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$\def\PP{\mathbb{P}}$Let $z_1$, $z_2$, ..., $z_n$ be points in $\PP^{k-1}$. I am interested in equations for when the $z_i$ lie on a rational normal curve (or degeneration thereof.)

Specifically, let $s: \PP^{k-1} \to \PP^{\binom{k+1}{2} - 1}$ be the $2$-uple Veronese. If the $z_i$ lie on a rational normal curve, then the $n \times \binom{k+1}{2}$ matrix $(s(z_1), s(z_2), \ldots, s(z_n))$ has rank $2k-1$.

Let $X$ be the subscheme of $(\PP^{k-1})^n$ where $\mathrm{rank}(s(z_1), s(z_2), \ldots, s(z_n)) \leq 2k-1$. Let $V$ be the subscheme of $(\PP^{k-1})^n$ obtained by taking the closure of those points that lie on a degree $k-1$ rational curve. So $V \subset X$. If $k \leq 2$, this is trivial. If $k=3$, this says that $n$ points lie on a conic if and only if the $6 \times n$ matrix $(s(z_1), s(z_2), \ldots, s(z_n))$ has a right kernel -- again, this is obvious. So the first hard situation is $k=4$.

I would love it if $V=X$. That seems too good to be true, once $k \geq 4$, although I don't actually have a counter-example. Here are weaker things that would make me happy:

  • Is $V$ an irreducible component of $X$?

  • Is $X$ reduced at a generic point of $V$?

  • Is there some explicit description of the other components of $V$?

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3 Answers 3

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Let me try the $\mathbb P^3$ case, and only look for a non-degenerate rational normal curve. Apply a linear map to move the first four of your points to be the standard four points ($[1,0,0,0]$,...) (if the four are coplanar, they're not on an rnc). Then apply the standard Cremona involution inverting each coordinate. Do the remaining points become collinear? If yes, they were on a rational normal curve (pre-Cremona). If no, they weren't. (For example, with six points, the remaining two are always collinear; not so with seven!)

I think this works the same in any dimension, at least generically. If we're lucky, the closure of this relation (to include linearly degenerate configurations) will be the right equations to check if the points are on a degenerate rnc.

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  • $\begingroup$ Whoa, that's great! Probably you need to intersect that over all choices of $k+1$ points, if you want to exclude other spurious components. But that is much cleaner than anything I had thought of. $\endgroup$
    – David E Speyer
    Commented Dec 12, 2014 at 22:43
  • $\begingroup$ Ah, that seems like a good idea. There is also an issue that one of the other points might be in the indeterminacy locus of the Cremona map if they are linearly degenerate somehow -- maybe your suggestion takes care of that? $\endgroup$
    – user47305
    Commented Dec 12, 2014 at 22:47
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I would like to suggest the following paper, where my coauthors and I try to give a partial answer to this question

https://arxiv.org/abs/1711.06286

Roughly speaking, the idea is to use the Gale transform to reduce to the planar situation. In fact, one has that (d+4) points in P^d lie on a rational normal curve iff their Gale duals lie on a conic in P^2.

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  • $\begingroup$ Your recent paper arxiv.org/abs/1903.00460 seems also relevant. $\endgroup$
    – Zach Teitler
    Commented Mar 7, 2019 at 11:23
  • $\begingroup$ Yes, that paper is also related. However, there we focus more on some geometric consequences of knowing explicit equations for the parameter space of points supported on a r.n.c. I think that the first one is more concerned with Speyer's question. Anyway thank you for pointing it out, and sorry for my late answer! $\endgroup$
    – Alessio
    Commented May 20, 2019 at 16:54
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Uggh, I lose. $7$ generic points in $\mathbb{P}^3$ don't lie on a rational normal curve, but apparently they DO lie on $3$ quadratics. (Because we are looking at the kernel of a $10 \times 7$ matrix.)

That means there is some fun Chasles theorem stuff here: Any quadratic which passes through $7$ of the intersection points of three quadratics should pass through the $8$th.

So my specific suggestion of using quadratics was wrong. I'd be glad to hear any suggestions as to what is right.

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    $\begingroup$ Maybe something like this will be helpful: If $V$ is a vector space of dimension $n$, and $d\ge 2n{+}1$ points $[v_i]\in\mathbb{P}(V)$ are in general position while the span of their squares $[{v_i}^2]\in \mathbb{P}\bigl(S^2(V)\bigr)$ lies in a projective subspace of dimension at most $2n-2$, then these $d$ points lie on a (unique) rational normal curve in $\mathbb{P}(V)$. This is in a paper by Chern and Griffiths (Abel's Theorem and Webs), and I always thought it was a neat fact. $\endgroup$
    – Robert Bryant
    Commented Dec 13, 2014 at 12:44

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