1
$\begingroup$

Let $i:\mathbb F\hookrightarrow\mathbb P^4$ be a cubic scroll i.e. $\mathbb F\simeq \mathbb P(\mathcal O_{\mathbb P^1}(1)\oplus\mathcal O_{\mathbb P^1}(2))\overset{n}{\rightarrow} \mathbb P^1$ with $\mathcal O_{\mathbb P(\mathcal O_{\mathbb P^1}(1)\oplus\mathcal O_{\mathbb P^1}(2))}(1)\simeq \mathcal O_{\mathbb P^4}(1)_{|\mathbb F}$. The main component of the space of conics on it should be $|i^*\mathcal O_{\mathbb P^4}(1)\otimes n^*\mathcal O_{\mathbb P^1}(-1)|\simeq \mathbb P^2$.
But how does it sit inside the Hilbert scheme $\mathbb P({\rm Sym}^2\mathcal E_3^*)\overset{t}{\rightarrow}Gr(3,5)$ of conics in $\mathbb P^4$ (namely what are the degrees on $\mathbb P^2$ of the restriction of the main classes on the Hilbert scheme)?

A natural idea that seems not working: as $\mathbb F$ in $\mathbb P^4$ is given by the $2\times 2$-minors of a $2\times 3$-matrix with linear entries, say $M_{\mathbb F}\in |M_{2\times 3}\otimes\mathcal O_{\mathbb P^4}(1)|$, one can use the universal conic on the Hilbert scheme (and the fact that it is a divisor in $\mathbb P(t^*\mathcal E_3)$ ) to get a section $M_{C(\mathbb F)}\in H^0(M_{2\times 3}\otimes t^*\mathcal E_3)$ whose degeneracy locus, when seen as an injective morphism $\varphi_{M_{C(\mathbb F)}}:\mathcal O_{\mathbb P({\rm Sym}^2\mathcal E_3^*)}^{\oplus 2}\rightarrow t^*\mathcal E_3^{\oplus 3}$, $\{rk(\varphi_{M_{C(\mathbb F)}})\leq 1\}$ should be the space of conics contained in $\mathbb F$. But unfortunately it has dimension $3$ (codimension $(2-1)(9-1)$).

$\endgroup$
2
  • $\begingroup$ What are "the main classes on the Hilbert scheme" for you? $\endgroup$
    – Sasha
    Sep 17 '21 at 18:26
  • $\begingroup$ Thank you for your answer. By "main classes" I mean the chern classes $t^*c_i(\mathcal E_3)$ coming from $Gr(3,5)$ and $c_1(\mathcal O_t(1))$. $\endgroup$
    – pi_1
    Sep 17 '21 at 18:44
1
$\begingroup$

Let me first describe the pullback of the tautological bundle of $\mathrm{Gr}(3,5)$. Let $V$ be a 3-dimensional vector space; then we can take $$ \mathbb{F} = \mathrm{Bl}_{[f]}(\mathbb{P}(V^\vee)) \subset \mathbb{P}(S^2V^\vee/ \langle f^2 \rangle) =: \mathbb{P}(W), $$ where $0 \ne f \in V^\vee$, and then the main component of the Hilbert scheme of conics on $\mathbb{F}$ is $\mathbb{P}(V)$ (because lines on $\mathbb{P}(V)$ go to conics on $\mathbb{F}$). Consider the Euler sequence $$ 0 \to \mathcal{O}(-1) \to V \otimes \mathcal{O} \to T(-1) \to 0 $$ on $\mathbb{P}(V)$ and its symmetric square $$ 0 \to V \otimes \mathcal{O}(-1) \to S^2V \otimes \mathcal{O} \to S^2T(-2) \to 0. $$ The morphism $W^\vee \otimes \mathcal{O} \hookrightarrow S^2V \otimes \mathcal{O} \to S^2T(-2)$ is not surjective; its cokernel is isomorphic to the cokernel of $V \otimes \mathcal{O}(-1) \to S^2V \otimes \mathcal{O} \to \mathcal{O}$ (where the second arrow is given by $f^2$), which allows one to check that the cokernel si the structure sheaf of the line $L \subset \mathbb{P}(V)$ corresponding to $f$. Consider the kernel of the corresponding map $$ \mathcal{E} := \mathrm{Ker}(S^2T(-2) \to \mathcal{O}_L). $$ The construction implies that there is an epimorphism $$ W^\vee \otimes \mathcal{O} \to \mathcal{E} $$ which induces a morphism $\mathbb{P}(V) \to \mathrm{Gr}(3,W)$. It is easy to see that this is the morphism from the above description, hence the pullbacks of the tautological classes of the Grassmannian are the Chern classes of $\mathcal{E}$.

If this is satisfying I can write a description of the remaining line bundle class later.

$\endgroup$
1
  • $\begingroup$ Thank you very much. I still need to completely check some details but it is satisfying (even if I would like to understand also where my reasoning is false). $\endgroup$
    – pi_1
    Sep 18 '21 at 17:58

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.