The set of lines in $\mathbb{R}^2$ is a Möbius band?

Question

I encountered a hard-to-believe and hard-to-understand statement in this problem:

Let $\mathcal{L} = \{\textrm{lines in}\ \mathbb{R}^2\}$. Consider the 2-to-1 map $f: S^1 \times \mathbb{R} \rightarrow \mathcal{L}$ given by $(\theta,x) \mapsto L_{\theta,x} = \{t(\mathrm{cos}\,\theta,\mathrm{sin}\,\theta)+x(-\mathrm{sin}\,\theta,\mathrm{cos}\,\theta): t\in\mathbb{R}\}$.

Show that for almost every $(\theta,x) \in S^1\times\mathbb{R}$, we have that $f^{-1}(L_{\theta,x})$ is a smooth submanifold of $X$.

Remark: The set of lines in $\mathbb{R}^2$ is a Möbius band.

So my question is, why the set is a Möbius band? And I really hope to get some help to understand the function $f$ - how it work: why $t(\mathrm{cos}\,\theta,\mathrm{sin}\,\theta)+x(-\mathrm{sin}\,\theta,\mathrm{cos}\,\theta)$ gives lines in $\mathbb{R}^2$? And why it is 2-to-1? Thank you very much for your guidance.

Answer 1

First, the remark (I think it is more important and also simpler to see). "The set of lines in $\mathbb{R}^2$ should immediately make us think of the projective plane. This previous answer shows why the space of unoriented lines is just a punctured projective plane.

Depending on your exposure to the classification of surfaces, you may declare the problem finished at this stage. After all, one construction of the projective plane is to glue a disc along the boundary of a möbiusband. Or you can try to convert a diagram for the punctured projective plane into a diagram for an (open) möbiusband. I have uploaded a series of diagrams depicting this here. You should be able to see that there is a slightly more efficient way of manipulating the diagram than what I did.

Now we just need to see why that particular map gives a parametrisation of the unoriented lines. I think a more sophisticated mind than mine should be able to explicitly see how this is a copy of $\mathbb{R}$ twisting along the circle, but I don't know the formalism to explain that.

We wish to understand how $x$ and $\theta$ determine the line $L_{\theta, x}$ geometrically. The simplest point is when $t = 0$, and this gives us $(-x \sin\theta, x \cos\theta)$. Then the part of the line with $t$ in front clearly says that the line emanates from this point at an angle of $\theta$. So modifying $x$ with $\theta$ fixed gives us all the lines in the equivalence class of parallel lines to the $t=0$ one. You can then show that we get all lines, and we get the same line when we do the involution $\theta \mapsto \pi + \theta, x \mapsto -x$.

Answer 2

Here is a geometric proof that the open subset $L:=\mathbb P^2\setminus \{[1:0:0]\}\subset \mathbb P^2$ of the projective plane is isomorphic to the total space of the bundle $\mathcal O_{\mathbb P^1}(1) $, where $\mathbb P^1$ is identified with the line $P\subset \mathbb P^2$ given by $z_0=0$ and the projection morphism of the bundle is $L\to \mathbb P^1:[z_0:z_1:z_2]\mapsto [0:z_1:z_2]$ .

Let $U_i\subset L \;(i=1,2)$ be the open subset $z_i\neq 0$.
The trivializations of $L$ are then given by : $$f_1:U_1\to \mathbb (P^1\cap U_1)\times \mathbb R: [z_0:z_1:z_2]\mapsto ([0:z_1:z_2], z_0/z_1)$$$$f_2:U_2\to (\mathbb P^1\cap U_2)\times \mathbb R: [z_0:z_1:z_2]\mapsto ([0:z_1:z_2], z_0/z_2) $$

The transition cocycle is determined by $g_{12}([z_0:z_1:z_2])=z_2/z_1$, which shows that our bundle is indeed $\mathcal O_{\mathbb P^1}(1)$.

The above proof is valid on a completely arbitrary field and has an obvious generalization to $\mathcal O_{\mathbb P^n}(1)$, whose total space is thus shown to be isomorphic to $\mathbb P^{n+1}\setminus \{[1:0:\cdots:0]\}$.
Over $\mathbb R$ the total space of $\mathcal O_{\mathbb P^1}(1) $ is the Möbius strip: this may be given as the definition of the Möbius strip or necessitates a proof if the Möbius strip is defined in another way.

Answer 3

I feel like the existing answers, while mathematically correct, may be a bit much for someone new to topology. Let me try a less formal explanation why "The Möbius strip is equivalent (homeomorphic) to the set of non-oriented lines in the plane".

Consider the standard construction of a Möbius strip by gluing the vertical ends of a flat strip of paper of width $[0,2\pi)$ and height $(-1,1)$ "with a 180° twist". Clearly you can specify any point on the strip uniquely by the "angle" $\theta\in [0,2\pi)$ and the "height" $x\in (-1,1)$; this is simply the parametrization of the paper strip before you glue it.

Similarly you can specify any line in the plane uniquely by an angle $\theta \in [0,\pi)$ (where does the line point?) and a height $(-\infty,\infty)$ (how far do we translate the line from the origin towards where it points?). This is what your $f$ does, except it does it "twice" by allowing $\theta \in [0,2\pi)$.

Note that $[0,\pi)$ and $(-\infty, \infty)$ are homeomorphic (can be stretched or shrunk in)to $[0,2\pi)$ and $(-1,1)$. Since they have the same parametrizations (up to homeomorphism) it seems the Möbius strip and the set of lines in the plane are really the same. However, to really be able to conclude that, we still need to talk about the "edges" of our parametrizations. To define the Möbius strip by gluing the edges of a strip means that we map the limiting angle $\theta\to2\pi$ to the same points as $\theta=0$ but with the heights $x\in(-1,1)$ inverted.

Is that property true for our parametrization of the lines too? Yes! The horizontal lines that you get for $\theta=0$ are exactly those that you get if you let $\theta\to\pi$, but the former diverge upwards as you increase the "height" parameter, while the latter diverge downwards.