# Problem finding the normal bundle

Consider the quadric hypersurface $$C=V(x_0x_3+x_1x_4+x_2x_5)\subset\mathbb{P}^5$$. It is obvious $$C$$ containes two $$\mathbb{P}^2$$'s given by the equation $$\mathbb{P}^2_a=V(x_3,x_4,x_5)$$ and $$\mathbb{P}^2_b=V(x_0,x_1,x_2)$$.

I'd like to compute the normal bundle $$\mathcal{N}_{\mathbb{P}^2_a\mid C}$$, so I thought I could use the sequence

$$\mathcal{N}_{\mathbb{P}^2_a\mid Y}\to \mathcal{N}_{\mathbb{P}^2_a\mid C} \to \mathcal{N}_{Y\mid C}|_{\mathbb{P}^2_a}$$

where $$Y$$ is a subvariety of $$C$$ containing $$\mathbb{P}^2_a$$. The problem is that I can't find a suitable subvariety, and I'm quite stuck since I don't know much on how to compute the normal bundle, outside this formula (and the one for the case of an hyperplane). Thanks in advance!

• For starters, this is a quadratic hypersurface, not a curve. Sep 10 '20 at 15:25
• You don't need $Y$, just use $0 \to N_{\mathbb{P}^2/C} \to N_{\mathbb{P}^2/\mathbb{P}^5} \to N_{C/\mathbb{P}^5}\vert_{\mathbb{P}^2} \to 0$. Sep 10 '20 at 15:30
• $N_{C | \mathbb{P}^5} = \mathcal{O}_C(2)$ because it is a quadratic hypersurface Sep 10 '20 at 15:53
• The closed embeddings $\mathbb{P}^2 \to C \to \mathbb{P}^5$ compose to give a standard (i.e. degree 1) embedding $\mathbb{P}^2 \to \mathbb{P}^5$ so $\mathcal{O}_{\mathbb{P}^5}(1) |_{\mathbb{P}^2} = \mathcal{O}_{\mathbb{P}^2}(1)$. This implies that $\mathcal{O}_C(2)|_{\mathbb{P}^2} = \mathcal{O}_{\mathbb{P}^2}(2)$ Sep 10 '20 at 16:09
• well, it is some rank 2 (probably nonsplit) vector bundle. I am not sure what more explicit description you have in mind. Now for this particular case, something interesting does happen which is that you have the Euler sequence $0 \to \Omega \to \mathcal{O}(-1)^{\oplus 3} \to \mathcal{O} \to 0$ so $0 \to \Omega(2) \to \mathcal{O}(1)^{\oplus 3} \to \mathcal{O}(2) \to 0$ is exact (you should check the maps are the same) so you probably get $N_{\mathbb{P}^2 | C} = \Omega_{\mathbb{P}^2}(2)$ Sep 10 '20 at 16:38

