# Quotient distance on $\mathbb{C}P^n$ is equivalent to distance induced by the Fubini-Study metric

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?

• I am not following what $p_i$ and $q_i$ are. – Ben McKay May 20 '20 at 13:37
• Sorry, I clarified the notation. – BLBA May 20 '20 at 13:40
• It is the quotient metric, so I think the answer is affirmative, but I have to think about your sequences. – Ben McKay May 20 '20 at 13:42
• Exactly, it's the quotient metric. – BLBA May 20 '20 at 13:43
• equivalent, as in the identity map is Bi-Lipschitz. – 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 $$\sigma_{w}(u,v)=\left-\left\left$$, 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 $$X=S^{2n+1}$$: $$\delta(x,z)=\inf_{s,t\in\mathbb{R}}\|e^{is}x-e^{it}z\|=\inf_{s,t\in\mathbb{R}}\sqrt{2-2 Re~ e^{i(s-t)}\left}=\sqrt{2-2 |\left|}.$$ 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 $$X$$.