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This is quite possibly a stupid question, but it is pretty far from what I normally do, so I wouldn't even know where to look it up.

If $X$ is a projective variety over an algebraically closed field of arbitrary characteristic and $Y\subset X$ a smooth divisor. Under which conditions can I contract $Y$ to a point, i.e. under which conditions is there a projective (smooth!?) variety $V$, and a morphism $f:X\rightarrow V$, such that $f$ is an isomorphism away from $Y$, and $Y$ is mapped to a point. What can one say if $Y$ is a strict normal crossings divisor?

Hints and references are very appreciated!

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    $\begingroup$ I don't think there is hope for any reasonable and useful condition in that generality... Think about the very special case of a projective line on a smooth proper surface: Already in this situation the answer to your question is quite sophisticated. $\endgroup$
    – Xandi Tuni
    Commented May 13, 2010 at 13:30
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    $\begingroup$ In the case when $Y$ is a curve on a smooth surface $X$, the most basic requirement is that $Y$ have negative self-intersection. If (and only if) $Y$ is in fact a copy of $\mathbb P^1$ with self-intersection -1, you can contract it to a point to again get a smooth surface. (Castelnuovo's criterion; cf. Ch. V of Hartshorne.) Otherwise you will get a singular surface, although I forget whether in general you can always contract to get a variety, as opposed to an algebraic space. Your general question has been studied by Artin in the context of algebraic spaces; see e.g. his 1970 ICM Talk. $\endgroup$
    – Emerton
    Commented May 13, 2010 at 14:25
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    $\begingroup$ Is Y smooth, or just X? $\endgroup$
    – Allen Knutson
    Commented May 14, 2010 at 13:35
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    $\begingroup$ I wouldn't even mind if X and Y were smooth. $\endgroup$
    – Lars
    Commented May 14, 2010 at 17:11

3 Answers 3

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For a smooth $Y$, a necessary condition for contractibility is that the conormal line bundle $N_{Y,X}^*$ is ample. It is also sufficient for contracting to an algebraic space. The reference is Algebraization of formal moduli. II. Existence of modifications. by M. Artin.

$Y$ can be contracted to a point on an algebraic (projective) variety if in addition $Y=\mathbb P^{n-1}$, $n=\dim X$. You can prove this easily by hands. Start with an ample divisor $H$ and then prove that an appropriate linear combination $|aH+bY|$ is base point free and is zero exactly on $Y$. You will find the argument in Matsuki's book on Mori's program for example.

So if $X$ is a surface and $Y=\mathbb P^1$ with $Y^2<0$ then it is contractible to a projective surface. For a reducible divisor $Y=\sum Y_i$ a necessary condition (which is also sufficient in the category of algebraic spaces) is that the matrix $(Y_i.Y_j)$ is negative definite. The strongest elementary sufficient condition for contractibility to a variety is that $\sum Y_i$ is a rational configuration of curves. This is contained in On isolated rational singularities of surfaces by M. Artin.

This paper also contains an example of an elliptic curve $Y$ with $Y^2=-1$ which is not contractible to an algebraic surface. The surface $X$ is the blowup of $\mathbb P^2$ at 10 sufficiently general points lying on a smooth cubic, $Y$ is the strict preimage of that cubic.

Finally, for an irreducible divisor $Y$ the resulting space $V$ is smooth iff $Y=\mathbb P^{n-1}$ and $N_{Y,X}=\mathcal O(-1)$. Indeed, $X\to V$ has to factor through the blowup of $V$ at a point by the universal property of the blowup. But then $X$ has to coincide with this blowup by Zariski main theorem. And on the blowup at a point the exceptional divisor is $\mathbb P^{n-1}$ with the normal bundle $\mathcal O(-1)$.

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  • $\begingroup$ @VA: So you are saying that if a birational morphism $f:X\to Y$ (say $X,Y$ smooth projective over $\mathbb C$) contracts a smooth divisor $E$ into a subvariety of codimension $\geq 2$ in $Y$, then $N^*_{E,X}$ is ample? Do you have a reference for this, that would interest me a lot! Thanks. $\endgroup$
    – Henri
    Commented Mar 25, 2012 at 22:04
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    $\begingroup$ I think that Artin proves that the ampleness of the conormal is sufficient. I don't think it is necessary though. $\endgroup$
    – Jorge Vitório Pereira
    Commented Jan 31, 2017 at 22:10
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    $\begingroup$ I remember reading somewhere the following example. Take a contractible rational curve with conormal bundle $O(2)\oplus O$ on a smooth 3-fold. The blow-up of such rational curve gives a contractible divisor with non-ample conormal bundle. $\endgroup$
    – Jorge Vitório Pereira
    Commented Feb 1, 2017 at 2:23
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In the case of smooth surfaces, this is already a tough question : If E is a rational curve, then E can be contracted to a point in a smooth variety if the self-intersection $E\cdotp E=-1$ (thm of Castelnuovo). If you don't impose the smoothness of the variety, then a sufficient condition is $E\cdotp E<0$.

Moreover, there exists a theorem, due to Grauert, which states that if $\sum E_i$ is a divisor on a smooth surface such that the matrix of intersections $(E_i\cdotp E_j)$ is negative, then there exists $f:S \to S'$ birational such that the exceptional locus of $f$ is exactly $\cup E_i$. But this theorem is valid only in the analytic setting, which means that $S'$ is a priori not an algebraic variety.

Concerning the general question of contraction of curves in an algebraic variety, the main results come from Mori's theory (cf the book of Kollar and Mori, Birational geometry of algebraic varieties)

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  • $\begingroup$ Assume that $\sum E_i$ is 1-connected and has arithmetic genus 0. Is it true that Grauert's theorem gives an algebraic surface in that case? $\endgroup$
    – quim
    Commented May 13, 2010 at 14:56
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    $\begingroup$ A compact analytic space is an algebraic space if and only if it is birational to an algebraic variety, so Grauert's use of analysis is not as restrictive as it may appear. $\endgroup$
    – Ben Wieland
    Commented May 13, 2010 at 17:03
  • $\begingroup$ What do you mean by 1-connected? Since it has arithmetic genus 0, it has to be a tree. $\endgroup$
    – VA.
    Commented May 13, 2010 at 23:56
  • $\begingroup$ Yeah, sorry. And thanks for the thorough answer. $\endgroup$
    – quim
    Commented May 14, 2010 at 8:30
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If you ask V to be smooth and if the divisor Y is irredicible then Y should be isomorphic to the projective space and an intersection on X of the divisor Y with a line L on Y have to be equal to $-1.$

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