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"!


1 Answer 1


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.

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

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