What's the $2$d-object that's in $1:1$ correspondence with all lines in $\Bbb R^2$?

Question

What's the 2d-object that's in 1:1 correspondence with all lines in $\Bbb R^2$?

(And what's the correct term for such an object? Locus? Fundamental domain?)

Writing a line as

$$y=ax+b \quad\text{ with }\quad (a,b)\in \Bbb R^2\tag 1$$

each point $(a,b)$ represents a different line, but this misses all vertical lines. The line for $a\to\infty$ is $x=0$ and the same as $a\to-\infty$. It can be included by representing lines as

$$y=\tan(a)\cdot x+b \quad\text{ with }\quad (a,b)\in \left[-\tfrac\pi2,\tfrac\pi2\right]\times\Bbb R\tag 2$$

However $(\pm\frac\pi2,b)$ will all yield the same line, so that these points have to be identified with each other. And the lines $x=c$, $c\neq0$ are still missing.

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For 2-dimensonal projective space $M=P\Bbb R^2$, I think all lines are represented by $M$ itself: For a line $g=(g_x:g_y:g_z)$ and a point $P=(x:y:z)$, the condition that $P\in g$ is $$g\cdot P = g_xx+g_yy+g_zz=0$$ which is symmetric in $g$ and $P$, and therefore the space of all lines is isomorphic to the space $M$ of all points.

To go from $M$ to $\Bbb R^2$ one can remove the line "at infinity", which is homeomorphic to $\Bbb S^1$. But $M\backslash \Bbb S^1$ has two connected components: One is a Möbius Band and one is a disc, and I have no idea how to interpret them in terms of unique affine lines.

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Addendum: Meanwhile I found a better representation of lines in the real $a$,$b$-plane: Let $g$ be the line

$$g:ax+by=a^2+b^2-r^2\tag 3$$ for some constant $r$, say $r=1$. Then $a^2+b^2 > r^2$ will yield all possible lines except the ones through the origin. Lines through the origin are realized by $a^2+b^2 = r^2$ where $(a,b)$ and $(-a,-b)$ yield the same line, i.e. we have to identify opposite points of the circle of radius $r$.

Points in the circle do not represent (new) lines, so we effectively poked a hole of radius $r$ into the plane and glued opposite sides of that hole together, yielding a Möbius band.

Here is a Desmos plot. The blue dot is $(a,b)$ in the $a$,$b$-plane which represents the blue line and can be dragged around. The red-ish circle represents the hole poked into the plane.

Actually, this can be used the other way round: All points in the circle represent lines, except the point at the origin which would represent the line at infinity. Points in the white area do not represent (new) lines. Again, opposite points of the circle are glued together because they represent the same line through the origin.

Moreover, $(3)$ generalizes nicely to higher dimensions: Each point $g\in\Bbb R^n$ with $|g| > r$ represents the unique $n-1$-dimensional hyper-plane

$$L_g: \langle g,x\rangle = |g|^2-r^2\tag 4$$ And the points with $|g|=r$ represent hyper-planes through the origin where $L_g = L_{-g}$, so that $g$ and $-g$ have to be indentified to achieve uniqueness. Hence the locus is $n$-dimensional projective space with one point removed.

$L_g$ has a distance of $(g^2-r^2)/|g|$ to the origin, and it is perpendicular to $g$ which means that for all $x,y\in L_g$ we have $\langle x-y,g\rangle=0$.

Answer 1

The brief answer to your question is: an open Möbius band.

In more detail, let's embed $\mathbb R^2$ into $\mathbb R^3$ by the map $(x,y) \mapsto (x,y,1)$. Using this embedding, I am going to abuse notation by identifying $$P = \{(x,y,z) \in \mathbb R^3 \mid z=1\} $$ Associated to each line $L \subset P$ is the unique plane $M_L \subset \mathbb R^3$ such that $L \subset M_L$ and $M_L$ passes through the origin. This plane $M_L$ is never the $x,y$-coordinate plane $\{(x,y,z) \in \mathbb R^n \mid z=0\}$.

Conversely, associated to each plane $M$ through the origin which is not the $x,y$-coordinate plane is a line $L_M = M \cap P$ in $P$.

This gives a one-to-one correspondence between lines in $P$ and planes in $\mathbb R^3$ through the origin that are not the $x,y$-coordinate plane.

The space of all planes through the origin is $\mathbb RP^2$, the real projective plane. Removing one point from $\mathbb RP^2$ --- any point, but in this example we are removing the one point that corresponds to the $x,y$-coordinate plane --- gives an open Möbius band.

Answer 2

Wanted to present a "twistorial" perspective. The oriented affine lines of $\mathbb{R}^2$ may be identified with the unit circle's tangent bundle $TS^1$: given any tangent vector $t$ to a point $u$ on the unit circle, we can extend $u$ to an oriented line and then displace by $t$. Conversely, given any oriented affine line, $u$ is the direction vector, and $t$ is the displacement of its closest point from the origin.

As $S^1$ is parallelizable (the bundle $TS^1$ is trivializable), we find the bundle $TS^1$ is equivalent to an infinite cylinder $\mathbb{R}\times S^1$. Explicitly, this amounts to "turning" each tangent space on the unit circle $90^\circ$ into the third dimension to make a vertical line. (Visualize the reverse process as "flattening" the infinite cylinder as a cotangent bundle as one does with fields to get crop circles.)

This double covers the space of unoriented lines, the covering map being "forgetting" orientations. The fibers of this covering are the orbits of the free $\mathbb{Z}_2$-action given by negating both coordinates of $\mathbb{R}\times S^1$. Thus, the space of unoriented affine lines of $\mathbb{R}^2$ is the quotient $(\mathbb{R}\times S^1)/\mathbb{Z}_2$. Since $\mathbb{R}\simeq(0,1)$ are homeomorphic, we can visualize the infinite cylinder as a band. We can cut this band in half for a rectangular fundamental domain, then identify opposite points on the opposite pair of closed edges, i.e. make a Mobius band.