I am trying to calculate the term: $$(t^a)_{ij} (t^a)_{kl}$$

In the book it's written that it equals to $$A\delta_{il}\delta_{kj}+B\delta_{ij}\delta_{kl}$$ and from using equation (18.40) $$tr[t^a](t^a)_{kl}=0$$ $$(t^a t^a)_{il}=4/3 \delta_{il}$$ we get: $$(t^a)_{ij}(t^a)_{kl}=1/2(\delta_{il}\delta_{kj}-1/3 \delta_{ij}\delta_{kl})$$

Now, I tried to prove the last equality, but I am not sure how to continue, we have: For $k=l, i=j , i\ne l, k\ne j$: $B= (t^a)_{jj}(t^a)_{ll}$. and for $i=l, k=j , i\ne j , k\ne l$ $A= (t^a)_{lj}(t^a)_{jl}$.

And as far as I can tell there's no summation convention here, am I right?

So how to get eq. (18.41) in the book?

  • $\begingroup$ Which one is 18.41? $\endgroup$ – user207455 Jul 7 '19 at 11:52
  • $\begingroup$ both $$tr[t^a](t^a)_{kl}=0$$ $$(t^a t^a)_{il}=4/3 \delta_{il}$$ are eq. (18.40) in the book. $\endgroup$ – MathematicalPhysicist Jul 7 '19 at 11:59

There is summation convention used in those formulae. In particular they are all summed over all values of $a$.

Using summation convention on both $a$ and other repeating indices: \begin{align} tr[t^a](t^a)_{kl} & = (t^a)_{ii} (t^a)_{kl} = 0, \\ (t^a)_{ii} (t^a)_{kl} & = A \delta_{il} \delta_{ki} + B \delta_{ii} \delta_{kl} = A \delta_{kl} + 3 B \delta_{kl} = 0. \end{align} Therefore: \begin{align} A = -3 B. \end{align}

For the second constraint: \begin{align} (t^a)_{ij} (t^a)_{jl} = A \delta_{il} \delta_{jj} + B \delta_{ij} \delta_{jl} = 3 A \delta_{il} + B \delta_{il} = \frac{4}{3} \delta_{il}, \end{align} or: \begin{align} 3A + B & = \frac{4}{3}. \end{align}

Solving the system of equations we get $A = \frac{1}{2}$, $B = -\frac{1}{6}$.


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