What is the projective bundle over $(\Bbb P^1\times\Bbb P^1)/(\Bbb Z/2)$ which isn't $\Bbb P(\mathcal{E})$ for some vector bundle $\mathcal{E}$?

I'm thinking about when a (Zariski-locally trivial) $$\Bbb P^n$$-bundle comes from the projectivization of a vector bundle, and I've found Sasha's answer here on MO stating that $$(\Bbb P^1\times\Bbb P^1)/(\Bbb Z/2)$$ has a $$\Bbb P^1$$-bundle which isn't the projectivization of a vector bundle. Unfortunately, it doesn't say what this bundle is. Does anyone know an explicit description of what the intended bundle is, and an easy way to see it's not the projectivization of a vector bundle?

• The variety $B= (\mathbb{P}^1 \times \mathbb{P}^1)/(\mathbb{Z}/2)$ is the base of the bundle, not the total space. Feb 5 '21 at 0:41
• @EricWofsey Yikes, thank you for the correction. Feb 5 '21 at 0:51
• @HankScorpio Have you read the paper that Sasha refers to in the comments of his answer? Feb 5 '21 at 6:15

Let $$X = (\mathbb{P}^1 \times \mathbb{P}^1)/(\mathbb{Z}/2)$$. This is a toric surface with 4 ordinary double points. Let $$\pi \colon Y \to X$$ be the blowup of one of these points and let $$E \subset Y$$ be the exceptional curve. Then $$E^2 = -2$$ and one has $$Ext^1(\mathcal{O}_Y, \mathcal{O}_Y(E)) = H^1(Y, \mathcal{O}_Y(E)) \cong H^1(Y, \mathcal{O}_E(E)) \cong \Bbbk.$$ Let the vector bundle $$F$$ on $$Y$$ be defined as the non-trivial extension $$0 \to \mathcal{O}_Y(E) \to F \to \mathcal{O}_Y \to 0.$$ Then it follows from the construction that $$F\vert_E \cong \mathcal{O}_E(-1)^{\oplus 2}.$$ Consider the projective bundle $$\mathbb{P}_Y(F)$$, the surface $$S = \mathbb{P}_E(F\vert_E) \subset \mathbb{P}_Y(F)$$, and the composition $$\mathbb{P}_Y(F) \to Y \to X.$$ One can check that this composition has another factorization $$\mathbb{P}_Y(F) \to Z \to X,$$ where the first morphism contracts the surface $$S$$ and the second morphism is a $$\mathbb{P}^1$$-bundle, which is not isomorphic to the projectivization of a vector bundle.
EDIT. Here is an explanation why $$Z \to X$$ is not a projectivization.
Let $$x_0 = \pi(E)$$ and let $$C \subset Z$$ be the fiber of $$Z \to X$$ over $$x_0$$. If $$Z \to X$$ is a projectivization, there is a line bundle $$L$$ on $$Z$$ such that $$L\vert_C \cong \mathcal{O}(1)$$. Then the pullback of $$L$$ to $$\mathbb{P}_Y(F)$$ restricts to $$S \cong E \times \mathbb{P}^1$$ as $$\mathcal{O}(0,1)$$. On the other hand, the Grothendieck bundle of $$\mathbb{P}_Y(F)$$ restricts to $$S$$ as $$\mathcal{O}(1,1)$$. From this we conclude that there is a line bundle on $$Y$$ that restricts to $$E$$ as $$\mathcal{O}(1)$$, but this is impossible, because $$Pic(Y)$$ is generated by $$\pi^*Pic(X)$$ (which restricts to $$E$$ trivially) and by $$\mathcal{O}_Y(E)$$ (which restrict to $$E$$ as $$\mathcal{O}(-2)$$).
• While I appreciate the answer, I don't really understand much of what's going on (maybe this is on me, I am still in my first course in AG). In particular, the linked section of the paper seems to treat computing some Brauer class map which I can't quite connect with the claims you make (how does this show that $Z\to X$ is not a projectivization, for instance?). Feb 5 '21 at 7:59