Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I was wondering about a generalization of the following property of surfaces of general type.

Let $X$ be a smooth projective surface of general type. Then there is no pencil of rational or elliptic curves on $X$, i.e., there are no pencils of varieties with non-maximal Kodaira dimension on $X$. (edit: see below what i mean by pencil)

Something like this will probably already fail in dimension three, but let me still be brave and ask.

Let $X$ be a smooth projective variety of general type satisfying some minimality condition, say it has ample canonical bundle. Then is it true that $X$ does not admit any pencil of varieties with non-maximal Kodaira dimension?

Slightly weaker, if $n$ is the dimension of $X$, is it true that $X$ does not admit any pencil of $(n-1)$-dimensional varieties with non-maximal Kodaira dimension?

If this fails, what extra conditions would make this work. Does it help if we assume the locally free sheaf of differentials to be ample as well?

The minimality condition is certainly necessary. If not you could just blow-up and your variety would contain a projective space. That would not be good.

Edit: When I say pencil on a variety, i mean the data of a variety $B$ and a non-isotrivial family of varieties $X\to B$. A pencil of abelian varieties on a variety $X$ is thus a base variety $B$ admitting a flat projective non-isotrivial family $X\to B$ of abelian varieties. If the base is a curve, this is a good definition. In general, this might not be a good definition. Note that the weaker version of this question is precisely about the case where the base variety is a curve.

share|improve this question
Your new definition of pencil seems problematic, because it implies for instance that there are no pencils of curves at all on $\mathbf{P^2}$. –  Artie Prendergast-Smith May 9 '13 at 16:30
I don't understand what you exactly mean with your question. Do you want a fiber space? In this case I don't understand why the blow up is a counterexample. Do you want just a pencil? For instance, does an elliptic surface embedded in a projective manifold of general type answer your question? –  diverietti May 9 '13 at 16:56
By the way, about your weaker question: if K_X is ample, I think it's easy to see by adjunction that X cannot have an algebraic family of codimension-1 subvarieties not of general type. –  Artie Prendergast-Smith May 9 '13 at 17:35
I'm a little confused by the question, however it is a well known fact that if "$X$" is a smooth complex projective variety of general type and "$x\in X$" is a very general point then any subvariety "$x\in V\subset X$" is of general type. (Idea of proof: If there is a covering family "$Y\to T$" then after cutting down $T$ we may assume that "$Y\to X$" is generically finite and hence "$Y$" is of general type. By the easy addition formula "$\kappa (Y)\leq \kappa (Y_t)+\dim T$" hence "$\kappa (Y_t)=\dim Y_t$". See arXiv:0812.3454 for this and related results.) –  hacon May 10 '13 at 14:22
Firstly, my apologies for being so unclear. I think I`m using some bad terminology. I will try to fix this. My second comment will be more substantial. –  Jonathan May 10 '13 at 16:37
show 5 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.