Let $CP(n)$ be the complex projective space with Fubini-Study metric with diameter $=\frac{\pi}{2}$. Fix a point say $p\in CP(n)$; my question is what is the set of points of maximum distance to the point $p$? (The hint given in the class is $CP(n-1)$.) I can't figure out why. Could anyone help me please? Or give any reference with a detailed study of this kind of properties?
antipodal map of complex projective space
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0@JasonDeVito, the $CP(n)$ should be written as $\mathbb{CP}^n$ the complex $n$-dimensional projective space. The Fubini-Study metric is the metric you get from Riemannian submersion form the round sphere of dimension $2n+1$ by Hopf fibration. – 2012-02-15
1 Answers
Here's a hint via a few lemmas.
The (identity component) of the isometry group of $S^{2n+1}$ is $SO(2n+2)$. The subgroup of isometries which preserve the Hopf fibration is $U(n+1)\subseteq SO(2n+2)$ (standard embedding). The subgroup $SU(n+1)$ inside of this, thus preserves the Hopf fibration, so descends to an isometry of $\mathbb{C}P^n$. $SU(n+1)$ acts transitively on $S^{2n+1}$, and hence transitively on $\mathbb{C}P^n$. (This is an outline of the proof of what Jim said in his comment). So, as Jim mentioned, we might as well focus on the point $p=[1:0:...:0]$.
Given any geodesic $\gamma$ in $\mathbb{C}P^n$ starting at $p$, and given any preimage $\tilde{p}$ of $p$ in $S^{2n+1}$, there is a unique geodesic $\tilde{\gamma}$ starting at $\tilde{p}$ which projects down to $\gamma$ and is everywhere orthogonal to the Hopf circles. In my head, I'm thinking of $\tilde{p}=(1,0...,0)\in S^{2n+1}$ where I'm thinking of $S^{2n+1} = \{(z_0,...,z_n)\in\mathbb{C}^{n+1}\mid \sum |z_i|^2=1\}$.
A geodesic $\gamma$ stops minimizing distance when $\tilde{\gamma}$ is half way around the sphere and the set of all such stopping points on $S^{2n+1}$ is $\{(0,z_1,...,z_n)\in S^{2n+1}\}$, diffeomorphic to $S^{2n-1}$.
The Hopf actions preserves this $S^{2n-1}$ and becomes $\mathbb{C}P^n$ in the quotient.
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0@Eduardo: As you phrased it, no, because $\tilde{\gamma}$ may not be unique. The issue is that $\tilde{\gamma}'(0)$ must project to $\gamma'(0)$, but there are many such choices. However, if you insist that $\tilde{\gamma}'(0)$ be a normal vector to $f^{-1}(p)$, then yes. See, for example, this MO question: https://mathoverflow.net/questions/194825/geodesics-and-riemannian-submersions – 2017-10-13