Let $X$ be a smooth variety and $D$ be a codimension 1 sub variety. Let $E$ be a rank 2 vector bundle on $X$. Let $L$ be a line bundle on $D$ which is quotient of $E|_D$. Then consider $Ker(E,L)$. Is it true that $\mathbb{P}(E)\cong \mathbb{P}(Ker(E,L))$?
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5$\begingroup$ What you describe is an "elementary modification". The associated projective bundles are typically isomorphic. Consider the case when $X$ is a projective line, $D$ is a rational point, and $E$ is a trivial rank $2$ bundle. Then $\text{Ker}(E,L)$ is the vector bundle whose sheaf of global sections is $\mathcal{O}(-1)\oplus \mathcal{O}$. Thus, the first projective bundle is the Hirzebruch surface $\Sigma_0$, yet the second projective bundle is $\Sigma_1$. $\endgroup$– Jason StarrCommented Jul 30, 2017 at 23:02
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$\begingroup$ Typo correction: "The associated projective bundles are typically NOT isomorphic". The rest of the comment gives an example where the projective bundles are NOT isomorphic. $\endgroup$– Jason StarrCommented Jul 31, 2017 at 0:00
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2$\begingroup$ . . . Also, I am not the downvoter. $\endgroup$– Jason StarrCommented Jul 31, 2017 at 0:01
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$\begingroup$ actually $\mathcal{O}(-1)\oplus \mathcal{O}(-1)$ is also an elementary modification of $\mathcal{O}\oplus \mathcal{O}$. But in this case the projective bundles are same. But thanks for the counterexample. $\endgroup$– user111251Commented Jul 31, 2017 at 8:51
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1$\begingroup$ In general the two projective bundles are related by a flip and accidentally the result of the flip can be isomorphic to the original bundle. The point is that the natural rational map given by the elementary modification is never and isomorphism and is resolved explicitly by a flip. As a bit of shameless self-promotion - take a look at Apppendix A of arxiv.org/abs/math/0008010 where this is explained and where you can find the references to the classical papers of Tyurin and Maruyama who analyzed this in detail. $\endgroup$– Tony PantevCommented Jul 31, 2017 at 10:15
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