#  Weakly convex biharmonic hypersurfaces in nonpositive curvature space forms
are minimal

Yong Luo111The author is supported by the DFG Collaborative Research Center
SFB/Transregio 71.

###### Abstract

A submanifold $M^{m}$ of a Euclidean space $R^{m+p}$ is said to have harmonic
mean curvature vector field if $\Delta\vec{H}=0$, where $\vec{H}$ is the mean
curvature vector field of $M\hookrightarrow R^{m+p}$ and $\Delta$ is the rough
Laplacian on $M$. There is a conjecture named after Bangyen Chen which states
that submanifolds of Euclidean spaces with harmonic mean curvature vector
fields are minimal. In this paper we prove that weakly convex hypersurfaces
(i.e. hypersurfaces whose principle curvatures are nonnegative) with harmonic
mean curvature vector fields in Euclidean spaces are minimal. Furthermore we
prove that weakly convex biharmonic hypersurfaces in nonpositive curved space
forms are minimal.

##  1 Introduction

Let $\vec{x}:M^{m}\to R^{m+p}$ be an immersion from a Riemmanian manifold $M$
of dimension $m$ to a Euclidean space of dimension $m+p$, $p\geq 1$. Denote by
$\vec{x},\vec{H},\Delta$ respectively the position vector of $M$, the mean
curvature vector field of $M$ and the Laplacian operator with respect to the
induced metric $g$ on $M$. Then it is well known that (see for example [5])

$\Delta\vec{x}=-n\vec{H}.$

This shows that $M$ is a minimal submanifold if and only if its coordinates
functions are harmonic functions. According to this equation, a submanifold in
a Euclidean space with harmonic mean curvature vector field, i.e.

$\displaystyle\Delta\vec{H}=0,$ (1.1)

if and only if

$\displaystyle\Delta^{2}\vec{x}=0.$ (1.2)

Therefore a submanifold with harmonic mean curvature vector field is called a
biharmonic submanifold.

There is also a variational description of biharmonic submanifolds as follows.
Assume that $\vec{x}:M\to N^{m+p}$ is an immersion to a Riemnnian manifold,
then the biharmonic energy of $\vec{x}$ is defined by

$\displaystyle E_{2}(\vec{x})=\int_{M}|\tau(\vec{x})|^{2}dM,$ (1.3)

where $\tau(\vec{x})$ is the tension field of $\vec{x}$. The critical points
of the functional $E_{2}$ satisfy the following E-L equation (see [10])

$\displaystyle-\Delta\vec{H}=\sum_{i=1}^{m}R^{N}(e_{i},\vec{H})e_{i},$ (1.4)

where $\\{e_{i}\\}$ is a local orthonormal frame on $M$. In particular, when
$N$ is a Euclidean space, this equation is just $\Delta\vec{H}=0$.

######  Definition 1.1.

A Submanifold satisfying equation (1.4) is called a biharmonic submanifold.

As is easy to see, minimal submanifolds are biharmonic submanifolds. It is
nature to arise the question whether the space of biharmonic submanifolds is
strictly larger than the space of minimal submanifolds. Concerning this
problem, when the ambient manifold is a Euclidean space Chen conjectured that
the answer is no.
Chen’s Conjecture. Suppose that $\vec{x}:M^{m}\to R^{m+p}$, $p\geq 1$,
satisfies

$\displaystyle\Delta\vec{H}=0.$ (1.5)

Then $\vec{H}=0$.

Since Chen’s conjecture arose, it remains to be open with very little progress
even for hypersurfaces with dimensions grater that 4. But in recent years it
attracts many attentions and there are some partial answers to this
conjecture. Now we give an overview of them, as to the author’s knowledge.
Chen’s conjecture is proved:

$\bullet$ For Surfaces in $R^{3}$ in [6] and [10] and in an unpublished work
by Chen himself, as reported in [4].

$\bullet$ For hypersurfaces in $R^{4}$ in [9] and a different proof in [7].

$\bullet$ For hypersurfaces which admit at most 2 distinct principle
curvatures in [8].

$\bullet$ For curves [8].

$\bullet$ For submanifolds $M^{m}$ which are pseudo-umbilic and $m\neq 4$ [8].

$\bullet$ For submanifolds of finite type [8].

