# When are two elementary transforms isomorphic?

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 self-intersection 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?

• Are you asking about an isomorphism over $C$, or about an abstract isomorphism? Dec 27, 2018 at 4:21
• @Sasha I mean an isomorphism over $C$, finding $L\in \mathrm{Pic}(C)$ such that $E_1\cong E_2\otimes L$ where $elm_{x_1}X=X_1=\mathbb{P}_C(E_1)$, $elm_{x_2}X=X_2=\mathbb{P}_C(E_2)$. Dec 27, 2018 at 6:40
• 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 $\mathrm{Aut}(C)$? Dec 27, 2018 at 7:57
• See an example of a non-trivial 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). Dec 27, 2018 at 10:04

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 2-torsion 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.