I would like to ask something i don't understand. In my textbook of manifolds, it says that the subset $M$ of $\mathbb R P^{n} \times\mathbb R P^{n} $, made from the pairs $(D,D')$ of the orthogonal directions of $\mathbb R^{n+1}-\{0\}$, is a submanifold of the manifold $\mathbb R P^{n} \times\mathbb R P^{n} $.
So, I thought of something, based on what the inner product function that Nick wrote on his answer, but I'm not sure about it. (thinking of $\mathbb R P^{n} $ as $S^{n}$ by identifying the antipodal points). Let $g:\mathbb R P^{n}\times\mathbb R P^{n}\to\mathbb R$, with $f=g\circ \pi$, where $f:S^{n}\times S^{n}\to\mathbb R$ is the inner product function, and $\pi:S^{n}\times S^{n}\to\mathbb R P^{n} \times \mathbb R P^{n} $ is the projection map. Then, $M=g^{−1}(0)$, and because $f$ is a submersion, so is $g$. So $M$ will be a submanifold of $\mathbb R P^{n} \times\mathbb R P^{n} $.
What do you think about it?? Thank you in advance!
Have you learned the theorem about "regular level sets"?
If you have coordinates $(x_0,\dots,x_n)$ and $(y_0,\dots,y_n)$ for two points in $\Bbb{R}^{n+1}$, then the subset $M$ is defined by the equation
$$ x_0 y_0 + x_1y_1 + \cdots + x_n y_n = 0 $$
Let $f$ be the function $f = \sum_i x_i y_i$. Then $M$ is the level set $f=0$. You can check that $0$ is a "regular value" of $f$ (by writing down the gradient of $f$ and seeing that it's non-zero on this set), and the theorem says that the level set is an embedded submanifold. The way I've worded it is for $\Bbb{R}^{n+1} \times \Bbb{R}^{n+1}$, but you can then project to $\Bbb{P}^n \times \Bbb{P}^n$.