4
$\begingroup$

Let $X$ be a compact Kahler manifold with first Chern class $c_1(X)>0$ (i.e. positive). Consider a family $\pi\colon \mathcal{X} \to \mathcal{D}$ over the unit disc $\mathcal{D}$, with $X_0=X$. Do we know that $c_1(X_t)>0$ for $t \neq 0$?

Easy example: let $Y$ be a compact Kahler manifold with $H^{2,0}(Y)=0$, then "Deform projective Kahler to projective Kahler"!

$\endgroup$

1 Answer 1

8
$\begingroup$

By Kodaira embedding theorem, if $X$ is a compact Kahler manifold and $c_1(X)$ is positive, then $X$ is projective and $-K_X$ is ample, i.e. some power $(-K_X)^{\otimes n}$ gives an embedding $$\phi \colon X \to \mathbb{P}^{N}.$$ Now, given the family $\pi \colon \mathcal{X} \to \mathcal{D}$ we can consider the relative canonical line bundle $\mathcal{K}$ on $\mathcal{X}$.

The restriction of $\mathcal{K}^{-1}$ to the central fiber $X_0=X$ is precisely $-K_X$ and, since ampleness is an open condition in families ([Lazarsfeld, Positivity in Algebraic Geometry I, Proposition 1.2.17 pag. 29]), we can conclude that $-K_{X_t}$ is also ample if $t$ is small enough.

In other words, $c_1(X_t)$ remains positive for $t$ close enough to $0$.

Notice that this is not necessarily true for large $t$. For instance, take a smooth cubic surface $X \subset \mathbb{P}^3$, which is a Del Pezzo surface, and consider a $1$-parameter degeneration to a cubic surface with a node. Then take the simultaneous resolution of singularities, which exists for Rational Double Points.

In this way we obtain a family $\pi \colon \mathcal{X} \to \mathcal{D}$ whose central fiber $X_0$ is isomorphic to $X$ and such that the fibre $X_{\tilde{t}}$ contains a $(-2)$ curve for some $\tilde{t} \in \mathcal{D}$. Therefore the first Chern class of $X_{\tilde{t}}$ is zero when restricted to this curve, in particular it is not positive.

Of course, by the previous considerations the surface $X_t$ does not contain any $(-2)$-curve if $t$ is small enough.

$\endgroup$
2
  • 1
    $\begingroup$ I guess you mean $-K_X$ is ample, right? $\endgroup$
    – Henri
    Jul 4, 2011 at 9:34
  • $\begingroup$ Yes, of course you are right. Thank you. $\endgroup$ Jul 4, 2011 at 9:47

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.