You have embeddings $$\mathbb{P}^2 \to C \to \mathbb{P}^5$$ which compose to give a standard embedding of $$\mathbb{P}^2 \to \mathbb{P}^5$$ which is the vanishing of coordinate functions: $$V(x_3, x_4, x_5)$$ where $$\mathbb{P}^5 = \mathrm{Proj}(k[x_0, \dots, x_5])$$. Let $$Z = V(x_3, x_4, x_5) = \mathbb{P}^2$$.
Because everything is regular (and thus the closed immersions are regular immersions) all normal sheaves are vector bundles (finite locally free) and there is a short exact sequence, $$0 \to \mathcal{N}_{Z/C} \to \mathcal{N}_{Z/\mathbb{P}^5} \to (\mathcal{N}_{C/\mathbb{P}^5})|_{Z} \to 0$$ we should compute these terms. First, since $$C$$ is a degree two hypersurface its sheaf of ideals is $$\mathcal{I} = \mathcal{O}_{\mathbb{P}^5}(-2)$$ explcitly there is an exact sequence, $$0 \to \mathcal{O}_{\mathbb{P}^5}(-2) \xrightarrow{F} \mathcal{O}_{\mathbb{P}^5} \to \mathcal{O}_{C} \to 0$$ where $$F = x_0 x_3 + x_1 x_4 + x_2 x_5$$. Therefore, $$\mathcal{C}_{C/\mathbb{P}^5} = \mathcal{O}_{\mathbb{P}^5}(-2) \otimes \mathcal{O}_{C} = \mathcal{O}_C(-2)$$ so the normal bundle is, $$\mathcal{N}_{C/\mathbb{P}^5} = \mathcal{C}_{C/ \mathbb{P}^5}^\vee = \mathcal{O}_C(2)$$ Furthermore, since $$\mathcal{O}_C(2)$$ is the pullback of $$\mathcal{O}_{\mathbb{P}^5}(2)$$ under $$C \to \mathbb{P}^5$$ and $$Z \to C \to \mathbb{P}^5$$ is a degree one hyperplane embedding so $$\mathcal{O}_{\mathbb{P}^5}(2)|_{Z} = \mathcal{O}_Z(2)$$ and thus, $$(\mathcal{N}_{C/\mathbb{P}^5})|_{Z} = \mathcal{O}_Z(2)$$
Next, to compute $$\mathcal{N}_{Z/\mathbb{P}^5}$$ we can use the Kozul complex for the ideal $$\mathcal{I} = (x_3, x_4, x_5)$$, $$0 \to \mathcal{O}_{\mathbb{P}^5}(-3) \to \mathcal{O}_{\mathbb{P}^5}(-2)^{\oplus 3} \to \mathcal{O}_{\mathbb{P}^5}(-1)^{\oplus 3} \to \mathcal{O}_{\mathbb{P}^5} \to \mathcal{O}_Z \to 0$$ then pulling back to $$Z$$ gives a right-exact sequence, $$\mathcal{O}_{Z}(-2)^{\oplus 3} \to \mathcal{O}_{Z}(-1)^{\oplus 3} \to \mathcal{C}_{Z/\mathbb{P}^5} \to 0$$ but the coordinate functions vanish on $$Z$$ so the first map is zero giving, $$\mathcal{N}_{Z/\mathbb{P}^5} = \mathcal{C}_{Z/\mathbb{P}^5}^\vee = \mathcal{O}_Z(1)^{\oplus 3}$$ Therefore we have an exact sequence, $$0 \to \mathcal{N}_{Z/C} \to \mathcal{O}_Z(1)^{\oplus 3} \to \mathcal{O}_Z(2) \to 0$$ Compare this with the Euler sequence, $$0 \to \Omega_{\mathbb{P}^2} \to \mathcal{O}_{\mathbb{P}^2}(-1)^{\oplus 3} \to \mathcal{O}_{\mathbb{P}^2} \to 0$$
Notice that maps $$\mathcal{O}_{\mathbb{P}^2}^{\oplus 3} \to \mathcal{O}_{\mathbb{P}^2}(1)$$ are classified by triplets of sections, $$\mathrm{Hom}(\mathcal{O}_{\mathbb{P}^2}^{\oplus 3}, \mathcal{O}_{\mathbb{P}^2}(1)) = \Gamma(\mathbb{P}^2, \mathcal{O}_{\mathbb{P}^2}(1))^{\oplus 3}$$ which are $$3 \times 3$$ matrices with surjective maps corresponding to nonsingular matrices. Therefore, for every pair of surjective maps $$\mathcal{O}_{\mathbb{P}^2}^{\oplus 3} \to \mathcal{O}_{\mathbb{P}^2}(1)$$ there is an automorphism of $$\mathcal{O}_{\mathbb{P}^2}^{\oplus 3}$$ compatible with them so there exists a diagram, $$\require{AMScd}$$ $$\begin{CD} 0 @>>> \mathcal{N}_{Z/C} @>>> \mathcal{O}_Z(1)^{\oplus 3} @>>> \mathcal{O}_Z(2) @>>> 0 \\ @. @VVV @VVV @| \\ 0 @>>> \Omega_{\mathbb{P}^2}(2) @>>> \mathcal{O}_Z(1)^{\oplus 3} @>>> \mathcal{O}_Z(2) @>>> 0 \\ \end{CD}$$
where the vertical maps are isomorphisms. Therefore, $$\mathcal{N}_{\mathbb{P}^2/\mathbb{P}^5} \cong \Omega_{\mathbb{P}^2}(2)$$.