Every point lies on a unique secant through $C$

Question

Let $C \subset \mathbb{P}^3$ be the twisted cubic (i.e., $C=\{(X_0^3:X_0^2 X_1:X_0 X_1^2:X_1 ^3) : (X_0,X_1) \in \mathbb{P}^1\}$). I need to show that every point $Q \in \mathbb{P}^3 \setminus C$ lies on a unique secant of $C$. Where's what I have so far:

A line in $\mathbb{P}^3$ joining $A=(a_0^3:a_0^2a_1,a_0a_1^2,a_1^3)$ and $B=(b_0^3:b_0^2b_1:b_0b_1^2:b_1^3)$ on $C$ is of the form $L=(ua_0^3+vb_0^3:ua_0^2a_1+vb_0^2b_1:ua_0a_1^2+vb_0b_1^2:ua_0^3+vb_0^3)$, $u,v \in \mathbb{C} - \{0\}$ (for an explanation on this see this). Hence, we want to show that for every $Q=(q_0:q_1:q_2:q_3)$ not in $C$ there exists $a_0,a_1,b_0,b_1,u,v$ such that $uA+vB=Q$ in $\mathbb{P}^3$.

Also, as $Q$ is not in $C$, we can suppose that $a_0,b_0 \neq 0$, and then divide by them and then we only want to find 4 unknowns $u,v,\frac{a_1}{a_0},\frac{b_1}{b_0}$. The problem is that the equation $uA+vB=Q$ gives rise to a (non-linear) system on these unknowns that I can't solve.

Can anyone help me? Thank you!

Answer 1

$\newcommand{\PSP}{{\mathbb P}}$

Uniqueness can be proven this way: Let $C \subseteq \PSP^3_k$ be the twisted cubic curve. Furthermore let $P_1,P_2$ and $Q_1,Q_2$ be four points on $C$ with $P_1 \neq P_2$ and $Q_1 \neq Q_2$. Let $l_1$ be the line through $P_1, P_2$ and $l_2$ be the line through $Q_1,Q_2$. Now let $R$ be a point on $l_1 \cap l_2$, apart from $C$. Then $l_1, l_2 \subseteq H$ for a certain plane $H \subseteq \PSP^3_k$. This plane intersects $C$ in at most three different points. So one must have $P_i = Q_j$ for certain $i,j$. As $R$ is a second common point of $l_1,l_2$ they must coincide, $l_1 = l_2$.

The proof of $\mathrm{Sec} C = \PSP^3_k$ can probably be done via the following gröbner base calculation (to be honest, the problem is, what is actually computed is only the defining ideal of the closure of $\mathrm{Sec} C$). Note that:

1) I used your parametrization of the secant variety in the variable vmat. 2) ideal() stands for ideal(0), it is the definining ideal of the closure of the image of the parametrization of $\mathrm{Sec} C$.

i1 : R=QQ[x_0..x_3]

o1 = R

o1 : PolynomialRing

i2 : S=QQ[s_0,s_1,t_0,t_1,u,v]

o2 = S

o2 : PolynomialRing

i3 : vmat={u*s_0^3+v*t_0^3,u*s_0^2*s_1+v*t_0^2*t_1,u*s_0*s_1^2+v*t_0*t_1^2,u*s_1^3+v*t_1^3}

       3     3    2       2        2       2    3     3
o3 = {s u + t v, s s u + t t v, s s u + t t v, s u + t v}
       0     0    0 1     0 1    0 1     0 1    1     1

o3 : List

i4 : phi=map(S,R,vmat)

               3     3    2       2        2       2    3     3
o4 = map(S,R,{s u + t v, s s u + t t v, s s u + t t v, s u + t v})
               0     0    0 1     0 1    0 1     0 1    1     1

o4 : RingMap S <--- R

i5 : ker phi

o5 = ideal ()

Note added: How to avoid using Bezout's theorem at the moment I have no simple idea, but I think for $\mathrm{Sec} C =\PSP^3_k$ a better proof is possible, which sheds light on the question if really every point of $\PSP^3_k-C$ lies on a secant of $C$:

If one takes an arbitrary point $P \notin C$ and projects from $P$, giving $\PSP^3_k - P \to \PSP^2_k$ and so $g:C \to \PSP^3_k - P \to \PSP^2$ then $g(C)$ should be a cubic curve in $\PSP^2_k$. As the genus of $g(C)$ is zero, the curve must be nodal or cuspidal. If it is nodal, the two points $R_1$ and $R_2$ of $C$ over the node $Q$ of $g(C)$ furnish a secant containing $P$. But in case $P$ lies on $\mathrm{Tan} C$, the tangent variety, then $g(C)$ is cuspidal with the cusp being the image of the tangent to $C$ through $P$. But then $g(C)$ can not be nodal too, because of the genus formula

$$p = \frac{1}{2}(3-1)(3-2) - n_c - n_n$$

where $n_c$ is the number of cusps and $n_n$ the number of nodes.

So in effect we see, that not every point of $\PSP^3_k-C$ lies on a secant of $C$, some lie only on a tangent - but the closure of the space of secants is indeed $\PSP^3_k$.