M Theory Lesson 262
Using only integer values for the squares of entries in 27 $F^{\dagger} V F$, 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
$3I = 26 + 2 \sqrt{697}$
$3R = \sqrt{53} - 2 \sqrt{2}$
and so the up-down entry $V_{ud}$ must equal $0.9744 = \sqrt{I^2 + R^2}$. Fortunately, according to wikipedia, this value is $0.9742 \pm 0.0002$. 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.
Let us just look at the up-down entry, $I + iR$. The solution is given by
$3I = 26 + 2 \sqrt{697}$
$3R = \sqrt{53} - 2 \sqrt{2}$
and so the up-down entry $V_{ud}$ must equal $0.9744 = \sqrt{I^2 + R^2}$. Fortunately, according to wikipedia, this value is $0.9742 \pm 0.0002$. 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.
posted by Kea | 5:13 PM
2 Comments:
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.
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
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