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May 20, 2022 at 1:26 comment added Will Sawin You'll want the map from the base to the moduli space of the fiber to be constant, for example if the fiber has no deformations, or if the moduli space and base have contrasting geometric properties (frequently if the base is a rational curve). After that, it becomes a question about the automorphism group of the fiber.
May 19, 2022 at 23:07 comment added R. van Dobben de Bruyn What do you mean by "fibre bundle in usual complex topology": a topologically locally trivial fibre bundle, or a holomorphically locally trivial fibre bundle? In the former case, the fibres need not even be isomorphic as complex manifolds (see e.g. Ehresmann's theorem, which implies that any smooth projective morphism of (smooth) varieties is a $C^\infty$ fibre bundle, but of course the holomorphic/algebraic structure can vary).
May 19, 2022 at 21:55 history edited Sean Lawton CC BY-SA 4.0
Minor edits.
May 19, 2022 at 21:09 comment added LSpice Name of @DanielLitt's reference: Serre - Éspaces fibrés algébriques.
May 19, 2022 at 21:06 history edited LSpice CC BY-SA 4.0
Typo in title; deleted "thanks"
May 19, 2022 at 19:23 comment added Daniel Litt For principal G-bundles, this is related to G being a "special group" in the sense of Serre; the original reference is here: numdam.org/item/SB_1951-1954__2__305_0
May 19, 2022 at 19:19 comment added Daniel Loughran @tota: The function field of $Y$ is not algebraically closed. This is the relevant field, not $\mathbb{C}$.
May 19, 2022 at 18:19 answer added Sean Lawton timeline score: 4
May 19, 2022 at 16:37 comment added abx You might have a look at this question and the answers there.
May 19, 2022 at 15:04 history asked tota CC BY-SA 4.0