Let $M$ be the space of pairs $\{(l,P)|l \subset P \subset R^3\}$ where $l$ is a one-dimensional subspace and $P$ is a two-dimensional subspace of $R^3$. Define a injection $M \rightarrow RP^2 \times RP^2$ by $(l,P) \rightarrow (l,P^{\perp})$ and use this map to define a topology on $M$. Prove or disprove:
(1) $M$ is a compact smooth submanifold.
(2) $M$ is non-orientable.
(3) The universal covering of $M$ is diffeomorphic to the set of unite quaternions in $\mathbb{H}$.
(4) The fundamental group $\pi(M)$ is isomorphic to the standard 8-element quaternion group $\{\pm 1, \pm i, \pm j, \pm k\}$
For (1), I think I maybe need to show $RP^2 \times RP^2 - M$ is open, then $M$ is closed and hence compact?
For (3) and (4), I cannot find a covering map.
I am sure there must be a more elegant solution, but my trial:
Note that $l\subseteq P$ iff $l\perp P^\perp$, so $M$ is a set of all pairs of lines $(l, p)$ such that $p\perp l$. There is a natural $4:1$ map $\pi: SO(3)\to M$ that sends a positive orthonormal basis $(a,b,c)$ to $\langle a \rangle, \langle b\rangle$. Identifying $M$ with a subset of $\mathbb{RP}^2\times\mathbb{RP}^2$, $\pi$ is smooth, which can be verified in local coordinates; by construction, it is a covering map, so $M$ is a smooth manifold. Further, it is the image of a compact manifold $SO(3)$, hence it is compact.
As a homogenous space, $M\simeq SO(3)/\{1,R_{z}(\pi),R_x(\pi), R_y(\pi)\}$ where $R_{xyz}(\pi)$ is a rotation by $\pi$ around the coordinate axes. The universal cover of $M$ is the universal cover of $SO(3)$, which is the sphere $S^3$ of unit quaternions, $SO(3)\simeq S^3/\{\pm 1\}$. The sphere $S^3$ acts on $\mathbb{R}^3$ by rotations $R_h: v\mapsto hvh^*$ where $v$ is a purely imaginary quaternion; $\pm k$ acts by $180^\circ$ rotation around the $ij$-plane (identified with the $xy$ plane). So, $\pm k$ may be identified with the preimage of $R_z(\pi)$ and similarly, $\pm i$ resp. $\pm j$ with the preimages of $R_x(\pi)$ and $R_y(\pi)$. Hence $M\simeq S^3/\{\pm 1, \pm i, \pm j, \pm k\}$.
(2) is not correct, $M$ is orientable. To show this, I will define the orientation of each tangent space directly. Let $m:=(p,q)$ be a pair of orthogonal lines and choose a positive basis $(a,b,c)\in SO(3)$ so that $\langle a\rangle =p$ and $\langle b\rangle =q$. I will define a basis $(w_1, w_2, w_3)$ of $T_m M$ as follows. The first tangent vector $w_1\in T_m M$ will be induced by an infinitesimal rotation of $(b,c)$ around $a$ in the "positive" direction (if your right hand thumb is in direction $a$, you fingers shows the positive rotation of $\langle b,c\rangle$). That is, $w_1$ is induced by a curve that leaves $p$ unchanged and rotates $q$ in the direction indicated. Similarly, $w_2$, $w_3$ are induced by infinitesimal rotation of the frame $(a,b,c) $around $b$ resp. $c$. A basis of $T_m M$ will be positively oriented, if it has the same orientation as $(w_1, w_2, w_3)$. It remains to show that this construction is independent of the choice of $(a,b,c)$. However, the three other choices are $(a,-b,-c)$, $(-a,b,-c)$ and $(-a,-b,c)$ which all have two minus-signs, so the same rule would induce the bases $(w_1, -w_2, -w_3)$, $(-w_1, w_2, -w_3)$ resp. $(-w_1, -w_2, w_3)$ which have all the same orientation as bases of $T_m M$. So, this choice of the orientation of $T_m M$ is well defined.
(I will be happy if somebody shows me a nicer, shorter and less-explicit solution. Here is a statement thath the full real flag manifold is always orientable -- their notation corresponds to $n_1=n_2=n_3=1$ in our case)