Infinitesimal deformations and moving cycles - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:18:37Z http://mathoverflow.net/feeds/question/116885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116885/infinitesimal-deformations-and-moving-cycles Infinitesimal deformations and moving cycles LMN 2012-12-20T17:30:26Z 2012-12-20T21:38:45Z <p>The wonderful responses to an earlier question <a href="http://mathoverflow.net/questions/111464/self-intersection-and-the-normal-bundle" rel="nofollow">http://mathoverflow.net/questions/111464/self-intersection-and-the-normal-bundle</a> motivated me to ask the following question:</p> <p>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)</p> <p>$\textbf{Question:}$ 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.</p> <p>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.</p> http://mathoverflow.net/questions/116885/infinitesimal-deformations-and-moving-cycles/116916#116916 Answer by Francesco Polizzi for Infinitesimal deformations and moving cycles Francesco Polizzi 2012-12-20T21:08:13Z 2012-12-20T21:30:02Z <p>The answer is <strong>no</strong>, since first-order deformations can be obstructed, so that they do not give necessarily global embedded deformations of the subscheme. </p> <p>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 $C$.</p> <p>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.</p> <p>For further details, see Shafarevich's book <em>Classical algebraic geometry 2: schemes and complex manifolds</em>, page 111 and Mumford's <em>Lectures on curves on algebraic surfaces</em>, Lecture 22.</p> http://mathoverflow.net/questions/116885/infinitesimal-deformations-and-moving-cycles/116920#116920 Answer by Daniel Litt for Infinitesimal deformations and moving cycles Daniel Litt 2012-12-20T21:38:45Z 2012-12-20T21:38:45Z <p>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.</p> <p>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.</p> <p>This will give a family of subschemes <em>algebraically</em> equivalent to $Z$--but unless the curve is rational, they may not be <em>rationally</em> 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).</p> <p>The question you are asking is thus not a <em>local</em> question about the Hilbert scheme, but rather one about its global geometry---namely, is there a <em>rational curve</em> 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 <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.kjm/1250523940" rel="nofollow">here</a>. </p>