Prescribing the discriminant locus of fiber spaces

Question

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 $0 < \dim_{\mathbb{C}} Y < \dim_{\mathbb{C}} X$.

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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 $D$?

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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?

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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.

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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.

Answer 1

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 $\dim(V) \ge 3$.