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The wonderful responses to an earlier question Self-intersection and the normal bundle motivated me to ask the following question:

Let $Y \subset X$ be a subvariety of a variety $X$. Infinitesimal deformations of $Y$ in $X$ are subschemes of $X \times \textrm{Spec } k[\epsilon]/(\epsilon^2)$ flat over $\textrm{Spec } k[\epsilon]/(\epsilon^2)$ and with closed fiber $Y$. Such subschemes correspond bijectively to sections of the normal bundle $\mathcal{N}_{Y/X}$. (Hartshorne, III.9)

$\textbf{Question:}$ Do infinitesmal deformations of a regularly embedded subvariety $Y \subset X$ of codimension $d$ naturally determine cycles in $X$ (rationally equivalent to $Y$)? This seems like a bit of a long shot, but comments of Charles and Donu in the linked question seem to suggest that something like this is true.

If this were true, it would be important for both the linked question, and in it's own right. References where I can learn the relevant material be greatly appreciated.

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The answer is no, since first-order deformations can be obstructed, so that they do not give necessarily global embedded deformations of the subscheme.

For instance, Mumford gives an example of a smooth surface $X$ containing a curve $C$ such that $h^0(N_{C/X}) \neq 0$ (and so $C$ can be deformed in $X$ infinitesimally to the first order) but $C$ cannot be moved globally; in other words, no effective cicle different from $C$ arises in $X$ from the first-order deformations of $C$.

The reduced subscheme of the Hilbert scheme $\textrm{Hilb}_X^{C}$ in a neighborhood of the point $\xi$ corresponding to $C$ consists only of $\xi$, but the tangent space of $\textrm{Hilb}_X^C$ at $\xi$ is $1$-dimensional. This means that the local ring of $\textrm{Hilb}_X^C$ at $\xi$ contains nilpotents elements, and shows that the appearance of nilpotents is unavoidable also in entirely "classical" questions of algebraic geometry.

For further details, see Shafarevich's book Classical algebraic geometry 2: schemes and complex manifolds, page 111 and Mumford's Lectures on curves on algebraic surfaces, Lecture 22.

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Francesco, Thanks for your answer! At the end of the second paragraph, you mention that "...there is no effective cycle arising from a first order deformation of $C$". What if you don't require the cycles to be effective - does the same result hold? Requiring the relevant cycle to be effective seems to be too much - I would (perhaps naively) expect to find curves $C$ on a surface $X$ so that $\mathcal{O}_X(C)$ has a one dimensional space of global sections - while the normal bundle has more global sections. – LMN Dec 20 '12 at 21:24
Sorry, I do not understand well your question. An embeddeed deformation of $C$ is a flat family and $h^0$ of the structure sheaf of the fibres must remain constant. So if $C$ is reduced and irreducible, the fibres must be effective and connected, since $h^0(C, O_C) =1$. So the only resonable cycle to expect is an effective, connected cycle (maybe non-reduced and with embedded components). – Francesco Polizzi Dec 20 '12 at 21:37
Ok, I see. I think you answered my question. – LMN Dec 20 '12 at 21:47
Notice that you can have $h^0(\mathcal{O}_X(C))=1$ and $h^0(C, N_C)>0$ even if the deformations are not obstructed. For instance, take an elliptic $C$ curve on an abelian surface $A$: it does not move in a linear system, but the normal bundle of $C$ in $A$ has a $1$-dimensional family of sections: in fact, you can move $C$ into its algebraic equivalence class, using the translations in $A$. The translates of $C$ are a $1$-dimensional family of effective curves on $A$ that are NOT linearly equivalent to $C$, and that come from the $1$-dimensional family of first-order deformations! – Francesco Polizzi Dec 20 '12 at 21:49
Great, thanks ! – LMN Dec 20 '12 at 22:05

I'd like to expand a bit on Francesco Polizzi's excellent answer, with a couple of examples. He's right of course that deformations may be obstructed, but there's another issue which prevents infinitesimal deformations from being realized as families of rationally equivalent cycles.

As Francesco observes, the existence of obstructed deformations of a subscheme $Z\subset X$ is equivalent to the non-smoothness of the Hilbert $\operatorname{Hilb}_X^{p(Z)}$ at the point corresponding to $Z$. Now suppose one has some unobstructed deformation of $Z$, say a flat family $\mathcal{Z}\to k[t]/t^n$ with closed fiber $Z$. If $X$ is projective and $[Z]\in\operatorname{Hilb}_X^{p(Z)}$ is a smooth point, then by the projectivity of $\operatorname{Hilb}^{p(Z)}_X$ we can slice $\operatorname{Hilb}^{p(Z)}_X$ by hypersurfaces tangent to the family $\mathcal{Z}$ and thus realize $\mathcal{Z}$ inside a family over some curve.

This will give a family of subschemes algebraically equivalent to $Z$--but unless the curve is rational, they may not be rationally equivalent to $Z$. As a simple example, consider deforming points in an elliptic curve. Such deformations are always unobstructed, since elliptic curves are smooth by definition. But no two distinct points in an elliptic curve are rationally equivalent (because e.g. any map from a rational curve to an elliptic curve is constant, or because distinct points on an elliptic curve give distinct line bundles).

The question you are asking is thus not a local question about the Hilbert scheme, but rather one about its global geometry---namely, is there a rational curve in the Hilbert scheme realizing some given germ of a deformation. This is very hard in general, even for zero cycles on surfaces--see e.g. Bloch's conjecture, and the related beautiful paper by Mumford here.

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