Let $X$ be a stable curve consisting of two components meeting at three points. Let $M$ be its versal deformation space. The locus in $M$ parametrizing singular curves is a divisor with three components $D_1, D_2, D_3$, each corresponding to a node on $X$. The question is, which curve does a general point on $D_1\cap D_2$ correspond to? More precisely, does it correspond to an irreducible nodal curve, or a curve with two components meeting at two points?


1 Answer 1


A general point on $D_1 \cap D_2$ is given by smoothing the third node, which produces an irreducible curve with two nodes.

  • $\begingroup$ Your other possibility would have the wrong arithmetic genus, unless two of the nodes collided, but I guess that's not in the versal deformation space at all. $\endgroup$ Jun 22, 2013 at 5:24
  • $\begingroup$ Thank you very much. May I ask for some more detail explanation of it? $\endgroup$
    – marker
    Jun 22, 2013 at 15:10
  • $\begingroup$ Perhaps you just need to distinguish deformations from resolutions. I.e. a curve with two components meeting at two nodes is obtained by partially resolving the original curve by separating the branches at the third node, but this is a not a "deformation". Indeed the only way to deform a node non trivially is to smooth it. A deformation preserves the arithmetic genus, as Allen Knutson referred to, but a resolution reduces it. I always recall the maxim of Brieskorn, something like that analysis of singularities has three key aspects: resolution, deformation, and monodromy. $\endgroup$
    – roy smith
    Jun 23, 2013 at 20:19

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