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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?

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    $\begingroup$ largest in what sense? $\endgroup$ – Will Sawin May 8 '17 at 0:00
  • $\begingroup$ Say, in diameter. $\endgroup$ – Eduardo Longa May 8 '17 at 3:49
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    $\begingroup$ I think we can make the diameter (measured internally) arbitrarily large by making the surface crinkly. Measured externally, the maximal diameter would be between two antipodal points. Those both lie in the same fiber, but it's easy to build sets getting arbitrarily close to that. $\endgroup$ – Will Sawin May 8 '17 at 5:03

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