Problem with SU(3) generators's trace

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    Su(3) Trace
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SUMMARY

The forum discussion centers on the trace of the commutator of SU(3) generators, specifically the expression $$Tr\left(\left[ T^a_8,T^b_8\right] T^c_8\right)=i\frac{3}{2}f^{abc}$$. The user questions the correctness of this relation, noting that the Dynkin index for the adjoint representation is 3, leading to the conclusion that $$Tr\left(T^a_8T^b_8\right)=3\delta^{ab}$$. The user's calculations suggest a discrepancy involving a factor of 1/2, prompting a request for clarification on the professor's assertion.

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  • Understanding of SU(3) group theory
  • Familiarity with Lie algebra and its generators
  • Knowledge of Dynkin indices and their significance
  • Basic grasp of commutators and traces in linear algebra
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  • Learn about the structure constants $$f^{abc}$$ in Lie algebras
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This discussion is beneficial for theoretical physicists, graduate students studying quantum field theory, and anyone interested in the mathematical foundations of gauge theories.

Einj
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Hi everyone. I'm not sure this is the correct section for this topic and if not my apologiez.
I'm studying SU(3) and my professor wrote down the following equality:

$$Tr\left(\left[ T^a_8,T^b_8\right] T^c_8\right)=i\frac{3}{2}f^{abc}$$

where Ts are generators of the adjoint representation. I'm not sure this relation is correct and I would like to have your opinion. The Dynkin index of the adjoint representation is 3 so:

$$Tr\left(T^a_8T^b_8\right)=3\delta^{ab}$$

Now, my reasoning is:

$$Tr\left(\left[T^a_8,T^b_8\right]\right)=if^{abd}Tr(T^d_8T^c_8)=if^{abd}3\delta^{dc}=3if^{abc}$$

The difference is just a 1/2 factor but I would like to know if I'm doing something wrong.

Thanks everybody
 
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The index for SU(3) should be 3 in the adjoint representaion (and N generally). You should probably ask him to clarify.
 

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