Pullback of a Hamiltonian

I understand that a Hamiltonian vector field $H$ creates a Hamiltonian flow $\phi_t$. Now, in order to prove that the Hamiltonian is conserved one uses the following \begin{eqnarray*} \frac{d}{dt}\phi_t^{*} H &=& \phi_t^* \mathcal{L}_{X_H} \\ &=& \phi_t^* i_{X_H}dH \\ &=& \phi_t^* i_{X_H}^2 \omega \\ &=& 0\ \end{eqnarray*} but I do not quite have clear what this $\phi_t^*$ is. It seems to be some kind of pullback of $H$. But to what? Also, why is the first equality true?

Finally, I was wondering how if the fact that the Hamiltonian is conserved related to the fact that the symplectic volume is constant under symplectomorphisms. These symplectomorphisms drive the the flow, that is different points in the integral curve are related via symplectic transformations. So how is the Hamiltonian conversation related to the invariance of the symplectic volume (Liouville's theorem)?

The pullback of a function (thought of as a 0-form) is just pre-composition. That is, $\phi_t^*H = H \circ \phi_t$. So $\frac{d}{dt} \phi_t^*H = \frac{d}{dt}(H \circ \phi_t)$. But (pretty much by definition of how vector fields act on functions) this is equal to $vH$ where $v$ is tangent to the flow lines of $\phi_t$. By definition of $\phi_t$ as the flow of $X_H$, this means $v= X_H$, and so you have

$$\frac{d}{dt}(H \circ \phi_t) = X_H(H) = dH \, (X_H)$$

But remember that the Hamiltonian vectors fields are characterized by $\omega(\cdot, X_H) = dH$. So you finally get

$$\frac{d}{dt}(H \circ \phi_t) = dH(X_H) = \omega(X_H,X_H) = 0$$

The last equality is just the skew-symmetry of $\omega$. This derivative being zero means $H$ is constant along the flow lines of $\phi_t$.

• A question about convention: You state that "$\frac{d}{dt}(H\circ \phi)$..is equal to $vH$." Should that be $v(H)$, or is the identification of this with $vH$ conventional? Mar 27 '16 at 15:07
• By $vH$ I mean $v(H)$, the derivation $v$ applied to the function $H$.
– Nick
Mar 28 '16 at 18:47