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I am studying an example of symplectic Lefschetz fibrations. As far as I know, given a Weinstein manifold $F$ and a collection $V_1,\ldots,V_k$ of exact framed Lagrangian spheres of $F$, there exists a unique up to deformation Lefschetz fibration $f:W\rightarrow\mathbb{C}$ whose regular fiber is $F$ and whose vanishing cycles are $V_1,\ldots,V_k$.

Now if we apply this fact to the case where the fiber $F$ is the plumbing of a collection of $T^*S^n$ according to any tree $T$ and the collection $V_1,\ldots,V_k$ are the zero sections, then we get the corresponding Lefschetz fibration $f:W\rightarrow\mathbb{C}$.

My question is: Is it possible to write $W$ as an explicit complex affine variety (i.e. defined by some polynomials that we can write down) and $f$ as a map defined on the ambient space $\mathbb{C}^N$ (e.g. linear map)?

I have been looking at the example where our tree is $A_k$. I know the fiber is $F=A_k^{2n}=\{x_1^2+\cdots+x_n^2+t^{k+1}-1=0\}$, but I think we cannot take $f$ to be $\mathbb{C}^{n+1}\rightarrow\mathbb{C}:(\mathbf{x},t)\mapsto x_1^2+\cdots+x_n^2+t^{k+1}-1$ because it is not Morse. I don't know if we can perturb this polynomial so that it becomes Morse, but then will the fiber still be $A_k$?

Thanks in advance.

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Yes, it's possible for the specific case that you are looking at. For $k=1$, this is obvious. For $k\geq2$, such a Lefschetz fibration can be constructed by applying a standard construction to the standard Lefschetz fibration on $\mathbb{C}^{n+1}$, which is called stabilization. Concretely, a stabilization to a Lefschetz fibration $\pi$ means to replace the smooth fiber of $\pi$ by attaching a critical handle along an unknot (in this case, a standard Legendrian $S^{n-1}$ on the boundary of the unit cosphere bundle), and then adding a critical point to the base. Notice that after attaching a critical value to the $T^\ast S^n$ you get an $A_2$ Milnor fiber since this amounts to attaching a critical handles to $D^{2n}$ along the Legendrian hopf link in $S^{2n-1}$. You can also start with the standard Lefschetz fibration on $\mathbb{C}^2$, so as to reduce the higher-dimensional stabilization construction to its original definition in four dimensions, and then viewing $\mathbb{C}^{n+1}$ as the result of a sequence of stabilizations applied to $\mathbb{C}^2$ (which means adding quadratic terms to the defining equation, which is linear, of $\mathbb{C}^2$).

In conclusion, for any $k$, there is a Lefschetz fibration on $\mathbb{C}^{n+1}$ whose fiber is symplectomorphic to the $A_k$ Milnor fiber. It's easy to generalize the above construction to other sphere plumbings. I hope this answers your question.

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  • $\begingroup$ Thank you for your answers which indeed solved my question. In fact, I just wanted to see why $W$ is (the completion) of the standard symplectic ball. I have searched the key construction you mentioned, the stabilization, and I find the fact that after doing a stabilization, say we get $W'$ from $W$, the former is symplectomorphic to the latter. I would like to ask, how this symplectomorphism changes the Liouville structures? I don't know what this map looks like. I have just heard that it comes from the cancellation of two handlebodies. Thanks again. $\endgroup$ – ChiHong Chow Oct 18 '17 at 9:39
  • $\begingroup$ @ChiHongChow Handle cancellation does not change the Weinstein structure. Probably the following paper due to Casals-Murphy is a good reference for illustrating the real Morse theoretical view point of symplectic Lefschetz fibrations: arxiv.org/abs/1610.06977 $\endgroup$ – YHBKJ Oct 18 '17 at 17:37

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