I'm a bit stuck with this exercise from a script I'm reading, and I'm not very familiar with projective $n$-space yet. The problem:
Let $L_1$ and $L_2$ be two disjoint lines in $\mathbb{P}^3$, and let $p\in\mathbb{P}^3\smallsetminus(L_1\cup L_2)$. Show that there is a unique line $L\subseteq\mathbb{P}^3$ meeting $L_1$, $L_2$, and $p$ (i.e. such that $p\in L$ and $L\cap L_i\neq\varnothing$ for $i=1,2$).
To be honest, I already have a problem with the term 'line'. As I take it, a line in $\mathbb{P}^3$ should be something cut out by two degree-1 polynomials (homogeneous, since it wouldn't be well defined otherwise, right?). But what are disjoint lines in projective space? As far as I understood it, two distinct lines should always intersect in exactly one point there, so how can they be disjoint? Can it be that these two statements mean a different kind of 'line'?
Any explanation of this, hints, or even a solution would be very appreciated. Thanks in advance!
Best Answer
Two lines always intersect in the projective plane $\mathbb{P}^2$. But in $\mathbb{P}^3$, two lines may be disjoint (otherwise called skew): for example the lines $x_0 = x_1 =0$ and $x_2 = x_3 = 0$ are disjoint in $\mathbb{P}^3$. Another perspective: lines in $\mathbb{P}^3$ correspond to planes through the origin in $\mathbb{C}^4$. Two lines in $\mathbb{P}^3$ are disjoint if and only if their corresponding planes intersect only at the origin.
As a hint for your problem, consider the projective planes $\Pi_i$ that contain $p$ and $L_i$. Show that these two projective planes meet in a projective line that contains $p$. Conversely, any line satisfying your conditions would lie in $\Pi_1$ and $\Pi_2$.