# deformation of curves with three nodes

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

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

• 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. – Allen Knutson Jun 22 '13 at 5:24
• Thank you very much. May I ask for some more detail explanation of it? – marker Jun 22 '13 at 15:10
• 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. – roy smith Jun 23 '13 at 20:19