On coverings of the complex sphere

Question

Here, everything takes place in $\mathbb{C}^d$ for some $d$, and the sphere $\mathcal{S} = \{\mathbf{x}\in\mathbb{C}^d:\|\mathbf{x}\| = 1\}$.

Given $\delta > 0$, consider a collection of vectors $\mathcal{Y}_{\delta}\subset\mathcal{S}$ such that for every $\mathbf{x}\in\mathcal{S}$ there is a $\mathbf{y}\in\mathcal{Y}_{\delta}$ such that $|\langle \mathbf{x},\mathbf{y}\rangle| > 1-\delta$. Consider now two vectors $\mathbf{v},\mathbf{w}\in\mathcal{S}$ such that $|\langle \mathbf{v},\mathbf{w}\rangle| < \epsilon$ for some small (but fixed) $\epsilon$ (i.e. the vectors $\mathbf{v}$ and $\mathbf{w}$ are "far away."). I would like to show that for every sufficiently small $\delta$ (independent of $\epsilon$) there is a $\mathbf{z}\in\mathcal{Y}_{\delta}$ such that $|\langle \mathbf{v},\mathbf{z} \rangle| > 1-\delta$ (this can be $1$ minus a constant times $\delta$ if that makes it easier) and $|\langle \mathbf{w},\mathbf{z}\rangle| \geq |\langle \mathbf{v},\mathbf{w}\rangle|.$

[Start Later Comment:] To clarify, I would like to see that the result holds for any collection $\mathcal{Y}_{\delta},\,\delta>0$ with the above property (designing your own $\mathcal{Y}_{\delta}$s is not allowed.). Although if one can design such $\mathcal{Y}_{\delta}$s with finite cardinalities so that the above result holds, that would also be interesting. [End Later Comment]

I feel I do not have the necessary tools to approach these kind of problems so any general suggestion would be great.

For some intuition, suppose everything were real and $d=2$ (i.e. we work in $\mathbb{R}^2$). Then $\mathcal{Y}_{\delta}$ could be chosen to be the vectors on the unit circle that are $O(\delta)$-apart. Given $\mathbf{v}$ and $\mathbf{w}$ that are sufficiently far away, I can find a vector $\mathbf{z}$ in $\mathcal{Y}_{\delta}$ that is $1-\delta$ close to $\mathbf{v}$, but at the same time, $\mathbf{z}$ can be chosen to be closer to $\mathbf{v}$ than it is to $\mathbf{w}$.

Edit: It seems there is some confusion regarding the quantifiers. It is fine if you can prove it for a given $\epsilon$ (you can pick $\epsilon = 0.1$, for example.). The problem is then: Given an arbitrary collection of set of vectors $\mathcal{Y}_{\delta},\,\delta > 0$ that satisy $\exists \delta_0 > 0,\,\forall \delta \in (0,\delta_0),\,\forall \mathbf{x}\in\mathcal{S},\,\exists\mathbf{y}\in\mathcal{Y}_{\delta},\,|\langle \mathbf{x},\mathbf{y}\rangle| > 1-\delta.$ Then show, (with $\epsilon = \frac{1}{10}$ for example), that $\exists \delta_0 > 0,\,\forall \delta \in (0,\delta_0),\,\forall \mathbf{v},\mathbf{w}\in\mathcal{S}$ with $|\langle \mathbf{v},\mathbf{w}\rangle| < \frac{1}{10}$ there exists $\mathbf{z}\in\mathcal{Y}_{\delta}$ with $|\langle \mathbf{v},\mathbf{z} \rangle| > 1-\delta$ and $|\langle \mathbf{w},\mathbf{z}\rangle| \geq |\langle \mathbf{v},\mathbf{w}\rangle|$.

Answer 1

We can make the obvious strategy work: push $v$ a little bit towards $w$ and then approximate this vector by a $z\in Y_{\delta}$.

Let me write $\langle v, w\rangle = s = |s|e^{i\alpha}$. Next, let's introduce $$ u = \frac{1}{L} (v+\sigma w) , \quad L = \|v+\sigma w\| ; $$ here, I'm going to choose $\sigma = O(\delta^{1/2})$ later on.

I can now pick a $z\in Y_{\delta}$ with $|\langle z, u \rangle|> 1-\delta$. Observe that this implies that $\|e^{i\beta}z-u\|< (2\delta)^{1/2}$ for suitable $\beta$; for convenience, I will assume that $\beta=0$ (replace $z$ by $e^{i\beta}z$ in the calculations below for the general case). Thus $$ \langle z, w \rangle = \langle u, w \rangle + O(\delta^{1/2}) = \frac{s+\overline{\sigma}}{L} + O(\delta^{1/2}) ; \quad\quad\quad\quad (1) $$ the implied constant in $O(\delta^{1/2})$ is $\sqrt{2}$. This means that if I now set $\sigma=A\delta^{1/2}e^{-i\alpha}$ with a sufficiently large $A>0$, then I have satisfied your second condition. Indeed, with this choice of $\sigma$, we have that $$ L^2 = 1 + A^2\delta + 2A|s|\delta^{1/2} , \quad\quad\quad\quad (2) $$ so if we take another look at (1), we see that I have made the scalar product $s$ larger by adding $\overline{\sigma}$, and the error terms coming from $L-1$ and $O(\delta^{1/2})$ can't completely destroy this achievement. More explicitly, notice that $|s+\overline{\sigma}|=|s|+A\delta^{1/2}$, and expand (2): $$ L = 1 + A|s|\delta^{1/2} + O(\delta)\quad\quad\quad\quad (3) $$ Recall that the $O(\delta^{1/2})$ term from (1) was really bounded by $(2\delta)^{1/2}$. Thus $$ \textrm{RHS of (1)} \ge (|s|+A\delta^{1/2})(1-A|s|\delta^{1/2}-O(\delta)) - (2\delta)^{1/2}\\ = |s| +A(1-|s|^2)\delta^{1/2} - (2\delta)^{1/2} - O(\delta) , $$ and this will be $\ge |s|$ for small $\delta>0$, as desired (provided we took $A>0$ large enough).

