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Let $\pi: X \to \mathbb{P}^n$ be a conic bundle over an (algebraically closed) field $k$. Let $g \in Aut(X)$ so that $g$ preserves the fibres of $\pi$. Clearly $g$ lives inside $PGL_3(k(\mathbb{P}^n))$ since it it determined by the action on the generic fibre. I think that in fact, $g$ must live inside $PGL_3(k)$. Is this true? If not, is it true if we further assume that $g$ has finite order?

My attempt is to look at the morphism $\mathbb{P}^n \to \mathbb{P}^8$ assigning to each point $x \in \mathbb{P}^n$ of the base, the corresponding matrix $g_x$ acting on the conic $\pi^{-1}(x)$. Then if the image has dimension greater than $0$ (I.E. $g$ is not from $PGL_3(k)$) it must intersect the divisor $D \subset \mathbb{P}^8$ cut out by the determinant equation, so that $g$ has vanishing determinant somewhere. Is this sufficient? I was worried about the existence of a family of projective matrices over $\mathbb{P}^n$ which are invertible everywhere.

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    $\begingroup$ The fiber of a conic bundle at the generic point $\eta $ is not in general isomorphic to $\mathbb{P}^1_{k(\eta )}$, hence its automorphism group is not $\operatorname{PGL}(2,k(\eta )) $. $\endgroup$
    – abx
    Commented Jun 28, 2023 at 10:01
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    $\begingroup$ For a proper, geometrically connected, smooth curve of arithmetic genus $0$, the least degree of a very ample invertible sheaf is at most $2$, and the dual of the dualizing sheaf is a very ample invertible sheaf of degree $2$ whose associated closed immersion is in $\mathbb{P}^2$. Thus, the automorphism group is a closed subgroup of $\text{Aut}(\mathbb{P}^2)$, but this is $\textbf{PGL}_3$, not $\textbf{PGL}_2$. Perhaps that is your mistake. $\endgroup$
    – Jason Starr
    Commented Jun 28, 2023 at 10:53
  • $\begingroup$ I'm sorry, I meant to use $PGL_3$ instead. Also, if the discriminant curve is smooth and sufficiently high genus, it looks like this group is either trivial or $C_2$. Is that correct? $\endgroup$
    – TCiur
    Commented Jun 29, 2023 at 5:55
  • $\begingroup$ Even if the discriminant is smooth of high degree, there could still be nontrivial automorphisms: think about blowing up the restriction over a high degree hypersurface of a (constant) cross-section of a product conic bundle. If you add the hypothesis that the family is minimal, that will probably reduce the automorphism group to a finite group (by forcing the global sections of the tangent bundle to vanish). $\endgroup$
    – Jason Starr
    Commented Jun 30, 2023 at 10:48

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