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The following matter should be widely known (if true). I am sorry for my ignorance!

For the natural $n$, let $E(n)$ be the corresponding elliptic surface.

In the analytic world, there exists a well-known surgery

$$(E(n-2) \setminus N_f) \cup_{T^3} (E(2) \setminus N_F) \simeq E(n).$$

$N_F$ is the neighborhood of the general fiber.

See p.23, for example.

Does any analogue of this hold holomorphically? Of course, instead of the surgery, one expects some form of the degeneration.

One may see (cf. Section 5) that $E(n)$ degenerates to $E(n) \cup_{T^2} E(0)$. Is this the best one can do?

For example, is there no way to degenerate $E(n)$ into some union of $E(m)$, $m < n$? Thank you!

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    $\begingroup$ What exactly is $E(n)$? $\endgroup$
    – Sasha
    Commented Oct 14 at 6:51
  • $\begingroup$ @Sasha Dear Sasha, topologically, this is a fiber sum of $n$ copies of $E(1)$, i.e., of $n$ rational elliptic surfaces. I admit that there are several complex structures; I would be glad to have the degeneration for any of them. For example, as in the book of Gompf-Stipsitz, you can take $\pi: E(1) \to \mathbb CP^1$, and, then, pull-back along $z^n: \mathbb CP^1 \to \mathbb CP^1$. $\endgroup$
    – Ivan Karpov
    Commented Oct 14 at 11:41

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You can degenerate the covering $z^n \colon \mathbb{P}^1 \to \mathbb{P}^1$ to a morphism $$ \mathbb{P}^1 \cup \mathbb{P}^1 \to \mathbb{P}^1 $$ equal to $z^{n-2}$ and $z^2$ on the first and second copies of $\mathbb{P}^1$ in the left-hand side, respectively. Pulling back $\pi \colon E(1) \to \mathbb{P}^1$ along this degeneration, you obtain a degeneration of $E(n)$.

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    $\begingroup$ I don't think this is true. I agree you can do this with some maps of degrees $n,n-2,2$ but why those specifically? When I calculate I get that $z^n$ doesn't degenerate unless the two branch points collide, in which case it degenerates into a union of $n$ degree $1$ maps glued using a single degree $0$ map. $\endgroup$
    – Will Sawin
    Commented Oct 14 at 13:30
  • $\begingroup$ Sorry, I should confess I accepted the answer without enough reflection if I understand its contents properly. Now I see that I am not able to reproduce the sought-for degeneration of the base as well as @WillSawin Probably, you could kindly add some more explanations? $\endgroup$
    – Ivan Karpov
    Commented Oct 14 at 15:02
  • $\begingroup$ @IvanKarpov To me the most natural algebraic geometry interpretation of what you mean by $E(n)$ is any elliptic surface over a rational curve with arithmetic Euler characteristic $n$. So the question would be: does there exist a degeneration of an elliptic surface with Euler characteristic $n$ to the union of an elliptic surface with Euler characteristic $m$ and an elliptic surface with Euler characteristic $n-m$? $\endgroup$
    – Will Sawin
    Commented Oct 14 at 15:06
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    $\begingroup$ To this question, the answer is positive, and one just has to pullback any elliptic surface of arithmetic Euler characteristic $1$ by a map from the blowup of $\mathbb P^1 \times \mathbb A^1$ at $(\infty,0)$ (i.e. a degeneration of $\mathbb P^1$ to $\mathbb P^1 \cup \mathbb P^1$ to $\mathbb P^1$ that has degrees $m$ and $n-m$ on the two special fibers. The map $(t,z) \mapsto (z/(z-1))^m (1- tz )^{n-m}$ does the trick for that. $\endgroup$
    – Will Sawin
    Commented Oct 14 at 15:09
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    $\begingroup$ @WillSawin: Consider a stable map from a 2-component rational curve to $\mathbb{P}^1$ equal to $z^{n-2}$ on the first component and $z^2$ on the second. In the space of stable maps it can be deformed to a map of degree $n$ from $\mathbb{P}^1$ to $\mathbb{P}^1$, and then in the space of these maps (which is an open subset of a projective space) to the map $z^n$. $\endgroup$
    – Sasha
    Commented Oct 14 at 15:09

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