All the lines on the Segre quadric

Question

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?

Answer 1

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!

Answer 2

Let $\mathbb{K}^2_2$ be the vector space of $2\times2$ matricies with values in the field $\mathbb{K}$, let $\mathbb{P}=\mathbb{P}\left(\mathbb{K}^2_2\right)\cong\mathbb{P}^3_{\mathbb{K}}$ the projectivized space of $\mathbb{K}^2_2$; by definition, the Segre quadric is $$ \Sigma=\left\{[M]\in\mathbb{P}\mid rank(M)=1\right\}. $$ Let $[M]=P\neq Q=[N]\in\Sigma$ and let $L$ be the line passing through $P$ and $Q$; by definition $$ L\subset\Sigma\iff\forall[A]\in L,\,rank(A)=1, $$ because $$ L=\left\{[sM+tN]\in\mathbb{P}\mid[s:t]\in\mathbb{P}^1_{\mathbb{K}}\right\} $$ then $$ L\subset\Sigma\iff\forall[s:t]\in\mathbb{P}^1_{\mathbb{K}},\,[sM+tN]\in\Sigma\iff rank(sM+tN)=1. $$ Let $[M]=\left[m_i^j\right]\neq[N]=\left[n_i^j\right]\in\Sigma$; by condition $rank(sM+tN)=1$ for any $[s:t]\in\mathbb{P}^1_{\mathbb{K}}$, one has $$ \det\begin{pmatrix} sm_1^1+tn_1^1 & sm_1^2+tn_1^2\\ sm_2^1+tn_2^1 & sm_2^2+tn_2^2 \end{pmatrix}=\dots=st(m_1^1n_2^2+m_2^2n_1^1-m_1^2n_2^1-m_2^1n_1^2)=0; $$ that is, the line $L$ passing through $P$ and $Q$ is in $\Sigma$ if and only if $m_1^1n_2^2+m_2^2n_1^1-m_1^2n_2^1-m_2^1n_1^2=0$!

In other words, a plane $\pi$ in $\mathbb{P}$ of Cartesian equation $a_0x_0+a_1x_1+a_2x_2+a_3x_3=0$ intersects $\Sigma$ in a pair of two lines if and only if $a_0a_3-a_1a_2=0$.

Otherwise, if $$ \exists[s:t]\in\mathbb{P}^1_{\mathbb{K}}\mid rank(sM+tN)=2 $$ $L$ is a secant line to $\Sigma$!

Remark: This reasoning is generalizable for the generic Segre variety $\Sigma_{m,n}$ in $\mathbb{P}^{(m+1)(n+1)-1}_{\mathbb{K}}$.

Answer 3

Note that all the structure on $\mathbb{P}^1\times \mathbb{P}^1$ comes from the Segre embedding, given by \begin{align*} \Sigma_1^1:\mathbb{P}^1\times \mathbb{P}^1 &\rightarrow \mathbb{V}(X_0X_3-X_1X_2)\subset \mathbb{P}^3\\ ([x:y],[z:w])&\mapsto [xz:xw:yz:yw]. \end{align*} This is definitionally an isomorphism. The line through the points $([x:y],[z:w])$ and $([x':y']:[z':w'])$ gets mapped under $\Sigma^1_1$ to $$\ell = \{[xz:xw:yz:yw]\cdot s+[xz:xw:yz:yw]\cdot t\;|\; [s:t]\in \mathbb{P}^1\}.$$ If this lies entirely in the surface, we have \begin{align*} \forall [s:t]\in \mathbb{P}^1, \quad (xzs+x'z't)(yws+y'w't)-(xws+x'w't)(yzs+y'w't)&\equiv 0 \\ \Rightarrow st(xy'-x'y)(zw'-w'z)&\equiv 0 \end{align*} Hence the line lies in the surface if and only if one of the following holds:

• $xy'-x'y=0$, i.e. $[x:y]=[x':y']=[\lambda:\mu]$. Then the line is equal to $\{[\lambda:\mu]\}\times \mathbb{P}^1 $, and $\Sigma^1_1$ maps it to $\mathbb{V}(\lambda X_2 - \mu X_0, \lambda X_3-\mu X_1)$.
• $zw'-w'z=0$, i.e. $[z:w]=[z':w']=[\lambda:\mu]$. Then the line is equal to $\mathbb{P}^1\times\{[\lambda:\mu]\}$, and $\Sigma^1_1$ maps it to $\mathbb{V}(\lambda X_1-\mu X_0, \lambda X_3-\mu X_2)$.

This shows that the two families above together constitute all the lines on the Segre quadric.