# Most Economical Description of Two Lines in $\mathbb{P}^{3}$

I have two lines $L_{1}$ and $L_{2}$ in $\mathbb{P}^{3}$, and in one case they intersect while in the other they are skew. I am wanting to do a computation which involves writing them out explicitly in coordinates and I'm hoping to use projective linear transformations to describe the two lines as economically as possible. I have an idea that I was hoping someone may critique or fix!

Let $(x_{0}:x_{1}:x_{2}:x_{3})$ be coordinates on $\mathbb{P}^{3}$. Now, we can most certainly use the automorphisms of $\mathbb{P}^{3}$, namely $\rm{PGL}_{4}(\mathbb{C})$, to describe one of the two lines as

$$L_{1}: \{x_{2}=0, x_{3}=0\}.$$

Now, in general, the second line will be the complete intersection of two hyperplanes like $a_{0}x_{0}+a_{1}x_{1}+a_{2}x_{2}+a_{3}x_{3}=0$ and $b_{0}x_{0}+b_{1}x_{1}+b_{2}x_{2}+b_{3}x_{3}=0$. Ideally, I would like to make these a lot simpler if at all possible!

My idea is that maybe $\rm{PGL}_{4}(\mathbb{C})$ contains a subgroup like $\rm{PGL}_{2}(\mathbb{C})$ which actually fixes the first line $L_{1}$. Then, maybe I can use this subgroup to make at least some of the $a_{i}$ and $b_{i}$ zero. For example, I think I should at least be able to make $L_{2}$ be given by $\{x_{1}=0\}$ plus another hyperplane vanishing.

So my question is, can this actually be done, and maybe it can even be done nicer than I described above? Does the two lines intersecting or not play any role in this simplification? My main worry is that I know a lot of intuition which comes from three-(real) dimensions is completely "accidental" so maybe in four-dimensions ($\mathbb{P}^{3} = \mathbb{P}(\mathbb{C}^{4})$), we can't make sense of a rotation around a fixed plane, leaving that plane fixed.

## 3 Answers

Yes. A simple way to do this amounts to choosing a basis for the $\mathbb{C}^4$.

Case 1: You have two skew lines $L_1, L_2$. Have the first line be spanned by $p_1, p_2$ and the second by $p_3, p_4$. These give a basis for the $\mathbb{C}^4$, relative to which the lines are

$$L_1 = [* : * : 0 : 0], \qquad L_2 = [0 : 0 : * : *].$$

Case 2: You have two intersecting lines. Let the point of intersection be $p$. Pick $q_1, q_2$ so that $L_i$ is spanned by $p,q_i$. Finally, pick some last point $r$ not on the plane spanned by $L_1,L_2$.

In the basis $p,q_1,q_2,r$, the two lines are:

$$L_1 = [* : * : 0 : 0], \qquad L_2 = [* : 0 : * : 0].$$

• Thanks Jake! Your result for the non-intersecting case I would regard as "ideal as possible." I convinced myself it wasn't possible in general, because I thought you could use PGL(4) to orient one line however you like and maybe some components of the other. However, I thought you wouldn't be able to change the relative orientation of the lines. So your result in Case #1 is general for skew lines? – Benighted Nov 22 '16 at 4:52
• Yes, it is general. This is easy to prove: if $p_1, p_2, p_3, p_4$ did not form a basis, then there would be some equation $a_1 p_1 + a_2 p_2 = a_3 p_3 + a_4 p_4$ giving a common point of $L_1, L_2$. – Jake Levinson Nov 22 '16 at 19:41

The simplest way to describe a line $l\subset \mathbb P^3$ in the problem at hand is to give two distinct points on it, say $a=(a_0:a_1:a_2:a_3)$ and $b=(b_0:b_1:b_2:b_3)$.
If you have a second line $m$ joining $c=(c_0:c_1:c_2:c_3)$ to $d=(d_0:d_1:d_2:d_3)$ the condition that the lines $l,m$ meet (i.e. are not skew) is simply $$\det [a^T b^T c^T d^T]=0$$ where you take the determinant of the matrix whose columns are obtained by transposing the four vectors $a,b,c,d$.

• Thanks a lot George. I should have been clearer with my question, but the computation I want to do requires directly using coordinates, so I was wondering if my idea about using the projective linear transformations to set as many unknowns to zero was correct. For example, it clearly suffices to give one line by x2=x3=0, can we, in enough generality, say the other line at least live in the hyperplane, say, x1=0? – Benighted Nov 22 '16 at 2:52
• Dear spietro: yes, you certainly can. Actually you can send any two skew lines to any two skew lines you want by a projective transformation. For example any two skew lines can be sent to the lines $x_2=x_3=0$ and $x_0=x_1=0$ by a projective transformation. Beware however that you can never send two intersecting lines to two non-intersecting lines (or vice versa) by a projective transformation. – Georges Elencwajg Nov 22 '16 at 8:17

A useful way to think about these questions is to reinterpret them in terms of linear algebra. $\mathbb{P}^3$ is $\mathbb{C}^4 - 0$ modulo scaling, so a point in $\mathbb{P}^3$ corresponds to a line through the origin in $\mathbb{C}^4$, and a line in $\mathbb{P}^3$ corresponds to a two dimensional subspace of $\mathbb{C}^4$.

An alternate way to express your conclusion "we can most certainly use the automorphisms of $\mathbb{P}^3$, namely $\text{PGL}_4(\mathbb{C})$, to describe one of the two lines as $L_1: \{x_2 = 0,\text{ } x_3 = 0\}$" is that for any two-dimensional subspace of $\mathbb{C}^4$, we may pick a basis $\{v_i\}$ such that the subspace is $\text{span}(v_0, v_1)$ is that subspace. Acting by $\text{GL}_4(\mathbb{C})$ corresponds to changing the basis.

If you have two two-dimensional subspaces of $\mathbb{C}^4$ that intersect only at $0$, you can pick a basis $\{v_i\}$ such that the first is $\text{span}(v_0,v_1)$ and the second is $\text{span}(v_2,v_3)$. Then your lines in $\mathbb{P}^3$ are given by $x_0=x_1 =0$ and by $x_2 = x_3 =0$. This is the case of skew lines.

You can treat the cases where the lines intersect similarly.

Your thoughts about looking at the subgroup that fixes a line is also a fruitful avenue to pursue. Subgroups of $\text{GL}_n$ that fix a subspace are examples of parabolic subgroups, which are conjugate to elements of $\text{GL}_n$ that are block upper triangular. People usually think about parabolic subgroups by thinking about the action of $\text{GL}_n$ on $\mathbb{C}^n$, similar to the idea I sketched above.

You might want to look up what a flag is. A parabolic subgroup of $\text{GL}_n(\mathbb{C})$ is the stabilizer of a flag.

• Thanks a lot :) Wish I could accept all these answers. It now makes perfect sense that given two planes in $\mathbb{C}^{4}$, you can choose a basis which will achieve what I was hoping. I guess it's nearly identical reasoning to how given two lines through the origin in $\mathbb{C}^{2}$, you can choose a basis such that the lines are in these directions. – Benighted Nov 22 '16 at 16:48