Let $n$ be an even, positive integer. Then, is the metric (in the metric-space sense) on $\mathbb{C}P^n$ induced by the Fubini-Study (Riemannian) metric equivalent to the quotient (pseudo?)-metric

$$ d_Q([x],[y]) = \inf\{d(p_1,q_1)+d(p_2,q_2)+\dotsb+d(p_{n},q_{n})\} > , $$ where the $\inf$ is taken over all finite sequences $(p_1, p_2, \dots, p_n)$ and $(q_1, q_2, \dots,q_n)$ with $[p_1]=[x]$, $[q_n]=[y]$, $[q_i]=[p_{i+1}]$, for $i=1,2,\dots, n-1$

and where $d$ is the great circle distance on $S^{2n+1}$, where we identify $\mathbb{C}P^n \cong S^{2n+1}/U(1)$. If this is indeed true, does someone have a reference?

  • $\begingroup$ I am not following what $p_i$ and $q_i$ are. $\endgroup$ – Ben McKay May 20 '20 at 13:37
  • $\begingroup$ Sorry, I clarified the notation. $\endgroup$ – BLBA May 20 '20 at 13:40
  • $\begingroup$ It is the quotient metric, so I think the answer is affirmative, but I have to think about your sequences. $\endgroup$ – Ben McKay May 20 '20 at 13:42
  • $\begingroup$ Exactly, it's the quotient metric. $\endgroup$ – BLBA May 20 '20 at 13:43
  • 2
    $\begingroup$ equivalent, as in the identity map is Bi-Lipschitz. $\endgroup$ – BLBA May 20 '20 at 13:52

This is a partial answer: we show that Fubini-Study metric does not exceed the quotient metric(and some ideas for other direction).

Let $(X,d)$ and $(Y,h)$ be metric spaces and let $q:X\to Y$ be a bijection. This map generates an equivalence relation on $X$: $x\sim z\Leftrightarrow q(x)=q(z)$. Moreover, we can view $Y=X/ \sim$.

Let $d_q$ be the quotient metric in the sense of the definition given in the question and the equivalence relation above. Note that $q$ generates one more semi-metric on $Y$: $$d'_q(y,w)=\inf \{d(x,z),~q(x)=y,~ q(z)=w\}.$$ The triangle inequality does not always hold. In fact, $d_q$ is the largest pseudo-metric such that $d_q\le d'_q$.

Proposition. $q$ is $\alpha$-Lipschitz iff $h\le \alpha d_q$.

If $h\le \alpha d_q$, then for any $x,z \in X$ we have $h(q(x), q(z))\le \alpha d_q(x,z)\le \alpha d(x,z)$. If $q$ is $\alpha$-Lipschitz, then for $y,w\in Y$, any chain $y=y_0,...,y_n=w$, we have $$h(y,w)\le h(y_0,y_1)+...+h(y_{n-1},y_n)\le \alpha(d(x_1,z_1)+...+d(x_n,z_n)),$$ where $q(x_k)=y_{k-1}$ and $q(z_k)=y_k$. Since the chain was chosen arbitrarily, we conclude $h\le \alpha d_q$. $\square$

Now let $X=S^{2n+1}$ with the length distance $d$ induced by the Euclidean metric $\left<\cdot,\cdot\right>$, and let $Y=\mathbb{C}P^n$ with the distance $h$ induced by Fubini-Study metric. Let $q$ be the standard quotient map.

Note that the pull-back $\sigma$ of the Fubini-Study (Hermitean) metric with respect to $q$ is $\sigma_{w}(u,v)=\left<u,v\right>-\left<u,w\right>\left<w,v\right>$, where $w\in S^{2n+1}$ and $u,v\in \mathbb{C}^{n+1}$. Hence, $\sqrt{\sigma_{w}(u,u)}\le\|u\|$, but if we take $u\perp w$, then $\sqrt{\sigma_{w}(u,u)}=\|u\|$. Therefore, the norm of the tangent map $Tq_w:T_wX\to T_{q(w)}Y$ is $1$. Since this is true for all $w$, it follows that $q$ is $1$-Lipschitz with respect to $d$ and $h$, from where $h\le d_q$.

I don't have a complete proof of the other direction, but I think it should be possible to show that $d'_q$ is dominated by $h$ (we may need to switch to the Euclidean distance on $X$ instead of the length distance, but they are equivalent).

For that it may be useful to consider the following metric on $X=S^{2n+1}$: $$\delta(x,z)=\inf_{s,t\in\mathbb{R}}\|e^{is}x-e^{it}z\|=\inf_{s,t\in\mathbb{R}}\sqrt{2-2 Re~ e^{i(s-t)}\left<x,z\right>}=\sqrt{2-2 |\left<x,z\right>|}.$$ Then, $d'_q(q(x),q(z))=\delta(x,z)$. It is possible to show that if $\gamma$ is a smooth curve in $X$, then $$\lim_{t\to 0} \frac{\delta(\gamma(t),w)}{|t|}=\sqrt{\sigma_{w}(u,u)},$$ where $w=\gamma(0)$ and $u=\gamma'(0)$. This means that the length induced by $\sigma$ and $\delta$ coincide on $X$. The problem is that this is for the curve in $X$, not in $Y$, and I don't think any curve in $Y$ can be lifted to $X$.


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