Translation Hypersurfaces with Constant S_r Curvature in the Euclidean Space

Abstract

The main goal of this paper is to present a complete description of all translation hypersurfaces with constant $r$-curvature $S_r$, in the Euclidean space ${ℝ}^{n+1}$, where $3\le r\le n-1$.

Key words
Euclidean space; Scherk's surface; Translation hypersurfaces; r-Curvature

INTRODUCTION

It is well known that translation hypersurfaces are very important in Differential Geometry, providing an interesting class of constant mean curvature hypersurfaces and minimal hypersurfaces in a number of spaces endowed with good symmetries and even in certain applications in Microeconomics. There are many results about them, for instance, Chen et al. (²⁰⁰³), Dillen et al. (¹⁹⁹¹), Inoguchi et al. (²⁰¹²), Lima et al. (²⁰¹⁴), Liu (¹⁹⁹⁹), López (²⁰¹¹), López and Moruz (²⁰¹⁵), López and Munteanu (²⁰¹²), Seo (²⁰¹³) and Chen (²⁰¹¹), for an interesting application in Microeconomics.

Scherk (¹⁸³⁵) obtained the following classical theorem: Let $M:=\{(x,y,z):z=f(x)+g(y)\}$ be a translation surface in ${ℝ}^3$, if is minimal then it must be a plane or the Scherk surface defined by

$z(x,y)=\frac{1}{a}\ln \left|\frac{\cos (ay)}{\cos (ax)}\right|,$

where $a$ is a nonzero constant. In a different aspect, Liu (¹⁹⁹⁹) considered the translation surfaces with constant mean curvature in $3$-dimensional Euclidean space and Lorentz-Minkowski space and Inoguchi et al. (²⁰¹²) characterized the minimal translation surfaces in the Heisenberg group $Ni{l}_3$, and López and Munteanu, the minimal translation surfaces in $So{l}_3$.

The concept of translation surfaces was also generalized to hypersurfaces of ${ℝ}^{n+1}$ by Dillen et al. (¹⁹⁹¹), who obtained a classification of minimal translation hypersurfaces of the $(n+1)$-dimensional Euclidean space. A classification of the translation hypersurfaces with constant mean curvature in $(n+1)$-dimensional Euclidean space was made by Chen et al. (²⁰⁰³).

The absence of an affine structure in hyperbolic space does not permit to give an intrinsic concept of translation surface as in the Euclidean setting. Considering the half-space model of hyperbolic space, López (²⁰¹¹), introduced the concept of translation surface and presented a classification of the minimal translation surfaces. Seo (²⁰¹³) has generalized the results obtained by Lopez to the case of translation hypersurfaces of the $(n+1)$-dimensional hyperbolic space.

Definition 1. We say that a hypersurface $M^n$ of the Euclidean space ${ℝ}^{n\mathrm{+}\mathrm{1}}$ is a translation hypersurface if it is the graph of a function given by

$F(x_1,\mathrm{\dots },x_n)=f_1(x_1)+\mathrm{\dots }+f_n(x_n)$

where $\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{)}$ are cartesian coordinates and each $f_i$ is a smooth function of one real variable for $i\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}n$.

Now, let $M^n\subset {ℝ}^{n+1}$ be an oriented hypersurface and ${\lambda }_1,\mathrm{\dots },{\lambda }_n$ denote the principal curvatures of $M^n$. For each $r=1,\mathrm{\dots },n$, we can consider similar problems to the above ones, related with the $r$-th elementary symmetric polynomials, $S_r$, given by

In particular, $S_1$ is the mean curvature, $S_2$ the scalar curvature and $S_n$ the Gauss-Kronecker curvature, up to normalization factors. A very useful relationship involving the various $S_r$ is given in the [Proposition 1, Caminha (²⁰⁰⁶)]. This result will play a central role along this paper.

Recently, some authors have studied the geometry of translational hypersurfaces under a condition in the $S_r$ curvature, where $r>1$. Namely, Leite (¹⁹⁹¹) gave a new example of a translation hypersurface of ${ℝ}^4$ with zero scalar curvature. Lima et al. 2014 presented a complete description of all translation hypersurfaces with zero scalar curvature in the Euclidean space ${ℝ}^{n+1}$ and Seo 2013 proved that if $M$ is a translation hypersurface with constant Gauss-Kronecker curvature $GK$ in ${ℝ}^{n+1}$, then $M$ is congruent to a cylinder, and hence $GK=0$.

In this paper, we obtain a complete classification of translation hypersurfaces of ${ℝ}^{n+1}$ with $S_r=0$. We prove the following

Theorem 1. Let $M^n\mathrm{(}n\mathrm{\ge }\mathrm{3}\mathrm{)}$ be a translation hypersurface in ${ℝ}^{n\mathrm{+}\mathrm{1}}$. Then, for , $M^n$ has zero $S_r$ curvature if, and only if, it is congruent to the graph of the following functions

• $F(x_1,\mathrm{\dots },x_n)={\displaystyle \sum _{i=1}^{n-r+1}}{a}_i{x}_i+{\displaystyle \sum _{j=n-r+2}^n}{f}_j(x_j)+b$ ,

on ${ℝ}^{n\mathrm{-}r\mathrm{+}\mathrm{1}}\mathrm{\times }{J}_{n\mathrm{-}r\mathrm{+}\mathrm{2}}\mathrm{\times }\mathrm{\cdots }\mathrm{\times }{J}_n$, for certain intervals $J_{n\mathrm{-}r\mathrm{+}\mathrm{2}}\mathrm{,}\mathrm{\dots }\mathrm{,}{J}_n$, and arbitrary smooth functions $f_i\mathrm{:}{J}_i\mathrm{\subset }ℝ\mathrm{\to }ℝ$. Which defines, after a suitable linear change of variables, a vertical cylinder, or

