# 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?

## 1 Answer

The question is not really well posed as it stands. First, the open hemisphere doesn't contain "exactly one representative of each equivalence class," because it doesn't contain any representatives of the equivalence classes of points on the equator. The best we can say is that it contains at most one representative of each equivalence class.

Second, the hemisphere isn't "the largest" such open set; it's merely a maximal one, meaning that it's not properly contained in any larger open set that contains at most representative of each class. There are plenty of other maximal open sets that are not hemispheres, such as one whose boundary is a deformation of the equator, as pictured in the drawing below.

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 $W$.