Finally, consider $$ |\langle v, z\rangle | =\left| L\langle u, z \rangle - \overline{\sigma}\langle w, z\rangle \right| \ge L (1-\delta) - A\delta^{1/2}\left( \frac{|s|+A\delta^{1/2}}{L}+O(\delta^{1/2})\right) . $$ We have one potentially dangerous term on the RHS, namely $-A|s|\delta^{1/2}/L$ (everything else we're subtracting is $O(\delta)$). However, (3) shows that everything will work out fine after multiplying out $L(1-\delta)$.

Answer 2

Intuitive Remark: for two unit vectors $ x \cdot y = |x| |y| \cos \theta = \cos \theta$, so are measuring the angle between the vectors.

• $|x \cdot y| > 1 - \delta$ is a precise way of saying $x \approx y$.

• Your $Y_\delta$ is defined so that for any $z \in S$, there is $z \approx y \in Y_\delta$.

• $|v \cdot w| < \epsilon$ mean essentially $v \perp w$ since their inner product is close to $0$.

• We would like to find $z \approx v$ with $z$ closer to $w$ then the original $v$.

These types of problems appear in coding theory, such as the Hamming bound.

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This seems rather difficult. The condition $w \cdot z = v \cdot w$ is equivalent to $ w \cdot (z - v) = 0$ which defines a plane. All points $z$ on one size of the plane is more similar to $v$ than to $w$.

Then consider the cone $ z \cdot v = 1 - \delta$. The cone, the plane and the sphere define a semi-circle (or hemisphere) and we looking for a $Y_\delta$ point in there.

For every point in that half-sphere, we can find a $Y_\delta$ point approximating it, and will be at most angle $2\delta$ from $v$ by Triangle inequality.

Answer 3

The statement you are hoping to prove (at least in the form you wrote it in the edits)
is false (already for $d=2$). I will work over the real numbers, but with a minor modification, this example works over the complex numbers as well.

As an example, consider the net $Y_\delta\subset S^1\subset R^2={\mathbb C}$ consisting of roots of unity of the order $4N$. Let $$ \delta= 1- \cos(\frac{\pi}{2N}). $$ By taking $N$ large we can get $\delta$ as small as you wish. Let $$ w=i, v=e^{\frac{i\pi}{10N}}.$$ Then for every $z\in Y_\delta\setminus \{\pm 1\}$, $$ 1- |<z, v>| >\delta. $$ On the other hand, for both $z=\pm 1$, we have $$ |<v, w>| > |<w,z>|=0, $$ which is the opposite inequality of the one you are hoping for.

Edit.

In what follows, I will use the notation $S=S^{2d-1}$ for the unit sphere in the complex vector space ${\mathbb C}^d$, equipped with the standard hermitian metric $\langle \cdot , \cdot \rangle$.

You are interested in the following function on $S\times S$: $$ D(v, w)= 1- |\langle v, w\rangle|. $$ This function is not a metric, but it is comparable to the chordal metric $d$ on the complex-projective space ${\mathbb C} P^{d-1}$, which is given by the formula: $$ d([v], [w])= 1- |\langle v,w\rangle|^2, \quad v, w\in S. $$ Here $[z]$ denotes the projection of a unit vector $z$ in ${\mathbb C}^d$ to ${\mathbb C} P^{d-1}$: $$ z=(z_1, z_2, \ldots, z_d)\mapsto [z]=[z_1: z_2: \ldots : z_d]. $$

Definition. A subset $Y\subset S$ will be called a $\delta$-pet if for every $x\in S$ there exists $y\in Y$ such that $D(x, y)\le \delta$.

Note. Recall that a subset $Y$ in a metric space $(X,d)$ is called an $\eta$-net if for every $x\in X$ there exists $y\in Y$ such that $d(x,y)\le \eta$. Define the function $\Theta(D)$ $$ \Theta(D)=\arccos(1-D)\in [0, \pi/2], \quad D\in [0, 1]. $$ Then a subset $Y\subset S$ is a $\delta$-pet if and only if the projection of $Y$ to the complex-projective space ${\mathbb C} P^{d-1}$ is a $\Theta(\delta)$-net for the angular (Fubini-Study) metric on ${\mathbb C} P^{d-1}$. The latter is given by $$ \angle([v], [w])= \Theta(D(v, w)), \quad v, w\in S. $$

Now, I can state a theorem.

Theorem. There exists $\delta_0>0$ such that for all $\delta\in (0, \delta_0)$, every $\delta$-pet $Y\subset S$ and every $v, w\in S$, there exists $z\in Y$ such that $$ D(z,w) \le D(z,v), $$ while $$ D(z,v)\le 10\delta. $$ One can, in principle, estimate the number $\delta_0$ and reduce the constant $10$ a bit, but I am too lazy for that. As I explained above, you cannot replace the inequality $$ D(z,v)\le 10\delta $$ with $$ D(z,v)\le \delta. $$

I do not know if this theorem is what you are actually asking for.If you are, I can write down a proof. It does use some elementary Riemannian geometry though.