• A generalized periodic Enneper hypersurface given by

  $F(x_1,\mathrm{\dots },x_n)$
  $=$
  ${\displaystyle \sum _{i=1}^{n-r-1}}{a}_i{x}_i$
  $+$
  ${\displaystyle \sum _{k=n-r}^{n-1}}{\displaystyle \frac{\sqrt{\beta }}{a_k}}\ln \left|{\displaystyle \frac{\cos \left(-{\displaystyle \frac{a_{n-r}\mathrm{\dots }{a}_{n-1}}{{\sigma }_{r-1}(a_{n-r},\mathrm{\dots },a_{n-1})}}\sqrt{\beta }{x}_n+b_n\right)}{\cos (a_k\sqrt{\beta }{x}_k+b_k)}}\right|+c$

on ${ℝ}^{n\mathrm{-}r\mathrm{-}\mathrm{1}}\mathrm{\times }{I}_{n\mathrm{-}r}\mathrm{\times }\mathrm{\cdots }\mathrm{\times }{I}_n$, with $a_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_{n\mathrm{-}r}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_{n\mathrm{-}\mathrm{1}}\mathrm{,}{b}_{n\mathrm{-}r}\mathrm{,}\mathrm{\dots }\mathrm{,}{b}_n$ and $c$ are real constants where $a_{n\mathrm{-}r}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_{n\mathrm{-}\mathrm{1}}$ and ${\sigma }_{r\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}{a}_{n\mathrm{-}r}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_{n\mathrm{-}\mathrm{1}}\mathrm{)}$ nonzero, $\beta \mathrm{=}\mathrm{1}\mathrm{+}{\displaystyle \mathrm{\sum }_{i\mathrm{=}\mathrm{1}}^{n\mathrm{-}r\mathrm{-}\mathrm{1}}}{a}_i^{\mathrm{2}}$, $I_k\mathrm{(}n\mathrm{-}r\mathrm{\le }k\mathrm{\le }n\mathrm{-}\mathrm{1}\mathrm{)}$ are open intervals defined by the conditions while $I_n$ is defined by .

Theorem 2. Any translation hypersurface in ${ℝ}^{n\mathrm{+}\mathrm{1}}\mathrm{(}n\mathrm{\ge }\mathrm{3}\mathrm{)}$ with $S_r$ constant, for , must have $S_r\mathrm{=}\mathrm{0}$.

Finally, we observe that, when one considers the upper half-space model of the $(n+1)$-dimensional hyperbolic space ${ℍ}^{n+1}$, that is,

${ℝ}_{+}^{n+1}=\{(x_1,\mathrm{\dots },x_n,x_{n+1})\in {ℝ}^{n+1}:x_{n+1}>0\}$

endowed with the hyperbolic metric $d{s}^2={\displaystyle \frac{1}{x_{n+1}^2}}\left(d{x}_1^2+\mathrm{\dots }+d{x}_{n+1}^2\right)$ then, unlike in the Euclidean setting, the coordinates $x_1,\mathrm{\dots },x_n$ are interchangeable, but the same does not happen with the coordinate $x_{n+1}$ and, due to this observation, López 2011 and Seo 2013 considered two classes of translation hypersurfaces in ${ℍ}^{n+1}$:

A hypersurface $M\mathrm{\subset }{ℍ}^{n\mathrm{+}\mathrm{1}}$ is called a translation hypersurface of type I (respectively, type II) if it is given by an immersion $X\mathrm{:}U\mathrm{\subset }{ℝ}^n\mathrm{\to }{ℍ}^{n\mathrm{+}\mathrm{1}}$ satisfying

$X(x_1,\mathrm{\dots },x_n)=(x_1,\mathrm{\dots },x_n,f_1(x_1)+\mathrm{\dots }+f_n(x_n))$

where each $f_i$ is a smooth function of a single variable. Respectively, in case of type II,

$X(x_1,\mathrm{\dots },x_n)=(x_1,\mathrm{\dots },x_{n-1},f_1(x_1)+\mathrm{\dots }+f_n(x_n),x_n)$

Seo proved

Theorem 3(Theorem 3.2, Seo 2013). There is no minimal translation hypersurface of type I in ${ℍ}^{n\mathrm{+}\mathrm{1}}$.

and with respect to type II surfaces he proved

Theorem 4(Theorem 3.3, Seo 2013). Let $M\mathrm{\subset }{ℍ}^{\mathrm{3}}$ be a minimal translation surface of type II given by the parametrization $X\mathit{}\mathrm{(}x\mathrm{,}z\mathrm{)}\mathrm{=}\mathrm{(}x\mathrm{,}f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{+}g\mathit{}\mathrm{(}z\mathrm{)}\mathrm{,}z\mathrm{)}$. Then the functions $f$ and $g$ are as follows:

$f(x)$

$=$

$ax+b,$

$g(z)$

$=$

$\sqrt{1+a^2}{\displaystyle \int \frac{c{z}^2}{\sqrt{1-c^2{z}^4}}𝑑z},$

where $a$, $b$, and $c$ are constants.

We emphasize that the result proved by Seo, Theorem 3.2 of Seo 2013, implies that our result (Theorem 2) is not valid in the hyperbolic space context.

