When is a surface in a threefold contractible to a curve? - MathOverflow most recent 30 from http://mathoverflow.net2013-05-25T06:55:55Zhttp://mathoverflow.net/feeds/question/31426http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/31426/when-is-a-surface-in-a-threefold-contractible-to-a-curveWhen is a surface in a threefold contractible to a curve?JME2010-07-11T16:11:56Z2010-08-14T18:39:11Z
<p>Given a threefold $Y$ containing a surface $S$. Under which conditions can I contract $S$ so that I still end up with a smooth variety?
In other words</p>
<blockquote>
<p>what are the conditions for the
existence of a smooth variety $X$ and
a morphism $Y\rightarrow X$ such that
the image of $S$ under the morphism is
a curve and the morphism is an
isomorphism away from $S$?</p>
</blockquote>
<p>What are the conditions when $Y$ is a fourfold and $S$ is still a surface?</p>
http://mathoverflow.net/questions/31426/when-is-a-surface-in-a-threefold-contractible-to-a-curve/31470#31470Answer by SandeepJ for When is a surface in a threefold contractible to a curve?SandeepJ2010-07-11T21:42:05Z2010-07-11T21:42:05Z<p>You want a divisorial contraction. This paper may be the answer.</p>
<ol>
<li>MR2041612 (2005c:14019) Tziolas,
Nikolaos . Terminal 3-fold
divisorial contractions of a surface
to a curve. I. Compositio Math. 139
(2003), no. 3, 239--261.</li>
</ol>
<p>from the paper "This paper studies divisorial contractions of a surface to a curve, i.e. when
dim $\Gamma = 1$ and X has only index 1 terminal singularities along $\Gamma$. It is not always true that given $\Gamma \subset X$, there is a terminal contraction of a surface to $\Gamma$. We investigate when there is one, give criteria for existence or not and in the case that there is a terminal contraction we also describe the singularities of Y."</p>
http://mathoverflow.net/questions/31426/when-is-a-surface-in-a-threefold-contractible-to-a-curve/31472#31472Answer by Donu Arapura for When is a surface in a threefold contractible to a curve?Donu Arapura2010-07-11T22:03:30Z2010-07-11T22:12:11Z<p>I think there are birational geometers lurking around who would do a better job. But let
me make a small attempt for now. If you blow up a smooth curve on a smooth threefold,
you would get a ruled surface with normal bundle restricting to $O(-1)$ along the rulings.
I think you can do the converse if you allow yourself to work in the category of smooth algebraic spaces [see Artin, Cor. 6.11, Algebraization of formal moduli..., Annals 1970],
but such a statement is generally be false for varieties. I remember learning this from
Moishezon long ago.</p>
<p>I think that if in addition to the above conditions, the fibre of the ruling is an extremal ray in Mori's sense, you can probably use his stuff to get a projective contraction. Take a look at Kenji Matsuki's book on the Intro. to the Mori program for more about that.</p>
http://mathoverflow.net/questions/31426/when-is-a-surface-in-a-threefold-contractible-to-a-curve/31518#31518Answer by Francesco Polizzi for When is a surface in a threefold contractible to a curve?Francesco Polizzi2010-07-12T09:39:05Z2010-07-12T09:39:05Z<p>You want a divisorial contraction $Y \to X$ on a smooth 3-fold Y. Such contractions have been classified by Mori in his paper "3-folds whose canonical bundles are not numerically effective", thm 3.3. There are exactly the following possibilities:</p>
<ol>
<li><p>the smooth blow-up of a point; in this case $S$ is isomorphic to $\mathbb{P}^2$ with normal bundle $\mathcal{O}(-1)$;</p></li>
<li><p>the smooth blow-up of a curve, in this case $S$ is a ruled surface whose normal bundle restricted to the ruling has degree $(-1)$; this is the situation described by Donu Arapura in his answer;</p></li>
<li><p>the contraction of a plane $S$ with normal bundle $\mathcal{O}(-2)$; in this case the
surface $X$ has an isolated singularity isomorphic to the quotient of $\mathbb{A}^3$ by the involution $(x,y,z) \to (-x, -y, -z)$;</p></li>
<li><p>the contraction of a smooth quadric $S$ whose rulings are numerically equivalent; in this case the image of $S$ is a single point, which is a singular point for $X$;</p></li>
<li><p>the contraction of a singular quadric $S$; again, the image of $S$ is a point in $X$.</p></li>
</ol>
<p>Summing up, if you want that $X$ is smooth and the image of $S$ is a curve, the only possibility is 2. </p>
<p>In the case where $Y$ is a smooth 4-fold and $S$ is a smooth surface, the answer can be found in the paper of Kawamata "Small contractions of four-dimensional algebraic manifolds": in this case the only possibility is that $S$ is the disjoint union of copies of $\mathbb{P}^2$, with normal bundle $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$</p>