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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
                             
\title{The zero set of conformal vector fields}
\author{Andrei Moroianu}
\author{Liviu Ornea}
\thanks{This work was accomplished in the framework of the
  Associated European Laboratory ``MathMode". L.O. is partially supported by
CNCSIS PNII  grant code 8, 525/2009.}

\address{Centre de Math{\'e}mathiques, Ecole Polytechnique, 91128
  Palaiseau Cedex, France} 
\email{am@math.polytechnique.fr}
\address{Univ. of Bucharest, Faculty of Mathematics,
14 Academiei str., 70109 Bucharest, Romania, and 
Institute of Mathematics ``Simion Stoilow" of the Romanian Academy,
21, Calea Grivitei str., 
010702-Bucharest, Romania.}
\email{lornea@gta.math.unibuc.ro, Liviu.Ornea@imar.ro}

\begin{document}


\begin{abstract}

We show that every connected component of the zero set of a conformal
vector field on a Riemannian manifold is totally umbilical. 

\vs


\noindent

2000 {\it Mathematics Subject Classification}: Primary 53C15, 53C25.

\medskip
\noindent{\it Keywords:} conformal vector field, Killing vector field,
totally umbilical manifolds. 
\end{abstract}

\maketitle

\section{Introduction}

A well-known result of S. Kobayashi, \cite{kob}, states that the
connected components of the 
set of zeros of a Killing vector field on a Riemannian manifold 
are totally geodesic submanifolds of even codimension (the case of 
of isolated singular points is included). This is a purely local result.

In \cite{bla}, D.E. Blair proved the similar statement for conformal
vector fields, the zero set being now a totally umbilical
submanifold. However, his proof only works on {\em compact} manifolds
and makes explicit use of the Obata theorem.  

In this short note we prove that, in fact, the result is local, like
that of Kobayashi. However, we were not able to prove it in full
generality. Precisely, we could not prove that each connected
component of the zero set of a conformal vector field is always a
submanifold. For Killing fields, this follows by identifying this set
with the image through the exponential map of a vector subspace of the
tangent space. In the conformal case, we could not find a connection
whose exponential map play this r\^ole. We thus need to assume that
the zero set contains a differentiable curve. We do not know whether
this assumption can be removed or not. 


\section{Main result}


Let $\zeta$ be a conformal vector field on a Riemannian manifold
$(M^n,g)$ ($n\ge 3$). We denote by $\xi$ its metric dual 1-form,
and by $\f:=-\d\xi/n$. Then $\xi$ satisfies:
\beq\label{1} \n_Y\xi=\frac12 Y\i d\xi+\f Y^\flat,\qquad \forall Y\in TM.
\eeq  

In \cite[Theorem 2.1]{C}, Capocci proved the following result:

\begin{ath} \cite{C}\label{cap} Let $x$ be a zero of the conformal
  field $\zeta$. Then $\zeta$ is a homothetic vector field with
  respect to some conformal metric $\tilde g$ defined on a
  neighbourhood of $x$ if and only if the gradient of 
$\f$ with respect to $g$ belongs to the image of $\n \zeta$ at
$x$; moreover, $\zeta$ is Killing with respect to $\tilde g$ if, in
addition, $\f(x)=0$. 
\end{ath}

Hence, if \ref{cap} applies, the connected component of the zero set
of $\zeta$ is totally geodesic with respect to $\tilde g$, and hence
totally umbilical with respect to $g$ (see, {\em e.g.} \cite[pp.2--3]{bla}).  

We can now state our result:

\begin{ath}
 Let $\zeta$ be a conformal vector field on a Riemannian manifold
$(M^n,g)$ ($n\ge 3$). If a connected component of its zero set
contains a smooth curve, then this connected component is a totally
umbilical submanifold of $M$. 
\end{ath}


We show that we can apply Theorem \ref{cap}.
 
