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


1 Answer 1


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 $H_t$.

  • $\begingroup$ 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, 2016 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, 2016 at 12:54
  • $\begingroup$ I get your point, thank you very much. $\endgroup$
    – trick1234
    Dec 9, 2016 at 2:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.