\begin{document}


\newcommand{\pp}[2]{\ensuremath{\frac{\partial #1}{\partial #2}}}
\newcommand{\cinfty}{\ensuremath{\mathcal{C}^\infty}}
\newcommand{\pinfty}{\ensuremath{\mathcal{P}^\infty}}
\newcommand{\piv}{\ensuremath{\pi_V}}
\newcommand{\pih}{\ensuremath{\pi_H}}
\newcommand{\Lie}{\ensuremath{\mathcal{L}}}
\renewcommand{\theenumi}{\arabic{enumi}}


\setcounter{page}{1}
\thispagestyle{empty}


\markboth{Benjamin Lent Davis
 and A\"{\i}ssa Wade }{Nonlinearizability
 of certain Poisson structures near a symplectic leaf}

\label{firstpage}
$ $
\bigskip

\bigskip


\centerline{{\Large 
Nonlinearizability
 of certain Poisson structures near a symplectic leaf
  }}


\bigskip
\bigskip
\centerline{\large by Benjamin Lent Davis 
\footnote[1]{Research partially supported by  the Faculty Development Fund of Saint Mary's College.}
and  A\"{\i}ssa Wade
\footnote[2]{Research partially supported by the Shapiro Funds}
 }


\vspace*{.7cm}

\begin{abstract}
We give an intrinsic proof that Vorobjev's first approximation of a
 Poisson manifold near a symplectic leaf is a Poisson manifold.
We also show that Conn's linearization results cannot
 be extended in Vorobjev's setting
\end{abstract}


\pagestyle{myheadings}

\section{Introduction}

A  Poisson structure on a smooth $n$-dimensional manifold $M$ is a bivector
 field $\pi \in \Gamma(\Lambda ^2 TM)$  such that the Schouten bracket
 $[\pi, \pi]=0$.It is known that every Poisson structure $\pi$ on $M$ 
 gives rise to a foliation by symplectic leaves. The study of the local structure of a Poisson manifold near a zero-dimensional leaf 
(i.e., a point  $m$ for which $\pi_m=0$ ) leads to the linearization problem,
 which we now describe. Recall that the {\em linear approximation}
 of a Poisson structure $\pi$ at a zero-dimensional leaf $m$ is determined
 by a Lie algebra $\mathfrak g$, called the {\em  transverse Lie algebra or isotropy Lie algebra} at $m$.
More precisely,  the linear approximation coincides with the 
canonical linear Poisson structure $\pi^{(1)}$ defined on the dual $\mathfrak{g}^*$ of $\mathfrak g$.
In \cite{weinstein83}, Weinstein  showed  that, if $\pi$ is a Poisson structure  on $M$ which vanishes at a point $m$ and whose transverse Lie algebra at $m$ is semi-simple, then $\pi$
 is formally isomorphic to its linear approximation 
$\pi^{(1)}$. The analytic and the smooth versions of this result were
 proved by Conn in \cite{conn84} and \cite{conn85}. Other
 partial results on the linearization of Poisson structures can be found in
 \cite{molinier93} and  \cite{dufour-zung}.  A survey of the literature on linearization of Poisson brackets may be found in \cite{fm04}.

Less studied is  the local structure of a Poisson
 manifold near a symplectic leaf with a nonzero dimension.
 Results on the Poisson topology of neighborhoods of symplectic leaves have been obtained by Ginzburg and Golubev \cite{gg}, Crainic \cite{crainic}, and
Fernandes \cite{fernandes}, \cite{fern02}.
The study of linearization near symplectic leaves of nonzero dimension was initiated by Vorobjev  (see \cite{vorobjev}),
 who defined the anologue of the  linear approximation $(\mathfrak g^*, \pi^{(1)})$
  for a symplectic leaf of dimension
 $d>0$. His goal was to generalize Conn's local linearization result to nonzero dimensional symplectic leaves.

In the present work, we give complete proofs  

of some results (see Theorems \ref{Main 1}
and  \ref{Lie alg}  below)
which were stated in \cite{vorobjev} without proofs.  We also give examples 
 of linearizable and nonlinearizable Poisson structures near a symplectic leaf
  having a nonzero dimension.  In particular, we show that Conn's linearization result cannot be extended in a straightforward way to the semi-local context.  To our knowledge, Example 3 (see below) is the first counter-example to
 the semi-local linearization question.


 
  


\section{Integrable geometric data}
Let $S$ be an embedded  submanifold of a smooth $n$-dimensional manifold $P$. Fix a tubular neighborhood  ${\cal N}$ of  $S$.
 This corresponds to a  vector bundle $p : {\cal N} \rightarrow S$. We 
 identify $S$ with the zero section of ${\cal N}$. Moreover, we denote $Vert=\ker p_*$. 
 An \emph{Ehresmann connection} on ${\cal N}$ is a projection map
 $\Gamma: T {\cal N} \rightarrow Vert$. Equivalently, we have 
 a smooth vector subbundle $Hor \subset {\cal N}$ such that
$$T_x{\cal N}= Hor_x \oplus Vert_x \quad \forall \ x \in {\cal N}.$$
 For any vector field $X \in \chi(S)$, there is a unique horizontal
 vector field $\overline{X} \in \chi(Hor)$ which is called the \emph{horizontal lift}
 of $X$, and  satisfies $$ p_*(\overline{X})=X.$$ 
 Define the curvature of $\Gamma$ by the formula
$$Curv_{\Gamma}(X,Y)=[\overline{X}, \overline{Y}]- \overline{[X,Y]}, \quad \mbox{ for any } \ X, Y \in \chi(S) .$$
\noindent  Suppose that $S$ is equipped with a symplectic form $\omega \in \Omega^2(S)$.
  Now, we consider a triple $(\Gamma, \nu, \varphi)$ called  {\em geometric data}
and formed by
 
\smallskip
$\bullet$ an Ehresmann connection $\Gamma$,

\smallskip
$\bullet$  a vertical bivector field $\nu \in \Gamma(\Lambda^2 Vert)$,

\smallskip
$\bullet$  and a nondegenerate 2-form $\varphi \in \Omega^2(S) \otimes \cinfty({\cal N})$ given by
 $$\varphi=\omega \otimes 1 + R,$$
 \noindent where  $\omega$ is the symplectic form on the zero section $S$, and $R$ vanishes on $S$, i.e.
$$\varphi_x(X,Y)=\omega_x(X,Y), \quad \forall \ x \in S, \quad \forall X, Y \in \chi(S).$$

