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? 



Projective bundle
Here is one method of constructing many such examples.



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> **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 $X$.
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**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.


