Theorem. One square of contexts has no global point [fdip-009E]
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  • November 3, 2026

A context in \(M_n(\mathbb C)\) is a unital commutative *-subalgebra. A character of a context is a unital *-homomorphism from that algebra to \(\mathbb C\). For a finite family of contexts, a global point is a choice of one character of each context such that the chosen characters agree on every matrix that lies in two of them. Such a point exists if and only if there is a function on the union of the contexts whose restriction to each context is a character. This equivalence is the Kochen-Specker formulation for a finite family.

Let \(\sigma _1,\sigma _2,\sigma _3\) be the Pauli matrices, and let \(I\) be the \(2\times 2\) identity. In \(M_4(\mathbb C)\), the 9 Kronecker products \[\begin {array}{ccc} \sigma _1\otimes I & I\otimes \sigma _1 & \sigma _1\otimes \sigma _1\\ I\otimes \sigma _2 & \sigma _2\otimes I & \sigma _2\otimes \sigma _2\\ \sigma _1\otimes \sigma _2 & \sigma _2\otimes \sigma _1 & \sigma _3\otimes \sigma _3 \end {array}\] are Hermitian involutions. Matrices in one row commute, and matrices in one column commute. The product of each row, in the displayed order, is the \(4\times 4\) identity \(I\otimes I\). The three column products are \(I\otimes I\), \(I\otimes I\) and \(-I\otimes I\). The unital *-algebras generated by the three rows and the three columns are 6 contexts. That family has no global point. Each of the 6 contexts does have a character.

The equivalence does not show that the family of every commutative subalgebra of \(M_n(\mathbb C)\) has no global point. Nothing proved here identifies a context with a selector. Whether global points of this kind organise those selectors remains open. The accounts of Mermin, Peres, Heunen, Landsman and Spitters, and Doering and Isham are not premises of the multiplication.

Proof. 

  • November 3, 2026

The two conditions are equivalent. From a global point, define a function by evaluating the character of any chosen context that contains the matrix. Agreement on shared matrices makes the value independent of the choice, and the restriction to each context is the chosen character. Conversely, the restrictions of such a function are characters, and they agree on shared matrices.

On an elementary tensor, \((A\otimes B)(e_j\otimes f_l)=Ae_j\otimes Bf_l\). Extending that rule by linearity and applying it twice gives \((A\otimes B)(C\otimes D)=(AC)\otimes (BD)\). The matrix entry of \(A\otimes B\) from \(e_j\otimes f_l\) to \(e_i\otimes f_k\) is \(A_{ij}B_{kl}\). The adjoint entry, placed from \(e_i\otimes f_k\) back to \(e_j\otimes f_l\), is the conjugate \(\overline {A_{ij}B_{kl}}=A^*_{ji}B^*_{lk}\), which is the matching entry of \(A^*\otimes B^*\). Thus \((A\otimes B)^*=A^*\otimes B^*\).

The Pauli matrices are Hermitian and square to \(I\). Multiplying the matrices gives \[\begin {aligned} \sigma _1\sigma _2&= \begin {pmatrix}0&1\\1&0\end {pmatrix} \begin {pmatrix}0&-i\\i&0\end {pmatrix} =\begin {pmatrix}i&0\\0&-i\end {pmatrix}=i\sigma _3,\\ \sigma _2\sigma _3&= \begin {pmatrix}0&-i\\i&0\end {pmatrix} \begin {pmatrix}1&0\\0&-1\end {pmatrix} =\begin {pmatrix}0&i\\i&0\end {pmatrix}=i\sigma _1,\\ \sigma _3\sigma _1&= \begin {pmatrix}1&0\\0&-1\end {pmatrix} \begin {pmatrix}0&1\\1&0\end {pmatrix} =\begin {pmatrix}0&1\\-1&0\end {pmatrix}=i\sigma _2. \end {aligned}\] Since the factors are Hermitian, \((\sigma _1\sigma _2)^*=\sigma _2\sigma _1\) and \((i\sigma _3)^*=-i\sigma _3\), so \(\sigma _2\sigma _1=-i\sigma _3\). The same adjoint step reverses the other two products. Distinct Pauli matrices therefore anticommute.

Each of the 9 Kronecker products is Hermitian, and its square is \(I\otimes I\). If the left factors commute and the right factors commute, the two orders of the Kronecker product agree. The identity commutes with every matrix, and every matrix commutes with itself, so this covers every pair in the first two rows and the first two columns. In the third row, and throughout the third column, each pair is a Kronecker product of two distinct Pauli matrices on the left and two distinct Pauli matrices on the right. Both sides anticommute, and the two signs cancel, so those tensors commute as well.