PRELIMINARIES AND BASIC RESULTS

Let ${\overline{M}}^{n+1}$ be a connected Riemannian manifold. In the remainder of this paper, we will be concerned with isometric immersions, $\mathrm{\Psi }:M^n\to {\overline{M}}^{n+1}$, from a connected, $n$-dimensional orientable Riemannian manifold, $M^n$, into ${\overline{M}}^{n+1}$. We fix an orientation of $M^n$, by choosing a globally defined unit normal vector field, $\xi$, on $M$. Denote by $A$, the corresponding shape operator. At each $p\in M$, $A$ restricts to a self-adjoint linear map $A_p:T_{p}M\to T_{p}M$. For each $1\le r\le n$, let $S_r:M^n\to ℝ$ be the smooth function such that $S_r(p)$ denotes the $r$-th elementary symmetric function on the eigenvalues of $A_p$, which can be defined by the identity

(1) $det(A_p-\lambda I)=\sum _{k=0}^n{(-1)}^{n-k}{S}_k(p){\lambda }^{n-k}.$

where $S_0=1$ by definition. If $p\in M^n$ and $\{e_l\}$ is a basis of $T_{p}M$, given by eigenvectors of $A_p$, with corresponding eigenvalues $\{{\lambda }_l\}$, one immediately sees that

$S_r={\sigma }_r({\lambda }_1,\mathrm{\dots },{\lambda }_n),$

where ${\sigma }_r\in ℝ[X_1,\mathrm{\dots },X_n]$ is the $r$-th elementary symmetric polynomial on $X_1,\mathrm{\dots },X_n$. Consequently,

In the next result we present an expression for the curvature $S_r$ of a translation hypersurface in the Euclidean space. This expression will play an essential role in this paper.

Proposition 1. Let $F\mathrm{:}\mathrm{\Omega }\mathrm{\subset }{ℝ}^n\mathrm{\to }ℝ$ be a smooth function, defined as $F\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{)}\mathrm{=}{\mathrm{\sum }}_{i\mathrm{=}\mathrm{1}}^n{f}_i\mathit{}\mathrm{(}{x}_i\mathrm{)}$, where each $f_i$ is a smooth function of one real variable. Let $M^n$ be the graphic of $F$, given in coordinates by

(2) $\phi (x_1,\mathrm{\dots },x_n)=\sum _{i=1}^n{x}_i{e}_i+F(x_1,\mathrm{\dots },x_n)e_{n+1}.$

The $S_r$ curvature of $M^n$ is given by

(3)

where the dot represents derivative with respect to the corresponding variable, that is, for each $j\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathrm{,}n$, one has ${\dot{f}}_j\mathrm{=}{\displaystyle \frac{d\mathit{}{f}_j}{d\mathit{}{x}_j}}\mathit{}\mathrm{(}{x}_j\mathrm{)}\mathrm{=}{\displaystyle \frac{\mathrm{\partial }\mathit{}F}{\mathrm{\partial }\mathit{}{x}_j}}\mathit{}\mathrm{(}{x}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{x}_n\mathrm{)}$ and $W^{\mathrm{2}}\mathrm{=}\mathrm{1}\mathrm{+}{\mathrm{|}\mathrm{\nabla }\mathit{}F\mathrm{|}}^{\mathrm{2}}$

Proof. Let $F$ be as stated in the Proposition, denote by $\nabla F={\sum }_{i=1}^n{\displaystyle \frac{\partial F}{\partial x^i}}{e}_i$ the Euclidean gradient of $F$ and $⟨,⟩$ the standard Euclidean inner product. Then, we have

$\nabla F=\sum _{i=1}^n{\dot{f}}_i{e}_i$

and the coordinate vector fields associated to the parametrization given in (2) have the following form

$\frac{\partial \phi }{\partial x_m}=e_m+{\dot{f}}_m{e}_{n+1},m=1,\mathrm{\dots },n.$

Hence, the elements $G_{ij}$ of the metric of $M^n$ are given by

implying that the matrix of the metric $G$ has the following form

$G=I_n+{(\nabla F)}^t\nabla F,$

where $I_n$ is the identity matrix of order $n$. Note that the $i$-th column of $G$, which will be denoted by $G^i$, has the expression given by the column vector

(4) $G^i=e_i+{\dot{f}}_i\nabla F.$

An easy calculation shows that the unitary normal vector field $\xi$ of $M^n$ satisfies

$W\xi =e_{n+1}-\nabla F,$

where $W^2=1+{|\nabla F|}^2$. Thus, the second fundamental form $B_{ij}$ of $M^n$ satisfies

implying that the matrix of $B$ is diagonal

$B=\frac{1}{W}\cdot \text{diag}({\ddot{f}}_1,\mathrm{\dots },{\ddot{f}}_n),$

with $i$-th column given by the column vector

(5) $B^i=\frac{{\ddot{f}}_i}{W}{e}_i.$

If $A$ denotes the matrix of the Weingarten mapping, then $A=G^{-1}B$. In (1), changing $\lambda$ by ${\lambda }^{-1}$ gives

$det(\lambda A-I)=\sum _{i=1}^n{(-1)}^{n-i}{S}_i{\lambda }^i.$

Thus, we conclude that the expression for curvature $S_r$ can be found by the following calculation

${(-1)}^{n-r}{S}_r=\frac{1}{r!}{\frac{d^r}{d{\lambda }^r}}_{|\lambda =0}det(\lambda A-I).$

Note that

${(-1)}^{n-r}detG\cdot S_r=\frac{1}{r!}detG\cdot {\frac{d^r}{d{\lambda }^r}}_{|\lambda =0}det(\lambda A-I)=\frac{1}{r!}{\frac{d^r}{d{\lambda }^r}}_{|\lambda =0}det(\lambda B-G).$

Due to the multilinearity of function $det$, on its $n$ column vectors, it follows immediately that

