Let $X$ be a quadric hypersurface in $\mathbb P^4$ and let $Z\subset X$ be a algebraic subset of pure dimension $1$ on $X$. Consider $Z$ smooth.
I want to find the $c_3(I_{Z/X})$.
I want somebody to help me verify if the calculations below are OK, also I need some help, hind or reference when $Z$ is singular.
Take the Euler characteristic of the following short exact sequence
\begin{equation} 0\rightarrow I_{Z/X}\rightarrow \mathcal O_X\rightarrow \mathcal O_{Z/X}\rightarrow 0 \end{equation} we have $\chi(I_{Z/X})=1-\chi(\mathcal O_{Z/X})=g$, where $g$ is the arithmetic genus of $Z$.
By the Hirzebruch-Riemann-Roch theorem \begin{align} \chi(I_{Z/X})&=\deg\left[Ch(I_{Z/X})\cdot td(TX)\right]_3\\ &=\deg\left[(1+c_1+\frac{1}{2}(c_1^2-2c_2)+\frac{1}{6}(c_1^3-3c_1c_2+3c_3))\cdot (1+\frac{1}{2}c_1^{'}+\frac{1}{12}((c_1^{'})^2+c_2^{'})+\frac{1}{24}(c_1^{'}c_2^{'}))\right]_3 \end{align}
where $[\ ]_3$ is the term of degree 3, $c_i=c_i(I_{Z/X})$ and $c_i^{'}=c_i(TX)$ for $i=1,2,3$.
By the other hand $c_1=0$, $c_2=[Z]$ (the fundamental cycle of $Z$), $c_1^{'}=3H$, $c_2^{'}=4H^2$ and $H=c_1(\mathcal O_X(1))$.
So \begin{align} \chi(I_{Z/X})&=\deg\left[(1-[Z]+\frac{1}{2}c_3)\cdot (1+\frac{3}{2}H+\frac{13}{12}H^2+\frac{1}{2}H^3)\right]_3\\ g &=\deg\left(\frac{1}{2}H^3+\frac{1}{2}c_3-\frac{3}{2}[Z]\cdot H\right). \end{align}
finally, we have the identity $$ (g-1)H^3-3[Z]\cdot H =c_3$$