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Let $f : X \to D$ be a proper flat holomorphic family of complete algebraic curves over a disk and the central fiber is a nodal curve with a unique singular point $p \in X_0$. Suppose that every fiber is smooth except the central fiber. I am interested in the topology of $X$, specifically $\pi_1(X)$. How much information about the homotopy type of $X$ can be computed in terms of the smooth punctured fibration $f : X^* \to D^*$.

There is an exact sequence,

$$ 1 \to \pi_1(\Sigma_g) \to \pi_1(X^*) \to \mathbb{Z} \to 1 $$

where $g$ is the genus of the fiber. This presents the fundamental group as a semi-direct product,

$$ \pi_1(X) = \pi_1(\Sigma_g) \rtimes \mathbb{Z} $$

via the monodromy action. My guess is there is some way to compute $\pi_1(X)$ using ideas in the vein of vanishing cycles and Milnor fibers of the singularity.

In the case $g = 1$, I am guessing there is a unique topological type of $X$ given the monodromy. However, for $g > 1$ it is not clear if $f : X^* \to D^*$ contains enough information to determine $\pi_1(X)$.

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    $\begingroup$ $X$ is homotopy equivalent to $X_0$. See for example mathoverflow.net/questions/264940/… $\endgroup$
    – Angelo
    Commented Apr 11, 2023 at 10:20
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    $\begingroup$ @Angelo The OP should specify that $X$ is smooth and $f$ is proper, otherwise there are silly counterexamples. $\endgroup$ Commented Apr 11, 2023 at 14:18
  • $\begingroup$ I do mean for $f$ to be proper and flat. @Angelo I was worried this might not be true if $X$ is not smooth. Is it still true? $\endgroup$
    – Ben C
    Commented Apr 11, 2023 at 17:15
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    $\begingroup$ If you allow $X$ to be singular, then there exists a pair of families that have $X_t$ the same for all $t\neq 0$, but where $X_0$ is either a cuspidal rational curve (thus simply connected) or a smooth elliptic curve (not simply connected). $\endgroup$ Commented Apr 11, 2023 at 17:49

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