Let us assume $X$ is a smooth, projective and unirational variety of dimension $n$ over $\mathbb{C}$.

Given a conic bundle $\pi: Y\rightarrow X$ such that $\omega_{\pi}^{-1}$ is relatively very ample with respect to $\pi$, here $\omega_{\pi}=\omega_Y\otimes \pi^{*}\omega_X^{-1}$. Assume all three possible types of fibers occur, that is we have either smooth conics $(\cong \mathbb{P}^1)$, pairs of intersecting lines $(\cong \mathbb{P}^1$ v $\mathbb{P}^1)$ or nonreduced double lines $(\cong 2\mathbb{P}^1)$

Is anything known about the cohomology of the relative tangent bundle of $\pi$, i.e. what is $H^i(Y,\mathcal{T}_{\pi})$ for $0\leq i\leq n+1$, where $\mathcal{T}_{\pi}=\omega_{\pi}^{-1}$?

I am interested in any information about these groups. I am especially interested in the groups for $i=0,1,2$. If it helps, one can assume that $\pi$ is a standard conic bundle, that is $Y$ is also smooth and we have $Pic(Y)=\mathbb{Z}K_Y\oplus\pi^{*}Pic(X)$ and $\pi^{-1}(C)$ is irreducible for any irreducible curve $C\subset X$. Here $K_Y$ is the canonical divisor of $Y$.


1 Answer 1


Such conic bundle is given by a rank three vector bundle, say $E$ on $X$, and a line subbundle $L \subset Sym^2E$ (just take $E$ to be the pushforward of $\omega^{-1}_\pi$, and $L$ corresponds to the equation of the conic bundle). Then the pushforward of $T_\pi$ to $X$ is isomorphic to $L \otimes E \otimes \det E^\vee$ (and derived pushforwards vanish). A priori this vector bundle on $X$ may have more or less any cohomology (except, of course, for the vanishing of the cohomology in degree $n+1$).

  • $\begingroup$ Thank you. I was hoping that something was known because this is a line bundle in this case...but if this question is equivalent to finding the cohomology of a rank three vector bundle on $X$, I guess it is really hard in this generality. $\endgroup$
    – Bernie
    Feb 26, 2016 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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