${\frac{d}{d\lambda }}_{|\lambda =0}det[\lambda B^1-G^1,\mathrm{\cdots },\lambda B^n-G^n]=\sum _{i=1}^n{(-1)}^{n-1}det[G^1,\mathrm{\cdots },\underset{\text{i-th term}}{\underbrace{B^i}},\mathrm{\cdots },G^n],$

leading to the conclusion

and thus

(6)

Now, applying the expressions (4) and (5) in (6) we reach to the expression

(7)

Calculating the determinant on the right in the equality above, we get

$det[e_1+{\dot{f}}_1\nabla F,$

$\mathrm{\dots }$

$,e_{i_1},\mathrm{\dots },e_{i_r},\mathrm{\dots },e_n+{\dot{f}}_n\nabla F]=$

$=$

$1+{\displaystyle \sum _{i\ne i_1,\mathrm{\dots },i_r}}{\dot{f}}_{i}det[e_1,\mathrm{\dots },e_{i_1},\mathrm{\dots },\underset{i\text{-th term}}{\underbrace{\nabla F}},\mathrm{\dots },e_{i_r},\mathrm{\dots },e_n]$

$=$

$1+{\displaystyle \sum _{\begin{array}{c}1\le i\le n\\ i\ne i_1,\mathrm{\dots },i_r\end{array}}}{\dot{f}}_i^2.$

Consequently, the expression for $S_r$ in (7) assumes the following form

Finally, using that $detG=W^2$ we obtain the desired expression

RESULTS

In order to prove Theorem 1 we need the following lemma.

Lemma 1. Let $f_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{f}_r$ be smooth functions of one real variable satisfying the differential equation

(8) $\sum _{k=1}^r\ddot{f_1}(x_1)\mathrm{\dots }\widehat{\ddot{f_k}(x_k)}\mathrm{\dots }\ddot{f_r}(x_r)(\beta +{\dot{f_k}}^2(x_k))=0,$

where $\beta$ is a positive real constant and the big hat means an omitted term. If $\ddot{f_i}\mathrm{\ne }\mathrm{0}$, for each $i\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathit{}r$ then

$\sum _{k=1}^r{f}_k(x_k)=\sum _{k=1}^{r-1}\frac{\sqrt{\beta }}{a_k}\ln \left|\frac{\cos \left(-{\displaystyle \frac{a_1\mathrm{\dots }{a}_{r-1}}{{\sigma }_{r-2}(a_1,\mathrm{\dots },a_{r-1})}}\sqrt{\beta }{x}_r+b_r\right)}{\cos (a_k\sqrt{\beta }{x}_k+b_k)}\right|+c$

where $a_i\mathrm{,}{b}_i\mathrm{,}c$, $i\mathrm{=}\mathrm{1}\mathrm{,}\mathrm{\dots }\mathit{}r$ are real constants with $a_i\mathrm{,}{\sigma }_{r\mathrm{-}\mathrm{2}}\mathit{}\mathrm{(}{a}_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{a}_{r\mathrm{-}\mathrm{1}}\mathrm{)}\mathrm{\ne }\mathrm{0}$.

Since the derivatives ${\ddot{f}}_i\ne 0$ it follows that $\ddot{f_1}(x_1)\mathrm{\dots }\ddot{f_r}(x_n)\ne 0$. Thus dividing (8) by this product we get the equivalent equation:

$\sum _{k=1}^r\frac{\beta +{\dot{f_k}}^2(x_k)}{\ddot{f_k}(x_k)}=0,$

which implies, after taking derivative with respect to $x_l$ for each $l=1,\mathrm{\dots }r$, that ${\left({\displaystyle \frac{\beta +{\dot{f_l}}^2(x_l)}{\ddot{f_l}(x_l)}}\right)}^{\prime }=0$, thus ${\displaystyle \frac{\beta +{\dot{f_l}}^2(x_l)}{\ddot{f_l}(x_l)}}={\tilde{a}}_l$ for some non null constant ${\tilde{a}}_l$. Thus, setting $a_l={\displaystyle \frac{1}{{\tilde{a}}_l}}$

$\frac{\ddot{f_l}(x_l)}{\beta +{\dot{f_l}}^2(x_l)}=a_l\mathit{}\text{for each}l=1,\mathrm{\dots },r$

which can be easily solved to give:

$\arctan \left(\frac{\dot{f_l}(x_l)}{\sqrt{\beta }}\right)=a_l\sqrt{\beta }x+b_l\mathit{}\text{for some constant}{b}_l$

and consequently

(9) $f_l(x_l)=-\frac{1}{a_l}\sqrt{\beta }\ln |\cos (a_l\sqrt{\beta }{x}_l+b_l)|+c_l,l=1,\mathrm{\dots },r.$

Now, since ${\sum }_{k=1}^r{\displaystyle \frac{1}{a_k}}=0$ it implies that ${\displaystyle \frac{1}{a_r}}=-{\displaystyle \frac{{\sigma }_{r-2}(a_1,\mathrm{\dots },a_{r-1})}{a_1\mathrm{\dots }{a}_{r-1}}}$, from (9) it follows that

$f_r(x_r)=\frac{{\sigma }_{r-2}(a_1,\mathrm{\dots },a_{r-1})}{a_1\mathrm{\dots }{a}_{r-1}}\sqrt{\beta }\ln |\cos (a_r\sqrt{\beta }{x}_r+b_r)|+c_r.$

Consequently

$\sum _{k=1}^r{f}_k(x_k)=\sum _{k=1}^{r-1}\frac{1}{a_k}\sqrt{\beta }\ln \left|\frac{\cos \left(-{\displaystyle \frac{a_1\mathrm{\dots }{a}_{r-1}}{{\sigma }_{r-2}(a_1,\mathrm{\dots },a_{r-1})}}\sqrt{\beta }{x}_r+b_r\right)}{\cos (a_k\sqrt{\beta }{x}_k+b_k)}\right|+c,$

where $c=c_1+\mathrm{\dots }+c_r$. ∎

With this lemma at hand we can go to the proof of Theorem 1.

