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

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.

• 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.
– Arty
Nov 29 '21 at 20:06
• 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. Nov 29 '21 at 20:41
• 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?
– Arty
Nov 30 '21 at 11:29
• 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 $Y\rightarrow W$.
– Arty
Nov 30 '21 at 11:32
• No, I am not saying this. There are examples of quadric surface bundles with a relative hyperplane but without points. Nov 30 '21 at 18:36