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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 $(2-1)(9-1)$).

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  • $\begingroup$ What are "the main classes on the Hilbert scheme" for you? $\endgroup$
    – Sasha
    Commented Sep 17, 2021 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
    Commented Sep 17, 2021 at 18:44

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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.

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  • $\begingroup$ 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
    Commented Sep 18, 2021 at 17:58

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