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I agree entirely

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

***

Here is an example of the poor quality of argument:

In fact, this is not even an argument. At best, it is a tacit admission by the critic that he is not able to carry out the derivation himself.

Moreover, at least one interested reader, namely Jay R. Yablon, has carried out the derivation himself (without using GA) and has found no mistake.

***

Here is an example of the poor quality of argument:

Richard D. Gill wrote:

The interested reader may search for the mistake themselves, it is hidden in the derivation (34)–(40) [in https://arxiv.org/abs/1911.11578 ].

In fact, this is not even an argument. At best, it is a tacit admission by the critic that he is not able to carry out the derivation himself.

Moreover, at least one interested reader, namely Jay R. Yablon, has carried out the derivation himself (without using GA) and has found no mistake.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:***

Here is an example of the poor quality of argument:Richard D. Gill wrote:

The interested reader may search for the mistake themselves, it is hidden in the derivation (34)–(40) [in https://arxiv.org/abs/1911.11578 ].

In fact, this is not even an argument. At best, it is a tacit admission by the critic that he is not able to carry out the derivation himself.

Moreover, at least one interested reader, namely Jay R. Yablon, has carried out the derivation himself (without using GA) and has found no mistake.

***

Good. We are back to talking about physics and related math. Let's stick to it, please. I also concur that there is no mathematical mistake in the derivation. It is fairly simple.

.

- FrediFizzx
- Independent Physics Researcher
**Posts:**1992**Joined:**Tue Mar 19, 2013 7:12 pm**Location:**N. California, USA

Joy Christian wrote:***

Here is an example of the poor quality of argument:Richard D. Gill wrote:

The interested reader may search for the mistake themselves, it is hidden in the derivation (34)–(40) [in https://arxiv.org/abs/1911.11578 ].

In fact, this is not even an argument. At best, it is a tacit admission by the critic that he is not able to carry out the derivation himself.

Moreover, at least one interested reader, namely Jay R. Yablon, has carried out the derivation himself (without using GA) and has found no mistake.

***

Indeed, that was not an argument. It was laziness. But it is a fact that the error which I claim is hidden in (34)–(40) is not new. I converted a bug into a feature by giving an exercise to my reader: find it for themselves. The solution to my exercise is in my earlier (published) work.

Boole gave Bell's inequality as an exercise to his readers!

Last I heard from Jay Yablon was that he had not succeeded in patching the work. He has written an unfinished manuscript in which he announces the results which he thinks he can prove, but he hasn't finished it yet.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

gill1109 wrote:

But it is a fact that the error which I claim is hidden in (34)–(40) is not new. I converted a bug into a feature by giving an exercise to my reader: find it for themselves.

This is complete nonsense. The truth is that Gill has not been able to carry out the derivation in eq. (34)-(40). Any undergraduate can do this calculation and see that there is no error in it.

Here is my challenge to Gill: Point out what his claimed error is, or admit that he has been making false claims about my local model for the past eight years to protect his vested interests.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:gill1109 wrote:

But it is a fact that the error which I claim is hidden in (34)–(40) is not new. I converted a bug into a feature by giving an exercise to my reader: find it for themselves.

This is complete nonsense. The truth is that Gill has not been able to carry out the derivation in eq. (34)-(40). Any undergraduate can do this calculation and see that there is no error in it.

Here is my challenge to Gill: Point out what his claimed error is, or admit that he has been making false claims about my local model for the past eight years to protect his vested interests.

***

You're absolutely right! No error in (34)-(40). Indeed, . In fact, that is exactly what *I* have been saying for the past eight years. I have to revise my paper. Will do that later this weekend.

You have one proof that , and another proof that . They can't both be correct, unless .

There is a much easier proof of the first statement, as I wrote in https://arxiv.org/abs/1203.1504v1, and as any decent undergraduate could figure out for themselves. The proof of the second statement is wrong.

