Consider a complex $t$ dimensional unit sphere.
Can we have $t$ sets of $2^t$ vectors $v_{ij}\in \Bbb C^t$ on the sphere where $i=1$ to $t$ and $j=1$ to $2^t$ on this with inner products satisfying $\sum_{i=1}^tv_{ij}\overline{v_{ij'}}=0$ where $j\neq j'$ ($\overline{v_{ij}}$ is complex conjugate transpose of $v_{ij}$)? How do you find such pairs of points?
The sum of $t$ inner products of vectors have to be $0$ and not the inner products themselves.
We want $\{V_j\}_{j=1}^t$ be $2^t \times t$ complex matrices with rows of $V_j$ being the $2^t$ vectors $v_{ij}$ such that:$$V_1\overline{V_1}+V_2\overline{V_2}+\dots+V_t\overline{V_t}=tI$$ where $I$ is the identity matrix.
In the above $\overline{V_j}$ refers to complex conjugate transpose of matrix $V_j$.
Let's just work on the real sphere.
Hint: If $(\sum v_{ij}) \cdot (1,1,\ldots 1) = \sqrt{nm}$, then $ \sum v_{ij} \times \sum v_{ij} = mn$ and thus $ \sum v_{ij} \times v_{ij'} = mn - mn = 0$.
It remains to show that some set of $mn$ vectors satisfy this condition. This follows by a continuity argument, and looking at the plane intersecting the unit sphere.
Note that the above is not a complete classification. For example, we could have used that the sum of absolute values of the terms is $\sqrt{mn}$, but this doesn't allow for the easy geometric interpretation of a sphere.