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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 $\phi_{\Omega'}=f$.

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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 $H_t$.

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  • $\begingroup$ Suppose that $\rho_t: F \to F$ is a family of diffeomorphism such that $\frac{d \rho_t}{dt}=X_t$ and $\rho_t \circ \phi_{\Omega} =f$. Do I understand correctly that the function $H_t$ is a function such that $dH_t = X_t \lrcorner\Omega \vert_F$? $\endgroup$ – trick1234 Dec 8 '16 at 12:07
  • $\begingroup$ I fixed a glitch in my formulas. To answer your question: yes, what you want to say is that $H_t$ should be a choice of time dependent Hamiltonian on $F$ such that $\rho_\epsilon \circ \phi_\Omega=f$. Following my sign conventions I think that there should be a minus sign in your last formula (but check this carefully). $\endgroup$ – Nikolaki Dec 8 '16 at 12:54
  • $\begingroup$ I get your point, thank you very much. $\endgroup$ – trick1234 Dec 9 '16 at 2:18

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