# Largest elementary neighbourhood

The real projective space $\mathbb{R}P^n$ can be defined as the quotient space of $\mathbb{S}^n$by the equivalence relation that identifies antipodal points. The largest open set of $\mathbb{S}^n$ that contains exactly one representative of each equivalence class is easily seen to be an open hemisphere.

Consider now the complex projective $\mathbb{C}P^n$ as the quotient space of $\mathbb{S}^{2n+1} \subset \mathbb{C}^{n+1}$ by the $\mathbb{S}^1$ action given by

$\lambda \cdot (z_1, \dots, z_{n+1}) = (\lambda z_1, \dots, \lambda z_{n+1}), \qquad \lambda \in \mathbb{S}^1, \, (z_1, \dots, z_{n+1}) \in \mathbb{S}^{2n+1}.$

In analogy with the real case, what is now the largest hypersurface of $\mathbb{S}^{2n+1}$ that contains exactly one representative of each equivalence class?

That said, it does make sense to ask for a maximal hypersurface in $\mathbb S^{2n+1}$ that contains at most one representative of each class. One such hypersurface is the set $W$ consisting of all $z = (z^1,\dots,z^{n+1})\in \mathbb S^{2n+1}$ such that $z^{n+1}$ is real and positive. Its image in $\mathbb C\mathbb P^n$ is the affine subspace consisting of all points $[z^1:\dots:z^{n+1}]$ such that $z^{n+1}\ne 0$.
To see that it's maximal, the basic idea is to check that any hypersurface $\widetilde W$ that strictly contains $W$ must contain at least one point where $\operatorname{Re} z^{n+1}=0$, and a neighborhood of that point in $\widetilde W$ will also contain points where $\operatorname{Re} z^{n+1}<0$; but the equivalence class of any such point is already represented by a point in $W$.