I am looking for (an example of) a minimal set of Gell-Mann matrices such that their closure under the Lie bracket is all of $\mathfrak{su}(3)$. By minimal I mean the set should be as small as possible.

Even though this seems quite simple, I can't make much sense of any of the sources that I have consulted. For example, here I found that this set should only contain two elements. Any help is much appreciated.

  • $\begingroup$ Look at this question and determine the generators for the Lie algebra. $\endgroup$ – Dietrich Burde Apr 16 at 16:52
  • $\begingroup$ Yes, exactly, but what are they? $\endgroup$ – Georg Apr 16 at 16:54

Well, you know it is not a set of two generators from elementary I,U,V-spin considerations in the eightfold way. Below, you see by inspection a set of just three will suffice. Avoid taking your three from one of the numerous su(2) subalgefbras, of course!

So take the set $\lambda_1,\lambda_2,\lambda_4$, for the sake of argument. Commuting them pairwise generates $\lambda_3,\lambda_7,\lambda_6$. Commuting the first and second set nets $\lambda_5,\lambda_8$. This is straightforward given the sparseness of the structure constants in this basis.

It is easier to see from the root diagram, or the meson octet, in physics. I hope you are not asking for isomorphisms among sufficient sets of threes. I gave you one.

  • $\begingroup$ You are right that the set you gave generates whole of su(3). But what are the elementary considerations that probe that it is impossible to generate su(3) with only two elements? $\endgroup$ – Georg Apr 17 at 11:06
  • $\begingroup$ There is another elementary way to see that two generators do not suffice: take any two Gell-Mann matrices, and take the commutator. This will yield a third matrix (or even zero). By the anti-symmetry of the structure constants, it is impossible to generate a fourth distinct element using the commutator between any of the three matrices you now have at your disposal. $\endgroup$ – Georg Apr 17 at 11:23
  • $\begingroup$ Basically yes, but in a few cases you get a sum of two generators on the right, e.g starting with 4,5, or 6,7. However, further commutation with the lin.combs still leads to closure to su(2)s. This is also evident in the standard "F" basis. $\endgroup$ – Cosmas Zachos Apr 17 at 14:25

Start with the two simple roots $\alpha_1$ and $\alpha_2$, corresponding to $E_{\alpha_1}$ and $E_{\alpha_2}$ respectively. The commutator will give you the third root $\alpha_3=\alpha_1+\alpha_2: E_{\alpha_3}=[E_{\alpha_1},E_{\alpha_2}]$. Take the negative of these (corresponding to transpose conjugate of the respective matrices). The commutators of $[E_{\alpha_1},E_{-\alpha_1}]$ and $[E_{\alpha_2},E_{-\alpha_2}]$ will give two linearly independent Cartan elements.

  • 1
    $\begingroup$ If I start with $E_{\alpha_1}$ and $E_{\alpha_2}$, by construction I am only allowed to use commutators to reach other elements. Thus we cannot "Take the negative of these (corresponding to transpose conjugate of the respective matrices)". $\endgroup$ – Georg Apr 16 at 21:07

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