There is a unique quadric through three disjoint lines

There is a classical exercise that three disjoint lines in $\mathbb{P}^3$ are contained in a quadric surface $Q$. The existence is trivial. Every quadric in $\mathbb{P}^3$ is determined by nine coefficients and if three distinct points of a line lie on $Q$ then the whole line lie on $Q$. Thus we obtain the system of 9 linear conditions on $Q$. But why such a quadric is unique? To be more precise, why these linear equations are linearly independent?

• It's at least geometrically obvious. Take a quadric surface $Q$ through the three lines. Since the lines are disjoint, they all lie in the same ruling of the quadric. Since all nondegenerate quadrics are equivalent, you can apply an automorphism of $\mathbb{P}^3$ taking $Q$ to some standard quadric and moreover taking one of the lines to some specific line. Moreover, if they lay in some other quadric $Q'$, they'd also all be in the same ruling of $Q'$. Now think about what a quadric surface looks like... – Daniel McLaury Mar 25 '15 at 16:40

There exists exactly one quadric containing three mutually disjoint lines in $$\mathbb P^3$$:
Indeed you may assume that these lines are

$$L_1: x_0=x_1=0\quad \quad L_2:x_2=x_3=0\quad\quad L_3: x_0=x_2 ,\:x_1=x_3$$

and then the unique quadric containing these three line $$L_i$$ has the equation

$$x_0x_3-x_1x_2=0$$

This is easy to check by writing that the polynomial

$$ax_0^2+bx_0x_1+\dots+jx_3^2$$

vanishes on the $$L_i$$'s.

• Surely, using some appropriate automorphism of $\mathbb{P}^3$ we can assume that the first line is given by $\{x_0=x_1=0\}$. But why can we assume that the other two are $L_2:x_2=x_3=0,L_3: x_0=x_2 ,\:x_1=x_3$? – vitaliy Mar 25 '15 at 21:20
• The automorphism group of $\mathbb P^3(k)$ acts transitively on triples of disjoint lines. This boils down to the fact that in the vector space $k^4$ if you fix two transversal planes, say $P$ and $Q$, you can find an automorphism of $k^4$ sending any plane $S$ transversal to both of them to any other such plane $S'$. You can prove that by using block-diagonal invertible $4\times 4$ matrices. It is the sort of calculations done when introducing charts on the grassmannian: cf. Griffiths-Harris page 193 and following ones. – Georges Elencwajg Mar 25 '15 at 22:10
• Georges Elencwajg, I have checked your assertion. Thank you! – vitaliy Mar 26 '15 at 21:12
• С удовольствием, дорогой Виталий ! – Georges Elencwajg Mar 26 '15 at 23:49
• @vitaliy. There is also a slick way to send one of the transverse planes to another one: consider them as graphs of isomorphisms $u,v:k^2\to k^2$ and use the transformation $v\circ u^{-1}$. If you are not convinced you might ask a new question and hope that a kind soul will answer :-) – Georges Elencwajg Mar 27 '15 at 9:41

Mutually skew lines $a,b$ and $c$ lie in the same ruling, and the quadric is the union of the lines of the other ruling. There exists exactly one line containing a given point $p \in a$ and intersecting $b$ and $c$, so the quadric is unique as the union of all the lines intersecting with $a,b$ and $c$.

• Took me a second to see how to do that without Schubert calculus: the lines through $p$ and meeting $b$ sweep out a plane $L$. Since $c$ doesn't meet $b$ it can't lie in $L$ (all pairs of lines in a projective plane meet), so it must intersect $L$ at a single point, $q$ (every line and plane in $\mathbb{P}^3$ meet). The line $\overline{pq}$ is then the unique line through $p$ meeting both $b$ and $c$. – Daniel McLaury Mar 25 '15 at 20:10

You already have two good answers; let me give a bad one. Here "bad" means "uses more macinery than necessary"; I am writing it anyway in case you like the approach.

Let $Q$ be the quadric containing our three lines $L_1, L_2, L_3$. Now for any other quadric $Q' \subset \mathbf P^3$, the intersection $Q \cap Q'$ is a quartic curve.

Suppose now there exists another $Q'$ which also contains $L_1$, $L_2$, $L_3$. Then $Q \cap Q' = L_1 \cup L_2 \cup L_3 \cup L$ for some other line $L$.

Now we use some machinery: intersection numbers. (Assuming we're working over $\mathbf C$, you can argue in terms of singular cohomology.) Fix one of our lines, say $L_1$. In $\mathbf P^3$ we know $L_1 \cdot Q' =2$, so on $Q$ we know that $L_1 \cdot (L_1 \cup L_2 \cup L_2 \cup L)=2$. But $L_1 \cdot L_j = 0$ for $j=1,2,3$ since these three lines are all in one ruling, and $L_1 \cdot L \leq 1$. This is a contradiction.