4
$\begingroup$

Let $X$ be a projective Calabi-Yau threefold with a single ordinary double point at $x \in X$, and smooth elsewhere. Is $X$ necessarily factorial?

I suspect that the answer is "yes", for the following reason. Some neighbourhood of $x$ is analytically isomorphic to the 'conifold' geometry $$ xy - wz = 0 ~~\mathrm{in}~~\mathbb{C}^4~. $$ This is not factorial; $x = w = 0$ is a non-Cartier divisor, and blowing up along this gives a small projective resolution of the conifold. However, analytic isomorphisms do not preserve the local class group, and $\mathcal{O}_{X,x}$ may nevertheless be a UFD. I suspect that this is necessarily true, since if not, we could blow up along the non-Cartier divisor, but one-nodal projective Calabi-Yau threefolds do not admit projective resolutions. I would like to know whether this argument can be made rigorous, or replaced by a simpler one.


Edit: It may be important to assume that $X$ is smoothable, and this is the case which matters to me.

$\endgroup$
6
$\begingroup$

Let $X$ be a Calabi-Yau three-fold with only ordinary double points. One can always find a small resolution $Y\rightarrow X$ where $Y$ is a (not necessarily Kaehler) complex manifold. Let $C_1,\ldots,C_n$ be the exceptional curves. Friedman proved in his paper "Simultaneous Resolution of Threefold Double Points" that $X$ has an infinitesimal smoothing if and only if there is a linear dependence relation $\sum_i a_i[C_i]=0$ in $H_2(Y,{\mathbb R})$ with all $a_i$ non-zero. Furthermore, by unobstructedness of the deformation theory of Calabi-Yau varieties with isolated ordinary double points (proved by Ran and Tian), having an infinitesimal smoothing is equivalent to having a smoothing.

In your case, $X$ just has a single ODP, so there is one exceptional curve $C$. So there is a smoothing if and only if $[C]=0$ in $H_2(Y,{\mathbb R})$. This is in turn equivalent to $X$ being factorial (there is no algebraic small resolution).

So what you want is true: if $X$ is smoothable, it is factorial. However, there are examples of non-smoothable $X$ with one ODP, and these are not factorial.

$\endgroup$
  • $\begingroup$ Thanks Mark, that's exactly what I was after. I was actually reading Friedman's paper, but with my poor grasp of deformation theory, I hadn't managed to extract what I needed! $\endgroup$ – Rhys Davies Aug 7 '13 at 15:25
  • $\begingroup$ Blowing-up the ODP one obtains as exceptional divisor a smooth quadric in $\mathbb{P}^3$. If there is no algebraic small resolution then the two rulings of this quadric must be numerically equivalent in the blown-up threefold, right? $\endgroup$ – Francesco Polizzi Aug 7 '13 at 15:25
  • 1
    $\begingroup$ Francesco, that's right. $\endgroup$ – Mark Gross Aug 7 '13 at 15:36
2
$\begingroup$

At the moment, I do not have in mind the answer for the general case. However, it is yes when the threefold is a complete intersection in some $\mathbb{P}^n$. In fact, in this situation a more general result holds.

Let us call a threefold singularity an ordinary $m$-ple point if the corresponding tangent cone is a cone over a smooth surface in $\mathbb{P}^3$. When $m=2$ we obtain precisely the ordinary double points. Then we have the following

Proposition. Let $Y \subset \mathbb{P}^n$ be a smooth, complete intersection fourfold and $X \subset Y$ be a reduced, irreducible threefold, which is complete intersection of $Y$ with a hypersurface of degree $d$. Assume that the singular locus of $X$ consists of $k$ ordinary multiple points $p_1, \ldots, p_k$ of multiplicity $m_1, \ldots, m_k$. If \begin{equation} \sum_{i=1}^k m_i < d \end{equation} then $X$ is factorial.

In particular, since a complete intersection Calabi-Yau threefold has always degree at least $5$, you get your result.

A proof of this proposition and related references can be find in my recent preprint with A. Rapagnetta and P. Sabatino On factoriality of threefolds with isolated singularities, arXiv:1305.4371.

$\endgroup$
  • $\begingroup$ Thank-you for the answer, Francesco. I will have a look at your preprint. However, the case of a complete intersection in $\mathbb{P}^n$ won't actually help me directly. $\endgroup$ – Rhys Davies Aug 7 '13 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.