Suppose that $\pi:E \to D$ is a 4-dimensional Lefschetz fibration over a disk,(more general, Lefschetz fibration over a surface with boundary ) and let $\Omega$ be a closed 2-form on $E$ such that it is non-degenerate fiberwise. 


For any $x \in E$, there is a decomposition $TE\_x=TE\_x^h \oplus TE\_x^v$, $TE^v=\ker d\pi$ and $TE\_x^h=\{v \in TE\_x: \Omega(v,w)=0,\forall w \in TE\_x^v\}$. Let $\partial\_t$ be the coordinate vector field on $S^1=\partial D$, $R$ be the horizontal lift of $\partial\_t$ and $\phi\_{\Omega}$ be the flow generated by $R$. Taking base point $1 \in S^1$, then we get a map $\phi\_{\Omega}: \pi^{-1}(1)\to\pi^{-1}(1)$, so the boundary $Y=\pi^{-1}(\partial D)$ is a mapping torus $Y\_{\phi\_{\Omega}}$.



> 
> My Question: Suppose that $f: \pi^{-1}(1)\to\pi^{-1}(1)$ is a Hamiltonian perturbation of $\phi\_{\Omega}$, can we find a symplectic form $\Omega'$ on $\pi:E \to D$ such that $\phi\_{\Omega'}=f$ ?
> 
> 
> 


I tried to reason as follows, but I am not sure. 


Let $\omega$ be restriction of $\Omega$ on $Y=\pi^{-1}(\partial D)$ and $\lambda=\pi^\*(dt)$, then $(\omega,\lambda)$ form a stable Hamiltonian structure on $Y\_{\phi\_{\Omega}}$. In a neighborhood of $\partial E$, we can identify $(E,\Omega)$ with $(Y\times[0,-\epsilon), \omega+ds\wedge\lambda)$.


Suppose that $f: \pi^{-1}(1)\to\pi^{-1}(1)$ is a Hamiltonian perturbation of $\phi\_{\Omega}$, we can define a mapping tours $Y\_f$. Let $\omega\_f$ be unique closed two form on $Y\_f$ such that $\omega\_f=\omega$ on fiber. Since $f$ Hamiltonian isotopic to $\phi\_{\Omega}$, so $[\omega\_f]=[\omega] \in H^2(Y ,\mathbb{R})$. 


Finding a family of close two form $\omega\_s$ on $Y$ such that $\omega\_s=\omega\_f$ near $s=0$ and $\omega\_s=\omega$ near $s=-\epsilon$. Replacing $(Y\times[0,-\epsilon), \omega+ds\wedge\lambda)$ by $(Y\times[0,-\epsilon), \omega\_s+ds\wedge\lambda)$, then we obtain a new symplectic form $\Omega'$ on $E$ and I hope that $\phi\_{\Omega'}=f$.



Perburb the Monodromy of Lefschetz fibration over a disk
The answer to your question is yes, given that the Hamiltonian perturbation indeed is sufficiently small. Conceptually the idea of the construction is better formulated as follows: fix the symplectic form $\Omega$ and instead deform the *fibration* by a smooth (in general non-symplectic) isotopy in order to obtain a different monodromy. (If you instead insist on fixing the fibration, the symplectic form will be deformed by the pull-back under this smooth isotopy.)


First, construct a standard symplectic neighbourhood of $F=\pi^{-1}(1) \subset (E,\Omega)$ being the product $$(F \times [-\epsilon,\epsilon]^2,\Omega|\_F \oplus dx \wedge dy).$$
We can moreover use an identification in which all $F \times \{(x,0)\}$ are fibres $\pi^{-1}(z)$ for $z \in \partial D$ in the base. However, all fibres for $z \notin \partial D$ can only be assumed to be $C^\infty$-close to the symplectic surfaces $F \times \mathrm{pt}$ (equality for all fibres would imply flatness of the fibration).


Second, take a smooth isotopy which deforms $(u,(x,y))$ to $(u,(x,H\_x(\phi^t\_{H\_t}(u))+y))$ , $u \in F$, for a smooth function $H\_t \colon F \to [-\epsilon,\epsilon]$. It is readily checked that the monodromy around $\partial D$ is deformed by the corresponding Hamiltonian diffeomorphim $\phi^\epsilon\_{H\_t}$. Note that you have to interpolate this smooth isotopy to make it defined on all of $E$, and also do this in away so that the image of a fibre is still symplectic. The latter can be done, but I am a bit brief at this point. Also, you need to be careful concerning the support of $H\_t$.


