Timeline for Is the trace of group generators a representation invariant?
Current License: CC BY-SA 4.0
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Oct 5, 2022 at 15:58 | comment | added | Cosmas Zachos | WP reviews, doesn't introduce. You seem as though you'd greatly profit from Wu-Ki Tung's classic book... | |
Oct 5, 2022 at 15:51 | comment | added | user346886 | @CosmasZachos that's a very comprehensive Wiki page. I think I should also start reading lecture notes on Lie groups and algebras in parallel to my introductory field theory books. I read also on the wiki page you sent that spinor representations are a result of lack of simple connectedness. For me at least that's a rather amazing / deep result which I'd like to know more about. | |
Oct 5, 2022 at 13:24 | comment | added | Cosmas Zachos | @osolitonmio The proof in MSE is fine and applies here. In fact, all you need do is trace $\pi_{(m,n)}(J_i) = J^{(m)}_i \otimes 1_{(2n+1)}+1_{(2m+1)} \otimes J^{(n)}_i $ in WP which is how all representations of the type you are discussing are defined in the first place! Make sure you understand why both direct product summands are traceless! | |
Oct 4, 2022 at 22:36 | history | edited | ZeroTheHero | CC BY-SA 4.0 |
added 347 characters in body
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Oct 4, 2022 at 19:50 | comment | added | ZeroTheHero | @osolitonmio in the case of finite-dim of Lorentz, the rep is still the exponential of generators (just not hermitian). I’ve never thought about this case and it might still work. Not sure if it would work for - say - indecomposables. maybe irreducible to enough. | |
Oct 4, 2022 at 19:05 | comment | added | user346886 | I think I can follow your proof for this case. But for the finite-dim Lorentz group irreps I suppose it doesn't hold? I did find this question and answer on MSE which might be the full answer math.stackexchange.com/questions/1483570/… ? I don't know for sure. | |
Oct 4, 2022 at 18:38 | history | answered | ZeroTheHero | CC BY-SA 4.0 |