[EM] Exact spatial model probabilities?

[Richard Lung]

Roger Penrose did a proof of super-luminal connections, using "magic dodecahedra" popularised in his book, Shadows of the Mind. Previously, as in the Bell theorem, the proof had only been statistical. Penrose offered a deterministic proof.

I offered my own explanation of super-luminal connections, in my free e-book, Statistical Relativity Elections. Essentially, I argued it was a geometric mean of super-luminal connections and sub-luminal connections. This comes out of the special relativity situation, which allows, in theory, for tachyons, as well as tardyons. (Quantum theory does not allow for individual tachyons.)

I call myself a naive physicist (an amateur, I scarcely pretend that much).


Regards,
Richard Lung.




On 22 Jan 2022, at 5:53 am, Forest Simmons <forest.simmons21 at gmail.com> wrote:

These are fantastic questions.

The spherical geometry makes sense for Universal Domain methods because, as Saari pointed out in his Geometry of Voting, the natural distance between two permutations is the number of transpositions it takes to convert one into the other... the Kemeny distance.

There are 12 permutations of three candidates ... which arrange themselves naturally on the face of a clock.  Any "Gray Code" where successive permutations differ by only one transposition will match the twelve hours with the twelve permutations:

For example, ABC, ACB, CAB, CBA, BCA, BAC, and back to  ABC. Each permutation will be diametrically opposite to its reverse order.

For permutations of four candidates you get 24 points on the surface of a sphere.

Each permutation is adjacent to three others. For example, ABCD is next to BACD, ACBD, and ABDC. When you arrange all 24 of these permutations symmetrically on the surface of a sphere, so that the distance between any two adjacent permutations is the same, you will find that reversed permutations are polar opposites.

It's harder to visualize the arrangement of the 120 permutations of five symbols, but topologically they  reside symmetrically on the surface of a 4-dimensional ball. The surface of a four dimensional compact manifold is in general a three dimensional manifold, in this case a 3-sphere, just as  the surface of a 3-dimensional ball is an ordinary 2-dimensional sphere ... 2D manifolds have area, while 3D manifolds have volume. A circle, the boundary of a disc, is a "1-sphere" because it has length, namely its circumference.

General Relativity (without black holes or worm holes) models the Big Bang as a four dimensional space-time manifold whose temporal cross sections are 3-spheres ... precisely the kind of spheres on which the 120 distinct complete rankings of five candidates must be arranged if you require adjacent rankings to be the same distance apart.

So the space of permutations of n symbols consists of n factorial points distributed evenly on the boundary of an (n-1) dimensional ball. A Euclidean ball of radius R consists of all vectors v having L_2 norms ||v|| less than or equal to R.The set where equality holds is its boundary. If the vector space has a basis of n vectors, the ball of radius R will be n dimensional, and its boundary will be (n-1) dimensional.

These high dimensional vector spaces are ubiquitous in science, especially in quantum mechanics and other applications of partial differential equations ... the general mathematical setting being "Functional Analysis."

So Saari had the right idea topologically, but not metrically ...the same topological space can have many different metrics that yield the same topology, and the Kemeny metric is not quite right; by making all adjacent permutations the exact same distance apart, the Kemeny metric introduces clone dependence into methods based on it ... most notably the Kemeny-Young method, and Borda as related to Saari's geometry of voting.

In my de-cloned Kemeny-Young posts, I have shown how to modify the Kemeny metric to make it respect the natural clustering of clone sets, while preserving the natural Kemeny generated topology. 

Topology is all about local stretching and shrinking without ripping apart or smashing together either the natural proximities or the large scale connections.

Now we start to see the difficulty of trying to work within Universal Domain rules: UD requires ballots that are permutations of the candidates, so if we identify voters with their ballots, voters are distributed on spheres. But issue spaces are products of line segments, like rectangles, blocks, etc.

If there are two issues, then issue space is a rectangle. If there are four candidates, then ballot space is a two dimensional sphere, so according to UD rules, that only allow voter influence through their ballot rankings, there is a topological mismatch of voter/ballot space and issue space.

For example, the Borsuk-Ulam theorem shows that any continuous map from an n-sphere into a rectangle (or any other lower dimensional product of intervals) must have a pair of polar opposite points that get mapped to the same point; i.e. this requires that two polar opposite rankings correspond to the exact same point in issue space.

That's one reason I take Universal Domain rules with a grain of salt, and feel free to supplement rankings with ratings, approval cutoffs, etc.

I know that's not any help with your main question, but it may give some insight into Durand's spherical distribution suggestion.


I'll try to give a practical suggestion next time!


El vie., 21 de ene. de 2022 7:02 a. m., Kristofer Munsterhjelm <km_elmet at t-online.de> escribió:
> My minimum-manipulability optimization system can find exact
> probabilities that a voter will vote according to a particular ranking,
> under impartial culture.
> 
> It's easy to approximately find such probabilities for spatial models as
> well by just creating an instance and randomly placing voters on it to
> determine what they vote, and hence in the limit of number of trials
> going to infinity, get the probability.
> 
> But I was wondering, is it possible to do so exactly? Let's say that in
> a spatial model, first c candidates are placed uniformly at random in an
> unit d-cube. Then voters are also placed uniformly at random in this
> cube and they prefer candidates closer to them to candidates further away.
> 
> Now I would suppose that the probability that a voter votes A first is
> equal to the volume of the Voronoi cell that belongs to A. (And the
> probability that a voter will vote A>B>C is the volume of the
> intersection of the closest-neighbor Voronoi cell for A with the
> second-closest-neighbor for B and the third-closest-neighbor for C.)
> 
> Well, that eliminates one source of randomness -- assuming I can exactly
> calculate the vertices of these regions. But there's still the second
> source in the randomness of the candidates. Do you know of any calculus
> tricks to get the probabilities over every possible candidate position,
> or is this simply too hairy?
> 
> One benefit of the spatial model over impartial culture is that it's
> much more clear how one may assign utilities to each ranking, so the
> optimization system could also be used to investigate tradeoffs between
> VSE and manipulability. (Durand suggested sampling on a sphere for
> associating utilities with impartial culture, but I haven't read the
> suggestion in detail. https://arxiv.org/pdf/1511.01303.pdf)
> 
> -km
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