My apologies for my mistake. I wrote the paper in a bit of a hurry. That's no excuse. Qui s'excuse, s'accuse.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

gill1109 wrote:

You're absolutely right! No error in (34)-(40). Indeed, . In fact, that is exactly what *I* have been saying for the past eight years. I have to revise my paper. Will do that later this weekend.

Good. Revise your paper. And if you are an honest debater, then don't forget to acknowledge in your revised version that I forced you to recognize, admit, and acknowledge your error.

gill1109 wrote:

You have one proof that , and another proof that . They can't both be correct, unless .

Your mistake here is to not understand that both and are correct because they hold under two different conditions. The proof of the second statement can only be wrong in the flat geometry of. It is absolutely correct in . But you have not understood this simple fact for the past eight years, so you are unlikely to understand it now.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Indeed there are things in your work which I never succeeded in understanding. Oh well, I'm just a statistician, and definitely not a physicist.

I try to be an honest debater and I don't like hurting anyone's feelings. Thanks for pointing out the mistake. Naturally, I have acknowledged your contribution in my revision, which arXiv tells me "is scheduled to be announced at Tue, 18 Feb 2020 01:00:00 GMT"

You write "note, however, that kappa is not a hidden variable. It is a part of the properties of a physical space Alice and Bob live in." So can Alice and Bob (or anybody else) find out what value it took in any particular trial of their experiment? After they have done the experiment they can get together, and then they will know which of the four possible pairs of outcomes were observed and which settings they both used, in each trial which they performed. According to your model, kappa is fluctuating between +/-1 completely at random. How hidden is it, actually?

I try to be an honest debater and I don't like hurting anyone's feelings. Thanks for pointing out the mistake. Naturally, I have acknowledged your contribution in my revision, which arXiv tells me "is scheduled to be announced at Tue, 18 Feb 2020 01:00:00 GMT"

You write "note, however, that kappa is not a hidden variable. It is a part of the properties of a physical space Alice and Bob live in." So can Alice and Bob (or anybody else) find out what value it took in any particular trial of their experiment? After they have done the experiment they can get together, and then they will know which of the four possible pairs of outcomes were observed and which settings they both used, in each trial which they performed. According to your model, kappa is fluctuating between +/-1 completely at random. How hidden is it, actually?

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

gill1109 wrote:

You write "note, however, that kappa is not a hidden variable. It is a part of the properties of a physical space Alice and Bob live in." So can Alice and Bob (or anybody else) find out what value it took in any particular trial of their experiment? After they have done the experiment they can get together, and then they will know which of the four possible pairs of outcomes were observed and which settings they both used, in each trial which they performed. According to your model, kappa is fluctuating between +/-1 completely at random. How hidden is it, actually?

The actual value of kappa is forever unknown to Alice and Bob, as well as to Charlie. All they can tell, even after the experiment, is whether kappa has been an odd number or even number for a given run of the experiment. If it had been an even number for both of them, then they would find that AB = -1. If it had been an odd number for either of them, then they would find that AB = +1. If it had been an odd number for both of them, then they would find that AB = -1.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:gill1109 wrote:

You write "note, however, that kappa is not a hidden variable. It is a part of the properties of a physical space Alice and Bob live in." So can Alice and Bob (or anybody else) find out what value it took in any particular trial of their experiment? After they have done the experiment they can get together, and then they will know which of the four possible pairs of outcomes were observed and which settings they both used, in each trial which they performed. According to your model, kappa is fluctuating between +/-1 completely at random. How hidden is it, actually?

The actual value of kappa is forever unknown to Alice and Bob, as well as to Charlie. All they can tell, even after the experiment, is whether kappa has been an odd number or even number for a given run of the experiment. If it had been an even number for both of them, then they would find that AB = -1. If it had been an odd number for either of them, then they would find that AB = +1. If it had been an odd number for both of them, then they would find that AB = -1.

