Equations for points to lie on a rational normal curve $\def\PP{\mathbb{P}}$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 $V$?
 A: 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.
A: 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.
A: 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.
