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:



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



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