The projection $p:{\cal N}\rightarrow S$ induces a map $p^*:\Omega^2(S)\otimes \cinfty({\cal N})\rightarrow \Omega^2({\cal N})$  by pulling back the first factor of the tensor product.
Define the \emph{coupling 2-form} $\widehat{\omega} \in \Omega^2({\cal N})$ by $\widehat{\omega}=p^*\varphi.
$
\newcommand{\ann}{\ensuremath{\mathrm{ann}}}
In a sufficiently small tubular neighborhood we have that, for any $x \in {\cal N}$, 
the linear map $\widehat{\omega}_{|Hor}(x): Hor_x \rightarrow \ann(Vert_x)$
is an isomorphism, where $\ann(Vert_x)$ is the annihilator of $Vert_x$.. If necessary, we may work with a smaller tubular neighborhood.
Define the \emph{horizontal coupling bivector} by  
\begin{equation}
\label{hcb}
\mu=\widehat{\omega}_{|Hor}^{-1}.
\end{equation}
 This gives  the identity
$$\widehat{\omega}(\mu\alpha,\mu\beta)=-\mu(\alpha,\beta).$$
\noindent With respect to the above notations
 we have the following theorem:
\begin{theorem}
\label{Main 1}
The  bivector field $\pi=\mu +\nu$ is a Poisson bivector field if and only if the 
 following \emph{integrability conditions} are satisfied
\begin{enumerate}
\item $\nu$ is a Poisson  bivector field (i.e. $[\nu, \nu]=0$);
 \item $[\overline{X}, \ \nu]=0,$ for all $X \in \chi(S)$;
 \item $Curv_{\Gamma}(X,Y)= \nu(d\widehat{\omega}(\overline{X}, \overline{Y}))$;
 \item $d \widehat{\omega}(\overline{X}_1, \overline{X}_2, \overline{X}_3)=0$, for all
 $X_1, X_2, X_3 \in \chi(S)$.
\end{enumerate}
\end{theorem}
 
\noindent The proof relies on the following three lemmas.
\begin{lemma}
\label{formula 1}
Let $M$ be a smooth manifold and let $\Lambda$ be a bivector field  on $M$.
 Then, 
$$ -\frac{1}{2}[\Lambda,\Lambda](\alpha,\beta,\gamma)=
\Big(\Lambda(d\Lambda(\alpha,\beta),\gamma)+\langle\alpha,[\Lambda \beta,\Lambda\gamma]\rangle
 \Big)+\mbox{c.p.},$$
\noindent for any $\alpha,\beta,\gamma\in\Omega^1(M))$. 
\end{lemma}
 
\begin{proof}
 We have the following formula (see~\cite{ks-m}):
$$\Lambda({\cal L}_{\Lambda \alpha}\beta - {\cal L}_{\Lambda \beta}\alpha 
-d(\Lambda(\alpha, \beta)))
 = [\Lambda \alpha, \ \Lambda \beta] + \frac{1}{2}[\Lambda,\Lambda](\alpha,\beta, \cdot).$$
\noindent There follows
$$
\frac{1}{2}[\Lambda,\Lambda](\alpha,\beta, \gamma)  =
 - {\cal L}_{\Lambda \alpha} \langle \beta, \ \Lambda \gamma \rangle 
- \langle \alpha, \ [\Lambda \beta, \ \Lambda \gamma] \rangle + \mbox{c.p.}
$$
\end{proof}

\begin{lemma}\label{jacobi} 
 For any horizontal 1-forms $\alpha,\beta,$ and $\gamma$, we have
$$ -\frac{1}{2}[\mu,\mu](\alpha,\beta,\gamma)= d\widehat{\omega}(\mu\alpha,\mu\beta,\mu\gamma).$$
\end{lemma}

\begin{proof} Indeed,
\begin{eqnarray*}
d\widehat{\omega}(\mu\alpha,\mu\beta,\mu\gamma)&=&
\Lie_{\mu\alpha}\Big(\widehat{\omega}(\mu\beta,\mu\gamma)\Big)+\widehat{\omega}(\mu\alpha,[\mu\beta,\mu\gamma])
+\mbox{c.p.}\\
&=&\mu(d\mu(\beta,\gamma),\alpha)+\langle\alpha,[\mu\beta,\mu\gamma]\rangle+\mbox{c.p.}\\
&=& -\frac{1}{2}[\mu,\mu](\alpha,\beta,\gamma)\mbox{ by  Lemma \ref{formula 1}.}
\end{eqnarray*}
\noindent There follows the lemma.
\end{proof}

\begin{lemma}
\label{prep}
For any horizontal 1-form $\alpha$ and for any vertical
 1-forms $\beta, \gamma$, we have
$$-\frac{1}{2}[\pi,\pi](\alpha,\beta,\gamma)= [\overline{X}, \nu](\alpha,\beta),$$
\noindent where $\overline{X}= \mu \alpha$. 
\end{lemma}


\begin{proof}
By Lemma~\ref{formula 1}, we have that
\begin{eqnarray*}
 - \frac{1}{2}[\pi,\pi](\alpha,\beta,\gamma)&=&
\pi\Big(d(\pi(\alpha,\beta)),\gamma\Big)+\pi\Big(d(\pi(\beta,\gamma)),\alpha \Big)
+\pi\Big(d(\pi(\gamma,\alpha)),\beta \Big)\\
&&+\langle
\alpha,[\pi\beta,\pi\gamma]\rangle+\langle\beta,[\pi\gamma,\pi\alpha]\rangle+\langle\gamma,[\pi\alpha,\pi\beta]\rangle.\\
&=& \pi(d\pi(\beta,\gamma),\alpha)+\langle \beta,[\pi\gamma,\pi\alpha]\rangle+\langle \gamma,[\pi\alpha,\pi\beta]\rangle\\
&=& -\mathcal{L}_{\overline{X}}(\pi(\beta,\gamma))+\langle \beta,[\pi\gamma,\pi\alpha]\rangle+\langle \gamma,[\pi\alpha,\pi\beta]\rangle\,\, (\star)
\end{eqnarray*}
since $\pi(\alpha,\beta)=0$ and $\pi(\gamma,\alpha)=0$, and the Lie
 bracket of  two vertical vector fields is again vertical implying that 
$\langle \alpha,[\pi\beta,\pi\gamma]\rangle=0$.
On the other hand, by the product rule,
$$\mathcal{L}_{\overline{X}}(\pi(\beta,\gamma))=(\mathcal{L}_{\overline{X}}\pi)(\beta,\gamma)+\pi(\mathcal{L}_{\overline{X}}\beta,\gamma)+\pi(\beta,\mathcal{L}_{\overline{X}}\gamma)$$
and so
$$(\mathcal{L}_{\overline{X}}\pi)(\beta,\gamma)=\mathcal{L}_{\overline{X}}(\pi(\beta,\gamma))-\pi(\mathcal{L}_{\overline{X}}\beta,\gamma)-\pi(\beta,\mathcal{L}_{\overline{X}}\gamma).\, (\star\star)$$
Concentrating now on the second term of $(\star\star)$, we set $Z=\pi\gamma$ and compute using the product rule that
\begin{eqnarray*}
-\pi(\mathcal{L}_{\overline{X}}\beta,\gamma)&=&(\mathcal{L}_{\overline{X}}\beta)(Z)\\
&=&\overline{X} \cdot \pi(\gamma,\beta)-\beta([\pi\alpha,\pi\gamma])\\
&=&-\mathcal{L}_{\overline{X}}(\pi(\beta,\gamma))+\beta([\pi\gamma,\pi\alpha]).
\end{eqnarray*}   
Performing a similar computation on the third term of $(\star\star)$, we find that
$$(\mathcal{L}_{\overline{X}}\pi)(\beta,\gamma)=-\mathcal{L}_{\overline{X}}(\pi(\beta,\gamma))+\langle\beta,[\pi\gamma,\pi\alpha]\rangle+\langle\gamma,[\pi\alpha,\pi\beta]\rangle.\,\,(\star\star\star)$$