$\bullet$ For submanifolds which are proper, i.e. any preimage of compact
subsets are compact in [1].

In this paper we prove Chen’s conjecture for weakly convex biharmonic
hypersurfaces in Euclidean spaces.

######  Theorem 1.2.

Assume that $\vec{x}:M^{m}\to R^{m+1}$ is a weakly convex biharmonic
submanifold in $R^{m+1}$, i.e. $\Delta\vec{H}=0.$ Then $\vec{H}=0$.

Chen’s conjecture is generalized to the so called generalized Chen’s
conjecture (see [2] [4] [5] [16] [17] and [15] etc.) allowing the ambient
manifolds to be nonpositive curved.
Generalized Chen’s Conjecture. Suppose that $f:M\to(N,h)$ is an isometric
immersion and the section curvature of $(N,h)$ nonpositive. Then if $f$ is
biharmonic, it is minimal.

The generalized Chen’s curvature turned out to be false by a counter example
constructed by Y-L. Ou and L. Tang (see [17]). But it remains interesting to
find out sufficient conditions which make it be true. It is easy to see from
the proof of theorem 1.2 that our argument is also applicable to find out that
for hypersurfaces weakly convex is a sufficient condition to guarantee the
generalized Chen’s conjecture to be true when the ambient manifold is a space
form $N(c)$ of constant curvature $c\leq 0$. Thus we have

######  Theorem 1.3.

Assume that $\vec{x}:M^{m}\to N^{m+1}(c)$ is a weakly convex biharmonic
hypersurfaces in a spcae form $N^{m+1}(c)$ with $c\leq 0$, then $\vec{H}=0$.

Other sufficient conditions have also been found out to guarantee the
generalized Chen’s conjecture to be true. For recent development in this
direction, we refer to [13] [11] etc..

We would like to point out that for closed (compact without boundary) bi-
harmonic submanifolds the Chen’s conjecture and the generalized Chen’s
conjecture is easily proved to be true by an argument due to Jiang (see [10])
or by directly using an integration by parts argument.

The Chen’s and generalized Chen’s conjectures belong to the research field of
classification of biharmonic submanifolds in Riemannian or pseudo-Riemannian
manifolds. Nowadays it is an active research field and for readers who have
interest with it we refer to a survey paper [12] by Montaldo and Oniciuc and
references therein.

Organization. In section 2 we give a brief sketch of submanifolds geometry.
Theorem 1.2 and theorem 1.3 are proved in section 3.

##  2 Preliminaries

Assume that $\vec{x}:M^{m}\to N^{m+p}$ is an immersion to a Riemannian
manifold $N$ with Riemannian metric $\langle,\rangle$, which is a bilinear
form on $TN\otimes TN$, the tensor of the tangent vector space of $N$. Then
$M$ inherits a Riemannian metric from $N$ by
$g_{ij}:=\langle\partial_{i}\vec{x},\partial_{j}\vec{x}\rangle$ and a volume
form by $\sqrt{\det g_{ij}}dx$. The second fundamental form of
$M\hookrightarrow N$, $h:TM\otimes TM\to NM$, is defined by

$h(X,Y):=D_{X}Y-\nabla_{X}Y,$

for any $X,Y\in TM$, where $D$ is the covariant derivative with respect to the
Levi-Civita connection on $N$, $\nabla$ is the Levi-Civita connection on $M$
with respect to the induced metric and $NM$ is the normal bundle of $M$. For
any normal vector field $\eta$ the Weingarten map $A_{\eta}:TM\to TM$ is
defined by

$\displaystyle D_{X}\eta=-A_{\eta}X+\nabla^{\bot}_{X}\eta,$ (2.1)

where $\nabla^{\bot}$ is the normal connection and as is well known that $h$
and $A$ are related by

$\displaystyle\langle h(X,Y),\eta\rangle=\langle A_{\eta}X,Y\rangle.$ (2.2)

For any $p\in M$, let $\\{e_{1},e_{2},...,e_{m},e_{m+1},...,e_{m+p}\\}$ be a
local orthonormal basis of $N$ such that $\\{e_{1},...,e_{m}\\}$ is an
orthonormal basis of $T_{p}M$. Then $h$ is decomposed at $p$ as

