# Lower bound on a norm of $\mathbb{CP}^2$ inducing a lower bound on the Euclidean norm of $\mathbb{C}^3$

Let $$|\cdot|$$ denote the usual Euclidean norm on $$\mathbb{C}^3$$ and fix some arbitrary metric $$\rho$$ on $$\mathbb{CP}^2$$. For $$\delta > 0$$ and any set $$\hat{P} \subset \mathbb{CP}^2$$, define the $$\delta$$-neighborhood of $$\hat{P}$$ by

$$N_{\delta}(\hat{P})= \{\hat{x} \in \mathbb{CP}^2: \rho(\hat{s},\hat{x})<\delta \text{ for some } \hat{s} \in \hat{P}\}.$$

Here, the hat notation refers to the obvious projection that sends nonzero vectors in $$\mathbb{C}^3$$ to their equivalence class in complex projective space via $$v = \lambda w$$ for some $$\lambda \in \mathbb{C}-\{0\}$$.

Now let's fix some nonzero $$y\in \mathbb{C}^3$$ with unit modulus and fix a complex 2-plane $$\Sigma \subset \mathbb{C}^3$$ going through the origin. We can write $$y = v +w$$ with $$v\in \Sigma$$ and $$w\in \Sigma^{\perp}$$. I'm reading a paper that uses the following fact without proof:

If $$\hat{y} \notin N_{\delta}(\hat{\Sigma})$$, then there is a constant $$K>0$$, depending only on the metric $$\rho$$, so that $$|w| \geq K \delta$$.

Essentially the fact is saying that if a vector lies far away from a plane at the projectivized level, then the component orthogonal to the plane can't be too small (up to a multiplicative constant that depends on the details of how you measure a vector being 'far away' from a plane).

I'd love to see some sketch of a proof for this.

I can sort of buy it for particular choices of metric like Fubini-Study, but the statement is made for arbitrary metric, so I imagine there is a slick proof using only general facts about how/whether metrics on $$\mathbb{CP}^2$$ "distort" if lifted to $$\mathbb{C}^3-\{0\}$$. I suppose there may be an equivalent formulation of this statement for $$\hat{y} \in N_{\delta}(\hat{\Sigma})$$ (with the inequality switched and strict), but let me leave it as written in the paper.

Suppose that there is no such uniform constant. Then for each $$L > 0$$, we may find some $$\hat{y} \notin N_{\delta}(\hat{P})$$ with $$|w| < L \delta$$, where $$w$$ denotes the perpendicular component of a representative $$y\in \hat{y}$$ having unit modulus. Indeed, for each positive integer $$n$$ there exists $$\hat{y}_n \notin N_{\delta}(\hat{P})$$ with $$|w_n| < \delta/n$$. This gives us a sequence $$\{y_n\}$$ on the unit sphere. Choose a convergent subsequence, again denoted $$\{y_n\}$$ with $$y_n \to y_*$$. Clearly the perpendicular component of $$y_*$$ must be zero. Hence $$y_* \in P$$, and thus $$\hat{y}_* \in N_{\delta}(P)$$.
On the other hand, $$\hat{y}_n \to \hat{y}_* \notin N_{\delta}(P)$$ since the quotient map is continuous and the complement of $$N_{\delta}(P)$$ is closed, hence containing all of its limit points. We have obtained our contradiction.