# Is a stabilizer subgroup a symmetric subalgebra?

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 $$\mathfrak{su}(n)$$.

• 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? – Torsten Schoeneberg Apr 23 at 16:39
• 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. – Georg May 6 at 9:37