Show that the set of lines in $\mathbb{R}^n$ is a (smooth) manifold of dimension $2(n-1)$ I was recently made aware of the result in the title. It's easy to show for $\mathbb{R}^2$, but I'm having trouble coming up with a generalization for $\mathbb{R}^n$.
There are a couple of ways to represent the set of lines in $\mathbb{R}^n$. One way is to let $X=\mathbb{R}^n \times (\mathbb{R}^n \setminus \lbrace 0\rbrace)$ with the usual topology, viewed as the space of points and direction vectors defining lines through those points. Then define a relation $\sim$ on $X$ by $(p_1,v_1)\sim (p_2,v_2) \iff \text{$v_1 \parallel v_2$ and $p_1 - p_2 \parallel v_1$}$. So $Y=X/\sim$ with the quotient topology is the space of lines in $\mathbb{R}^n$, up to homeomorphism.
We can also let $Y=\mathbb{R}^n \times \mathbb{R}P^{n-1}$ with the product topology, viewed as the set of translated lines through the origin.
Clearly both definitions of $Y$ the set of lines are well-defined and homeomorphic, but how do we come up with a smooth atlas of charts onto $\mathbb{R}^{2(n-1)}$? I tried considering $n$ open sets $\lbrace U_i \rbrace_{1\leq i\leq n}$ where $U_i$ is the set of all the lines not parallel to $e_i$ the $i$-th standard basis vector, but could not get very far.


I tried considering $n$ open sets $\lbrace U_i \rbrace_{1\leq i\leq n}$ where $U_i$ is the set of all the lines not parallel to $e_i$ the $i$-th standard basis vector, but could not get very far.

You have to exclude more than a single point for this to work cleanly; it's best to exclude a subspace of codimension $1$.
The atlas: Let $U_i$ be the set of all lines not contained in a hyperplane $x_i=c$ parallel to the $i$th coordinate hyperplane. For coordinates on $U_i$, associate each such line to the ordered pair $(u,v)$ of its intersections with the hyperplanes $x_i=0$ and $x_i=1$ respectively.
The dimension is $2(n-1)$, of course.
