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 ].
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.
***
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.
***
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.
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.
***
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.
gill1109 wrote:
You have one proof that , and another proof that . They can't both be correct, unless .
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?
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.
***
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.
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.
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.
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.
***
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.
>> 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
>>
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.
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.
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