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 $\mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$.
Best Answer
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.