I’m considering the disjoint sets in projective space $\mathbb{P}^3(\mathbb{C})$ defined as follows:
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$L_1$ given by the equations $X_1=0$ and $X_2=0$
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$L_2$ given by the equations $X_0=X_1$ and $X_2=X_3$
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$L_3$ given by the equations $X_0=0$ and $X_2=2X_3$
I know each of them are projective lines and any two of them are disjoint. I also know there’s a quadric containing these three lines, but I’m not sure how to find it.
My idea was to write the most general equation for a quadric in $\mathbb{P}^3(\mathbb{C})$ as a starting point, but I’m not sure how to incorporate the restrictions given by the equations of the $L_i$.
I also tried trying to start from the restrictions themselves: for instance $(X_0-X_1)(X_2-X_3)=0$ is a quadric containing $L_2$, but I can’t keep adding factors corresponding to the other conditions, as this would give a polynomial of degree greater than $2$…
Is there a general way to solve this kind of problems?
Best Answer
Brute force works without a ton of fuss, since it turns this in to a linear algebra problem: write $$\varphi = ax_0^2+bx_0x_1+cx_0x_2+dx_0x_3+ex_1^2+fx_1x_2+gx_1x_3+hx_2^2+ix_2x_3+jx_3^2$$ and plug in.
After a bit of linear algebra, you'll find that $a=b=d=e=h=i=j=0$ and $f=c$ and $g=-2c$. So your quadric is a multiple of $x_0x_2+x_1x_2-2x_1x_3$, which does indeed vanish on all three lines and is nondegenerate because it's nonsingular (check the Jacobian).
It is true that there's a unique quadric through any triple of disjoint lines in $\Bbb P^3$, but I don't know how to prove that result without doing essentially this again. Maybe someone else will have something smarter to say.