Combining ($\star$) and ($\star\star\star$), we see that 
\begin{displaymath}
\begin{array}{rcl}
- \frac{1}{2}[\pi,\pi](\alpha,\beta,\gamma) &=& (\mathcal{L}_{\overline{X}}\pi)(\beta,\gamma),\\
&=& (\mathcal{L}_{\overline{X}}(\mu+\nu))(\beta,\gamma,)\\
&=& (\mathcal{L}_{\overline{X}}\mu)(\beta,\gamma),\\
&=& 0,
\end{array}
\end{displaymath}
since $\beta$ and $\gamma$ are vertical.
\end{proof}

\medskip

\noindent{\bf Proof of Theorem \ref{Main 1}:} 
The splitting $T{\cal N}=Hor\oplus Vert$ induces the splittings
$$\bigwedge^k T{\cal N}=\bigoplus_{i+j=k} Hor^i \wedge Vert^j,$$
where $Hor^i \wedge Vert^j$ is the wedge product of $i$ copies of the vector bundle $Hor$ with
$j$ copies 
of the vector bundle $Vert$.  A section of $Hor^i \wedge Vert^j$ is said to be 
a \emph{multivector field of degree $(i,j)$}. 

 Thus, the trivector field $[\pi,\pi]$ is a sum of component trivector fields of degrees (3,0), (2,1), (1,2), and (0,3).   We will show that   conditions 1-4
  given in Theorem \ref{Main 1} are
  equivalent to the vanishing of 
 $[\pi,\pi]$ in degree $(n, 3-n)$, where $n=0,..,3$.
We have $$[\pi,\pi]=[\mu+\nu,\mu+\nu] = [\mu,\mu]+2[\mu,\nu]+[\nu,\nu].$$

  In general, the horizontal  distribution  is not necessarily integrable, but 
$Vert$ is always integrable. 

\noindent \textbf{Degree (0,3):} 
\noindent Thus, the degree (0,3) component of $[\pi,\pi]$ is exactly $[\nu,\nu]$. 
\medskip

\noindent \textbf{Degree (1,2):}
\noindent Let $\alpha$ be a horizontal 1-form, and
 let $\beta, \gamma$ be two vertical 1-forms.
By lemma \ref{prep}, we get
\begin{equation}
\label{Eq 2} 
[\pi, \pi](\alpha,\beta,\gamma)=-2 [\overline{X}, \nu](\alpha,\beta),
\end{equation}
\noindent where $\overline{X}= \mu \alpha$. 

\medskip
\noindent \textbf{Degree (2,1):}

 
\noindent Suppose $\alpha,\beta$ are horizontal 1-forms and $\gamma$ is a vertical 1-form. 
By Lemma~\ref{formula 1}, we have that
\begin{eqnarray*}
- \frac{1}{2}[\pi,\pi](\alpha,\beta,\gamma)&=&
\pi\Big(d\pi(\alpha,\beta),\gamma\Big)+\pi\Big(d\pi(\beta,\gamma),\alpha \Big)
+\pi\Big(d\pi(\gamma,\alpha),\beta \Big)\\
&&+\langle \alpha,[\pi\beta,\pi\gamma]\rangle+\langle\beta,[\pi\gamma,\pi\alpha]\rangle+\langle\gamma,[\pi\alpha,\pi\beta]\rangle.
\end{eqnarray*}
Several terms vanish automatically.  In particular $\pi(\beta,\gamma)=0$ and $\pi(\gamma,\alpha)=0$ since $\pi$ vanishes in degree (1,1). 
 Without loss of generality, we can suppose that $\overline{X}=\mu \alpha$, $\overline{Y}=\mu \beta$
 are horizontal lifts of vector fields $X, Y \in \chi(M)$. Setting
$Z:=\pi\gamma$, we get
$$\langle \alpha,[\pi\beta,\pi\gamma]\rangle=\langle\alpha,\Lie_{\overline{Y}} Z\rangle=0$$
and 
$$\langle \beta,[\pi\gamma,\pi\alpha]\rangle=\langle\beta,-\Lie_{\overline{X}} Z\rangle=0$$
since the Lie derivative of a vertical vector field by the horizontal lift of a vector field on $S$ is again vertical.
Thus,
\begin{eqnarray*}
- \frac{1}{2}[\pi,\pi](\alpha,\beta,\gamma)
&=&\pi(d(\pi(\alpha,\beta)),\gamma)+\langle\gamma,[\pi\alpha,\pi\beta]\rangle\cr
&=& \nu(d(\mu(\alpha, \beta)), \gamma))+\langle\gamma, [\overline{X}, \overline{Y}]\rangle\cr
&=&-\nu(d(\widehat{\omega}(\overline{X}, \overline{Y})), \gamma)
+\langle\gamma, [\overline{X}, \overline{Y}]\rangle
\end{eqnarray*}
\noindent But
$$[\overline{X}, \overline{Y}]=\overline{[X,Y]} +Curv_{\Gamma}(X,Y) \quad
{\rm and} \quad \langle \gamma, \overline{[X,Y]}\rangle=0.$$
\noindent It follows
\begin{equation}
\label{Eq 3}
- \frac{1}{2}[\pi,\pi](\alpha,\beta,\gamma)= -
\nu(d(\widehat{\omega}(\overline{X}, \overline{Y})), \gamma)+ 
\langle \gamma, Curv_{\Gamma}(X,Y)\rangle.
\end{equation}

\noindent 
\textbf{Degree (3,0):} 
\noindent We have seen earlier that the degree (3,0) components of $[\pi,\pi]$ and $[\mu,\mu]$ are equal, thus
$$[\pi,\pi](\alpha,\beta,\gamma)=[\mu,\mu](\alpha,\beta,\gamma).$$ 
By Lemma~\ref{formula 1}, 
\begin{equation}
\label{Eq 4}
d\widehat{\omega}(\mu\alpha,\mu\beta,\mu\gamma)=-\frac{1}{2}[\mu,\mu](\alpha,\beta,\gamma).
\end{equation}

\noindent This completes the proof of Theorem \ref{Main 1}.
\hfill $\Box$

\medskip
\noindent{\bf Definition} A geometric data $(\Gamma, \nu, \varphi)$ satisfying conditions 1-4 of Theorem \ref{Main 1}
 is said to be  {\em integrable}. In  this case, the corresponding  bivector field $\pi=\mu + \nu$ 
  is called the {\em  coupling Poisson bivector field associated with} $(\Gamma, \nu, \varphi)$. 
\medskip

