In a $d$ dimensional vector space defined over $\mathbb{C}$, how do I calculate the largest number $N(\epsilon, d)$ of vectors $\{V_i\}$ which satisfies the following properties. Here $\epsilon$ is small but finite compared to 1.
$$\langle V_i, V_i\rangle = 1$$
$$|\langle V_i, V_j\rangle| \leq \epsilon, i \neq j$$
Some examples are as follows.
$N(0, d)$ = d
$N\left(\frac{1}{2}, 2\right)$ = 3
$N\left(\frac{1}{\sqrt{2}}, 2\right) = 6$
How do I obtain any general formula for $N(\epsilon, d)$. Even an approximate form for $N(\epsilon, d)$ in the large $d$ and small $\epsilon$ limit works fine for me.
EDIT: The problem is now resolved. See the cross-posted answer to the same question at Math Overflow.
Consider the $n+1$ elementary basis vectors in $\mathbb{R}^{n+1}$. The endpoints form the regular simplex, ie the tetrahedron equivalent in $n$ dimensions and all lie in a $n$-dim hyperplane. The center of this simplex is the point $(\frac{1}{n+1}, .., \frac{1}{n+1})$. Hence the vectors from the center to the corners of this $n$-dim tetrahedron have pairwise equal angles to each other. I believe this construction gives the arrangement of $d+1$ vectors in $\mathbb{R}^d$ with angles closed to $\frac{\pi}{2}$. Hence this construction should provide the threshhold for $\varepsilon$ where $N(\varepsilon, d)=d+1$.
To compute $\varepsilon$, pick two of the vectors, say $v_1=(1-\frac{1}{n+1}, -\frac{1}{n+1}, .., -\frac{1}{n+1})$ and $v_2=(-\frac{1}{n+1}, .., -\frac{1}{n+1}, 1-\frac{1}{n+1})$. I get $\varepsilon=\frac{3d-1}{d(d+1)}$ so for $\varepsilon < \frac{3d-1}{d(d+1)}$ we have $N(\varepsilon, d)=d$ and $N(\frac{3d-1}{d(d+1)}, d)=d+1$.