Explain answer on Lie algebras from mathoverflow I was reading this question on mathoverflow click me and the answer it received. Although I understand the question, the answer is too short for me, so I would like to have some clarification.
So the Lie algebra is $\mathfrak{su}(N)$ and then the author says WLOG the subspace is again a Lie subalgebra. 
Question 1: How do we know that each subspace is also invariant under the commutator?
Then he says that the action leaves the inner product invariant. So he means the action 
\begin{align*}
\Phi: SU(N) \times \mathbb{C}^n &\rightarrow \mathbb{C}^n &(A,x) &\mapsto Ax?
\end{align*}
Question 2: What does reducibility mean in this case and what does he mean with non-isomorphic modules (aren't these modules vector spaces here)? But still, I don't know why this decomposition is possible.
Question 3: Then he explains how the centralizers must look like (where does this conclusion come from) and he additionally states a trace-condition. I don't see where this condition comes from, since $\mathfrak{su}(N)$ is the same as $\mathfrak{u}(N).$
If anything is unclear, please let me know.
 A: (1) We don't know that an arbitrary subspace is a Lie subalgebra, but the statement is that $C_{H_0}(S)= C_{H_0}(\mathfrak{m})$ for some Lie subalgebra $\mathfrak{m}$ (indeed, $\mathfrak{m}$ is just the Lie subalgebra generated by $S$).
(2) Complete reduciblity means that $W$ decomposes into a direct sum of simple submodules. The reason this works is that $\mathfrak{su}(N)$ leaves the inner product on $W$ invariant (that is, $\langle xv,w\rangle=-\langle v,xw\rangle$ for all $v,w\in W$ and $x\in\mathfrak{su}(N)$). From this, it is easy to prove that if $V\leq W$ is a submodule, then the orthogonal compliment $V^{\perp}$ is also a submodule.
As far as the modules being non-isomorphic, you must note that an isomorphism $\phi:V_i\to V_j$ of $\mathfrak{su}(N)$-module must satisfy $\phi(xv)=x\phi(v)$ for all $v\in V_i$ and $x\in\mathfrak{su}(N)$. In particular, it is not enough that they have the same dimension.
[An easy exercise to see what is going on is to look at the $\mathbb{C}[x]$-modules $U=\mathbb{C}[x]/(x)$ and $V=\mathbb{C}[x]/(x-1)$. Both are 1-dimensional $\mathbb{C}$-vector spaces, but they are non-isomorphic as $\mathbb{C}[x]$-modules. To see this, note  that $xu=0$ for all $u\in U$, while $xv=v$ for all $v\in V$. In particular, if $\phi:U\to V$ is an isomorphism, then $\phi(1)\neq 0$, but $0=\phi(0)=\phi(x1)=x\phi(1)=\phi(1)$, a contradiction.]
(3) Given the decomposition $W=\bigoplus_i V_i^{k_i}$ as irreducible $\mathfrak{m}$-modules, we have that any element $y$ of the centralizer of $\mathfrak{m}$ defines an endomorphism
$$\phi_y:\bigoplus_iV_i^{k_i}\to\bigoplus_iV_i^{k_i}$$
by $\phi_y(v)=yv$. Since $[x,y]=0$ for all $x\in\mathfrak{m}$, we have 
$$\phi_y(xv)=y(xv)=(yx)v=(xy)v=x(yv)=x\phi_y(v).$$
In particular, $\phi_y$ is an element of $$\mathrm{End}_{\mathfrak{m}}\left(\bigoplus_iV_i^{k_i}\right),$$
which is unitary because $y$ is unitary.
By Schur's lemma, $\mathrm{Hom}_{\mathfrak{m}}(V_i,V_j)=0$ for $i\neq j$, and a little linear algebra shows that $\mathrm{End}_{\mathfrak{m}}(V_i^{k_1})\cong M_{k_i}(\mathbb{C})$. Therefore, you get that the centralizer is the set of unitary matrices in $\bigoplus_i M_{k_i}(\mathbb{C})$. That is, as a Lie algebra, the centralizer is $\bigoplus_i\mathfrak{u}(k_i)$.