***

Thanks! I was thinking of lambda, and getting mixed up as usual.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

***

In response to Richard D. Gill's latest update, I have had to revise my existing 2012 reply to him on the arXiv: https://arxiv.org/abs/1203.2529 (see Appendix C).

For the convenience of potential readers, I have also made a separate PDF file of the new appendix here: http://libertesphilosophica.info/blog/w ... plyapp.pdf

The only change in the revised version of Appendix C is the following new line:

***

In response to Richard D. Gill's latest update, I have had to revise my existing 2012 reply to him on the arXiv: https://arxiv.org/abs/1203.2529 (see Appendix C).

For the convenience of potential readers, I have also made a separate PDF file of the new appendix here: http://libertesphilosophica.info/blog/w ... plyapp.pdf

The only change in the revised version of Appendix C is the following new line:

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

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 are easily seen too square to -1, not to +1; what I wrote (that they square to +1) was wrong. But the trivector (pseudoscalar) M = e1e2e3 does square to +1, another easy computation. 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.

Remember: the real Clifford algebra Cl(p, q) is defined by having n = p + q elements e1, e2, ..., en, of which p square to +1 and the remaining q square to -1. The only other rules you have to remember is: they anti-commute; they commute with the scalar 1; the algebra is associative. This gives us a real vector space of dimension 2^n = 2^(p+q) spanned by the products of a finite number of e1, ... , en, taken in order. There are 2^n subsets of n elements e1, ..., en including the empty set (an empty product is by definition equal to the scalar 1).

Remember: the real Clifford algebra Cl(p, q) is defined by having n = p + q elements e1, e2, ..., en, of which p square to +1 and the remaining q square to -1. The only other rules you have to remember is: they anti-commute; they commute with the scalar 1; the algebra is associative. This gives us a real vector space of dimension 2^n = 2^(p+q) spanned by the products of a finite number of e1, ... , en, taken in order. There are 2^n subsets of n elements e1, ..., en including the empty set (an empty product is by definition equal to the scalar 1).

Last edited by gill1109 on Tue Feb 18, 2020 5:09 am, edited 1 time in total.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

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

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

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 the 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 must do the algebra or check the references, or both.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

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 the 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 must 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.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

Joy Christian wrote:gill1109 wrote:

You write "note, however, that kappa is not a hidden variable. It is a part of the properties of a physical space Alice and Bob live in." So can Alice and Bob (or anybody else) find out what value it took in any particular trial of their experiment? After they have done the experiment they can get together, and then they will know which of the four possible pairs of outcomes were observed and which settings they both used, in each trial which they performed. According to your model, kappa is fluctuating between +/-1 completely at random. How hidden is it, actually?

The actual value of kappa is forever unknown to Alice and Bob, as well as to Charlie. All they can tell, even after the experiment, is whether kappa has been an odd number or even number for a given run of the experiment. If it had been an even number for both of them, then they would find that AB = -1. If it had been an odd number for either of them, then they would find that AB = +1. If it had been an odd number for both of them, then they would find that AB = -1.

***

Let's analyze that a bit further. It is a bit strange as the "even" or "odd" has to be on both A or B to get -1 but the "odd" is on A or B to get +1. This seems to be like an unbalanced coin toss. Is the probability to get AB = +1, 1/3? If so, that doesn't match the QM probability. Perhaps the S^3 action is a non-linear function for "even" and "odd".

.

- FrediFizzx
- Independent Physics Researcher
**Posts:**1992**Joined:**Tue Mar 19, 2013 7:12 pm**Location:**N. California, USA

Joy Christian wrote:gill1109 wrote:Joy Christian wrote:

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 * e4`

ans = 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.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

***

Here is your mistake:

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.

***

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.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

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.

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.

- gill1109
- Mathematical Statistician
**Posts:**1958**Joined:**Tue Feb 04, 2014 10:39 pm**Location:**Leiden

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.

***

- Joy Christian
- Research Physicist
**Posts:**2377**Joined:**Wed Feb 05, 2014 4:49 am**Location:**Oxford, United Kingdom

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