Projective bundle 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? 
 A: 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.
