Adjoint representation of SU(3)

[nigelscott]

Not sure if this is the correct forum but here goes.

I am trying to prove [T^a,T^b] = if^abcT^c

Where (T^a)^bc = -if^abc and f^abcare the structure constants for SU(3).

I picked f¹²³ and generated the three 8 x 8 matrices .. T¹, T² and T³.
The matrices components are all 0 except for,

(T¹)²³ = -i
(T(¹)⁴⁷ = -i/2
(T¹)⁵⁶ = i/2

(T²)⁴⁶ = -i/2
(T²)⁵⁷ = -i/2

(T³)⁴⁵ = -i/2
(T³)⁶⁷= i/2

When I compute [T¹,T²] I get 0. What am I missing?

 

[fresh_42]

All your generators $T^i$ seem to be nilpotent which can't be. They have to be regular, i.e. invertible.

 

[nigelscott]

Yes, so how does one build the correct generators? This worked fine for SU(2) so something else is missing. I suspect the Jacobi identities play a role but I'm not sure how to proceed. Thanks

 

[fresh_42]

Have a look on the Gell-Mann matrices: (try $\lim_{t→0} \exp(t \lambda_i)$, or simply $1+\lambda_i$)

https://en.wikipedia.org/wiki/Gell-Mann_matrices

Remember that the adjoint representation $Ad$ is a group homomorphism $SU(3) \longrightarrow GL(\mathfrak{su}(3))$ with $Ad(\exp(\lambda_i)) = \exp(ad (\lambda_i)).$

 

[nigelscott]

Thanks. I am familiar with the G-M matrices. I am after the 8 x 8 adjoint versions of the generators. I think I am making progress. When I add indeces:

(T¹)^32,
(T¹)^74,
(T¹)^65,

(T²)^64,
(T²)^75,

(T³)^54,
(T³)^76,

into the mix I almost get the right answer. I am wondering if interchanging the second and third indeces changes the sign of the constant.

Forgive me if I am making this hard work but I no expert in GT.

 

[nikkkom]

Adjoint representation of SU(3) has 8 matrices, not 3.