\section{Parametrizations}



In this section, we first recall how a Euclidean hypersurface of constant
rank can be locally parametrized by the use of the Gauss parametrization.
Then, we  discuss a class of envelopes of hyperplanes depending on parameters
as well how they can be  described in terms of the Gauss parametrization.


\subsection{The Gauss parametrization}


Let $f\colon M^n\to\mathbb{R}^{n+1}$ be an isometric immersion of constant rank 
$k$ for $1\leq k\leq n-1$.  By that we mean that its second fundamental 
form $A$ has constant rank $k$ or, equivalently, that the relative nullity
subspaces, i.e., the kernels of its second form  $\Delta(x)=\mbox{ker } A(x)$, 
satisfy $\dim\Delta(x)=n-k$  at any $x\in M^n$. 
In this situation, it is a standard fact that the tangent distribution 
$x\in M^n\mapsto\Delta(x)$ is integrable and that its totally geodesic 
leaves are mapped by $f$ into open subsets of affine subspaces of $\mathbb{R}^{n+1}$

A hypersurface of constant rank can be locally parametrized in terms of 
the image of its Gauss map $N$ and its support function $\gamma=\<f,N\>$.  
This parametrization is known as the  \emph{Gauss parametrization} 
and was described in \cite{dg} but it was already used by Sbrana  
in \cite{sb1} and \cite{sb2} long before.
\vspace{1ex}

Let  $(g,\gamma)$ be a pair formed by an isometric immersion 
$g\colon L^k\to\mathbb{S}^n$ into the unit sphere and a function 
$\gamma\in C^\infty(L)$. Denote by $\bar\pi\colon\Lambda\to L^k$ the normal 
bundle of $g$ and set $h=i\circ g$ where $i\colon\mathbb{S}^n\to\mathbb{R}^{n+1}$ 
is the standard inclusion. It was shown in \cite{dg} that the map 
$\psi\colon \Lambda\to\mathbb{R}^{n+1}$ given by
$$
\psi(x,w)=\gamma(x)h(x)+h_*\mbox{grad\,}\gamma(x)+i_*w
$$
parametrizes (at regular points) a hypersurface of constant rank $k$ such
that the fibers of $\Lambda$ are identified with the leaves of the 
relative nullity foliation of $\psi$.

Conversely, any  hypersurface $f\colon M^n\to\mathbb{R}^{n+1}$ of constant  rank $k$ can be 
locally parametrized as above.   In fact, let  $U\subset M^n$ be 
an open saturated subset  of leaves of relative nullity
and let $\pi\colon U\to L^{n-k}$ denote the projection onto the quotient space.
The Gauss map $N$ of $f$  induces an immersion $g\colon L^{n-k}\to\mathbb{S}^n$ 
given by $g\circ\pi = N$. Moreover, since the support function  
$\<f,N\>$ is  constant along the relative nullity leaves, hence it induces 
a function $\gamma\in C^\infty(L)$. Now the Gauss parametrization allows 
to recover $f$ by means of the pair $(g,\gamma)$. 
\vspace{1,5ex}


The next statement presents some basic properties of the 
Gauss parametrization.



\begin{proposition}{\hspace*{-1ex}}{\bf .  }\label{gp} The following assertions hold:
\begin{itemize}
\item[(i)] The map $\psi\colon \Lambda\to\mathbb{R}^{n+1}$ is regular at $(x,w)$ 
if and only if  the self adjoint operator  
$$
P_w(x)=\gamma(x)I+\emph{Hess}\,\gamma(x)-A_w
$$ 
on $T_xL$ is nonsingular. Here $A_w$ is the shape operator of $g$ 
with respect to $w$. 
\item[(ii)]  The map $\psi$ when restricted to the open subset $V$ of regular 
points is an immersed hypersurface having the map $N\colon\Lambda\to\mathbb{S}^n$ 
given by $N(x,w)=g(x)$ as a Gauss map of rank $k$.
\item[(iii)] If $(x,w)\in V$ there is $j(x,w)\colon T_xL\to T_{(x,w)}\Lambda$ 
so that $j\colon T_xL\to\Delta^\perp(x,w)\subset T_{(x,v)}\Lambda$ is
an isometry such that
\begin{equation} \label{eq}
h_*=\psi_*\circ j,\;\;\;P_w^{-1}=\bar\pi_*\circ j\;\;\mbox{and}\;\;
A\circ  j=-j\circ P_w^{-1}
\end{equation} 
where $A$ is the shape operator of $\psi$ at $(x,w)$ with respect to $N$.
\end{itemize}
\end{proposition}

\noindent\emph{Proof: } See \cite{dg}.\qed


\subsection{Envelopes of hyperplanes}


Let $P_u\colon\mathbb{R}^n\to\mathbb{R}^{n+1}$, $n\geq 2$, denote a smooth $k$-parameter 
family of affine hyperplanes parametrized by $u=(u_1,\ldots,u_k)$ on an 
open subset $U\subset\mathbb{R}^k$ with $1\leq k\leq n-1$. 
\vspace{1ex}

We say that a hypersurface  $f\colon M^n\to\mathbb{R}^{n+1}$ is the 
\emph{envelope of hyperplanes} of $(P_u)_{u\in U}$  if there exists 
a smooth totally geodesic foliation of $M^n$ by leaves 
$(L_u)_{u\in U}$ of dimension  $n-k$ parametrized by an embedding 
$h\colon U\to M^n$ transversal to the foliation and embeddings 
$j_u\colon L_u\to P_u(\mathbb{R}^n)$ as  open subset of $(n-k)$-dimensional 
affine subspaces of $\mathbb{R}^{n+1}$ such that $j_u=f\vert_{L_u}$ and 
$P_u(\mathbb{R}^n)=f_*T_{h(u)}M$ for any $u\in U$. 
\vspace{1ex}


Clearly, the leaves of $(L_u)_{u\in U}$ are contained in the relative nullity
of $f$. Also notice that any hypersurface $f\colon M^n\to\mathbb{R}^{n+1}$ with constant 
index of  relative nullity $\nu=n-k$ is the envelope of the $k$-parameter family of 
tangent hyperplanes.
\vspace{1ex} 

A $k$-parameter family of affine hyperplanes $(P_u)_{u\in U}$ can be given 
in terms of a smooth family  of equations of the form
$$
G(u)=\varphi_1x_1+\cdots +\varphi_{n+1}x_{n+1}-\varphi_0=0
$$
where $\varphi_j\in C^\infty(U)$, $0\leq j\leq n+1$, and  
$x=(x_1,\ldots,x_{n+1})$ are coordinates in $\mathbb{R}^{n+1}$ with respect 
to a canonical base.  

Assume that the map 
$\varphi=(\varphi_1,\ldots,\varphi_{n+1})\colon U\to\mathbb{R}^{n+1}$
is an immersion and without loss of generality that 
$0\not\in\varphi(U)$.  Let $g\colon U\to\mathbb{S}^n\subset\mathbb{R}^{n+1}$ 
and $\gamma\in C^\infty(U)$ be given by
$$
g=\frac{1}{\phi}(\varphi_1,\ldots,\varphi_{n+1})\;\;\mbox{and}
\;\;\gamma=\frac{\varphi_0}{\phi}
$$
where $\phi^2=\sum_{j=1}^{n+1}\varphi_j^2$. 

