Question about the paper "Infinitely many monotone Lagrangian tori in del Pezzo surfaces"

Question

In the paper "Infinitely many monotone Lagrangian tori in del Pezzo surfaces" by Renato Vianna, the author constructs an infinite amount of non-symplectomorphic monotone Lagrangian tori in del Pezzo surfaces.

From what I understand, the basic idea is the following. Let's focus on the case of $\mathbb{C}\mathbb{P}^2$, so the Delzant polytope is a triangle and the central fiber is known to be monotone. Then we apply nodal trades to the vertices and let them slide through this monotone fiber. Then we want to visualize the fibration in another way, so that the new created fiber does not lie on the eigenline, and we apply a mutation so that we are able to visualize the desired fiber.

What I fail to understand is why will the fiber that will replace the previous monotone fiber be monotone? Is this a straightforward thing to see, or is there some depth to it? Any insight is appreciated.

Answer 1

There are two ways:

1. Show that doing a nodal slide across the central fiber is same as performing a zero-area Lagrangian surgery of the fiber with a disc given by parallel a circle transporting to the singular fiber.
2. I would guess that this is what Vianna had in mind. Vianna shows that performing three rational blowdowns (loosely, consider rational blowdown to be a nodal trade, but this actually changes the manifold since you're doing it at a non-Delzant corner) $\mathbb P (a^2,b^2,c^2) $ is a standard $\mathbb P ^2$ when $(a,b,c)$ is a Markov triple (Corr. 2.5 in the paper). Let $T(a,b,c)$ be the central torus fiber in the orbifold $\mathbb P (a^2,b^2,c^2)$, thus it is monotone. Now, by abuse of notation, call the torus obtained in $\mathbb P^2$, after performing the three rational blowdowns, $T(a,b,c)$, which is the Vianna tori corresponding to the Markov triple $(a,b,c)$. Now you check that performing a rational blowdown preserves monotonicity to finish off.