# canonical representation of three skew lines in $\mathbb{P}^3$

Consider three skew (non-intersecting) lines $L,M,N$ in $\mathbb{P}^3$. Each line is given by two equations of the form $\alpha_{i,j}^\top z = 0, \, i=1,2,3, j = 1,2$, where $z=(z_0,z_1,z_2,z_3)$ are the homogeneous coordinates of $\mathbb{P}^3$ and $\alpha_{i,j} \in \mathbb{P}^3$.

What i want is to show that up to projective equivalence the equations for the three lines can be written as (this is possible according to the hint of exercise 2.12 in Harris, Algebraic Geometry - a first course)

$L: \, z_0=z_1=0 \\ M: \, z_2=z_3 = 0 \\ N: \, z_0=z_3, \, z_1=z_2.$

Since say $L, M$ are non-intersecting, this implies that $\alpha_{i,j}$ for every $i=1,2, j=1,2$ are in general position. So they are projectively equivalent to $(1,0,0,0), (0,1,0,0),(0,0,1,0),(0,0,0,1)$. This gives us the description that we want for $L$ and $M$. But how about $N$?

Note here that the most general result that we have is projective equivalence between two sets of 5 points in general position. In the context of the present question, i am concerned about two things:

1) The fact that the lines $L,M,N$ are skew does not imply in general that every group of 5 points among the $\alpha_{i,j}$ will be in general position.

2) The $\alpha_{i,j}$ are six!

Question: What am i missing here? How can the above two issues be addressed?

Let usthink in terms of linear algebra. You have $3$ $2$-dimensional subspaces $S$, $T$ and $U$ in $V=k^4$ which intersect pairwise trivially.

Since $S\cap T=0$, we have $V=S\oplus T$. Let $\{u_1,u_2\}$ be a basis of $U$, and suppose that $u_1=s_1+t_1$ and $u_2=s_2+t_2$ with the $s_i$ in $S$ and the $t_i$ in $T$. Since $U\cap S=U\cap T=0$, the four of $s_1$, $s_2$, $t_1$, $t_2$ are non-zero.

• If $s_1$ and $s_2$ are linearly dependent, so that for example $u_2=\alpha u_1$, then $\alpha u_1-u_2=\alpha t_1-t_2\in T\cap U=0$ so that in fact also $t_2=\alpha t_1$, which is absurd because it tells us that $u_1=u_2$.

• If $t_1$ and $t_2$ are linearly dependent, we can do the same.

• If neither of these two cases occur, then $\{s_1,s_2,t_1,t_2\}$ is a basis of $V$ and if $x_1,x_2,x_3,x_4$ are coordinates with respect to it, the equations of our subspaces are \begin{align} &x_3=x_4=0 && \text{for $S$,} \\ &x_1=x_2=0 && \text{for $T$ and} \\ &x_1-x_3=x_2-x_4=0 && \text{for $U$.} \end{align}

• I just found it :-) How did I come up with it? I just explored the possibilities. – Mariano Suárez-Álvarez Apr 7 '15 at 19:02
• Dear Mariano, there seems to be some mix-up in your first bullet. It is perfectly possible for $s_1$ and $s_2$ to be dependent without having $u_2=\alpha u_1$. For example take $u_1=s_1+t_1, u_2=2s_1+t_1$. So your conclusion " $\{s_1,s_2,t_1,t_2\}$ is a basis of $V$ " seems to me unwarranted. – Georges Elencwajg Apr 7 '15 at 21:58
• @GeorgesElencwajg, but in that example $U\ni 2u_1-u_2=2(s_1+t_1)-(2s_1+t_1)=t_1\in T$, so that $t_1=0$, from which we see that $u_1=s_1\in S\cap T=0$, which is impossible. SOmething more is indeed needed, though :-) I'll try to fix it later. – Mariano Suárez-Álvarez Apr 8 '15 at 0:23
• You are right Mariano: my objection was not valid:sorry. I have upvoted your answer! – Georges Elencwajg Apr 8 '15 at 7:25
• A geometer should post a geometric answer! – Mariano Suárez-Álvarez Apr 8 '15 at 7:29