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There are a number of techniques in algebraic geometry that can be used to show that a given reducible (often genus-zero) curve $C$ in a smooth variety $X$ becomes smooth and irreducible after a (generic) deformation. Such techniques are used, for instance, to show that (for smooth varieties in characteristic zero) rational chain-connectedness implies rational connectedness.

However, suppose you have a genus-zero reducible curve that you suspect cannot be deformed (under a generic deformation) to a smooth rational curve--or even an irreducible rational curve. (More precisely, you suspect that for a given irreducible component of the Kontsevich moduli of stable maps, the general point corresponds to a reducible curve.) Are there any obvious criteria that would allow you to show this?

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    $\begingroup$ the obvious technique is to show the tangent space of the component of moduli has the same dimension as the subset parametrizing reducible curves. E.g. the extreme case is if the reducible curve is "rigid", i.e. does not deform at all. $\endgroup$
    – roy smith
    Commented Dec 3, 2012 at 2:42
  • $\begingroup$ Roy: Thanks for the excellent suggestion. Unless I misunderstand, the tangent space dimension will typically be harder to compute than the "expected dimension" (tangent space dimension minus obstruction dimension), since the latter is often accessible via Riemann-Roch. $$ $$ One advantage of this technique is that it is (in principle) capable of working even if we are looking at an irreducible component of the moduli space, some of whose points lie in the closure of the moduli of irreducible curves. $\endgroup$
    – Charles Staats
    Commented Dec 3, 2012 at 17:05

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I agree with Roy that the main technique is to prove that a subscheme of the moduli space paramterizing reducible curves is the entire moduli space using an infinitesimal argument. Another argument is numerical, e.g., consider a Hirzebruch surface with directrix curve $C_0$. Consider a reducible, genus $0$ curve in the surface with negative intersection number with $C_0$, yet whose class is not a multiple of $[C_0]$. This curve cannot be deformed to an irreducible curve, or else it would have nonnegative intersection number with $[C_0]$.

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  • $\begingroup$ I'm familiar with this technique in the case of surfaces--a more basic example is to take $X$ to be the blowup of $\mathbb P^2$ at a point, and consider reducible curves that have negative intersection number with the exceptional divisor $E$. But I have two questions: 1. Can this sort of argument be applied if the ambient $X$ is not a surface? 2. Is there any hope for this sort of argument if we want to show that the general curve is reducible on an irreducible component of the moduli space (but not the whole connected component)? $\endgroup$
    – Charles Staats
    Commented Dec 3, 2012 at 16:58
  • $\begingroup$ Edit: Probably, I mean to say that the general curve cannot be infinitesimally deformed to be irreducible. (But some points of the moduli component might lie in the closure of the space of irreducible curves.) $\endgroup$
    – Charles Staats
    Commented Dec 3, 2012 at 17:04
  • $\begingroup$ Additional note: I've just looked up the definition of "Hirzebruch surface," and actually my "more basic example" is a special case of your example. $\endgroup$
    – Charles Staats
    Commented Dec 3, 2012 at 18:59

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