# Normal bundle to complete intersection in $\mathbb{P}^n$

Let $X\subset\mathbb{P}^n$ be a complete intersection defined by irreducible polynomials $f_1,...,f_k$ of degrees $d_1,...,d_k$. How to show that the normal bundle of $X$ is isomorphic to $\bigoplus\limits_{i=1}^k\mathcal{O}_X(d_i)$?

By definition, the normal bundle is $(\mathscr I/\mathscr I^2)^\vee$, where $\mathscr I$ is the ideal sheaf of $X$.

Since $X$ is a complete intersection, the ideal sheaf $\mathscr I$ is resolved by the Koszul complex. I.e., we have an exact sequence of the form $$0 \to \mathscr K \to \bigoplus_{i,j} \mathscr O(-d_i-d_j) \xrightarrow{\varphi} \bigoplus \mathscr O(-d_i) \to \mathscr I \to 0.$$

Where $\varphi$ is multiplication by the vector $(f_1,\ldots, f_n)$. We have that $\mathscr I / \mathscr I^2 \simeq \mathscr I \otimes_{\mathscr O_{\mathbb P^n}} \mathscr O_X$. Thus, tensoring the above sequence with $\mathscr O_X$ we get

$$\mathscr K \otimes \mathscr O_X \to \bigoplus_{i,j} \mathscr O_X(-d_i-d_j) \xrightarrow{\varphi} \bigoplus \mathscr O_X(-d_i) \to \mathscr I/\mathscr I^2 \to 0.$$

But the middle map $\varphi$ is zero in $\mathscr O_X$. Hence $$\bigoplus \mathscr O_X(-d_i) \simeq \mathscr I/\mathscr I^2.$$

Dualizing, we get the result.

• This argument with Koszul complex seems nice. Thank you. Dec 6, 2015 at 23:04
• Dear @Fredrik Meyer, sorry for my late reply, could you kindly tell me where can I find the fact that you have mentioned $\mathscr I / \mathscr I^2 \simeq \mathscr I \otimes_{\mathscr O_{\mathbb P^n}} \mathscr O_X$? I really appreciate it. Nov 22, 2020 at 14:54
• @Steve Hi! This is based on the local fact that if $B=A/I$, then $I/I^2 \simeq I \otimes_B B$ as $B$-modules. To prove it, looks at $I \otimes B \to I/I^2$ given by $i \otimes b \mapsto ib$. First, check that it is well-defined. Nov 24, 2020 at 15:25
• Dear @FredrikMeyer, thanks for your help. Nov 25, 2020 at 0:24

Since every $$f_i$$ is a section in $$H^{0}(\mathbb{P}^n,\mathcal{O}(d_i))$$, denote $$F=\oplus{f_i}$$ the section in $$\oplus\mathcal{O}(d_i)$$, obviously $$Z(F)=X$$. Now restric $$dF$$ to $$\mathcal{T}_{\mathbb{P}^n}|_X$$, we get an exact sequence: $$0\to\mathcal{T}_X\to\mathcal{T}_{\mathbb{P}^n}|_X\to\oplus\mathcal{O}_X(d_i)\to0$$ Compare with

$$0\to\mathcal{T}_X\to\mathcal{T}_{\mathbb{P}^n}|_X\to N\to0$$