## Fake Critiques by Gill, Moldoveanu, and Weatherall Debunked

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### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:You have a nasty surprise coming your way. Hold your breath.

I'm looking forward to your next production!

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
Joy Christian wrote:
gill1109 wrote:You explicitly describe the subalgebra in Equation (13) of https://arxiv.org/pdf/1908.06172.pdf. It includes the element $\epsilon$, which satisfies $\epsilon^2 = 1$. This is the same element as $\lambda I_3{\bf e}_\infty$ in the multiplication table, Table 1. $( \epsilon - 1)( \epsilon + 1) = 0$, so $( \epsilon - 1)$ and $( \epsilon + 1)$ are a pair of zero-divisors.

You have a nasty habit of keep repeating your mistakes despite my having corrected them dozens of times before. What you claim holds only if you assume +1 = -1, or equivalently, 2 = 0.

As you know, I believe that it follows from your mathematical assumptions that +1 = -1, or equivalently, 2 = 0. You seem to claim that there is no contradiction because your assumptions are physical, not mathematical. Hm, next time I write a mathematical paper I'll make that assumption up front, at the beginning of the paper. I'll then quickly be able to prove the Riemann hypothesis. Problem is, I could also equally quickly disprove it. Well, I'll write two papers and just send them to different journals!

Quite a few people believe that the contradiction between QM and Bell's theorem is a consequence of inconsistency of the ZFC axioms of mathematics. Han Geurdes, Alessandro de Castro, are several. Itamar Pitowsky saw issues in measurability. Tim Palmer thinks that Bell's theorem is false and that he has a counter-example using chaos theory, fractals, and p-adic analysis. Andrei Khrennikov also had ideas in that direction.

Stop waffling. You have a nasty surprise coming your way. Hold your breath.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:
gill1109 wrote:You explicitly describe the subalgebra in Equation (13) of https://arxiv.org/pdf/1908.06172.pdf. It includes the element $\epsilon$, which satisfies $\epsilon^2 = 1$. This is the same element as $\lambda I_3{\bf e}_\infty$ in the multiplication table, Table 1. $( \epsilon - 1)( \epsilon + 1) = 0$, so $( \epsilon - 1)$ and $( \epsilon + 1)$ are a pair of zero-divisors.

You have a nasty habit of keep repeating your mistakes despite my having corrected them dozens of times before. What you claim holds only if you assume +1 = -1, or equivalently, 2 = 0.

As you know, I believe that it follows from your mathematical assumptions that +1 = -1, or equivalently, 2 = 0. You seem to claim that there is no contradiction because your assumptions are physical, not mathematical. Hm, next time I write a mathematical paper I'll make that assumption up front, at the beginning of the paper. I'll then quickly be able to prove the Riemann hypothesis. Problem is, I could also equally quickly disprove it. Well, I'll write two papers and just send them to different journals!

Quite a few people believe that the contradiction between QM and Bell's theorem is a consequence of inconsistency of the ZFC axioms of mathematics. Han Geurdes, Alessandro de Castro, are several. Itamar Pitowsky saw issues in measurability. Tim Palmer thinks that Bell's theorem is false and that he has a counter-example using chaos theory, fractals, and p-adic analysis. Andrei Khrennikov also had ideas in that direction.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
You explicitly describe the subalgebra in Equation (13) of https://arxiv.org/pdf/1908.06172.pdf. It includes the element $\epsilon$, which satisfies $\epsilon^2 = 1$. This is the same element as $\lambda I_3{\bf e}_\infty$ in the multiplication table, Table 1. $( \epsilon - 1)( \epsilon + 1) = 0$, so $( \epsilon - 1)$ and $( \epsilon + 1)$ are a pair of zero-divisors.

You have a nasty habit of keep repeating your mistakes despite my having corrected them dozens of times before. What you claim holds only if you assume +1 = -1, or equivalently, 2 = 0.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:
gill1109 wrote:What do you mean by the term "even algebra"?