Being $g$ is an immersion, the pair $(g,\gamma)$  gives a 
hypersurface $f\colon M^n\to\mathbb{R}^{n+1}$ by means of the Gauss 
parametrization. Clearly $f$ is the envelope of $(P_u)_{u\in U}$ 
and the leaves $(L_u)_{u\in U}$ of the envelope coincide with the
relative nullity foliation of $f$.  Moreover,  the envelope of
$(P_u)_{u\in U}$ can be locally given as the solution of the 
system of equations
$$
(R)\begin{cases}
G(u)=0\\
G_{u_j}(u)=0,\;\;j=1,\ldots,k.
\end{cases}
$$  

We have shown the following fact.

\begin{proposition}{\hspace*{-1ex}}{\bf .  } Any hypersurface $f\colon M^n\to\mathbb{R}^{n+1}$
of constant rank $k$ can be locally  given as the envelope
of a smooth family of affine hyperplanes
$$
G(u)=\varphi_1x_1+\cdots +\varphi_{n+1}x_{n+1}-\varphi_0=0
$$
where $(\varphi_1,\ldots,\varphi_{n+1})\colon U\to\mathbb{R}^{n+1}$
is an immersion of an open subset $U$ of $\mathbb{R}^k$. 
Then $f$ is locally the solution of the  system of equations $(R)$.
\end{proposition}


\section{A class of surfaces}


A surface $g\colon L^2\to\mathbb{S}^n$ in the unit sphere is called
\emph{hyperbolic} (respectively, \emph{elliptic}) if there exists a
tensor $J$ on $L^2$ satisfying $J^2=I$ and $J\neq I$ 
(respectively, $J^2=-I$) and such that the second fundamental form 
$\alpha_g\colon TL\times TL\to N_gL$ of $g$ satisfies
\begin{equation} \label{eq:aJ}
\alpha_g(JX,Y)=\alpha_g(X,JY)
\end{equation} 
for all vector fields $X,Y\in\mathfrak{X}(L)$. 
Local coordinates $(u,v)$ on $L^2$  are called
\emph{real-conjugate} for $g$ if the condition
$$
\alpha_g(\partial_u,\partial_v)=0
$$
holds where $\partial_u=\partial/\partial u$ and $\partial_v=\partial/\partial v$.
They are called \emph{complex-conjugate} if the condition
$\alpha_g(\partial,\bar\partial)=0$
holds where  $\partial=\partial_z=(1/2)(\partial_u-i\partial_v)$, that is, if we have that
$$
\alpha_g(\partial_u,\partial_u) + \alpha_g(\partial_v,\partial_v)=0.
$$

A simple argument (see \cite{dft}) gives the following result.

\begin{proposition}{\hspace*{-1ex}}{\bf .  }\label{coordinates}
Let $g\colon L^2\to\mathbb{S}^n$ be a hyperbolic (respectively, elliptic) surface.
Then  there exists locally a real-conjugate (respectively, complex-conjugate) 
system of coordinates on $L^2$ for $g$. Conversely, if there exists real-conjugate
(respectively, complex-conjugate) coordinates on $L^2$, then
$g$ is a hyperbolic (respectively, elliptic) surface.
\end{proposition}

Let $g\colon L^2\to\mathbb{S}^n$ be a simply-connected surface that carries a
real-conjugate system of coordinates $(u,v)$. Equivalently, the isometric 
immersion $h=i\circ g\colon L^2\to\mathbb{R}^{n+1}$ satisfies
\begin{equation} \label{eq:guv}
h_{uv}-\gg h_u-\Gamma^2  h_v + Fh=0
\end{equation} 
where $\gg,\Gamma^2 $ are the Christoffel symbols given by
$$
\nabla_{\partial_u}\partial_v=\gg\partial_u+\Gamma^2 \partial_v
$$
and $F=\<\partial_u,\partial_v\>$.

We are interested in surfaces for which, in addition, the following system of 
differential equations admits solution:
\begin{equation} \label{eq:siti}
d\mu+2\mu\o=0\;\;\;\mbox{where}\;\;\; \o=\Gamma^2  du+\gg dv.
\end{equation} 
This is the case if and only if the integrability condition 
\begin{equation} \label{integcon}
\gg_u=\Gamma^2 _v
\end{equation} 
is satisfied.

\begin{proposition}{\hspace*{-1ex}}{\bf .  } Let $g\colon L^2\to\mathbb{S}^n$ be a hyperbolic surface 
with real conjugate coordinates $(u,v)$ such that the induced metric satisfies
condition (\ref{integcon}).  Then there is a positive function 
$\mu\in C^\infty(L)$ such that $\varphi$ is a solution of (\ref{eq:guv})
if and only if $\psi=\sqrt{\mu}\varphi$ is a solution of 
$$
\psi_{uv}+M\psi=0
$$
where  $M\in C^\infty(L)$ is given by
\begin{equation} \label{Mm}
M=F-\frac{\mu_{uv}}{2\mu}+\frac{\mu_u\mu_v}{4\mu^2}\cdot
\end{equation} 
In particular, the immersion 
$k=\sqrt{\mu}\,h\colon L^2\to\mathbb{R}^{n+1}$ satisfies
\begin{equation} \label{form}
k_{uv}+Mk=0.
\end{equation} 
Conversely, let $k\colon L^2\to\mathbb{R}^{n+1}$ be an isometric immersion 
that for a system of coordinates  $(u,v)$ satisfies (\ref{form})
where  $M\in C^\infty(L)$. Then $(u,v)$ are real conjugate coordinates
for the immersion $g=(1/\|k\|)\,k\colon L^2\to\mathbb{S}^n$ and
condition (\ref{integcon}) is satisfied for the induced metric.
\end{proposition}

\noindent\emph{Proof: } If $g$ satisfies the integrability condition, then
$$
\gg=-\frac{\mu_v}{2\mu},\;\;\;\Gamma^2 =-\frac{\mu_u}{2\mu}
$$
where $\mu=c\,e^{-2\int\o}$ for any $c\in\mathbb{R}_+$. Hence (\ref{eq:guv}) 
becomes
\begin{equation} \label{eq:guv2}
h_{uv}+\frac{\mu_v}{2\mu}h_u+\frac{\mu_u}{2\mu}h_v+Fh=0.
\end{equation} 
It follows easily that $k=\sqrt{\mu}\,h$ takes the form (\ref{form}) 
where $M$ is given by (\ref{Mm}).
The converse is a straightforward computation.\vspace{1,5ex}\qed

Let $g\colon L^2\to\mathbb{S}^n$ be a simply-connected surface endowed
with  complex-conjugate coordinates $(z,\bar{z})$.  
Equivalently, the isometric immersion $h=i\circ g\colon L^2\to\mathbb{R}^{n+1}$
satisfies
\begin{equation} \label{eq:guz}
h_{z\bar z}-\Gamma h_z-\bar\Gamma h_{\bar z}+Fh=0
\end{equation} 
where the Christoffel symbols, obtained using the $\mathbb{C}$-linear extensions 
of the metric of $L^2$ and the corresponding connection, are given by
$$
\nabla_{\partial}\bar\partial=\Gamma\partial+\bar \Gamma\bar\partial
$$ 
and $F=\<\partial,\bar\partial\>$.

We are interested in surfaces for which, in addition, the following 
system of differential equations for $\mu$ real admits solutions:
\begin{equation}  \label{eq:comp}
\mu_{\bar{z}}+2\mu\Gamma=0.
\end{equation} 
This is the case if and only if the integrability condition 
\begin{equation} \label{integcon2}
\Gamma_z=\bar{\Gamma}_{\bar z}, 
\end{equation} 
that is, $\Gamma_z\;\mbox{is real}$, is satisfied.

\begin{proposition}{\hspace*{-1ex}}{\bf .  } Let $g\colon L^2\to\mathbb{S}^n$ be an elliptic surface
with complex  conjugate coordinates $(z,\bar{z})$  such that the induced 
metric satisfies condition (\ref{integcon2}). 
Then, there is a positive solution  $\mu\in C^\infty(L)$ 
of (\ref{eq:comp}) such that $\varphi$ is a solution of (\ref{eq:guz})
if and only if $\psi=\sqrt{\mu}\varphi$ is a solution of 
$$
\psi_{z\bar z}+M\psi=0
$$
where  $M\in C^\infty(L)$ is given by
\begin{equation} \label{Mm2}
M=F-\frac{\mu_{z\bar z}}{2\mu}+\frac{\mu_z\mu_{\bar z}}{4\mu^2}\cdot
\end{equation} 
In particular, the immersion 
$k=\sqrt{\mu}\,h\colon L^2\to\mathbb{R}^{n+1}$ satisfies
\begin{equation} \label{form2}
k_{z\bar z}+Mk=0.
\end{equation} 

Conversely, let $k\colon L^2\to\mathbb{R}^{n+1}$ be an isometric immersion that 
 for a system of coordinates  $(z,\bar{z})$ satisfies (\ref{form2})
where  $M\in C^\infty(L)$. Then $(z,\bar{z})$ are complex conjugate
coordinates for the immersion $g=(1/\|k\|)k\colon L^2\to\mathbb{S}^n$  and 
condition  (\ref{integcon2}) is satisfied for the induced metric.
\end{proposition}

