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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!

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    $\begingroup$ For starters, this is a quadratic hypersurface, not a curve. $\endgroup$ Sep 10 '20 at 15:25
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    $\begingroup$ You don't need $Y$, just use $0 \to N_{\mathbb{P}^2/C} \to N_{\mathbb{P}^2/\mathbb{P}^5} \to N_{C/\mathbb{P}^5}\vert_{\mathbb{P}^2} \to 0$. $\endgroup$
    – Sasha
    Sep 10 '20 at 15:30
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    $\begingroup$ $ N_{C | \mathbb{P}^5} = \mathcal{O}_C(2) $ because it is a quadratic hypersurface $\endgroup$
    – Ben C
    Sep 10 '20 at 15:53
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    $\begingroup$ 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 $\mathcal{O}_C(2)|_{\mathbb{P}^2} = \mathcal{O}_{\mathbb{P}^2}(2)$ $\endgroup$
    – Ben C
    Sep 10 '20 at 16:09
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    $\begingroup$ 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 $N_{\mathbb{P}^2 | C} = \Omega_{\mathbb{P}^2}(2)$ $\endgroup$
    – Ben C
    Sep 10 '20 at 16:38
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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, $\mathcal{N}_{\mathbb{P}^2/\mathbb{P}^5} \cong \Omega_{\mathbb{P}^2}(2)$.

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