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

In summary, the conversation discusses the concept of a "scalar (under rotation) 1-chain" in the context of differential geometry and Lie algebras. It is explained that the Jacobi identity requires the transformation properties of the Lie product to be derived solely from the other generators, and that this observation leads to the conclusion that only scalar 1-cochains need to be considered in finding the cohomology of a Lie algebra. There is confusion about the notation and coefficients used in the paper's explanation, particularly with regards to the term \alpha_1 \phi_{JJ} and how it relates to the overall scalar 1-cochain. The speaker asks for clarification on this topic.
  • #1
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 [itex]\mu: V\times V \rightarrow V[/itex] that satisfies the Jacobi identity,
[tex]\mu(x,\mu(y,z))=\mu(\mu(x,y),z)+\mu(y,\mu(x,z)).[/tex]
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, [itex]J_{\mu\nu}[/itex], and [itex]y,z[/itex] are other generators carrying Lorentz indices, say [itex]Y_\rho, Z_\sigma[/itex] respectively. Suppose [itex]\mu(y,z)=\mu(Y_\rho,Z_\sigma)=W[/itex]. Substituting this into the l.h.s above, one finds that the Jacobi identity requires that the transformation properties of [itex]W[/itex] under the Lorentz group are derived solely from those of [itex]Y_\rho,Z_\sigma[/itex], i.e. in this case [itex]W[/itex] ought to transform as a second-rank covariant tensor. Another way of saying this is that [itex]\mu[/itex] 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:

[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:

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
[tex]\phi = \alpha_1 \phi_{JJ} +\alpha_2 \phi_{KJ} +\alpha_3 \phi_{JK} +\alpha_4 \phi_{KK}[/tex]
with [itex]\phi_{JJ}=\Pi^a\otimes J_a[/itex] etc.

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

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 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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  • #2


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

What is a scalar 1-chain?

A scalar 1-chain is a mathematical concept used in topology and differential geometry. It represents a one-dimensional object, or a curve, that can be continuously deformed into a point without breaking or tearing.

How does a scalar 1-chain behave under rotation?

Under rotation, a scalar 1-chain remains unchanged. This is because a scalar 1-chain is a one-dimensional object and does not have any orientation, so it is not affected by rotations.

What is the difference between a scalar 1-chain and a vector 1-chain?

A scalar 1-chain only represents the curve itself, while a vector 1-chain also includes the direction and magnitude of the curve. This means that a scalar 1-chain can be thought of as a simplified version of a vector 1-chain.

How are scalar 1-chains used in physics?

In physics, scalar 1-chains are used to describe the path of a particle as it moves through space. They are also used in the study of fields, such as electric and magnetic fields, to map out the behavior of these fields in a given space.

Can a scalar 1-chain be higher-dimensional?

No, a scalar 1-chain is always one-dimensional. It cannot have any higher dimensions, as it represents a curve that can be continuously deformed into a point, which is a zero-dimensional object.

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