Suppose $S$ is a compact complex surface, $C\subset S$ is a one dimensional irreducible subvariety (a curve). Suppose further, there exists a family of biholomorphism of $S$ nearby the identity map. In other words, there exist $F(t,s),t\in\mathbb C,|t|<1$ which is holomorphic in all variables and biholomorphisms for fixed t and $F(0,\cdot)=Id$. If the image of the curve $C$ under $F(t,\cdot)$ is not fixed (sweep out on open set in $S$). Does that mean that $C$ is not rigid and the self-intersetion number of $C$ is not negative?
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ By definition your curve moves in the family $(C_t)$, with $C_t:= F_t^*C$, so it is not rigid, and $(C)^2=C\cdot C_t\geq 0$. What's the problem? $\endgroup$– abxJan 25, 2016 at 16:53
-
$\begingroup$ i thought the family should be flat, isn't it?and $C$ may not be smooth, so i don't know it the family is flat or not. $\endgroup$– user42804Jan 25, 2016 at 17:11
-
$\begingroup$ Coluld you explain a little bit more why $(C)^2=C⋅C_t$? $\endgroup$– user42804Jan 25, 2016 at 17:17
-
$\begingroup$ The family is automatically flat. And, flat or not, all the $C_t$ have the same cohomology class — how could it vary? $\endgroup$– abxJan 25, 2016 at 18:30
-
1$\begingroup$ View $F$ as an automorphism of $S\times D$ ($D=$ unit disk); it maps isomorphically $C\times D$ to the family $\mathcal{C}=\bigcup C_t\times \{t\} $. Since the constant family is obviously flat over $D$, so is $\mathcal{C}$. $\endgroup$– abxJan 25, 2016 at 21:51
|
Show 1 more comment