What is a scalar (under rotation) 1-chain ?

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The discussion centers on the concept of a "scalar (under rotation) 1-chain" within the context of differential geometry and Lie algebras. The Lie product map, denoted as μ, adheres to the Jacobi identity and is crucial for understanding the transformation properties of cochains. The example provided illustrates how the scalar nature of cochains simplifies the analysis of the cohomology of a Lie algebra, particularly in the case of the galilean algebra. The user seeks clarification on the representation of scalar 1-cochains and the nature of coefficients involved in their expressions.

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ianhoolihan
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What is a "scalar (under rotation) 1-chain"?

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:

...Lie product map \mu: V\times V \rightarrow V that satisfies the Jacobi identity,
\mu(x,\mu(y,z))=\mu(\mu(x,y),z)+\mu(y,\mu(x,z)).
This is usually written as a cyclic sum, a form that, in the case at hand, obscures its content. To clarify the latter, take as an example the case where [\itex]x[/itex] is a Lorentz group generator, J_{\mu\nu}, and y,z are other generators carrying Lorentz indices, say Y_\rho, Z_\sigma respectively. Suppose \mu(y,z)=\mu(Y_\rho,Z_\sigma)=W. Substituting this into the l.h.s above, one finds that the Jacobi identity requires that the transformation properties of W under the Lorentz group are derived solely from those of Y_\rho,Z_\sigma, i.e. in this case W ought to transform as a second-rank covariant tensor. Another way of saying this is that \mu itself is a Lorentz scalar, an observation that we use later on.

Now things begin with finding the cohomology of a Lie algebra. The galilean algebra is taken as an example, and the Lie product is given in terms of differential forms:

\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

where barred indices refer to boosts. The paper then goes on to say:

By an argument based on the observation made [above]...we conclude that only scalar (under rotations) cochains need to be considered.

I do not see how this applies. I assume it somehow helps to simplify "the most general scalar 1-cochain":

...the most general scalar 1-cochain is given by
\phi = \alpha_1 \phi_{JJ} +\alpha_2 \phi_{KJ} +\alpha_3 \phi_{JK} +\alpha_4 \phi_{KK}
with \phi_{JJ}=\Pi^a\otimes J_a etc.

Now if someone could clarify this all to me, that'd be great. More specifically:

1. does \phi_{JJ}=\Pi^a\otimes J_a mean \phi_{JJ}=\Pi^a\otimes J_a=\Pi^1 J_1 + \Pi^1 J_2+ ... +\Pi^3 J_3 i.e. with nine terms (remember the unbarred indices are rotation only, so three generators)?

2. Are the \alpha_i real coefficients, or arrays? I.e. I would have thought
\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}}
where I have let A=\{\{a\},\{\bar{a}\}\}. This is the closest I can get to the given expression, but here I have \alpha_1 \phi_{JJ} = \alpha_1 \Pi^a \otimes T_a = \phi ^a_b\Pi^b\otimes T_a, which doesn't seem to work. I am assuming the the \alpha_i are simple scalars, which somehow is to do with \phi being a "scalar 1-cochain".

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

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