Let $C$ be a smooth projective curve and $X=\mathbb{P}_C(E)$ be a ruled surface over $C$. Let $x_1,\ x_2\in X$ be closed points and define $X_1,\ X_2$ to be elementary transforms of $X$ at $x_1,\ x_2$, respectively. Then when $X_1\cong X_2$ holds? Even if $x_1,\ x_2$ are in the same fiber $\pi^{1}(p)$ where $\pi: X\to C$ and $p\in C$, $X_1$ and $X_2$ might be different, because there might be a minimal section passing $x_1$ but no one passing $x_2$ for instance. Then the selfintersection numbers of a minimal section for $X_1$ and $X_2$ become different. It seems to me very hard to classify the transforms. Is there any result about this question?

1$\begingroup$ Are you asking about an isomorphism over $C$, or about an abstract isomorphism? $\endgroup$– SashaDec 27, 2018 at 4:21

$\begingroup$ @Sasha I mean an isomorphism over $C$, finding $L\in \mathrm{Pic}(C)$ such that $E_1\cong E_2\otimes L$ where $elm_{x_1}X=X_1=\mathbb{P}_C(E_1)$, $elm_{x_2}X=X_2=\mathbb{P}_C(E_2)$. $\endgroup$– user190964Dec 27, 2018 at 6:40

$\begingroup$ Oh.. If I restrict the case into isomorphisms over $C$, then $X_1$ and $X_2$ never be isomorphic if $\pi(x_1)\neq\pi(x_2)$, right? Then what is the difference for abstract isomorphism, is there more information than $\mathrm{Aut}(C)$? $\endgroup$– user190964Dec 27, 2018 at 7:57

$\begingroup$ See an example of a nontrivial isomorphism over $C$ in my answer. In general the difference, indeed, is in automorphisms of $C$ (unless $g(C) = 0$ and so the structure of a projective bundle on $C$ might be not unique). $\endgroup$– SashaDec 27, 2018 at 10:04
1 Answer
Let $c_1,c_2 \in C$ be a pair of points. The existence of an isomorphism $X_1 \cong X_2$ (for some choices of points $x_1$, $x_2$ in $X$ over $c_1$ and $c_2$) over $C$ is equivalent to the existence of a line bundle $L$ on $C$ of degree 1 such that $$ L^2 \cong O_C(c_1 + c_2) $$ and the existence of a morphism $E \otimes L^{1} \to E$ such that $$ 0 \to E \otimes L^{1} \to E \to O_{c_1} \oplus O_{c_2} \to 0. $$ Indeed, the elementary transformation $X'$ of $X$ at some point $x \in X$ over $c \in C$ comes with a distinguished point $x'$ over $c$ such that the elementary transformation of $X'$ at $x'$ is isomorphic to $X$. Thus, if $X_1 \cong X_2$ then the elementary transformation of $X_1$ at the point $x'_2$ is isomorphic to $X$, and this is isomorphic to the projectivization of the kernel of an surjective appropriate morphism $E \to O_{c_1} \oplus O_{c_2}$. The isomorphism $L^2 \cong O_C(c_1 + c_2)$ is obtained by taking the determinant of the above exact sequence.
As an example of a nontrivial situation of that type, let me assume that $L$ is a line bundle such that $L^2 \cong O_C(c_1 + c_2)$ (note that it is determined by the points $c_1$ and $c_2$ up to 2torsion in the Jacobian of $C$) and take $$ E = O_C \oplus L^{1}, $$ so that $E \otimes L^{1} \cong L^{1} \oplus L^{2}$ and consider the morphism $$ L^{1} \oplus L^{2} \to O_C \oplus L^{1} $$ defined by the matrix $$ \begin{pmatrix} 0 & s \\ 1 & 0 \end{pmatrix}, $$ where $s$ is the section of $L^2$ with zeroes at $c_1$ and $c_2$.
In general, for a given vector bundle $E$ the existence of such isomorphisms depends very much on the properties of $E$. For instance, if $\deg(E)$ is even and $E$ is stable, no such nontrivial isomorphisms exist.

$\begingroup$ In the last statement, which "iso"morphism are you saying? $\endgroup$ Jan 3, 2019 at 1:53