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Modular class

Let $(M, \Pi)$ be a Poisson manifold. Let us assume, for simplicity that

is orientable. Let $\Om$ be a volume form on $\Om$ . Consider for any $f\in\Coo(M)$ , $L_{X_f}\Om\in\Om^n M$ . There exists a function $\phi_{\Om}(f)$ such that $L_{X_f}\Om = \phi_{\Om}(f)\Om$ .
$\begin{fact} $\phi_{\Om}$\ is a vector field. \end{fact}$

$\begin{proof} We have to prove $\phi_{\Om}(fg)=\phi_{\Om}(f)g +f\phi_{\Om}(g)$. ... ...g\phi_{\Om}(f)\Om + f\phi_{\Om}(g)\Om \end{displaymath}hence thesis. \end{proof}$

$\begin{defn} $\phi_{\Om}$\ is called the \textbf{modular vector field} of $(M, \Pi)$ with respect to $\Om$. \end{defn}$

$\begin{fact} The modular vector field is an infinitesimal Poisson field. \end{fact}$

$\begin{proof} We have seen that this is equivalent to $\phi_{\Om}\in\Der(\Coo(M)... ...poiss{\phi_{\Om}(f)}{g} + \poiss{f}{\phi_{\Om}(g)}. \end{displaymath}\end{proof}$

$\begin{fact} $L_{\phi_{\Om}}\Om = 0$. \end{fact}$

$\begin{proof} \begin{displaymath} L_{\phi_{\Om}}\Om=d i_{\phi_{\Om}}\Om + i_{\ph... ...{\Pi}\Om), \end{displaymath}because $di_{\Pi}\Om=i_{\phi_{\Om}}\Om$. \end{proof}$
Take another volume form $\Om'=a\Om$ . Then

$\begin{displaymath} L_{X_f}\Om'=\phi_{\Om'}(f)\Om' = \phi_{\Om'}(f)a\Om \end{displaymath}$

$\begin{displaymath} L_{X_f}(a\Om)=aL_{X_f}\Om + X_f(a)\Om=a\phi_{\Om}(f)+X_f(a) \end{displaymath}$

Furtermore

$\begin{displaymath} a\phi_{\Om'}(f)=a\phi_{\Om}(f)+X_f(a), \end{displaymath}$

$\begin{displaymath} \phi_{\Om'}(f)=\phi_{\Om}(f)+\inv{a}X_f(a), \end{displaymath}$

and

$\begin{displaymath} \inv{a}X_f(a)=\inv{a}\poiss{f}{a}=\poiss{f}{\log{\vert a\vert}}= -\poiss{\log{\vert a\vert}}{f} \end{displaymath}$

Hence the modular vector fields with respect to different volume forms differ for a hamiltonian vector field.

$\begin{displaymath} \phi_{\Om'}=\phi_{\Om}+X_{-\log{\vert a\vert}}. \end{displaymath}$

$\begin{defn} The vector field $\phi_{\Om}$\ defines a class $[\phi_{\Om}]\in\rH^... ...ndent of $\Om$, and is called the \textbf{(Poisson) modular class}. \end{defn}$

$\begin{defn} Let $(M, \Pi)$\ be a Poisson manifold such that $[\phi_{\Om}]=0$. Then $(M, \Pi)$\ is called \textbf{unimodular}. \end{defn}$

$\begin{exer} On $(\bR^2, f(x,y)dx\land dy)$\ compute the modular class. \end{exer}$

$\begin{examples} \mbox{} \begin{enumerate} \item Let $(M, \om)$\ be a compact sy... ...t for vector fields tangent to leaves \cite{abb}. \end{enumerate}\end{examples}$
Let $(M, \Pi)$ be compact unimodular Poisson manifold. Then there exists $\Om$ such that $L_{\phi_{\Om}}\Om=0$ . Then
$\begin{align*} \int_M\poiss{f}{g}\Om &= \int_M (L_{x_f}g)\Om \\ &= \int_M L_{X_... ...}d(g\Om)}_{=0}-\int_M gL_{X_f}\Om \\ &= -\int_Mg\phi_{\Om}(f)\Om. \end{align*}$
This is called also infinitesimal KMS condition. Being $(M, \Pi)$ unimodular, we can choose a volume form $\Om$ such that $\phi_{\Om}\equiv 0$ , so $\int_M\poiss{f}{g}\Om=0$ , i.e.

$\begin{displaymath} \int_M \Om\:\Coo(M)\to \bR \end{displaymath}$

is a Poisson trace.

Next: Computation for Poisson cohomology Up: Poisson cohomology Previous: Poisson cohomology Contents

Pawel Witkowski 2006-06-26