It is possible to construct (in many ways) a family of Calabi-Yau quintics $\mathcal{X}\rightarrow \Delta$, over disk, such that the fiber over $0$ has a singularities locally given by the equation $x_1^2+x_2^2+x_3^2+x_4^2=0$ and the others are smooth quintics. In such families there is vanishing cycle in generic fiber which is a Lagrangian isomorphic to $S^3$ and collapses to the node at central fiber. This Lagrangian $S^3$ is obtained by the deformation of the local equation: $$x_1^2+x_2^2+x_3^2+x_4^2=0 \rightarrow x_1^2+x_2^2+x_3^2+x_4^2=\epsilon$$

Here is my questions: Consider the Fermat quintic $z_1^5+\cdots+z_5^5=0$ and its real locus which is a Lagrangian isomorphic to $S^3 /\mathbb{Z}_2$. Is there any one parameter family of C.Y 3-folds with one fiber isomorphic to Fermat quintic, such that real locus of quintic apears as vanishing cycle for this family ?

Comment: I think if there is such family then the singular fiber has to have orbifold singularities given by local equation: ${x_1^2+x_2^2+x_3^2+x_4^2=0}/\mathbb{Z}_2$ and I think its impossible to write a family of degree 5 equations with this type local singularities!

  • $\begingroup$ aren't those $A_1$ singularities? $\endgroup$ Jan 9, 2011 at 19:18
  • $\begingroup$ Which one do you mean? The one with Z_2 quotient or without ? There might be a mistake with naming but I think my question is clear. Is not it? :) $\endgroup$ Jan 9, 2011 at 21:36
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    $\begingroup$ In the first part of the question it seems that you just ask for a Lefschetz pencil? I guess if you take a generic pencil of hyperplane sections of $P^4$ in the 5-fold Veronese embedding it is Lefschetz (all singularities of fibers are ODP's). Then you can choose one singular fiber and a small neighborhood of the point of the base of the pencil. $\endgroup$
    – Sasha
    Jan 10, 2011 at 6:09
  • $\begingroup$ This is a great question, I don't understand why no one apart from me seems to like it :)... To Sasha and Sandor -- it seems to me that you misread the question. The real question start after: "Here is my questions..." , prior to this Mohammad just explains basic things about vanishing cycles, is not he? $\endgroup$ Jan 10, 2011 at 9:23
  • $\begingroup$ @ Sasha: Please look at Dmitri's comment. $\endgroup$ Jan 10, 2011 at 15:10

1 Answer 1


Let me say first, that I really like this question. Very unusual question about such well known things (in fact I did not know even that the real quintic $\sum_i x_i^5=0$ is $\mathbb RP^3$).

This is not an answer, but more like my interpretation of this question (hopefully correct one). The first comment is about vanishing cycles. Usually, when we speak about vanishing cycles in symplectic geometry, we speak about lagrangan spheres in a fiber of a Lefshetz fibraton constructed using parallel transport given by the symplectic connection on the fibers. But Lefshetz pencil is just the most simple object of algebraic geometry. We can consider instead other one-parameter families of algebraic varieties, that can have singularities more complicated than double points $A_1$. In this case again we can ask what will be the shape of the vanishing cycle constructed via symplectic parallel transport? How does it look like? The point that I want to make is that in a large case of situations, this vanishing cycle will not look at all like a manifold.

Namely, we will consider the example coming from a function on $\mathbb C^n$ with isolated singularity. I.e., we have an analytic function $F:\mathbb C^n\to \mathbb C$ with isolated singularity at $0$ and consider its level sets: hyper-surfaces $F_t:=F^{-1}(t)$ intersected with $B_\varepsilon(0)$ (the ball of radius $\varepsilon$). Then we know that the homotopy type of $F_t\cap B_{\varepsilon}$ is a bucket of $k$ spheres, where $k$ is the Milnor number of the singularity. Now, I think (and here I can not provide the proof), that the vanishing cycle is a deformation retract of $F_t\cap B_{\varepsilon}$. Hence it can not be diffeomorphic to a manifold unless $k=1$. In this case it has to be homotopy equivalent to a sphere, i.e., homeomoerphic to it (by Poincare).

In the case when the singularity of $F$ is non-degenerate (double point), the vanishing cycle is of course diffeomorphic to $S^n$, but maybe if we consider Brieskorn singularities we will be able to get vanishing cycles that are exotic spheres as well?

The above situation concerns the case when the total space of the fibration is smooth. In this case we does not seem to be able to get $\mathbb RP^{n-1}$ as a vanishing cycle. But if we allow singularities in the total space of the fibration, we can get it. Indeed if we consider the function $\sum_i x_i^2=0$ on the total space $\mathbb C^n/\mathbb Z_2$, where $\mathbb Z_2$ is acting by $z\to -z$, the vanishing cycle will be diffomorphic to $\mathbb RP^{n-1}$. In sum, it seems to me that the following question is interesting: can we describe the class of manifolds what appear as vanishing cycles of algebraic one-parameter families?

As for the question itself, I don't know what is the answer...

  • $\begingroup$ All you said is correct, and my question is some how exactly what you asked at the end: Can a family of Quintics develop a singulary given by the local equation $x_1^2+\cdots+x_4^2$ on $\mathbb{C}/\mathbb{Z}_2$ ? $\endgroup$ Jan 10, 2011 at 15:16

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