Let $X$ be a smooth cubic threefold over $\mathbb C$. Let $L_1,L_2$ be two lines (can be the same line) on $X$, and let $I_{L_1},I_{L_2}$ be their ideal sheaves. For simplicity, we denote them as $I_1$ and $I_2$.
According to the proof of page 9, Lemma 4.3 of this paper, $\chi(I_{1},I_2)=-1$. It didn't provide the details of computation. However, I got a different number by my own computation. I would like to know where I was wrong.
By Hirzebruch Riemann-Roch, we have $$\chi(I_{1},I_2)=\int_Xch(I_1,I_2)\cdot td(X),$$ where the $td(X)$ is the Todd class of $X$, which is well known to be $(1,H,\frac23H^2,\frac13H^3)=(1,H,2L,P)$. $H$ is the hyperplane class, $L$ is the line class and $P$ is the point class.
$ch(I_1,I_2)=ch^{\vee}(I_1)\cdot ch(I_2)$ and $ch^{\vee}(I_1)$ should be interpreted as Chern character $ch(I_1^{\vee})$ of $I_1^{\vee}$, which the derived dual according to this post. The second term is easy to find: $ch(I_2)=1-L$. To compute $ch^{\vee}(I_1)$, we should take locally free resolution of $I_1$. By Serre's construction, $\dim\text{Ext}^1(I_1,\mathcal{O}_X)=1$ and there is an exact sequence
$$ 0\to \mathcal{O}_X\to E\to I_1\to 0,\tag{1}\label{1} $$ such that $E$ is locally free of rank 2 and $E$ is unique up to isomorphism.
Apply $\mathcal{Hom}(-,\mathcal{O}_X)$ to the exact sequence, we have
$$0\to \mathcal{Hom}(I_1,\mathcal{O}_X)\to \mathcal{Hom}(E,\mathcal{O}_X)\to \mathcal{Hom}(\mathcal{O}_X,\mathcal{O}_X)\to \mathcal{Ext}^1(I_1,\mathcal{O}_X)\to 0.$$
The derived dual $I_1^{\vee}$ should be the two-term complex $[\mathcal{Hom}(E,\mathcal{O}_X)\to \mathcal{Hom}(\mathcal{O}_X,\mathcal{O}_X)]$, which is just $[E^{\vee}\to \mathcal{O}_X]$. From the exact sequence $(\ref{1})$, $c_1(E)=0$, so $E$ is self-dual, and $ch(E)=2-L$, so it follows that $$ch^{\vee}(I_1)=ch(E^{\vee}\to \mathcal{O}_X)=(2-L)-1=1-L.$$ Therefore $ch(I_1,I_2)=(1-L)(1-L)=1-2L+P$. Multiply it with the Todd class, we have its zero order term being $$P-2L\cdot H+P=P-2P+P=0.$$ So $\chi(I_1,I_2)=0.$
Is there anything wrong with my computation?
There is mistake in squaring $(1-L)$, because $L^2 = 0$ (by degree reasons), not $P$. Also some steps could be simplified.