A topology on the set of lines?

Of course any set $X$ can have a topology, but are there more natural topologies, metrics or similar on the set of straight lines in $\mathbb R^2$?

• Instead of constantly editing more questions in, you should revert back to the original question pertaining to lines in the plane and ask about lines in $\mathbb{R}^3$ in a different question. Jul 4, 2016 at 15:16
• @AlexProvost. You are right!
– Lehs
Jul 4, 2016 at 15:21
• You might be interested in the Grassmanian manifold. I don't really have the expertise to describe it, usefully but one can certainly give topology (and more) to the affine lines in a vector space using it. Jul 4, 2016 at 20:40
• Not enough for an answer: You might be interested in reading about the Hough transform (which turns lines into points, making options for a topology fairly straightforward). Jul 5, 2016 at 4:19

There is in fact a natural smooth structure on the set $L$ of lines in $\mathbb{R}^2$. The atlas is made up of two sets $U,V$ corresponding to the non-vertical (resp. non-horizontal) lines. Note that a non-vertical line $y = mx + c$ is given by two uniquely determined parameters, and so the map $U \to \mathbb{R}^2$ which maps a non-vertical line to $(m,c)$ is a chart. Similarly, a non-horizontal line $x = my + c$ yields two parameters for the chart $V \to \mathbb{R}^2$. Finally, the composition of these two charts in any order is smooth: for instance, the elements of $U \cap V$ are lines $y = mx + c$ with $m \neq 0$, and this implies $x = m^{-1}y - cm^{-1}$. So the composition of the two charts is given by $(m^{-1},-cm^{-1})$, which is smooth with a smooth inverse.

In fact this smooth manifold is well-known: it is the punctured projective plane $\mathbb{R}P^2 \setminus \{[0:0:1]\}$. Indeed there is a well-defined smooth embedding $L \to \mathbb{R}P^2$ given by $\{ ax+by+c = 0 \} \mapsto [a:b:c]$. It is almost surjective, except that we cannot have $a=b=0$ and $c \neq 0$. Upon removal of the point $[0:0:1]$, we obtain a diffeomorphism $L \cong \mathbb{R}P^2 \setminus \{[0:0:1]\}$.

Note that the projective plane minus a point is a well-known and easily visualisable surface: it is the Möbius strip with its boundary removed! To make the correspondence between $L$ and the Möbius strip clearer, notice that a line in $L$ is almost uniquely determined by the two following quantities: its angle $\in \mathbb{R}P^1 \cong S^1$ (where $0^\circ = 180^\circ$), and its distance from the origin $\in [0,\infty)$. However this correspondence fails: if $d > 0$ and $\theta \in S^1$ are fixed, there are precisely two lines with angle $\theta$ with distance $d$ from the origin. Fine; let's fix this by allowing distances to be negative! Then if $(d,\theta)$ corresponds to a line, $(-d,\theta)$ corresponds to the other line across the origin. Perfect! But wait... if this were true, this would establish a correspondence between $L$ and the product $S^1 \times \mathbb{R}$, aka the open cylinder, aka the trivial line bundle over $S^1$. But this cannot be the case: any simple closed curve on the open cylinder disconnects it (Jordan curve property for $\mathbb{R}^2 \setminus \{0\} \cong \text{open cylinder}$), but this doesn't hold in $L$. Indeed, remove the loop in $L$ that consists of all lines through the origin. Then it is clear that you can still go from any line in $L$ to any other using a path that never uses a line through the origin.

So what goes wrong? The problem is that it is impossible to consistently assign a choice of "positive direction" to the copy of the real line attached to each angle $\theta \in S^1$. It is possible locally, in a neighborhood of each line, but these local pictures cannot be glued together consistently because of a topological obstruction. For instance, if we start with a line through the origin and rotate this line $180^\circ$, we end up with the same line, but the notion of "distance to the origin" has been reversed! This is a manifestation of the nonorientability of the Möbius bundle over the circle.