$h(X,Y)=\sum_{\alpha=m+1}^{m+p}h_{\alpha}(X,Y)e_{\alpha}.$

The mean curvature vector field is defined as

$\displaystyle\vec{H}:=\frac{1}{m}\sum_{i=1}^{m}h(e_{i},e_{i})=\sum_{\alpha=m+1}^{m+p}H_{\alpha}e_{\alpha},$
(2.3)

where

$\displaystyle H_{\alpha}:=\frac{1}{m}\sum_{i=1}^{m}h_{\alpha}(e_{i},e_{i}).$
(2.4)

##  3 Proof of theorem 1.2

It is well known that for a submanifold $M^{m}$ in a Euclidean space to be
biharmonic, i.e. $\Delta\vec{H}=0$ if and only if ([4])

$\displaystyle\Delta^{\bot}\vec{H}-\sum_{i=1}^{m}h(A_{\vec{H}}e_{i},e_{i})$
$\displaystyle=$ $\displaystyle 0,$ (3.1) $\displaystyle
m\nabla|\vec{H}|^{2}+4\sum_{i=1}^{m}A_{\nabla^{\bot}_{e_{i}}\vec{H}}e_{i}$
$\displaystyle=$ $\displaystyle 0,$ (3.2)

where $\Delta^{\bot}$ is the (nonpositive) Laplace operator associated with
the normal connection $\nabla^{\bot}$.

Assume that $\vec{H}=H\nu$, where $\nu$ is the unit normal vector field on
$M$. Note that by the assumption that $M$ is weakly convex, we have $H\geq 0$.
Define

$\displaystyle B:=\\{p\in M:H(p)>0\\}$ (3.3)

We will prove that $B$ is an empty set by a contradiction argument, and so $M$
is minimal and we are done.

If $B$ is not empty, we see that $B$ is an open subset of $M$. We assume that
$B_{1}$ is a nonempty connect component of $B$. We will prove that $H\equiv 0$
in $B_{1}$, thus a contradiction.

We prove it in two steps.

Step 1. $H$ is a constant in $B_{1}$.

Let $p\in B_{1}$ be a point. Around $p$ we choose a local orthonormal frame
$\\{e_{k},k=1,...,m\\}$ such that $\langle h,\nu\rangle$ is a diagonal matrix
at $p$, where $\nu$ is the unit normal vector field of $M$.

For any $1\leq k\leq m$ we have at $p$

$\displaystyle\langle\sum_{i=1}^{m}A_{\nabla^{\bot}_{e_{i}}\vec{H}}e_{i},e_{k}\rangle$
$\displaystyle=$ $\displaystyle\sum_{i=1}^{m}\langle
h(e_{i},e_{k}),\nabla^{\bot}_{e_{i}}\vec{H}\rangle$ $\displaystyle=$
$\displaystyle\langle h(e_{k},e_{k}),\nabla^{\bot}_{e_{k}}\vec{H}\rangle.$

By equation (3.2) we have

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
m\nabla_{e_{k}}|\vec{H}|^{2}+4\langle\sum_{i=1}^{m}A_{\nabla^{\bot}_{e_{i}}\vec{H}}e_{i},e_{k}\rangle$
$\displaystyle=$ $\displaystyle m\nabla_{e_{k}}|\vec{H}|^{2}+4\langle
h(e_{k},e_{k}),\nabla^{\bot}_{e_{k}}\vec{H}\rangle$ $\displaystyle=$
$\displaystyle
2mH\nabla_{e_{k}}H+4\lambda_{k}\langle\nu,\nabla^{\bot}_{e_{k}}\vec{H}\rangle$
$\displaystyle=$ $\displaystyle(2mH+4\lambda_{k})\nabla_{e_{k}}H,$

where $\lambda_{k}:=h(e_{k},e_{k})$ is the $k$th principle curvature of $M$ at
$p$, which is nonnegative by the assumption that $M$ is weakly convex.

By $2mH+4\lambda_{k}>0$ at $p$, one gets

$\displaystyle\nabla_{e_{k}}H=0\,\,\,\,at\,\,\,\,p,$ (3.4)

for any $k=1,...,m$, which implies that

$\displaystyle\nabla H=0\,\,\,\,at\,\,\,\,p.$ (3.5)

Because $p$ is an arbitrary point in $B_{1}$, we see that

$\displaystyle\nabla H=0\,\,\,\,in\,\,\,\,B_{1}.$ (3.6)

Therefore we get that $H$ is a constant in $B_{1}$.

Step 2. $H$ is zero in $B_{1}$.

Let $p\in B_{1}$, by step 1, we see that

$\displaystyle\Delta|\vec{H}|^{2}(p)=0.$ (3.7)

On the other hand, by equation (3.1), we have

$\displaystyle\Delta|\vec{H}|^{2}$ $\displaystyle=$ $\displaystyle
2|\nabla^{\bot}\vec{H}|^{2}+2\langle\vec{H},\Delta^{\bot}\vec{H}\rangle$ (3.8)
$\displaystyle\geq$ $\displaystyle 2\sum_{i=1}^{m}\langle
h(A_{\vec{H}}e_{i},e_{i}),\vec{H}\rangle$ $\displaystyle=$ $\displaystyle
2\sum_{i=1}^{m}\langle A_{\vec{H}}e_{i},A_{\vec{H}}e_{i}\rangle$
$\displaystyle=$ $\displaystyle 2\sum_{i=1}^{m}H^{2}\langle
A_{\nu}e_{i},A_{\nu}e_{i}\rangle$ $\displaystyle\geq$ $\displaystyle 2nH^{4}.$

From (3.7)-(3.8), we get $H(p)=0$. Because $p$ is an arbitrary point in
$B_{1}$, we see that $H\equiv 0$ in $B_{1}$, a contradiction.

This completes the proof of theorem 1.2 $\hfill\Box$

A sketch of proof of theorem 1.3. If $M$ is a submanifolds in a space form
$N(c)$, then it is biharmonic if and only if

$\displaystyle\Delta^{\bot}\vec{H}-\sum_{i=1}^{m}h(A_{\vec{H}}e_{i},e_{i})+cm\vec{H}$
$\displaystyle=$ $\displaystyle 0,$ (3.9) $\displaystyle
m\nabla|\vec{H}|^{2}+4\sum_{i=1}^{m}A_{\nabla^{\bot}_{e_{i}}\vec{H}}e_{i}$
$\displaystyle=$ $\displaystyle 0.$ (3.10)

The same as the step 1 in the proof of theorem 1.2, by using equation (3.10),
we can prove that if $H(p)\neq 0$ at a point $p\in M$, then it is a constant
around a neighborhood of the point $p$. Thus we have at $p$,
$\Delta|\vec{H}|^{2}=0.$

On the other hand at $p$

$\displaystyle\Delta|\vec{H}|^{2}$ $\displaystyle=$ $\displaystyle
2|\nabla\vec{H}|^{2}+2\langle\vec{H},\Delta\vec{H}\rangle$ $\displaystyle=$
$\displaystyle
2|\nabla\vec{H}|^{2}-2\sum_{i=1}^{m}R^{N}(e_{i},\vec{H},e_{i},\vec{H})$
$\displaystyle=$ $\displaystyle 2|\nabla\vec{H}|^{2}-2cm|\vec{H}|^{2}$
$\displaystyle\geq$ $\displaystyle 2|\nabla\vec{H}|^{2}.$

Therefore $\nabla\vec{H}(p)=0$. Now we choose an orthogonal basis
$\\{e_{i},i=1,...,m\\}$ of $T_{p}M$. Computing directly one gets at $p$

$\displaystyle
0=\langle\nabla_{e_{i}}\vec{H},e_{j}\rangle=H\langle\nabla_{e_{i}}\nu,e_{j}\rangle=-H\langle\nu,\nabla_{e_{i}}e_{j}\rangle=H\langle
h(e_{i},e_{j}),\nu\rangle,$ (3.11)

for any $1\leq i,j\leq m$. Taking trace over this equality we get

$H^{2}(p)=0.$

Therefore $H(p)=0$. This is a contradiction to $H(p)\neq 0$.

This completes the proof of theorem 1.3. $\hfill\Box$

Acknowledgement. Special thanks due to my supervisor professor Guofang Wang
for his constant encouragement and support.

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Yong Luo

Mathematisches Institut,

Albert-Ludwigs-Universität Freiburg,

Eckerstr. 1, 79104 Freiburg, Germany.

<EMAIL_ADDRESS>