A question on the manifold $ \{n\otimes n-m\otimes m:n,m\in S^2,(n,m)=0\} $

Question

Consider a manifold $ N $ defined as follows $$ N=\{n\otimes n-m\otimes m:n,m\in S^2,\quad(n,m)=0\}\subset M^{3\times 3}, $$ 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. For a closed curve $ \alpha:S^1\to N $, we can consider the homotopy class of it. Define maps $$ p_1:N\to\mathbb{R}P^2,\quad p_2:N\to\mathbb{R}P^2 $$ as follows. For any $ A\in N $, we can represent $ A=n\otimes n-m\otimes m $ with $ (n,m)\in S^2\times S^2 $ and $ (n,m)=0 $, we define $ p_1(A)=n\otimes n $ and $ p_2(A)=m\otimes m $. I guess that if $ [\alpha] $ corresponds to the element of order $ 4 $ in the fundamental group, then either of $ [p_i(\alpha)] $, $ i=1,2 $ is non-trivial. However I do not know how to prove this or give a conterexample. I have already know the covering space and fundamental group of $ N $ but I do not know how to go on. Can you give me some hints or references?

Answer 1

I'm not sure exactly what you were claiming, but here is a correct claim.

$\pi_1(N)$ has six order four elements $\{\pm i, \pm j, \pm k\}$, and there are three natural maps $p_1, p_2, p_3: N \to \Bbb{RP}^2$. Each order-four element of $\pi_1(N)$ survives under two of these maps, and is annihalated under one.

In particular, given any order-four $x \in \pi_1(N)$, at least one of $(p_1)_*x$ or $(p_2)_* x$ are nonzero, which is one way to interpret the question you posed. (It may survive under only one, or it may survive under both. If you use all of $p_1, p_2, p_3$, then all elements survive under exactly two.)

---

It seems from your comments you understand that your manifold $N$ is diffeomorphic to the quotient of $SO(3)$ by the subgroup $D = \Bbb Z/2 \times \Bbb Z/2$, where this is the subgroup of diagonal matrices in $SO(3)$. I will just think of $N$ as this quotient. In particular, there are three projection maps $p_i: N \to \Bbb{RP}^2$, sending a matrix in $SO(3)$ (up to multiplication by a diagonal matrix in $SO(3)$) to its $i$'th column (up to multiplication by $\pm 1$). The first two of these are exactly what you described; in your formulation, the third is $p_3(n \otimes n^T - m \otimes m^T) = \{o, -o\}$, where $o$ is a unit vector in the kernel of $n \otimes n^T - m \otimes m^T$; equivalently, $o$ is a unit vector perpendicular to both $m$ and $n$.

Consider the homomorphism $p: S^3 \to SO(3)$, given by sending a unit quaternion to the rotation it induces on $\text{span}(i,j,k)$. This is a double cover, and it's easy to verify that $p^{-1}(D) = Q = \{\pm 1, \pm i, \pm j, \pm k\}$ is the order-8 group of unit quaternions; all elements other than $\pm 1$ have order $4$. Thus we have $$N \cong SO(3)/D \cong S^3/p^{-1}(D) = S^3/Q.$$ This diffeomorphism provides an isomorphism $\pi_1(N) = Q$, justifying the claim that $\pi_1(N)$ has six elements of order four.

Now for each map $p_i: N \to \Bbb{RP}^2$, using the relationship between the fundamental group and deck transformations of the universal covering space, we can determine the corresponding map on fundamental groups $\epsilon_i: Q \to \Bbb Z/2$ as follows. Write $P_i: S^3 \to S^2$ for the lift of $p_i$ to the universal covers. For each $q \in Q$, we have $$P_i(q) = P_i(1)(-1)^{\epsilon_i(q)}.$$

Because $i$ acts by conjugation as $\begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ and similarly for $j, k$, it is then straightforward to compute that $$\epsilon_1(\pm i) = 0, \quad \epsilon_2(\pm i) = \epsilon_3(\pm i) = 1,$$ $$\epsilon_2(\pm j) = 0, \quad \epsilon_1(\pm j) = \epsilon_3(\pm j) = 1,$$ $$\epsilon_3(\pm k) = 0, \quad \epsilon_1(\pm k) = \epsilon_2(\pm k) = 1.$$

Answer 2

I'm not too comfortable with the description of $N$ in terms of matrix tensor products, but the manifold $M/(\mathbb{Z}_2\times\mathbb{Z}_2)$ you describe in your comment is the manifold of pairs of orthogonal lines in $\mathbb{R}^3$, which can be viewed as the projectivized tangent bundle of $\mathbb{R}P^2$. This makes it clear that there is a fibre bundle $$ \mathbb{R}P^1 \to N\stackrel{p_1}{\to} \mathbb{R}P^2, $$ whose long exact sequence in homotopy groups looks like $$ \cdots \to \pi_2(\mathbb{R}P^2) \to \pi_1(\mathbb{R}P^1) \to \pi_1(N) \xrightarrow{(p_1)_\sharp}\pi_1(\mathbb{R}P^2)\to \pi_0(\mathbb{R}P^1) \to \cdots . $$ Since $\mathbb{R}P^1$ is connected the homomorphism induced by $p_1$ from $\pi_1(N)$ to $\pi_1(\mathbb{R}P^2)\cong\mathbb{Z}_2$ is surjective. If you know $\pi_1(N)\cong\mathbb{Z}_4$ by other means, this should imply that the generator is mapped non-trivially.