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The Bell-Wigner polytope has the following extreme points

$(1,0,0,0,0,0)$
$(0,1,0,0,0,0)$
$(0,0,1,0,0,0)$
$(1,1,0,1,0,0)$
$(1,0,1,0,1,0)$
$(0,1,1,0,0,1)$
$(1,1,1,1,1,1)$
$(0,0,0,0,0,0)$

I checked the largest set of independent position vectors consists of 7 points thus it's a six dimensional polytope. I wanted to find all the facet inequalities. I was thinking of the following method

For each selection of six points out of the given eight points do the following : see if they are linearly independent, if they are then obtain five independent vectors from them and form the 5 dimensional hyperplane they define and check if all the extreme eight points lie on one side of this hyperplane, if yes it defines a facet.

Am I correct ? Also can any kind of symmetry or some other thing help me simplify my method ?

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Yes that will work, if you change the word "extreme eight points" to "all 8 points" in your pseudo-code, and since you only have 8 points in 6 dimensional space, the number of facets will be very small and at any rate, you only have ${8 \choose 2} = 28$ subsets of points to check for facet inequalities. If you ever have a situation where you want the inequality description for a convex hull/polytope where the number of points minus the dimension is considerably higher than just 2, update your question and I can give you some sophisticated methods (that use computers, of course). However your case is so trivial with 8 points in 6 dimensions, brute force is the easiest way to go, especially with a computer.

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  • $\begingroup$ By extreme I mean all 8 given points ( as they are extreme points of the polytope ). $\endgroup$ – sashas Jun 26 '15 at 19:01
  • $\begingroup$ thanks, sure if I deal with a complex polytope I will surely update the question. $\endgroup$ – sashas Jun 26 '15 at 19:02
  • $\begingroup$ As a side note, I should have added one more thing. It looks like the "all 1" point and the "all 0" point are on different sides of the hyperplane defined by the other 6 points. If that is true, and these two extreme points cannot see each other except through of the convex hull of the middle 6 points, and those 6 points are not linearly dependent, then you can get the facets very easily: The facets will be precisely when you choose either the "all 1" or "all 0" point (but not both), and then you choose 5 out of the "middle" 6 points. So you would have 12 facets, all easily described. $\endgroup$ – user2566092 Jun 26 '15 at 19:12
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    $\begingroup$ @sasha Basically, a union of two simplexes along a common facet that glues them together. $\endgroup$ – user2566092 Jun 26 '15 at 19:16

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