I'm not quite sure why one would use a sphere, unless you were specifically trying to get a version of Arrow's theorem.
If anything it looks like it fails precisely because the space is not homologically trivial, but I'm a bit unsure how to make that precise. A similar set up with just [0,1]^n as preference space works perfectly fine just by averaging all the scores for each candidate.
I kind of sense that requiring a function X^k -> X to exist is somehow hard if X is not 'simple', but I'm not yet sure what the obstruction is.
I thought about averaging the scores, which gives you a point inside the circle, and then projecting onto the circle with a ray from the centre, which is continuous everywhere apart from where the average is at the centre (e.g. for two voters this is when they have exactly opposite views).
So if you have a continuous probability distribution on the domain the probability of undecidability has measure zero.
It's not the undecidability that is a problem, it's the discontinuity. Undecidable answers are manageable, random answers however are very annoying to deal with.
Yea I think one reason to restrict to spheres is because the voting function takes as input the relative preferences (like in [0,1]^n how does all 0s differ from all 1s), which implies the vectors should be normalized
As it turns out choosing a simplex instead doesn't change things much from the hypercube. I think the arithmetic mean also still works. In stark contrast to the sphere.
If anything it looks like it fails precisely because the space is not homologically trivial, but I'm a bit unsure how to make that precise. A similar set up with just [0,1]^n as preference space works perfectly fine just by averaging all the scores for each candidate.
I kind of sense that requiring a function X^k -> X to exist is somehow hard if X is not 'simple', but I'm not yet sure what the obstruction is.