\noindent\emph{Proof: } We have $\mu=c\,e^{-2\int\o}$ for any $c\in\mathbb{R}_+$ where
$\o=\Gamma d\bar{z}$. Then (\ref{eq:guz}) takes the form
$$
h_{z\bar z}+\frac{\mu_z}{2\mu}h_z+\frac{\mu_{\bar z}}{2\mu}h_{\bar z}+Fh=0.
$$
It follows easily that $k=\sqrt{\mu}\,h$ is as in  (\ref{form2}) where 
$M$ is given by (\ref{Mm2}).
The converse is a straightforward computation.\qed



\section{The main result}



After introducing the necessary terminology and definitions, we  present 
the main result of the paper in terms of the Gauss parametrization, as 
is the case in the paper by Sbrana.  The proof of the alternative version 
of the theorem in terms  of envelopes of hyperplanes given in the introduction 
can easily be obtained from this version using results from the preceding 
sections. 
\vspace{1,5ex}


By a \emph{variation} $F$ of an isometric immersion $f\colon M^n\to\mathbb{R}^{n+1}$ 
we mean a smooth map $F\colon (-\epsilon, \epsilon)\times M^n\to\mathbb{R}^{n+1}$ 
such that $f_t=F(t,\cdot)$ is an immersion for each $t\in I=(-\epsilon,
\epsilon)$ and  $f=f_0$.
The variational vector field of $F$ is the section ${\cal T}\in\Gamma(f^*(T\mathbb{R}^{n+1}))$
of the Riemannian vector bundle $f^*(T\mathbb{R}^{n+1})$ defined as
$$
{\cal T}(x)=F_*\partial/\partial t|_{t=0}(x).
$$

A variation $F$ of a given isometric immersion
$f\colon M^n\to\mathbb{R}^{n+1}$ is called an \emph{isometric bending} if 
$f_t$ is an isometric immersion for any $t\in I$.
The variational vector field of an isometric bending satisfies
$$
\<\tilde\nabla_X{\cal T},f_*Y\>+\<f_*X,\tilde\nabla_Y{\cal T}\>=0
$$
for any $X,Y\in\mathfrak{X}(M)$. Equivalently, it satisfies that
$$
\<\tilde\nabla_X{\cal T},f_*X\>=0
$$
for any $X\in\mathfrak{X}(M)$.
\vspace{1ex}

An isometric bending $F$ is called \emph{trivial} if it is produced by a smooth
one-parameter family of isometries of $\mathbb{R}^{n+1}$, that is, if there
exist a smooth family ${\cal C}\colon I\to O(n+1)$ of orthogonal transformations
of $\mathbb{R}^{n+1}$ and a smooth map $v\colon I\to \mathbb{R}^{n+1}$ such that
$$
F(t,x)={\cal C}(t)f(x)+v(t).
$$
For a trivial isometric bending the variational vector field is of the form
$$
{\cal T}(x)={\cal D}f(x)+w
$$
where ${\cal D}={\cal C}'(0)$ is a skew-symmetric linear endomorphism of $\mathbb{R}^{n+1}$
and $w=v'(0)$ a vector in  $\mathbb{R}^{n+1}$.
Conversely, given a skew-symmetric linear endomorphism ${\cal D}$ of $\mathbb{R}^{n+1}$
and a vector $w\in\mathbb{R}^{n+1}$, the map
$$
F(t,x)=e^{t{\cal D}}f(x)+tw
$$
defines a  trivial isometric bending that has ${\cal T}={\cal D}f+w$ as variational
vector field.
\vspace{1ex}

By an \emph{infinitesimal bending} ${\cal T}$ of an isometric immersion
$f\colon M^n\to\mathbb{R}^{n+1}$ we mean an element of $\Gamma(f^*(T\mathbb{R}^{n+1}))$
that satisfies
\begin{equation} \label{iif0}
\<\tilde\nabla_X{\cal T},f_*Y\>+\<f_*X,\tilde\nabla_Y{\cal T}\>=0
\end{equation} 
for any $X,Y\in\mathfrak{X}(M)$.
An infinitesimal bending is said to be \emph{trivial} if
$$
{\cal T}(x)={\cal D}f(x)+w
$$
where ${\cal D}$ is a skew-symmetric linear
endomorphism of $\mathbb{R}^{n+1}$ and $w\in\mathbb{R}^{n+1}$.
\vspace{1ex}

An isometric immersion $f\colon M^n\to\mathbb{R}^{n+1}$ is called \emph{infinitesimally
bendable} if it admits a nontrivial infinitesimal bending.  Otherwise, 
it is said that $f$ is \emph{infinitesimally rigid}.
\vspace{1ex}

Multiplying a given infinitesimal bending by a real constant and adding a trivial
infinitesimal bending yields a new infinitesimal bending. In the sequel,
we identify two infinitesimal bendings ${\cal T}_1$ and ${\cal T}_2$ if
${\cal T}_2={\cal T}_0 + c\,{\cal T}_1$ where ${\cal T}_0$ is a trivial infinitesimal bending and 
$0\neq c\in\mathbb{R}$.
\vspace{1ex}


We have already observed  that hypersurfaces of rank at least three at any point 
are infinitesimally rigid.  Therefore, the interesting case to be considered is 
the one of constant rank two.  We see next that even in this special case  
hypersurfaces are ``generically" infinitesimally rigid.
\vspace{1ex}

We call the pair $(g,\gamma)$ a \emph{special hyperbolic pair}
(respectively, \emph{special elliptic pair}) if  $g\colon L^2\to\mathbb{S}^n$
is a hyperbolic (respectively, elliptic) surface so that system (\ref{eq:siti})
(respectively, system (\ref{eq:comp})) has solution and $\gamma\in C^\infty(L)$ 
satisfies (\ref{eq:guv}) (respectively,  (\ref{eq:guz})).

\begin{theorem}{\hspace*{-1ex}}{\bf .  }\label{main} Let $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, be an
infinitesimally bendable hypersurface of constant rank two that is neither
surface-like nor ruled on any open subset of $M^n$. Then, there is
an open and dense subset of $M^n$ such that along any connected
component $f$ is parametrized in terms of the Gauss parametrization
by a special hyperbolic or a special elliptic pair.

Conversely, any hypersurface parametrized in terms of the Gauss
parametrization by a special hyperbolic or special elliptic pair
admits locally a  unique infinitesimal bending.
\end{theorem}

The case of ruled hypersurfaces that has been excluded from consideration 
in the above result is rather simple and will be treated separately in 
Section $5$.



\section{Existence and uniqueness}



 We study the system of differential equations of  an infinitesimal 
bending of a Euclidean hypersurface and discuss its integrability
conditions. This yields a kind of fundamental theorem for 
infinitesimal bendings that is,  basically, contained in Sbrana's 
paper  \cite{sb1}.  In fact, the case of arbitrary codimension 
was later taken on by Schouten \cite{sc} but presented in a rather difficult 
terminology.  We point out that in this section some  long but straightforward 
computations are only indicated.
\vspace{1,5ex}

Given a hypersurface $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, in the sequel 
we associate to any infinitesimal bending ${\cal T}$ of $f$ the variation
$F\colon \mathbb{R}\times M^n\to\mathbb{R}^{n+1}$ with variational vector field 
${\cal T}$ given by
$$
F(t,x)=f(x)+t{\cal T}(x).
$$
It is usually said that  $f_t=F(t,\cdot)$ is isometric to $f$ up to first
order for if
\begin{equation} \label{var}
\|f_{t*}X\|^2=\|f_*X\|^2+t^2\|\tilde\nabla_X{\cal T}\|^2
\end{equation} 
for all $X\in\mathfrak{X}(M)$.

