SU(n) - conjugate representation

[tom.stoer]

Very simple question, but I can't find the answer.

Taking an su(n) Lie algebra with hermitean generators we have

$$[T^a, T^b] = if^{abc}T^c$$

One immediately finds that the new generators

$$\tilde{T}^a = (-T^a)^\ast$$

define the same algebra, i.e. fulfil the same commutation relations

$$[\tilde{T}^a, \tilde{T}^b] = if^{abc}\tilde{T}^c$$

One can show easily that for n=2 the two sets of generators are equivalent, i.e. related by a transformation

$$\tilde{T}^a = S T^a S^{-1}$$

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

$$\tilde{T}^a = S T^a S^{-1}$$

 

[fzero]

In general for $$su(n)$$ the fundamental representation $$V$$ and the antifundamental representation $$V^*=\text{Hom}(V,\mathbb{C})$$ are complex conjugate representations with opposite complex structures, so are distinct. In the case of $$su(2)= sl(2,\mathbb{R})$$ the fundamental representation is actually real, hence the equivalence you found.

 

[tom.stoer]

Thanks; I know this. But what does it mean on the level of the matrices T and S? How can I show using simply the T's that the above mentioned S does not exist for su(n) with n>2.

 

[Dickfore]

You are requiring the representations to be real, if I am not mistaken. There was some requirement that the eigenvalues of the generators should come in pairs of opposite sign.

 

[tom.stoer]

Again I know that, but why? Let's make a simple example. I give you the Gell-Mann matrices and let you calculate an S that does the job. Where do you fail to determine S? And why?

 

[tom.stoer]

I think I figured it out.

One can check that iff S exists then it maps eigenvalues to eigenvalues. But using the conjugate generators one sees that it maps eigenvalues to "minus eigenvalues". These two mappings can only coincide iff the eigenvalues come in pairs.

One can check for the Gell-Mann matrices that for a=1..7 the eigenvalues come in pairs +1, -1. But for a=8 the eigenvalues are ~ 1, 1, -2 and are not paired. Therefore assuming S to do the job leads to a contradiction, whcih means that for SU(3) no such S can exist.

 

[Dickfore]

One can check that iff S exists then it maps eigenvalues to eigenvalues.

The characteristic polynomials for two similar matrices (meaning they are connected via $B = S A S^{-1}$) are:

$$\mathrm{det} \left(B - \lambda \, 1 \right) = \mathrm{det} \left[ S (A - \lambda \, 1) S^{-1} \right] = \mathrm{det}{S} \, \mathrm{det}\left(A - \lambda 1\right) \, \mathrm{det}{S^{-1}} = \mathrm{det}\left(A - \lambda 1\right)$$

the same, so you are right there.

But using the conjugate generators one sees that it maps eigenvalues to "minus eigenvalues".

Because the generators are hermitian, they have real eigenvalues. Thus, the eigenvalues of $\tilde{T^{a}} = -(T^{a})^{\ast}$ should be the set $\{-\lambda^{\ast}\} = \{-\lambda\}$.

These two mappings can only coincide iff the eigenvalues come in pairs.

True, however, we have just proven that the condition is a necessary one. I am not sure whether it is easy to show that it is a sufficient one as well, or even if that is true.

One can check for the Gell-Mann matrices that for a=1..7 the eigenvalues come in pairs +1, -1. But for a=8 the eigenvalues are ~ 1, 1, -2 and are not paired. Therefore assuming S to do the job leads to a contradiction, whcih means that for SU(3) no such S can exist.

Yes, I wanted to reply to your first reply with the 8th Gell-Mann matrix, but you beat me to it.

 

[tom.stoer]

Thanks for responsing and for LaTeX.

True, however, we have just proven that the condition is a necessary one. I am not sure whether it is easy to show that it is a sufficient one as well, or even if that is true.

My original problem was to understand why the fundamental and the conjugate rep. of SU(3) are not equivalent. This is what we have solved now. I agree that we only discussed a necessary condition, we did not show whether it's suffcient.

 

[fzero]

Thanks for responsing and for LaTeX.
My original problem was to understand why the fundamental and the conjugate rep. of SU(3) are not equivalent. This is what we have solved now. I agree that we only discussed a necessary condition, we did not show whether it's suffcient.

The fundamental and antifundamental representations are related by the $$\mathbb{Z}_2$$ outer automorphism of $$SU(n)$$. You can either see this from the weight diagram or by labeling the Dynkin diagram with the highest weights. This means that $$SU(n)$$ has inequivalent actions on them. For $$SU(2)$$, there are no outer automorphisms, so there was an inner automorphism that related the conjugate representations.

For $$SU(3)$$, what distinguishes the two representations is the choice of $$\lambda_{2,5,7}$$. If we take these to be the fundamental representation, then the antifundamental matrices $$-\lambda_{2,5,7}$$ correspond to an inequivalent complex structure.