Proving A_μ' in Lie(SU(N)) for U in SU(N)

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Homework Help Overview

The discussion revolves around proving a property related to the Lie algebra of the special unitary group SU(N). The original poster seeks to establish that if a matrix \( A_\mu \) belongs to the Lie algebra of SU(N), then the transformed matrix \( A_\mu' \) also belongs to the same Lie algebra for any \( U \) in SU(N). The context involves properties of matrices and their transformations under group actions.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the properties of matrices in the context of Lie groups, questioning the definitions and relationships between the matrices involved. There is a focus on verifying the defining properties of the Lie algebra, particularly concerning hermitian and anti-hermitian conditions. Some participants also seek clarification on the meaning of the coupling constant \( g \) in the context of non-abelian gauge theory.

Discussion Status

The discussion is active, with participants sharing their thoughts on the properties of the matrices and their transformations. Some have made progress in their reasoning, while others are still clarifying foundational concepts. There is no explicit consensus yet, but various interpretations and approaches are being explored.

Contextual Notes

Participants note the importance of properties such as hermiticity and the relationship between the Lie algebra and the group itself. There is an ongoing examination of the implications of these properties for the proof being discussed.

latentcorpse
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How would you go about proving that if A_\mu \in \text{Lie}(SU(N)) then A_\mu' \in \text{Lie}(SU(N)) \forall U \in SU(N)

where A_\mu'=U A_\mu U^{-1}-\frac{1}{g} ( \partial_\mu U) U^{-1}?

Presumably we need to check some defining property of being in the Lie Group?

Thanks.
 
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Could tell me what g is?
 
praharmitra said:
Could tell me what g is?
Thanks for your reply.
Well this is to do with the physics of non abelian gauge theory so I think we just take it as the dimensionless coupling constant?
In short, I think we just treat it as a constant.
 
Ah! well, here is what I have done

Property for Lie(SU(N)) : A^\dagger = A

Property for SU(N) : A^\dagger = A^{-1}

What I am able to show is that

<br /> A&#039;_\mu^\dagger = U A_\mu ^\dagger U^{-1}+\frac{1}{g} ( \partial_\mu U) U^{-1}<br />

There is a plus sign where there should be a minus. I am trying to figure that out.EDIT: Ahh! Got it. I forgot that \partial_\mu^\dagger = -\partial_\mu that fixes it.
 
praharmitra said:
Ah! well, here is what I have done

Property for Lie(SU(N)) : A^\dagger = A

Property for SU(N) : A^\dagger = A^{-1}

What I am able to show is that

<br /> A&#039;_\mu^\dagger = U A_\mu ^\dagger U^{-1}+\frac{1}{g} ( \partial_\mu U) U^{-1}<br />

There is a plus sign where there should be a minus. I am trying to figure that out.


EDIT: Ahh! Got it. I forgot that \partial_\mu^\dagger = -\partial_\mu that fixes it.

Shouldn't the Lie group, \text{Lie}(SU(N)) be anti hermtian matrices i.e. A^\dagger=-A?
 
Well,
<br /> U^\dagger = U^{-1} \Rightarrow exp(iA)^\dagger = exp(iA)^{-1}<br />

<br /> \Rightarrow exp(-iA^\dagger) = exp(-iA) \Rightarrow A^\dagger = A<br />
 
praharmitra said:
Well,
<br /> U^\dagger = U^{-1} \Rightarrow exp(iA)^\dagger = exp(iA)^{-1}<br />

<br /> \Rightarrow exp(-iA^\dagger) = exp(-iA) \Rightarrow A^\dagger = A<br />

I think in my notes, we have the i absorbed into the matrix A since we define the lie algebra to be the vector space of traceless, anti-hermitian matrices...
 

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