Any irreducible component of the HIlbert scheme contains an irreducible element - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T21:27:24Z http://mathoverflow.net/feeds/question/101722 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101722/any-irreducible-component-of-the-hilbert-scheme-contains-an-irreducible-element Any irreducible component of the HIlbert scheme contains an irreducible element Naga Venkata 2012-07-09T02:22:17Z 2012-07-16T17:17:38Z <p>Consider $Hilb_{d,g}$ the Hilbert scheme of curves in $\mathbb{P}^3$ of degree $d$ and genus $g$. Is it true that if $L$ is an irreducible component of $Hilb_{d,g}$ then there exists a curve $C \in L$ such that $C$ is irreducible? Is there some criterion under which this is true?</p> <p>If the former statement is true, can we further say that a generic element of $L$ is an irreducible curve?</p> http://mathoverflow.net/questions/101722/any-irreducible-component-of-the-hilbert-scheme-contains-an-irreducible-element/101759#101759 Answer by Jack Huizenga for Any irreducible component of the HIlbert scheme contains an irreducible element Jack Huizenga 2012-07-09T09:46:28Z 2012-07-09T09:46:28Z <p>The Hilbert scheme is pretty horribly behaved, and positive results of this nature are quite rare.</p> <p>You probably mean to ask if there exists $C\in L$ such that $C$ is both reduced and irreducible. If you don't demand reducedness, the answer is yes. For suppose $H$ is a component of a Hilbert scheme of curves in $\mathbb{P}^n$, and let $C\in H$ be a general member of $H$. Let $M_t\in PGL(n+1)$ be a family of linear transformations which limits to the projection from a general codimension 2 plane in $\mathbb{P}^n$ as $t\to 0$ (so that it projects onto a line). Then $M_t(C)$ gives a curve in $H$, but the limiting curve in $H$ is supported entirely on a line, hence is irreducible.</p> <p>Whenever you have a component with a reduced and irreducible member, it is in fact true that the general member is reduced and irreducible, as being integral is open in flat families.</p> <p>In general, the Hilbert scheme is always connected (a theorem of Hartshorne); however, irreducible components typically meet at highly non-reduced members instead of the "nice" objects we're actually trying to parameterize.</p> <p>It is not reasonable to expect that every component of a Hilbert scheme of curves has an irreducible member. For instance, the Hilbert scheme parameterizing twisted cubic curves also has a separate component parameterizing unions of plane cubic curves and an isolated point. You may object this latter object is not a "curve," but part of the problem with Hilbert schemes is that the Hilbert scheme of curves doesn't even only see things of pure dimension 1.</p> <p>Similarly, the Hilbert scheme corresponding to the Hilbert polynomial of two skew lines does not have any reduced and irreducible curves in it.</p> <p>To get more satisfying counterexamples, one only needs to increase the degree and genus--things get really nasty really fast.</p> http://mathoverflow.net/questions/101722/any-irreducible-component-of-the-hilbert-scheme-contains-an-irreducible-element/102373#102373 Answer by Vladimir Baranovsky for Any irreducible component of the HIlbert scheme contains an irreducible element Vladimir Baranovsky 2012-07-16T17:17:38Z 2012-07-16T17:17:38Z <p>You may want to look at curves in three dimensional projective space. I vaguely remember, although I may be wrong, that there was a component with general member non-reduced. </p>