9 ms·

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 prec

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.

dylwil3 2y ago
Yep, see Eckmann for a generalization and precise characterization: https://core.ac.uk/download/pdf/82385648.pdf https://core.ac.uk/download/pdf/82385648.pdf
- contravariant 2y ago
  Awesome, always nice to see my mathematical intuition still works. Also an interesting piece of mathematics. My main takeaway was the following conclusion > [E]xcept for the contractible case either no social choice function can exist on P, or if it exists for all n then unexpected properties turn up.
- ykonstant 2y ago
  The notion of "space with mean" from that paper seems to be of independent interest; nice.
vcdimension 2y ago
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.
- yjk 2y ago
  In addition, I would argue undecidability is a feature, not a bug. I'm not sure why any other answer would be desired in that case.
  - contravariant 2y ago
    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.
rtolsma 2y ago
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
- contravariant 2y ago
  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.

dylwil3 2y ago

Yep, see Eckmann for a generalization and precise characterization: https://core.ac.uk/download/pdf/82385648.pdf https://core.ac.uk/download/pdf/82385648.pdf

contravariant 2y ago
Awesome, always nice to see my mathematical intuition still works. Also an interesting piece of mathematics. My main takeaway was the following conclusion > [E]xcept for the contractible case either no social choice function can exist on P, or if it exists for all n then unexpected properties turn up.
ykonstant 2y ago
The notion of "space with mean" from that paper seems to be of independent interest; nice.

vcdimension 2y ago

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.

yjk 2y ago
In addition, I would argue undecidability is a feature, not a bug. I'm not sure why any other answer would be desired in that case.
- contravariant 2y ago
  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.

rtolsma 2y ago

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

contravariant 2y ago
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.