Is a homogeneous symplectic isotopy always Hamiltonian?

Question

The following statements are written in the Guillermou-Kashiwara-Shapira's paper "Sheaf quantization of Hamiltonian isotopies and applications to non-displaceability".

Let $M$ be a smooth manifold and $I$ an open interval of $\mathbb{R}$ containing the origin. Consider a homogeneous symplectic isotopy $\Phi:\dot{T}^*M\times I\rightarrow\dot{T}^*M$ where $\dot{T}^*M=T^*M-0_M$, that is, for each $t\in I$, $\phi_t=\Phi(\cdot,t)$ is a homogeneous symplectomorphism and $\phi_0=id_{\dot{T}^*M}$. Set $v_\Phi=\frac{\partial\Phi}{\partial t}:\dot{T}^*M\times I\rightarrow T\dot{T}^*M$, $f=\alpha_M(v_\Phi):\dot{T}^*M\times I\rightarrow\mathbb{R}$ where $\alpha_M$ is the canonical Liouville 1-form of $\dot{T}^*M$ and $f_t=f(\cdot,t)$. Then $v_\Phi=X_{f_t}$ where $X_{f_t}$ is a Hamiltonian vector field in $\dot{T}^*M$.

In this statement, I cannot understand why the identity $v_\Phi=X_{f_t}$ holds. In particular, I don't know how to use the homogenity condition.

Answer 1

From your other question: Any homogeneous symplectomorphism of tangent bundle $\dot{T}^*M=T^*M-0_M$ preserves the canonical Liouville form? you know that a homogeneous symplectomorphism is exact: $$ \forall t, \phi_t^* \alpha = \alpha. $$ Hence by differentiating with respect to t you get $$\mathcal{L}_{X_t} \alpha = 0$$ where the vector field $X_t$ is defined by $(d/dt)\phi_t = X_t\circ \phi_t$. By the Cartan formula, $\mathcal{L}_{X_t} \alpha = d(\iota_{X_t} \alpha) + \iota_{X_t}d\alpha$, and hence $$ \iota_{X_t}d\alpha = -d(\iota_{X_t} \alpha).$$ This means that $X_t$ is the hamiltonian vector field associated with the function $\iota_{X_t} \alpha$.