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Let $P$ denote a parabolic subgroup scheme of $\operatorname{Sp}(2n;F)$, where $F$ is a field (I am interested in $K=\mathbb{Q}_p$ so possibly okay to assume local with characteristic $0$ if it makes life easier).

Let $X$ be a smooth quasi-projective scheme over $F$, and let $T\longrightarrow X$ denote a principal $P$-bundle over $X$, i.e. a $P$-torsor over $X$. By definition, $P$ is étale locally trivial.

Question: Is $P$ also Zariski locally trivial?

A reference or a proof would be appreciated!

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    $\begingroup$ Considering the Levi decomposition reduces you to a Levi. But, the Levis of symplectic groups are themselves symplectic groups times general linear groups. But, both symplectic groups and general linear groups are special (see en.m.wikipedia.org/wiki/Special_group_(algebraic_group_theory) ). $\endgroup$ Commented Oct 14 at 12:14
  • $\begingroup$ Amazing, thanks! Is there an easy way to see what you said about the structure of the Levi? $\endgroup$
    – kindasorta
    Commented Oct 14 at 13:01
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    $\begingroup$ Re, the isogeny type of a Levi subgroup can be read off from the Dynkin diagram by deleting nodes, and derived subgroups of Levi subgroups of simply connected groups are simply connected. This almost pins down the structure; to get the rest of the way there I think you have to actually sit down and compute with root lattices. (Or you can look at the centraliser of a split torus: its eigenspaces are either self-dual, in which case you get a symplectic factor, or come in dual pairs, in which case you get a general linear factor.) $\endgroup$
    – LSpice
    Commented Oct 14 at 14:51
  • $\begingroup$ I wish I could accept an amalgam of both your answers $\endgroup$
    – kindasorta
    Commented Oct 14 at 15:01

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Here is a more general answer than that which I suggested in the comments.

Throughout the following let us fix a field $F$.

Definition (Serre): A linear algebraic group $G$ over $F$ is special if for every finite type reduced $F$-scheme $X$, any $G$-torsor $T\to X$ is trivializable Zariski locally on $X$.

Example: For any symplectic space $(V,\psi)$ over $F$, the group $G=\mathrm{Sp}(V,\psi)$ is special. Let me sketch the proof. Symplectic vector bundles on $X$ (i.e., a vector bundle $\mathscr{E}$ together with a symplectic pairing $\omega\colon \mathscr{E}\otimes_{\mathscr{O}_X}\mathscr{E}\to\mathscr{L}$) form a stack for the fppf topology. As $G=\mathrm{Aut}(V\otimes_F\mathscr{O}_X,\psi\otimes 1)$ we see that $G$-bundles on $X$ correspond to isomorphism classes of symplectic bundles $(\mathscr{E},\omega)$ which are fppf locally isomorphic to $(V\otimes_F\mathscr{O}_X,\psi\otimes 1)$. Thus, it suffices to show such an isomorphism actually happens Zariski locally.

By a standard spreading out argument (or something more general like [Č1, Lemma 2.1]) it suffices to treat the case when $X=\mathrm{Spec}(R)$ for a local ring $R$. In this case both $\mathscr{E}$ and $\mathscr{L}$ are trivial, and so we’re essentially claiming that (up to isomorphism) the only symplectic pairing on $R\otimes_F V$ is $\psi\otimes 1$. This is a simple exercise which I leave to you. $\blacksquare$

Thus, the answer to your question is yes by the following observation.

Proposition: If $G$ is a special reductive group, then so is any parabolic $P\subseteq G$.

Proof: Let $\mathcal{Q}$ be a $P$-torsor on some reduced finite type $F$-scheme $X$. As in the example we may assume that $X=\mathrm{Spec}(R)$ where $R$ is a local ring. By assumption $\mathcal{Q}\times^P G$ is trivializable. Thus, we are reduced to showing that the kernel of $$H^1(X_\mathrm{fppf},P)\to H^1(X_\mathrm{fppf},G)$$ is trivial. But, observe that $P_X$ is a parabolic group scheme of $G_X$, and so this always holds by the following lemma (which should be more well-known). $\blacksquare$

Lemma (see [Č2, §1.3.5]): Let $S=\mathrm{Spec}(R)$ with $R$ semi-local, and $H$ a reductive group $S$-scheme. Then, for any parabolic subgroup $S$-scheme $Q\subseteq H$ the map $$H^1(S_\mathrm{fppf},Q)\to H^1(S_\mathrm{fppf},H)$$ is injective.

Up to injectivity vs. trivial kernel issues, this amounts to the questions of why $H(R)\to (H/Q)(R)$ is surjective, but this follows from Bruhat-like decompositions (see the references in loc. cit.).

——

Of course, if one wants to avoid such general/abstract results, one can apply the ideas from the comments. Namely, if $L$ is the Levi factor of $P$, then for affine $X$ the natural map

$$H^1(X_\mathrm{fppf},L)\xrightarrow{\sim} H^1(X_\mathrm{fppf},P).$$

This is essentially because the difference between $P$ and $L$ is the (split) unipotent group $R_u(P)$ which is an iterative extension of $\mathbb{G}_a$’s, which (as being just the structure sheaf), have vanishing cohomology on affines (see loc. cit. for precise references).

But then, as Loren explains in the comments, when $G=\mathrm{Sp}(V,\psi)$ then the only possibility for $L$ is that it is of the form $\mathrm{GL}(W)\times \mathrm{Sp}(V’,\psi’)$ for some $W$ and some $(V’,\psi’)$. But, obviously for the general linear group, and by the example above for the symplectic group, both of these groups are special, and thus evidently so is their product.

References:

[Č1] Česnavičius, K., 2015. Poitou–Tate without restrictions on the order. Mathematical Research Letters, 22(6), pp.1621-1666.

[Č2] Česnavičius, K., 2022. Problems about torsors over regular rings: With an appendix by Yifei Zhao. Acta Mathematica Vietnamica, 47(1), pp.39-107.

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  • $\begingroup$ The spreading out argument is beautiful! Thank you very much, I had no idea such a thing existed. $\endgroup$
    – kindasorta
    Commented Oct 16 at 8:10

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