# Restriction of vector bundles

I am trying to compute the Chern classes of the restriction of a rank two vector bundle on $\mathbb{P}^3$, denoted by $E$, with fixed Chern classes, $c_1(E) = c_1$ and $c_2(E) = c_2$, to a hyperplane $H \subset \mathbb{P}^3$. For this let $s$ be the equation of the hiperplane $H$ and consider the exact sequence:

$$0 \to \mathcal{O}_{\mathbb{P^3}}(-1) \to^{s} \mathcal{O}_{\mathbb{P^3}} \to \mathcal{O}_{\mathbb{P^3}}|_{H} \to 0$$

tensoring this sequence by $E(k)$ where $k$ is some integer, one has :

$$0 \to E(k-1) \to^{s} E(k) \to E(k)|_{H} \to 0$$

and computing the Chern classes one sees that $c_1(E(k)|_{H}) = 2$ and $c_2(E(k)|_{H}) = 3-2k-c_1$

But this seems to be very strange to me, once the first Chern class of the restriction does not change with the Chern classes of the bundle $E$, and the second Chern class of $E$ does not play any role in it. Moreover, if this was true, one will have that $c_1(E(k-1)|_{H}) = c_1(E(k)|_{H})=2$ for a rank two vector bundle on $\mathbb{P}^2$ which is clearly an absurd.

Since I triple checked my computations, I would like to know if the multiplicative property of the Chern classes on the exact sequences does not hold here.

• They do, but not in the way you think. If $i: H\hookrightarrow \mathbb{P}^3$ is the inclusion, what you will get using the multiplicative property are the Chern classes of $i_*(E_{|H})$ on $\mathbb{P}^3$, which are not the Chern classes of $E_{|H}$ on $H$: they are related in a more complicated way, through the Grothendieck-Riemann-Roch theorem. – abx Jun 13 '18 at 14:21
OK, I'll write my comment as an answer. The computation of the OP using the multiplicative property of exact sequences gives the Chern classes of the coherent sheaf $i_*(E_{|H})$ on $\mathbb{P}^3$, where $i:H\hookrightarrow \mathbb{P}^3$ is the inclusion map. They are different from the Chern classes of the vector bundle $E_{|H}$; they are related, but in a nontrivial way, through the Grothendieck-Riemann-Roch theorem.