Let $X$ be a smooth cubic threefold over $\mathbb{C}$ and let $L \subset X$ be a line.

The projection from $L$ yields a rational map $X \dashrightarrow \mathbb{P}^2$. Resolving the indeterminacy by blowing up $L$, we obtain a conic bundle morphism $\pi: \tilde{X} \to \mathbb{P}^2$.

Consider the relative anticanonical bundle $\omega_{\pi}^{-1}$. By standard properties of conic bundles we obtain an embedding $$\tilde{X} \hookrightarrow \mathbb{P}(\mathcal{E})$$ which respects $\pi$. Here $\mathcal{E} := \pi_*\omega_{\pi}^{-1}$ is a vector bundle of rank $3$ on $\mathbb{P}^2$ and $\mathbb{P}(\mathcal{E}) \to \mathbb{P}^2$ is the associated $\mathbb{P}^2$-bundle over $\mathbb{P}^2$.

Is there is an explicit description of the vector bundle $\mathcal{E}$? For example, is it independent of $X$ or $L$? Is it determined by the normal bundle of $L$? Is it a direct sum of lines bundles?

Note that $\mathbb{P}(\mathcal{E}) \cong \mathbb{P}(\mathcal{E} \otimes \mathcal{O}(k))$ for any $k \in \mathbb{Z}$. In particular, I am quite happy with determining $\mathcal{E}$ up to twist by a line bundle.


Choose coordinates $(U,V,X,Y,Z)$ in $\mathbb{P}^4$ so that $L$ is the line $X=Y=Z=0$. The equation of $X$ is of the form $$ AU^2+2BUV+CV^2+2D U + 2EV +F=0\ ,$$where $A,B,\ldots F$ are homogeneous forms in $X,Y,Z$ of degree $1,1,1,2,2,3$. You can view this equation as a section $s \in H^0(\mathbb{P}^2,\mathrm{Sym}^2\mathcal{E} \otimes \mathcal{O}_{\mathbb{P}^2}(1))$, where $\mathcal{E}$ is the vector bundle $\mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}(1)$. Put $P:=\mathbb{P}_{\mathbb{P}^2}(\mathcal{E})$, and let $p:P\rightarrow \mathbb{P}^2$ be the structure map. Then $s$ defines a section of the line bundle $\mathcal{O}_{P}(2)\otimes p^*\mathcal{O}_{\mathbb{P}^2}(1)$ on $P$, whose zero locus is $\tilde{X} $. So up to a twist, your vector bundle is just $\mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}\oplus \mathcal{O}_{\mathbb{P}^2}(1)$. You may then compute $\omega _{\pi }$ by the adjunction formula and see which twist is needed to get $\pi _*\omega _{\pi }$.

  • 6
    $\begingroup$ Just to add a bit to abx answer. You can apply the same procedure to the ambient space $\mathbb{P}^4$. If you blowup a line in it, you get a map from the blowup to $\mathbb{P}^2$ which is a $\mathbb{P}^2$-fibration. By definition of the blowup, it contains the blowup of the cubic, so it provides the ambient bundle for your conic fibration. Let me also add, that in his answer abx uses the Grothendieck's convention for the projectivization (so $\mathbb{P}(E)$ parameterizes 1-dimensional quotients of $E$). $\endgroup$
    – Sasha
    Jan 15 '17 at 19:06

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