Let $g_t$ be the metric on $M^n$ induce by $f_t$. Then,
$$
\partial/\partial t\vert_{t=0}\,g_t(X,Y)=0
$$
for all $X,Y\in\mathfrak{X}(M)$.  Consequently, we have that the  associated
one-parameter family of Levi-Civita connections and the corresponding
family of curvature tensors satisfy 
$$
\partial/\partial t\vert_{t=0}\,f_*\nabla^t_XY=0
$$
and
\begin{equation} \label{curvature}
\partial/\partial t\vert_{t=0}\,\<R^t(X,Y)Z,W\>=0
\end{equation} 
for all $X,Y,Z,W\in\mathfrak{X}(M)$.
\vspace{1ex}

 Let $N(t)$ denote a Gauss map of $f_t$ and $A(t)$ the second fundamental 
form of $f_t$ with respect to $N(t)$ so that the map $t\in\mathbb{R}\mapsto N(t)$ 
is smooth.  Then  $N=N(0)$ is the Gauss map and $A=A(0)$ is the second 
fundamental form of $f$. Moreover, let us define
$L\in\Gamma(\mbox{End}(TM,f^*(T\mathbb{R}^{n+1}))$ by
$$
LX=\tilde\nabla_X{\cal T}={\cal T}_*X.
$$
Then (\ref{iif0}) can be written as
\begin{equation} \label{iif}
\<LX,f_*Y\>+\<f_*X,LY\>=0
\end{equation} 
for all $X,Y\in\mathfrak{X}(M)$.


\begin{lemma}{\hspace*{-1ex}}{\bf .  } We have that 
${\cal Y}=\partial/\partial t\vert_{t=0}\, N(t)\in\Gamma(f^*(T\mathbb{R}^{n+1}))$ satisfies
\begin{equation} \label{4}
\<{\cal Y},N\>=0
\end{equation} 
and
\begin{equation} \label{3}
\<{\cal Y},f_*X\>+\<LX,N\>=0
\end{equation} 
for all $X\in\mathfrak{X}(M)$.
\end{lemma}

\noindent\emph{Proof: } The derivative with respect to $t$ at $t=0$ of $\<N(t),N(t)\>=1$ 
gives (\ref{4}) whereas  of $\<N(t),f_{t*}X\>=0$ yields (\ref{3}).\qed


\begin{lemma}{\hspace*{-1ex}}{\bf .  }  
We have that $B=\partial/\partial t\vert_{t=0}\, A(t)\in\Gamma(\mbox{End}(TM))$ 
is symmetric and satisfies 
\begin{equation} \label{1}
(\tilde\nabla_XL)Y=\<BX,Y\>N+\<AX,Y\>{\cal Y}
\end{equation} 
and
\begin{equation} \label{2}
{\cal Y}_*X=-f_*BX-LAX
\end{equation} 
for all $X,Y\in\mathfrak{X}(M)$.
\end{lemma}

\noindent\emph{Proof: } The derivative with respect to $t$ at $t=0$ of the Gauss formula
$$
\tilde\nabla_Xf_{t*}Y=f_{t*}\nabla^t_XY + g_t(A(t)X,Y)N(t)
$$
 easily gives (\ref{1}).
As for the Weingarten formula
$$
\tilde\nabla_XN(t)=-f_{t*}(A(t)X)
$$
we have that its derivative at $t=0$ yields (\ref{2}).
\vspace{1,5ex}\qed

If ${\cal T}={\cal D}f+w$ is a trivial infinitesimal bending then 
$L={\cal D}\circ f_*$.  It follows that 
${\cal Y}={\cal D}N$  and that $B=0$ since 
$$
\<BX,Y\>=\<(\tilde\nabla_XL)Y,N\>=\<(\tilde\nabla_X{\cal D})Y,N\>=0
$$
for all $X,Y\in\mathfrak{X}(M)$.

\begin{proposition}{\hspace*{-1ex}}{\bf .  }  The tensor $B$ is a symmetric Codazzi 
tensor, i.e.,
\begin{equation} \label{6}
\left(\nabla_XB\right)Y-\left(\nabla_YB\right)X=0
\end{equation} 
such that
\begin{equation} \label{5}
BX\wedge AY-BY\wedge AX=0
\end{equation} 
for all $X,Y\in\mathfrak{X}(M)$.
\end{proposition}

\noindent\emph{Proof: }  The derivative at $t=0$ of the Codazzi equation
$$
\left(\nabla^t_XA(t)\right)Y=\left(\nabla^t_YA(t)\right)X
$$
gives (\ref{6}).
To obtain (\ref{5}) we compute the derivative at $t=0$ of the
Gauss equation
$$
R^t(X,Y)Z=g(t)(A(t)Y,Z)A(t)X-g(t)(A(t)X,Z)A(t)Y
$$
and use (\ref{curvature}).\vspace{1,5ex}\qed

The next result is to be expected bearing in mind the nature of the Gauss
and Codazzi equations as the integrability conditions for the system of
differential equations associated to an isometric immersion as a hypersurface.

\begin{lemma}{\hspace*{-1ex}}{\bf .  }\label{sys} Equations (\ref{6}) and (\ref{5}) are the
integrability conditions of the system of  differential equations
(\ref{1}) and (\ref{2}) for $L$ and ${\cal Y}$, that is,
$$
(S)\begin{cases}
{\cal Y}_*X=-LAX-f_*BX\\
(\tilde\nabla_XL)Y=\<BX,Y\>N+\<AX,Y\>{\cal Y}.
\end{cases}
$$
\end{lemma}

\noindent\emph{Proof: } For the first equation, we have to show that
\begin{equation} \label{condi1}
\tilde\nabla_X{\cal Y}_*Y-\tilde\nabla_Y{\cal Y}_*X-\tilde\nabla_{[X,Y]}{\cal Y}=0
\end{equation} 
for all $X,Y\in\mathfrak{X}(M)$. One has that
$$
\tilde\nabla_X{\cal Y}_*Y=-(\tilde\nabla_XL)AY-L(\nabla_XA)Y-LA\nabla_XY-f_*\nabla_XBY-\<AX,BY\>N.
$$
Then (\ref{condi1}) is equivalent to
$$
(\tilde\nabla_XL)AY-(\tilde\nabla_YL)AX+f_*((\nabla_XB)Y-(\nabla_YB)X)
+(\<AX,BY\>-\<AY,BX\>)N=0.
$$
Replacing the first two terms by the use of the second equation in $(S)$ 
it is easily seen that (\ref{condi1}) follows from (\ref{6}).

It is easy to see that the integrability condition for the second
equation is
\begin{equation} \label{condi2}
(\tilde\nabla_X\tilde\nabla_YL-\tilde\nabla_Y\tilde\nabla_XL-\tilde\nabla_{[X,Y]}L)Z=-LR(X,Y)Z
\end{equation} 
for all $X,Y,Z,W\in\mathfrak{X}(M)$.
A straightforward computation using (\ref{1}) gives
\begin{eqnarray*} 
(\tilde\nabla_X\tilde\nabla_YL)Z\!\!\!&=&\!\!\!\<(\nabla_XB)Y,Z\>N+\<B\nabla_XY,Z\>N
-\<BY,Z\>f_*AX+\<(\nabla_XA)Y,Z\>{\cal Y}\\
\!\!\!&&\!\!\!+\; \<A\nabla_XY,Z\>{\cal Y}
-\<AY,Z\>LAX-\<AY,Z\>f_*BX.
\end{eqnarray*} 
That $A$ is a Codazzi tensor together with (\ref{6}) yields
\begin{eqnarray*} 
(\tilde\nabla_X\tilde\nabla_YL-\tilde\nabla_Y\tilde\nabla_XL-\tilde\nabla_{[X,Y]}L)Z
\!\!\!&=&\!\!\!-\<BY,Z\>f_*AX-\<AY,Z\>(LAX+f_*BX)\\
\!\!\!&&\!\!\!+\,\<BX,Z\>f_*AY-\<AX,Z\>(LAY+f_*BX).
\end{eqnarray*} 
On the other hand,  we have
$$
LR(X,Y)Z=\<AY,Z\>LAX-\<AX,Z\>LAY,
$$
and (\ref{condi2}) follows using (\ref{5}).\vspace{1,5ex}\qed

Next we consider the case of hypersurfaces of constant rank two.

\begin{corollary}\label{kerB}{\hspace*{-1ex}}{\bf .  } If $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, 
is an infinitesimally bendable hypersurface of constant rank two, 
then $\Delta\subset\ker B$.
\end{corollary}

\noindent\emph{Proof: }  This follows easily from (\ref{5}).

\begin{theorem}{\hspace*{-1ex}}{\bf .  }\label{mains}  Let $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, be a 
simply-connected  hypersurface of constant rank two.  
Then, the set of all symmetric Codazzi tensors
$B\in\Gamma(\mbox{End}(TM))$  such that $\Delta\subset\ker B$ and
$$
BX\wedge AY-BY\wedge AX=0
$$
for all $X,Y\in\mathfrak{X}(M)$ is in one-to-one correspondence with 
the set of all infinitesimal bendings of $f$ so that $B=0$ corresponds 
to the trivial one.
\end{theorem}

\noindent\emph{Proof: } Given  $B\in\Gamma(\mbox{End}(TM))$ as in the statement,  
we first prove that there exists a solution ${\cal Y}$  and $L$ of system 
$(S)$ such that (\ref{iif}), (\ref{4}) and (\ref{3}) are satisfied. 
In particular, this gives the existence of an infinitesimal bending 
${\cal T}$ such that $L={\cal T}_*$. To see this, observe that by (\ref{1}) the 
one-form $\o=\<L,v\>$ is closed for any $v\in\mathbb{R}^{n+1}$.

