The $\mathfrak{su}(3)$ structure constants $f^{abc}$ are defined by $$[T^a,T^b] = i f^{abc} T^c,$$ with $T^a$ being the generators of the group $\mathrm{SU}(3)$. They are usually written out in a very non-transparent way as $$f^{123} = 1, \ f^{147} = - f^{156} = f^{246} = f^{257} = f^{345} = - f^{367} = \frac{1}{2}, \ f^{458} = f^{678} = \frac{\sqrt{3}}{2}.$$

The question: Is it possible to write these structure constants in some covariant way so that they don't appear as random?

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    $\begingroup$ Yes of course. In the standard Cartan approach one decomposes the Lie algebra in Cartan generators and roots $\alpha$, $\beta$ etc. The structure constants are then given by the $N_{\alpha,\beta}$ in $[E_\alpha,E_\beta]=N_{\alpha,\beta}E_{\alpha+\beta}$. $\endgroup$ – user178876 Sep 11 '18 at 12:11
  • $\begingroup$ BTW, people did come up with other bases of SU(3), e.g. here. It's just unclear if this is worthwhile, in practice you only need some quantities like the Dynkin index or the quadratic Casimir, hardly ever the explicit matrices. Another very convenient basis can be found in the Susyno package. $\endgroup$ – user178876 Sep 11 '18 at 13:33
  • $\begingroup$ Yes of course. J J Sylvester introduced "nonions" in 1882, p 649. $\endgroup$ – Cosmas Zachos Sep 11 '18 at 15:59
  • $\begingroup$ This clock-and-shift language, of course, applies to *all SU(N)*s. $\endgroup$ – Cosmas Zachos Sep 11 '18 at 16:01

The numerical values of the components of the structure constants can be not only "written down". They may be explicitly calculated from the explicit form of the generators. The values that you referred to correspond to a particular basis that is the most widespread convention, the Gell-Mann matrices:


They're the basis of all 3x3 Hermitian traceless matrices. Two of them, the 3rd and 8th, are taken to be diagonal. The remaining 6 are off-diagonal matrices that only have $+1,+1$; or $+i,-i$ at the $ij$- and $ji$-location of the matrix. The order is so that we do $ij=12,13,23$ in this order, and for each value of $ij$, we first write the matrix with $1,1$ and then with $+i,-i$, with the $-i$ above the diagonal.

The first three Gell-Mann matrices are just the three Pauli matrices for $SU(2)$. Then we're adding the remaining five matrices – out of them, the last one is the diagonal one, up to the overall sign, it's the unique diagonal matrix that is traceless and orthogonal to all previous matrices (including and especially the diagonal, third Gell-Mann or Pauli matrix).

All of the Gell-Mann matrices are normalized so that the trace of their square equals two.

If you just substitute $T^a = \lambda^a/2$ for your generators, you may simply calculate all the commutators and identify them as another Gell-Mann matrix.

Any basis of the 8-dimensional space of $SU(3)$ generators will contain either "random numbers" or entries of different kinds. For example, in a comment, marmot mentioned that you may choose the basis from the Cartan algebra and then the eigenstates corresponding to some roots etc. In this basis, the structure constants will still look somewhat random and you will have to distinguish two types of the indices $a,b,c=1,2,\dots 8$. But there will be a nicer logic in them than in the Gell-Mann basis.

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