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

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

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

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

Thanks.
 
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praharmitra said:
Could tell me what [tex]g[/tex] 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)) : [tex]A^\dagger = A[/tex]

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

What I am able to show is that

[itex] A'_\mu^\dagger = U A_\mu ^\dagger U^{-1}+\frac{1}{g} ( \partial_\mu U) U^{-1}[/itex]

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 [tex]\partial_\mu^\dagger = -\partial_\mu[/tex] that fixes it.
 
praharmitra said:
Ah! well, here is what I have done

Property for Lie(SU(N)) : [tex]A^\dagger = A[/tex]

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

What I am able to show is that

[itex] A'_\mu^\dagger = U A_\mu ^\dagger U^{-1}+\frac{1}{g} ( \partial_\mu U) U^{-1}[/itex]

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 [tex]\partial_\mu^\dagger = -\partial_\mu[/tex] that fixes it.

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

[tex] \Rightarrow exp(-iA^\dagger) = exp(-iA) \Rightarrow A^\dagger = A[/tex]
 
praharmitra said:
Well,
[tex] U^\dagger = U^{-1} \Rightarrow exp(iA)^\dagger = exp(iA)^{-1}[/tex]

[tex] \Rightarrow exp(-iA^\dagger) = exp(-iA) \Rightarrow A^\dagger = A[/tex]

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...