# Ideals of semisimple Lie algebras

Let $\mathfrak{g} \subset gl(V)$ be a semisimple Lie algebra. I already know that symmetric bilinear form $f(x,y)=\mathbf{Trace}(XY)$ is nonsingular on $\mathfrak{g}$. And I've read that any ideal $\mathfrak{g_1}$ of $\mathfrak{g}$ is also semisimple, from which it also follows that $\mathfrak{g_1}$ is nonsingular space with respect to this bilinear form.

Could someone help me to show that $\mathfrak{g_1}$ is actually nonsingular?

I suppose that $f(x,y)=\mathbf{Trace}(XY)$ means in fact $$f(x,y)=\mathbf{Trace}(ad_xad_y)$$ since I don't get otherwise what $X,Y$ mean since $f$ is function of $x,y$... in other words I assume that we are speaking about the Killing form. Then the thing is quite easy: suppose $\mathfrak{g_1}$ is an ideal, then consider $\mathfrak{g}=\mathfrak{g_1}\oplus\mathfrak{g}_{2}$, find a base for $\mathfrak{g}$ made by vector of $\mathfrak{g}_1$ followed by vector of $\mathfrak{g}_2$ and then in this base consider the block matrix form of the adjoint matrix $ad_x$. Since $\mathfrak{g_1}$ is an ideal you have that $[x,\mathfrak{g}]\in\mathfrak{g_1}$ whenever $x\in\mathfrak{g_1}$ so you have $$ad_x=[x,.]=\begin{array}{c} \begin{array}{cc} \mathfrak{g}_{1} & \mathfrak{g}_{2}\end{array}\\ \left(\begin{array}{cc} * & *\\ 0 & 0 \end{array}\right) \end{array}$$ Which means that for every $y \in \mathfrak{g}$ you have $ad_xad_y$ of the form $$ad_xad_y=\begin{array}{c} \begin{array}{cc} \mathfrak{g}_{1} & \mathfrak{g}_{2}\end{array}\\ \left(\begin{array}{cc} * & *\\ 0 & 0 \end{array}\right) \end{array}$$ So for every vector $v \in \mathfrak{g}$ let us decompose it as $v=v_1+v_2$ where $v_1 \in \mathfrak{g}_{1}$ and $v_2 \in \mathfrak{g}_{2}$. Then, given the form of the matrix above, you have that for every vector $x \in \mathfrak{g}_{1}$ the ${Trace}(ad_xad_v)=0$ if and only if ${Trace}(ad_xad_{v_1})=0$. This means that $$ker\left(f_{|\mathfrak{g}_{1}}\right)\subseteq ker\left(f\right)$$ and since $ker\left(f\right)=0$ therefore $ker\left(f_{|\mathfrak{g}_{1}}\right)=0$
• I imagine that $f(x,y)$ should have been $f(X,Y)$, as $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}(V)$ by assumption. Also, shouldn't it be $[X,\mathfrak{g}_2]\in \mathfrak{g}_2$, which changes the block form? – Jason DeVito May 23 '18 at 19:51
• Oh, I see, you are assuming $X,Y\in \mathfrak{g}_1$. Then in fact $[X,\mathfrak{g}_2] \in \mathfrak{g}_1\cap\mathfrak{g}_2 = 0$, so the top left $*$ in each matrix is actual $0$. – Jason DeVito May 23 '18 at 19:55
• @TorstenSchoeneberg Suppose that the form is singular on $\mathfrak{g}_1$ the subspace that is the kernel on $\mathfrak{g}_1$ is also a subspace on which the the form is null on $\mathfrak{g}$ since all the other blocks of the matrix $ad_xad_y$ are zeros. So since the form is not degenerate on $\mathfrak{g}$ you deduce it can't be degenerate on $\mathfrak{g}_1$ – Dac0 May 23 '18 at 21:17
• But you show that $ad_x ad_y$ has this block form only if both $x,y \in \mathfrak{g}_1$. I do not think you can conclude $ker(f_{|\mathfrak{g}}) \subseteq ker(f_{|\mathfrak{g}_1})$ if $\mathfrak{g}_2$ is any complementary subspace to $\mathfrak{g}_1$. But if $\mathfrak{g}_2$ is an ideal as well, then for all $x,y \in \mathfrak{g}$, $ad_x ad_y$ will be in diagonal (block) form, and from there one should get a full proof. (Also, your last sentence before the edit still makes no sense grammatically, and some other sentences could also need some punctuation and grammar improvement.) – Torsten Schoeneberg May 24 '18 at 1:40