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?



Rationality of Hilbert schemes?
[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.)


