- #1
- 5,779
- 172
Very simple question, but I can't find the answer.
Taking an su(n) Lie algebra with hermitean generators we have
[tex][T^a, T^b] = if^{abc}T^c[/tex]
One immediately finds that the new generators
[tex]\tilde{T}^a = (-T^a)^\ast [/tex]
define the same algebra, i.e. fulfil the same commutation relations
[tex][\tilde{T}^a, \tilde{T}^b] = if^{abc}\tilde{T}^c[/tex]
One can show easily that for n=2 the two sets of generators are equivalent, i.e. related by a transformation
[tex]\tilde{T}^a = S T^a S^{-1}[/tex]
I know that for n>2 this is no longer true, i.e. that the two representations are not equivalent. That means that for n>2 this S cannot exist. My question is, how can one show algebraically that for n=3, 4, ... no S can exist such that
[tex]\tilde{T}^a = S T^a S^{-1}[/tex]
Taking an su(n) Lie algebra with hermitean generators we have
[tex][T^a, T^b] = if^{abc}T^c[/tex]
One immediately finds that the new generators
[tex]\tilde{T}^a = (-T^a)^\ast [/tex]
define the same algebra, i.e. fulfil the same commutation relations
[tex][\tilde{T}^a, \tilde{T}^b] = if^{abc}\tilde{T}^c[/tex]
One can show easily that for n=2 the two sets of generators are equivalent, i.e. related by a transformation
[tex]\tilde{T}^a = S T^a S^{-1}[/tex]
I know that for n>2 this is no longer true, i.e. that the two representations are not equivalent. That means that for n>2 this S cannot exist. My question is, how can one show algebraically that for n=3, 4, ... no S can exist such that
[tex]\tilde{T}^a = S T^a S^{-1}[/tex]