Given a solution  ${\cal Y}$ and $L$ of $(S)$, we define a smooth function by
$$
\tau=\<{\cal Y},N\>,
$$
a smooth one-form by
$$
\theta(X)=\<{\cal Y},f_*X\>+\<LX,N\>
$$
and a smooth symmetric bilinear tensor by
$$
\beta(X,Y)=\<LX,f_*Y\>+\<LY,f_*X\>.
$$
A straightforward calculation gives that
\begin{equation} \label{tau}
d\tau=-\theta\circ A,
\end{equation} 
\begin{equation} \label{theta}
(\nabla_X\theta)Y=-\beta(AX,Y)+2\tau\<AX,Y\>
\end{equation} 
and
\begin{equation} \label{M}
(\nabla_Z\beta)(X,Y)=\<AX,Y\>\theta(Z)+\<AX,Z\>\theta(Y)
\end{equation} 
for all $X,Y,Z\in\mathfrak{X}(M)$.


We claim that the system of differential equations formed by (\ref{tau}), 
(\ref{theta}) and (\ref{M}) is completely integrable.  The integrability 
condition for the first equation is easy to verify. For the second equation, 
we have to see that
\begin{equation} \label{cal1}
(\nabla_X\nabla_Y\theta-\nabla_Y\nabla_X\theta-\nabla_{[X,Y]}\theta)Z
=-\theta(R(X,Y)Z)
\end{equation} 
holds. Using (\ref{tau}) and (\ref{theta}) we obtain 
$$
(\nabla_X\nabla_Y\theta)Z=-(\nabla_X\beta)(AY,Z)
-\beta(\nabla_XAY,Z)-2\theta(AX)\<AY,Z\>
+2\tau\<\nabla_XAY,Z\>.
$$
Hence
\begin{align*}
 (\nabla_X\nabla_Y\theta-\nabla_Y\nabla_X\theta-\nabla_{[X,Y]}\theta)Z
=&-(\nabla_X\beta)(AY,Z)+(\nabla_Y\beta)(AX,Z)\\
&\,-2\theta(AX)\<AY,Z\>+2\theta(AY)\<AX,Z\>.
\end{align*}
Using (\ref{M}) we obtain that
$$
(\nabla_X\nabla_Y\theta-\nabla_Y\nabla_X\theta-\nabla_{[X,Y]}\theta)Z
=-\theta(AX)\<AY,Z\>+\theta(AY)\<AX,Z\>.
$$
On the other hand, we have from the Gauss equation that
$$
\theta(R(X,Y)Z)=\<AY,Z\>\theta(AX)-\<AX,Z\>\theta(AY),
$$
and (\ref{cal1}) follows.

Finally, the integrability condition for the last equation, namely, that  
\begin{equation} \label{cal2}
\!\!(\nabla_X\nabla_Y\beta-\nabla_Y\nabla_X\beta-\nabla_{[X,Y]}\beta)(Z,W)
=-\beta(R(X,Y)Z,W)-\beta(R(X,Y)W,Z)
\end{equation} 
can be verified by a similar computation, and this
proves the claim.

Start with a solution $L^*$ and ${\cal Y}^*$ of system $(S)$ with 
corresponding tensors $\theta^*$,  $\beta^*$ and function $\tau^*$.
Fix a point $p_0\in M^n$ and let $L_0$ and ${\cal Y}_0$ be a solution of the 
integrable system
$$
(S_0)\begin{cases}
{\cal Y}_*X=-LAX\\
(\tilde\nabla_XL)Y=\<AX,Y\>{\cal Y}
\end{cases}
$$
with  initial conditions $\theta_0(p_0)=\theta^*(p_0)$, 
$\beta_0(p_0)=\beta^*(p_0)$ and $\tau_0(p_0)=\tau^*(p_0)$.
Then $L=L^*-L_0$ and
${\cal Y}={\cal Y}^*-{\cal Y}_0$ are a solution of $(S)$ such that 
$\theta=\theta^*-\theta_0$,  $\beta=\beta^*-\beta_0$ and 
$\tau=\tau^*-\tau_0$. Clearly $\theta(p_0)=\beta(p_0)=\tau(p_0)=0$.
Since $\theta, \beta$ and $\tau$ solve the homogeneous integrable system
(\ref{tau}), (\ref{theta}) and (\ref{M}), hence $\theta=\beta=\tau=0$. 


Given any two pairs  $L_j,{\cal Y}_j$, obtained as above, let ${\cal T}_j$, $1\leq j\leq 2$, 
be the associated infinitesimal bendings. 
It remains to show that ${\cal T}={\cal T}_1-{\cal T}_2$ is a trivial infinitesimal bending.

We have that the pair $L=L_1-L_2$, ${\cal Y}={\cal Y}_1-{\cal Y}_2$ satisfies $(S_0)$
as well as  (\ref{iif}), (\ref{4}) and (\ref{3}). Fix $p_0\in M^n$ and
define  a skew-symmetric linear endomorphism ${\cal C}$ of $\mathbb{R}^{n+1}$ by
$$
{\cal C}f_*(p_0)X=L(p_0)X\;\;\mbox{and}\;\;{\cal C}N(p_0)={\cal Y}(p_0)
$$
and a vector $v\in\mathbb{R}^{n+1}$ by $v={\cal T}(p_0)-{\cal C}f(p_0)$.
Consider the  trivial infinitesimal bending $\tilde{{\cal T}}={\cal C}f+v$ and
$\tilde{{\cal Y}}={\cal C}N$.  Then, the pair $\tilde{L}$ and $\tilde{{\cal Y}}$ satisfies
$(S_0)$. Thus, also the pair $L^*=L-\tilde{L}$, ${\cal Y}^*={\cal Y}-\tilde{{\cal Y}}$
solves system $(S_0)$. Moreover, ${\cal T}^*(p_0)=0$, ${\cal Y}^*(p_0)=0$ and
$L^*(p_0)=L(p_0)-\tilde{L}(p_0)=0$.
Thus ${\cal T}^*=0$ and hence ${\cal T}=\tilde{{\cal T}}$.\qed





\section{The proof of Theorem \ref{main}}


In the sequel, let $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, be a 
hypersurface of constant  rank two. Recall that the \emph{splitting tensor} 
$C\colon\Gamma(\Delta)\to\Gamma(\mbox{End}(\Delta^\perp))$
is defined by 
$$
C_TX=-(\nabla_XT)_{\Delta^\perp}
$$
for any $T\in\Gamma(\Delta)$ and $X\in\mathfrak{X}(M)$. 
 From the Codazzi equation, it follows that
\begin{equation} \label{codazzi}
\nabla_TA=AC_T=C_T^tA
\end{equation} 
for any $T\in\Gamma(\Delta)$.

\begin{proposition}{\hspace*{-1ex}}{\bf .  }  Assume that the splitting tensor at any point
satisfies $C_T\in\mbox{span}\{I\}$ for any $T\in\Delta$, where $I$ denotes the 
identity section of $\mbox{End}(\Delta^\perp)$. Then $f$ is surface-like.
\end{proposition}

