Contracting a rational curve in a Calabi-Yau threefold

Question

Let $X$ be a Calabi-Yau threefold and $C \subset X$ be a rational curve with $N_{C/X}\cong \mathcal{O}\oplus \mathcal{O}(-2)$. Can one contract the curve $C$? Assuming the answer is yes, what kind of singularity does one get? Explicit equation?

Answer 1

Given a rational curve $C$ in a threefold $X$ with normal bundle $N_{C/X}\cong{\mathcal O}\oplus{\mathcal O}(-2)$, what Miles Reid proves is that, at least locally, "$C$ either contracts or moves". $C$ certainly moves infinitesimally to order one, since $\dim H^0(N_{C/X})=1$. Reid proves that if $C$ moves to order $k$ but no further, then locally at least it is contractible to the hypersurface singularity in Francesco's answer. Alternatively, if $C$ deforms to arbitrary order, then there is a small analytic disc over which $C$ deforms in a ruling inside $X$. In this case, any contraction that contracts $C$ will also have to contract all other fibres of the ruling, and locally the singularity is an $A_1$ surface double point times the small disc, given by the equation $x^2+y^2+z^2=0$ inside affine four-space.

Whether a contraction exists inside a projective Calabi-Yau threefold, in either case, is a global question, and seems quite subtle. I don't really have anything else to say in the isolated case. It is very easy to produce non-isolated examples: let $\bar X$ be a singular Calabi-Yau threefold containing a smooth curve $S$ of $A_1$ singularities; many such examples exist. Blowing up $S$ gives a smooth Calabi-Yau threefold $X$ with an exceptional divisor which is a ruled surface over $S$; the fibres of the ruling have the normal bundle you want.

Just to see that the problem is really global, consider the following, related example. Let $\bar Y$ be a singular Calabi-Yau threefold which contains a smooth curve $Z$ of $A_2$ singularities, in such a way that when you blow up $Z$ inside $\bar Y$, you get an irreducible exceptional divisor $D$ which fibres over $Z$. Over any point $p\in Z$ the fibre in $D$ will of course be a line pair, but it can happen that $D$ is irreducible because of monodromy. This can also be achieved inside a projective Calabi-Yau threefold $Y$. Either one of the lines in the fibre still has the normal bundle you want, and the numerical class of the curve can be contracted, but only by contracting both fibres along the whole of $Z$, so the contraction has to give $\bar Y$ back, with a curve of $A_2$ singularities.

Answer 2

This is only a partial answer.

Threefolds $V$ with an isolated singular point $p$ such that there is a small resolution $\pi \colon X \to V$ with exceptional set isomorphic to $\mathbb{P}^1$ were investigated by H. B. Laufer in the paper

On $\mathbb{CP}^1$ as an exceptional set, Recent Developments in Several Complex Variables 100, Annals of Mathematics Studies (1981).

Laufer proves the following

Theorem. Let $V$ be an analytic space of dimension $n \geq 3$ with an isolated singularity at $p$. Suppose that there exists a non-zero holomorphic form $\omega$ on $V - \{p\}$ (i.e., that $p$ is a Gorenstein singularity). Assume moreover that there exists a small resolution $\pi \colon X \to V$ with irreducible exceptional locus $C := \pi^{-1}(p)$.

Then $C$ is isomorphic to $\mathbb{P}^1$, $n=3$ and the possible bidegrees for the normal bundle $N_{C/X}$ are $(-1, \, -1)$, $(-2, \, 0)$ and $(-3, \, 1)$.

In the $(-2, \, 0)$ case the local form for $V$ around $p$ is given by a hypersurface in $\mathbb{C}^4$ of equation $$x^2+y^2+z^2+w^{2k}=0,$$ with $k \geq 2$ (whereas for $k=1$ we have normal bundle of type $(-1, -1)$).

I did not have the time to check if for some of these example the small resolution $X$ is a Calabi Yau threefold. If so, this would provide an affermative answer to your question.

Answer 3

This is actually a comment on Francesco's answer, but too long for a comment. The examples given by Laufer are local; one way to make them global would be to start with a Calabi-Yau $\,X\,$ with one singular point of type $A_{2k-1}$ -- for instance the quintic in $\mathbb{P}^4$, with coordinates $(x_0,x_1,x_2,t,u)$, given by $u^3\sum x_i^2+ut^4+\sum x_i^5=0$. The point $p:=(0,\ldots ,0,1)$ is a singular point of type $A_{3}$. Laufer gives a resolution $\hat{V}\rightarrow V$ of an analytic neighborhood $V$ of $p$; glueing $X\smallsetminus \{p\} $ and $\hat{V}$ along $V\smallsetminus \{p\} $ provides a compact manifold $\hat{X}$ wih trivial canonical bundle, and a map $\pi :\hat{X}\rightarrow X$ such that $\pi ^{-1}(p)$ is a $\mathbb{P}^1$ with the required normal bundle. What is not clear to me is whether $\,\hat{X}\,$ is Kähler, hence Calabi-Yau.