I am going through Variations on a Theme by Kepler, Guillemin and Sternberg.
It is stated that a symplectic transformation on a manifold $\phi:X \rightarrow X$ induces a transformation on function. $$f \rightsquigarrow f \circ \phi^{-1} $$
They say: It turns out that every function $f$ on phase space generates a (local) one-parameter group, $\phi_t$, of symplectic transformations such that $$\{f, g\} = \dfrac{d}{dt}g \circ \phi_{-t}\rvert_{t=0} $$
My understanding is that they are referring to the flow of the Hamiltonian vector field $X^f = \{f, \cdot\}$ $$\phi_t = \exp(tX^f) $$
I am however confused as to how everything fits together. We have for the flow $\rho$ of a complete vector field $X$ that: $$\dfrac{d\rho_t(p)}{dt} = X(\rho_t(p)) $$
and so $$X^f(g) = \{f, g\} = \dfrac{d}{dt}g \circ \phi_{-t}\rvert_{t=0} $$
Intuitively, it seems to me that there is a minus sign error, indeed we are considering the change in $g$ as we travel against the flow, and I would expect this to be $-X^f(g)$
I am asking this because I am trying to understand the statement "left Poisson bracket by $f$ is the infinitesimal symmetry associated to $f$" and how the above statements fit together
EDIT:
According to Wikipedia, some authors define the Hamiltonian vector field with the opposite sign. My best answer is this must be the case here.
$$-\{f, \cdot\} = X^f
$$
Then indeed
$$X^f(g) = -\{f, g\} = -\dfrac{d}{dt}g \circ \phi_{-t}\rvert_{t=0}
$$
This can also be checked with the fact the Lie derivative for functions is $\mathcal{L}_X f = X(f)$ along with the general definition of the Lie derivative of a k-form:
$$\mathcal{L}_X \omega := \dfrac{d}{dt}(\exp(tX))^*\omega \Big \rvert_{t=0} $$
giving $$\mathcal{L}_X f = \dfrac{d}{dt}f \circ(\phi_t)) \Big \rvert_{t=0} = X(f) $$
In this way, for a function $f$, $-\{f, \cdot\}$ = $X^f$ is the infinitesimal generator for the one-parameter family of symmetry transformations on the manifold $$\phi_t:X\rightarrow X$$ $$\phi_t = \exp^{tX^f}$$
while $\phi^*_{-t}$ is the induced transformation on a function $g$
Is it thus correct to say the following ? $$\dfrac{d}{dt}\exp(tX)\Big\rvert_{t=0}f = X(f) = \dfrac{d}{dt}f\circ\exp(-tX)\Big\rvert_{t=0} $$