Twisted cubic as a plane section of the Segre threefold

Question

As usual, I have not a clue on how to tackle excercises our instructor assigns from Harris. The fact is that, at least to me, Harris is very concise and do not really builds the ''mindset'' to do solve this problems.

It is asked to show that the twisted cubic $C\subset\mathbb{P}^3$ can be realized as the itersection of the Segre threefold $\Sigma_{1,2}\subset\mathbb{P}^5$ with a 3-plane $\mathbb{P}^3\subset\mathbb{P}^5$

I know that $\Sigma_{1,1}=V(x_0x_3-x_1x_2)$ and that I can write $\Sigma_{2,1}=V(x_0x_3-x_1x_2;x_2x_5-x_4x_3;x_0x_5-x_1x_4)$
while the twisted cubic as
$C=V(x_0x_3-x_1x_2; x_1x_3-{x_2}^2; x_0x_2-{x_1}^2)$.

So both surfaces are in the zero locus of the quadric $x_0x_3-x_1x_2$. What I would neeed then is a 3-plane embodying the two constraints $x_1x_3={x_2}^2; x_0x_2={x_1}^2$ do these two equations define a 3-plane in $\mathbb{P}^5$?

Answer 1

You're really quite close to the solution now that you've written things down.

$\Sigma_{2,1}=V(x_0x_3-x_1x_2;x_2x_5-x_4x_3;x_0x_5-x_1x_4)$
while the twisted cubic as
$C=V(x_0x_3-x_1x_2; x_1x_3-{x_2}^2; x_0x_2-{x_1}^2)$.

Can we impose some linear conditions so that the last equations match the first equations?

Hint:

Intersecting with a 3-plane is equivalent to enforcing two linear conditions. We notice that $x_4$ and $x_5$ appear in the first equations but not the second, so enforcing two linear relations of the form $x_4=\sum_{i=0}^{i=3} c_ix_i$ and $x_5=\sum_{i=0}^{i=3} d_ix_i$ would let us eliminate $x_4$ and $x_5$ via intersecting with a 3-plane. What relations should we impose?

Full solution:

If we enforce $x_4=x_1$ and $x_5=x_2$, we get that the equations for $\Sigma_{2,1}$ turn in to $V(x_0x_3-x_1x_2, x_2^2-x_1x_3, x_0x_2-x_1^2)$ which after flipping the sign on the second equation exactly match the equations of the twisted cubic. Since $V(x_1=x_4,x_2=x_5)$ determines a copy of $\Bbb P^3$, we see that our intersection $V(x_0x_3-x_1x_2;x_2x_5-x_4x_3;x_0x_5-x_1x_4)\cap V(x_1=x_4,x_2=x_5)$ exactly gives us a copy of the twisted cubic.