On Complete Conformally flat submanifolds with nullity in Euclidean space

TL;DR

This paper classifies complete conformally flat submanifolds in Euclidean space with positive nullity index, showing they are cylinders over flat or constant curvature submanifolds under various curvature conditions.

Contribution

It provides a complete classification of such submanifolds based on nullity index and scalar curvature, extending previous results in conformal geometry.

Findings

Nullity index at least two implies the submanifold is flat and cylindrical.

Non-negative scalar curvature with positive nullity implies a cylindrical structure over a constant curvature submanifold.

Non-zero scalar curvature with nullity one implies a cylindrical structure over a submanifold with non-zero constant sectional curvature.

Abstract

In this note, we investigate conformally flat submanifolds of Euclidean space with positive index of relative nullity. Let $M^n$ be a complete conformally flat manifold and let $f\colon M^n\to \R^m$ be an isometric immersion. We prove the following results: (1) If the index of relative nullity is at least two, then $M^n$ is flat and $f$ is a cylinder over a flat submanifold. (2) If the scalar curvature of $M^n$ is non-negative and the index of relative nullity is positive, then $f$ is a cylinder over a submanifold with constant non-negative sectional curvature. (3) If the scalar curvature of $M^n$ is non-zero and the index of relative nullity is constant and equal to one, then $f$ is a cylinder over a $(n-1)$-dimensional submanifold with non-zero constant sectional curvature.

Equations

$$\Delta (x)=\{X\in T_{x}M:\alpha (X,Y)=0\text{mbox}forallY\in T_{x}M\}.$$

$$⟨{\nabla }_{X}Y,T⟩=⟨X,Y⟩⟨\delta ,T⟩$$

$$R(X,Y,Z,W)=L(X,W)⟨Y,Z⟩-L(X,Z)⟨Y,W⟩+L(Y,Z)⟨X,W⟩-L(Y,W)⟨X,Z⟩$$

$$L(X,Y)=\frac{1}{n-2}\left(\mathrm{Ric}(X,Y)-\frac{s}{2(n-1)}⟨X,Y⟩\right)$$

$$K(X,Y)=L(X,X)+L(Y,Y)$$

$$R(X,Y,Z,W)=⟨\alpha (X,W),\alpha (Y,Z)⟩-⟨\alpha (X,Z),\alpha (Y,W)⟩$$

$$\mathrm{Ric}(X,Y)=⟨nH,\alpha (X,Y)⟩-\sum _{j=1}^n⟨\alpha (X,X_j),\alpha (Y,X_j)⟩$$

$$L(X,X)+L(Y,Y)=⟨\alpha (X,X),\alpha (Y,Y)⟩-\Vert \alpha (X,Y){\Vert }^2$$

$$\mathrm{Ric}(X,X)+\mathrm{Ric}(Y,Y)=\frac{s}{n-1}+(n-2)(⟨\alpha (X,X),\alpha (Y,Y)⟩-\Vert \alpha (X,Y){\Vert }^2)$$

$$\mathrm{Ric}(Y,Y)=\frac{s}{n-1}$$

$$L(Y,Y)=-L(X,X)=\frac{s}{2(n-1)(n-2)}=:h\text{and}L(X,Y)=0$$

$$L(Y,Z)=0$$

$$L(Y+Z,Y+Z)=2h,$$

$$({\nabla }_{Y}L)(X,X)=({\nabla }_{X}L)(Y,X)$$

$$Y(h)=0,$$

$$h=h(t)=h(\gamma (t)),t\in I,$$

$$({\nabla }_{{\gamma }^{\prime }(t)}L)(Y,Z)=({\nabla }_{Y}L)({\gamma }^{\prime }(t),Z)$$

$$⟨Y,Z⟩⟨\mathrm{grad}\log \sqrt{|h(t)|},{\gamma }^{\prime }(t)⟩=⟨{\nabla }_{Y}Z,{\gamma }^{\prime }(t)⟩$$

Full text

††footnotetext: 2010 Mathematics Subject Classification. Primary 53B25, 53C40, 53C42.††footnotetext: Keywords. Conformally flat submanifolds, index of relative nullity, scalar curvature

1. Introduction

A Riemannian manifold $M^{n}$ is said to be conformally flat if each point lies in an open neighborhood conformal to an open subset of the Euclidean space $\mathord{\mathbb{R}}^{n}$. The geometry and topology of such Riemannian manifolds have been investigated by several authors from the intrinsic point of view. Some of the many papers are [cat16, ku49, ku50, cadjnd11, cahe06, no93, sy88].

