themselves obey the following algebra \[3.4\] {Y\_(f),Y\_(g)}=0, {Y\_(f),Y\_(g)}=’ \^\_[S\^1]{}dx (f\_’ g\_- f\_g\_’) From the geometrical point of view the $Y_\pm$ are one forms on $S^1$ and the $f$ are scalars while the $u,v$ are vector fields. In one dimension, $p-$times covariant and $q-$times contravariant tensors are the same thing as scalar densities of weight $p-q$, hence all integrals are over scalar densities of weight one which are spatially diffeomorphism invariant. However, the Hamiltonian vector fields corresponding to the $V_\pm(\xi)$ only act on the $Y_\pm$ not on the $u,v,f,f'$. Specifically we have \[3.5\] {V\_(u),Y\_(f)}=0, {V\_(u),Y\_(f)}=Y\_(u f’) Hence the Virasoro generators $V_\pm(u)$ act on $Y_\pm$ by infinitesimal diffeomorphisms while they leave $Y_\mp$ invariant. Consider the Hamiltonian flow of $V_\pm(u)$ defined by \[3.6\] \[\^\_u(t)\](F):=\_[n=0]{}\^ {V\_(u),F}\_[(n)]{} where $F$ is any smooth function on phase space and the repeated Poisson bracket is inductively defined by $\{G,F\}_{(0)}=F,\;
\{G,F\}_{(n+1)}=\{G,\{G,F\}_{(n)}\}$. It is easy to check that \[3.7\] \[\^\_u(t)\](Y\_(x))=Y\_(x), \[\^\_u(t)\](Y\_(x))=(\[\^u\_t\]\^Y\_)(x) where $t\mapsto \varphi^u_t$ is the one parameter group of diffeomorphisms of $S^1$ defined by the integral curves $t\mapsto c^u_x(t)$ through $x$ of the vector field $u$ on $S^1$, that is, $\varphi^u_t(x)=c^u_x(t)$. Here $\varphi^\ast$ denotes the pull – back of $p-$forms.

Since the Hamiltonian flow of a smooth phase space 