Suppose that $\pi:E \to D$ is a 4-dimensional Lefschetz fibration over a disk, 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 $\partial E=\pi^{-1}(\partial D)$ is a mapping torus $Y_{\phi_{\Omega}}$.

For the mapping torus, we can define a periodic orbit to be an integral curve of vector field $R$. I don't know whether the periodic orbit are non degenerate.

Can I do a perturbation $\Omega'$ of $\Omega$ such that the periodic orbits with period $\le d$ on the mapping torus $Y_{\phi_{\Omega'}}=\pi^{-1}(\partial D)$ are non-degenerate?

  • $\begingroup$ Just a comment that might spark a thought: $\phi_\Omega$ is a symplectomorphism w.r.t. $\Omega$, and periodic orbits of period $p$ are fixed points of $\phi^p_\Omega$. For a generic time-dependent Hamiltonian $H$ on $\pi^{-1}(1)$, the perturbed map $\psi_H\circ \phi_\Omega$ has only nondegenerate fixed points (where $\psi_H$ is the time-1 flow of the Hamiltonian isotopy). However, we can't recast a change $\Omega\mapsto\Omega'$ as an existence of $H$, because $\psi_H\circ\phi_\Omega$ is still a symplectomorphism w.r.t. the original $\Omega$. $\endgroup$ Aug 1, 2015 at 0:29

1 Answer 1


This is very rough and probably not too precise but here is what I think.

For a two-form $\omega$ on a manifold $M$ denote by $\text{null}(\omega_x)$ the vector subspace of the tangent space $T_xM$ defined as follows: $$\text{null}(\omega_x) = \left\{ \, W \in T_xM \,\, | \,\, \omega_x(W,v) = 0 \, \text{ for all } \, v \in T_xM \, \right\},$$ i.e. the null-space of $\omega$ contains all tangent vectors that are orthogonal to the whole tangent space $T_xM$. In the case when $M=E$ and the assumption that $\Omega$ is non-degenerate fiber-wise, $\dim{\text{null}(\Omega)} = 2$ or $0$. When the dimension of the null-space is $0$ then $\Omega$ is non-degenerate on $E$ and hence it is symplectic. Otherwise, it is symplectic on the fibers only. However, when restricted to the even-dimensional (more precisely three-dimensional) fibration over the unit circle, the null-space is one-dimensional in both cases, so it defines a line field.

Let $z = y_1 + i y_2$ be a notation for the points on the disc $D = \{ z \in \mathbb{C} \, : \, |z| < 1 + \varepsilon \}$. In polar coordinates we have $z = R e^{i\theta}$ where $R = |z|$ and $\theta = \text{Arg}(z) \in [0,2\pi)$. Define the functions $R^2(x) = |\pi(x)|^2$ and $\theta(x) = \text{Arg}(\pi(x))$. For $z \in D$ let $E_z = \pi^{-1}(z)$ be the fiber over $z$. Now, the lift of the standard symplectic form $2 \, dy_1 \wedge dy_2$ can be written as $$\pi^*(2 \, dy_1 \wedge dy_2) = d (R^2) \wedge d \theta.$$ Denote by $M_1 = \{ x \in E \, : \, |\pi(x)|^2 = 1 \}$. Take $$l(x) = \{ W \in T_xM_1 \, : \, \Omega_x(W,v) = 0 \, \text{ for all } \, v \in T_x E_{\pi(x)} \, \}.$$ This is a line field and is equal to $T_xE^h \cap T_xM_1 = l(x)$, i.e. $l(x)$ is a one dimensional line field tangent to $M_1$ and $\Omega$-orthogonal to the fibers. Observe that you can also write it as the null-space of $\omega_x = \Omega_x - d R^2 \wedge d \theta$. The lift of the vector field tangent to the unit circle in the $D$ lies on $l(x)$. Take any function $H : E \to \mathbb{R}$ and restrict it to $M_1$. If you look at the form $\Omega'_x = \Omega_x - d(R^2) \wedge \theta - d H \wedge d \theta$ restricted to $M_1$, then take its one dimensional null-space $l_H(x)$. It gives you a line field on $M_1$. The vector field on the unit circle lifts to a vector field whose span defines $L_H(x)$. Furthermore, its tangent trajectories give rise to a large family of symplecotmorphisms (parametrized by $H$) on the fiber $\pi^{-1}(1)$ by using the constriction of first return-map. Then, using the null-space of $\omega$, you can trivialize $M_1$ by creating the cover $p : E_1 \times \mathbb{R} \to M_1$ and as a result you obtain more or less something like a time-dependent (actually, $\theta$-dependent) Hamiltonian $\tilde{H}(y,\theta)$ where $(y,\theta) \in E_1 \times \mathbb{R}$ that is $2\pi-$periodic and gives rise to a symplectomorphism on $E_1$ (a version of the Poincare first return map). Then it looks like generically, it is possible that its periodic points are isolated.

