7
$\begingroup$

It is an apparently well known result that if one has a 1-parameter family of smooth Calabi-Yau 3-folds which acquires a node at a boundary point, then there is a vanishing lagrangian 3-sphere, and the geometrical monodromy is a Dehn-twist along this sphere whose effect on the homology of our base fiber is given by the classical Picard-Lefschetz formula $$\alpha \mapsto \alpha-(\delta\cdot \alpha) \delta,$$ where $\delta$ is the homology class of the vanishing 3-sphere. (This formula or an analogue of it actually holds in much more general families than just CY 3-folds, but that's the case I'm interested in). For a modern reference for the above result, see Chapter 3 of Looijenga's $\textit{Isolated Singular Points on Complete Intersections}$.

My question is whether there is a known analogue for the case when the family in question acquires more than one $A_1$-singularity. For example, if the smooth family acquires 4 nodes, is there a known formula for the monodromy action around such a fiber for Calabi-Yau 3-folds?

If no formula exists, is there any sense for what it's matrix might look like, roughly? Do we know, for example, that $Rank(S-1)=1$, where $S$ is the monodromy action and $1$ is the identity matrix?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

In general, I think that the monodromy should be $$\alpha\mapsto \alpha - \sum_i (\delta_i\cdot \alpha)\delta_i $$ where $\delta_i$ are the vanishing cycles corresponding to each node. I'll see whether I can write down an outline of an argument later.

... Meanwhile later... Let me complement Tim's excellent comment with a more obscure explanation. The so called variation map $S-I: H^i(X_t)\to H^i(X_t)$ of the cohomology of the nearby fibre factors through a the cohomology of a (complex of) sheaf(ves) $\mathbb{R}\phi\mathbb{Z}$ supported on singular locus of the special fibre $X_0$. In the case that $X_0$ has isolated singularities, this sheaf decomposes into a sum of the corresponding sheaves supported at each of the critical points, and maps decomposes as well. The formula I stated above would follow this together with the standard Picard-Lefschetz formula.

Improve this answer
$\endgroup$
5
  • 4
    $\begingroup$ Indeed: in a nearby fibre, the vanishing cycles will be disjoint. Far from the critical points, the whole family over the disc is trivialized by radial parallel transport into the central fibre; near the critical points, one has a standard quadratic model. Show in this way that the smooth monodromy is a simultaneous Dehn twist along the vanishing spheres $V_i$. Then prove Picard-Lefschetz by observing that for any $n$-cycle $x$, $(S-1)(x)$ is alway homologous to a cycle supported in neighborhoods $T^*V_i$ of the vanishing cycles, hence to a linear combination of the $\delta_i$. $\endgroup$
    – Tim Perutz
    Commented Feb 12, 2013 at 20:17
  • $\begingroup$ Is $\phi$ the map defining the family or something in the radial direction? Is there an easy way to see that the variation map factors in this way, or a reference for it? Thank you both for your good answers. I'm starting to get the picture, and I think this will definitely help with what I'm working on. A follow-up question is if the vanishing spheres are independent in homology. More to the point, do you know what $Rank(S-I)$ is? It seems to depend on the number of nodes that appear, unless there's some relation between these spheres. $\endgroup$
    – HNuer
    Commented Feb 13, 2013 at 15:15
  • $\begingroup$ Dimca has a book on Sheaves in Topology which would be a good source some of this. I'll say more later. Sorry, I have to run again. $\endgroup$
    – Donu Arapura
    Commented Feb 13, 2013 at 15:22
  • $\begingroup$ @Tim Perutz How do you know the vanishing cycles will be disjoint? $\endgroup$
    – Harry Reed
    Commented Mar 23, 2017 at 10:11
  • 1
    $\begingroup$ @Harry Reed The $i$th vanishing cycle can be defined, by means of a suitable connection for the family and a specified "vanishing path" to the critical value, as the points in the regular fiber such that limiting parallel transport takes you to the $i$th node. Since we can use the same path for all the nodes, these loci are disjoint. $\endgroup$
    – Tim Perutz
    Commented Mar 23, 2017 at 13:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .