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For plane curve singularities most questions have been answered, in large part due to the Newton-Puiseux expansion. I've heard that there are a number of open problems regarding space curve singularities but have not found any clear statements. (One problem or set of problems seems to be around the characterization of which semigroups can arise.) Could someone point me to some of these problems? I am primarily interested in local singularities, over the complex numbers or reals, and to start with, unibranched singularities.

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Community wiki, since this seems not to expect/admit a "unique" right answer? – Yemon Choi Sep 13 '10 at 6:56
Tags: open-problem and open-problems-list are appropriate – Alexander Chervov Dec 10 '12 at 7:57

Edited. As Richard points out this is not really an answer, but rather a comment, that even for plane sinuglarities there are still some open questions. There is a facinating, completely (new and) open conjecture in the following article

The Hilbert scheme of a plane curve singularity and the HOMFLY polynomial of its link

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This answer does not address my question, but the reference is certainly nice. Thank you. – Richard Montgomery Sep 11 '10 at 18:13

I personally have a question I want answered, although it is not limited to space curves singularities. All unibranch curve singularities have a natural degeneration to the monomial singularity with the same semigroup; in particular, plane curve singularities with more than one Puiseux pair will degenerate to space(or higher) curve singularities. For example, the plane curve singularity $k[t^4,t^6 + t^7]$ degenerates to the space curve singularity $k[t^4, t^6, t^{13}]$.

I would like to understand how the (local contribution to the) relative compactified Jacobian degenerates in this family. The question I really want answered is: which torsion free sheaves on the central fibre deform out? In particular, it is known that the compactified Jacobian is irreducible if and only if the singularity is planar. Does the deformation to the monomial curve select out a component of the compactified Jacobian of the monomial curve singularity and, if so, is this component homeomorphic to the compactified Jacobian of the original planar curve singularity?

Let me also mention an obvious question following upon the work of Campillo, Delgado, and Gusein-Zade, who observe that the semigroup of a plane curve singularity "is" its Alexander polynomial. So: what is the topological meaning of the semigroup of a space curve singularity? An answer would of course bear upon the question of which such semigroups occur.

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Dear Vivek: (A) Could you specify the degeneration of $(t^4, t^6 + t^7, 0)$ degenerates to $(t^4, t^6, t^{13})$? (I do see how the space curve {\it projects} onto the plane curve.) (B) I recall reading a criteria which the semigroup of a plane curve must satisfy, and that some space curves fail this criterion. Do you recall that that criterion? – Richard Montgomery Nov 9 '10 at 5:37

I think the following is still open:

"Does there exist an isolated space curve singularity that is rigid?"

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Could you tell me what you mean by rigid' and isolated', or give me a reference? I come from the Arnol'd-Mather-Thom world of singularity theory, where rigid' would mean, roughly, that any small perturbation of the curve (perhaps with finite jet restrictions imposed) is locally diffeomorphic to the curve. Rigid' would then imply `simple' a la Arnol'd. (Simple unibranched space curve singularities, and so it would seem, rigid ones have been classified by Bruce and Gaffney. There are many.) – Richard Montgomery Sep 20 '10 at 20:01
@Richard Montgomery: Thanks for the response. Here is an (imprecise) explanation of terminology. "Isolated" means that singular locus of the curve is $0$-dimensional, so one excludes things like $z^2=y=0$. If you are coming from the Arnol'd-Mather-Thom world, then I would guess you are already assuming this. "Rigid" means any small perturbation of the singularity is analytically equivalent to the trivial perturbation of the singularity. For example, the node $x y =0$ (a simple singularity) is not rigid because it admits the perturbation $x y= t$. – jlk Sep 21 '10 at 3:27
One reference is Introduction to singularities and deformations by Greuel, Lossen, and Shustin. The conjecture is stated as Conjecture 1.11 on page 236. I don't think they were the first ones to state the conjecture, but I can't think of another reference off-hand. – jlk Sep 21 '10 at 3:28
Here I am working over the complex numbers $\mathbb{C}$. I just noticed you are also interested in real singularities. – jlk Sep 21 '10 at 3:29

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