Yesterday we didn't write out the $(W^{+}, W^{-}, Z)$ bosons as a Koide triplet, so let us do that now. This triplet is clearly of the form
$(1,1, \frac{\alpha}{2})$
where $\alpha \cos \theta_{W} = 2$, for the Weinberg angle $\theta_{W}$. The eigenvalue set at the Koide phase $\phi = 0$ is
$\lambda \in \{ 1 + \alpha, 1 - \frac{\alpha}{2} \}$
with multiplicity $2$ for the second eigenvalue. At $\theta_{W} = \pi /6$ we obtain
$\lambda \in \{ 1 + \frac{4}{\sqrt{3}}, 1 - \frac{2}{\sqrt{3}} \}$.
Squaring and summing the rest mass triplet for the Higgs, we obtain $m_{H} = 11/2$ exactly, in the units chosen, which just happen to be the Koide scale for the top quark triplet.
14 years ago
So this post is probably junk, because the second triplet is some random triplet of form $(1,1,x)$. But note what happened: the $(W,W,Z)$ triplet is used to write down a real Koide matrix, which itself has a triplet. So one can define a whole sequence of triplets by feeding each set into a (real) Koide matrix. Not sure why each triplet in the sequence should be able to define the Higgs scale, but perhaps $(W,W,Z)$ is not the only one.
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