# Largest angular distance from an orthonormal basis

Let $S^{n-1}$ be the unit sphere in $\mathbb{R}^{n}$.

Let $\{e_1,\ldots,e_n\}$ be an orthonormal basis for $\mathbb{R}^{n}$.

Let $\Sigma=\{e_1,-e_1,\ldots,e_n,-e_n\}$ be the set of $2n$ points thought of as a subset of $S^{n-1}$.

For any $p\in S^{n-1}$, let $$c_n(p) = \mathrm{dist}(p,\Sigma)$$ where dist is the intrinsic distance of $S^{n-1}$.

Let $$c_n = \sup_{p\in S^{n-1}} c_n(p).$$

Questions

(a) Is the formula for $c_n$ written down somewhere? If so, what is it? If not, can you figure one out?

(b) What is $\lim_{n\rightarrow\infty}c_n$?

• Your $\Sigma$ is also known as the orthoplex code (=generalized octahedron). Neil Sloane has a database of spherical codes, and I would search that first. I also expect SPLAG to discuss the $c_n$ sequence in the early chapters, but my copy is in my office, and I don't remember this detail. – Jyrki Lahtonen Jul 10 '13 at 18:54
• $c_n = \cos^{-1}\frac{1}{\sqrt{n}}$, it is attained by vectors of the form $\frac{1}{\sqrt{n}}(\pm e_1 \pm e_2 \pm \ldots\pm e_n)$ – achille hui Jul 10 '13 at 19:00
• Correct, @achille (I calculated it as I didn't remember). Wanna type it out as an answer or should I? – Jyrki Lahtonen Jul 10 '13 at 19:01
• @Jyrki I'll do it. – achille hui Jul 10 '13 at 19:05
• Thanks, I guess it's not surprising (b) is $\pi/2$...or is it? – Mud Jul 10 '13 at 19:29

Let $\theta(\vec{p},\vec{q})$ be the intrinsic distance/angle between any two unit vectors $\vec{p}, \vec{q} \in S^{n-1}$. Notice $1 - \cos \theta(\vec{p},\vec{q}) = 1 - \vec{p}\cdot\vec{q}$ is a stricting increasing function of $\theta$, this means for any finite set of unit vectors $\Sigma \subset S^{n-1}$, we have:

$$1 - \cos\theta(\vec{p},\Sigma) = 1 - \cos(\min_{\vec{q}\in\Sigma}\theta(\vec{p},\vec{q})) = \min_{\vec{q}\in\Sigma}(1 - \cos \theta(\vec{p},\vec{q})) = 1 - \max_{\vec{q}\in\Sigma}\vec{p}\cdot\vec{q}$$

When $\Sigma = \{ \pm \vec{e}_1, \ldots, \pm \vec{e}_n \}$, this reduces to:

$$\cos\theta(\vec{p},\Sigma) = \max_{i=1..n} |p_i|$$

where $p_i = \vec{p}\cdot\vec{e}_i$ is the coordinates of $\vec{p}$ with respect to the orthonormal base $\{ \vec{e}_1, \ldots \vec{e}_n \}$.

Finding $c_n$ is equivalent to maximizing $1 - \cos\theta(\vec{p},\Sigma)$. This is equivalent to minimizing $\cos\theta(\vec{p},\Sigma)$ which in turn equivalent to minimizing $$(\cos \theta(\vec{p},\Sigma))^2 = \max_{i=1..n}p_i^2$$

Since $\sum_{i=1}^n p_i^2 = 1$, the minimum of $\{ \max_{i=1..n}p_i^2 : \vec{p} \in S^{n-1} \}$ is clearly equal to $1/n$ and attained when $\vec{p}$ has the form $\frac{1}{\sqrt{n}}(\pm \vec{e}_1 \pm \cdots \pm \vec{e}_n )$. From this, we can conclude:

$$c_n = \sup_{\vec{p} \in S^{n-1}} \theta(\vec{p},\Sigma) = \cos^{-1}\frac{1}{\sqrt{n}} \quad\text{ and }\quad \lim_{n\to\infty} c_n = \cos^{-1} 0 = \frac{\pi}{2}.$$