\noindent{\bf Remark 1}
Let $X_1, X_2 , X_3$ be vector fields on the base manifold $S$,
 and let $\overline{X}_1, \overline{X}_2, \overline{X}_3$ be their  horizontal lifts respectievely.
 Using the above notations, we have
$$
d\widehat{\omega}(\overline{X}_1, \dots, \overline{X}_3)=
 \Lie_{\overline{X}_1}\Big(\widehat{\omega}(\overline{X}_2, \overline{X}_3)\Big)-
\widehat{\omega}([\overline{X}_1, \overline{X}_2], \overline{X}_3)+c.p.  $$
\noindent The curvature $Curv_{\Gamma}$ takes values in the vertical bundle $Vert$.
It follows that
$$\widehat{\omega}\Big([\overline{X}_1, \overline{X}_2], \overline{X}_3\Big)=
\widehat{\omega}\Big(\overline{[X_1, X_2]}, \overline{X}_3\Big).$$
\noindent Consequently, 
\begin{eqnarray*}
d\widehat{\omega}(\overline{X}_1, \dots, \overline{X}_3)&=&
\Lie_{\overline{X}_1}\Big(\widehat{\omega}(\overline{X}_2, \overline{X}_3)\Big)-
\widehat{\omega}\Big(\overline{[X_1, X_2]}, \overline{X}_3\Big)+c.p. \cr
&=& \Lie_{\overline{X}_1}\Big(\varphi(X_2, X_3)\Big)-\varphi\Big([X_1, X_2], X_3\Big)+c.p.\cr 
&=& (\partial_{\Gamma}\varphi)(X_1,X_2, X_3),
\end{eqnarray*}

\noindent where $$\partial_{\Gamma}:\Omega^k(S)\otimes\cinfty({\cal N})
\rightarrow\Omega^{k+1}(S)\otimes\cinfty({\cal N})$$ is the operator defined by the formula
\begin{eqnarray*}
\partial_{\Gamma}\mathbb F(X_0,\ldots,X_k)&=&\sum_{i=0}^n (-1)^i
 \Lie_{\overline{X}_i}\Big(\mathbb F(X_0,\ldots,\hat{X_i},\ldots,X_k)\Big)\\
&&+\sum_{i<j} (-1)^{i+j}\mathbb F([X_i,X_j],X_0,\ldots,\hat{X_i},\ldots,\hat{X_j},\ldots,X_k).\end{eqnarray*}
\noindent Hence integrability condition 4 in Theorem \ref{Main 1} can be replaced by $\partial_{\Gamma}\varphi=0$.

\medskip
\noindent{\bf Remark 2}
We have shown that, given a manifold $P$, any  integable geometric data $(\Gamma, \nu,\varphi)$ with respect
to a tubular  neighborhood ${\cal N}$ of a symplectic submanifold $(S,\omega)$ of $P$
induces a Poisson bivector field $\pi$ on ${\cal N}$,
 which admits $S$ as a symplectic leaf. The converse is also true.
Indeed, suppose $\pi$ is a Poisson structure on the total space
${\cal N}$ of a vector bundle over a symplectic manifold $(S, \omega)$,
 where $S$ is identified with the zero section of ${\cal N}$, and  $S$
 is a symplectic leaf of ${\cal N}$.
 Let $Vert=\ker p_*$, where $p$ is the natural projection of ${\cal N}$ onto $S$. In addition,
 we assume that $\pi$ is {\em horizontally nondegenerate}, that is
$\pi_{|\ann(Vert)}$ is nondegenerate.
 Then, there is a natural Ehresmann connection
 associated with $\pi$ which is determined by the horizontal subbundle
 $$Hor= \pi(\ann(Vert)).$$
 \noindent The bivector field $\pi$ can be decomposed into
$\pi=\mu+\nu$, where $\mu$ and $\nu$ are horizontal and
 vertical, respectively. Furthermore, we can define
$$\varphi(X,Y)= \pi^{-1}_{|Hor} (\overline{X}, \overline{Y}),$$
\noindent where $\overline{X}, \overline{Y}$ are the lifts of the vector
 fields $X, Y \in \chi(S)$. Using Equations (1)--(4), we
  conclude that integrability conditions 1--4 are satisfied.


\medskip
\noindent {\bf Remark 3.}
Brahic \cite{brahic} and Vaisman \cite{vaisman04} have each given independent proofs of  Theorem \ref{Main 1}.    The proof given here
 is intrinsic, and  has the advantage of clearly relating Vorobjev's linearization formula to geometric identities for transitive Lie algebroids
 (see below).


\section{Transitive Lie algebroids}

A \emph{Lie algebroid} over a manifold $S$ is a real vector bundle
  $p: E \rightarrow S$ together with a bundle map 
$\rho: E\rightarrow TS$ and a real Lie algebra structure 
$[\cdot,\cdot]_E$ on $\Gamma(E)$ such that the following 
 \emph{Leibniz rule} holds
$$[v,fw]_E=f[v,w]_E+(\rho(v)\cdot f)w,$$
\noindent for any $f\in\cinfty(S)$ and $v,w\in\Gamma(E)$.
The map $\rho$ is called the \emph{anchor} of the Lie algebroid.  
Using the Jacobi identity for $[ \cdot , \cdot]_E$
 and the Leibniz identity, one can show that
 the induced map $\rho:\Gamma(E)\rightarrow\chi(S)$ is a Lie algebra 
 homomorphism, i.e.
$$\rho[u,v]_E=[\rho(u), \rho(v)].$$

\medskip

Given any Lie algebroid, the distribution spanned by the image of the anchor is integrable,
 and the leaves of the resulting foliation are called the \emph{orbits} of the Lie algebroid.
  When the anchor is surjective, we call the Lie algebroid \emph{transitive}.

If $E\stackrel{\rho}{\rightarrow} S$ is a transitive Lie algebroid  with anchor map $\rho$,
then we obtain a short exact sequence of vector bundles
$$0\rightarrow I\stackrel{v}{\rightarrow} E\stackrel{\rho}{\rightarrow} TS\rightarrow 0,$$
where $I=\ker\rho$ is called the \emph{isotropy bundle}.

\medskip
 Recall that a \emph{connection} for a transitive Lie algebroid $E\stackrel{\rho}{\rightarrow} S$ 
 is a  splitting $\sigma:TS\rightarrow E$ of the sequence above.
Its  \emph{curvature } $R_\sigma\in\Omega^2(S)\otimes\Gamma(I)$ is given  by
$$R_\sigma (X,Y)=[\sigma X,\sigma Y]-\sigma [X,Y], \quad \mbox{ for all $X,Y\in\chi(S)$}.$$

Any connection $\sigma$ for a transitive Lie algebroid induces a covariant derivative 
for the isotropy bundle $I$ defined by
$$\nabla^{\sigma}_X s=[\sigma X,s], \quad \mbox{ for all $X\in\chi(S)$ and $s\in\Gamma(I)$.}$$
By a  \emph{covariant derivative} for a vector bundle ${\cal N}\rightarrow S$,
 we mean a linear map $\nabla:\Gamma({\cal N})\rightarrow\Omega^1(S)\otimes\Gamma({\cal N})$ satisfying 
the following properties:
$$\nabla_X fs= X(f)\cdot s+ f\nabla_X s \quad
 {\rm and} \quad \nabla_{fX} s=  f\nabla_X s \,$$ for all $X\in\chi(S)$, $f\in\cinfty(S)$, 
and $s\in\Gamma({\cal N})$.

