# Commutator of SU(2) Casimir operators in su(3)

I have a $\mathfrak{su}(3)$ Lie algebra spanned by 8 generators, $$\left\{J_1,J_2,J_3,J_4,J_5,J_6,J_7,J_8 \right\}$$

Now, I can choose infinitely many $\mathfrak{su}(2)$ sub-algebras composed of three operators (does not have to be generators but also linear combinations). For every such sub-algebra I can define Casimir operator. I want to know if Casimir operators from different sub-algebras commute with each other or not? Maybe there is some theorem I don't know about?

• – Qmechanic Dec 12 '14 at 13:57
• Why don't you just test if they commute? You know all the commutation relations. (Spoiler: As $\mathfrak{su}(3)$ is semisimple, it has only one quadratic Casimir, meaning they won't.) – ACuriousMind Dec 12 '14 at 13:58
• possible duplicate of Definition of Casimir operator and its properties – Brandon Enright Dec 13 '14 at 0:21
• Perhaps it is a better question than ACuriousMind gave it credit for? The OP is probably impressed that the quadratic Casimirs of the I,V, and U-spin subalgebras are proportional to diag(1,1,0), diag(1,0,1), and diag(0,1,1) in the triplet, 3: He does not care for SU(3) Casimirs. – Cosmas Zachos Mar 7 '16 at 17:12

So here are two SU(2)s I am considering, in the Gell-Mann basis. The first one is just I-spin, comprised of $\vec{I}=(\lambda_1,\lambda_2,\lambda_3)$, and the 2nd one the formally equivalent one, $U \vec{I} U^{-1}$, where, for simplicity, take $U\equiv \exp(i\theta \lambda_4)$. The corresponding quadratic SU(2) Casimirs are, un-normalized, C=diag(1,1,0), idempotent, and $UCU^{-1}$, respectively.
You are asking if $[C,UCU^{-1} ]$vanishes or not. It suffices to check that the term linear in θ fails to vanish (the zeroth order term vanishing, of course). That is, you are inspecting $$i\theta (C\lambda_4 C -C\lambda_4 -\lambda_4 C+C\lambda_4C).$$ Since $C\lambda_4 C=0$, and the other two terms in the parenthesis add up to $-\lambda_4$, the commutator fails to vanish, $=-i\theta \lambda_4$ plus higher orders in θ, and the answer to your question is: not.