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

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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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