the proof of Theorem 8. $\square$

The following example produces submanifolds satisfying the conditions in part (ii) of Theorem 8 or 11.

Example 24

Let $F=g\times id_{{\mathbb {R}}^{n-2}}:U\times {\mathbb {R}}^{n-2}\rightarrow {\mathbb {R}}^{n+1}$ be a cylinder over a rotational surface $g(x,\theta )=(x\cos \theta , x\cos \theta , \varphi (x)), (x,\theta )\in U,$ where $\varphi (x)$ is a smooth function. Consider a cylinder $h=\gamma \times id_{{\mathbb {R}}^n}$ in ${\mathbb {R}}^{n+2}$ over a unit speed plane curve $\gamma$ with curvature k. Then the isometric immersion $f=h\circ F$ satisfies the conditions in part (ii) of Theorems 8 and 11, with constant curvature H and $a=(1,0,\dots ,0)$, if the function $\varphi (x)$ solves the ordinary differential equation

$$\begin{aligned} \varphi \varphi ^{\prime \prime }-1-\varphi ^{\prime \, 2}=\pm \varphi \sqrt{(1+\varphi ^{\prime \, 2})\left( n^2H^2(1+\varphi ^{\prime \, 2})^2-k^2\right) }. \end{aligned}$$

In particular, g can be chosen as a Delaunay surface and $\gamma$ as the curve with curvature $k=c_0(1+\varphi ^{\prime \, 2})$ for a constant $c_0$ such that $0<|c_0|<n|H|$.