Choosing vectors on a complex sphere Consider a complex $t$ dimensional unit sphere. 
Say we pick $n$ points on this with inner products in the set $\{a_1,a_2,\dots,a_r\}$ (we have $n$ inner products with value $1$).  Note the set elements do not have to be distinct.
Is it possible to pick a second set of $n$ points with inner products in the set $\{-a_1,-a_2,\dots,-a_r\}$ (we have $n$ inner products with value $1$)? That is if $v_iv_j = a_k$, I would like to have $w_iw_j=-a_k$.
How do you find such pairs of points?
 A: I'm interpreting this question as "Given $\{v_i\}$ a set of $n$ vectors in $\mathbb{C}^t$, does there exist $\{w_i\}$, a set of $n$ vectors such that the inner products are negative ..."
I will assume that $ t \geq 2 $, which is the more interesting case. $t=1$ is easily dealt with.
In general, no. 
For $n = 2$, if the points are $v_1, v_2$, then we can take $w_1 = v_1, w_2 = -v_2$, which satisfies the conditions.
For $n > 2$, if we pick the points $ v_i = (1,0,\ldots, 0)$, then we want $n$ points such that $w_i \overline{w_j} = -1$.
This implies that $0 \leq ( \sum w_i) (\sum \overline{w_i}) = n - n(n-1) =n(2-n)< 0 $
If you want $v_i$ to be distinct, then take a small perturbation of the above.
This gives that $|\sum w_i \overline{w_j} | \leq n$ is a necessary condition. It is not clear if this is a sufficient condition.

If the question is asking to find a set of $\{v_i\} $, then for $n \leq t$, simply take $v_i = e_i$, which is the unit vector in direction $i$. Then $v_i \overline {v_j} = \delta_{i,j}$, and hence we may set $w_i = v_i$ which satisfies $ w_i w_j = - 0 $ if $ i \neq j$.
I do not know if there are solutions for $n > t$.
