most recent 30 from http://mathoverflow.net 2013-05-20T20:29:46Z http://mathoverflow.net/feeds/question/69450 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69450/deform-projective-kahler-to-projective-kahler SergY 2011-07-04T08:47:38Z 2012-12-26T08:56:41Z

<p>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 $t \neq 0$? </p> <p>Easy example: let $Y$ be a compact Kahler manifold with $H^{2,0}(Y)=0$, then "Deform projective Kahler to projective Kahler"!</p>

http://mathoverflow.net/questions/69450/deform-projective-kahler-to-projective-kahler/69451#69451 Francesco Polizzi 2011-07-04T09:19:32Z 2011-07-05T06:41:51Z

<p>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 $\mathcal{X}$. </p> <p>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. </p> <p>In other words, $c_1(X_t)$ remains positive for $t$ close enough to $0$. </p> <p>Notice that this is <strong>not</strong> 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. </p> <p>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 <strong>not</strong> positive.</p> <p>Of course, by the previous considerations the surface $X_t$ does not contain any $(-2)$-curve if $t$ is small enough.</p>