Unique line in $\mathbb{P}^3$ through a point $p$ and intersecting two disjoint lines

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!

Solution 1:

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 $\Pi_2$.

Solution 2:

Let $L_1$ and $L_2$ be two disjoint lines in $\mathbb{P}^3$, and let $p\in\mathbb{P}^3\setminus(L_1\cup L_2)$. Show that there is a unique line $L\subseteq\mathbb{P}^3$ such that $p\in L$ and $L\cap L_i\neq\varnothing$ for $i=1,2$.

One possibility is to view this as a linear algebra problem in $k^4$.

Recall (cf. Lemma 6.17 in Gathmann's notes) that we have a one-to-one correspondence

\begin{align} \{\text{Cones in $\mathbb{A}^{n+1}$}\}&\longleftrightarrow \{\text{Projective varieties in $\mathbb{P}^n$}\}\\ Y&\longmapsto \mathbb{P}(Y):=\{(a_0:\cdots:a_n) : (a_0,\ldots,a_n)\in Y\setminus\{\mathbf{0}\}\} \end{align}

with inverse given by $X\mapsto C(X):=\{\mathbf{0}\}\cup\{(a_0,\ldots, a_n) : (a_0:\cdots:a_n)\in X\}$ for $X\subseteq\mathbb{P}^n$.

Using this correspondence, the problem can be translated from the language of projective geometry to the language of linear algebra as follows:

• The lines $L_1$ and $L_2$ correspond to two-dimensional linear subspaces $U=\mathrm{span}_k\{\mathbf{u}_1,\mathbf{u}_2\}=C(L_1)$ and $V=\mathrm{span}_k\{\mathbf{v}_1,\mathbf{v}_2\}=C(L_2)$ of $k^4$, respectively.

• The point $p$ corresponds to a one-dimensional linear subspace $W=\mathrm{span}_k\{\mathbf{w}\}=C(\{p\})$.

• $L_1\cap L_2=\varnothing$ corresponds to $U\cap V=\{\mathbf{0}\}$.

• $p\not\in L_1\cup L_2$ corresponds to $U\cap W=V\cap W=\{\mathbf{0}\}$.

• We want to show that there is a unique two-dimensional subspace $Z$ of $k^4$, such that $W\subseteq Z$, $U\cap Z\neq \{\mathbf{0}\}$ and $V\cap Z\neq \{\mathbf{0}\}$; the desired claim then follows with $L=\mathbb{P}(Z)$.

One way to do this (I leave it for you to check the details!) is to form the hyperplanes $H_1=\mathrm{span}_k\{\mathbf{u}_1,\mathbf{u}_2,\mathbf{w}\}$ and $H_2=\mathrm{span}_k\{\mathbf{v}_1,\mathbf{v}_2,\mathbf{w}\}$, and then set $Z=H_1\cap H_2$.