\noindent\emph{Proof: } See Lemma $6$ in \cite{dft}.\vspace{1,5ex}\qed

Assume further that $f$ is infinitesimally bendable. 
Locally and because of the rank assumption,  there is an orthonormal
tangent frame spanning $\Delta^\perp$ such that
\begin{equation} \label{matrix}
A\vert_{\Delta^\perp}=\begin{bmatrix}
\lambda_1&0\\
0&\lambda_2
\end{bmatrix}.
\end{equation} 

\begin{lemma}{\hspace*{-1ex}}{\bf .  } If $B\neq 0$ at any point of $M^n$, then 
\begin{equation} \label{matrix2}
B\vert_{\Delta^\perp}=\begin{bmatrix}
\lambda\lambda_1&b\\
b&-\lambda\lambda_2
\end{bmatrix}.
\end{equation} 
\end{lemma}

\noindent\emph{Proof: } By Corollary \ref{kerB} we have that $\Delta\subset\ker B$.
Now (\ref{matrix2}) follows easily from (\ref{5}).\qed

\begin{lemma}{\hspace*{-1ex}}{\bf .  }\label{conditions} We have that 
$D=(A\vert_{\Delta^\perp})^{-1}B\vert_{\Delta^\perp}\in\Gamma(\mbox{End}(\Delta^\perp))$ 
satisfies:
\begin{itemize}
\item[(i)] $[D,C_T]=0$ for all $T\in\Delta$,
\item[(ii)] $\nabla_TD=0$ for all $T\in\Delta$,
\item[(iii)] $\mbox{tr\,} D=0$,
\item[(iv)] $T(\det D)=0$ for all $T\in\Delta$.
\end{itemize}
\end{lemma}

\noindent\emph{Proof: } We denote $A=A\vert_{\Delta^\perp}$ and $B=B\vert_{\Delta^\perp}$.   
From (\ref{codazzi}) we obtain $\nabla_TB=BC_T.$
Hence
$$
BC_T=C_T^tB.
$$
We have using  (\ref{codazzi}) that
$$
ADC_T=BC_T=C_T^tB=C_T^tAD=AC_TD,
$$
and $(i)$ follows. We have
\begin{eqnarray*} 
A\nabla_TD\!\!\!&=&\!\!\!\nabla_T(AD)-(\nabla_TA)D=\nabla_TB-(\nabla_TA)D
=BC_T-AC_TD\\
\!\!\!&=&\!\!\!BC_T-C_T^tAD=BC_T-C_T^tB=0,
\end{eqnarray*} 
and this yields $(ii)$. We obtain from (\ref{matrix}) and (\ref{matrix2}) that
$$
D=\begin{bmatrix}
\lambda&b/\lambda_1\\
b/\lambda_2&-\lambda
\end{bmatrix},
$$
which gives $(iii)$. Now part $(iv)$ follows
from  $(ii)$ and $(iii)$.\qed


\begin{proposition}{\hspace*{-1ex}}{\bf .  }\label{ruled} Assume that $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$,
is not surface-like on any open subset of $M^n$. Then $f$ is ruled
along any open subset where $D\neq 0$ satisfies $\det D=0$.
\end{proposition}

\noindent\emph{Proof: } By Lemma \ref{conditions} there is an orthogonal frame $X,Y$
of $\Delta^\perp$ with $Y$ of unit length such that $DY=0$ and $DX=Y$.
We claim that $f$ is ruled by the integral leaves of
the distribution
$\Delta\oplus\mbox{span}\{Y\}$. To see this, we have to show that
$$
(i)\; \<AY,Y\>=0,\;(ii)\;\nabla_TY=0,\;
(iii)\;\<\nabla_YT,X\>=0\;\;\mbox{and}\;\;(iv)\;\<\nabla_YY,X\>=0
$$
for all $T\in\Delta$. We have
$$
\<AY,Y\>=\<ADX,Y\>=\<BX,Y\>=\<BY,X\>=\<ADY,X\>=0.
$$
Condition $(ii)$ follows easily using $\nabla_TD=0$.
Since  $[D,C_T]=0$, we obtain
$$
\<\nabla_YT,X\>=-\<C_TY,X\>=-\<C_TDX,X\>=-\<DC_TX,X\>=0.
$$
We have that
$$
BY=ADY=0\;\;\mbox{and}\;\;BX=ADX=AY=\lambda X,\;\;\lambda\neq 0,
$$
and condition $(iv)$ follows easily using (\ref{6}).
\vspace{1,5ex}\qed

In the sequel, we consider the case $\det D\neq 0$. By the above, this 
is always the case under the assumptions of Theorem \ref{main}.  
\vspace{1ex}

By part $(iii)$ of Lemma \ref{conditions} the eigenvalues of $D$ are 
the solutions of  $t^2+\det D=0$. Therefore, on each connected component 
of an open subset of $ M^n$ either $\det D<0$ and thus $D$ has two 
smooth real eigenvalues $\{\mu,-\mu\}$ or $\det D>0$ and thus $D$ has a 
pair of smooth complex eigenvalues $\{i\mu,-i\mu\}$. Then 
$J\in\Gamma(\mbox{End}(\Delta^\perp))$ defined by 
\begin{equation} \label{jota}
D=\mu J
\end{equation} 
satisfies $J^2=I$ in the first case and $J^2=-I$ in the second case.

\begin{lemma}{\hspace*{-1ex}}{\bf .  }\label{para}  The eigenspaces of $D$ are parallel and the
eigenvalues constant along the leaves of $\Delta$.
\end{lemma}

\noindent\emph{Proof: } Follows from parts $(ii)$ and $(iv)$ of Lemma \ref{conditions}.
\vspace{1,5ex}\qed

A hypersurface $f\colon M^n\to\mathbb{R}^{n+1}$ of rank two is said to be
\emph{hyperbolic} (respectively, \emph{elliptic}) if there exists
$J\in\Gamma(\mbox{End}(\Delta^\perp))$ satisfying the following conditions:
\begin{itemize}
\item[(i)]  $J^2=I$ and $J\neq I$ (respectively, $J^2=-I$).
\item[(ii)] $\nabla_TJ=0$ for all $T\in\Gamma(\Delta)$.
\item[(iii)] $C_T\in\mbox{span}\{I,J\}$ for all $T\in\Gamma(\Delta)$.
\end{itemize}

\begin{proposition}\label{hypel}
{\hspace*{-1ex}}{\bf .  } Assume that $f\colon M^n\to\mathbb{R}^{n+1}$ is neither surface-like
nor ruled on any open subset $\tilde{M}^n$ of $M^n$. Then, there is an 
open and dense subset $\tilde M^n$ of $M^n$ such that the restriction of 
$f$ to any connected component of $\tilde{M}^n$ is either hyperbolic or 
elliptic.
\end{proposition}

\noindent\emph{Proof: } Let  $J\in\Gamma(\mbox{End}(\Delta^\perp))$ be defined by (\ref{jota}).
The subspace $S$ of all elements in $\mbox{End}(\Delta^\perp)$ that
commute  with $D$, i.e., that commute with $J$, is $S=\mbox{span}\{I,J\}$.
Thus condition $(iii)$ in the above definition follows from part 
$(i)$ of Lemma \ref{conditions}.\vspace{1,5ex}\qed


Given a submersion $\pi\colon M\to L$ between differentiable manifolds,
then  $X\in \mathfrak{X}(M)$ is said  to be \emph{projectable} if it is 
$\pi$-related to some  $\bar X\in \mathfrak{X}(L)$, that is, if there 
exists $\bar X\in \mathfrak{X}(L)$ such that $\pi_*X=\bar X\circ \pi$.
\vspace{1,5ex}

In the sequel, we denote by $\pi\colon M^n\to L^2$ the submersion onto the
(local) quotient space of leaves of $\Delta$, namely, onto $L^2=M^n/\Delta$.
A tensor $D\in\mbox{End}(\Delta^\perp)$  is said to be  \emph{projectable} with
respect to $\pi$ if it is the horizontal lift of some tensor $\bar D$ on $L$.
Clearly, $D$ is projectable with respect to  $\pi$ if and only if for all
$\bar{x}\in L$,  $x,y\in \pi^{-1}(\bar{x})$,
$v\in\Delta^\perp(x)$ and $w\in \Delta^\perp(y)$ with  $\pi_*v=\pi_*w$,
we have  that $\pi_*Dv=\pi_*Dw$.

