Timeline for Is Sp(1).Sp(1).Sp(1) the homotopy-fixed locus of Triality?
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| Apr 12, 2019 at 21:21 | answer | added | Theo Johnson-Freyd | timeline score: 5 | |
| Apr 12, 2019 at 19:58 | comment | added | Urs Schreiber | Thanks, I see. One just needs to beware that the triple dot-product above is not two nested binary central products -- at least not without changing along the way which central subgroups are understood. Anyway, the notation is not so important here. | |
| Apr 12, 2019 at 19:16 | comment | added | Theo Johnson-Freyd | I didn't know those papers. The open circle is used for the central product in the literature I've seen on finite groups, eg the Atlas. | |
| Apr 12, 2019 at 16:20 | comment | added | Urs Schreiber | @TheoJohnson-Freyd regarding the dot-notation, there is a list of references using it this way here: ncatlab.org/nlab/show/Sp%28n%29.Sp%281%29#References . What's an example of a reference using the open ∘ instead? | |
| Apr 12, 2019 at 16:18 | comment | added | Urs Schreiber | @TheoJohnson-Freyd sure, the three Sp(1) "dot-factors" want to be thought of as associated with the three outer nodes of the Dynkin diagram, but does that alone serve to settle the question? Probably, if one chases through details carefully. Hopefully somebody has written that out? Or can write it out on the spot, if it's really easy? | |
| Apr 12, 2019 at 15:38 | comment | added | Theo Johnson-Freyd | Incidentally, in the group theory papers I have read, the central product $(G \times H) / Z$, where $Z = Z(G) = Z(H)$, is usually written $G \circ H$. The group theorists seem to usually use $G \cdot H$ for a (usually non-split) extension with normal subgroup $G$ and quotient $H$. | |
| Apr 12, 2019 at 15:37 | comment | added | Theo Johnson-Freyd | The non-homotopy fixed points of triality (in its usual manifestation) is the exceptional group $G_2$. This is not contained in $Sp(1)^3$, so $Sp(1)^3$ definitely is not the anything fixed points of triality. It probably doesn't matter, but $G_2$ does contain an $Sp(1)^2$ centralizing an element of order $2$. I write "in its usual manifestation" because implicitly I splitting of the map $Aut(Spin(8)) \to Out(Spin(8)) = Aut(Spin(8))/Inn(Spin(8))$. IIRC, there are inequivalent conjugacy classes of such a splitting, but there is a standard splitting coming from the Chevalley presentation. | |
| Apr 12, 2019 at 15:32 | comment | added | Theo Johnson-Freyd | Isn't this $Sp(1)^3$ just the one spanned by the three outer nodes on the Dynkin diagram? That $Sp(1)^3$ is acted on nontrivially by triality. The action is by outer automorphisms, so remains highly nontrivial even if you look at the classifying space $BSp(1)^3$. Or do you have a different $Sp(1)^3$ in mind? | |
| Apr 12, 2019 at 14:45 | history | asked | Urs Schreiber | CC BY-SA 4.0 |