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

Richard David Gill

doi:10.3390/e23050631

Mathematical Institute, Leiden University, Rapenburg 70, 2311 EZ Leiden, The Netherlands

Entropy2021, 23(5), 631;https://doi.org/10.3390/e23050631

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) [] discussed the even sub-algebra of

C ℓ_{4, 0}

, isomorphic to

C ℓ_{0, 3}

:

One can take as basis for the eight-dimensional real vector space $C ℓ_{0, 3}$ 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 $∥ v - 1 ∥ . ∥ v + 1 ∥ = 0$ hence either $∥ v - 1 ∥ = 0$ or $∥ v + 1 ∥ = 0$ (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.

Reference

Gill, R.D. Does Geometric Algebra Provide a Loophole to Bell’s Theorem? Entropy 2020, 22, 61. [Google Scholar] [CrossRef] [PubMed]

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

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