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Tony Pantev
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For a hyperplane section Z of X, when there exists its etale X-neighbourhood such that Z is a fibre of its morphism to a curve
I am not sure I understand the question. If you had such an etale neighborhood of $Z$, then the normal bundle of $Z$ in $U$ will be trivial. But the normal bundle to $Z$ in $U$ is isomorphic to the normal bundle of $Z$ in $X$, and $N_{Z/X}$ can never be trivial for a hyperplane section.
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Singular fibers of an elliptic fibered K3 surface.
If $S$ is elliptic, then the fiber and the section generate a sublattice in $Pic(S)$ which is isomorphic to $U$. So if $S$ is elliptic and $Pic(S) \cong U(k)$ you will have $k = 1$. Did you mean to ask for $S$ to be only genus one fibered and $Pic(S) \cong U(k)$ ?
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Quasi-unipotent monodromy for general families
Two classical references are the paper "Periods of integrals on algebraic manifolds III", Publ. Math. IHES 38 (1970) by Griffiths and the paper "Variation of Hodge structure: the singularities of the period mapping" Invent. math. 22 (1973) by Schmid. They in particular explain Borel's proof of the quasi-unipotency theorem that I mentioned above. There are many other modern references. For instance, you may want to take a look at the excellent book "Period mappings and period domains" by Carlson, Mueller-Stach, and Peters.
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Quasi-unipotent monodromy for general families
Ah, I noticed I missed an adjective in the comment - I was defining what it means for $G$ to be quasi-unipotent. I edited the answer to reflect this correctly. The unipotency of $mon$ is defined in a similar manner - we say that $mon$ is unpotent, when $G$ is a connected unipotent algebraic group, i.e. when when it coincides with its unipotent radical. The references are numerous and the applications are usually Hodge theoretic. The quasi-unipotency of a local systen can is also very useful when we compute cohomology.
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Question on K3 Surface
@alex24: You should accept Remke's answer. It correctly and completely answers your question.
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Question on K3 Surface
As Remke Kloosterman points out - I got my Hurwitz formula wrong. To get the canonical class of the double cover to be trivial, you need to take a branch divisor that is in the twice the anticanonical linear system of the rational elliptic surface. So you simply need to take two smooth fibers as your branch divisor. This gives a smooth K3. It is the fiber product of the rational elliptic surface and a $\mathbb{P}^{1}$ doubly covering your original $\mathbb{P}^{1}$ with branching at the two points over which the two smooth fibers sit.
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Question on K3 Surface
Of course, you are right! I don't know what I was thinking. Somehow I did the calculation in my head without writing and, of course, I got it wrong.
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