Correction: Gill, R.D. Does Geometric Algebra Provide a Loophole to Bell’s Theorem? Entropy 2020, 22, 61

Corrections are made to my paper “Gill, R.D. Does Geometric Algebra Provide a Loophole to Bell’s Theorem? Entropy 2020, 22, 61”. Firstly, there was an obvious and easily corrected mathematical error at the end of Section 6 of the paper. In the Clifford algebra under consideration, the basis bivectors $M{e}_i$ do not square to the identity, but to minus the identity. However, the trivector M does square to the identity and hence non-zero divisors of zero, $M-1$ and $M+1$, can be found by the same argument as was given in the paper.

Secondly, in response to a complaint about ad hominem and ad verecundam arguments, a number of scientifically superfluous but insulting sentences have been deleted, and other disrespectful remarks have been rendered neutral by omission of derogatory adjectives. I would like to apologize to Dr. Joy Christian for unwarranted offence.

The end of Section 6 of Gill (2020) [1] discussed the even sub-algebra of $\mathcal{C}{\ell }_{4,0}$, isomorphic to $\mathcal{C}{\ell }_{0,3}$:

One can take as basis for the eight-dimensional real vector space $\mathcal{C}{\ell }_{0,3}$ the scalar 1, three anti-commuting vectors $e_i$, three bivectors $v_i$, and the pseudo-scalar $M=e_1{e}_2{e}_3$. The algebra multiplication is associative and unitary (there exists a multiplicative unit, 1). The pseudo-scalar M squares to $-1$. Scalar and pseudo-scalar commute with everything. The three basis vectors $e_i$, by definition, square to $-1$. The three basis bivectors $v_i=M{e}_i$ square to $+1$. Take any unit bivector v. It satisfies $v^2=1$ hence $v^2-1=(v-1)(v+1)=0$. If the space could be given a norm such that the norm of a product is the product of the norms, we would have $\parallel v-1\parallel .\parallel v+1\parallel =0$ hence either $\parallel v-1\parallel =0$ or $\parallel v+1\parallel =0$ (or both), implying that either $v-1=0$ or $v+1=0$ (or both), implying that $v=1$ or $v=-1$, neither of which are true.

But the bivectors $v_i$ square to $-1$ and the trivector M squares to $+1$. Still, it then follows that $(M+1)(M-1)=0$, and by the argument originally given, it follows that $M=1$ or $M=-1$, a contradiction.