If you have to ask this, then you shouldn't be doing this stuff. I would stick to statistics if I were you.

My question was rhetorical, your answer is revealing

Enough said.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
What do you mean by the term "even algebra"?

If you have to ask this, then you shouldn't be doing this stuff. I would stick to statistics if I were you.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:The even sub-algebra of $Cl_{(4,0)}$ is an even algebra. But the algebra $Cl_{(3,0)}$ is not even an even algebra.

What do you mean by the term "even algebra"?

The even sub-algebra of the real Clifford algebra $Cl_{(4,0)}$ is the sub-algebra consisting of real linear combinations of all products of an even number of the generating elements e_1, e_2, e_3, e_4. It is well known (and easily checked) to be isomorphic to $Cl_{(0,3)}$, not to $Cl_{(3,0)}$

You explicitly describe the subalgebra in Equation (13) of https://arxiv.org/pdf/1908.06172.pdf. It includes the element $\epsilon$, which satisfies $\epsilon^2 = 1$. This is the same element as $\lambda I_3{\bf e}_\infty$ in the multiplication table, Table 1. $( \epsilon - 1)( \epsilon + 1) = 0$, so $( \epsilon - 1)$ and $( \epsilon + 1)$ are a pair of zero-divisors

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
Joy Christian wrote:
gill1109 wrote:The even sub-algebra of the real Clifford algebra Cl(4, 0) is isomorphic to the real Clifford algebra Cl(0, 3), as you said yourself.

No, I never said that, either orally or in writing. That is your claim.

You said something that implies this immediately. From your RSOS paper:
Joy Christian wrote:In this higher dimensional space, ${\bf e}_\infty$ is then a unit vector,
$\|{\bf e}_\infty \|^2={\bf e}_\infty\cdot {\bf e}_\infty=1 \iff {\bf e}_\infty^2=1,$ (2.32)
and the corresponding algebraic representation space (2.31) is nothing but the eight-dimensional even sub-algebra of the $2^4=16$-dimensional Clifford algebra $Cl_{(4,0)}$.

You said that your 8-dimensional algebra is the even sub-algebra of $Cl_{(4,0)}$. Now please consult some standard textbooks or do the algebra yourself.

Alternatively, we can go directly to the nub of the matter. Fire up GAviewer and check that in $Cl_{(4,0)}$, (e1 e2 e3 e34)^2 = 1 (or do the algebra yourself). Note that e1e2e3e4 is an element of the even sub-algebra of $Cl_{(4,0)}$. Because (e1 e2 e3 e34)^2 - 1 = 0, it follows that (e1 e2 e3 e34 - 1)(e1 e2 e3 e34 + 1) = 0.

We have a pair of zero-divisors of the even sub-algebra of $Cl_{(4,0)}$: elements x = e1 e2 e3 e34 - 1, and y = e1 e2 e3 e34 = 1 such that xy = 0, but neither x nor y = 0.

If your algebra could be normed, it would mean that ||x|| ||y|| = 0, hence ||x|| = 0 or ||y|| = 0, hence x = 0 or y = 0, hence e1e2e3e4 = +/- 1.

The even sub-algebra of $Cl_{(4,0)}$ is an even algebra. But the algebra $Cl_{(3,0)}$ is not even an even algebra. So they cannot be isomorphic to each other. Period. I never claimed they were.

The 8-dimensional even subalgebra of the algebra $Cl_{(4,0)}$ has been normed without zero divisors. See Eq. (46) of my paper and do the math yourself: https://arxiv.org/abs/1908.06172.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:
gill1109 wrote:The even sub-algebra of the real Clifford algebra Cl(4, 0) is isomorphic to the real Clifford algebra Cl(0, 3), as you said yourself.

No, I never said that, either orally or in writing. That is your claim.

You said something that implies this immediately. From your RSOS paper:
Joy Christian wrote:In this higher dimensional space, ${\bf e}_\infty$ is then a unit vector,
$\|{\bf e}_\infty \|^2={\bf e}_\infty\cdot {\bf e}_\infty=1 \iff {\bf e}_\infty^2=1,$ (2.32)
and the corresponding algebraic representation space (2.31) is nothing but the eight-dimensional even sub-algebra of the $2^4=16$-dimensional Clifford algebra $Cl_{(4,0)}$.

You said that your 8-dimensional algebra is the even sub-algebra of $Cl_{(4,0)}$. Now please consult some standard textbooks or do the algebra yourself.

Alternatively, we can go directly to the nub of the matter. Fire up GAviewer and check that in $Cl_{(4,0)}$, (e1 e2 e3 e34)^2 = 1 (or do the algebra yourself). Note that e1e2e3e4 is an element of the even sub-algebra of $Cl_{(4,0)}$. Because (e1 e2 e3 e34)^2 - 1 = 0, it follows that (e1 e2 e3 e34 - 1)(e1 e2 e3 e34 + 1) = 0.

We have a pair of zero-divisors of the even sub-algebra of $Cl_{(4,0)}$: elements x = e1 e2 e3 e34 - 1, and y = e1 e2 e3 e34 = 1 such that xy = 0, but neither x nor y = 0.

If your algebra could be normed, it would mean that ||x|| ||y|| = 0, hence ||x|| = 0 or ||y|| = 0, hence x = 0 or y = 0, hence e1e2e3e4 = +/- 1.

Recently you also wrote
Joy Christian wrote:Quantum mechanics cannot predict individual event-by-event outcomes for any phenomena. A good example is the decay of a radioactive element. Quantum mechanics can only predict probabilities for such a phenomenon. It is a statistical theory. And, in my opinion, no theory, including classical mechanics, can predict individual event-by-event outcomes for complex phenomena such as the weather or an outcome of a coin toss.

That is also one standpoint which is fully consistent with Bell's theorem. As John Bell remarked, Niels Bohr would have found Bell's theorem uninteresting because he already knew that, it was exactly what he had been saying all the time.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
I must say, that I think that the editorial process at "Entropy" is too rapid. This may work in many fields but not in mathematics. I would have liked more time to polish the paper, but I was pushed very hard to submit a "final version" which could be published in 2019 even if it is officially in the year 2020 volume. Anyway, I have now been informed of the procedure for publishing a correction note, and I will take my time in order to check everything throughout the paper, yet again.

You can try to improve your fake critique as much as you like. Your strawman is not going to suddenly come to life.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

I must say, that I think that the editorial process at "Entropy" is too rapid. This may work in many fields but not in mathematics. I would have liked more time to polish the paper, but I was pushed very hard to submit a "final version" which could be published in 2019 even if it is officially in the year 2020 volume. Anyway, I have now been informed of the procedure for publishing a correction note, and I will take my time in order to check everything throughout the paper, yet again.

The editors involved in handling my paper were Kevin H. Knuth (editor-in-chief) and Andrei Khrennikov (editor of special issue "Quantum Information Revolution: Impact to Foundations"). That's the name of last year's Växjö conference. Of course, I agree that J Christian should have been asked to review the paper, too.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
That was a very good report. I was able to make use of it to much improve the paper.

Funny, because there is absolutely no improvement in your paper.

In fact, it is easy to see that the report I have posted was completely ignored by the editors. Your paper was accepted immidiately without your responses to the second round of reviews.

Here are the dates of the review process, published online: "Received: 21 October 2019 / Revised: 27 December 2019 / Accepted: 30 December 2019 / Published: 31 December 2019."

Also, I confirm that I was never asked or informed by Entropy about reviewing your paper. That is not how good journals operate. A good journal would have asked me to publish a rejoinder along with your paper, if not review it. That is a common courtesy offered by any good journal.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

That was a very good report. I was able to make use of it to much improve the paper. The editors handling the paper were very pleased with the result.

