Let $U_{d,n}\subseteq(\mathbb{P}^d)^n$ denote the locus of $n$-distinct points in projective space $\mathbb{P}^d$ that lie on a rational normal curve of degree $d$, and let $V_{d,n}$ denote its topological closure. I would like to know if, for $d \le n-3$, the spaces $V_{d,n}$ are normal. If they are, this implies that their GIT quotients by $\text{SL}_{d+1}$ are normal, and hence that the birational morphisms from $\overline{M}_{0,n}$ to these GIT quotients (constructed in http://arxiv.org/abs/1012.4835) are contractions. This would clarify some results in this preprint and would be useful in future work as well.

Note that these loci $V_{d,n}$ can be constructed in other ways than the topological closure one given above. For instance, one can take the Kontsevich stable map space $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^d,d)$ and use the product of evaluation maps $\overline{\mathcal{M}}_{0,n}(\mathbb{P}^d,d) \rightarrow \mathbb{P}^d$ to cut out this locus. Similarly, if $\mathcal{H}$ denotes the closed component of the Hilbert scheme parametrizing rational normal curves of degree $d$ then there is an incidence locus in $\mathcal{H}\times(\mathbb{P}^d)^n$ and the projection to $(\mathbb{P}^d)^n$ also defines $V_{d,n}$. Perhaps one of these constructions helps address normality?


Hi Noah,

Consider the forgetful morphism $(\mathbb P^d)^{n+1}\to (\mathbb P^d)^n$. This restricts to a forgetful morphism $\pi:V_{d,n+1}\to V_{d,n}$. This is kind of like the map $\overline{\mathscr M}_{g,n+1}\to \overline{\mathscr M}_{g,n}$. The general fiber of $\pi$ is a $\mathbb P^1$ (the original degree $d$ rational normal curve).

Now let's see normality. Usually the best way to prove normality is via Serre's criterion: normal is equivalent to $R_1$ and $S_2$. So, you want to prove these conditions.

For $n=d+3$ it is easy since $V_{d,n}=(\mathbb P^d)^n$. So it is normal (actually smooth of course). In particular it is both $R_1$ and $S_2$, in fact it is Cohen-Macaulay (CM).

Now consider $\pi:V_{d,n+1}\to V_{d,n}$. If $V_{d,n}$ is $R_1$, then I think (likely) so is $V_{d,n+1}$: the singular set of $V_{d,n+1}$ is contained in the union of the pre-image of the singular set of $V_{d,n}$ and the locus where $\pi$ is not smooth. The former is of at least codimension $2$ by induction. I am not entirely certain about the latter, but in any case the non-smooth locus has two parts: the locus of fibers with multiple components and the locus of singular set of the reduced singular fibers. The latter of these is definitely of at least codimension $2$ since the locus of these fibers is of at least codimension $1$ and the singular part is at least codimension $1$ in that. So, you need that the locus of non-reduced fibers is at least codimension $2$. I think this seems OK, but this is one of those things I did not carefully work out. It seems to me that (any component of this) ought to be contained in (a component of) the reduced singular locus. The reason I think that is that it seems that there should be a partial "smoothing" of a non-reduced fiber when you just separate the components. If this is true, then the locus is at least of codimension $2$. Then again, if this fails then normality fails, so at least you got a necessary condition.

This should take care of condition $R_1$. Now on to $S_2$.

In some sense this is harder, because we need this condtion at the singular set as well while above we just needed to show that the singular set is not too big.

For $d=2$ this is easy, because a conic (including degenerate ones) is defined by a single equation. So we can prove that $V_{d,n}$ is actually CM (=Cohen-Macaulay) and hence $S_2$. This follows by induction: If $V_{d,n}$ is CM, then so is $V_{d,n}\times \mathbb P^d$ and $V_{d,n+1}$ is a Cartier divisor in $V_{d,n}\times \mathbb P^d$ and hence itself is CM.

The same proof does not work for $d>2$. Although a smooth curve in a smooth total space is a local complete intersection, as pointed out by mdeland in the remarks this does not remain true for all degenerations. (I should have realized this as this is a famous example...) At the same time I feel that one might still be able to do something like this. After all we do not need the individual curves to be lci or even S_2. (If we did, this example would lead to non-normality).

It seems this definitely works for $d=2$ but not for other $d$'s. However, at least this might give you some ideas on how one might go about proving something like this. I would add that if this does not turn out to be normal, then perhaps this is not the "right" compactification to consider. At the least you should assume that your degenrate curves have no embedded points, but I could imagine that the best to do is to demand that degenerations be stable, i.e., consider the pointed curve degenerating to a pointed degenerate curve that is stable. This would lead to a partial resolution of $V_{d,n}$ and the above proof would essentially prove that it is normal. Of course, this may be what you're doing in general or it does not work for some reason, I am not familiar with the relevant results, so this is just the first thing I would try.

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    $\begingroup$ I believe if you take a twisted cubic (d = 3), degenerating to a plane nodal cubic with a spatial embedded point, then it is not lci at the embedded point? $\endgroup$ – mdeland Apr 2 '11 at 15:49
  • $\begingroup$ @mdeland: you're absolutely right. I guess that guess did not pan out. :( I'll edit the answer when I have a little time. $\endgroup$ – Sándor Kovács Apr 3 '11 at 1:04
  • $\begingroup$ Thanks, Sándor! I'm at AGNES right now so don't have much time to check this out, but it looks very helpful and informative! $\endgroup$ – Noah Giansiracusa Apr 3 '11 at 12:04

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