# How does one find vanishing algebraic cycles?

I have a question, related to what I asked before. Let's consider a smooth hyperplane section $X$ of a smooth projective variety $Y$ over $\mathbb C$. According to Weak Lefschetz theorem, cohomology groups of $X$ coincide with those of $Y$ in all dimensions except for the middle one. In the middle dimensions the pull-back $i^\*: H^d(Y) \to H^d(X)$ is injective, but not surjective, and the "new" cycles on $X$ are called vanishing cycles. (The reason for such a name is that these "new" cycles will vanish when we approach singular fibers on the Lefschetz pencil.) Vanishing cycles also can be described as the ones that live in the kernel of $i_*: H^d(X) \to H^{d+2}(Y)$.

Let's consider the case when $X$ is even-dimensional, so that we can hope that the vanishing cycles are algebraic.

For example, for a smooth even-dimensional quadric in $\mathbb {CP}^n$ there exist one vanishing cycle - it is a difference $[E_1] - [E_2]$ of two maximal linear subspaces from different classes.

Another example I thought about is a smooth cubic surface in $\mathbb {CP}^3$, vanishing cycles here are generated by differences of pairs of lines $[l_1] - [l_2]$ lying on the cubic.

Now I wanted to ask, what are other examples people have in mind? I'm interested in the case, when vanishing cycles actually are algebraic.

Is there a general method to describe such cycles in concrete situations (like MG(3,6))?

Thanks

EDIT: Probably winter break is not a best time to start a bounty...

It seems to me, that on a generic hypersurface of $CP^n$ of degree more than 3 if $n=2$ and of degree more then $2$ when $n>3$ there are no algebraic vanishing cycles. This is surely the case for generic hypersurfaces in $\mathbb CP^3$ of deree $4$ and more. For them there is only one integral class in $H^{1,1}$, this is the class given by the hyperplane section. So the two examples that you give are in a certain sence very exceptional.

• Thanks for this interesting remark. Do you know any examples of vanishing cycles being algebraic and how people construct them when the ambient space $Y$ is not a projective space? – Evgeny Shinder Dec 22 '09 at 3:15
• Evgeny, I never thought about such examples before, because for me like for Tim, a vanishing cycle is more an object of symplectic geometry, we think of them as of little half-dimesnional lagrangean spheres. But in the case of $MG(3,6)$ your approach seems to be completely reasonable. Usually, when you deform a variety, its complex structure and Hodge classes change. But not in the case of $MG(3,6)$ -- the complex structure is unique here. I tried to comment on this varitety in your previous question. – Dmitri Panov Dec 22 '09 at 12:14

Evgeny Shinder firmly locates his question in the algebraic world. Nonetheless, I'm going to advertise how cleanly Picard-Lefschetz theory works in symplectic topology (cf. Seidel, "A long exact sequence for Floer cohomology"). As partial justification, I'll point out that the algebraic situation is far more complicated, as evidenced not only by Dmitri's answer but by the fact that the Hodge conjecture is open!

A symplectic Lefschetz pencil on a manifold $X$ of real dimension $2n+2$, together with a path from a regular value $b$ to a critical value, gives rise to a geometric vanishing cycle $V$. This is an embedded Lagrangian submanifold in $X_b$. It comes with a preferred isotopy class of diffeomorphisms $S^n\to V$, and hence (by the Lagrangian neighbourhood theorem) a symplectomorphism between a neighbourhood of $V$ in $X_b$ and a neighbourhood of $S^n$ in $T^*S^n$. The description is more-or-less reversible: the existence of a Lagrangian sphere in $M$ implies that, symplectically (but not necessarily algebraically, even when $M$ is algebraic), an ordinary-double-point degeneration of $M$ exists.

It might be interesting to think about the algebraic symplectic case (but I haven't).

Here's the method I had in mind looking at the examples I had. Let $X \subset Y$ be a smooth hyperplane section. Our goal is to detect cycles on $X$ that are not complete intersections of $X$ with cycles on $Y$.

Let's consider a cycle $Z$ on $Y$ such that $Z \cap X$ is reducible, say $$Z \cap X = A + B$$ for some cycles $A$, $B$ on $X$. Then we may hope $A - B$ or some similar combination can be a vanishing.

For quadrics and cubics above we take $Z$ to be tangent linear subspace of appropriate dimension.

Has anybody seen something like that applied in other cases?

The sad thing is that I sort of can make it work for $MG(3,6)$ to describe a vanishing cycle, but it's unclear from the description I get whether the cycle is rational or not.