Proof of the Theorem 1 From Proposition 1, we have that $M^n$ has zero $S_r$ curvature if, and only if,

(10)

In order to ease the analysis, we divide the proof in four cases.

Case 1: Suppose ${\ddot{f}}_i(x_i)=0$, $\forall i=1,\mathrm{\dots },n-r+1$. In this case, we have no restrictions on the functions $f_{n-r+2},\mathrm{\dots },f_n$. Thus

$\mathrm{\Psi }(x_1,\mathrm{\dots },x_n)=(x_1,\mathrm{\dots },x_n,\sum _{i=1}^{n-r+1}{a}_i{x}_i+\sum _{j=n-r+2}^n{f}_j(x_j)+b)$

where $a_i,b\in ℝ$ and for $l=n-r+2,\mathrm{\dots },n$, the functions $f_l:I_l\subset ℝ\to ℝ$ are arbitrary smooth functions of one real variable. Note that the parametrization obtained comprise hyperplanes.

Case 2: Suppose ${\ddot{f}}_i(x_i)=0$, $\forall i=1,\mathrm{\dots },n-r$, then, there are constants ${\alpha }_i$ such that ${\dot{f}}_i={\alpha }_i$, for $i=1,\mathrm{\dots },n-r$. From (10) we have

${\ddot{f}}_{n-r+1}\mathrm{\dots }{\ddot{f}}_n(1+{\alpha }_1^2+\mathrm{\cdots }+{\alpha }_{n-r}^2)=0,$

from which we conclude that ${\ddot{f}}_k=0$ for some $k\in \{n-r+1,\mathrm{\dots }n\}$ and thus, this case is contained in the Case $1$.

Case 3: Now suppose ${\ddot{f}}_i(x_i)=0$, $\forall i=1,\mathrm{\dots },n-r-1$ and ${\ddot{f}}_k(x_k)\ne 0$, for every $k=n-r,\mathrm{\dots },n$. Observe that if we had ${\ddot{f}}_k(x_k)=0$ for some $k=n-r,\mathrm{\dots },n$ the analysis would reduce to the Cases 1 and 2. In this case, there are constants ${\alpha }_i$ such that ${\dot{f}}_i={\alpha }_i$ for any $1\le i\le n-r-1$. From (10) we have

$\sum _{k=n-r}^n{\ddot{f}}_{n-r}\mathrm{\dots }\widehat{{\ddot{f}}_k}\mathrm{\dots }{\ddot{f}}_n(\beta +{\dot{f}}_k^2)=0$

where $\beta =1+{\displaystyle \sum _{k=1}^{n-r-1}}{\alpha }_k^2$ and the hat means an omitted term. Then, from Lemma 1 we have that

$\sum _{k=n-r}^n{f}_k(x_k)=\sum _{k=n-r}^{n-1}\frac{\sqrt{\beta }}{a_k}\ln \left|\frac{\cos \left(-{\displaystyle \frac{a_{n-r}\mathrm{\dots }{a}_{n-1}}{{\sigma }_{r-1}(a_{n-r},\mathrm{\dots },a_{n-1})}}\sqrt{\beta }{x}_n+b_n\right)}{\cos (a_k\sqrt{\beta }{x}_k+b_k)}\right|+c$

where $a_{n-r},\mathrm{\dots },a_{n-1},b_{n-r},\mathrm{\dots },b_n$ and $c$ are real constants, and $a_{n-r},\mathrm{\dots },a_{n-1}$, and ${\sigma }_{r-1}(a_{n-r},\mathrm{\dots },a_{n-1})$ are nonzero.

Case 4: Finally, suppose ${\ddot{f}}_i(x_i)=0$, where $1\le i\le k$ and $n-k\ge r+2$, and ${\ddot{f}}_i(x_i)\ne 0$ for any $i>k$. We will show that this case cannot occur. In fact, note that for any fixed $l\ge k+1$

$+$

${\displaystyle \sum _{\begin{array}{c}1\le m\le n\\ m\ne i_1,\mathrm{\dots }{i}_r\end{array}}}{\dot{f}}_m^2)$

$=$

$+$

Derivative with respect to the variable $x_l(l\ge k+1)$, in the above equality, gives

$+$

${\displaystyle \sum _{\begin{array}{c}1\le m\le n\\ m\ne l,i_1,\mathrm{\dots },i_{r-1}\end{array}}}{\dot{f}}_m^2)$

(11) $+$

That is, if we set

$A_l$

$=$

$B_l$

$=$

then, it follows that $A_l,B_l$ do not depend on the variable $x_l$ and we can write

(12) $A_l{\stackrel{\dot{}\dot{}\dot{}}{f}}_l+2B_l{\dot{f}}_l{\ddot{f}}_l=0.$

We have two possible situations to take into account: Case I.$A_l\ne 0,\forall l\ge k+1$, and Case II. there is an $l\ge k+1$ such that $A_l=0$.