Let $\sigma$ be a connection for a transitive Lie algebroid  $E$.
For simplicity, the corresponding covariant derivative
 will be denoted by $\nabla$ instead of $\nabla^{\sigma}$ when there is no ambiguity.
Define the \emph{curvature of} $\nabla$  by 
$$R_{\nabla}(X_1,X_2)=[\nabla_{X_1} , \ \nabla_{X_2}] 
-\nabla_{[X_1,X_2]} ,$$
\noindent for all $X_1,X_2\in\chi(S)$. 

The Jacobi identity yields geometric identities when evaluated on isotropic and coisotropic sections.
\begin{theorem}
\label{connection}
 Let $\sigma: TS\rightarrow E$ be a connection for a transitive Lie algebroid 
and let $\nabla$ be the induced covariant derivative for the isotropy bundle $I$. 
Then, for all $X, X_1, X_2, X_3\in\chi(S)$ and for all  $s, s_1, s_2\in\Gamma(I)$, we have
\begin{enumerate}
\item []{\bf (i)} The isotropy bundle $I$ is a Lie algebroid.
\item []{\bf (ii)} $\nabla_X[s_1,s_2]=[\nabla_X s_1,s_2]+[s_1,\nabla_X s_2]$,
\item []{\bf (iii)} $[R_{\sigma}(X_1,X_2),s]-R_{\nabla}(X_1,X_2)(s)=0$,
\item []{\bf (iv)} $\nabla_{X_1}R_{\sigma}(X_2,X_3)+R_{\sigma}(X_1,[X_2,X_3])+\mbox{c.p.}=0.$
\end{enumerate}
\end{theorem}
\begin{proof}
Simply use the definitions of the connection, 
the covariant derivative and their respective curvatures, and the fact that 
$$\mathcal{J}(\sigma X,s_1,s_2)=0, \quad \mathcal{J}(\sigma X_1, \sigma X_2,s)=0,
 \quad{and} \quad \mathcal{J}(\sigma X_1, \sigma X_2, \sigma X_3)=0,$$
 where $\mathcal{J}$ is the jacobiator of the bracket on sections of $E$.
\end{proof}
\noindent Versions of this theorem have been noted by several authors
  (see~\cite{mackenzie},\cite{fernandes},\cite{vorobjev}).\medskip
\medskip

 Any connection $\sigma$ for a transitive Lie algebroid $E \rightarrow S$
 gives rise to a {\em homogeneous} Ehresmann connection $\Gamma$ on the dual $I^*$ of the isotropy
bundle, i.e. the horizontal lift of every vector field $X \in \chi(S)$ preserves
 the space $\cinfty_{\mbox{lin}}({I^*})$ of fiberwise linear functions on $I^*$.
 To define the horizontal lift of $X$, we introduce 
 the natural isomorphism 
\begin{equation}
\label{ell1}
\ell: \Gamma(I)\rightarrow\cinfty_{\mbox{lin}}(I^*)
\end{equation}
given by the natural pairing. The horizontal lift 
 $\overline{X}$ of a vector $X\in \chi(S)$ satisfies
$$ \Lie_{\overline{X}}(\ell(s))=\ell(\nabla_X s).$$
Let $Hor \subset TI^*$ be the corresponding horizontal subbundle. Then
 $$TI^*=Hor \oplus Vert,$$
\noindent where $Vert=\ker  p_*$  with $p_*: TI^* \rightarrow TS$.
Define the \emph{Lie-Poisson bivector} $\nu$  by the following formula:
\begin{equation}
\label{Eq 6}
\nu(d u_1, du_2)=\ell([s_1, s_2]), \quad \mbox{where}  \ \ell(s_i)=u_i.
\end{equation}
The Lie-Poisson bivector defines a Poisson structure on $TI^*$ 
 (see Section 16.5 of \cite{weinstein}).
 Now, we suppose that $S$ is equipped with a symplectic form $\omega$.
 Let 
 \begin{equation}
 \label{Eq 7}
 \varphi= \omega\otimes 1 + \ell\circ R_{\sigma}.
 \end{equation}
\begin{theorem}
\label{Lie alg}
 Given a transitive Lie algebroid $E$ over a symplectic manifold $S$ 
 together with
a connection $\sigma: TS \rightarrow E$,
the geometric data $(\Gamma, \nu, \varphi)$ defined as above is integrable.
\end{theorem}

\begin{proof}
We will show that geometric identities \textbf{(i)--(iv)}
  in Theorem \ref{connection} are equivalent to integrability
  conditions 1--4 in Theorem \ref{Main 1}.

\noindent \textbf{1.}  Immediate by Equation \ref{Eq 6}.

\noindent \textbf{2.} By Theorem \ref{connection}\textbf{(ii)}, we have 
$$\nabla_X[s_1,s_2]-[\nabla_X s_1,s_2]-[s_1,\nabla_X s_2]=0.$$
\noindent But
$$\ell(\nabla_X[s_1,s_2])=\langle \overline{X}, d \ell([s_1, s_2])
\rangle=\Lie_{\overline{X}}(\nu(du_1, du_2)),$$
where $\ell(s_i)=u_i$. Similarly
$$\ell([\nabla_X s_1,s_2]+[s_1,\nabla_X s_2])
=\nu(d( \Lie_{\overline{X}}  u_1), du_2)+
\nu(du_1,d( \Lie_{\overline{X}}  u_2)).$$
\noindent It follows that 
$$\ell(\nabla_X[s_1,s_2]-[\nabla_X s_1,s_2]-[s_1,\nabla_X s_2])=
 [\overline{X}, \nu] (du_1, du_2).$$
\noindent Hence,
$$
\nabla_X[s_1,s_2]-[\nabla_X s_1,s_2]+[s_1,\nabla_X s_2]=0
\iff [\overline{X}, \nu]=0.
$$

