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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!

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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 $\Pi_2$.

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  • $\begingroup$ Hello @Michael, and thanks! I'm still a bit confused though, could you have a look at [Remark 3.1.4 here], especially the last part of (ii). Is it correct that this kind of 'line' is not the same as the one the exercise is about (as otherwise, the discrepancy would make no sense)? $\endgroup$ – InvisiblePanda Mar 3 '12 at 14:43
  • $\begingroup$ Rand, I cannot access your link. If you repost it, I'd be glad to look at it. $\endgroup$ – Michael Joyce Mar 3 '12 at 15:59
  • $\begingroup$ Oh, just noticed it, I'm sorry! Seems like I forgot to put the (link) behind it. It's here. The one I meant is Remark 3.1.4. $\endgroup$ – InvisiblePanda Mar 3 '12 at 17:44
  • $\begingroup$ That last sentence looks like an oversight by the author. When saying that two lines always intersect in a point, he is referring to the fact that two lines in the projective plane always intersect in a point. The corresponding statement in $\mathbb{P}^n$ is that a linear space of dimension $k$ and a linear space of dimension $n-k$ always intersect, and if they are in general position, they will intersect in a point (otherwise, they can intersect in a higher dimensional linear space). $\endgroup$ – Michael Joyce Mar 3 '12 at 18:05
  • $\begingroup$ Hello @Michael, thank you for your answer and comments, they have been very helpful! $\endgroup$ – InvisiblePanda Mar 4 '12 at 7:39
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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 $Z=H_1\cap H_2$.

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