Consider the Lie algebra $\mathfrak{su}(n)$ and the set of operators that do not change the direction of the vector $\psi$, $$ K:=\{ s\in \mathfrak{su}(n)\ :\ s \psi \propto \psi \}. $$

Let $P$ be the orthogonal complement of $K$, $\mathfrak{su}(n)=K\oplus P$.

I would like to show that $K$ is a symmetric (Lie) subalgebra, $$ [K,K]\subset K, \quad [K,P]\subset P, \quad [P,P]\subset K. $$

The first two relations are shown by just acting with the commutator on $\psi$ (after choosing arbitrary elements from the respective sets). I checked the third relation for $\mathfrak{su}(2)$ and $\mathfrak{su}(3)$, but I can't find a way to show it for $\mathfrak{su}(n)$.

  • $\begingroup$ By "orthogonal" you mean with respect to the Killing form, right? -- Also, maybe that's a clear convention in your setting, but what vector space exactly do we let the Lie algebra act on here, i.e. $\psi$ is an element of what? I presume $\Bbb C^2$ but seen as four-dimensional real vector space? $\endgroup$ – Torsten Schoeneberg Apr 23 at 16:39
  • $\begingroup$ Yes, I mean with respect to the Killing form. The vector $\psi$ is an element of $\mathbb C^n$, if you want it is represented as an $n$ dimensional coulmn vector with complex entries. $\endgroup$ – Georg May 6 at 9:37

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