I don't know if this referee was one of the referees whom I had suggested to the editors to review the paper. I proposed J.J. Christian as one of the referees, and also some other "opponents" of Bell's theorem, as well as qualified persons with more conventional views.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
Joy Christian wrote:
gill1109 wrote:Anyway, it is easy to check immediately that (e1 e2 e3 e4) (e1 e2 e3 e4) = 1. From M^2 = 1 we find (M – 1)(M + 1) = 0 and that leads to a contradiction with the claim that the even sub-algebra can be normed.

There is no contradiction, and you know it. I have demonstrated your mistake on the RSOS page of my paper at least a dozen times. You have a habit of keep repeating your false claims. In particular, your claimed "contradiction" is constructed by assuming +1 = -1, or equivalently, 2 = 0.

Whatever. I have added a correction note at the end of the arXiv paper http://arxiv.org/abs/1203.1504, I have asked the editors of Entropy (and of the special issue in which my paper is included) to have it added to the paper, too.

No "whatever." You don't know what you are talking about. As I noted, your claimed "contradiction" is constructed by assuming +1 = -1, or equivalently, 2 = 0. But you are unable to see that.

Here is my actual paper again that you are misrepresenting: https://arxiv.org/abs/1908.06172.

Also, you say that you have added a correction note about your mistake. But you fail to mention in that note that without my pointing out your mistake to you in my rebuttal below you would have never seen that mistake yourself. Indeed, you did not see it for months while repeating it all over the Internet as I remained silent and let you repeat it until your paper was published in Entropy: https://www.academia.edu/41843329/Point ... _s_Theorem.

You say you have asked the editors of the journal Entropy to add the correction note. What you really mean is that you have asked yourself, because you are an editor of that journal.

But despite you being an editor of the journal in which your paper is published, at least one of the reviewers of your paper was brutally honest about your attempt to criticize my work. I reproduce here from the reviewer reports of your paper, published online by the journal. It is important to note that this report was ignored by the editors and your paper was accepted:

https://www.mdpi.com/1099-4300/22/1/61/review_report.

I would describe all of your attempts of the past eight years to criticize my work exactly the same way. I am now pleased to note that people do not have to believe me. They can just look at this report of an anonymous reviewer, chosen by you, an editor of Entropy, to review your own paper!

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:
gill1109 wrote:Anyway, it is easy to check immediately that (e1 e2 e3 e4) (e1 e2 e3 e4) = 1. From M^2 = 1 we find (M – 1)(M + 1) = 0 and that leads to a contradiction with the claim that the even sub-algebra can be normed.

There is no contradiction, and you know it. I have demonstrated your mistake on the RSOS page of my paper at least a dozen times. You have a habit of keep repeating your false claims. In particular, your claimed "contradiction" is constructed by assuming +1 = -1, or equivalently, 2 = 0.

Whatever. I have added a correction note at the end of the arXiv paper http://arxiv.org/abs/1203.1504, I have asked the editors of Entropy (and of the special issue in which my paper is included) to have it added to the paper, too.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
Anyway, it is easy to check immediately that (e1 e2 e3 e4) (e1 e2 e3 e4) = 1. From M^2 = 1 we find (M – 1)(M + 1) = 0 and that leads to a contradiction with the claim that the even sub-algebra can be normed.

There is no contradiction, and you know it. I have demonstrated your mistake on the RSOS page of my paper at least a dozen times. You have a habit of keep repeating your false claims.

In particular, your claimed "contradiction" is constructed by assuming +1 = -1, or equivalently, 2 = 0.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

gill1109 wrote:
The even sub-algebra of the real Clifford algebra Cl(4, 0) is isomorphic to the real Clifford algebra Cl(0, 3), as you said yourself.

No, I never said that, either orally or in writing. That is your claim.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Yes, I learnt a great deal from studying your papers, and also from your talk in Berlin at the meeting of "Die Junge Akademie an der Berlin-Brandenburgischen Akademie der Wissenschaften und der Deutschen Akademie der Naturforscher Leopoldina" April 28 - 30, 2008, and also from our meeting in Oxford, February 2012. I am deeply indebted to you.