Case I. $A_l\mathrm{\ne }\mathrm{0}$: Under this assumption, there are constants ${\alpha }_l(l=k+1,\mathrm{\dots },n)$ such that equation (12) becomes ${\stackrel{\dot{}\dot{}\dot{}}{f}}_l+2{\alpha }_l{\dot{f}}_l{\ddot{f}}_l=0$. Furthermore, it can be shown that for $\{l_1,\mathrm{\dots },l_{r+1}\}\subset \{k+1,\mathrm{\dots },n\}$

(13) $\frac{{\partial }^{r+1}{G}_r(f_1,\mathrm{\dots },f_n)}{\partial x_{l_1}\mathrm{\cdots }\partial x_{l_{r+1}}}=2\sum _{k=1}^{r+1}\left({\dot{f}}_{l_k}{\ddot{f}}_{l_k}\prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}\right)$

where

Since $S_r=0$ it follows that $G_r=0$, and using that ${\displaystyle \prod _{k=1}^{r+1}}{\dot{f}}_{l_k}{\ddot{f}}_{l_k}\ne 0$ we obtain

(14) ${\displaystyle \sum _{k=1}^{r+1}}\left({\displaystyle \prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}}{\displaystyle \frac{{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}}{{\dot{f}}_{l_m}{\ddot{f}}_{l_m}}}\right)$

$=$

${\displaystyle \frac{{\displaystyle \sum _{s=1}^{r+1}}\left({\dot{f}}_{l_s}{\ddot{f}}_{l_s}{\displaystyle \prod _{\begin{array}{c}m=1\\ m\ne s\end{array}}^{r+1}}{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}\right)}{{\displaystyle \prod _{k=1}^{r+1}}{\dot{f}}_{l_k}{\ddot{f}}_{l_k}}}=0.$

Now, for $l=l_1,\mathrm{\dots }{l}_{r+1}$, substitute ${\stackrel{\dot{}\dot{}\dot{}}{f}}_l+2{\alpha }_l{\dot{f}}_l{\ddot{f}}_l=0$ in (14) to obtain the identity

(15) ${\sigma }_r({\alpha }_{l_1},\mathrm{\dots },{\alpha }_{l_r},{\alpha }_{l_{r+1}})=0$

for any $l_1,\mathrm{\dots },l_r,l_{r+1}\in \{k+1,\mathrm{\dots },n\}$. Hence we conclude that,

${\sigma }_r({\alpha }_{k+1},\mathrm{\dots },{\alpha }_n)$

$=$

$0$

${\sigma }_{r+1}({\alpha }_{k+1},\mathrm{\dots },{\alpha }_n)$

$=$

$0.$

These equalities, from [Proposition 1, Caminha (²⁰⁰⁶)], imply that at most $r-1$ of the constants ${\alpha }_l(l\ge k+1)$ are nonzero. If ${\alpha }_{l_1}\ne 0,\mathrm{\dots },{\alpha }_{l_m}\ne 0$ with $m\le r-1$, in the expression obtained for $B_l$, making $l\ne l_1,\mathrm{\dots },l_m$ and taking derivatives with respect to the variables $x_{l_1},\mathrm{\dots },x_{l_m}$ we get

$\prod _{j=l_1}^{l_m}{\stackrel{\dot{}\dot{}\dot{}}{f}}_j\cdot {\sigma }_{r-m}({\ddot{f}}_{k+1},\mathrm{\dots },\widehat{{\ddot{f}}_l},\mathrm{\dots },{\widehat{\ddot{f}}}_{l_1},\mathrm{\dots },{\widehat{\ddot{f}}}_{l_m},\mathrm{\dots },{\ddot{f}}_n)=0$

for all $l\in \{k+1,\mathrm{\dots },n\}\setminus \{l_1,\mathrm{\dots },l_m\}$. As ${\stackrel{\dot{}\dot{}\dot{}}{f}}_j\ne 0$ for all $j\in \{l_1\mathrm{\dots },l_m\}$, we obtain that

${\sigma }_{r-m}({\ddot{f}}_{k+1},\mathrm{\dots },{\widehat{\ddot{f}}}_l,\mathrm{\dots },{\widehat{\ddot{f}}}_{l_1},\mathrm{\dots },{\widehat{\ddot{f}}}_{l_m},\mathrm{\dots },{\ddot{f}}_n)=0$

for all $l\in \{k+1,\mathrm{\dots },n\}\setminus \{l_1,\mathrm{\dots },l_m\}$. Consequently,

${\sigma }_{r-m}({\ddot{f}}_{k+1},,\mathrm{\dots },{\widehat{\ddot{f}}}_{l_1},\mathrm{\dots },{\widehat{\ddot{f}}}_{l_m},\mathrm{\dots },,{\ddot{f}}_n)$

$=$

$0$

${\sigma }_{r-m+1}({\ddot{f}}_{k+1},,\mathrm{\dots },{\widehat{\ddot{f}}}_{l_1},\mathrm{\dots },{\widehat{\ddot{f}}}_{l_m},\mathrm{\dots },,{\ddot{f}}_n)$

$=$

$0.$

Since $(n-k-m)-(r-m)=n-k-r\ge 2$, at most $r-m-1$ of the functions ${\ddot{f}}_l$ are nonzero, for $k+1\le l\le n$ and $l\ne l_1,\mathrm{\dots },l_m$, leading to a contradiction. So, ${\alpha }_j=0$ for all $j\in \{l_1\mathrm{\dots },l_{r-1}\}$, which implies that ${\ddot{f}}_l$ is constant for all $l\in \{k+1,\mathrm{\dots },n\}$. Now, again from equation (11) we get

From which, we conclude that

${\sigma }_r({\ddot{f}}_{k+1},\mathrm{\dots },{\ddot{f}}_n)$

$=$

$0$

${\sigma }_{r+1}({\ddot{f}}_{k+1},\mathrm{\dots },{\ddot{f}}_n)$

$=$

$0.$

Therefore, at most $r-1$ of the functions ${\ddot{f}}_l(k+1\le l\le n)$ are nonzero, leading to a contradiction. Thus, it follows that Case 4 cannot occur, if $A_l\ne 0$ for every $l$.

