Upper bound on the dimension of the Hilbert scheme of space cuves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:29:30Z http://mathoverflow.net/feeds/question/111166 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111166/upper-bound-on-the-dimension-of-the-hilbert-scheme-of-space-cuves Upper bound on the dimension of the Hilbert scheme of space cuves Naga Venkata 2012-11-01T15:56:46Z 2013-03-08T04:22:00Z <p>Denote by $H_{P,Q}$ the flag Hilbert scheme parametrizing a pair $(C,X)$ such that $X$ is a degree $d$ surface in $\mathbb{P}^3$ with Hilbert polynomial $Q$ and $C \subset X$ is a curve with Hilbert polynomial $P$. Let $p_1$ be the natural projection map from $H_{P,Q}$ to $H_P$.</p> <p>Then the questions are:</p> <p>1) Is there an upper bound on the dimension of $Im(P_1)$ in terms of the degree $d$?</p> <p>2) If the dimension of $H_{P,Q}$ is large (say greater than $d^2$) then if we replace $Q$ by the Hilbert polynomial of degree $d-1$ surfaces in $\mathbb{P}^3$ then is the corresponding dimension of $Im(P_1)$ is equal to the one before? This is equivalent to saying that if dimension of $H_{P,Q}$ is large then for a generic curve in $Im(P_1)$, do we have that $I_{d-1}(C) \not= \emptyset$? </p> <p>3) Can we also say that the degrees of the defining equations of a generic curve in $Im(P_1)$ is the same? That is to say is there a fixed $r$-tuple of integers $(a_i)$ such that a generic curve in $Im(P_1)$ is defined by $r$ equations $Q_i$ of degree $a_i$ respectively? (of course, I do not expect $Q_i$ to be fixed)</p> <p>Partial/known results and ideas of approaching this problem are most welcome.</p> http://mathoverflow.net/questions/111166/upper-bound-on-the-dimension-of-the-hilbert-scheme-of-space-cuves/111201#111201 Answer by Will Sawin for Upper bound on the dimension of the Hilbert scheme of space cuves Will Sawin 2012-11-01T19:53:17Z 2012-11-01T19:53:17Z <p>Since every degree $d$ hypersurface in $\mathbb P^3$ has the same Hilbert polynomial, <code>$Q(n)=\left(\begin{array}{c} n+3 \\ 3 \end{array}\right) - \left(\begin{array}{c} n+3-d \\ 3 \end{array}\right)$</code>, the introduction of $Q$ is unnecessary.</p> <p>1) I agree with Charles Staats. In fact his example can be extended to show that, for each degree $d$, not just $1$, the dimension of the image is unbounded. Set</p> <p><code>$P(n)=\left(\begin{array}{c} n+3 \\ 3 \end{array}\right) - \left(\begin{array}{c} n+3-d \\ 3 \end{array}\right)-\left(\begin{array}{c} n+3-e \\ 3 \end{array}\right) + \left(\begin{array}{c} n+3-d-e \\ 3 \end{array}\right)$</code></p> <p>the Hilbert polynomial of the complete intersection of a degree $d$ hypersurface and a degree $e$ hypersurface. Than the image of $P_1$ certainly includes each complete intersection of a degree $d$ hypersurface and a degree $e$ hypersurface, and the dimension of the space of such intersections clearly goes to $\infty$ as $e$ does.</p> <p>2) Clearly, the image of $H_{P,Q(d)}$ inside $H_P$ is chain of ascending closed sets. It eventually encapsulates the whole space because every curve is contained inside some hypersurface, including the curves corresponding to the generic points of $H_P$.</p> <p>Thus, if there is a lower bound for the dimension that applies to $H_{P,Q(d)}$, it also applise to $H_{P,Q(d+1)}$, so the dimension of the image of the projection never increases. So it is a necessary condition that the dimension of $H_{P,Q(d-1)}$ is already equal to the dimension of $H_P$. This condition is obviously also sufficient.</p> <p>I do not expect there to be a lower bound on $H_{P,Q(d)}$ in terms of $d$ that implies $H_{P,Q(d)}\geq H_P$, because there is no compelling reason for there to be, but I cannot think of an example.</p> <p>3) If you have some scheme parameterizing curves, you can take a generic point of an irreducible component and view its fiber as a curve over a larger field. Take the defining equations of the curve. For most points in the irreducible component, these defining equations still make sense, and for most of those points, those defining equations reproduce the fiber, since the set of points where it fails to reproduce the fiber is constructible.</p> <p>Obviously one cannot expect generic conditions on different irreducible components to be the same.</p>