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.  
 A: 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 $a,b,c,d$.
A: 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:
$$L_1 = [* : * : 0 : 0], \qquad L_2 = [* : 0 : * : 0].$$
A: 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.
