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Correction to Entropy 2020, 22(1), 61.
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# Correction: Gill, R.D. Does Geometric Algebra Provide a Loophole to Bell’s Theorem? Entropy 2020, 22, 61

Mathematical Institute, Leiden University, Rapenburg 70, 2311 EZ Leiden, The Netherlands
Entropy 2021, 23(5), 631; https://doi.org/10.3390/e23050631
Received: 21 April 2021 / Accepted: 8 May 2021 / Published: 19 May 2021
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

1. 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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