2
$\begingroup$

I have two "rookie questions" about elliptic surfaces:

  1. Let $S$ be an elliptic surface over $\mathbb{C}$, i.e. a smooth, projective algebraic surface equipped with a morphism $f: S \to C$ to a curve $C$ such that the generic fibre is an elliptic curve over $\mathbb{C}(C)$. The Kodaira dimension of an elliptic surface is at most $1$ (but can be $0$ or $-\infty$). If one asks that $g(C) \geq 2$, does this force the Kodaira dimension to be equal to $1$, or can it still be $0$ or $-\infty$?

  2. A minimal elliptic surface is usually defined to be an elliptic surface which does not contain any vertical $(-1)$-curves, i.e. $(-1)$-curves contained in the fibres of the morphism $f: S \to C$. However there can be horizontal $(-1)$-curves. Does contracting such a curve always give another elliptic surface? If the Kodaira dimension is $1$ then this is certainly the case since it is a birational invariant. How do the fibres change in this case? I mean, they still have to be elliptic curves, but I don't have a very clear picture of what the relationship between the "new" and the "old" fibres is. I guess $C$ will not change, but I just can't imagine how the picture looks.

$\endgroup$
1
  • 4
    $\begingroup$ Answer to 1 is given by Francesco. Concerning the new question 2 two things can be said. First, if the base is a curve of genus $>0$ there can not be a section that is a rational curve (this is a simple exercise). Second, the answer to the beginning of your second question is NO. Indeed, the simplest such elliptic surface is obtained by blowing up $\mathbb CP^2$ in $9$ points - the base points of an elliptic pencil. If you blow down these curves you get back $\mathbb CP^2$, which is not elliptic. $\endgroup$ Apr 16, 2012 at 23:11

1 Answer 1

4
$\begingroup$

If $g(C) \geq 2$ then $\textrm{kod}(S)=1$.

The same holds also if $g(C)=1$ and $f$ is not locally trivial.

See [Barth-Hulek-Peters-Van de Ven, Compact Complex Surfaces], Proposition 12.5 page 215 (Chapter V).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.