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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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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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  • $\begingroup$ 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? $\endgroup$
    – diverietti
    Commented Feb 13, 2012 at 13:28
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    $\begingroup$ 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). $\endgroup$
    – Francesco Polizzi
    Commented Feb 13, 2012 at 13:42
  • $\begingroup$ 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. $\endgroup$
    – Francesco Polizzi
    Commented Feb 13, 2012 at 13:47
  • $\begingroup$ On the other hand... I think that in the analytic category this should always work, isn't it? $\endgroup$
    – diverietti
    Commented Feb 14, 2012 at 10:27
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    $\begingroup$ Nevermind, found it!. $\endgroup$
    – diverietti
    Commented Feb 14, 2012 at 17:01

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