What is a "scalar (under rotation) 1chain"?by ianhoolihan Tags: None 

#1
Mar612, 06:09 PM

P: 145

Hi all,
I am trying to make sense of a paper involving differenital geometry and Lie algebras. Here's the part I am confused about: [tex]\mu=\frac{1}{2}\epsilon_{ab}^c\Pi^a\Pi^b\otimes J_c +\epsilon_{ab}^c\Pi^a\Pi^{\bar{b}}\otimes K_c[/tex] where barred indices refer to boosts. The paper then goes on to say: 1. does [itex]\phi_{JJ}=\Pi^a\otimes J_a[/itex] mean [itex]\phi_{JJ}=\Pi^a\otimes J_a=\Pi^1 J_1 + \Pi^1 J_2+ ... +\Pi^3 J_3[/itex] i.e. with nine terms (remember the unbarred indices are rotation only, so three generators)? 2. Are the [itex]\alpha_i[/itex] real coefficients, or arrays? I.e. I would have thought [tex]\phi=\phi^A_B\Pi^B \otimes T_A = ... = \phi ^a_b\Pi^b\otimes T_a+\phi ^a_{\bar{b}}\Pi^{\bar{b}}\otimes T_a+\phi ^{\bar{a}}_b\Pi^b\otimes T_{\bar{a}}+\phi ^{\bar{a}}_{\bar{b}}\Pi^{\bar{b}}\otimes T_{\bar{a}}[/tex] where I have let [itex]A=\{\{a\},\{\bar{a}\}\}[/itex]. This is the closest I can get to the given expression, but here I have [itex]\alpha_1 \phi_{JJ} = \alpha_1 \Pi^a \otimes T_a = \phi ^a_b\Pi^b\otimes T_a[/itex], which doesn't seem to work. I am assuming the the [itex]\alpha_i[/itex] are simple scalars, which somehow is to do with [itex]\phi[/itex] being a "scalar 1cochain". I have a few more questions, but that will suffice for now  hopefully this gets the ball rolling, and I can work them out myself, once I understand what's going on here. Cheers, Ianhoolihan 



#2
Mar712, 01:20 PM

P: 145

Any help, or even incomplete hints in the right direction? Cheers



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