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

Question

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.

Answer 1

Hmm, I think I've worked my way to exactly such a 'slick' proof:

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.