From now on we identify 1-forms and vectors via the metric $g$
whenever there is no risk of confusion.
We fix some point $x$ on $M$, and a local orthonormal frame $\{e_i\}$
parallel at $x$. We also assume that all vector fields $X$, $Y$,... are
parallel at $x$ in order to simplify the computations. Differentiating
\eqref{1} in the direction of some vector field $X$ yields
\beq\label{2} \n_X\n_Y\xi=\frac12 Y\i\n_X d\xi+X(\f) Y,\qquad \forall
X,Y\in TM. 
\eeq  
On the other hand, we also have 
\beq\label{3}
\n_X\n_Y\xi=R_{X,Y}\xi+\n_Y\n_X \xi
\stackrel{\eqref{1}}{=}R_{X,Y}\xi+\n_Y(\frac12 X\i d\xi+\f X). 
\eeq 
We use Einstein's summation convention on repeated subscripts. From
\eqref{2} and \eqref{3} we get
\bea
\n_X d\xi&=& \frac12 e_i\wedge (e_i \i\n_X
d\xi)\stackrel{\eqref{2}}{=}
e_i\wedge\n_X\n_{e_i}\xi\\
&\stackrel{\eqref{3}}{=}&e_i\wedge (R_{X,e_i}\xi+\frac12 X\i
\n_{e_i}d\xi+e_i(\f) X)\\
&=&e_i\wedge R_{X,e_i}\xi+\frac12 \n_X d\xi+ d\f\wedge X.
\eea
To compute the curvature term we can use the first Bianchi identity:
\bea
e_i\wedge R_{X,e_i}\xi&=&e_i\wedge e_j R(X,e_i,\xi,e_j)=e_i\wedge
e_j(R(X,e_j,\xi,e_i)+R(X,\xi,e_i,e_j))\\
&=&-e_i\wedge e_j R(X,e_i,\xi,e_j)+2R_{X,\xi},
\eea
whence $e_i\wedge R_{X,e_i}\xi=R_{X,\xi}$. The previous equation yields
\beq\label{4}
\n_X d\xi= 2R_{X,\xi}+2d\f\wedge X.
\eeq

\begin{elem} Let $Z$ denote the zero set of $\zeta$. Assume that $Z$
  contains a smooth curve $c(t)$ (that we may suppose parametrized by
  arc length, {\em i.e.} $g(\dot 
  c,\dot c)=1$). 

Then  $\nabla_{\dot
    c}\zeta=0$, $\f(c(t))=0$ and $\dot c(\f)=0$. 
\end{elem}
\bp Extend $\dot c$ to a vector field $Y$ on $M$. Since the flow
$\f_s$ of
$Y$ through any point of the curve $c$ stays (locally) on the same
curve, the Lie derivative of $\zeta$ with respect to $Y$ vanishes at
every point of $c$:
$$(\LL_Y\zeta)_{c(t)}=-\frac{d}{ds}\res_{s=0}((\f_s)_*\zeta)_{c(t)}
=-\frac{d}{ds}\res_{s=0}(\f_s)_*(\zeta_{\f_s^{-1}c(t))})=0.$$
We thus get at $c(t)$
$$\nabla_{\dot
    c}\zeta=\n_Y\zeta=\n_\zeta Y=0.$$
Moreover, since $g(\n_{\dot c}\zeta,\dot c)=\f g(\dot c,\dot c)=\f$, we
must have $\f(c(t))=0$. Differentiating this with respect to $t$
yields the last relation.

\r 


Taking \eqref{1} into account we get $\dot
c\i d\xi=0$. We differentiate this relation with respect to $\dot c$
and obtain at any point of $c$:
\bea 0&=&\n_{\dot c}(\dot c\i d\xi)={\n_{\dot c}\dot c}\i d\xi+\dot
c\i\n_{\dot c} d\xi \\
&\stackrel{\eqref{4}}{=}&4{\n_{\dot c}\dot c}\i \n\xi+2R_{\dot
  c,\zeta}\dot c+\dot c(\f)\dot c-2d\f g(\dot c,\dot c)\\
&=&4{\n_{\dot c}\dot c}\i \n\xi- 2d\f.
\eea
This shows in particular that $d\f$ (more precisely, the gradient of
$\f$ with respect to $g$) belongs to the image of $\n\zeta$ at
$c(0)$. 

According to Theorem \ref{cap}, this completes the proof.





\begin{thebibliography}{22}

\bibitem{bla} D.E. Blair, {\em On the zeros of a conformal vector
    field}, Nagoya Math. J. {\bf 55} (1974), 1--3. 

\bibitem{C} M.S. Capocci, {\em Essential conformal vector fields},
  Class. Quantum Grav. {\bf 16} (1999), 927--935. 

\bibitem{kob} S. Kobayashi, {\em Fixed points of isometries}, Nagoya
  Math. J. {\bf 13} (1958) 63--68. 
\end{thebibliography}


\end{document}