Does a simple subalgebra have to be an ideal of the larger algebra?

Solution 1:

The following facts hopefully clear the fog.

• It is possible for a simple Lie algebra $L_1$ to be a sub Lie algebra, but not an ideal, of a bigger simple Lie algebra $L_2$. For example, consider $L_1=\mathfrak{sl}_2$ as a subalgebra of $L_2=\mathfrak{sl}_3$. Obviously $L_1$ is not an ideal, and both the Lie algebras are simple. In fact, a fundamental tool when developing the structure theory of simple Lie algebras is to observe that to each root there is a corresponding copy of $\mathfrak{sl}_2$.
• However, when we talk about a direct sum of Lie algebras, $L_1\oplus L_2$, then this alone implies that $[L_1,L_2]=0$. It follows that both $L_1\oplus 0$ and $0\oplus L_2$ are automatically ideals of $L_1\oplus L_2$.
• So when we decompose a semi-simple Lie algebra into a direct sum of simple ones, those summands are automatically also ideals by the previous bullet.

Solution 2:

The answer is negative. Take, for instance$$\mathfrak{so}(3,\Bbb R)=\left\{\begin{bmatrix}0&a&c\\-a&0&b\\-c&-b&0\end{bmatrix}\,\middle|\,a,b,c\in\Bbb R\right\},$$which is a simple Lie algebra. Then $\mathfrak{so}(3,\Bbb R)$ is a subalgebra of$$\mathfrak{sl}(3,\Bbb R)=\{M\in\mathfrak{gl}(3,\Bbb R)\mid\operatorname{tr}M=0\}.$$But, for instance, if you take$$M=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix},$$then $M\in\mathfrak{sl}(3,\Bbb R)$, but$$(\forall a,b,c\in\Bbb R):\left[\begin{bmatrix}0&a&c\\-a&0&b\\-c&-b&0\end{bmatrix},M\right]=\begin{bmatrix}c & 0 & 0 \\ b & 0 & 0 \\ 0 & -a & -c\end{bmatrix},$$which only belongs to $\mathfrak{so}(3,\Bbb R)$ when $a=b=c=0$.