Consider a manifold $ N $ defined as follows $$ N=\{n\otimes n-m\otimes m:n,m\in S^2,\,\,(n,m)=0\}\subset M^{3\times 3}, $$ where $ S^2 $ denotes the two dimensional sphere, $ (\cdot,\cdot) $ represents the inner product of the space $ \mathbb{R}^3 $ and $ M^{3\times 3} $ denotes the set of $ 3\times 3 $ matrices. Here $$ (n\otimes n-m\otimes m)=(n_in_j-m_im_j)_{1\leq i,j\leq 3}. $$ I want to determine the fundamental group and universal covering space of $ N $.
I note that the manifold $ \{n\otimes n:n\in S^2\} $ is actually $ \mathbb{R}P^2 $ and want to use same arguments on $ N $ here. However, I do not how to go on. Can you give me some hints or references?
Edit:
A simple way to determine the fundamental group and the universal cover is the following. We will show that $N$ is homeomorphic to $S^3/Q_8$, where $Q_8$ is the quaternion group.
Define $M=\{(n,m)\colon n,m\in S^2, n\cdot m=0\}$. It is easy to see that $M$ is the total space of the unit circle bundle of $TS^2$, so $M\cong \Bbb RP^3\cong SO(3)$.
Note that $\Bbb Z_2\times\Bbb Z_2$ acts on $M$ freely, where the action is generated by $(n,m)\mapsto (-n,m)$ and $(n,m)\mapsto(n,-m)$. This implies that $M/(\Bbb Z_2\times\Bbb Z_2)\cong N$ by sending the class of $(n,m)$ to $n\otimes n-m\otimes m$.
Under the homeomorphism $M\to SO(3),\ (n,m)\mapsto \{n,m,n\times m\}$. these actions get transferred to $\{n,m,n\times m\}\mapsto \{-n,m,-n\times m\}$ and $\{n,m,n\times m\}\mapsto \{n,-m,-n\times m\}$ on $SO(3)$. So the orbit space $SO(3)/(\Bbb Z_2\times\Bbb Z_2)$ is homeomorphic to $N$. Consider the action of $Q_8$ on unit quaternions given by right multiplication. This action is compactible with the double cover $S^3\to SO(3)$, so we have homeomorphisms $S^3/Q_8\cong SO(3)/(\Bbb Z_2\times\Bbb Z_2)\cong N$. Hence, $\pi_1(N)\cong Q_8$, and $S^3$ is its universal covering space.
Alternative descriptions of $N$:
We will see that $N$ is homeomorphic to the total space of the projectivized tangent bundle of $\Bbb RP^2$, $P(T\Bbb RP^2)$.
First, we determine how elements of $N$ acts on $\Bbb R^3$.
Summation convention is assumed, and we work under any orthonormal basis. Let $v\in\Bbb R^3$, $v_i$ is its $i$th compoenent. Then $$((n\otimes n-m\otimes m)v)_i=(n\otimes n-m\otimes m)_{ij}v_j=n_in_jv_j-m_im_jv_j=(n\cdot v)n-(m\cdot v)m$$
Observe that the linear map above is a projection onto the plane spanned by $n$ and $m$ (denoted $\langle n,m\rangle$) followed by a reflection (in this plane) about the line spanned by $n$ (or equivalently by orthogonality, reflection about the plane that is normal to $m$). More rigorously, working with the orthonormal basis $\{n,m,n\times m\}$, we have $v=v_1n+v_2m+v_3(n\times m)$ and \begin{align}(\text{ref}\circ\text{proj})(v)&=\text{ref}(v_1n+v_2m)\\ &=v_1n-v_2m=(v\cdot n)n-(v\cdot m)m\end{align} This shows that each element of $N$ is uniquely characterized by the plane $\langle n,m\rangle$ (and hence the orthogonal complement $\langle n\times m\rangle$) and the line $\langle n\rangle$ (axis of reflection)
Define the following map $$f:N\to P(T\Bbb RP^2),\ (n\otimes n-m\otimes m)\mapsto(\langle n\times m\rangle,\langle n\rangle)$$ This map is a homeomorphism ($N$ is compact, as it's a closed and bounded subset of $\Bbb R^9$; so this is a continuous bijection from compact space to Hausdorff space).
Another observation is that $N$ can be identified with the set of sequences of subspaces $\langle n\rangle\le\langle n,m\rangle\le\langle n,m,n\times m\rangle=\Bbb R^3$, which is the (complete) flag space $F(\Bbb R^3)$. For more information, see this wikipedia article.