The row products are \[\begin {aligned} (\sigma _1\otimes I)(I\otimes \sigma _1)(\sigma _1\otimes \sigma _1) &=(\sigma _1\otimes \sigma _1)^2=I\otimes I,\\ (I\otimes \sigma _2)(\sigma _2\otimes I)(\sigma _2\otimes \sigma _2) &=(\sigma _2\otimes \sigma _2)^2=I\otimes I,\\ (\sigma _1\otimes \sigma _2)(\sigma _2\otimes \sigma _1) (\sigma _3\otimes \sigma _3) &=(i\sigma _3\otimes (-i\sigma _3))(\sigma _3\otimes \sigma _3) =I\otimes I. \end {aligned}\] The column products are \[\begin {aligned} (\sigma _1\otimes I)(I\otimes \sigma _2)(\sigma _1\otimes \sigma _2) &=(\sigma _1\otimes \sigma _2)^2=I\otimes I,\\ (I\otimes \sigma _1)(\sigma _2\otimes I)(\sigma _2\otimes \sigma _1) &=(\sigma _2\otimes \sigma _1)^2=I\otimes I,\\ (\sigma _1\otimes \sigma _1)(\sigma _2\otimes \sigma _2) (\sigma _3\otimes \sigma _3) &=(i\sigma _3\otimes i\sigma _3)(\sigma _3\otimes \sigma _3) =-(I\otimes I). \end {aligned}\] Within a row or a column the three matrices commute. If \((AB)C=I\otimes I\) and \(C^2=I\otimes I\), then \(AB=(AB)C^2=((AB)C)C=C\). If instead \((AB)C=-(I\otimes I)\), then \(AB=(AB)C^2=((AB)C)C=(-(I\otimes I))C=-C\). The first case is every row and the first two columns. The second case is the third column.

A character sends a Hermitian involution to a complex number \(t\) with \(t^2=1\), hence to \(\pm 1\), and it sends a product of commuting elements to the product of the values. The three values on a row therefore multiply to \(+1\). The three column products force the values \(+1\), \(+1\) and \(-1\). The product of the three row constraints equals the product of all 9 values and equals \(+1\). The product of the three column constraints is the same product of 9 values and equals \(-1\). No choice of signs satisfies both. A global point would supply such signs, so the 6 contexts have no global point.

Each context separately has a character. The trace multiplies over a Kronecker product, \(\operatorname {Tr}(A\otimes B)=\operatorname {Tr}(A)\operatorname {Tr}(B)\), because the diagonal entries are \(A_{ii}B_{kk}\). Thus \(\operatorname {Tr}\bigl ((\sigma _a\otimes \sigma _b)(\sigma _c\otimes \sigma _d)\bigr ) =\operatorname {Tr}(\sigma _a\sigma _c)\operatorname {Tr}(\sigma _b\sigma _d)\), where \(\sigma _0\) means \(I\). The square of each of these 4 matrices is \(I\), so its trace against itself is \(\operatorname {Tr}(I)=2\). Each displayed product \(i\sigma _k\) has trace zero, so each \(\sigma _k\) has trace zero. The product of two distinct matrices among the 4 is either one of them or \(\pm i\) times one of them, and in every case the trace is zero. Distinct tensors \(\sigma _a\otimes \sigma _b\) are therefore orthogonal for \(\operatorname {Tr}(P^*Q)\). The square of such a tensor is \(I\otimes I\), so \(\operatorname {Tr}(P^*P)=4\). The 4 matrices of any one row or column are the identity tensor together with three generators. They are 4 distinct tensors of this form, each with nonzero trace against itself, so they are linearly independent.

Write \(A\) and \(B\) for the first two matrices of a context and \(C\) for the third. They commute. Assign \(\pm 1\) freely to \(A\) and \(B\). Where \(AB=C\), set the value of \(C\) equal to the product of the other two values. Then \(AC=A(AB)=B\) and \(BC=B(AB)=A\), and the values match. Where \(AB=-C\), which is the third column, set the value of \(C\) equal to the opposite of that product. Then \(AC=A(-AB)=-B\). Since \(A\) and \(B\) commute, \(BC=B(-AB)=-A\). Each value carries the same minus sign. Every square equals \(I\otimes I\) and has value \(+1\). Opposite orders agree because the matrices commute and the values commute. These 4 matrices therefore span the algebra they generate, and the linear extension of the assignment is a unital algebra homomorphism on that basis. Every basis matrix is Hermitian and every value is real, so the extension preserves the *-operation.