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Bell polytopes with nontrivial symmetries

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Take $N$ parties, each of which receives an input $s_i \in {1, \dots, m_i}$ and produces an output $r_i \in {1, \dots, v_i}$, possibly in a nondeterministic manner. We are interested in joint conditional probabilities of the form $p(r_1r_2\dots r_N|s_1s_2\dots s_N)$. Bell polytope is the polytope spanned by the probability distributions of the form $p(r_1r_2\dots r_N|s_1s_2\dots s_N) = \delta_{r_1, r_{1, s_1}}\dots\delta_{r_N, r_{N, s_N}}$ for all possible choices of numbers $r_{i,s_i}$ (in other words, each input $s_i$ produces a result $r_{i,s_i}$ either with probability 0 or 1, regardless of other players' inputs).

Every Bell polytope has a certain amount of trivial symmetries, like permutation of parties or relabelling of inputs or outputs. Is it possible to give an explicit Bell polytope with nontrivial symmetries? (e.g. transformations of the polytope into itself that takes faces to faces and is not trivial in the above sense) In other words, I'm interested whether a specific Bell scenario can possess any "hidden" symmetries

Bell polytopes in literature are usually characterized by their faces, given by sets of inequalities (Bell inequalities), which, however, usually do not have any manifest symmetry group.

This post has been migrated from (A51.SE)
asked Sep 19, 2011 in Theoretical Physics by Marcin Kotowski (405 points) [ no revision ]
At first glance, it seems any further symmetry would necessarily take you out of the regime of product states, and that such a state would necessarily produce correlations outside of the polytope. That said, this is just speculation, but perhaps it is one way to prove that there aren't any.

This post has been migrated from (A51.SE)

2 Answers

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Matty Hoban pointed me to a paper (PDF here) by Itamar Pitowsky from 1991 which looks the geometry of correlation polytopes and their symmetries. I haven't read the paper in full, but glancing through it, on page 400 (page 6 of the actual paper) under the statement of results the author seems to say that the cardinality of the symmetry group is $n! 2^n$ which would be consistent with just the bit flips and permutations, and with the existence of only the trivial symmetries you mention.

This post has been migrated from (A51.SE)
answered Sep 23, 2011 by Joe Fitzsimons (3,555 points) [ no revision ]
Thanks for the reference. Though, the author doesn't actually seem to prove that there are no symmetries but the $n!2^n$ trivial ones - he only identifies these symmetries and claims that they generate the full group (maybe it follows from some facts about polytopes, I don't know).

This post has been migrated from (A51.SE)
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Any symmetry of the local hidden variable polytope must map a vertex of the polytope to another other vertex (or trivially to itself). This is true in general by convexity. By the duality between vertex representation and facet representation we only need consider vertices. I have modified the way you write vertices to obtain $p(r_{1} r_{2} ... r_{N}|s_{1} s_{2} ... s_{N})=\delta^{r_{1}}_{f_{1}(s_{1})}\delta^{r_{2}}_{f_{1}(s_{2})}...\delta^{r_{N}}_{f_{N}(s_{N})}$ where $f_{j}(s_{j})$ is the image of $s_{j}$ under a single-site function $f_{j}:\mathbb{Z}_{m_{j}}\rightarrow\mathbb{Z}_{v_{j}}$.

Therefore, a symmetry will map from the product of single-site maps $\delta^{r_{1}}_{f_{1}(s_{1})}\delta^{r_{2}}_{f_{1}(s_{2})}...\delta^{r_{N}}_{f_{N}(s_{N})}$ to other products of single-site maps $\delta^{r_{1}}_{f'_{1}(s_{1})}\delta^{r_{2}}_{f'_{1}(s_{2})}...\delta^{r_{N}}_{f'_{N}(s_{N})}$ with $f_{j}$ not necessarily equal to $f'_{j}$. Of course, one can reorder the products by permuting the parties and still produce a product of delta functions. Locality prevents us from allowing delta functions of the form $\delta^{r_{j}}_{f_{j'}(s_{j'})}$ with $j\neq j'$. Therefore, other than permutations the only symmetry transformations that are allowed will be transformations on the maps $f_{j}(s_{j})\rightarrow f'_{j}(s_{j})$.

We only need to consider each site's marginal probability distribution $p(r_{j}|s_{j})$ which can be written as a $m_{j}v_{j}$ length real vector. The vertices have $m_{j}$ non-zero elements which have unity value for each value of $s_{j}$. In order to conserve these two conditions of the vertex probability distributions, the only allowed transformations on the real vectors that are allowed are a restricted class of permutations of row elements. The restricted class of permutations of row elements is naturally generated by relabelling a measurement outcome for each value of $s_{j}$ and relabelling values of $s_{j}$.

This applies for the full probability distribution polytope. However, for other forms of correlations such as joint outcome statistics, e.g. $p(\sum_{j}^{n}r_{j}|s_{1} s_{2} ... s_{N})$ there are other subtle forms of symmetry outside of the 'trivial' classes. If you want me to elaborate, I can.

This is my first post to the TP.SE. I'm sorry if it is not detailed enough.

This post has been migrated from (A51.SE)
answered Oct 3, 2011 by Matty Hoban (435 points) [ no revision ]
No worries, Joe, I'm glad you put it up anyway. Sorry, I have been slow to reply to this question. Have been a bit busy.

This post has been migrated from (A51.SE)
Apparently I need to wait 24 hours to award the bounty, but I'll do so then.

This post has been migrated from (A51.SE)
Welcome to TP.SE. It's good to see you here. Sorry to have stolen your reference! You deserve the rep from my answer too, so I'll use a bounty to transfer the rep.

This post has been migrated from (A51.SE)

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