Case $A_l\mathrm{=}\mathrm{0}$:In this case, we have $B_l\dot{f_l}\ddot{f_l}=0$ implying

$A_l$

$=$

$B_l$

$=$

Derivative of $A_l$ with respect to variable $x_s$, for $s=k+1,\mathrm{\dots },n$ and $s\ne l$, gives

(16)

Now, for $i_1,\mathrm{\dots },i_r\in \{k+1,\mathrm{\dots },n\}$ with $i_1,\mathrm{\dots },i_r,l$ distinct indices, taking the derivatives of $B_l$ with respect to $x_{i_1},\mathrm{\dots },x_{i_r}$ gives

${\stackrel{\dot{}\dot{}\dot{}}{f}}_{i_1}\mathrm{\dots }{\stackrel{\dot{}\dot{}\dot{}}{f}}_{i_r}=0.$

Consequently, for at most $r-1$ indices, say $i_1,\mathrm{\dots },i_{r-1}$, we can have ${\stackrel{\dot{}\dot{}\dot{}}{f}}_{i_m}\ne 0$, $(m=1,\mathrm{\dots },r-1)$, and ${\stackrel{\dot{}\dot{}\dot{}}{f}}_j=0$ for every $j=k+1,\mathrm{\dots },n$, with $j\ne l,i_1,\mathrm{\dots },i_{r-1}$. Thus ${\stackrel{\dot{}\dot{}\dot{}}{f}}_{i_m}\ne 0$, with $i_m\ne l$, together with equation $\frac{\partial B_l}{\partial x_{i_m}}=0$ implies that the sum

Now, if ${\stackrel{\dot{}\dot{}\dot{}}{f}}_j=0$ we have by equation (16) that

Therefore,

From which, we conclude that

${\sigma }_{r-1}({\ddot{f}}_{k+1},\mathrm{\dots },{\widehat{\ddot{f}}}_l,\mathrm{\dots },{\ddot{f}}_n)$

$=$

$0$

${\sigma }_r({\ddot{f}}_{k+1},\mathrm{\dots },{\widehat{\ddot{f}}}_l,\mathrm{\dots },{\ddot{f}}_n)$

$=$

$0.$

Thus, for at most $r-2(r\ge 3)$ indices we must have ${\ddot{f}}_j\ne 0$, for every $j=k+1,\mathrm{\dots },n$, and $j\ne l$. This contradicts the hypothesis assumed in Case 4. Hence, $A_l=0$ cannot occur. Since the case $A_l\ne 0$, cannot occur as well, it follows that Case 4 is not possible. This completes the proof of the theorem. ∎

[Proof of the Theorem 2] Let $M^n\subset {ℝ}^{n+1}$ be a translation hypersurface with constant $S_r$ curvature. First, note that

(17) ${\displaystyle \frac{{\partial }^m{W}^{r+2}}{\partial x_{i_1}\mathrm{\cdots }\partial x_{i_m}}}$

$=$

${\displaystyle \prod _{j=1}^m}(r+4-2j)\cdot {\displaystyle \prod _{k=1}^m}{\dot{f}}_{i_k}{\ddot{f}}_{i_k}\cdot W^{r+2-2m}.$

We have as a consequence of the proof of Theorem 1, see (13), the identity

$\frac{{\partial }^{r+1}{G}_r(f_1,\mathrm{\dots },f_n)}{\partial x_{l_1}\mathrm{\cdots }\partial x_{l_{r+1}}}=2\sum _{k=1}^{r+1}\left({\dot{f}}_{l_k}{\ddot{f}}_{l_k}\prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}\right)$

where . With this we conclude, by Proposition 1, that

${\displaystyle \prod _{j=1}^{r+1}}(r+4-2j)\cdot {\displaystyle \prod _{k=1}^{r+1}}{\dot{f}}_{l_k}{\ddot{f}}_{l_k}\cdot W^{-r}{S}_r$

$=$

${\displaystyle \frac{{\partial }^{r+1}(W^{r+2}{S}_r)}{\partial x_{l_1}\mathrm{\cdots }\partial x_{l_{r+1}}}}$

(18) $=$

$2{\displaystyle \sum _{k=1}^{r+1}}\left({\dot{f}}_{l_k}{\ddot{f}}_{l_k}{\displaystyle \prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}}{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}\right).$

Now, we have two cases to consider: $r$ odd and $r$ even.

Case $r$ odd: Suppose that there are $l_1,\mathrm{\dots },l_{r+1}$ such that ${\displaystyle \prod _{k=1}^{r+1}}{\dot{f}}_{l_k}{\ddot{f}}_{l_k}\ne 0$. Then,

$Q_r:={\displaystyle \prod _{j=1}^{r+1}}(r+4-2j)\cdot W^{-r}{S}_r$

$=$

$2{\displaystyle \frac{{\displaystyle \sum _{s=1}^{r+1}}\left({\dot{f}}_{l_s}{\ddot{f}}_{l_s}{\displaystyle \prod _{\begin{array}{c}m=1\\ m\ne s\end{array}}^{r+1}}{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}\right)}{{\displaystyle \prod _{k=1}^{r+1}}{\dot{f}}_{l_k}{\ddot{f}}_{l_k}}}$

$=$

$2{\displaystyle \sum _{k=1}^{r+1}}\left({\displaystyle \prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}}{\displaystyle \frac{{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}}{{\dot{f}}_{l_m}{\ddot{f}}_{l_m}}}\right).$

Therefore,

$\frac{{\partial }^{r+1}{Q}_r}{\partial x_{l_1}\mathrm{\cdots }\partial x_{l_{r+1}}}=0.$

On the other hand, using (17) we obtain

$\frac{{\partial }^{r+1}{Q}_r}{\partial x_{l_1}\mathrm{\cdots }\partial x_{l_{r+1}}}=\prod _{j=1}^{r+1}(r+4-2j)\prod _{i=1}^{r+1}(-r+2-2i)\prod _{k=1}^{r+1}{\dot{f}}_{l_k}{\ddot{f}}_{l_k}{W}^{-3r-2}{S}_r.$

Since $r$ is odd, we conclude that $r+4-2j\ne 0$ and $-r+2-2j\ne 0$, for any $j\in \mathbb{N}$ and, therefore, $S_r=0$.

