I am interested in degree 6 genus 4 curves in $\mathbb{P}^{3}$, in other words, the complete intersection curve of a quadric and a cubic. Is there any result on the Hilbert scheme of such curves concerning:

(1) Its number of irreducible components;

(2) Singularities of the component containing smooth degree 6 genus 4 curves (is this component smooth?);

(3) How could one degenerate a smooth degree 6 genus 4 curve to other components? Like in twisted cubics case one can degenerate a twisted cubic to a plane nodal cubic with a spatial nonreduced structure on the node, I am not sure if there is similar degenerations in degree 6 genus 4 curve case since in general degenerations can be bad.


  • $\begingroup$ 1) There are probably tons of "junk" irreducible components. For example, like in the case of the twisted cubic you reference, there is another irreducible component parametrizing a degree 6 plane curve together with 6 additional points. You can check this is indeed a separate component: we know smooth canonical curves are general in the main component, since they are smooth points of the Hilbert scheme, but you cannot degenerate connected canonical curves to these disconnected plane curves union 6 points (and you can check these fill a full component by calculating H^0(N_C)). (Continued...) $\endgroup$ – Aaron Landesman Apr 5 '17 at 7:05
  • $\begingroup$ 2 and 3) It is true that a smooth canonical curve is a smooth point of the Hilbert scheme (you can use the normal bundle sequence, the euler sequence, and Serre duality to show H^1(N_C) = 0). But, the connected component of the Hilbert scheme cannot be smooth. For example, any point in the intersection of the two components described in the previous comment will not be smooth. For example, I think slow projections of canonical curves onto a plane will lie in the intersection of the two components. These will be singular points of the Hilbert scheme because they lie on two components. $\endgroup$ – Aaron Landesman Apr 5 '17 at 7:09
  • $\begingroup$ Yeah, I understand for the main irreducible component, smooth curves corresponds to smooth points, so the question is really asking along the intersection of the main component with other components, are the points on the main irreducible component smooth? This is true for twisted cubic case, since the two irreducible component themselves are smooth. $\endgroup$ – B.X. Apr 16 '17 at 21:46
  • $\begingroup$ I'm not sure how to answer that part of the question off the top of my head. If that's what the question is really asking, maybe you could edit the question to clarify that? It seemed to me like you were also asking several other questions (notably (1) and (3)) the way I read it. $\endgroup$ – Aaron Landesman Apr 18 '17 at 16:43

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