So a deformation of $\Omega$ that gives rise to non-degenerate periodic orbits can be obtained by adding $d\theta \wedge \big(R^2 + H \big)$. Observe that although $\theta$ is not well-defined on the unit circle, its differential $d \theta$ is, because $d \theta = d(\theta + 2\pi)$. Furthermore, if you feel uncomfortable about $d \theta$ around the center of $D$, then multiply your whole perturbation $d\theta \wedge \big(R^2 + H \big)$ by an infinitely smooth bump-function which is constantly $1$ everywhere in $D$ except for a very small disc neighborhood around its center and it is zero at the center.

  • $\begingroup$ Thank you very much for your answer. Please allow me to ask one more question. Suppose that $R_H$ is the horizontal lift of $\partial \theta$ over $M_1$ in your construction, $\phi_H$ is the flow generated by $R_H$. By directly computation, I show that $R_H$ $C^k$ converges to the original one if $H$ $C^{k+1}$ converges to 0. Does this implies that flow $\phi_H$ $C^k$ converges to the original one, and the periodic points of time-1 map will be close to the original periodic points? $\endgroup$
    – Trick1234
    Aug 4, 2015 at 13:42
  • $\begingroup$ It looks like it. You can also have a one-parameter family, by taking $\varepsilon H$... By the way, since I do not know what kind of fibration you have, what its topology is and whether it is locally trivializable (something like Ehresmann's theorem), you have to make sure that the trajectories of the lifted vector fields make one full circle to return to the fiber $E_1$. If the fibers are compact, you have no troubles. Would you like to share some details about the fibration and what you need it for? There might be alternative ways for working with it... $\endgroup$ Aug 5, 2015 at 10:23
  • $\begingroup$ Just as I described above, a Lefschetz fibration $E$ over a disk with one singularity point and fiber $\Sigma$, where $\Sigma$ is a closed surface. The boundary $Y=\partial E$ is a fibration over $S^1$. Given $d>0$, I want to know that whether all the periodic orbit are non-degenerate, your answer basically solved my problem. Big thanks! $\endgroup$
    – Trick1234
    Aug 6, 2015 at 12:29
  • $\begingroup$ In fact, I am reading Michael Usher's paper arXiv:math/0603128. Suppose $F:E_d(\pi) \to D$ is relative Hilbert scheme of $E$, which carries some symplectic form $\tilde{\Omega}$, $Y_d(\pi)=F^{-1}(\partial D)$. I am just curious that can we establish a correspondence between non-degenerate periodic orbit of $Y_{\phi_{\Omega}}$ with degree $d$ and the constant section of $Y_d(\pi)$. Since $(Y_d(\pi),\tilde{\Omega})$ is isomorphic to a mapping torus $Y_{\Phi}$, where $\Phi$ is a $C^0$ perturbation of $Sym^d(\phi_{\Omega}):Sym^d(\Sigma) \to Sym^d(\Sigma)$, so intuitively it seems true. $\endgroup$
    – Trick1234
    Aug 6, 2015 at 13:24
  • $\begingroup$ I see, so if the fibers are compact, then everything is ok. This means, as I expected, that the fibration is locally smoothly trivializible, by Ehresmann's theorem. By the way, you can obtain the fibration as a mapping torus of a more general smooth map, not necessarily a symplectomorphism. It requires only constructions from differential topology, without reference to symplectic forms. Of course, if you need your isotopy class to be symplectic, then you use the form. But in general, if I am not wrong, for a generic singularity, I think the gluing map is isotopic to a Dehn twist. $\endgroup$ Aug 7, 2015 at 13:26

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