# Conics on the complete intersection of two quadrics

Let $$Q_1, Q_2\subset \mathbb{P}^4$$ be two degree two smooth hypersurfaces, and $$S:=Q_1\cap Q_2$$ be their complete intersection. If $$S$$ is smooth, then it is known that $$S$$ is a del Pezzo surface of degree 4. Now let $$c_1, c_2\subset C$$ be two conics.

I'm wondering that is there any criterion for $$\operatorname{Hom}(I_{c_1/S}, I_{c_2/S})\neq 0$$? Here $$I_{c_i/S}$$ is the ideal sheaf of $$c_i$$ in $$S$$.

When $$S$$ is smooth, then using some theory of curves on del Pezzo surface, we can prove that $$\mathcal{O}_S(-c_1)\cong \mathcal{O}_S(-c_2)$$. But when $$S$$ is singular, is there any similar result? For example, if there's a quadric $$Q_3$$ in the pencil spaned by $$Q_1$$ and $$Q_2$$, such that $$\langle c_1\rangle\cup \langle c_2\rangle\cup S\subset Q_3$$, where $$\langle c_i\rangle$$ is the projective plane spaned by $$c_i$$, then can we conclude that $$\operatorname{Hom}(I_{c_1/S}, I_{c_2/S})\neq 0$$?

• When $S$ is smooth it is not true in general that $\mathcal{O}_S(-c_1) \cong \mathcal{O}_S(-c_2)$. In fact, there are 10 linear systems of conics on $S$, and if you take conics from different linear systems, their ideal sheaves are not isomorphic. Nov 3, 2021 at 9:34
• Oh thanks! Maybe I need to double-check my computations. Is it true under the assumption that $<c_1>$ and $<c_2>$ are both in $Q_3\in <Q_1, Q_2>$?
– Kim
Nov 3, 2021 at 9:58
• Still not true in general --- in this case $Q_3$ must be singular, and if it is a cone over a smooth quadric surface $\bar{Q}_3$, planes on $Q_3$ correspond to lines on $\bar{Q}_3$, and if two lines belong to different rulings, conics are not linearly equivalent. Nov 3, 2021 at 10:07
• Oh I see, you mean if $<c_1>$ and $<c_2>$ in $Q_3$ that correspond to two non-linearly equivalent lines on $\overline{Q_3}$, then $<c_1>$ and $<c_2>$ are also not linearly equivalent on $Q_3$, and thus $c_1$ and $c_2$ are not linearly equivalent on $S$?
– Kim
Nov 3, 2021 at 10:16

If $$f \colon I_{c_1} \to I_{c_2}$$ is a non-trivial morphism then it is an isomorphism. Indeed, $$f$$ is injective because both $$I_{c_1}$$ and $$I_{c_2}$$ are torsion free, and since the Hilbert polynomials of $$I_{c_1}$$ and $$I_{c_2}$$ are equal, the Hilbert polynomial of the cokernel is zero, hence the cokernel vanishes.
Thus, $$\mathrm{Hom}(I_{c_1},I_{c_2}) \ne 0$$ if and only if $$c_1$$ and $$c_2$$ are linearly equivalent Weil divisors.