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Let $X\to \Delta$ be a projective family, smooth over $\Delta^*$, such that all fibers over $t\in \Delta^*$ are isomorphic. Does the monodromy representation factor through the algebraic automorphism group of the smooth fiber, $Aut(X_t)$?

This is certainly false for non-isotrivial families, since Dehn twists on curves are infinite order, whereas smooth curves of genus $>1$ have finite automorphism groups.

Example 1: The family $y^2 = x^3+t$ has monodromy of order 6, which is precisely the automorphism group of the Eisenstein elliptic curve.

Example 2: The family of smooth quadric surfaces $\mathbb P^1\times \mathbb P^1$ degenerating to a quadric cone has monodromy of order 2, which corresponds to swapping the factors.

In the symplectic category, there is a notion of parallel transport, so we have a monodromy map $\pi_1(\Delta^*)\to Symp(X_t)/Ham(X_t)$. Perhaps if the family is algebraically isotrivial, then the monodromy is valued in $Aut(X_t)$ and is homotopy invariant, so there is no quotient?

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  • $\begingroup$ I think you want also the total space $X$ to be smooth. Otherwise you can pull back your family by $t\mapsto t^n$ and get a monodromy of arbitrarily high order. $\endgroup$ – abx Aug 21 '18 at 6:53
  • $\begingroup$ It seems to me that such a base change lowers the order. $\endgroup$ – François Aug 21 '18 at 7:49
  • $\begingroup$ Oops! You are right. The "new" monodromy is $T^n$, which is indeed of lower order. $\endgroup$ – abx Aug 21 '18 at 8:01
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I think the answer is yes, depending on how you are defining the monodromy*. I take "isotrivial" to mean that you have a smooth fibre bundle over the punctured plane where the fibres have complex structures and any two fibres are biholomorphic. Take the covering space of $X/X_0$ corresponding to the subgroup of $\pi_1$ given by the kernel of the projection to $\pi_1$ of the base ($\mathbb{Z}$). This is now an isotrivial family over $\mathbb{C}$, which I claim is trivial (see below). The deck group is $\mathbb{Z}$ acting as a translation in $\mathbb{C}$ coupled with a biholomorphism of the fibre, and this automorphism is the monodromy (in particular, you see it's an automorphism).

In general, given a fibre bundle where the complex structure varies, you get a map from the base to $J/Diff$, where J is the space of all complex structures. In the isotrivial case, this is the constant map, so lifts to a fibre. The fibre is the orbit of a complex structure under the action of the diffeomorphism group, which is $Diff/Aut$ (as $Aut$ is the stabiliser). In particular, if the base is contractible then this map lifts to a map from the base to $Diff$, which you can use to pullback the complex structure in each fibre to get the trivial family of complex manifolds. If the base is $S^1$ (equivalently $\mathbb{C}^*$) then this map lifts to a map from the interval into $Diff$ whose endpoints map to $Aut$ (starting at the identity). Again, you can think of the automorphism at the endpoint 1 as the monodromy.

*For instance, symplectic monodromy will depend (up to Hamiltonian isotopy) on the precise choice of loop in the base and the choice of symplectic form.

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  • $\begingroup$ Thanks for your answer, Jonny Evans! I'm a little concerned that there is a detail missing because the second paragraph doesn't use compactness of the fibers at all. There are non-trivial holomorphic families of $\mathbb C^2$'s over a contractible base, for example. $\endgroup$ – François Aug 28 '18 at 18:17
  • $\begingroup$ Possibly I'm missing some assumption about the existence of a suitable fine moduli space? I'll have a think. $\endgroup$ – Jonny Evans Aug 28 '18 at 21:57
  • $\begingroup$ Ah ha, if your fibres are compact then there is a theorem of Grauert and Fischer tells you that your family is a locally trivial bundle (which I guess in my head is what I was assuming isotrivial meant, despite what I wrote). As your question was about projective families, this should be enough. Indeed, an easier way to say what I said would be: pick local trivialisations and your monodromy ends up being a composition of biholomorphic transition maps, hence it's biholomorphic. $\endgroup$ – Jonny Evans Aug 28 '18 at 22:08

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