点到直线的"有向距离"

hbghlyj

本帖对点到直线的"有向距离"提出疑问：

考虑2对1映射 $f: S^1 \times \mathbb{R} \rightarrow \mathcal{L}$，定义为$$f(\theta,x) =\text{直线} L_{\theta,x} = \{t(\cos\theta,\sin\theta)+x(-\sin\theta,\cos\theta): t\in\mathbb{R}\}$$
其中$(\cos\theta,\sin\theta) ∈S^1$，$θ$是倾斜角，$x$是原点到直线的"有向距离"。
$f$是2对1的，因为$f(\theta,x)=f(\pi+\theta,-x)$，这在上式把$t$换成$-t$就能看出。
我的问题是，为什么这个集合是一个Möbius带？非常感谢您的指导。

hbghlyj

关于$\mathbb{R}^2$，搜到一帖 The space of all lines on a plane is an open Möbius Band.

hbghlyj

又见于Exercise 5. The space of lines in $\Bbb R^2$.
下方有提示：For example, lines which are not vertical can be parametrized by 2 numbers: the angle $θ ∈
(−π/2, π/2)$ which tells you how much the line is tilted, and $r ∈ \Bbb R$ which is the signed distance of the line from the origin $0 ∈ \Bbb R^2$ (using the + sign if the line passes above 0, and the − sign if it passes below 0).

hbghlyj

那么$\mathbb{R}^3$中的直线组成什么？sword.cit.ie/cgi/viewcontent.cgi?article=1003 … ;context=dpttem_kpre
$\mathbb{R}^n$中的直线呢？