Suppose we consider all the pairs of line segments: $X = \{ (\overline{AB}, \overline{CD}) : \overline{AB} \cap \overline{CD} \neq \varnothing \} $ can we describe this as a polygon in some high dimensional Euclidean space? We have that $A,B,C,D \in \mathbb{R}^2$ so that $X \subseteq \mathbb{R}^8$. Also, line intersection does not change under translations, or dilations or perspective transforms. So there is an action of the affine linear group, $\text{ASL}_2(V) = \mathbb{R}^2 \ltimes \text{GL}_2(\mathbb{R}^2)$.

The translation action could be as simple as setting $A = \vec{0}$ the zero vector. Rescaling and rotating is like setting $B = \vec{1} = (1,0) \in \mathbb{R}^2$. So the result would be a four dimensional space. Have I missed anything? The affine linear group is the indirect product of the group of translations and the special linear group, so it should be a $2 + 3 = 5$ dimensional manifold. The space of intersecting pairs of line segments should be a $8 - (2 + 3) = 3$ dimensional space.

So we are describing a 3-manifold? Is this the just the 3-sphere? The quotient space of $$\mathbb{R}^8 / \big(\mathbb{R}^2 \ltimes \text{GL}_2(\mathbb{R}^2) \big) \simeq \mathbb{R}^6 / \text{GL}_2(\mathbb{R}^2)$$ as if it were a division problem. I wonder what these quotient spaces are like. If I use coordinates, we are saying there exists $s \geq 0$ and $t \geq 0$ such that: \begin{eqnarray*} s\vec{A} + (1-s)\vec{B} = t\vec{C} + (1-t)\vec{D} \quad\text{for some}\quad s,t \geq 0 \end{eqnarray*} So in addition to the four points $\vec{A}, \vec{B}, \vec{C}, \vec{D} \in \mathbb{R}^2$ there are two scalars $s,t \in \mathbb{R}$ for a total of 10 degrees of freedom if we include every single point. We have also shown that the space $X$ is a variety, since we've gotten it to look like $\{ f(s,t, \vec{A}, \vec{B}, \vec{C}, \vec{D}) = 0 \} / G$ for an appropriate group of symmetries $G$.

  • $\begingroup$ From the mathematical description the elements of $X$ are ordered pairs of directed line segments. We may view the set of ordered pairs of (undirected) line segments in $\Bbb R^2$, for example, as $(\Bbb R^2)^4 / \sim \cong \Bbb R^8 / \sim$, where $\sim$ is the equivalence relation generated by $(AB, CD) \sim (AB, DC)$ and $(AB, CD) \sim (BA, CD)$. This quotient has an orbifold structure, but at a glance I would expect it to have orbifold singularities along the sets $\{A = B\}$ and $\{C = D\}$, that is, on the set where one segment or the other is a just a point. $\endgroup$ Aug 9, 2019 at 20:13
  • $\begingroup$ ...and I think you mean $AGL(\Bbb R^2) = \Bbb R^2 \ltimes GL(\Bbb R^2)$, which is a $4$-dimensional group. We can then identify the open, dense subset $\{(AB, CD) : A \neq B, C \not\in \overleftrightarrow BC\}$ with $\Bbb R^2$ using the normalizations $A = (0, 0), B = (1, 0), C = (0, 1)$ you provided (the element of $\Bbb R^2$ is just the location of $D$), and so identify the intersection of $X$ with that open, dense set with the "infinite trapezoid" $\{(x, y) : y \leq 0, 0 \leq x \leq 1 - y\}$. $\endgroup$ Aug 9, 2019 at 20:36

2 Answers 2


Consider the pair of segments $((0,0), (2,0)), ((1,0), (3,0))$. They intersect. But the pair $((0,0), (2,0)), ((1,\epsilon), (3,0))$ does not no matter how small $\epsilon$ is. Therefore this is not an open set.

Now consider the pair of segments $((0,0), (1,0)), ((1,0), (2,0))$. Do we consider them to intersect?

If not, then it is trivial to construct a sequence $S_n = ((0,0), (\frac{n+1}{n},0)), ((1,0), (2,0))$ of intersecting segments whose limit is that one and it wouldn't be closed.

But if we have closed segments, then it is closed. This can be seen easily as follows. Topologically, the set of pairs of line segments is an 8-dimensional Euclidean space. Therefore being closed is equivalent to the statement that all Cauchy sequences converge to a point in the set.

So let $S_n$ be a Cauchy sequence of intersecting line segments. It is easy to verify that its limit exists and is a pair of line segments. But does the limit converge? Each in the sequence may intersect at one or an infinite number of points. Pick one point per intersection. From this we can define two new sequences, $a_n$ and $b_n$ such that for all $n$, $0 \leq a_n, b_n \leq 1$ and the point $a_n$ of the way along line segment 1 in $S_n$ intersects the point $b_n$ of the way along line segment 2 in $S_n$.