Now, if for at most $r$ indices we have ${\ddot{f}}_j\ne 0$ for example $j=l_1,\mathrm{\dots },l_r$ then

$W^{r+2}{S}_r={\ddot{f}}_{l_1}\mathrm{\cdots }{\ddot{f}}_{l_r}\alpha ,$

for some constant $\alpha \ne 0$. Thus,

$(r+2)W^r{\dot{f}}_{l_1}{\ddot{f}}_{l_1}{S}_r={\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_1}{\ddot{f}}_{l_2}\mathrm{\cdots }{\ddot{f}}_{l_r}\alpha .$

If ${\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_1}=0$, then $S_r=0$. Otherwise,

$(r+2)W^{r+2}{\dot{f}}_{l_1}{\ddot{f}}_{l_1}{S}_r={\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_1}{\ddot{f}}_{l_2}\mathrm{\cdots }{\ddot{f}}_{l_r}{W}^2\alpha \mathit{}\Rightarrow \mathit{}(r+2){\dot{f}}_{l_1}{({\ddot{f}}_{l_1})}^2={\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_1}{W}^2.$

As $r>1$ implies that $W$ does not depend on the variables $x_{l_2},\mathrm{\dots },x_{l_n}$, it follows that ${\ddot{f}}_{l_2}=\mathrm{\cdots }={\ddot{f}}_{l_n}=0$ leading to a contradiction.

Case $r$ even: In this case, there is a natural $q\ge 2$ such that $r=2q$. Then $r+1\ge q+2$ and consequently

$\prod _{k=1}^{r+1}(r+4-2k)=0.$

Therefore, by (18) we get

$\sum _{k=1}^{r+1}\left({\dot{f}}_{l_k}{\ddot{f}}_{l_k}\prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}\right)=0.$

Suppose that there are $l_1,\mathrm{\dots },l_{r+1}$ such that ${\displaystyle \prod _{k=1}^{r+1}}{\ddot{f}}_{l_k}\ne 0$. In this case,

$\sum _{k=1}^{r+1}\left(\prod _{\begin{array}{c}m=1\\ m\ne k\end{array}}^{r+1}\frac{{\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_m}}{{\dot{f}}_{l_m}{\ddot{f}}_{l_m}}\right)=0.$

We conclude that for each $l_i$ there is a constant ${\alpha }_{l_i}$ such that ${\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_i}={\alpha }_{l_i}{\dot{f}}_{l_i}{\ddot{f}}_{l_i}$. Now, it is easy to verify (see (11)) that

$(r+2){\dot{f}}_{l_{r+1}}{\ddot{f}}_{l_{r+1}}{W}^r{S}_r$

$=$

${\displaystyle \frac{\partial G_r(f_1,\mathrm{\dots },f_n)}{\partial x_{l_{r+1}}}}$

$=$

${\stackrel{\dot{}\dot{}\dot{}}{f}}_{l_{r+1}}{G}_{r-1}(f_1,\mathrm{\dots },{\widehat{f}}_l,\mathrm{\dots },f_n)$

$+$

Therefore,

Differentiating this identity with respect to the variable $x_{l_{r+1}}$, gives

$(r+2)r{\dot{f}}_{l_{r+1}}{\ddot{f}}_{l_{r+1}}{W}^{r-2}{S}_r=0\mathit{}\text{implying that}\mathit{}{S}_r=0.$

Finally, suppose that for any $(r+1)$-tuple of indices, say $l_1,\mathrm{\dots },l_{r+1}$ it holds that ${\displaystyle \prod _{k=1}^{r+1}}{\ddot{f}}_{l_k}=0$. Then,

${\sigma }_{r+1}({\ddot{f}}_1,\mathrm{\dots },{\ddot{f}}_n)$

$=$

$0$

${\sigma }_{r+2}({\ddot{f}}_1,\mathrm{\dots },{\ddot{f}}_n)$

$=$

$0.$

Implying that at least $n-r$ derivatives ${\ddot{f}}_l$ vanish, i.e., there are at most $r$ functions such that ${\ddot{f}}_j\ne 0$ for example $j=l_1,\mathrm{\dots },l_r$. Thus, by Proposition 1

$W^{r+2}{S}_r={\ddot{f}}_{l_1}\mathrm{\cdots }{\ddot{f}}_{l_r}\alpha$

for some constant $\alpha \ne 0$. We conclude that $S_r=0$ analogously to the way it was presented for the case $r$ odd. ∎

Acknowledgments

The authors would like to express their gratitude to referee for valuable suggestions on the improvement of the paper. The first, second and third author are partially supported by Coordenação de Aperfeiçoamento de Pessoal de Nível Superior (CAPES). The fourth author is partially supported by CAPES and Conselho Nacional de Desenvolvimento Científico e Tecnológico (CNPq).

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Publication Dates

• Publication in this collection
  01 Dec 2016
• Date of issue
  Oct-Dec 2016

History

• Received
  12 May 2015
• Accepted
  20 Oct 2015