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 $\mathbb R^2$?

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    $\begingroup$ 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. $\endgroup$ Jul 4, 2016 at 15:16
  • $\begingroup$ @AlexProvost. You are right! $\endgroup$
    – Lehs
    Jul 4, 2016 at 15:21
  • $\begingroup$ 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. $\endgroup$ Jul 4, 2016 at 20:40
  • $\begingroup$ 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). $\endgroup$ Jul 5, 2016 at 4:19

3 Answers 3


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.

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  • $\begingroup$ 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? $\endgroup$
    – snulty
    Jul 4, 2016 at 15:54
  • $\begingroup$ @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.) $\endgroup$ Jul 4, 2016 at 16:12
  • $\begingroup$ @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. $\endgroup$ Jul 4, 2016 at 16:12
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    $\begingroup$ @JanDvorak It appears to have been rendered in real life! Source. $\endgroup$ Jul 4, 2016 at 22:57
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    $\begingroup$ @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 $\endgroup$ 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).

  • $\begingroup$ What is the subspace topology of $L_+$? $\endgroup$ Jul 5, 2016 at 15:50
  • $\begingroup$ @EduardoLonga As I said in my answer, $L_+$ is homeomorphic to the punctured plane $\mathbb{R}^2\setminus\{(0, 0)\}$. $\endgroup$ Jul 5, 2016 at 17:14
  • $\begingroup$ Ah, you impose the topology on this set of lines so that it is homeomorphic to the punctured plane. Now I got it. $\endgroup$ Jul 5, 2016 at 17:36
  • $\begingroup$ @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. $\endgroup$ Jul 5, 2016 at 18:06

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