Obstructions to being a hyperplane section or a fibre of a Lefschetz pencil

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Given a smooth projective variety $X$, when could $X$ fail to be a hyperplane section in some other variety $Y$, or fail to be the fibre of some Lefschetz pencil $\widetilde{Y} \rightarrow \mathbb{P}^{1}$?

Here, the variety $Y$ is not fixed, but simply required to exist.

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2 Answers

Your question is related to the problem of the existence of non-trivial extensions of subvarieties $X \subset \mathbb P^N$ to $\mathbb P^{N+1}$. An extension of $X$ is just a subvariety $Y$ of $\mathbb P^{N+1}$ such that $X= Y \cap \mathbb P^{N}$. It is called trivial if $Y$ is the join of $X$ and a point outside of $\mathbb P^N$.

This is a classical question that was studied by the Italian school of algebraic geometry. For instance, Scorza proved that the Veronese surface in $\mathbb P^5$ does not admit non-trivial extensions.

More recently, the problem has been studied by Zak, S. L'vovsky, L. Badescu, among many others. In Extensions of projective varieties and deformations by S. L'vovsky you will find the following result:

Theorem. Suppose $X$ is not $\mathbb P^N$ nor a quadric. If $\dim X \ge 2$ and $H^1(X,TX\otimes \mathcal O_{\mathbb P^N}(-1))=0$ then every extension of $X$ is trivial.

For a very nice introduction to this circle of ideas see the first chapter of the book Projective geometry and formal geometry by L. Badescu. Unfortunately, the relevant Chapter doesn't seem to be available from Google books.

Of course this does not answer your question as the embedding of $X$ into $\mathbb P^N$ is fixed.

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A generic curve over $\mathbb{C}$ of large genus, say at least $24$, will not be a fibre of a Lefschetz pencil or even a hyperplane section.

The reason is that by the theorems of Harris-Mumford and Eisenbud-Harris, the moduli space of curves of large genus is of general type, so there can be no rational curve passing through a generic point. To use this one needs to know that all the smooth fibres of the pencil are not isomorphic. Since the local monodromy around a singular fibre is infinite by the Picard-Lefschetz formula, it suffices to show that there must be at least one singular fibre. But if all fibres are smooth, then $\tilde{Y}$ (the total space of the Lefschetz pencil) must be isomorphic to $C \times \mathbb{P}^1$. This cannot happen since $C \times \mathbb{P}^1$ is not a blow up of any other surface.

Over a field of characteristic zero any very ample linear system contains a Lefschetz pencil, so it follows from the above that the smooth members of the linear system cannot all be isomorphic. But this would give rise to a unirational variety containing a generic point of the moduli space and this is not possible.

A similar argument should work for other classes of varieties whose moduli spaces are not uniruled.

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Since OP did not specify that Y be smooth, this is technically false.  Every X is a hyperplane section of a projective cone over X. – Jason Starr Nov 4 2011 at 17:03
Sandor -- I am afraid I do not understand the point you are trying to make.  The OP stated that X is smooth.  The projective cone over X is singular, but not along the generic hyperplane section X.  Rather it is singular at the vertex of the cone.  So I am afraid I do not understand your comment. – Jason Starr Nov 4 2011 at 18:18
Jason, sorry. I had something else in mind. :( – Sándor Kovács Nov 4 2011 at 18:25
Thanks. This is a very good answer. Perhaps I should have asked the converse: given a smooth projective variety $X$, when can and how can I hope to realize it as the fibre of a Lefschetz pencil?  – A. Pascal Nov 4 2011 at 18:30
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