Suppose we define a map L:R^n -> C = R^2 by sending the kth standard basis vector e_k |-> t^k, where t := exp(2k*pi/n), and extending linearly. L takes the origin O in the n-cube [0,1]^n to the origin in C, and the vertices edge-adjacent to O to the nth roots of unity. Then it's fairly straightforward to check that the vectors C := (cos(2k*pi/n)) (k = 1,...,n) and S := (sin(2k*pi/n)) (k = 1,...,n) define ker(L) via: ker(L) = (v in R^n | v*C = v*S = 0}, and so C and S form a basis for the orthogonal complement Perp(L) of ker(L). And since C*C = S*S (= n/2), L takes Perp(L) to C via a uniform scaling (of factor sqrt(n/2)), which ensures that L is an orthogonal projection R^n -> C. --Dan David W. wrote: << I had the following conjecture, probably nothing more than a corollary to projective geometers: Given an n-dimensional hypercube H, it is possible to orthogonally project H to some suitably coordinated complex plane P so that some vertex V of H maps to 0 and the vertices H-edge-adjacent to V map onto the n-th roots of unity. I leave it to the geometers to state this correctly, and if true, I imagine it is a fairly elementary theorem of projective geometry.

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Sometimes the brain has a mind of its own.