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Let $X$ be a projective manifold with $\dim_{\mathbb{C}} X \geq 3$. Assume $X$ is the total space of a fiber space, i.e., there is a proper surjective holomorphic map $f : X \to Y$ with connected fibers. The discriminant locus of $f$ is the set of all points $y \in Y$ such that $f^{-1}(y)$ is singular. We will assume $Y$ is a normal irreducible projective variety with $0 < \dim_{\mathbb{C}} Y < \dim_{\mathbb{C}} X$.


The following general question interests me:

Given a divisor $D \subset Y$, can we construct a fiber space $f : X \to Y$ whose discriminant locus is $D$?


This question is too big, so I will ask a more specific question:

Can we construct an example of a fiber space $f : X \to Y$ where the codimension-one part of the discriminant locus does not have normal crossings?


Notice that if we look at surfaces $\dim_{\mathbb{C}} X =2$, then the discriminant locus is a finite set of discrete points on a curve.


Edit: I would like to add that I would appreciate as many examples/references as one has. I think it is worth cultivating a bank of examples.

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    $\begingroup$ Just take any fiber bundle over $Y$ of fiber dimension $\geq 1$, then take a section over $D$, then form the blowing up of the total space along the image of $D$ under the section, and finally resolve any singularities of the new total space. $\endgroup$ Aug 20, 2020 at 9:28

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Let me just give an example. Let $V$ be a vector space, set $Y = \mathbb{P}(S^2V^\vee)$, and let $$ X \subset \mathbb{P}(V) \times \mathbb{P}(S^2V^\vee) $$ be the universal quadric, i.e., the natural divisor of bidegree $(2,1)$. The projection $X \to \mathbb{P}(V)$ is a projectivization of a vector bundle, hence $X$ is smooth. On the other hand, the projection $$ X \to Y = \mathbb{P}(S^2V^\vee) $$ is a quadric bundle, its discriminant divisor $D \subset \mathbb{P}(S^2V^\vee)$ is the symmetric determinantal hypersurface, and it is not a normal crossing divisor (it is normal, but singular) when $\dim(V) \ge 3$.

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  • $\begingroup$ Thank you for the example. I'm not familiar with the construction, do you have any references? Or perhaps, even a heuristic for constructing such examples? $\endgroup$
    – AmorFati
    Aug 20, 2020 at 6:15
  • $\begingroup$ Essentially, there is no reason for the discriminant divisor to be normal crossing. So, if you take a general morphism $f$ it will give you an example similar to the above. $\endgroup$
    – Sasha
    Aug 20, 2020 at 10:17

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