Are all exact Lagrangian spheres, vanishing cycles?

Question

Let $\pi: E \to D$ be an exact Lefschetz fibration with corners (fibers with boundary)over the disk. Fix a point $\theta \in \partial D$ and consider the fiber $F_\theta = \pi^{-1}(\theta)$ over that point. Then the vanishing cycles in $F_\theta$ are exact Lagrangian spheres. By vanishing cycle we mean a sphere embedded in the fiber that degenerates to a point along a path that joins $\theta$ with a critical point of $\pi$.

As pointed out by Paul Seidel in his book, the fact that vanishing cycles are exact Lagrangian spheres is trivial except when $\dim E = 4$ (see 16b, page 221 of "Fukaya categories and Picard-Lefschetz theory"). A bit later in that same chapter is pointed out that any exact Lagrangian sphere in an exact Symplectic manifold $F$ appears as the vanishing cycle in some Lefschetz fibration that has $F$ as one of its fibers.

Now my question: assume $\dim E = 4$ (I'm not sure if this is relevant but maybe it simplifies it in terms of the framings of the Lagrangian spheres). Assume also that we fix the Lefschetz fibration $\pi: E \to D$. Suppose that the set of vanishing cycles generates the homology $H_1(F_\theta)$. Let $V \subset F_\theta$ be an exact Lagrangian sphere (in this case, a simple closed curve).

Is $V$ a vanishing cycle? That is, does necessarily exist a path in $D$ joining $\theta$ with some critical point that has $V$ as its vanishing cycle?

Answer 1

No, this is not necessarily true. In fact, this is never the case if $E = B^4$, and $F$ has connected boundary and genus $g \ge 2$.

Since $H_1(E) = 0$, the vanishing cycles span $H_1(F)$. An Euler characteristic computation tells you that the Lefschetz fibration has $2g$ critical points. Choosing a set of paths in $D$ from a basepoint to the critical points, we get a set $C$ of $2g$ vanishing cycles, and Dehn twists along them generate a subgroup $\Gamma$ of the mapping class group of $F$. Every other vanishing cycle is in the orbit $\Gamma\cdot C$, so that in particular the subgroup generated by all vanishing cycles is $\Gamma$ itself. But Humphries proved that the mapping class group of a surface of genus $g \ge 2$ and with one boundary component cannot be generated by fewer than $2g+1$ Dehn twists, so $\Gamma$ is a proper subgroup of the mapping class group of $\Gamma$, so there has to be a simple closed curve which is not a vanishing cycle. (I believe that the reference for Humphries' result is: Humphries, Generators for the mapping class group, 1979.)

The torus knots $T(2,2g+1)$ display this phenomenon very concretely: there is a Lefschetz fibration on $B^4$ whose fibres have genus $g$ and whose boundary has an open book with binding $T(2,2g+1)$. (This comes from Morsifying the function $f\colon \mathbb{C}^2 \to \mathbb{C}$ mapping $(x,y)$ to $x^2+y^{2g+1}$.) The surface of genus $g$ is a plumbing of $2g$ bands, and the vanishing cycles can be taken to be the cores of these bands. But the subgroup they generate is a proper subgroup of the mapping class group: the proof by authority is that there is a well-known presentation that includes these generators but also has an extra Dehn twist.