Timeline for Is SO(4) a subgroup of SU(3)?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Nov 16, 2021 at 21:19 | comment | added | LSpice | It's the irreducibility that I was missing. Thanks! | |
| Nov 16, 2021 at 20:32 | comment | added | Robert Bryant | @LSpice: Because the Lie algebra so(4) in su(3) has to be a module over the Lie algebra of the maximal torus t (of any subalgebra, actually), and su(3)/t is the direct sum of three inequivalent irreducible modules of dimension 2 (the three weight spaces), so so(4)/t must be a sum of some number of these weight spaces. By dimension count, it would have to be two of these weight spaces, but the Lie bracket of any two contains the third. | |
| Nov 16, 2021 at 20:19 | comment | added | LSpice | Why would the Lie algebra have to contain the entire weight space if it intersected it non-trivially? | |
| Nov 16, 2021 at 18:25 | history | answered | Robert Bryant | CC BY-SA 4.0 |