Consider a collection of unit vectors $v\_1, \ldots, v\_n$ in $\mathbb{R}^d$ (we think of $n$ being much larger than $d$). I would like to minimize the sum:


$$\sum\_{i\neq j}|\langle v\_i,v\_j\rangle|.$$


Clearly, if $n=d$, the minimum is attained by taking $v\_i=e\_i$. Could it be that for $n>d$ in order to minimize the latter expression it is still best to take the vectors $v\_i=e\_{i\, \text{mod}\, d}$? 



The minimum of a sum of absolute values of inner products in $\mathbb{R}^d$
It is quite likely. At least, the proof for the case $d\mid n$ is easy. First of all, the restriction $i\ne j$ does not matter: adding $n$ ones changes nothing in the problem. Now notice that $|\langle v\_i,v\_j\rangle|\ge \langle v\_i,v\_j\rangle^2=\langle V\_i,V\_j\rangle$ where $V\_i=v\_i\otimes v\_i$. Now, $\langle V\_i,I\rangle=1$ for all $i$ ($I$ is the identity matrix, as usual), so $\langle\sum\_i V\_i,I\rangle=n$ and, by Cauchy-Schwarz, $\|\sum\_i V\_i\|^2\ge n^2/\|I\|^2=n^2/d$ (the norm here is the Frobenius norm, i.e., the square root of the sum of the squares of the matrix elements), which results in $\min\_{v\_i}\sum\_{i,j}\langle v\_i,v\_j\rangle^2\ge n^2/d$. For the conjectured minimizer, both this estimate and the crude inequalities $|\langle v\_i,v\_j\rangle|\ge \langle v\_i,v\_j\rangle^2$ become identities, whence the conclusion. 


I do not see off hand how to modify this argument for the case $d\not\mid n$ but it still makes the conjecture quite plausible. In the worst case scenario, you are off by at most $d/4$ from the true minimum with your system. 


