Arcadian Functor

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Marni D. Sheppeard

Saturday, February 14, 2009

M Theory Lesson 262

Using only integer values for the squares of entries in 27 FVF, it follows that the small (squares of) entries in the imaginary part must sum to 3. On solving for the variables I, J, K, R, G and B one has the freedom of signs in sorting out the CKM values.

Let us just look at the up-down entry, I+iR. The solution is given by

3I=26+2697
3R=53-22

and so the up-down entry Vud must equal 0.9744=I2+R2. Fortunately, according to wikipedia, this value is 0.9742±0.0002. It should not be difficult for the reader to find expressions for the other CKM entries, based on the Fourier transform.

Aside: Tommaso Dorigo continues with his excellent series of posts on fairy fields.

2 Comments:

Blogger Matti Pitkänen said...

Hi,

reading of some earlier post made me thinking again about the notion of one-element field. One-element field looks rather self-contradictory notion since 1 and 0 should be represented by same element.

I realized that the real units expressible as ratios of infinite rationals could however provide a well-defined realization of this notion.


a) The condition that same element represents the neutral element of both sum and product gives strong constraint on one-element field. Consider an algebra formed by reals with sum and product defined in the following manner. Sum, call it '+', corresponds to the ordinary product x*y for reals whereas product, call it '*', is identified as the non-commutative product x'*'y= y:th power of x. x=1 represents both the neutral element (0) of '+' and the unit of '*'. The sub-algebras generated by 1 and multiple powers x'*'x'*'...'*'x form commutative sub-algebras of this algebra.

b) When one restricts the consideration to x=1 one obtains one-element field as sub-field which is however trivial since '+' and '*' are identical operations in this subset.


c) One can get over this difficulty by keeping the operations'+' and '*', by assuming one-element property only with respect to the real and various p-adic norms, and by replacing ordinary real unit 1 with the algebra of real units formed from infinite primes by requiring that the real and various p-adic norms of the resulting numbers are equal to one. As far as real and various p-adic norms are considered, one has commutative one-element field. When number theoretic anatomy is taken into account, the algebra contains infinite number of elements and is non-commutative with respect to the product since the number theoretic anatomies of y:th power of x and x:th power of y are different.

February 15, 2009 2:27 AM  
Anonymous Anonymous said...

Great, Matti. Sounds like you're making progress, although I don't think one is 'supposed to' think in terms of zeroes and ones in this context, anyway.

Remote Kea

February 15, 2009 8:37 AM  

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