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I am interested in knowing the exact form of the anti-commutation of two generators of $su(N)$ lie algebra.

Let us denote $T^a$ to be the generator of $su(N)$ lie algebra in the defining representation. Since the number of generators is $n^2-1$, the index takes value in $a=1, ..., n-1$. The normalization of $T^a$ is $$Tr(T^aT^b)=\frac{1}{2} \delta^{ab}$$ The anti commutation of two such generators is $$\{T^a, T^b\}=T^a T^b+T^bT^a=\frac{1}{N}\delta^{ab}I+d^{abc}T^c$$ where $d^{abc}$ is a totally symmetric tensor in all the three indices. In https://pdfs.semanticscholar.org/1101/914fc76a36d4fb0ab0022f8c4ec6295d8d1f.pdf, it was shown that $$d^{abc}d^{abh}=\frac{N^2-4}{N}\delta^{ch}$$ where the repeated indices are to be summed over. In the above, we contract over two indices from each $d$-tensor.

My questions are:

1) Is there a simple expression for $$d^{abc}d^{agh}$$ where we only contract one index for each $d$-tensor? (in terms of $N$)

2) Is there a simple expression for $d^{abc}$ itself? (in terms of $N$)

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    $\begingroup$ for $N=3$ the elements of $d^{abc}$ are given in table 2 of these notes; the complexity of this result suggests a "simple" expression valid for any $N$ is not likely. $\endgroup$ – Carlo Beenakker Aug 12 '18 at 20:24
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It is only a partial answer so far: to the best of my understanding, the fully symmetrized version of $d^{abc}d^{agh}$ can be expressed via Kronecker deltas, see Example 5.1 here.

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