I also learnt a great deal from John Baez' paper, and even from Wikipedia too

The even sub-algebra of the real Clifford algebra Cl(4, 0) is isomorphic to the real Clifford algebra Cl(0, 3), as you said yourself.

Anyway, it is easy to check immediately that (e1 e2 e3 e4) (e1 e2 e3 e4) = 1. From M^2 = 1 we find (M – 1)(M + 1) = 0 and that leads to a contradiction with the claim that the even sub-algebra can be normed.

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

***
Here is your mistake:

Joy Christian wrote:
... the basis vectors e1, e2, and e3 in Geometric Algebra square to +1, not -1. And the basis bivectors square to -1, not +1.

So Cl(0, 3) is not even an even algebra.

And you have the audacity to throw references and ideas at me that you have learned from my papers. But what you have learned is still undigested gobbledygook. First, learn it correctly.

***

### Re: Fake Critiques by Gill, Moldoveanu, and Weatherall Debun

Joy Christian wrote:
gill1109 wrote:
Joy Christian wrote:
gill1109 wrote:Thanks. You are also right, at the very end of your Appendix C, that I screwed up in my argument that the real Clifford algebra Cl(0, 3) is not a division algebra. In that algebra the vectors e1, e2, e3 by definition square to -1, not to +1. The bivectors e1e2, e1e3, e2e3 square to -1, not to +1, as I wrote. But the trivector (pseudo scalar) M = e1 e2 e3 squares to +1. Thus 1 - M^2 = 0, hence (1 - M)(1 + M) = 0. We do have a zero divisor. Therefore, Cl(0, 3) is not a division algebra. I have to put out a correction.

Your entire argument is gobbledygook. I have never claimed that Cl(0, 3) is a division algebra. You have invented that claim and then criticized it, as you have done regarding many aspects of my work.

I agree, you never claimed that Cl(0, 3) is a division algebra, and I never said that you made that claim. You did, however, claim that your 8-dimensional algebra was the even subalgebra of Cl(4, 0). In the RSOS paper one can find your assertion: "the corresponding algebraic representation space (2.31) is nothing but the eight-dimensional even sub-algebra of the 2^4 = 16-dimensional Clifford algebra Cl4,0)". Now take a look at https://en.wikipedia.org/wiki/Clifford_algebra#Grading, close to the bottom of the section.
Of course, you don't have to believe everything you read on Wikipedia. One should do the algebra or check the references, or both.

Please. Don't insult me by citing Wikipedia. Read my paper instead: https://arxiv.org/abs/1908.06172.

I have read it, carefully. Here is my reading suggestion for you, if you distrust Wikipedia
The Octonions
John C. Baez
Bull. Amer. Math. Soc. 39 (2002), 145-205.
Errata: Bull. Amer. Math. Soc. 42 (2005), 213.
http://math.ucr.edu/home/baez/octonions/
See Theorems 1, 2 and 3 in Section 1.1 Preliminaries.

If one finds mistakes in Wikipedia and "reliable sources" confirm that they are mistakes, one can and should edit Wikipedia to correct it.

My claim is also proven by GAViewer.

Code: Select all
>> e1 * e2 * e3 * e4ans = 1.00*e1^e2^e3^e4>> (e1 * e2 * e3 * e4) * (e1 * e2 * e3 * e4)ans = 1.00>> (1 - e1 * e2 * e3 * e4)*(1 + e1 * e2 * e3 * e4)ans = 0>>

1 and e1 * e2 * e3 * e4 are both elements of the even subalgebra of the real Clifford algebra Cl(4, 0).
Hence (1 - e1 * e2 * e3 * e4) and (1 + e1 * e2 * e3 * e4) are both two elements of that subalgebra, too. They are evidently non-zero. Their product, however, is zero.
Too further "identify" Cl(0, 3) with even elements of this algebra, you can, for instance, take e1 e2, e1 e3, and e2 e3 to be the three *vectors*, and their products with M = e1 e2 e3 e4 as the three bi-vectors.

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