# Centralizer of semi-simple element in semi-simple Lie algebra

Let $L$ be a finite dimensional semi-simple Lie algebra, and $H$ a toral (maximal abelian) subalgebra. For any $h\in H$ I want to prove that $C_L(h)$ is reductive, i.e. its radical (=maximal solvable ideal) is equal to its center. How should I proceed?

What I did: Let $L=H\oplus (\bigoplus_{\alpha} L_a)$ be the root space decomposition of $L$ w.r.t. $H$. Let $L_{\alpha}=\langle x_{\alpha}\rangle.$ Then an element $h+\sum_{i=1}^k x_{\alpha_i}$ will commute with if and only if $h$ commutes with each $x_{\alpha_i}$. But $[h,x_{\alpha_i}]=\alpha_i(h)x_{\alpha_i}$ which forces that $\alpha_i(h)=0$ for $i=1,2,\cdots,k$.

After this I couldn't do anything. Any hint?

Hint: You have observed that $$C_L(h)=\mathfrak h\oplus_{\alpha:\alpha(h)=0}\mathfrak g_{\alpha}$$. Since this is the decomposition into joint eigenspaces for the adjoint action of $$\mathfrak h\subset C_L(h)$$, it follows from standard arguments (the projection onto an eigenspace of a linear map can be written as a polynomial in the map) that an $$\mathfrak h$$-invariant subspace of $$C_L(\mathfrak h)$$ must be the direct sum of a linear subspace of $$\mathfrak h$$ and some of the root spaces. In particular, this applies to any ideal in $$C_L(\mathfrak h)$$ and hence to the radical $$\mathfrak r$$. Now first argue directly that an ideal containing a root space cannot be solvable and second that an ideal contained in $$\mathfrak h$$ must actually be contained in the joint kernel of $$\{\alpha:\alpha(h)=0\}$$, which is exactly the center of $$C_L(h)$$.