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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.

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1 Answer 1

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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 $\mathbb P^n$

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