\noindent \textbf{3.} By Theorem \ref{connection}\textbf{(iii)}, we have 
\begin{eqnarray*}
0&=&\ell \Big([R_{\sigma}(X_1,X_2), s]-R_{\nabla}(X_1,X_2)(s)\Big)\cr
 &=&\nu\Big(d(\ell(R_{\sigma}(X_1, X_2)), \ d(\ell(s)) \Big)-
 \langle [ \overline{X}_1, \overline{X}_2]+\overline{[X_1, X_2]}, d \ell(s) \rangle.
\end{eqnarray*}
\noindent Since $\varphi(X_1, X_2)- \ell(R_{\sigma}(X,Y))$ is constant on each
 fiber, we have
$$\nu\Big(d(\ell(R_{\sigma}(X_1, X_2)), \ d(\ell(s)) \Big)=
\nu\Big(d(\varphi(X_1, X_2)), \ d(\ell(s)) \Big).$$
 Therefore,
 \begin{eqnarray*}
0&=& \ell \Big([R_{\sigma}(X_1,X_2), s]-R_{\nabla}(X_1,X_2)(s)\Big)\cr
 &=&\nu\Big(d(\varphi(X_1, X_2)), \ d(\ell(s)) \Big)-
 \langle Curv_{\Gamma}(X_1, X_2), \ d \ell(s) \rangle \cr
 &=&\nu\Big(d(\widehat{\omega}(X_1, X_2)), \ d(\ell(s)) \Big)-
 \langle Curv_{\Gamma}(X_1, X_2),  \ d \ell(s) \rangle .
 \end{eqnarray*}
We obtain that
$$
 ad_{R_{\sigma}(X_1,X_2)}-R_{\nabla}(X_1,X_2)=0 \iff
Curv_{\Gamma}(X_1, X_2)=
\nu^\sharp d(\widehat{\omega}(X_1, X_2)). 
$$

\noindent \textbf{4.} By Theorem \ref{connection}\textbf{(iv)}, 
\begin{eqnarray*}
\ell\Big(\nabla_{X_1}R_{\sigma}(X_2,X_3)+R_{\sigma}(X_1,[X_2,X_3])\Big)
&=&  \Lie_{\overline{X_1}}\Big( \ell \circ R_{\sigma}(X_2,X_3)\Big)\cr
&& +\ell \circ R_{\sigma}([X_1,X_2],X_3).\cr
\end{eqnarray*}
\noindent Since $\partial_{\Gamma} (\omega \otimes 1)= d \omega \otimes 1=0$, we get
\begin{eqnarray*}
\ell\Big(\nabla_{X_1}R_{\sigma}(X_2,X_3)+R_{\sigma}(X_1,[X_2,X_3])\Big)+c.p.
&=&\Lie_{\overline{X_1}}\Big( \varphi (X_2,X_3)\Big)\cr
 &&+\varphi([X_1,X_2],X_3) +c.p\cr
& & \cr
 &=& \partial_{\Gamma} \varphi (X_1, X_2, X_3).
\end{eqnarray*}
This completes the proof. \end{proof}

The resulting coupling Poisson structure on $I^*$ depends on the choice of connection, but is unique up to isomorphism by the following proposition of Vorobjev proved in \cite{vorobjev}.

 \begin{proposition}
Let  $E_1$ and $E_2$ be two isomorphic transitive Lie algebroids over the same 
 symplectic base manifold $(S,\omega)$,
 and let $\sigma_1: TS \rightarrow E$, $\sigma_1: TS \rightarrow E$ be 
 two connections.  There exists a diffeomorphism
  $\psi$ from a neighborhood ${\cal V}_1$ of the zero section
 $S \subset I_1^*$ onto a neighborhood ${\cal V}_2$ of the zero section
 $S \subset I_2^*$ such that $\psi_{|S}=id$ and
 $\psi_*\pi_1=\pi_2$, where $I_i$ is the isotropy bundle of $E_i$ and
 $\pi_i$ is the coupling Poisson bivector field associated with $\sigma_i$, for $i=1,2$.
  
\end{proposition}


\section{Vorobjev (non)linearizability}

In this section, we are interested in the particular case where the isotropy bundle is the conormal bundle $N^*S$
 to a symplectic leaf $S$ of a Poisson manifold.
Given a Poisson manifold $(P,\Lambda)$, the Poisson tensor
induces a natural map from the cotangent bundle to the tangent bundle by 
 the formula
\begin{eqnarray*}
\Lambda ^\sharp :T^*P&\rightarrow&TP\\
\alpha&\mapsto& \Lambda(\alpha,\cdot).\\
\end{eqnarray*}

It is known that  $T^*P \rightarrow P$ is a Lie algebroid,
 called the \emph{Poisson algebroid} (see \cite{weinstein},\cite{vaisman})
whose  anchor map is $\Lambda^\sharp$ and whose Lie bracket is given by
$$[\alpha, \beta]_{T^*P}= \Lie_{\Lambda^\sharp \alpha} \beta -
\Lie_{\Lambda^\sharp \beta} \alpha - d\Lambda(\alpha, \beta).$$

The restriction to a symplectic leaf $(S, \omega)$ give a transitive
 Lie algebroid $T^*P_{|S} \rightarrow S$. Furtheremore, the kernel of the anchor map
 $\rho$ of this transitive Lie algebroid
 coincides with the conormal bundle $N^*S$ of $S$. 
Let $E$ be a tubular neighborhood of $S$, and let $p: E \rightarrow S$ be  the corresponding
vector bundle. 
The derivative of $p$ gives a connection for the 
transitive Lie algebroid $T^*P_{|S} \rightarrow S$, namely,
$\sigma:TS\rightarrow T^*P|_S$ defined by
$$\sigma(X)=p^*\omega(X),$$
where $\omega\in\Omega^2(S)$ is symplectic form of $S$.
This connection is called the \emph{pullback connection} induced by the tubular neighborhood.
We have the splitting
$$TP_{|S}=TS \oplus NS.$$ 
\noindent Moreover, we know that the normal bundle $NS$ is endowed with a canonical Poisson
  structure
 $\nu$ (see Equation (5)).  Theorem \ref{Lie alg} says that, up to a shrinking of the tubular
  neighborhood $E$ of $S$,
 there is a coupling Poisson bivector field $\pi$ on $E$ having $S$ as a symplectic leaf.

\medskip
\noindent {\bf Definition.}
 The Vorobjev-Poisson structure $\pi$ is called
 the {\em first approximation} of $\Lambda$ at the symplectic leaf 
$(S, \omega)$ with respect to the neighborhood $E$. 

We say that $\Lambda$ is {\em Vorobjev linearizable at the symplectic leaf $S$} if 
 there is a tubular neighborhood $E \subset P$ of $S$ with fibers tangent to the
normal bundle $NS$ and a diffeomorphism $\psi : E \rightarrow U \subset P$ such
$\psi_{|S}=id$ and $\psi_* \pi= \Lambda$.

We now introduce a family of Poisson manifolds called Casimir-weighted products 
and use them to give some concrete examples of Vorobjev \\
(non)linearizability.

