4
$\begingroup$

Let $k$ be a field with char $k \neq 2$. For $a,b \in k^{\times}$, let $(a,b)$ denote the quaternion algebra with $i^2=a$ and $j^{2}=b$, and let $C(a,b)$ denote the projective plane conic given by $ax^2+by^2=cz^2$. A theorem of Witt (Theorem 1.4.2 in [1]) says that $(a_{1},b_{1})$ is $k$-isomorphic to $(a_{2},b_{2})$ iff $C(a_{1},b_{1})$ is $k$-isomorphic to $C(a_{2},b_{2})$.

On the other hand, according to [2], if $C(a,b)$ doesn't have rational points, then there exists an indecomposable vector bundle $S$ over $C(a,b)$ of rank two, and any other indecomposable rank two vector bundle over $C(a,b)$ is $S$ tensored with some power of the tangent bundle. By [Remark 3.7, 2], $End_{k}(S)$ is a quaternion algebra $(a',b')$, and by the fact just stated, the endomorphism ring of any indecomposable rank two vector bundle over $C(a,b)$ is $k$-isomorphic to $(a',b')$.

My question is the following:

Suppose that $C(a,b)$ does not have a $k$-rational point and suppose that $\mathcal{L}$ is an indecomposable vector bundle of rank two over $C(a,b)$. Is $End_{k}(\mathcal{L})$ $k$-isomorphic to $(a,b)$? Why?

References:

[1] P. Gille and T. Szamuely, Central Simple Algebras and Galois Cohomology, Cambridge Studies in Advanced Mathematics, 101. Cambridge University Press, Cambridge, 2006.

[2] I. Biswas and D.S. Nagaraj, Vector bundles over a nondegenerate conic, J. Aust. Math. Soc. 86 (2009), 145-154.

$\endgroup$
1
$\begingroup$

Let $S$ denote an indecomposable bundle over $C(a,b)$ of rank two and degree two. By the proof of Proposition 3.5 [2], $S \otimes \Omega^{1}$ corresponds to a nonsplit extension of $\mathcal{O}_{C(a,b)}$ by $\Omega^{1}$. The answer now follows from Bhargav's answer to

Explicit Bijection between Central Simple Algebras and twists of $\mathbb P^n$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.