Around 1919, Cartan [car17] initiated the investigation of such Riemannian manifolds from the submanifold point of view by studying the case of conformally flat Euclidean hypersurfaces (see also [mmf85, pin85]). In 1977, Moore [mo77] extended Cartan’s result in higher (but still low) codimension (see also [df96, df99, mm78]). Recently, the author, in collaboration with Dajczer and Vlachos, investigated in [17] the case of conformally flat submanifolds with flat normal bundle in arbitrary codimension (see also [dote11]).

In this short note, we address and deal with the following:

Problem. *Classify complete conformally flat submanifolds of Euclidean space with positive index of relative nullity and arbitrary codimension. *

Recall that the index of relative nullity at a point $x\in M^{n}$ of a submanifold $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ is defined as the dimension of the kernel of its second fundamental form $\alpha\colon TM\times TM\to N_{f}M$, with values in the normal bundle.

The first result provides a complete answer in the case where the index of relative nullity is at least two, and is stated as follows:

Theorem 1.1**.**

Let $M^{n}$ be a complete, conformally flat manifold and let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion with index of relative nullity at least two at any point of $M^{n}$. Then $M^{n}$ is flat and $f$ is a cylinder over a flat submanifold.

The next result provides a complete answer in the case where the scalar curvature is non-negative.

Theorem 1.2**.**

Let $M^{n}$ be a complete, conformally flat manifold with non-negative scalar curvature and let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion with positive index of relative nullity. Then $f$ is a cylinder over a submanifold with constant non-negative sectional curvature.

Observe that there are complete conformally flat manifolds such that the scalar curvature is non-negative while the sectional curvature is not. Easy examples are the Riemannian products $M^{n}=\mathord{\mathbb{S}}^{n-m}\times\mathord{\mathbb{H}}^{m},\ n\geq 2m,$ where $\mathord{\mathbb{S}}^{n-m}$ and $\mathord{\mathbb{H}}^{m}$ are the sphere and the hyperbolic space of sectional curvature $1$ and $-1$, respectively.

Finally, the next result provides a complete answer (both local and global) in the case where the scalar is non-zero and the index of relative nullity is constant and equal to one.

Theorem 1.3**.**

Let $M^{n}$ be a conformally flat manifold with non-zero scalar curvature and let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion with constant index of relative nullity equal to one. Then $f$ is locally either a cylinder over a $(n-1)$-dimensional submanifold with non-zero constant sectional curvature or a cone over a $(n-1)$-dimensional spherical submanifold with constant sectional curvature. Moreover, if $M^{n}$ is complete, then $f$ is globally a cylinder over a $(n-1)$-dimensional submanifold with non-zero constant sectional curvature.

Remarks**.**

(I) If the ambient space form in Theorem 1.2 is replaced by the sphere $\mathord{\mathbb{S}}_{c}^{m}$ of constant sectional curvature $c$, then an intrinsic classification can be obtained, provided that ${\rm{scal}}(M^{n})\geq c(n-1)$. This classification follows from a result of Carron and Herzlich [cahe06], since in this case $M^{n}$ turns out to have non-negative Ricci curvature. However, we do not obtain any (direct) information on the immersion $f$.

