# I Adjoint representation of SU(3)

1. Jul 24, 2016

### nigelscott

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

I am trying to prove [Ta,Tb] = ifabcTc

Where (Ta)bc = -ifabc and fabcare 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?

2. Jul 24, 2016

### Staff: Mentor

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

3. Jul 24, 2016

### 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

4. Jul 24, 2016

### Staff: Mentor

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 $Ad(\exp(\lambda_i)) = \exp(ad (\lambda_i)).$

Last edited: Jul 24, 2016
5. Jul 24, 2016

### 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.

6. Aug 2, 2016

### nikkkom

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