Let $X=Y\times Z$ be a product of complex manifolds $Y,Z$. Is it true that there exists no rigid curve on $X$? Here I mean by a rigid curve a curve which is not a member of any family of curves on $X$.
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$\begingroup$ This would be true if you were asking about rational curves. $\endgroup$– Tom GraberCommented Mar 4, 2013 at 17:03
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1$\begingroup$ Hi Tom. How do you see the assertion for rational curves? Thanks. $\endgroup$– Atsushi KanazawaCommented Mar 4, 2013 at 18:18
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3$\begingroup$ If you have a map from P^1, to the product, then you have one to each factor separately. If one of those maps is constant, then you just deform the map by moving the point. If both maps are nonconstant, then you get a copy of $P^1 \times P^1$ in the product, where your curve lives, and so it deforms inside that surface. $\endgroup$– Tom GraberCommented Mar 4, 2013 at 18:51
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$\begingroup$ I see. The argument works only for $\mathbb{P}^1$. Thanks. $\endgroup$– Atsushi KanazawaCommented Mar 4, 2013 at 20:41
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1 Answer
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These curves may actually exist, as the following example shows.
Let $C$ be a smooth curve of genus $g$. Then the diagonal $\Delta \subset C \times C$ is isomorphic to $C$ and has self intersection $\Delta^2= 2-2g$.
On the other hand, the tangent space of the Hilbert scheme $\mathscr{H}$ of $\Delta$ in $C \times C$ at the point $[\Delta]$ has dimension given by
$$ \dim _{[\Delta]} \mathscr{H} = h^0(\Delta, \, N_{\Delta/ C \times C}) = h^0(\Delta, \mathscr{O}_{\Delta}(\Delta))$$ `
and this quantity is $0$ for $g \geq 2$, because in that case $\Delta^2 < 0$.
This precisely means that $\Delta$ is rigid in $C \times C$ as soon as $g \geq 2$.
The cone of effective curves in $C \times C$ ($g \geq 2$) is quite subtle, and not fully understood in general. For further details, see [R. Lazarsfeld, Positivity in Algebraic Geometry I, Section 1.5].
answered Mar 4, 2013 at 11:05