(II) If $f\colon M^{n}\to\mathord{\mathbb{Q}}_{c}^{m}$ is an isometric immersion of a conformally flat manifold into a space form of constant sectional curvature $c$, then one can prove the following: (i) If the index of relative nullity is at least two, then $M^{n}$ has constant sectional curvature $c$. In particular, if $f$ is also minimal, then $f$ is totally geodesic. (ii) If the index of relative nullity is constant and equal to one, then $f$ is a $1$-generalized cone (for the definition, see [17]) over an isometric immersion $F\colon\Sigma^{n-1}\to\mathord{\mathbb{Q}}_{\tilde{c}}^{m-1}$ into an umbilical submanifold of $\mathord{\mathbb{Q}}_{c}^{m}$. **

Notes & Comments**.**

The special case of minimal conformally flat hypersurfaces $f\colon M^{n}\to\mathord{\mathbb{Q}}_{c}^{n+1},$ $n\geq 4,$ was treated by do Carmo and Dajczer in [DCD] (without any additional assumption on the index of relative nullity), where they showed that these are actually generalized catenoids, extending that way a previous result due to Blair [Blair] for the case $c=0$.

For the “neighbor” class of Einstein manifolds one can prove that: any minimal isometric immersion $f\colon M^{n}\to\mathord{\mathbb{Q}}_{c}^{m}$ of an Einstein manifold with positive index of relative nullity is totally geodesic. A related result of Di Scala [discala], in the case where the ambient space is the Euclidean one, states that: any minimal isometric immersion $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ of a Kähler-Einstein manifold is totally geodesic. However, it is not yet known if the assumption on Kähler can be dropped (this was conjectured by Di Scala in the same paper). Of course, in some special cases the conjecture is true, as have already been pointed out in [discala]. Finally, we note that Di Scala’s theorem still holds true if the Kähler (intrinsic) assumption is replaced by the (extrinsic) assumption on $f$ having flat normal bundle. This follows directly from Nölker’s theorem [no90], since, in this case, $f$ has homothetical Gauss map. **

2. Preliminaries

In this section we recall some basic facts and definitions. Let $M^{n}$ be a Riemannian manifold and let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion. The index of relative nullity $\nu(x)$ at $x\in M^{n}$ is the dimension of the relative nullity subspace $\Delta(x)\subset T_{x}M$ given by

[TABLE]

It is a standard fact that on any open subset where the index of relative nullity $\nu(x)$ is constant, the relative nullity distribution $x\mapsto\Delta(x)$ is integrable and its leaves are totally geodesic in $M^{n}$ and $\mathord{\mathbb{R}}^{m}$. Moreover, if $M^{n}$ is complete then the leaves are also complete along the open subset where the index reaches its minimum (see [13]). If $M^{n}$ splits as a Riemannian product $M^{n}=\Sigma^{n-k}\times\mathord{\mathbb{R}}^{k}$ and there is an isometric immersion $F\colon\Sigma^{n-k}\to\mathord{\mathbb{R}}^{m-k}$ such that $f=F\times{\rm id}_{\mathord{\mathbb{R}}^{k}}$, then we say that $f$ is a $k$-cylinder (or simply a cylinder) over $F$.

The following is due to Hartman [har70]; cf. [dt].

Theorem 2.1**.**

Let $M^{n}$ be a complete Riemannian manifold with non-negative Ricci curvature and let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion with minimal index of relative nullity $\nu_{0}>0$. Then $f$ is a $\nu_{0}$-cylinder.

A smooth tangent distribution $D$ is called totally umbilical if there exists a smooth section $\delta\in\Gamma(D^{\perp})$ such that

[TABLE]

for all $X,Y\in D$ and $T\in D^{\perp}$. The following is contained in [dt].

Proposition 2.2**.**

Let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion of a Riemannian manifold with constant index of relative nullity $\nu=1$. Assume that the conullity distribution $\Delta^{\perp}$ is totally umbilical (respectively, totally geodesic). Then $f$ is locally a cone over an isometric immersion $F\colon\Sigma^{n-1}\to\mathord{\mathbb{S}}^{m-1}\subset\mathord{\mathbb{R}}^{m}$ (respectively, a cylinder over an isometric immersion $F\colon\Sigma^{n-1}\to\mathord{\mathbb{R}}^{m-1}\subset\mathord{\mathbb{R}}^{m}$).

