TITLE: The minimum of a sum of absolute values of inner products in $\mathbb{R}^d$
QUESTION [14 upvotes]: 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 $v_i=e_{i\, \text{mod}\, d}$?

REPLY [4 votes]: In fact, the minimizers of the sum $\sum \langle v_i, v_j \rangle^2$ are precisely tight frames, i.e. sets such that for some $C>0$and for each $x\in \mathbb R^d$, one has $\| x \|^2 = C \sum \langle x, v_j \rangle^2$. This was proved in Benedetto, Fickus "Finite Normalized Tight Frames", Adv. Comp. Math., 18 (2003), pp. 357-385.
Thus for $n$ not a multiple of $d$, the set $v_i = e_{i\mod d}$ is not a minimizer, since it is not a tight frame. 
Although, I believe that for $\ell^1$ the answer may still be correct.