\begin{lemma}\label{prop*}{\hspace*{-1ex}}{\bf .  }
Let $f\colon M^n\to\mathbb{R}^{n+1}$ be a  hypersurface of rank two parametrized  
by a pair $(g,\gamma)$ in terms of the  Gauss parametrization.
If $f$ is  hyperbolic (respectively, elliptic) with respect to  
$J\in\Gamma(\mbox{End}(\Delta^\perp))$ and $D=\mu J$ satisfies  $(i)$--$(iv)$ 
in Lemma~\ref{conditions}, then $J$ and $D$ are the horizontal lifts of 
tensors $\bar{J}$ and $\bar D=\bar\mu\bar{J}$ on $L^2$ such that 
$\mu=\bar{\mu}\circ\pi$, $\bar{J}^2=\bar I$ 
(respectively, $\bar{J}^2=-\bar I$), the pair  $(g,\gamma)$ is hyperbolic  
(respectively, elliptic)  with respect to  $\bar{J}$ and $\bar D$ satisfies:
\begin{itemize}
\item[(a)] $\mbox{tr\,}\bar{D}=0$, 
\item[(b)] $\left(\nabla'_{\bar{X}}\bar{D}\right)\bar{Y}
-\left(\nabla'_{\bar{Y}}\bar{D}\right){\bar{X}}=0$
for all $\bar{X},\bar{Y}\in \mathfrak{X}(L)$ 
\end{itemize}
where
$\nabla'$ is the  Levi-Civita connection  of the metric induced by $g$.

Conversely, if the pair $(g,\gamma)$ is hyperbolic  (respectively, 
elliptic) with respect to a tensor $\bar{J}$ on $L^2$ satisfying 
${\bar J}^2=\bar I$ (respectively, ${\bar J}^2=-\bar I$), then the  
hypersurface  $f$ is hyperbolic (respectively, elliptic) with  respect 
to the horizontal lift $J$ of $\bar J$. In addition, the horizontal lift 
$D=\mu J$ of a tensor $\bar{D}=\bar{\mu}\bar{J}$, $\mu=\bar{\mu}\circ\pi$,  
satisfying $(a)$ and $(b)$ also fulfills the properties $(i)$--$(iv)$ in 
Lemma \ref{conditions}.
\end{lemma}

\noindent\emph{Proof: } We have from parts $(i)$ and $(ii)$ of Lemma \ref{conditions} 
and  Corollary $13$ in \cite{dft2} that the tensor $D$ is projectable. 
Then part $(iii)$ of Lemma \ref{conditions} gives 
$\mbox{tr\,}\bar{D}=\mbox{tr\,} D=0$.

From part $(iv)$ of Lemma \ref{conditions} we have that that $\det D$ 
is projectable and from Lemma~\ref{para} that also $J$ is projectable.
We have from the Gauss parametrization that
$$
f_*AX=-N_*X=-h_*\pi_*X
$$ 
where  $h=i\circ g$. Hence,
\begin{equation} \label{first}
f_*ADX=h_*\pi_*DX=-h_*\bar{D}\pi_*X
\end{equation} 
for any $X\in\mathfrak{X}(M)$. In particular,
\begin{equation} \label{bra}
f_*AD[X,Y]=-h_*\bar{D}\pi_*[X,Y]=-h_*\bar{D}[\pi_*X,\pi_*Y]
\end{equation} 
for any $X,Y\in\mathfrak{X}(M)$. Moreover,
\begin{eqnarray}\label{net}
f_*\nabla_XADY\!\!\!&=&\!\!\!\tilde\nabla_Xf_*ADY-\<AX,ADY\>N\nonumber\\
\!\!\!&=&\!\!\!-\tilde\nabla_{\pi_*X}h_*\bar{D}\pi_*Y
-\<h_*\pi_*X,h_*\bar{D}\pi_*Y\>h\circ\pi\nonumber\\
\!\!\!&=&\!\!\!-h_*\nabla'_{\pi_*X}\bar{D}\pi_*Y-\alpha_h(\pi_*X,\bar{D}\pi_*Y)
-\<\pi_*X,\bar{D}\pi_*Y\>h\circ\pi\nonumber\\
\!\!\!&=&\!\!\!-h_*\nabla'_{\pi_*X}\bar{D}\pi_*Y-\alpha_g(\pi_*X,\bar{D}\pi_*Y).
\end{eqnarray}
 From (\ref{6}) and the above, we have that
$$
0=f_*(\nabla_XB)Y-f_*(\nabla_YB)X=f_*\nabla_XADY-f_*\nabla_YADX-f_*AD[X,Y].
$$
We conclude that part $(b)$ holds as well as 
$$
\alpha_g(\pi_*X,\bar{D}\pi_*Y)=\alpha_g(\bar{D}\pi_*X,\pi_*Y).
$$
Since $\bar D\in\mbox{span}\{I,\bar J\}$ but $\bar D\not\in\mbox{span}\{I\}$, the preceding
equation is equivalent to
\begin{equation} \label{equiv}
\alpha_g(\pi_*X,\bar{J}\pi_*Y)=\alpha_g(\bar{J}\pi_*X,\pi_*Y)
\end{equation} 
and thus $g$ is hyperbolic (respectively, elliptic) with respect to  $\bar J$.

To deal with the function $\gamma$ we first show that condition (\ref{eq:aJ})
is equivalent to
$$
(\mbox{Hess\,} h^v+h^v I)J=J^t(\mbox{Hess\,} h^v+h^v I)
$$
where $\mbox{Hess\,} h^v$ is the endomorphism  of $TL$ associated to the Hessian
and $h^v=\<h,v\>$ for any $v\in\mathbb{R}^{n+1}$.
We have that the Hessian of $h^v$ satisfies
$$
\mbox{Hess\,} h^v(X,Y)=\<\alpha_h(X,Y),v\>=\<i_*\alpha_g(X,Y)-\<X,Y\>h,v\>
$$
for all $X,Y\in\mathfrak{X}(L)$. Thus
$$
\<i_*\alpha_g(JX,Y)-i_*\alpha_g(X,JY),v\>
=\<((\mbox{Hess\,} {h^v}+h^v I)J-J^t(\mbox{Hess\,} {h^v}+h^v I))X,Y\>
$$
for all $X,Y\in\mathfrak{X}(L)$.

It remains to prove that
\begin{equation} \label{prenda}
\left(\mbox{Hess\,}\gamma+\gamma I\right)\bar{J}
=\bar{J}^t\left(\mbox{Hess\,}\gamma+\gamma I\right).
\end{equation} 
By the Gauss parametrization, there exists a diffeomorphism 
$\theta\colon U\subset\Lambda\to M^n$ from an open neighborhood of the 
zero section of $\Lambda$ such that $\pi\circ\theta=\bar\pi$ and
$$
f\circ\theta(x,w)=\gamma(x)h(x)+h_*\nabla\gamma(x)+w
$$
for any $(x,w)\in\Lambda$.
Let $j\colon T_xL\to T_{(x,w)}\Lambda$ be the linear isometry in
Proposition \ref{gp}. Then,  for any $\bar{X},\bar{Y}\in T_xL$ we obtain 
using (\ref{eq}) and (\ref{first})   that
\begin{eqnarray}\label{addpw}
-\< AD\theta_* j\bar{X},\theta_*j\bar{Y}\>\!\!\!&=&\!\!\!
-\< f_ *AD\theta_* j\bar{X},f_* \theta_*j\bar{Y}\>\nonumber\\
\!\!\!&=&\!\!\!\<h_*\bar{D}\pi_* \theta_* j\bar{X},h_*\bar{Y}\>\nonumber\\
\!\!\!&=&\!\!\!\<\bar{D}\bar\pi_*j\bar{X},\bar{Y}\>'\nonumber\\
\!\!\!&=&\!\!\!\<\bar{D}P_w^{-1}\bar{X},\bar{Y}\>'.
\end{eqnarray}
It follows that $\bar{D}P_w^{-1}=P_w^{-1}\bar{D}^t$, or equivalently, that
$P_w\bar{D}=\bar{D}^tP_w$. And because $\bar D\in\mbox{span}\{I,\bar J\}$, this 
is equivalent to $P_w\bar{J}=\bar{J}^tP_w$.
Moreover, using that $A_w\bar{J}=\bar{J}^{t}A_w$ as follows from
(\ref{equiv}), we conclude that  (\ref{prenda}) is satisfied.\qed

\begin{lemma}\label{propr}{\hspace*{-1ex}}{\bf .  }
The following assertions on a surface  $g\colon L^2\to\mathbb{S}^n$ are equivalent:
\begin{itemize}
\item[(i)] The surface $g$ is hyperbolic (respectively, elliptic) with respect
to a tensor $\bar{J}$ on $L^2$ satisfying ${\bar J}^2=I$ (respectively,
${\bar J}^2=-I$), and there is $\bar{D}=\bar{\mu}\bar{J}$,  
$\bar\mu>0$, such that
\begin{itemize}
\item[(a)] $\mbox{tr\,}\bar{D}=0$,
\item[(b)] $\left(\nabla'_{\bar{X}}\bar{D}\right)\bar{Y}
-\left(\nabla'_{\bar{Y}}\bar{D}\right){\bar{X}}=0$
for all $\bar{X},\bar{Y}\in \mathfrak{X}(L).$
\end{itemize}
\item[(ii)] There exist real-conjugate (respectively, complex-conjugate) 
coordinates on $L^2$ such that system  (\ref{eq:siti}) 
(respectively, (\ref{eq:comp})) has solution.
\end{itemize}
\end{lemma}

