TITLE: Symmetric tensor of Lie algebra of $su(N)$
QUESTION [6 upvotes]: 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 $N$)

REPLY [2 votes]: 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.