Find all the lines on the quadric surface in $\mathbb P^3$ defined by the equation
$$xw=yz$$
(with the homogeneous coordinates $[x:y:z:w]$ of course).
Now it is well known that by the Segre embedding there is an isomorphism $\mathbb P^1 \times \mathbb P^1$ with our quadric. Moreover the images $\{[u_0,v_0]\}\times \mathbb P^1$ and $\mathbb P^1 \times \{[s_0,t_0]\}$ give two rulings of lines of our quadric. I have heard however from my teacher these are all the lines on the quadric. Is there any way to show this that is not too theoretically involved?
Best Answer
Any line in $\mathbb{P}^{3}$ is an intersection of two planes. In order to understand what lines are inside the quadric $xw=yz$, it is enough to understand when the hyperplane sections of $xw=yz$ in $\mathbb{P}^3$ contain a line.
Let $ax+by+cz+dw=0$ be a plane in $\mathbb{P}^3$. By Bezout's theorem, the intersection of $ax+by+cz+dw=0$ with $xw=yz$ is a curve of degree 2. If this curve is irreducible, then there is no hope of finding a line there! If this curve is a union of two lines, then we have successfully found a pair of lines.
So we need to understand what conditions need to be imposed on the coefficients $[a, b, c, d]\in\mathbb{P}^3$ such that the plane $ax+by+cz+dw=0$ intersects $xw=yz$ in a pair of lines.
Claim. $ax+by+cz+dw=0$ intersects $xw=yz$ in a pair of lines if and only if $ad=bc$.
Proof. $(\Leftarrow)$ Multiply $ax+by+cz+dw=0$ by $w$ and use $xw=yz$ to get $ayz+byw+czw+dw^2=0$. This is a plane curve in $\mathbb{P}^{2}$ (with coordinates $y, z, w$), unsurprisingly. Now, multiply both sides by $b$ to get $$ abyz+b^2yw+bczw+bdw^2=0 $$ Use the hypothesis $ad=bc$ to get $$ abyz+b^2yw + adzw + bdw^2 = 0 $$ which conveniently factors as $$ (az+bw)(by+dw)=0 $$ so we get a pair of lines.
$(\Rightarrow)$ I will leave this as an exercise. Try to factor the quadric equation, and show that such a factorization forces $ad=bc$. $\square$
So now we can answer the question "What are all the lines on the quadric surface $xw=yz$?" Well, the proof of the claim shows that the lines on $xw=yz$ are of the form $az+bw=0$ and $by+dw=0$ (viewed in $\mathbb{P}^2$ with coordinates $y,z,w$) such that $ad=bc$. Here $a, b, c, d$ come from the hyperplane $ax+by+cz+dw=0$. Now once you can fix any $[a, b]\in\mathbb{P}^{1}$, you get the line $az+bw=0$. And if you fix $[b, d]\in\mathbb{P}^{1}$, you get the line $by+dw=0$. I think these two families of lines are the desired rulings.
I am very interested in seeing a more concise and conceptual answer!