Consider the Hilbert scheme of curves in $P^3$ with genus $g$ and degree $d$, $H_{g,d}$. Is this rational for some $g$ and $d$?
Edit: For which $(g,d)$ is this rational?
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6$\begingroup$ Surely it's rational for some $g$ and $d$, e.g. $(g,d)=(0,3)$ (rational normal cubics). Not for all $(g,d)$, though, because for $g$ large enough ${\cal M}_g$ is of general type, and dominates this Hilbert scheme once $d$ is large enough given $g$. $\endgroup$– Noam D. ElkiesCommented Dec 29, 2016 at 5:14
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$\begingroup$ @NoamD.Elkies Thanks, I would be happy to accept that as an answer. Do you know a good reference for the higher genus case you describe? $\endgroup$– Elle NajtCommented Dec 29, 2016 at 19:39
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$\begingroup$ @NoamD.Elkies Specifically I mean ... a reference for why $M_g$ is of general type, and for why when $d >> g$ $M_g$ dominates the corresponding Hilbert scheme. I'm also curious about the low degree cases as well, but I'm not sure where to study this material. $\endgroup$– Elle NajtCommented Dec 29, 2016 at 19:45
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$\begingroup$ The fact that ${\cal M}_g$ is of general type for $g \geq 24$ is "a famous result due to Harris, Mumford and Eisenbud" as G.Farkas puts it at mathematik.hu-berlin.de/~farkas/m22.pdf (which proves the same result for $g=22$). $\endgroup$– Noam D. ElkiesCommented Dec 29, 2016 at 21:36
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$\begingroup$ For $g>0$, if $d \gg g$ then the sections of any divisor of degree $d$ embed $C$ as a curve of degree $d$ in a projective space of dimension $\geq 3$. Now project to a generic ${\bf P}^3$ and you're done. $\endgroup$– Noam D. ElkiesCommented Dec 29, 2016 at 21:43
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1 Answer
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[expanding on my comments above]
$H_{g,d}$ is rational for some but not all $(g,d)$.
A rational example is $H_{1,4}$: a quartic elliptic curve in ${\bf P}^3$ is the complete intersection of two quadrics, so $H_{1,4}$ is just the Grassmanian of $2$-dimensional subspaces of the $10$-dimensional space of quadrics. Similarly for other cases where the curve must be a smooth complete intersection; the simplest examples are $(g,d) = (0,1)$ and $(0,2)$ if you allow curves that do not span ${\bf P}^3$.
On the other hand, given $g$, for $d$ large enough the Hilbert scheme $H_{g,d}$ dominates the moduli space ${\cal M}_g$ of genus-$g$ curves, because every such curve can be embedded in ${\bf P}^3$ with degree $d$. (Use sections of a degree-$d$ divisor to embed in ${\bf P}^{d-g}$, and project down to a generic ${\bf P}^3$.) But ${\cal M}_g$ is of general type once $g \geq 24$ (Harris-Mumford-Eisenbud), so $H_{g,d}$ cannot be rational or even unirational for such $g$.
The question of characterizing all $(g,d)$ for which $H_{g,d}$ is rational, or even unirational, might be quite hard. (I gather that "unirational" can be more accessible; e.g. the Hilbert scheme $H_{0,3}$ of rational normal cubics is certainly unirational but it's not obvious to me that it is actually rational as I rashly claimed in my first comment.)
answered Dec 30, 2016 at 3:35
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2$\begingroup$ In fact $H_{0,3}$ is rational. This is easier to see if you use the space of stable maps of degree $e$, genus $0$ curves to $\mathbb{P}^n$, say $\overline{\mathcal{M}}_{0,0}(\mathbb{P}^n,e)$. Then the standard torus action on $\mathbb{P}^n$ induces a torus action on the space of stable maps. Using Bialynicki-Birula, you are reduced to showing rationality of certain fixed point loci; for $e=3$, the fixed point loci are just points. Herb Clemens has a paper about this. $\endgroup$ Commented Dec 30, 2016 at 9:25