Brauer group elements associated to conic bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T15:21:26Z http://mathoverflow.net/feeds/question/118449 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118449/brauer-group-elements-associated-to-conic-bundles Brauer group elements associated to conic bundles Daniel Loughran 2013-01-09T13:49:34Z 2013-01-11T11:32:45Z <p>Let $X$ be a non-singular projective variety over a field $k$ (perhaps not of characteristic $2$), and let $\pi:Y\to X$ be a conic bundle over $X$ i.e. a proper morphism all of whose fibres are isomorphic to plane conics. Let $Z \subset X$ denote the discriminant locus, i.e. the closed subset of $X$ consisting of those point whose fibres are singular (this is a divisor on $X$). Finally let $U = X \setminus Z$.</p> <p>Then, $\pi$ restricts to give a smooth morphism over $U$, which locally for the étale topology is a trivial bundle of $\mathbb{P}^1$'s, hence we obtain an element $A \in H^1(U,PGL_2)$ associated to $X$. A standard argument using the exact sequence defining $PGL_2$ gives rise to an injective map $H^1(U,PGL_2) \to H^2(U,\mathbb{G}_m)=\mathrm{Br}~U$ by which we obtain an element of $\mathrm{Br}~U[2]$ (which by abuse of notation we also denote by $A$).</p> <p>For any discrete valutation $v$ of the function field $k(U)$, we have a residue map $$\mathrm{res}_v:\mathrm{Br}~U \to H^1(k(v),\mathbb{Q}/\mathbb{Z}).$$ My first question is about how the geometric properties of $\pi$ are related to the algebraic properties of $A$.</p> <blockquote> <blockquote> <p>Question 1. Let $v$ be a valuation of $k(U)$ corresponding to an irreducible divisor $D$ supported on $Z$. What is the a relationship between $\mathrm{res}_v(A)$ and the fibre of $\pi$ over $D$?</p> </blockquote> </blockquote> <p>This is slightly vague so I will try to make more precise what I am after.</p> <p>Naively, one might expect that $\mathrm{res}_v(A)$ is non-zero since the fibre over $D$ is singular. However, this is not quite true as birational conic bundles should give rise to the same Brauer group element on sufficiently small open subsets of $X$. So there may be some another conic bundle also giving rise to $A$ for which the fibre over $D$ is non-singular. The following is a more precise version of the previous question which hopes to get around this issue (we use the same notation as Question 1).</p> <blockquote> <blockquote> <p>Question 2. Is $\mathbb{res}_v(A) =0$ if and only if the singular conic $\pi^{-1}(D)$ (considered over $k(v)=k(D)$) is split over $k(D)$?</p> </blockquote> </blockquote> <p>Here by split I mean that the two singular lines are defined over $k(D)$, rather than conjugate over some quadratic field extension. Also I am slightly abusing notation, as really it is the generic fibre of $\pi^{-1}(D)$ over $D$ which is a conic over $k(D)$.</p> <p>Finally, one might expect that conic bundles for which all singular conics $\pi^{-1}(D)$ are not split over $k(D)$ are in some sense "relatively minimal". Unfortunately I can't make this precise except in the case where $\mathrm{dim}~ Y = 2$, where relatively minimal means that one may not contract any of divisors lying in the fibres of $\pi$. I would like a higher dimensional analogue of this.</p> <blockquote> <blockquote> <p>Question 3. Suppose that for some divisor $D \subset Z$ the corresponding singular conic $\pi^{-1}(D)$ over $k(D)$ is split. Then is it possible to contract one of the components of $\pi^{-1}(D)$? More precisely, does there exist a conic bundle $\pi':Y'\to X$ and a morphism $f:Y \to Y'$ (respecting $\pi$ and $\pi'$) such that $Z' = Z \setminus D$? Here $Z'$ denotes the discriminant locus of $Y'$.</p> </blockquote> </blockquote>