There is an interesting paper by Peres which connects the Kochen-Specker theorem with the F_4 group, or the 24 cell. The 117 projectors with the original KS theorem in 3-dim Hilbert space is simplified by considering a four dimensional Hilbert space, or a system of 4 qubits. This involves only 18 projector operators. The space 24-cells is a system of root vectors for the F_4 group. Each root vector is paired with its negative to define a line through the origin in 4d space. These 24 lines are the 24 rays of Peres. The root vectors are
1 (2,0,0,0) 2 (0,2,0,0) 3 (0,0,2,0) 4 (0,0,0,2)
5 (1,1,1,1) 6 (1,1,-1,-1) 7 (1,-1,1,-1) 8 (1,-1,-1,1)
9 (-1,1,1,1) 10 (1,-1,1,1) 11 (1,1,-1,1) 12 (1,1,1,-1)
13 (1,1,0,0) 14 (1,-1,0,0) 15 (0,0,1,1) 16 (0,0,1,-1)
17 (0,1,0,1) 18 (0,1,0,-1) 19 (1,0,1,0) 20 (1,0,-1,0)
21 (1,0,0,-1) 22 (1,0,0,1) 23 (0,1,-1,0) 24 (0,1,1,0)
(I hope this table works out here) Consider these as 24 quantum states |ψ_i>, properly normalized, in a 4 dimensionl Hilbert Space e.g. it might be a system of two qubits. For each state we can define a projection operator
P_i = |ψ_i)(ψ_i|
P_i are are Hermitian operators with three eigenvlaues of 0 and one of 1. They can be considered as observables and we could set up an experimental system where we prepare states and measure these observables to check that they comply with the rules of quantum mechanics. There are sets of 4 operators which commute because the 4 rays they are based on are mutually orthogonal. An example would be the four operators P_1, P_2, P_3, P4.
Quantum mechanics tells us if we measure these commuting observables in any order we will end up with a state which is a common eigenvector i.e. one of the first four rays. The values of the observables will always be given by 1,0,0,0 in some order. This can be checked experimentally. There exist 36 sets of 4 different rays that are mutually orthogonal, but we just need 9 of them as follows:
{P2, P4, P19, P20}
{P10, P11, P21, P24}
{P7, P8, P13, P15}
{P2, P3, P21, P22}
{P6, P8, P17, P19}
{P11, P12, P14, P15}
{P6, P7, P22, P24}
{P3, P4, P13, P14}
{P10, P12, P17, P20}
At this point you need to check two things, firstly that each of these sets of 4 observables are mutually commuting because the rays are orthogonal, secondly that there are 18 observables each of which appears in exactly two sets.
Now assume there is some hidden variable theory which explains this system and which reproduces all the predictions of quantum mechanics. At any given moment the system is in a definite state, and values for each of the 18 operators are determined. The values must be 0 or 1. with the rules they are equal to 1 for exactly one observable in each of the 9 sets, the other three values in each set will be 0. Consequently, there must be nine values set to one overall. This leads to a contradiction, for each observable appears twice so which ever observables have the value of 1 there will always be an even number of ones in total, and 9 is not even.
The F_4 group is what Drey and Manogue have shown diagonalizes the Jordan matrix. This is a matrix of octonions in a 3x3. It is of the form
J^3(O) =
|x O O'|
|O* y O"|
|O'* O"* z |
with some determinant structure. The diagonalization of this matrix is by a group which also has quantum group structure.
Their papers on this are at
1
2
3
Cheers LC