What is topologically the set of all straight lines in $\mathbb{R}^d$? More structures on it?

Question

If we consider the set of all straight lines in $\mathbb{R}^d$, then what is it topologically? If it's topologically 'nice' i.e. a manifold, probably we could put a smooth manifold structure on it too. Can we put a Riemannian manifold structure?

For the first question, I'd think of it as a disjoint union over the points $p$ of all lines passing through a fixed point $p$. So I'm thinking it's something like $\mathbb{RP}^{d-1}\times \mathbb{R}^d$. But is it? If it is, I guess the other questions might be answered from it.

Answer 1

Let $L^n$ denote the set of lines in $\mathbb R^n$. It's not the disjoint union over $p$ of lines passing through $p$, because for $p\ne q$, the sets $\{\text{lines passing through $p$}\}$ and $\{\text{lines passing through $q$}\}$ intersect. It's also not equal to the Grassmannian $G(2,\mathbb R^{d+1})$, because it's not compact.

One way to topologize $L^n$ is as a vector bundle over $\mathbb R\mathbb P^{n-1}$. The fiber over each line $[x]\in \mathbb R\mathbb P^{n-1}$ is the set of all lines parallel to $[x]$, which obtains a vector space structure by viewing it as the quotient space $\mathbb R^n/\operatorname{span}(x)$. The projection maps each line $\ell$ in $\mathbb R^n$ to the unique line parallel to $\ell$ and passing through the origin.

It obtains a smooth structure as a homogeneous space: the affine group $\mathbb R^n \rtimes \operatorname{GL}(n,\mathbb R)$ acts transitively on $L^n$, with closed isotropy group $\mathbb R^{n-1}\rtimes \operatorname{GL}(n-1,\mathbb R)$.

Answer 2

As requested by the OP, here are more details. Consider a line $l$ in $\mathbb{R}^2$. This line is given by $ax+by=c$. If $c\not=0$, then the line does not pass through the origin. Now, consider the plane in $\mathbb{R}^3$ given by $ax+by=cz$. If you look at a slice of this plane at $z=1$, you are left with the line that you started with. Moreover, the plane $ax+by=cz$ is unique and uniquely determines the original line.

If you consider the normal vector to this plane, you have a line in $\mathbb{R}^3$ that determines the plane constructed above. The set of lines in $\mathbb{R}^3$ is given by $\mathbb{R}\mathbb{P}^2$.

The problem with this construction is that when $a$ and $b$ are zero, but $c$ is nonzero, such a plane does not correspond to a line. Therefore, the space that you're looking for is an open subset of $\mathbb{R}\mathbb{P}^2$.