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Let $C\subset X$ be a smooth irreducible curve of genus $g$, embedded in a smooth projective 3-fold $X$. So its homology class $\beta=[C]\in H_2(X)$ is an irreducible class. I want to compare two moduli schemes:

  • The Chow scheme $CH_1(X,\beta)$ of 1-cycles of degree $\beta$ on $X$, and
  • the Hilbert scheme $I^{CM}_{1-g}(X,\beta)$ of Cohen-Macaulay curves of arithmetic genus $g$ in class $\beta$.

The latter is an open subscheme of the full Hilbert scheme $I_{1-g}(X,\beta)$, parametrizing subschemes $Z\subset X$ such that $\chi(\mathscr O_Z)=1-g$ and $[Z]=\beta$. I do not know much about the former.

The notation for the Chow scheme does not include the genus, according to the fact that the arithmetic genus of a CM curve in class $\beta$ can change -- even for irreducible classes, I think. I cannot figure any (set-theoretic) difference between the Chow scheme and the space of all Cohen-Macaulay curves in class $\beta$ (although I never heard of a proper definition of the latter). So here is my question:

Are the moduli schemes $I^{CM}_{1-g}(X,\beta)$, with varying $g$, exactly the connected components of $CH_1(X,\beta)$? Are they at least open in the Chow scheme?

A comment: how many $g$? According to this post, the arithmetic genus of a CM curve is only bounded from above for fixed $\beta$, but since our $\beta$ is irreducible, I think all curves in class $\beta$ will be reduced, so we should get a lower bound as well.

Edit. By irreducible curve class, I mean that one cannot write $\beta=\beta_1+\beta_2$, with $\beta_i$ the class of a curve.

Thanks!

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    $\begingroup$ A Cohen-Macaulay curve can still be nonreduced, i.e., some primary ideals of minimal primes need not be prime. There are typically moduli of that nonreduced structure. The Hilbert scheme sees those moduli, but the Chow scheme does not. $\endgroup$
    – Jason Starr
    Commented Aug 12, 2016 at 16:23
  • $\begingroup$ I know there can be fat components on a CM curve, but can this happen for irreducible homology classes? My point was that if $\beta$ is irreducible, all curves in that class are reduced, and Hilb does not contain points corresponding to nonreduced curves. Is it wrong? $\endgroup$
    – Brenin
    Commented Aug 12, 2016 at 17:09
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    $\begingroup$ I was confused about the term irreducible class. If by irreducible class you mean a class that can be decomposed (nontrivially) as a sum of other effective curve classes, then you are correct that you cannot have nonreduced structure. However, a decomposable class can be represented by an irreducible curve. $\endgroup$
    – Jason Starr
    Commented Aug 12, 2016 at 17:39
  • $\begingroup$ @JasonStarr So at least as points Chow is the union of these finitely many Hilbert schemes of CM curves. Do you think they are in fact its connected components? It looks like the arithmetic genus is the only discrete invariant left. But do we even have an immersion $I_{1-g}^{CM}(Y,\beta)\to CH_1(Y,\beta)$? $\endgroup$
    – Brenin
    Commented Aug 15, 2016 at 14:51

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