Smooth manifold structure on space of all planes in $\mathbb{R}^3$? I am looking at the space of all planes in $\mathbb{R}^3$, and I am wondering if there exists any particular natural smooth manifold structure on it. I thought you could send the normal vector of each plane to its equivalence class $l$ in the real projective plane $\mathbb{R}\mathbb{P}^2$, and then sending its displacement along its normal vector to a number $r$, and then get a bijection to $\mathbb{R}\mathbb{P}^2\times\mathbb{R}$. But I realized that we don't really have a canonical choice of direction to measure the displacement in, so my idea fails there.
 A: Let $P$ be an affine plane in $\mathbb{R}^3$. There is a unique line $\ell$ through the origin which is normal to $P$. Another affine plane has the same normal line $\ell$ if and only if it is parallel to $P$. If $P\cap\ell = \{a\}$, then $P = a + P_0$ where $P_0$ is the plane through the origin with $\ell$ as its normal line. So every plane parallel to $P$ is given by a plane through the origin parallel to $P$, namely $P_0$, and a point on the normal line through the origin, namely $a$. Note that $(P_0, a) \in \gamma_2^{\perp}$ where $\gamma_2$ is the tautological line bundle over $\operatorname{Gr}_2(3) = \mathbb{RP}^2$. Conversely, for any point $(P', a') \in \gamma_2^{\perp}$, it defines the affine plane $a' + P'$.
More generally, I believe the set of affine $k$-planes in an $n$-dimensional real vector space $V$ is isomorphic to $\gamma_k^{\perp}$ where $\gamma_k \to \operatorname{Gr}_k(V)$ is the tautological line bundle.
A: Think about what happens for oriented planes - the set of oriented planes through the origin forms a copy of $S^2$ (since any such plane is identified by its oriented unit normal), and the translation along the normal vector is now a canonical way to get any plane. So the set of oriented affine planes has the natural structure $S^2 \times \mathbb R$, and the equivalence relation to forget orientation is $(p, s) \sim (-p, -s)$; so the space you're interested in has the smooth manifold structure coming from the corresponding quotient of $S^2 \times \mathbb R$ by this (free and proper) $\mathbb Z/2\mathbb Z$ action.
As for identifying this quotient as something more familiar, perhaps someone else knows a better approach than I.
A: Think of $R^3$ as the affine space $A: t=1$ in $R^4$. A plane is $R^3$ is then a 3 dimensional vector subspace of $R^4$ which meets this $A$ transversaly, ie which is not parallel to it.
In coordinate, the plane $ax+by+cz+dt=0$ meets $A$ along the plane $ax+by+cz+d=0$, which is a plane unless $a=b=c=0$.
Therefore, an answer is that the set of plane in $R^3$ is the complement of 1 point in $RP^3$ ; it appears to be the tautological line bundle over $RP^2$
A: Yes, this structure is called the Grassmannian
https://en.wikipedia.org/wiki/Grassmannian
