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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 $\bigoplus\limits_{i=1}^k\mathcal{O}_X(d_i)$?

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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.

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  • $\begingroup$ This argument with Koszul complex seems nice. Thank you. $\endgroup$ – mahavishnu Dec 6 '15 at 23:04
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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

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

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