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Let $\pi:X\rightarrow W$ be a morphism of smooth projective varieties over a field $k$ whose generic fiber is a smooth quadric, and let $r$ be the dimension of the fibers of $\pi$.

Does there always exists a rank $r+2$ vector bundle $f:\mathcal{E}\rightarrow W$ on $W$ such that $X$ can be embedded in $\mathbb{P}(\mathcal{E})$ as a divisor and $X\cap f^{-1}(w) = \pi^{-1}(w)$ for all $w\in W$?

When $r=1$ this is true. Indeed, $\omega_{X}^{-1}$ is relatively very ample and we can take $\mathcal{E} = \pi_{*}\omega_{X}^{-1}$. Does there exist a similar construction for $r\geq 2$?

In this paper

ARNAUD BEAUVILLE, Variétés de Prym et jacobiennes intermédiaires, Annales scientifiques de l’É.N.S. 4e série, tome 10, no 3 (1977), p. 309-391

A quadric bundle over $\mathbb{P}^2$ is defined as a smooth projective variety $X$ with a morphism $\pi:X\rightarrow\mathbb{P}^2$ whose fibers are isomorphic to $r$-dimensional quadrics.

According to Proposition $1.2$ there exists a vector bundle $\mathcal{E}\rightarrow\mathbb{P}^2$ of rank $r+2$ and a form $q\in H^0(\mathbb{P}^2,Sym^2\mathcal{E}(h))$ such that $X$ identifies with the zero locus of $q$ in the projective bundle $\mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$.

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No. For instance, take your favorite $\mathbb{P}^1$-bundle $Y \to W$ which is not a projectivization of a rank 2 vector bundle and set $$ X := Y \times \mathbb{P}^1. $$ This is a quadric surface bundle over $W$, but if there is an embedding $X \hookrightarrow \mathbb{P}(\mathcal{E})$, then its restriction to any fiber of $X$ over $\mathbb{P}^1$ would give a rational section of $Y \to W$, which is impossible.

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  • $\begingroup$ I just fund a paper by Beauville saying that the answer is positive for quadric bundles over $\mathbb{P}^2$, and his proof does not seem to be related to the fact that the base is $\mathbb{P}^2$. I added some explanation in my question. However, I do not understand why your example shouldn't fit Beauville's definition of quadric bundle. For $W = \mathbb{P}^2$ the quadric bundle $X\rightarrow W$ you considered is flat and seems to fit Beauville's definition. $\endgroup$
    – Arty
    Nov 29 '21 at 20:06
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    $\begingroup$ There are two possible interpretations of the word "quadric". One, is "an algebraic variety, isomorphic to a hypersurface of degree 2 in a projective space". The other is "an algebraic variety which after base change to an algebraic closure of the base field is isomorphic to a hypersurface of degree 2 in a projective space". Beauville is using the former interpretation, and I was using the latter. $\endgroup$
    – Sasha
    Nov 29 '21 at 20:41
  • $\begingroup$ If I understand correctly you are saying that the restriction of the relative hyperplane section of $\mathbb{P}(\mathcal{E})$ determines a point on each fiber of $Y\rightarrow W$ and hence gives a section or $W\rightarrow Y$. The existence of such section would imply that $Y\rightarrow W$ is locally trivial. Did I get it right? $\endgroup$
    – Arty
    Nov 30 '21 at 11:29
  • $\begingroup$ It seems that you construction is related to the fact that you can distinguish in family which one of the two $\mathbb{P}^1$'s giving the quadrics is the one coming from a fiber of $Y\rightarrow W$. $\endgroup$
    – Arty
    Nov 30 '21 at 11:32
  • $\begingroup$ No, I am not saying this. There are examples of quadric surface bundles with a relative hyperplane but without points. $\endgroup$
    – Sasha
    Nov 30 '21 at 18:36

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