如何证明扭三次曲线$ \nu(\mathbb{P}^1) $上任意有限点处于一般位置？

Alright, let's tackle these two related questions head-on—they both center on a key property of the twisted cubic curve, so we can unpack them together.

First, let's formalize the twisted cubic #

The twisted cubic is exactly the image of the Veronese map you defined:
$$\nu : \mathbb{P}^1 \to \mathbb{P}^3, \quad \nu([X_0:X_1]) = [X_0^3:X_02X_1:X_0X_1^2:X_13]$$
This is a degree-3 rational curve embedded in $\mathbb{P}^3$, and its core feature here is that it's non-degenerate (doesn't lie in any hyperplane of $\mathbb{P}^3$) and has no "special" collinear or coplanar point sets beyond what's forced by its structure.

Question 2: Prove any four points on $\nu(\mathbb{P}^1)$ span $\mathbb{P}^3$ #

Let's break this into concrete steps:

1. Parameterize points on $\mathbb{P}^1$
   Any point on $\mathbb{P}^1$ can be written as either $[1:t]$ for some scalar $t$ in our base field $k$, or the "point at infinity" $[0:1]$. Let's take four distinct points on $\mathbb{P}^1$:

   • Case 1: All four are affine points: $p_i = [1:t_i]$ where $t_1, t_2, t_3, t_4$ are distinct elements of $k$.
   • Case 2: One point is $[0:1]$ (the infinite point), and the other three are affine: $p_1=[0:1], p_i=[1:t_i]$ for $i=2,3,4$ with distinct $t_i$.

2. Compute their images under $\nu$

   • For affine points $[1:t_i]$, their image in $\mathbb{P}^3$ has homogeneous coordinates $[1, t_i, t_i^2, t_i^3]$.
   • For $[0:1]$, the image is $[0,0,0,1]$.

3. Check linear independence of the coordinate vectors
   To prove four points span $\mathbb{P}^3$, we just need to show their corresponding homogeneous coordinate vectors are linearly independent in $k^4$ (since linear independence in $k^4$ translates to spanning $\mathbb{P}^3$).

   • Case 1: The matrix formed by the four vectors is:

     1  t₁  t₁²  t₁³
     1  t₂  t₂²  t₂³
     1  t₃  t₃²  t₃³
     1  t₄  t₄²  t₄³
     

     This is a 4x4 Vandermonde matrix, whose determinant is $\prod_{1 \leq i < j \leq 4} (t_j - t_i)$. Since all $t_i$ are distinct, this determinant is non-zero—so the vectors are linearly independent.

   • Case 2: The matrix becomes:

     0  0  0  1
     1  t₂  t₂²  t₂³
     1  t₃  t₃²  t₃³
     1  t₄  t₄²  t₄³
     

     Calculating the determinant here (expanding along the first row) gives $(-1)^{1+4} \times 1 \times \text{3x3 Vandermonde determinant of } t_2,t_3,t_4$. Again, since $t_2,t_3,t_4$ are distinct, this determinant is non-zero—so vectors are linearly independent.

   In both cases, the four points span $\mathbb{P}^3$.

Question 1: Why does any finite point set on the twisted cubic lie in general position? #

Recall that a set of points in $\mathbb{P}^n$ is in general position if every subset of $n+1$ points spans $\mathbb{P}^n$. For $\mathbb{P}^3$, this means every 4-point subset must span the entire space.

From the proof above, we already know any four distinct points on the twisted cubic span $\mathbb{P}^3$. What about smaller subsets?

• Any 2 points: They define a unique line, and since the twisted cubic is a degree-3 curve, a line can intersect it at most 3 times (by Bezout's theorem)—but we're taking distinct points, so no three points are collinear (otherwise that line would intersect the curve at three points, but our parameterization shows a line would correspond to a quadratic equation in $t$, which can have at most 2 roots).
• Any 3 points: They span a plane, and again, a plane intersects the twisted cubic in at most 3 points (Bezout's theorem, degree 3 curve intersects degree 2 plane in 3 points). So no four points lie on a plane, which we already confirmed.

Since every subset of 4 points spans $\mathbb{P}^3$, and smaller subsets behave as expected (no unnecessary collinearity/coplanarity), any finite set of points on the twisted cubic is in general position.

内容的提问来源于stack exchange，提问作者YoYo