Is $\mathbb{R}P^6 = \mathbb{R}P^3 \times \mathbb{R}P^3$? Question is in the title. I know that $S^n$ is not $S^1 \times \cdots \times S^1$, but it is true that $\mathbb{R}^n = \mathbb{R}^1 \times \cdots \times \mathbb{R}^1$.
 A: No. The universal covers of the two spaces are, respectively, $\Bbb S^6$ and $\Bbb S^3 \times \Bbb S^3$, but these spaces are not homeomorphic, as, e.g., their $3$rd homology groups differ: $H_3(\Bbb S^3 \times \Bbb S^3) = \Bbb Z \oplus \Bbb Z$ but $H_3(\Bbb S^6) = 0$.
Another way to see that the proposed "product rule" for $\Bbb R \Bbb P^k$ fails is to observe that $\Bbb{RP}^1 \times \Bbb{RP}^1 \cong \Bbb S^1 \times \Bbb S^1$ is orientable but $\Bbb{RP}^2$ is not.
A: No. We have $$\pi_3(\Bbb RP^3\times \Bbb RP^3) = \pi_3(\Bbb RP^3)^2 = \pi_3(S^3)^2= \Bbb Z^2$$ but $\pi_3(\Bbb RP^6) = 0$, so they cannot be homeomorphic.
A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$If $k$ is a positive integer, then $\Reals\Proj^{2k-1}$ is orientable and $\Reals\Proj^{2k}$ is non-orientable.
Consequently, $\Reals\Proj^{6}$ is not orientable, but $\Reals\Proj^{3} \times \Reals\Proj^{3}$ is a product of orientable manifolds and hence orientable.
To address the question more generally: It's true that $\Reals^{n}$ is a Cartesian product $\Reals \times \dots \times \Reals$, but this decomposition in no way corresponds to sets of lines through the origin. In the present situation, for example, $\Reals\Proj^{6}$ is the set of lines through the origin in $\Reals^{7}$, while $\Reals\Proj^{3} \times \Reals\Proj^{3}$ is the set of ordered pairs of lines through the origin in $\Reals^{4}$, i.e., a certain set of pairs of lines in $\Reals^{8} \simeq \Reals^{4} \times \Reals^{4}$. There's no reason to expect these spaces to bear any resemblance aside from having the same dimension.
