Nice! That does it, then.

]]>That’s most of what I want; I also want the action on the pieces to be derivable from the overall action. (Obviously is nothing like a subgroup of — a rank 4 group doesn’t sit inside a rank 2 — so the derivation would have to be in a subtler sense I haven’t quite hit on.)

]]>Ah, right. Well if it’s true for the factors, it must be true for the whole group: a finite normal subgroup has to project to a finite normal subgroup in each component and so, by your argument, be contained in the center in each component.

]]>It’s nice—much better than the usual pictorial depictions of the Standard Model. But for me, at least, the animated picture is less useful than the discussion in words that explains it. For example, if you don’t know that the W can change left-handed quarks of one flavor to left-handed quarks of another flavor, but not the same for right-handed quarks, it’s not clear that playing with the animation will teach you that! But maybe other people learn differently than I do.

]]>https://www.quantamagazine.org/a-new-map-of-the-standard-model-of-particle-physics-20201022/

{Dated, as indicated, 2020-10-22.)

]]>When I tried to ask Sternberg about it he asked if I could compute the mass of the neutrino [….]

I don’t suppose the relevant SU(3) reps have any hope of combining to a G_2 rep?

If you can’t even compute the mass of the neutrino I’m not sure I want to talk to you. But anyway, the answer to your question is *yes*.

I think this is how interest in octonions and the Standard Model first started. In 1973, Murat Günaydin and Feza Gürsey published a paper “Quark structure and octonions” where they noticed that SU(3) is the subgroup of G_{2} fixing a unit imaginary octonion (this was surely known earlier), and for each quark there is a lepton such that together their representations of SU(3) combine to give the resulting representation of SU(3) on the complexified octonions. So, this rep extends to a rep of G_{2}.

That’s better.

]]>Gabriel:

It can’t be true that all normal subgroups are contained in the center. For example, there is the whole group, but also , etc..

Whoops! All normal subgroups of a so-called ‘simple’ Lie group like are contained in the center, but this is not true for or even . (Simple Lie groups are defined not to be Lie groups that are simple as groups, but Lie groups that are connected and have simple Lie algebras. Equivalently, a Lie group is simple if it’s connected and every normal subgroup is either discrete or the whole group.)

I feel that every finite normal subgroup of is a subgroup of the center, but I’m not going to try to prove this right now.

I like your solution to Puzzle 1 very much!

]]>Whoops! Thanks!

]]>