Jochen,
Your chart which I copy below, though it looks clumsy, is a sort of polytope. This is a tesseract in 4-dimensions. The z_A and z_B states form a bipartite entanglement. The same occurs for the states x_A and x_B, where the states z and x are in superposition quantum mechanically. We may then think of some additional spin state that for x_A and x_B these are replaced with the states Z_A and Z_B, say as some needle state. This means the bipartite system is replaced with a qu4-nit or quadpartite entanglement. This 4-entangled system then has the states (±1, ±1, ±1, ±1). A form of quantum teleportation may accomplish this.
With the identification of states that have Hamming distance 1 do not contribute to superposition and for this and with the separable states (-1, -1, -1, -1) and (1, 1, 1, 1) we have the Kirwan polytope. The Kirwan convexity theorem proves that the momentum map has image that is a convex set. This convex set or Kirwan polytope intersects the positive Weyl chamber. I have some more on this in my essay. For the 4-tangle case the states (1011), (1101), (1110) and (0111), where the remainder define with (0000) and (1111) the Kirwan polytope. This has 12 vertices and 12 faces. This has a relationship to the 24-cell of the F4 group.
Of course I have used the idea of quantum teleportation, but we might however drop the idea of this being a quantum teleportation and simply a classical replacement. The result will have a 4-tangle obstruction similar to the 3-tangle obstruction. In that case the cube defines the set of tripartite states and the Hamming 1-distance states removed (110), (101) and (0,1,1) define a double tetrahedron. The 12-cell above is constructed from 4 of these.
Jochen,
Your chart which I copy below, though it looks clumsy, is a sort of polytope. This is a tesseract in 4-dimensions. The z_A and z_B states form a bipartite entanglement. The same occurs for the states x_A and x_B, where the states z and x are in superposition quantum mechanically. We may then think of some additional spin state that for x_A and x_B these are replaced with the states Z_A and Z_B, say as some needle state. This means the bipartite system is replaced with a qu4-it or quadpartite entanglement. This 4-entangled system then has the states (±1, ±1, ±1, ±1). A form of quantum teleportation may accomplish this.
With the identification of states that have Hamming distance 1 do not contribute to superposition and for this and with the separable states (-1, -1, -1, -1) and (1, 1, 1, 1) we have the Kirwan polytope. The Kirwan convexity theorem proves that the momentum map has image that is a convex set. This convex set or Kirwan polytope intersects the positive Weyl chamber. I have some more on this in my essay. For the 4-tangle case the states (1011), (1101), (1110) and (0111), where the remainder define with (0000) and (1111) the Kirwan polytope. This has 12 vertices and 12 faces. This has a relationship to the 24-cell of the F4 group.
Of course, I have used the idea of quantum teleportation, but we might however drop the idea of this being a quantum teleportation and simply a classical replacement. The result will have a 4-tangle obstruction similar to the 3-tangle obstruction. In that case the cube defines the set of tripartite states and the Hamming 1-distance states removed (110), (101) and (0,1,1) define a double tetrahedron. The 12-cell above is constructed from 4 of these.
That this has some relationship to the 24-cell and the F4 group means this argument is similar to the Kochen-Specker theorem for 4-dimensions. So this insight should work.
Cheers LC
State xA zA xB zB P(λi)
λ1 1 1 1 1 p1
λ2 1 1 1 -1 p2
λ3 1 1 -1 1 p3
λ4 1 1 -1 -1 p4
λ5 1 -1 1 1 p5
λ6 1 -1 1 -1 p6
λ7 1 -1 -1 1 p7
λ8 1 -1 -1 -1 p8
λ9 -1 1 1 1 p9
λ10 -1 1 1 -1 p10
λ11 -1 1 -1 1 p11
λ12 -1 1 -1 -1 p12
λ13 -1 -1 1 1 p13
λ14 -1 -1 1 -1 p14
λ15 -1 -1 -1 1 p15
λ16 -1 -1 -1 -1 p16