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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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I understand from the physics analog of this question that you accept (2) is a basis-dependent subjective vortex. For (1), also subjective, yes and no. N is actually the least of your problems, and you might fix it by extrapolation of the generic results of MacFarlane, Sundry, and Weisz, 1968, (2.23), for SU(3), $$ 3d_{ijk}d_{pqk}= \delta_{ip} \delta_{jq} + \delta_{iq} \delta_{jp}- \delta_{ij} \delta_{pq} + f_{ipm} f_{jqm}+ f_{iqm} f_{jpm} ~. $$ Just don't try to eliminate the f bilinears by (2.10) because you are stuck.

I strongly urge you to use the more abstract invariants' structure leading up to them, instead of obsessing with parochial identities, however.

Back to (2), e.g. for su(3), the fully symmetric ds are properties of the Gell-Mann matrix basis, not the algebra, $$ 4 d^{ijk}=\operatorname {Tr} (\{ \lambda^i,\lambda^j\}\lambda^k). $$ You can read up in you standard text that they vanish when the number of indices from the set {2,5,7} is odd, from evident antisymmetry of the Gell-Mann matrices in that basis.

The nonvanishing independent elements are, naturally, (cf D B Lichtenberg, Unitary Symmetry & Elementary Particles):

$1/\sqrt 3$ for {118,228,338} ;

$-1/\sqrt 3$ for 888;

$-1/2\sqrt3$ for {448,558,668,778};

1/2 for {344,355,146,157,256};

-1/2 for {247,366,377}.

As indicated, simplification is somewhat subjective.

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  • $\begingroup$ Do you know math.stackexchange.com/q/3236324/79069 for semi-simple Lie algebra? $f^{abc}$ can be totally antisymmetric for semi-simple Lie algebra I think? $\endgroup$
    – wonderich
    Commented May 22, 2019 at 23:29
  • $\begingroup$ You may indeed arrange $-i2\operatorname{tr}(\lambda_c[\lambda_a,\lambda_b])=f_{abc}$ fully antisymmetric for semisimple ones. Will go to that question next. $\endgroup$ Commented May 23, 2019 at 0:04
  • $\begingroup$ As I comment there (2.10) of McF, S & W is derived for semi simple ones in their method. $\endgroup$ Commented May 23, 2019 at 0:27
  • $\begingroup$ I think your values for the elements of $d$ are incorrect, e.g. 344 See en.wikipedia.org/wiki/Special_unitary_group $\endgroup$
    – Rudyard
    Commented Jun 6, 2020 at 11:09
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    $\begingroup$ @Rudyard Thanks, you certainly appear to be right: the $\sqrt 3$s come from contamination by $\lambda_8$. Unfortunately I don't recall my source for what I have. Will fix in time. $\endgroup$ Commented Jun 6, 2020 at 11:44

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