Let $(P_1,\pi_1)$ and $(P_2,\pi_2)$ be Poisson manifolds with Casimir functions $f_1$ and $f_2$, respectively. 
 The \emph{Casimir-weighted product} is the Poisson manifold $(P_1\times P_2,
f_2\pi_1+f_1\pi_2)$. 
 That $f_2\pi_1+f_1\pi_2$ is Poisson is an easy computation using Schouten brackets. 
 The next proposition describes the symplectic leaves of a Casimir-weighted product.
\begin{proposition}
\label{productleaves}
Suppose that $(P_1,\pi_1)$ and $(P_2,\pi_2)$ are Poisson manifolds with smooth nowhere
vanishing
 Casimir functions $f_1$ and $f_2$, respectively.  
Then every symplectic leaf of the Casimir-weighted product is of the form
$(S_1\times S_2,\frac{1}{f_2}\omega_1+\frac{1}{f_1}\omega_2)$, where $(S_1,\omega_1)$ and
 $(S_2,\omega_2)$ are symplectic leaves of $(P_1,\pi_1)$ and $(P_2,\pi_2)$, respectively.
\end{proposition}


We now compute Vorobjev linearizations of  Casimir-weighted products of  symplectic Poisson manifolds by  Lie-Poisson manifolds.
\begin{theorem}\label{vorobcasimir}
Let $(S,\pi_S)$ be a symplectic Poisson manifold, and let $f$ be a Casimir for a Lie-Poisson
 manifold $(\mathfrak{g}^*,\pi_{\mathfrak{g}^*})$ such that $f(0)=1$.  Then the Vorobjev linearization 
of the Casimir-weighted product at the leaf $S\times\{0\}$ is
 $$\left(S\times\mathfrak{g}^*, \frac{1}{J^1_0f} \pi_S+\pi_{\mathfrak{g}^*}\right),
$$where $J^1_0f=1 + d_0f \in\cinfty(S\times \mathfrak{g}^*)$ denotes the first jet of $f$ at 0.
\end{theorem}
\begin{proof}
Projection to the first factor makes the Casimir-weighted product into a vector bundle 
$pr_1:S\times\mathfrak{g}^*\rightarrow S$.   Any vector bundle is canonically isomorphic to 
the normal bundle of the zero-section, thus $S\times\mathfrak{g}^*\tilde{\rightarrow} NS$
 and so we have  a canonical tubular neighborhood of the symplectic leaf $S\times {0}$.

We now find the coupling Poisson bivector $\pi=\mu+\nu$ induced by this tubular neighborhood. 
By Equation (\ref{Eq 6}), the Lie-Poisson bivector is the Poisson bivector of the
direct product of $(S,0)$ and the Lie-Poisson manifold $(\mathfrak{g}^*,\pi_{\mathfrak{g^*}})$.
Essentially this means that $\nu=\pi_{\mathfrak{g^*}}$.

By Equation (\ref{hcb}),  the computation of the horizontal coupling form $\mu$ is a two step processes: First we must compute the horizontal distribution $Hor$ of the connection, and then we must compute the coupling 2-form $\widehat{\omega}$.

We now show that the covariant derivative $\nabla$ on the normal bundle $NS$ is 
flat and that the horizontal distribution $Hor$ has leaves $S\times\{\xi\}$ for each
$\xi\in\mathfrak{g}^*$. Let $(q_1,\ldots,q_n,p_1,\ldots,p_n)$ be a Darboux chart on $S$. 
Let  $y_1,\ldots,y_k$ for $\mathfrak{g}$ be any basis.  Each $y_j$ is a linear function on $\mathfrak{g}^*$
 representing a section $dy_j$ of $N^*S$ by pulling-back by the composition of maps 
$S\times\mathfrak{g}^*\rightarrow\mathfrak{g}^*\tilde{\rightarrow}T_0\mathfrak{g}^*$.
Then $$(pr_1^* q_1, \ldots , pr_1^*q_n, pr_1^* q_1, \ldots , pr_1^*q_n, y_1 , \ldots , y_m)$$
 are coordinates on $S \times \mathfrak{g}^*.$
We see that each $dy_j$ is a $\nabla$-parallel section by the computation
\begin{displaymath}
\begin{array}{rcl}
\nabla_{X_{q_i}}dy_j&=&[\sigma(X_{q_i}),dy_j]\\
&=& [pr_1^*dq_i,dy_j]\\
&=& d\{pr_1^*q_i,y_j\}\\
&=& 0,
\end{array}
\end{displaymath}
since $q_i$ and $y_j$ are coordinates on the first and second factor of $S\times\mathfrak{g}^*$,
 respectively, and the bracket $\{\, ,\, \}$ is computed using the bivector $f\pi_S+\pi_{{\mathfrak{g}^*}}$.
Thus, $\{dy_1,\ldots,dy_k\}$ is a $\nabla$-parallel frame field for $N^*S$.
 The dual frame $\{\partial y_1,\ldots,\partial y_k\}$ is necessarily parallel for the dual covariant 
derivative on the dual bundle $NS$.
 Consequently, the horizontal distribution  for $NS$ is spanned by 
$\{\partial q_1,\ldots,\partial q_n,\partial p_1,\ldots,\partial p_n\}$, and so the leaves of the distribution are as claimed above.


Equation (\ref{Eq 7}) for the horizontal coupling form is $\varphi=\omega\otimes 1 + \ell\circ R_\sigma$.  Thus, we must compute the curvature of the pullback connection, $R_\sigma$. 
 In particular, 
\begin{displaymath}
\begin{array}{rcl}
\ell\circ R_\sigma(X_{q_i},X_{p_j})&=&[\sigma(X_{q_i}),\sigma(X_{p_j})]-\sigma([X_{q_i},X_{p_j}])\\
&=& \ell([pr_1^*dq_i,pr_1^*dp_j])\\
&=&\ell(d\{pr_1^*q_i,pr_1^*p_j\})\\
&=&\delta_{ij} \ell(df) \\
&=& \delta_{ij} \Big({\partial f \over \partial y_k} (0) \ell (d y_k) \Big),
\end{array}
\end{displaymath}
showing that $\ell \circ R_\sigma=
\omega\otimes (J^1_0f -1),$ where $J_0^1f$ is the 1-jet of $f$ at $0$.
Consequently, the coupling 2-form is $\varphi=\omega\otimes 1 +\ell \circ R_\sigma=\omega\otimes J^1_0f,$
and so the coupling bivector is
$\mu=\frac{1}{J^1_0f} \pi_S.$
\end{proof}


