by **gill1109** » Tue Feb 18, 2020 7:53 am

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 * 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.

[quote="Joy Christian"][quote="gill1109"][quote="Joy Christian"][quote="gill1109"]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.[/quote]

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.

[/quote]

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 [url]https://en.wikipedia.org/wiki/Clifford_algebra#Grading[/url], 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.[/quote]

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

[/quote]

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.

[url]http://math.ucr.edu/home/baez/octonions[/url]/

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]

>> 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

>> [/code]

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.