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Let X be a variety. Suppose $\mathcal{E_1}$ and $\mathcal{E_2}$ are two vector bundles on X. Is there an example such that $\mathbb{P}(\mathcal{E_1})$ and $\mathbb{P}(\mathcal{E_2})$ are isomorphic as varieties but not as $\mathbb{P}^n$-bundles over X?

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    $\begingroup$ To avoid silly examples, you want $X$ to be connected. $\endgroup$ Jul 27, 2017 at 23:32

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Here is one method of constructing many such examples.

Claim. Let $(X,H)$ be a projective variety, let $\phi \colon X \stackrel\sim\to X$ be an automorphism, and let $\mathscr L$ be a line bundle on $X$ with $P_H(\mathscr L,n) \neq P_H(\mathcal O_X,n)$ such that $\phi^* \mathscr L \not \cong \mathscr L$ and $\phi^* \mathscr L \not \cong \mathscr L^{-1}$. Define $\mathscr E_1 = \mathcal O_X \oplus \mathscr L$ and $\mathscr E_2 = \mathcal O_X \oplus \phi^* \mathscr L$. Then $\mathbb P(\mathscr E_1) \cong \mathbb P(\mathscr E_2)$ as varieties, but not as $\mathbb P^1$-bundles over $X$.

Example. Let $X = \mathbb P^1 \times \mathbb P^1$, $\phi$ the coordinate swap, and $\mathscr L = \mathcal O(1,0)$.

Example. Let $X = E$ be an elliptic curve, $\phi = [-1]$ the inversion, $P$ a point that is not $2$-torsion, and $\mathscr L = \mathcal O_X(P)$.

Proof of claim. Clearly $\phi$ can be enhanced to a commutative diagram (in fact, a pullback square) $$\begin{array}{ccc}\mathbb P(\mathcal O_X \oplus \phi^* \mathscr L) & \to & \mathbb P(\mathcal O_X \oplus \mathscr L)\\ \downarrow & & \downarrow \\ X & \stackrel \phi \to & X\end{array}$$ whose horizontal arrows are isomorphisms. This shows that $\mathbb P(\mathscr E_1) \cong \mathbb P(\mathscr E_2)$ as varieties.

Suppose that $\mathbb P(\mathscr E_1) \cong \mathbb P(\mathscr E_2)$ as $\mathbb P^1$-bundles over $X$ (equivalently, as varieties over $X$). The long exact sequence $$\ldots \to H^1_{\operatorname{\acute et}}(X,\mathbb G_m) \to H^1_{\operatorname{\acute et}}(X,\operatorname{GL}_2) \to H^1_{\operatorname{\acute et}}(X,\operatorname{PGL}_2) \to \ldots$$ shows that two vector bundles give the same $\operatorname{PGL}_2$-bundle (equivalently, the same projective bundle) if and only if they differ by a line bundle. Hence, there exists a line bundle $\mathscr M$ such that $$\mathcal O_X \oplus \mathscr L \cong \mathscr M \otimes \left(\mathcal O_X \oplus \phi^* \mathscr L\right) = \mathscr M \oplus (\mathscr M \otimes \phi^* \mathscr L).$$ By assumption, the Hilbert polynomials of $\mathscr L$ and $\mathscr O_X$ are not the same, so the factors in the decomposition $\mathcal O_X \oplus \mathscr L$ have different Gieseker slopes. Uniqueness of the Harder–Narasimhan filtration implies that $\mathscr M = \mathcal O_X$ and $\mathscr M \otimes \phi^* \mathscr L = \mathscr L$, or $\mathscr M = \mathscr L$ and $\mathscr M \otimes \phi^* \mathscr L = \mathscr O_X$. Both contradict our hypotheses. $\square$

I haven't thought about what conditions give $\mathscr L_1 \oplus \mathscr L_2 \cong \mathscr L_3 \oplus \mathscr L_4 \Rightarrow (\mathscr L_1, \mathscr L_2) = (\mathscr L_3, \mathscr L_4)$. My current proof uses stability, but there may well be an easier argument, perhaps dropping the hypothesis on Hilbert polynomials.

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  • $\begingroup$ Nice answer! Second paragraph of proof, should be suppose isomorphism $\endgroup$
    – Chen Jiang
    Jul 28, 2017 at 0:54
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    $\begingroup$ So, essentially you are saying that if we take a $\mathbb P^n$-bundle over $X$ and pull it back via an automorphism of $X$, then most likely we will get a different $\mathbb P^n$-bundle. However, this feels like that what we get is actually abstractly the same $\mathbb P^n$-bundle. In other words, I interpreted the original question, or perhaps I should say that the possibly more interesting question is whether there are $\mathbb P^n$-bundles whose ambient spaces are isomorphic, but they are essentially different, i.e., that there is no automorphism of $X$ that would take one to the other. $\endgroup$ Jul 28, 2017 at 6:11
  • $\begingroup$ @SándorKovács: that would be a very interesting question indeed. But I don't think it's what the OP asked. My interpretation comes from Grothendieck's philosophy of relative schemes; yours seems more in line with the philosophy surrounding MRC fibrations, starting from the total space $\mathbb P(\mathscr E)$. $\endgroup$ Jul 28, 2017 at 23:56
  • $\begingroup$ @R.vanDobbendeBruyn: I understood your interpretation and I wasn't saying that there is anything wrong with it. I also don't have any insight into what the OP has intended, I didn't mean to suggest that I did. Cheers. $\endgroup$ Jul 29, 2017 at 16:33
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    $\begingroup$ @R.vanDobbendeBruyn: I would actually suggest a different way to mark the difference between the two interpretations: There is an obvious action of the automorphism group of $X$ on the set of projective bundles over $X$. You showed that this action is non-trivial. There is also an equivalence relation given by being isomorphic as abstract schemes (and not as $X$-schemes). Clearly the equivalence given by the automorphisms of $X$ is finer. My suggested interpretation is asking whether they are indeed different equivalence relations. $\endgroup$ Jul 29, 2017 at 16:47

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