We also need the following two well-known results; cf. [dt].

Proposition 2.3**.**

A Riemannian product is conformally flat if and only if one of the following possibilities holds:

1. (i)

One of the factors is one-dimensional and the other one has constant sectional curvature. 2. (ii)

Both factors have dimension greater than one and are either both flat or have opposite constant sectional curvatures.

Proposition 2.4**.**

Let $M=M_{1}\times_{\rho}M_{2}$ be a warped product manifold. If $M_{1}$ has dimension one then $M$ is conformally flat if and only if $M_{2}$ has constant sectional curvature.

3. The proofs

Let $M^{n}$ be a conformally flat manifold and let $f\colon M^{n}\to\mathord{\mathbb{R}}^{m}$ be an isometric immersion. It is well-known that in this case the curvature tensor has the form

[TABLE]

in terms of the Schouten tensor given by

[TABLE]

where $s$ denotes the scalar curvature. In particular, the sectional curvature is given by

[TABLE]

where $X,Y\in TM$ are orthonormal vectors.

A straightforward computation of the Ricci tensor using the Gauss equation

[TABLE]

yields

[TABLE]

where $X_{1},\dots,X_{n}$ is an orthonormal tangent basis.

We obtain from (3.2) and (3.3) that

[TABLE]

for any pair $X,Y\in TM$ of orthonormal vectors. Using (3.1) it follows from (3.5) that

[TABLE]

for any pair $X,Y\in TM$ of orthonormal vectors. Now, assume that $\nu>0$ and choose a unit length $X\in\Delta$. Using (3.4), it follows from (3.6) that

[TABLE]

for all unit length $Y\perp X$.

Proof of Theorem 1.1: It follows from (3.4) and (3.7) that $s=0$. Thus, it follows from (3.7) that $M^{n}$ is Ricci flat. Since $M^{n}$ is conformally flat we obtain that $M^{n}$ is flat. The desired result follows from Theorem 2.1 and Proposition 2.3. ∎

Proof of Theorem 1.2: It follows from (3.7) that ${\rm{Ric}}\geq 0$. The desired result follows from Theorem 2.1 and Proposition 2.3. ∎

Proof of Theorem 1.3: It follows from (3.5), (3.1) and (3.4) that

[TABLE]

for any unit length vectors $X\in\Delta$ and $Y\in\Delta^{\perp}$. Moreover, we have that

[TABLE]

for any pair $Y,Z\in\Delta^{\perp}$ of orthonormal vectors. Indeed, if $Y$ and $Z$ are two such vectors then using (3.8) we get

[TABLE]

and (3.9) follows. Now, since $M^{n}$ is conformally flat we have that $L$ is a Codazzi tensor. Thus

[TABLE]

for all unit length $X\in\Delta$ and $Y\in\Delta^{\perp}$. It follows, using (3.8) and (3.9) that

[TABLE]

for all unit length $Y\in\Delta^{\perp}$. Therefore

[TABLE]

where $\gamma\colon I\subset\mathord{\mathbb{R}}\to M^{n}$ is a leaf of the nullity distribution $\Delta$, parametrized by arc length. Using again the fact that $L$ is a Codazzi tensor, we get

[TABLE]

for all unit length $Y,Z\in\Delta^{\perp}$, or equivalently,

[TABLE]

for all unit length $Y,Z\in\Delta^{\perp}$, where we have used again equations (3.8) and (3.9). Thus, if the scalar curvature is constant, then $\Delta^{\perp}$ is totally geodesic and the desired result follows from Propositions 2.2 and 2.3. On the other hand, if the scalar is not constant then $\Delta^{\perp}$ is totally umbilical and the desired result follows from Propositions 2.2 and 2.4. Finally, if $M^{n}$ is complete then the result is immediate and the proof is complete. ∎