• I've always seen it that the mobius strip was $\Bbb R P^2$ with a disk removed, but I suppose with the disk being contractible these are the same? Jul 4, 2016 at 15:54
• @snulty One has to be a little careful, because removing homotopic (or even homeomorphic) subspaces from a larger space might not result in homeomorphic spaces. (E.g, the plane is homeomorphic to the open disk, but removing both from the plane yields distinct spaces.) Jul 4, 2016 at 16:12
• @snulty But removing a small closed ball in a coordinate chart of a manifold amounts to removing a point. This amounts to the identity $\mathbb{R}^n \setminus \{0\} \cong \mathbb{R}^n \setminus B^n$, where $B^n$ is the closed unit $n$-ball. The identity follows from the homeomorphism $(0,\infty) \cong (1,\infty)$ provided by the exponential map. Then just expand each vector outward radially by exponentiating its modulus. Jul 4, 2016 at 16:12
• @JanDvorak It appears to have been rendered in real life! Source. Jul 4, 2016 at 22:57
• @PaulWintz It's pretty standard notation for homogeneous coordinates, i.e. equivalence classes of points in projective space. For example $[0:0:1] = [0:0:2]$. See en.wikipedia.org/wiki/Homogeneous_coordinates#Notation Sep 8, 2022 at 21:34

Intuitively we can say whether two lines are close: say, using an angle and distance between them. This defines a metric space, hence also a topology. I'll show another, more elegant way to achieve the same.

Any straight line in $\mathbb{R}^2$ can be non-uniquely specified by a point and a non-zero direction vector. So, let $X = \mathbb{R}^2 \times \left(\mathbb{R}^2 \setminus 0\right)$, where the first component is a point, and the second is a vector. Put the standard topology on $X$.

Now, define an equivalence relation on $X$ such that two pairs are equivalent iff they represent the same line: $(p_1, v_1) \sim (p_2, v_2) \Leftrightarrow v_1 \| v_2 \,\land\, (p_2-p_1) \| v_2$ (where $a\|b$ means vectors $a$ and $b$ are parallel).

We can now identify $X/\sim$ with the set of all lines in $\mathbb{R}^2$ and put the standard quotient topology on it.

Adding to the existing answers - which are quite correct - let me give yet another way of visualizing the space of lines.

Let's first look at the lines in $\mathbb{R}^2$ not passing through the origin; call this set $L_+$. An element $l$ of $L_+$ is specified by an element $\alpha_l$ of the punctured plane $\mathbb{R}^2\setminus\{(0, 0)\}$: namely, $\alpha_l$ is the point on $l$ closest to the origin. To put it another way, draw a line from the origin to $\alpha_l$; the perpendicular to this line, through $\alpha_l$, is $l$. It's easy to check that this is a homeomorphism between $L_+$ (with the subspace topology) and $\mathbb{R}^2\setminus\{(0, 0)\}$.

Now, what happens at the origin? Well, a line through the origin is specified by its angle: specifically, by a point on $S^1$ . . .

. . . Except that this winds up double-counting lines. E.g. the $x$-axis can be specified by either $(0, 1)$ or $(1, 0)$. So really, the set of lines through the origin looks like $S^1$ with opposite points identified - that is, the projective space $\mathbb{RP}^1$. (This is of course homeomorphic to $S^1$, but that's peculiar to the number $1$.)

So what description does this give of the space of all lines look like? Well, it looks like the plane $\mathbb{R}^2$, with the origin replaced by a copy of the projective space $\mathbb{RP}^1$. Making this precise is a bit tricky, but a good exercise. And this turns out to be a kind of degenerate example of a crucial construction in algebraic geometry - the blowup (see https://en.wikipedia.org/wiki/Blowing_up).

Incidentally, it's noting that - while this description of the space of lines is correct, and has a number of nice properties - it is misleading in certain ways. In particular, the description sounds non-homogeneous (the origin seems like a special point), while it is clear from thinking about lines in the plane that the space of lines has no distinguished points (and this is clearer from the other descriptions).

• What is the subspace topology of $L_+$? Jul 5, 2016 at 15:50
• @EduardoLonga As I said in my answer, $L_+$ is homeomorphic to the punctured plane $\mathbb{R}^2\setminus\{(0, 0)\}$. Jul 5, 2016 at 17:14
• Ah, you impose the topology on this set of lines so that it is homeomorphic to the punctured plane. Now I got it. Jul 5, 2016 at 17:36
• @EduardoLonga Sort of. The topology on the space of all lines in the plane is the same as in the other answers; I'm just giving a different description of that space. So, under that topology, $L_+$ happens to be homeomorphic to the punctured plane. Jul 5, 2016 at 18:06