\newcommand{\pu}{\ensuremath{\partial_u}}
\newcommand{\pv}{\ensuremath{\partial_v}}

\begin{example}{\textbf{1}}
Let $(T,\pi_T=\pu\wedge\pv)$ be the  \emph{unit symplectic torus} obtained by identifying
opposite sides 
of the unit square. Let $(\mathfrak{g}^*,0)=(\mathbb{R},0)$ be the 1-dimensional Lie-Poisson manifold, and let $f(z)=e^z$ be a Casimir.  The Casimir-weighted product 
$$\left(T\times\mathbb{R},e^z\pu\wedge\pv\right),$$
has Vorbjev linearization given by 
$$\left(T\times\mathbb{R},\left(\frac{1}{1+z}\right)\pu\wedge\pv\right),$$
A  linearizing isomorphism 
is given by $\psi(x,y,z)=(x,y,-1+e^{-z}).$
\end{example}

  \begin{example}{\textbf{2}}
Let $(T,\pi_T=\pu\wedge\pv)$ be the unit symplectic torus.
Let $(\mathfrak{g}^*,\pi_\mathfrak{g})=(\mathbb{R},0)$ be the 1-dimensional Lie-Poisson
manifold, and 
let $f(z)=1+z^2$ be a Casimir.  The Casimir-weighted product 
$$\left(T\times\mathbb{R},(1+z^2)\pu\wedge\pv\right),$$
has Vorbjev linearization given by 
$$\left(T\times\mathbb{R},\pu\wedge\pv\right).$$
There is no Poisson isomorphism from the Casimir-weighted product to the Vorobjev linearization.  To see this, note that any such isomorphism must induce isomorphisms of symplectic leaves.  Recall that an invariant of a compact 2n-dimensional symplectic manifold is the \emph{symplectic volume},
$$\mathrm{Vol}(S,\omega)=\int_S \omega^n.$$
The Casimir-weighted product  has leaves of non-constant symplectic volume $1+z^2$, but the Vorobjev linearization has leaves of constant symplectic volume 1.
\end{example}


The following example shows that a Poisson manifold is not necessarily Vorobjev linearizable at a
symplectic leaf possessing 
a semisimple transverse Lie algebra of compact type. In other words, Conn's theorem
 \cite{conn85} cannot be extended in this context.
\newcommand{\px}{\ensuremath{\partial_x}}
\newcommand{\py}{\ensuremath{\partial_y}}
\newcommand{\pz}{\ensuremath{\partial_z}}

\begin{example}{\textbf{3}}
Let $(T,\pi_T=\pu\wedge\pv)$ be the unit symplectic torus. Consider the Lie-Poisson manifold
$(\mathfrak{g}^*,\pi_{\mathfrak{g}^*})=(\mathfrak{so}(3)^*,x\py\wedge\pz+y\pz\wedge\px+z\px\wedge\py).$  Recall that the symplectic leaves of $\mathfrak{so}(3)^*$ are the origin together with the spheres centered at the origin, and so $f=1+x^2+y^2+z^2$ is a Casimir.  Moreover, the sphere at radius $r$ has symplectic form $\omega=\frac{1}{r}dA$, where $dA=dx\wedge dy+dy\wedge dz+dz\wedge dx$ is the standard area form on $\mathbb{R}^3$ (see\cite{mr}, pp.457-458).
By Proposition \ref{productleaves}, $S=T\times\{0\}$ is a symplectic leaf of the Casimir-weighted product
$$(P,\pi)=(T\times\mathfrak{so}(3)^*,f\pi_T+\pi_{\mathfrak{g}^*}).$$
 We will show that $(P,\pi)$ is not Vorobjev linearizable at $S$.
By Theorem \ref{vorobcasimir}, the Vorobjev linearization of $(P,\pi)$ at $S$ is the direct product
$$(T\times\mathfrak{so}(3)^*,\pi_T+\pi_{\mathfrak{g}^*}).$$
Suppose for a contradiction that $$\psi:(T\times\mathfrak{so}(3)^*,f\pi_T+\pi_\mathfrak{g})\tilde{\rightarrow}(T\times\mathfrak{so}(3)^*,\pi_T+\pi_{\mathfrak{g}^*})$$
is an isomorphism of Poisson manifolds.  The map $\psi$ induces isomorphisms of symplectic leaves.   By Proposition \ref{productleaves},
$$(S_1,\omega_1)=(T\times S^2_{r_1},\frac{1}{1+r_1^2}\omega_T+\frac{1}{r_1}dA)$$
is a 4-dimensional symplectic leaf of $(P,\pi)$ when $r_1>0$.
Again by Proposition \ref{productleaves}, the only 4-dimensional leafs of the direct product are of the form
$$(S_2,\omega_2)=(T\times S^2_{r_2},\omega_T+\frac{1}{r_2}dA)$$  
when $r_2>0$.
Since $\psi$ restricts to a symplectomorphism
$(S_1,\omega_1)\tilde{\rightarrow}(S_2,\omega_2),$ there must be leaves of  equal symplectic volume.  The symplectic volume of a product is the product of the symplectic volumes, thus,
$$\mathrm{Vol}(S_1,\omega_1)=\mathrm{Vol}\left(T,\frac{\omega_T}{1+r_1^2}\right)\mathrm{Vol}\left(S^2_{r_1},\frac{dA}{r_1}\right)=\left(\frac{1}{1+r_1^2}\right)\left(\frac{4\pi r_1^2}{r_1}\right)=\frac{4\pi r_1}{1+r_1^2},$$
$$\mathrm{Vol}(S_2,\omega_2)=\mathrm{Vol}\left(T,\omega_T\right)\mathrm{Vol}\left(S^2_{r_2},\frac{dA}{r_2}\right)=\left( 1 \right)\left(\frac{4\pi r_2^2}{r_2}\right)=4\pi r_2,$$
and so 
\begin{equation}
\label{Eq 9}
r_2=\frac{r_1}{1+r_1^2}.
\end{equation}

On the other hand, the second homotopy group $\pi_2(T\times S^2)=\mathbb{Z}$ with generator $\phi:S^2\rightarrow \{t\}\times S^2$, where $t\in T$ is any point.  The diffeomorphism $\psi$ of $T\times S^2$ induces an isomorphism $\psi_*$ of homotopy groups, and so $\psi_*\phi=\pm \phi$. By assumption, $\psi$ is a symplectomorphism $\psi^*\omega_2=\omega_1$, thus
$$
4\pi r_1=\int_{\phi} \omega_1 = \int_{\phi} \psi^*\omega_2= \int_{\psi_*\phi} \omega_2=\int_{\pm\phi} \omega_2=\pm 4\pi r_2.
$$
But this contradicts Equation (\ref{Eq 9}).
\end{example}


\noindent{\bf Remark 4}
 Let $\Lambda$ be a Poisson structure horizontally nondegenerate
  defined on the total space $E$ of a vector bundle $p: E \rightarrow S$
 over a compact base manifold $S$. We identify $S$ with the zero section 
and suppose that $S$ is a symplectic leaf of $(E, \Lambda)$. 
 We denote by $(\Gamma, \nu, \varphi)$ the geometric data
 associated with the first approximation $\pi$ of $\Lambda$.
 In \cite{brahic 2}, the author shows that the germ of $\Lambda$ along
 $S$ is isomorphic to a Poisson bivector field $\Lambda'$ admitting
 $(\Gamma, \nu, \varphi')$ as  associated geometric data.  The above example shows that  
$\varphi' = \varphi$ cannot occur in that case.  

\section{Acknowledgements}
Rui Loja Fernandes suggested a simplification to the proof of Example 3. Thanks also to Alan Weinstein for comments on an earlier draft of this paper.
{"config": "arxiv", "file": "math0406611.tex"}