Now we can take a subsequence of $a_n$ such that the subsequence converges. (In fact there is one that converges to the infimum of the set of $x$ such that $a_n < x$ an infinite number of times.) And then a subsequence of the corresponding subsequence of $b_n$ such that it converges as well.

But the limit of that subsequence of $S_n$ is the limit of the original sequence. It is straightforward to verify that it intersects at the limiting point found from the subsequence of $a_n, b_n$ that we just constructed. So it is in the set.

Therefore this is a closed set.

Incidentally its boundary is the set of pairs of intersecting line segments where one end of a line segment is in the other line segment.


First, note that if we're realizing our space of pairs as $(\Bbb R^2)^4 \cong \Bbb R^8$ via the map $(\overline{AB}, \overline{CD}) \leftrightarrow (A, B, C, D)$, then our space is really the set of ordered pairs of directed line segments. If we instead want to work with the space, e.g., of ordered pairs of (undirected) line segments, we're identifying $\overline{AB}$ and $\overline{BA}$ for every pair $(A, B)$, and so our space of pairs is $(\Bbb R^2)^4 / (\Bbb Z_2 \times \Bbb Z_2) \cong \Bbb R^8 / (\Bbb Z_2 \times \Bbb Z_2)$, where the actions of $(1, 0)$ and $(0, 1)$ respectively reverse the directions of the first and second segments in the ordered pair. In this case, our space inherits an orbifold structure, but at a glance it does not admit a manifold structure: It has singularities along the images under the quotient map $\Bbb R^8 \to \Bbb R^8 / (\Bbb Z_2 \times \Bbb Z_2)$ of the sets $\{A = B\}$ and $\{C = D\}$, that is, the pairs for which at least one of the segments is degenerate, i.e., just a point.

From now on, though, we'll work with the set of ordered pairs of directed line segments, though the story is similar if we work with unordered pairs and/or undirected segments. As you've pointed out, there is a natural action of $AGL(\Bbb R^2) = \Bbb R^2 \ltimes GL(\Bbb R^2)$ on $\Bbb R^*$ that (since affine maps preserve the property of whether two segments intersect) restricts to an action on $X$.

As you've already pointed out, on the (Zariski-)open, dense subset $Y = \{A \neq B, C \not\in \overline{AB}\}$ of $\Bbb R^8$, we can apply the $AGL(\Bbb R^2)$-action to normalize $A = (0, 0), B = (1, 0), C = (0, 1)$, but writing in coordinates a general affine transformation $T$ we find that imposing these three normalizations use all of the freedom of the group, or more precisely that imposing $TA = A, TB = B, TC = C$ forces $T = \operatorname{id}$. Thus, $AGL(\Bbb R^2)$-orbits of elements of $Y$ are parameterized exactly by $D = (x, y) \in \Bbb R^2$, and some simple plane geometry shows that $\overline{AB}$ intersects $\overline{CD}$ iff $y \leq 0$ and $0 \leq x \leq 1 - y$, a "closed, infinite trapezoid" in $\Bbb R^2$. The set of pairs $(\overline{AB}, \overline{CD})$ of points for which no three endpoints (i.e., no three of $A, B, C, D$) are collinear is the interior of this "trapezoid", which is homeomorphic to $\Bbb R^2$.

Compositions of the maps $(\overline{AB}, \overline{CD}) \mapsto (\overline{CD}, \overline{AB})$ and $(\overline{AB}, \overline{CD}) \mapsto (\overline{BA}, \overline{CD})$ map $Y$ to various other open, dense subsets of $\Bbb R^8$.

By definition, what is not covered by these open sets are the pairs for which the segments are collinear. In this case, by applying an affine transformation we may assume that the common line $Z$ is the $x$-axis $\Bbb R \times \{0\} \cong \Bbb R$, and the remaining freedom is exactly the $2$-dimensional group of affine transformations of $\Bbb R$. If $A \neq B$, we can use up all of this freedom to impose $A = 0, B = 1$, and $\overline{AB}$ intersects $\overline{CD}$ iff $0 \leq C \leq 1$ or $0 \leq D \leq 1$. The map $(\overline{AB}, \overline{CD}) \mapsto (\overline{CD}, \overline{AB})$ maps $Z$ to the set of pairs for which $C \neq D$. Again by definition what is not overed by $Z$ and its image under this map is the pairs of degenerate segments, that is, those of the form $(\overline{AA}, \overline{CC})$. There are two orbits of these, one comprising the pairs with $A \neq C$ and one comprising the pairs with $A = C$.


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