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Let $S$ be a smooth complex algebraic variety, let $b$ be a closed point of $X$, and let $f:X\to S$ be a polarized family of smooth projective varieties over $S$. This induces a monodromy representation $$\pi_1(S) \to \mathrm{Aut}(\mathrm{H}^i_{prim}(X_b,\mathbb Z))\subset \mathrm{GL}_N(\mathbb C).$$ (Here $N$ is the dimension of $\mathrm{H}^i$, and $\pi_1(S)$ denotes the topological fundamental group of $S(\mathbb C)$.)

Clearly, this representation is not "quasi-unipotent".

But can we find a set of generators, say $g_1,\ldots, g_n$ of $\pi_1(S)$, such that each $g_i$ acts quasi-unipotently?

The problem is maybe clearest to see when $S$ is projective. In this case, there is no "boundary", so how to get the set of generators to be quasi-unipotent?

PS. I could not extract what I'm asking from Quasi-unipotent monodromy for general families unfortunately.

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    $\begingroup$ The answer is no, in general. If S is a compact Shimura variety which is a closed subvariety of the moduli of ppav with "high" level structure, then you get a family of abelian variety parametrised by S. The level structure ensures that there are no torsion elements in the fundamental group of S, and compactness of S ensures there are no unipotent elements in $\pi _1(S )$ $\endgroup$
    – Venkataramana
    Commented Jun 14, 2017 at 12:00
  • $\begingroup$ @Venkataramana Thank you for your comment. I think I'm missing the argument for why "compactness of $S$ ensures $\pi_1(S)$ has no unipotent elements". How does one prove that? $\endgroup$
    – Randy
    Commented Jun 14, 2017 at 12:02
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    $\begingroup$ the Shimura variety S is of the form $\Gamma\backslash G/K$ where $G$ is a Hermitian type real semi-simple group and $K$ its maximal compact subgroup. Thus, the compactness is equivalent to the compactness of $\Gamma\backslash G$. If $\Gamma $ s a co-compact arithmetic subgroup of $G$, then $\Gamma$ cannot contain unipotent elements (this is the easier part of the Godement criterion; you can find it in Raghunathan's book on discrete subgroups of Lie groups). $\endgroup$
    – Venkataramana
    Commented Jun 14, 2017 at 15:55
  • $\begingroup$ @Venkataramana This is a really nice example! $\endgroup$
    – Daniel Litt
    Commented Jun 14, 2017 at 17:29
  • $\begingroup$ @Dan Litt Thank you! $\endgroup$
    – Venkataramana
    Commented Jun 14, 2017 at 20:16

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