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Let $X\subset\mathbb P^4$ a projective hypersurface with an ordinary double point at $o\in X$.

Blow-up $\mathbb P^4$ at $o$ and let $E\simeq\mathbb P^3$ the exceptional divisor of this blow-up. Consider the strict transform $\widetilde X$ of $X$. Then, it is easy to verify that $Y=\widetilde X\cap E$ is a quadric hypersurface in $E\simeq\mathbb P^3$, thus isomorphic to $\mathbb P^1\times\mathbb P^1$.

Now, take an hyperplane (if you want, generic) section $C$ of $Y$ in $E$, that is $C=Y\cap H(\simeq\mathbb P^1)$, where $H$ is a hyperplane in $E\simeq\mathbb P^3$. Obviously, the curve $C$ meets each line of the two rulings of $Y$ in exactly one point. So, choose one ruling and define the obvious morphism $\varphi\colon Y\to C$ obtained by collapsing any line of the choosen ruling to its intersection point with $C$.

Here is my question.

Does there exist a smooth projective variety $\overline X$, containing a curve isomorphic to $C$ (which I still call $C$), and a regular morphism $\Phi\colon\widetilde X\to\overline X$ such that:

  • $\Phi|_{\widetilde X\setminus Y}\colon\widetilde X\setminus Y\quad\overset{\simeq}\to\quad\overline X\setminus C$
  • $\Phi|_{Y}\colon Y\to C$ is the $\varphi$ above.

I guess this is very classical, but I wasn't able to find it.

Thanks in advance!

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up vote 6 down vote accepted

I guess the answer in no when $\deg(X) \geq 3$.

In fact, in this case Griffiths showed that the two rulings $L_1$ and $L_2$ in $E$ are numerically equivalent in $\widetilde{X}$ (although not algebraically equivalent in general), so the corresponding $K_{\widetilde{X}}$-negative rays $R_1$ and $R_2$ in the Mori cone of $\widetilde{X}$ coincide.

It follows that the contraction of $R_1$ is actually the contraction of the whole quadric, that is it is not possible to contract the two rulings separately.

The situation is very subtle: in fact, the completed local ring $\widehat{\mathcal{O}}_{X,o}$ is not factorial, but the fact that $L_1$ is numerically equivalent to $L_2$ implies that $\mathcal{O}_{X,o}$ is factorial.

See [Debarre, Higher Dimensional Algebraic Geometry, p. 160] and the references given therein.

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Dear Francesco, I don't really understand your answer... The problem is fairly local (around the isolated double point), so how can the degree of the hypersurface influence the answer? –  diverietti Feb 13 '12 at 13:28
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The (subtle) point is really this: the problem is not a local one. When you have a smooth threefold $Y$ containing a quadric $Q$ with normal bundle $\mathcal{O}_Q(-1, -1)$, then two things can happen: either the two rulings of $Q$ are numerically equivalent in $Y$ or they are not. Only in the latter case you can contract the ruling separately and it is impossible to tell, locally analytically, the difference between the two cases (see Debarre's book, top of page 161). –  Francesco Polizzi Feb 13 '12 at 13:42
    
Regarding your problem, Griffits showed that the two rulings in the exceptional divisor are numerically equivalent in $\widetilde{X}$ when $d \geq 3$, so (unless X is a quadric threefold) one cannot contracts separately the rulings. This is also explained by Debarre. –  Francesco Polizzi Feb 13 '12 at 13:47
    
I'll take a look, thank you! –  diverietti Feb 13 '12 at 13:49
1  
Nevermind, found it!. –  diverietti Feb 14 '12 at 17:01
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