Alg. Geo. Homework 1: Twisted Cubic

I’m trying to study a bit for the upcoming exams, one of them is for a weirdly structured “geometry” course made up by 2/3 intro Differential Geometry (smooth manifolds) and 1/3 Algebraic Geometry (disappointingly ‘800 style i.e. no schemes, just a ton of examples and definitions and boredom).

I’ll try to do some of the exercises, typing them up is more of a challenge to myself and a way to motivate me slightly more (such is life), this is just a disclaimer that this kind of post may be super boring even moreso than usual.

Let’s talk about the twisted cubic. It’s a curve that arises as a special case of the Veronese map

$v : \mathbb P^1 \to \mathbb P^3$
$[x, y] \mapsto [x^3, x^2y, xy^2, y^3]$

or, in affine coordinates, by $t \mapsto (t, t^2, t^3)$ . In other words we are talking about $C = v(\mathbb P^1)$ .

Exercise 1. Consider the following homogenous polynomials $F_0 = Z_0Z_2 - Z_1^2$ , $F_1 = Z_0Z_3 - Z_1Z_2$ and $F_2 = Z_1Z_3 - Z_2^2$ and the quadrics $Q_i = V(F_i)$ they define in $\mathbb P^3$ . Prove that $C = Q_0 \cap Q_1 \cap Q_2$ .

Ok, it’s straightforward to prove that a point of $C$ is in the zero locus of all three polynomials, let’s do the converse.
Let $p = [z_0, z_1, z_2, z_3] \in Q_0 \cap Q_1 \cap Q_2$ , since we are in projective space we know these coordinates cannot all be zero, thus we note that $z_0, z_3$ cannot be simultaneously zero otherwise $latex F_0(p) = z_0z_2 – z_1^2 = 0$ implies $z_1 = 0$ and $F_2(p) = z_1z_3 - z_2^2 = 0$ implies $z_2 = 0$ .

Suppose $z_0 \neq 0$ , then we find a preimage for $p$ through the Veronese map, namely $v(z_0, z_1) = [z_0^3, z_0^2z_1, z_0z_1^2, z_1^3] = [z_0, z_1, z_1^2/z_0, z_1^3/z_0^2]$ , indeed since $p$ must satisfy those polynomial relations we see $F_0(p) = z_0z_2 - z_1^2 = 0$ implies $z_2 = z_1^2/z_0$ and $F_1(p) = z_0z_3-z_1z_2 = 0$ implies $z_3 = z_1z_2/z_0 = z_1^3/z_0$ , thus $p = v(z_0, z_1) \in C$ .
(and similarly in the case $z_0 = 0, z_3 \neq 0$ ).

Exercise 2. Prove that all three quadrics are necessary, namely that any two of them intersect in the twisted cubic plus a line: $Q_i \cap Q_j = C \cup L$

This is only slightly more involved, I’m going to do it with the first two quadrics. Let $p = [z_0, z_1, z_2, z_3] \in Q_0 \cap Q_1$ .
If $z_0 = 0$ then, as before, $F_0(p) = 0\implies z_1 = 0$ , this time though $F_1$ becomes just the null polynomial giving no information and there’s no other relation. Thus any point of the form $[0, 0, z_2, z_3]$ belongs to the intersection (and this is the line).

If $z_0 \neq 0$ then we do the same trick as before deducing the relations $z_2 = z_1^2/z_0, z_3 = z_1^3/z_0^2$ from the polynomials and verifying that indeed $p = [z_0, z_1, z_1^2/z_0, z_1^3/z_0^2] = [z_0^3, z_0^2z_1, z_0z_1^2, z_1^3] \in C$ .

Partially Ordered Thoughts

Alg. Geo. Homework 1: Twisted Cubic

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Alg. Geo. Homework 1: Twisted Cubic

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