\noindent\emph{Proof: }  We make use of Proposition \ref{coordinates}. In the case of 
real coordinates $(u,v)$ and since we have $\bar{D}\partial u=\bar\mu\partial u$, 
$\bar{D}\partial v=-\bar\mu\partial v$, we easily see that
$$
\left(\nabla'_{\partial u}\bar{D}\right)\partial v
-\left(\nabla'_{\partial v}\bar{D}\right)\partial u=0
$$
is equivalent to the system (\ref{eq:siti}). The case of complex coordinates
is similar.\vspace{1,5ex}\qed

\noindent\emph{Proof of Theorem \ref{main}:} By Proposition \ref{hypel}, 
on each connected component of an  open and dense subset of $M^n$ the hypersurface $f$ 
is either hyperbolic or elliptic with respect to  $J\in\Gamma(\mbox{End}(\Delta^\perp))$. 
It follows from Lemma \ref{conditions} that there exists $D\in\Gamma(\mbox{End}(\Delta^\perp))$ 
satisfying the properties  $(i)$--$(iv)$. 

Let $f$ be parameterized by a pair $(g,\gamma)$ in terms of the Gauss 
parametrization.
By Lemma \ref{prop*} if $f$ is  hyperbolic (respectively, elliptic)
with respect to $J$, then $J$ and $D$ can be projected to  tensors $\bar{J}$ and
$\bar D\in\mbox{span}\{I,\bar{J}\}$ on $L^2$, with   $\bar{J}^2=I$ (respectively,
$\bar{J}^2=-I$).
Moreover,  the pair $(g,\gamma)$ is  hyperbolic (respectively, elliptic)  
with respect to  $\bar{J}$ and $\bar D$ satisfies $(a)$ and $(b)$ in 
Lemma~\ref{prop*}. Now the proof of the direct statement follows from 
Lemma \ref{propr}.

Conversely, let $f\colon M^n\to\mathbb{R}^{n+1}$ be a simply-connected 
hypersurface parameterized  in terms of the Gauss parametrization
by a special hyperbolic or special elliptic pair $(g,\gamma)$.
By Lemma~\ref{propr},  there exists $\bar{D}=\bar{\mu}\bar{J}$ satisfying 
equations $(a)$ and $(b)$. 
By Lemma~\ref{prop*} the hypersurface  $f$ is hyperbolic or elliptic 
with respect to the horizontal lift $J$ of $\bar J$, and the horizontal 
lift $D=\mu J$ of  $\bar{D}=\bar{\mu}\bar{J}$ satisfies $(i)$--$(iv)$ in  
Lemma \ref{conditions}. 

To conclude from Theorem \ref{mains} that $f$ admits a unique nontrivial 
infinitesimal bending it remains to show that $B\in\Gamma(\mbox{End}(TM))$ 
defined by $B\vert_{\Delta^\perp}=A\vert_{\Delta^\perp}D$ and $\Delta\subset\ker B$
is symmetric and satisfies equations (\ref{6}) and (\ref{5}). In fact, 
that $B$ is symmetric follows easily from (\ref{addpw}). 

Part $(b)$ of Lemma \ref{prop*} together with (\ref{bra}) and (\ref{net}) 
imply that
$$
f_*((\nabla_XB)Y-(\nabla_YB)X)=f_*\nabla_XADY-f_*\nabla_YADX-f_*AD[X,Y]=0
$$
for any $X,Y\in\Gamma(\Delta^\perp)$.  
Since  $D$ is projectable, we can use Corollary $13$ in \cite{dft2} and 
deduce that
$$
\nabla_TD=[D,C_T]
$$
for any $T\in\Delta$. Using (\ref{codazzi}) we obtain 
$$
(\nabla_TB)X=(\nabla_TAD)X=(\nabla_TA)DX+A[D,C_T]X=ADC_TX
$$
for any $T\in\Delta$ and $X\in\Delta^\perp$. It follows that
$$
(\nabla_XB)T-(\nabla_TB)X=0
$$
for any $T\in\Delta$ and $X\in\Delta^\perp$. Since
$$
(\nabla_SB)T-(\nabla_TB)S=B[S,T]=0
$$
for any $S,T\in\Delta$, we have shown that (\ref{bra}) holds.
Since (\ref{net}) is equivalent to $\mbox{tr\,} D=0$, the proof follows.\qed


\section{The ruled case}


In this section, we discuss the infinitesimal bendings of ruled 
hypersurfaces that have not been considered yet.
\vspace{1ex}

Let $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, be a ruled hypersurface without
flat points which is not surface-like on any open subset of $M^n$. Then $f$
has rank two and there exists locally an orthonormal frame $X,Y$ of
$\Delta^\perp$ such that the second fundamental form is the form
$$
A\vert_{\Delta^\perp}=\begin{bmatrix}
\lambda&\mu\\
\mu&0
\end{bmatrix}.
$$
Note that if $M^n$ is simply-connected then the set of all isometric 
immersions of $M^n$ into $\mathbb{R}^{n+1}$ consists of ruled immersions with 
the same rulings; see \cite{dft}. Moreover, this set
can be parametrized by the set of all smooth functions in an interval.
In fact, the second fundamental form of any other immersion must
be of the form
$$
A\vert_{\Delta^\perp}=\begin{bmatrix}
\lambda + \theta&\mu\\
\mu&0
\end{bmatrix}
$$
where $\theta\in C^\infty(M)$ is determined by choosing a smooth function
along an integral curve of $X$ and extending it to $M^n$ by requiring that
\begin{equation} \label{codazzi2}
Y(\theta)=\<\nabla_XX,Y\>\theta\;\;\mbox{and}\;\;T(\theta)
=\<\nabla_XX,T\>\theta
\end{equation} 
for any $T\in\Delta$. 


\begin{proposition}{\hspace*{-1ex}}{\bf .  }
 Let $f\colon M^n\to\mathbb{R}^{n+1}$, $n\geq 3$, be a simply-connected ruled
hypersurface of constant rank two that is not surface-like on any open 
subset of $M^n$.  Then, any infinitesimal bending is the variational 
vector field of an isometric bending.
\end{proposition}

\noindent\emph{Proof: } Since $M^n$ is simply-connected, there is a global orthonormal 
frame $\{X,Y\}$ of $\Delta^\perp$ as above. By Lemma \ref{conditions} and 
Proposition \ref{ruled}, the Codazzi tensor $B$ on $M^n$ is given by 
$B|_{\Delta^\perp}=A|_{\Delta^\perp}D$ and $\Delta\subset\ker B$, where 
$D=\theta J$ and  $J\in\Gamma(End(\Delta^\perp))$ is such that  $JX=Y$ and 
$JY=0$. Moreover, $\theta\in C^{\infty}(M)$ is arbitrarily prescribed along 
an integral curve of $X$ and  required to satisfy (\ref{codazzi2}). 
Therefore, the one-parameter family of Codazzi tensors $A(t)=A+tB$, $t\in\mathbb{R}$, 
gives rise to an isometric bending of $f$ having the infinitesimal bending 
determined by $B$ as its variational vector field. 




