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Let $f:X\rightarrow Y$ be a blow-up of a complex manifold $Y$ along a smooth submanifold. Is it easy to see that the exceptional locus contains a rational curve? Could anyone give a proof or suggest a reference for this fact?

Liu
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  • How can you say something "is easy to see", when you cannot devise a proof? Maybe you can give us some background why for you the statement is easy to see, and what is the problem with translating this "easy observation" into a correct proof. This will probably help you getting a better answer. – Nils Matthes Mar 03 '13 at 09:22
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    It is my typo. Of course "It is easy to..." should be "Is it easy to...?". – Liu Mar 03 '13 at 10:19
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    The fiber over any point of the smooth submanifold (assumed of codimension $\geq 2$) is a projective space of positive dimension and thus certainly contains a rational curve. – Georges Elencwajg Mar 03 '13 at 12:06

1 Answers1

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What says Georges in the comments is if $f$ is the blowing-up of $Y$ along a smooth closed submanifold $Z$, then

Claime: The exceptional divisor is the projective bundle $\mathbb P(J/J^2)$ over $Z$, where $J$ is the sheaf of ideals defining $Z$ in $Y$.

The quotient $J/J^2$ is locally free of rank $\mathrm{codim}(Z, Y)$. So if $\mathrm{codim}(Z, Y)\ge 2$, then the exceptional divisor contains full of rational curves.

Proof of the claim: the blowing-up $X\to Y$ is given by $$X=\mathrm{Proj} (\oplus_{n\ge 0} J^n). $$ So the exceptional divisor $E$ is $$E= X\times_Y V(J)=\mathrm{Proj} (\oplus_{n\ge 0} J^n)\otimes O_Y/J =\mathrm{Proj} (\oplus_{n\ge 0} J^n/J^{n+1}). $$ AS $Z$ is locally complete intersection in $Y$ (because both $Z$ and $Y$ are smooth), $J/J^2$ is locally free (it is the conormal bundle) and the canonical graded homomorphism $$ \mathrm{Sym}(J/J^2) \to \oplus_{n\ge 0} J^n/J^{n+1} $$ is an isomorphism. Thus $E\simeq \mathbb P(J/J^2)$.